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1500 | https://math.stackexchange.com/questions/4788015/necessary-and-sufficient-condition-for-line-of-curvature-to-be-parametric-curve | differential geometry - Necessary and sufficient condition for line of curvature to be parametric curve - Mathematics Stack Exchange
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Necessary and sufficient condition for line of curvature to be parametric curve
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Necessary and sufficient condition for line of curvature to be parametric curve is F=0, M=0 and EN−GL≠0
Let the line of curvature b parametric curve such that u= constant and v=constant.
So combined the differential equation of parameter curve is dudv=0.
Now differential equation of line of curvature is (EM-FL)du 2 2+(EN-GL)dudv+((FN-GM)dv 2 2 = 0
So from above equations EM-FL=0,FN-GM=0 and EN−GL≠0.
But as per my understanding since parametric curve has one parameter so other is constant so u=constant or v=constant or in another form parametric curve is r(u,c) or r(c,v) where c is constant.
Hence either du=0 or dv=0. So either EM-FL=0 or FN-GM=0. Further, Since dvdu=0 hence EN−GL≠0 .
Could you explain why book has both parameter of parametric curve u=constant and v=constant and why both term taken EM-FL and FN-GM of du 2 2 and dv 2 2 equal to 0.
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edited Oct 17, 2023 at 1:27
ajay jhajhariaajay jhajharia
asked Oct 16, 2023 at 17:02
ajay jhajhariaajay jhajharia
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1 You should actually learn to use MathJax if you're going to contribute to and post on this site. Then you can write actual mathematics so that it is legible for all of us. Then you write E N−G L≠0 E N−G L≠0. I think you are misinterpreting whatever book you're reading. It's clearly classical and "old-fashioned" in today's world. At any rate, the conditions F=M=0 F=M=0 tell you that both parameter curves are lines of curvature, not that one is. And the inequality tells you that only those are lines of curvature, i.e., that the point is not an umbilic point.Ted Shifrin –Ted Shifrin 2023-10-16 17:37:30 +00:00 Commented Oct 16, 2023 at 17:37
Ok understood. But why both term taken EM-FL and FN-GM of du2 and dv2 are taken equal to 0. Where as in case u=constant then du=0 so it should be EM-FL ≠0 similarly when v=u=constant then dv=0 so it should be EN-GM ≠0 ajay jhajharia –ajay jhajharia 2023-10-17 07:28:50 +00:00 Commented Oct 17, 2023 at 7:28
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1501 | https://emedicine.medscape.com/article/157961-medication | No Results
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Right Ventricular Infarction Medication
Medication Summary
The goals of pharmacotherapy for right ventricular infarction are to reduce morbidity and prevent complications. Cardiovascular agents included in treatment are reperfusion agents (ie, tissue plasminogen activators) and inotropic agents. [5, 9]
Inotropic Agents
Inotropic therapy is indicated for right ventricular failure when cardiogenic shock persists after right ventricular end-diastolic pressure has been optimized. Inotropes should be used until more data are available. Dobutamine is an inotropic agent used to improve right ventricular contractility and maintain cardiac output.
Dobutamine (Dobutrex)
Dobutamine (Dobutrex)
Dobutamine produces vasodilation and increases the inotropic state. At higher dosages, this agent may cause increased heart rate, exacerbating myocardial ischemia.
Milrinone
Milrinone
Milrinone is a bi-pyridine positive inotrope and vasodilator with little chronotropic activity. It is different in mode of action from both digitalis glycosides and catecholamines. It selectively inhibits phosphodiesterase type III (PDE III) in cardiac and smooth vascular muscle, resulting in reduced afterload, reduced preload, and increased inotropy.
Tissue Plasminogen Activators
Tissue plasminogen activators bind to fibrin and convert plasminogen to plasmin, which in turn initiates local fibrinolysis with limited systemic proteolysis. Thrombolytic therapy may contribute to an early survival benefit in patients presenting within 6 hours of onset of onset of inferior wall myocardial infarction with right ventricular involvement diagnosed by ECG or other noninvasive criteria.
Alteplase (Activase)
Alteplase (Activase)
Alteplase is a tissue plasminogen activator used in the management of acute myocardial infarction, acute ischemic stroke, and pulmonary embolism. Heparin or aspirin may be administered with and after alteplase infusions to reduce the risk of rethrombosis. The safety and efficacy of concomitant administration of heparin or aspirin during the first 24 hours after symptom onset have not been investigated.
Reteplase (Retavase)
Reteplase (Retavase)
Reteplase is a recombinant plasminogen activator that forms plasmin after facilitating cleavage of endogenous plasminogen. It is used in the management of acute myocardial infarction. Heparin or aspirin may be administered with and after reteplase infusions.
Adrenergic Agonists
Adrenergic agonists stimulate beta- and alpha-adrenergic receptors, causing increased contractility and heart rate, as well as vasoconstriction. These actions increase systemic blood pressure and coronary blood flow.
Norepinephrine (Levophed)
Norepinephrine (Levophed)
Norepinephrine is a naturally occurring catecholamine with potent alpha-receptor and mild beta-receptor activity. It stimulates beta1- and alpha-adrenergic receptors, resulting in increased cardiac muscle contractility, heart rate, and vasoconstriction. It increases blood pressure and afterload. Increased afterload may result in decreased cardiac output, increased myocardial oxygen demand, and cardiac ischemia.
Antidiuretic Hormone Analogs
Antidiuretic hormone analogs increase cyclic adenosine monophosphate (cAMP), increasing water permeability at the renal tubules. An example of this analog is vasopressin, which is a direct vasoconstrictor without inotropic or chronotropic effects.
Vasopressin (Pitressin)
Vasopressin (Pitressin)
Vasopressin increases water resorption at the distal renal tubular epithelium (ADH effect). It promotes smooth muscle contraction throughout the vascular bed of the renal tubular epithelium (vasopressor effects). Vasoconstriction is also increased in splanchnic, portal, coronary, cerebral, peripheral, pulmonary, and intrahepatic vessels.
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Tables
| | | |
---
| Leads | Sensitivity (%) | Specificity (%) |
| V1 | 28 | 92 |
| V3 R | 69 | 97 |
| V4 R | 93 | 95 |
Leads
Sensitivity (%)
Specificity (%)
V1
28
92
V3 R
69
97
V4 R
93
95
Contributor Information and Disclosures
Claudia Dima, MD, FACC Interventional Cardiology
Disclosure: Nothing to disclose.
David L Coven, MD, PhD Attending Physician (Interventional Cardiology) Jamaica Hospital Medical Center Northwell Health - Lenox Hill Hospital
David L Coven, MD, PhD is a member of the following medical societies: American College of Physicians, American Medical Association, Massachusetts Medical Society
Disclosure: Nothing to disclose.
Ashish Pershad, MD Consulting Staff, Heart and Vascular Center of Arizona
Ashish Pershad, MD is a member of the following medical societies: American College of Cardiology
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Received salary from Medscape for employment.
Richard A Lange, MD, MBA President, Texas Tech University Health Sciences Center, Dean, Paul L Foster School of Medicine
Richard A Lange, MD, MBA is a member of the following medical societies: Alpha Omega Alpha, American College of Cardiology, American Heart Association, Association of Subspecialty Professors
Disclosure: Nothing to disclose.
Eric H Yang, MD Associate Professor of Medicine, Director of Cardiac Catherization Laboratory and Interventional Cardiology, Mayo Clinic ArizonA
Eric H Yang, MD is a member of the following medical societies: Alpha Omega Alpha
Disclosure: Nothing to disclose.
George A Stouffer, III, MD Ernest and Hazel Craige Distinguished Professor of Medicine and Cardiology, Chief of Cardiology, University of North Carolina Medical Center
George A Stouffer, III, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Cardiology, American College of Physicians, American Heart Association, Phi Beta Kappa, Society for Cardiovascular Angiography and Interventions
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1502 | https://en.wikipedia.org/wiki/Generalized_Pochhammer_symbol | Generalized Pochhammer symbol - Wikipedia
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In mathematics, the generalized Pochhammer symbol of parameter α>0{\displaystyle \alpha >0} and partitionκ=(κ 1,κ 2,…,κ m){\displaystyle \kappa =(\kappa {1},\kappa {2},\ldots ,\kappa _{m})} generalizes the classical Pochhammer symbol, named after Leo August Pochhammer, and is defined as
(a)κ(α)=∏i=1 m∏j=1 κ i(a−i−1 α+j−1).{\displaystyle (a){\kappa }^{(\alpha )}=\prod {i=1}^{m}\prod {j=1}^{\kappa {i}}\left(a-{\frac {i-1}{\alpha }}+j-1\right).}
It is used in multivariate analysis.
References
[edit]
Dunkl, Charles F.; Xu, Yuan (2001), Orthogonal Polynomials of Several Variables, Encyclopedia of Mathematics and its Applications, vol.81, Cambridge University Press, p.308, ISBN9780521800433
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1503 | https://www.justintimemedicine.com/curriculum/4954 | Key Concept:Diarrhea and/or vomiting cause systemic disruptions, some of which create emergent situations. | JustInTimeMedicine
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25 | Constipation; Diarrhea; Vomiting
Key Concept:Diarrhea and/or vomiting cause systemic disruptions, some of which create emergent situations.
Key Concept:Diarrhea and/or vomiting cause systemic disruptions, some of which create emergent situations.
Updated 11/22/2023 5:09 PM - Kate Baird
For patients with chronic diarrhea, obvious systemic consequences stem from the GI tract’s inability to digest and absorb nutrients, with the specific nutrient deficiencies dictated by the portion of the GI tract affected by disease. While it can be easy to pigeonhole the GI tract in terms of its digestive and absorptive functions and think of it mostly in terms of how we get nutrients from our food into our bodies, the GI tract also plays a significant role in fluid and electrolyte regulation. In a previous Key Concept this week, we took a close look at absorption and secretion along different segments of the digestive tract, and that look included fluids and electrolytes, H+ and HCO 3− among them. In the words of Gennari and Weiss, “The gastrointestinal tract is a slumbering giant with regard to acid-base homeostasis. Large amounts of H+ and HCO 3− traverse the specialized epithelia of the various components of the gut every day, but under normal conditions, only a small amount of alkali (approximately 30 to 40 mmol) is lost in the stool.”1 This loss of bicarbonate is usually readily offset by renal acid excretion; losing a bit of each keeps their relative concentrations in balance and it's that balance that determines pH. That stool is usually slightly alkaline and urine is usually slightly acidic, is part of what keeps our bloodstream in its happy pH range of 7.35-7.45. But, disease of the digestive tract can cause fluid loss and electrolyte shifts that overwhelm normal regulatory mechanisms, causing hypovolemia, acid-base imbalance, and electrolyte disturbances.
While it would be nice at this point to discuss the one acid-base and one electrolyte disruption that occur as a result of vomiting and diarrhea, it’s not so simple. As you know, different segments of the digestive tract have different functions, as dictated by the cells and transporters in a given segment. We can, however, present a few generalizations that will be very useful clinically. This discussion references repetitive vomiting and diarrhea, not a single episode, that change blood volume, potassium concentration, chloride concentration, and pH. It also builds on your previous work with renal physiology in your CPR Intersession, and fluid and electrolyte imbalance in several places in the ECE and MCE, not to mention your developing GI expertise. As you know, any time volume and/or potassium reach critical lows, the patient needs emergent attention. Notice that, in both vomiting and diarrhea, the hypovolemia decreases glomerular filtration rate (GFR) which cripples the kidneys' ability to fully participate in renal compensation for the resulting electrolyte disturbances. This, of course, emphasizes the importance of volume replacement as a mainstay of treatment, with attention to replacing not just the quantity but also the type of fluid that is being lost.
Fig 1: Excess HCl Loss & Secretion
Causes Metabolic Alkalosis
Vomiting typically causes hypovolemia, hypokalemia, hypochloremia, and metabolic alkalosis. Essentially, these are manifestations that result from what’s lost in the emesis. Fluid that would have been absorbed during its transit through the GI tract is coming “up and out” rather than “down and through” causing net fluid loss, not to mention the fact that intake is probably decreased as well. Since that fluid contains potassium and HCl (gastric acid), it’s easy to see how repetitive vomiting can cause both a metabolic alkalosis (loss of H+) as well as hypokalemia and hypochloremia. However, there’s an additional mechanism at play that further explains the metabolic alkalosis of vomiting; you could even argue that it provides the primary rationale for this acid-base disturbance. Figure 1 reminds us that the hydrogen that goes into a molecule of HCl is generated in a gastric parietal cell by the carbonic anhydrase reaction. This reaction also creates a bicarbonate anion. As the hydrogen is pumped across the apical membrane into the stomach, bicarb’s basolateral exchange with chloride not only brings in chloride that will exit the other side via a CFTR channel, but it also causes bicarb to be secreted into the bloodstream. Take a minute here to follow those steps in Figure 1, starting with the carbonic anhydrase reaction and then tracing the paths of hydrogen, bicarb, and chloride. This basolateral secretion of HCO 3− into the bloodstream is sometimes called the “alkaline tide” which normally accompanies gastric acid secretion with eating. The metabolic alkalosis of vomiting is essentially an exaggeration of this response; HCl is lost from the stomach so more is made to replace it, causing even more bicarb to be secreted "out the back end" of the parietal cell and into the bloodstream. As long as the patient has sufficient respiratory function, respiratory compensation for the metabolic alkalosis can help bring pH back down towards normal. Think though what that compensation will be, and check your ideas here.
Diarrhea typically causes hypovolemia, hypokalemia, hyperchloremia, and metabolic acidosis. Read that one more time and identify the two main differences between the systemic disruptions of diarrhea and those of vomiting. You’ll be more likely to remember them if you identify them yourself. We’ll wait… You found them, didn’t you? Diarrhea and vomiting typically have opposite results when it comes to acid base and chloride disturbance, with diarrhea causing hyperchloremic metabolic acidosis. Moving on to mechanisms, you can probably think through why diarrhea causes hypokalemia and metabolic acidosis. As emphasized in this week's previous Key Concept on GI absorption and secretion,bicarb is secreted along most of the length of the intestines, creating stool that is just a bit alkaline. The greater the stool volume as with diarrhea, the more significant the alkali loss, and net loss of base leaves the bloodstream acidic. We also emphasized that the colon is the only segment that secretes potassium, and that this secretion is amped up at times of increased luminal flow and when the patient’s vascular volume is low. Fluid loss through diarrhea commonly causes dehydration and hypovolemia which, as you know, activates RAAS. This results in secretion of aldosterone by the adrenal glands which exacerbates potassium loss both through the kidneys via urine (in exchange for sodium) and through colonic secretion. Along the lines of acid-base compensation for the metabolic acidosis, respiratory compensation can help bring pH back down towards normal. Think though what that compensation will be, then check your ideas here.
Having checked hypokalemia and metabolic acidosis off our “to understand” list, we’re just left with hyperchloremia which actually relates back to the metabolic acidosis. Remember that thing called anion gap? It’s relevant here and is used clinically to help distinguish among causes of metabolic acidosis. Think about what the name tells you—anion gap. It means that there are anions that we don’t usually measure, but we know they’re there, helping to balance the positive charge of sodium. We call these “unmeasured anions” and they cause a “gap” between the main cation (sodium) and anions that we usually do measure (chloride and bicarb). We know that those anions are there, but they are not accounted for in a routine electrolyte analysis. In the US, this is how we calculate the serum anion gap which has a normal range hovering around 8 mEq/L (but will vary a bit among labs):
serum anion gap = [Na+] - ([Cl-] + [HCO 3−])
You can see that we add the concentrations of the anions and subtract that sum from the concentration of sodium—"the plus minus the minuses." Since that anion gap normally represents mostly albumin (which is abundant in the bloodstream and carries a negative charge), there are corrections for hypoalbuminemia that we won’t go into here.
Having briefly reviewed anion gap, we can get back to the task at hand—understanding the hyperchloremia that accompanies the metabolic acidosis caused by diarrhea. In general, there are three ways that a patient can develop a metabolic acidosis which, by definition, means low bicarb concentration and low pH:
through acid ingestion or creation (e.g. salicylate/aspirin overdose, lactic acidosis, ketoacidosis—those “MUDPILES” causes);
through GI bicarb loss (diarrhea); and
through renal acid retention (renal tubular acidosis).
The first mechanism of acid ingestion or creation causes anion gap to be high. This means that the acid added to the bloodstream has a negative charge that’s balancing the positive charge of sodium and replacing the low bicarb. So, pH is low and bicarb is low (by definition of metabolic acidosis) and what’s been added to the bloodstream is both acidic and carries a negative charge, causing acidosis and high anion gap. However, diarrhea directly causes bicarb loss and the loss of bicarb is balanced by a net gain of chloride, creating a normal anion gap metabolic acidosis. Here, the additional chloride anions balance the positive charge of sodium (as opposed to unmeasured anions like aspirin or lactic acid) and the patient is hyperchloremic. Many sources use hyperchloremic metabolic acidosis and normal anion gap metabolic acidosis as interchangeable terms. Why is chloride high with diarrhea or with renal tubular acidosis? As stated by Kraut and Madias, “gastrointestinal or urinary NaHCO 3 loss can lead to a deficit of Na+ and a reduction in extracellular fluid volume, which stimulates renal retention of Na+ and Cl−, the lost HCO 3− thereby being replaced by the retained Cl−.”4 Thus, with renal tubular acidosis, the kidneys create the hyperchloremia while working to correct the hypovolemia and sodium deficit.
Fig 2: Chloride Concentration in
High vs Normal Anion Gap
While we’re nowhere near a thousand words on this topic of chloride in normal vs high anion gap, we can still look at one of those pictures that’s worth a thousand words to drive our point home. Figure 2 shows a graphic representation of anion gap in metabolic acidosis with the “minuses” stacked up and visually balancing the “plus” of sodium. Analyze each ion's colored section and follow the changes from normal to each type of acidosis—first bicarb, then anion gap, then chloride. As defined by a metabolic acidosis, bicarb is decreased in both normal anion gap and high anion gap, though the cause of its decrease differs.In high anion gap, negatively charged, unmeasured molecules have been added to the bloodstream; they take the place of the low bicarb to balance the positive charge of sodium, thus increasing the gap. However, in a normal anion gap metabolic acidosis, it’s excess chloride that’s balancing the positive charge of sodium. Since we account for chloride in the anion gap equation, the gap remains normal. Whether your patient has diarrhea or diabetic ketoacidosis, next time you analyze the labs for a patient with metabolic acidosis, be sure tomind the gap.
References
Emmett, M., & Palmer, B. F. Acid-base and electrolyte abnormalities with diarrhea. UpToDate, Topic 2320 (Version 13.0), Waltham, MA. (Accessed on October 3, 2017, subscription needed. MSU does not have a site license for UpToDate.)
Emmett, M., & Szerlip, H. Approach to the adult with metabolic acidosis. UpToDate, Topic 2291 (Version 28.0), Waltham, MA. (Accessed on October 3, 2017, subscription needed. MSU does not have a site license for UpToDate.)
Gennari, F. J. & Weise, W. J. Acid Base Disturbances in Gastrointestinal Disease. Clinical journal of the american society of nephrology: CJASN, 3, 1861–1868, 2008. doi: 10.2215/CJN.02450508
Kraut, J. A., & Madias, N. E. (2012). Differential Diagnosis of Nongap Metabolic Acidosis: Value of a Systematic Approach. Clinical journal of the american society of nephrology: CJASN, 7(4), 71–679.
Kumar, V., Abbas, A. K., & Aster, J. C. (2021). Robbins &Cotran pathologic basis of disease (10th ed.). Philadelphia, PA: Elsevier/Saunders.
Marino, P. L. (2014). The ICU book (4th ed.). Philadelphia, PA: Wolters Kluwer Health/Lippincott Williams & Wilkins.
Mulroney, S. E., & Myers, A. K. (2016). Netter's essential physiology (2nd ed.). Philadelphia, PA: Elsevier.
Costanzo, L. S. (2022). Costanzo Physiology(7th ed.). Philadelphia, PA: Elsevier.
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Jana Simmons, Ph.D.
Department of Biochemistry & Molecular Biology
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1504 | https://terpconnect.umd.edu/~wbreslyn/chemistry/isotopes/isotopes-notation.html | Isotopes: Naming and Notation
Understanding Isotopes ----------------------
Naming and Notation
Intro to Isotopes
A = N + Z Calcs
Abundance
Notation & Naming
Isotopes vs Ions
Uses
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More Chem Help
Notation and Naming (AZE notation)
In chemistry naming and notation are essential for clear communication. There are three common ways we can represent an element.
Periodic TableHyphen NotationNuclear Notation
Carbon-12 or C-12
(6 neutrons) 12 6 C 6 12 C
Carbon-13 or C-13
(7 neutrons) 13 6 C 6 13 C
Carbon-14 or C-14
(8 neutrons) 14 6 C 6 14 C
Note: in hyphen notation, the number after the hyphen is the mass number (protons + neutrons).
For the Periodic Table, the Atomic Number is on top and the average atomic mass is on the bottom.
For nuclear notation, the mass number of the isotope goes on top and the atomic number goes on the bottom.
Sometimes the Atomic Number is omitted from the nuclear notation since we already know Carbon has six protons from the atomic number on the periodic table.
Exercise: Write the Nuclear Notation for the three stable isotopes of Neon which have 10, 11, and 12 neutrons. imgage Ne Neon-20 Neon-21 Neon-22 Show Answer Answer: 20 10 N e 10 20 N e, 21 10 N e 10 21 N e, 22 10 N e 10 22 N e
Exercise: Given the hyphen notation, write the nuclear notation. Refer to the periodic table as needed.
Chlorine-34, Li-7, and H-2
Show Answer
Answer: 35 17 C l 17 35 C l, 7 3 L i 3 7 L i, 2 1 L i 1 2 L i
Hydrogen is often given as an example of isotopes. Note that the three stable isotopes of Hydrogen are given special names. No other isotopes have special names.
Protium is the only isotope that exists without any neutrons. It has one proton and one electron.
Exercise 3: How is Hydrogen different from other elements?
Show Answer
Answer: It is given specific names (protium, deuterium, and tritium) and protium, Hydrogen-1, 1 1 H 1 1 H , has no neutrons.
Next Up: Isotopes vs Ions
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1506 | https://www.masterorganicchemistry.com/2010/09/22/five-key-factors-that-influence-acidity/ | Skip to content
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Home / Five Key Factors That Influence Acidity
Acid Base Reactions
By James Ashenhurst
Five Key Factors That Influence Acidity
Last updated: February 12th, 2025 |
Five Key Factors That Affect Acidity
Today we’ll talk about what’s behind the trends in acidity for different molecules and discuss the most important factors that determine these values.
I’ve written in schoolmarmish tones before about how pKa is one of the most important measures you can learn in organic chemistry, and not knowing some basic pKa values before an exam is a lot like walking up to a poker table without knowing the values of the hands: you’re going to lose your shirt. (See article: Know Your pKas)
Let’s quickly review the basics of acidity and basicity. Here’s the condensed version:
Bronsted acids are proton donors, Lewis acids are electron pair acceptors. Converse: Brønsted base = proton acceptor, Lewis base = electron pair donor.
A conjugate base is what you obtain when you remove a proton (H+) from a compound. For instance, HO– is the conjugate base of water. O2- is the conjugate base of HO–. Conversely, conjugate acids are what you obtain when you add a proton to a compound. The conjugate acid of water is H3O+.
Quick quiz: is pH 1 acidic or basic? pKa is similar to pH in that low (and even negative values) denote strong acids. That’s because pKa is based on the equilibrium:
According to this, anything which stabilizes the conjugate base will increase the acidity. Therefore pKa is also a measure of how stable the conjugate base is. Put another way, strong acids have weak conjugate bases, and vice versa.
With that out of the way, let’s get started.
Table of Contents
Factor #1 – Charge.
Factor #2 – The Role of the Atom
Factor #3 – Resonance
Factor #4 – Inductive effects
Factor #5 – Orbitals
Quiz Yourself!
Notes
1. Factor #1 – Charge.
Removal of a proton, H+ , decreases the formal charge on an atom or molecule by one unit. This is, of course, easiest to do when an atom bears a charge of +1 in the first place, and becomes progressively more difficult as the overall charge becomes negative. The acidity trends reflect this:
Note that once a conjugate base (B-) is negative, a second deprotonation will make the dianion (B 2-). While far from impossible, forming the dianion can be difficult due to the buildup of negative charge and the corresponding electronic repulsions that result.
2. Factor #2 – The Role of the Atom
This point causes a lot of confusion due to the presence of two seemingly conflicting trends.
Here’s the first point: acidity increases as we go across a row in the periodic table. This makes sense, right? It makes sense that HF is more electronegative than H2O, NH3, and CH4 due to the greater electronegativity of fluorine versus oxygen, nitrogen, and carbon. A fluorine bearing a negative charge is a happy fluorine.
But here’s the seemingly strange thing. HF itself is not a “strong” acid, at least not in the sense that it ionizes completely in water. HF is a weaker acid than HCl, HBr, and HI. What’s going on here?
You could make two arguments for why this is. The first reason has to do with the shorter (and stronger) H-F bond as compared to the larger hydrogen halides.
The second has to do with the stability of the conjugate base. The fluoride anion, F(–) is a tiny and vicious little beast, with the smallest ionic radius of any other ion bearing a single negative charge. Its charge is therefore spread over a smaller volume than those of the larger halides, which is energetically unfavorable: for one thing, F(–) begs for solvation, which will lead to a lower entropy term in the ΔG.
Note that this trend also holds for H2O and H2S, with H2S being about 10 million times more acidic.
3. Factor #3 – Resonance
A huge stabilizing factor for a conjugate base is if the negative charge can be delocalized through resonance. The classic examples are with phenol (C6H5OH) which is about a million times more acidic than water, and with acetic acid (pKa of ~4).
Watch out though – it isn’t enough for a π system to simply be adjacent to a proton – the electrons of the conjugate base have to be in an orbital which allows for effective overlap.
4. Factor #4 – Inductive effects
Electronegative atoms can draw negative charge toward themselves, which can lead to considerable stabilization of conjugate bases. Check out these examples:
Predictably, this effect is going to be related to two major factors: 1) the electronegativity of the element (the more electronegative, the more acidic) and the distance between the electronegative element and the negative charge.
5. Factor #5 – Orbitals
Again, the acidity relates nicely to the stability of the conjugate base. And the stability of the conjugate base depends on how well it can accomodate its newfound pair of electrons. In an effect akin to electronegativity, the more s character in the orbital, the closer the electrons will be to the nucleus, and the lower in energy (= stable! ) they will be.
Look at the difference between the pKa of acetylene and alkanes: 25! That’s 10 to the power of 25, as in, “100 times bigger than Avogadro’s number”.
Just to give you an idea of scale. That’s the amazing thing about chemistry – the sheer range in the power of different phenomena is awe-inspiring. (See article: pKa Values Span 60 Orders Of Magnitude)
There’s actually a mnemonic I’ve found that can help you remember these effects. This is credited to Dr. Christine Pruis, Senior Lecturer at Arizona State University Tempe.
Charge
Atom
Resonance
Dipole Induction
Orbitals
= CARDIO.
Tread carefully with mnemonics, but there you go.
Quiz Yourself!
Become a MOC member to see the clickable quiz with answers on the back.
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Notes
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Allylic Bromination
Bonus Topic: Allylic Rearrangements
In Summary: Free Radicals
Synthesis (2) - Reactions of Alkanes
Free Radicals Practice Quizzes
07 Stereochemistry and Chirality
Types of Isomers: Constitutional Isomers, Stereoisomers, Enantiomers, and Diastereomers
How To Draw The Enantiomer Of A Chiral Molecule
How To Draw A Bond Rotation
Introduction to Assigning (R) and (S): The Cahn-Ingold-Prelog Rules
Assigning Cahn-Ingold-Prelog (CIP) Priorities (2) - The Method of Dots
Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems
Assigning R/S To Newman Projections (And Converting Newman To Line Diagrams)
How To Determine R and S Configurations On A Fischer Projection
The Meso Trap
Optical Rotation, Optical Activity, and Specific Rotation
Optical Purity and Enantiomeric Excess
What's a Racemic Mixture?
Chiral Allenes And Chiral Axes
Stereochemistry Practice Problems and Quizzes
08 Substitution Reactions
Nucleophilic Substitution Reactions - Introduction
Two Types of Nucleophilic Substitution Reactions
The SN2 Mechanism
Why the SN2 Reaction Is Powerful
The SN1 Mechanism
The Conjugate Acid Is A Better Leaving Group
Comparing the SN1 and SN2 Reactions
Polar Protic? Polar Aprotic? Nonpolar? All About Solvents
Steric Hindrance is Like a Fat Goalie
Common Blind Spot: Intramolecular Reactions
Substitution Practice - SN1
Substitution Practice - SN2
09 Elimination Reactions
Elimination Reactions (1): Introduction And The Key Pattern
Elimination Reactions (2): The Zaitsev Rule
Elimination Reactions Are Favored By Heat
Two Elimination Reaction Patterns
The E1 Reaction
The E2 Mechanism
E1 vs E2: Comparing the E1 and E2 Reactions
Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings
Bulky Bases in Elimination Reactions
Comparing the E1 vs SN1 Reactions
Elimination (E1) Reactions With Rearrangements
E1cB - Elimination (Unimolecular) Conjugate Base
Elimination (E1) Practice Problems And Solutions
Elimination (E2) Practice Problems and Solutions
10 Rearrangements
Introduction to Rearrangement Reactions
Rearrangement Reactions (1) - Hydride Shifts
Carbocation Rearrangement Reactions (2) - Alkyl Shifts
Pinacol Rearrangement
The SN1, E1, and Alkene Addition Reactions All Pass Through A Carbocation Intermediate
11 SN1/SN2/E1/E2 Decision
Identifying Where Substitution and Elimination Reactions Happen
Deciding SN1/SN2/E1/E2 (1) - The Substrate
Deciding SN1/SN2/E1/E2 (2) - The Nucleophile/Base
SN1 vs E1 and SN2 vs E2 : The Temperature
Deciding SN1/SN2/E1/E2 - The Solvent
Wrapup: The Key Factors For Determining SN1/SN2/E1/E2
Alkyl Halide Reaction Map And Summary
SN1 SN2 E1 E2 Practice Problems
12 Alkene Reactions
E and Z Notation For Alkenes (+ Cis/Trans)
Alkene Stability
Alkene Addition Reactions: "Regioselectivity" and "Stereoselectivity" (Syn/Anti)
Stereoselective and Stereospecific Reactions
Hydrohalogenation of Alkenes and Markovnikov's Rule
Hydration of Alkenes With Aqueous Acid
Rearrangements in Alkene Addition Reactions
Halogenation of Alkenes and Halohydrin Formation
Oxymercuration Demercuration of Alkenes
Hydroboration Oxidation of Alkenes
m-CPBA (meta-chloroperoxybenzoic acid)
OsO4 (Osmium Tetroxide) for Dihydroxylation of Alkenes
Palladium on Carbon (Pd/C) for Catalytic Hydrogenation of Alkenes
Cyclopropanation of Alkenes
A Fourth Alkene Addition Pattern - Free Radical Addition
Alkene Reactions: Ozonolysis
Summary: Three Key Families Of Alkene Reaction Mechanisms
Synthesis (4) - Alkene Reaction Map, Including Alkyl Halide Reactions
Alkene Reactions Practice Problems
13 Alkyne Reactions
Acetylides from Alkynes, And Substitution Reactions of Acetylides
Partial Reduction of Alkynes With Lindlar's Catalyst
Partial Reduction of Alkynes With Na/NH3 To Obtain Trans Alkenes
Alkyne Hydroboration With "R2BH"
Hydration and Oxymercuration of Alkynes
Hydrohalogenation of Alkynes
Alkyne Halogenation: Bromination and Chlorination of Alkynes
Oxidation of Alkynes With O3 and KMnO4
Alkenes To Alkynes Via Halogenation And Elimination Reactions
Alkynes Are A Blank Canvas
Synthesis (5) - Reactions of Alkynes
Alkyne Reactions Practice Problems With Answers
14 Alcohols, Epoxides and Ethers
Alcohols - Nomenclature and Properties
Alcohols Can Act As Acids Or Bases (And Why It Matters)
Alcohols - Acidity and Basicity
The Williamson Ether Synthesis
Ethers From Alkenes, Tertiary Alkyl Halides and Alkoxymercuration
Alcohols To Ethers via Acid Catalysis
Cleavage Of Ethers With Acid
Epoxides - The Outlier Of The Ether Family
Opening of Epoxides With Acid
Epoxide Ring Opening With Base
Making Alkyl Halides From Alcohols
Tosylates And Mesylates
PBr3 and SOCl2
Elimination Reactions of Alcohols
Elimination of Alcohols To Alkenes With POCl3
Alcohol Oxidation: "Strong" and "Weak" Oxidants
Demystifying The Mechanisms of Alcohol Oxidations
Protecting Groups For Alcohols
Thiols And Thioethers
Calculating the oxidation state of a carbon
Oxidation and Reduction in Organic Chemistry
Oxidation Ladders
SOCl2 Mechanism For Alcohols To Alkyl Halides: SN2 versus SNi
Alcohol Reactions Roadmap (PDF)
Alcohol Reaction Practice Problems
Epoxide Reaction Quizzes
Oxidation and Reduction Practice Quizzes
15 Organometallics
What's An Organometallic?
Formation of Grignard and Organolithium Reagents
Organometallics Are Strong Bases
Reactions of Grignard Reagents
Protecting Groups In Grignard Reactions
Synthesis Problems Involving Grignard Reagents
Grignard Reactions And Synthesis (2)
Organocuprates (Gilman Reagents): How They're Made
Gilman Reagents (Organocuprates): What They're Used For
The Heck, Suzuki, and Olefin Metathesis Reactions (And Why They Don't Belong In Most Introductory Organic Chemistry Courses)
Reaction Map: Reactions of Organometallics
Grignard Practice Problems
16 Spectroscopy
Degrees of Unsaturation (or IHD, Index of Hydrogen Deficiency)
Conjugation And Color (+ How Bleach Works)
Introduction To UV-Vis Spectroscopy
UV-Vis Spectroscopy: Absorbance of Carbonyls
UV-Vis Spectroscopy: Practice Questions
Bond Vibrations, Infrared Spectroscopy, and the "Ball and Spring" Model
Infrared Spectroscopy: A Quick Primer On Interpreting Spectra
IR Spectroscopy: 4 Practice Problems
1H NMR: How Many Signals?
Homotopic, Enantiotopic, Diastereotopic
Diastereotopic Protons in 1H NMR Spectroscopy: Examples
13-C NMR - How Many Signals
Liquid Gold: Pheromones In Doe Urine
Natural Product Isolation (1) - Extraction
Natural Product Isolation (2) - Purification Techniques, An Overview
Structure Determination Case Study: Deer Tarsal Gland Pheromone
17 Dienes and MO Theory
What To Expect In Organic Chemistry 2
Are these molecules conjugated?
Conjugation And Resonance In Organic Chemistry
Bonding And Antibonding Pi Orbitals
Molecular Orbitals of The Allyl Cation, Allyl Radical, and Allyl Anion
Pi Molecular Orbitals of Butadiene
Reactions of Dienes: 1,2 and 1,4 Addition
Thermodynamic and Kinetic Products
More On 1,2 and 1,4 Additions To Dienes
s-cis and s-trans
The Diels-Alder Reaction
Cyclic Dienes and Dienophiles in the Diels-Alder Reaction
Stereochemistry of the Diels-Alder Reaction
Exo vs Endo Products In The Diels Alder: How To Tell Them Apart
HOMO and LUMO In the Diels Alder Reaction
Why Are Endo vs Exo Products Favored in the Diels-Alder Reaction?
Diels-Alder Reaction: Kinetic and Thermodynamic Control
The Retro Diels-Alder Reaction
The Intramolecular Diels Alder Reaction
Regiochemistry In The Diels-Alder Reaction
The Cope and Claisen Rearrangements
Electrocyclic Reactions
Electrocyclic Ring Opening And Closure (2) - Six (or Eight) Pi Electrons
Diels Alder Practice Problems
Molecular Orbital Theory Practice
18 Aromaticity
Introduction To Aromaticity
Rules For Aromaticity
Huckel's Rule: What Does 4n+2 Mean?
Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems
Antiaromatic Compounds and Antiaromaticity
The Pi Molecular Orbitals of Benzene
The Pi Molecular Orbitals of Cyclobutadiene
Frost Circles
Aromaticity Practice Quizzes
19 Reactions of Aromatic Molecules
Electrophilic Aromatic Substitution: Introduction
Activating and Deactivating Groups In Electrophilic Aromatic Substitution
Electrophilic Aromatic Substitution - The Mechanism
Ortho-, Para- and Meta- Directors in Electrophilic Aromatic Substitution
Understanding Ortho, Para, and Meta Directors
Why are halogens ortho- para- directors?
Disubstituted Benzenes: The Strongest Electron-Donor "Wins"
Electrophilic Aromatic Substitutions (1) - Halogenation of Benzene
Electrophilic Aromatic Substitutions (2) - Nitration and Sulfonation
EAS Reactions (3) - Friedel-Crafts Acylation and Friedel-Crafts Alkylation
Intramolecular Friedel-Crafts Reactions
Nucleophilic Aromatic Substitution (NAS)
Nucleophilic Aromatic Substitution (2) - The Benzyne Mechanism
Reactions on the "Benzylic" Carbon: Bromination And Oxidation
The Wolff-Kishner, Clemmensen, And Other Carbonyl Reductions
More Reactions on the Aromatic Sidechain: Reduction of Nitro Groups and the Baeyer Villiger
Aromatic Synthesis (1) - "Order Of Operations"
Synthesis of Benzene Derivatives (2) - Polarity Reversal
Aromatic Synthesis (3) - Sulfonyl Blocking Groups
Birch Reduction
Synthesis (7): Reaction Map of Benzene and Related Aromatic Compounds
Aromatic Reactions and Synthesis Practice
Electrophilic Aromatic Substitution Practice Problems
20 Aldehydes and Ketones
What's The Alpha Carbon In Carbonyl Compounds?
Nucleophilic Addition To Carbonyls
Aldehydes and Ketones: 14 Reactions With The Same Mechanism
Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones
Grignard Reagents For Addition To Aldehydes and Ketones
Wittig Reaction
Hydrates, Hemiacetals, and Acetals
Imines - Properties, Formation, Reactions, and Mechanisms
All About Enamines
Breaking Down Carbonyl Reaction Mechanisms: Reactions of Anionic Nucleophiles (Part 2)
Aldehydes Ketones Reaction Practice
21 Carboxylic Acid Derivatives
Nucleophilic Acyl Substitution (With Negatively Charged Nucleophiles)
Addition-Elimination Mechanisms With Neutral Nucleophiles (Including Acid Catalysis)
Basic Hydrolysis of Esters - Saponification
Transesterification
Proton Transfer
Fischer Esterification - Carboxylic Acid to Ester Under Acidic Conditions
Lithium Aluminum Hydride (LiAlH4) For Reduction of Carboxylic Acid Derivatives
LiAlH[Ot-Bu]3 For The Reduction of Acid Halides To Aldehydes
Di-isobutyl Aluminum Hydride (DIBAL) For The Partial Reduction of Esters and Nitriles
Amide Hydrolysis
Thionyl Chloride (SOCl2) And Conversion of Carboxylic Acids to Acid Halides
Diazomethane (CH2N2)
Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One
Making Music With Mechanisms (PADPED)
Carboxylic Acid Derivatives Practice Questions
22 Enols and Enolates
Keto-Enol Tautomerism
Enolates - Formation, Stability, and Simple Reactions
Kinetic Versus Thermodynamic Enolates
Aldol Addition and Condensation Reactions
Reactions of Enols - Acid-Catalyzed Aldol, Halogenation, and Mannich Reactions
Claisen Condensation and Dieckmann Condensation
Decarboxylation
The Malonic Ester and Acetoacetic Ester Synthesis
The Michael Addition Reaction and Conjugate Addition
The Robinson Annulation
Haloform Reaction
The Hell–Volhard–Zelinsky Reaction
Enols and Enolates Practice Quizzes
23 Amines
The Amide Functional Group: Properties, Synthesis, and Nomenclature
Basicity of Amines And pKaH
5 Key Basicity Trends of Amines
The Mesomeric Effect And Aromatic Amines
Nucleophilicity of Amines
Alkylation of Amines (Sucks!)
Reductive Amination
The Gabriel Synthesis
Some Reactions of Azides
The Hofmann Elimination
The Hofmann and Curtius Rearrangements
The Cope Elimination
Protecting Groups for Amines - Carbamates
The Strecker Synthesis of Amino Acids
Introduction to Peptide Synthesis
Reactions of Diazonium Salts: Sandmeyer and Related Reactions
Amine Practice Questions
24 Carbohydrates
D and L Notation For Sugars
Pyranoses and Furanoses: Ring-Chain Tautomerism In Sugars
What is Mutarotation?
Reducing Sugars
The Big Damn Post Of Carbohydrate-Related Chemistry Definitions
The Haworth Projection
Converting a Fischer Projection To A Haworth (And Vice Versa)
Reactions of Sugars: Glycosylation and Protection
The Ruff Degradation and Kiliani-Fischer Synthesis
Isoelectric Points of Amino Acids (and How To Calculate Them)
Carbohydrates Practice
Amino Acid Quizzes
25 Fun and Miscellaneous
A Gallery of Some Interesting Molecules From Nature
Screw Organic Chemistry, I'm Just Going To Write About Cats
On Cats, Part 1: Conformations and Configurations
On Cats, Part 2: Cat Line Diagrams
On Cats, Part 4: Enantiocats
On Cats, Part 6: Stereocenters
Organic Chemistry Is Shit
The Organic Chemistry Behind "The Pill"
Maybe they should call them, "Formal Wins" ?
Why Do Organic Chemists Use Kilocalories?
The Principle of Least Effort
Organic Chemistry GIFS - Resonance Forms
Reproducibility In Organic Chemistry
What Holds The Nucleus Together?
How Reactions Are Like Music
Organic Chemistry and the New MCAT
26 Organic Chemistry Tips and Tricks
Common Mistakes: Formal Charges Can Mislead
Partial Charges Give Clues About Electron Flow
Draw The Ugly Version First
Organic Chemistry Study Tips: Learn the Trends
The 8 Types of Arrows In Organic Chemistry, Explained
Top 10 Skills To Master Before An Organic Chemistry 2 Final
Common Mistakes with Carbonyls: Carboxylic Acids... Are Acids!
Planning Organic Synthesis With "Reaction Maps"
Alkene Addition Pattern #1: The "Carbocation Pathway"
Alkene Addition Pattern #2: The "Three-Membered Ring" Pathway
Alkene Addition Pattern #3: The "Concerted" Pathway
Number Your Carbons!
The 4 Major Classes of Reactions in Org 1
How (and why) electrons flow
Grossman's Rule
Three Exam Tips
A 3-Step Method For Thinking Through Synthesis Problems
Putting It Together
Putting Diels-Alder Products in Perspective
The Ups and Downs of Cyclohexanes
The Most Annoying Exceptions in Org 1 (Part 1)
The Most Annoying Exceptions in Org 1 (Part 2)
The Marriage May Be Bad, But the Divorce Still Costs Money
9 Nomenclature Conventions To Know
Nucleophile attacks Electrophile
27 Case Studies of Successful O-Chem Students
Success Stories: How Corina Got The The "Hard" Professor - And Got An A+ Anyway
How Helena Aced Organic Chemistry
From a "Drop" To B+ in Org 2 – How A Hard Working Student Turned It Around
How Serge Aced Organic Chemistry
Success Stories: How Zach Aced Organic Chemistry 1
Success Stories: How Kari Went From C– to B+
How Esther Bounced Back From a "C" To Get A's In Organic Chemistry 1 And 2
How Tyrell Got The Highest Grade In Her Organic Chemistry Course
This Is Why Students Use Flashcards
Success Stories: How Stu Aced Organic Chemistry
How John Pulled Up His Organic Chemistry Exam Grades
Success Stories: How Nathan Aced Organic Chemistry (Without It Taking Over His Life)
How Chris Aced Org 1 and Org 2
Interview: How Jay Got an A+ In Organic Chemistry
How to Do Well in Organic Chemistry: One Student's Advice
"America's Top TA" Shares His Secrets For Teaching O-Chem
"Organic Chemistry Is Like..." - A Few Metaphors
How To Do Well In Organic Chemistry: Advice From A Tutor
Guest post: "I went from being afraid of tests to actually looking forward to them".
Comments
Comment section
136 thoughts on “Five Key Factors That Influence Acidity”
Thank you for your work! From Brazil.
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3. Hello,
I believe I have told you this before, but it is well worth repeating: this is an amazing website! Thank you for all the work you have done preparing it.
I am the guy scouring the internet trying to clen up the confusion about the pKa of water and the pKa of H3O+ and the pKb of OH-. I see that you have the correct value of 14 for the pKa of water, but I have noticed on this page in section 1 and in the table on that you have the pKa of H3O+ as -1.74. It is actually 0.00 (Because the pKb (and pKa) of water is 14.00, then the pKa of water’s conjugate acid must be 0.00.) The pKb of OH- is also 0.00.
We just published an open access paper in Helvetica Chimica Acta ( in which we show that the operation of a combination pH electrode is based on the fact that the pKa of H3O+ is 0.00. This paper also gives a thorough description of many other aspects of the pKa of water discussion. It is worth a read!
Thanks for listening!
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4. Pingback: Rank the following solutions in order from most acidic to least acidic - studyhw
5. can you able to explain everything based on nature that around us?
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6. can you able to explain everything based on nature that around us
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7. Please let me know the reason for increase in acidity of carboxylic acid when a vinyl group directly attaches to it which is contrary to resonance effect.
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It isn’t resonance. It’s an inductive effect, essentially.
Alkene carbons are sp2 hybridized and the electrons are held slightly closer to the nucleus. This has the effect of making sp2 hybridized carbons effectively more electronegative than sp3 hybridized carbons.
The effect is even stronger for alkynes (sp hybridized carbons) which have even higher effective electronegativity. For example the pka of propiolic acid is about 1.8.
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8. One more question…How am i supposed to determine that which reason is correct for reasoning the acidity?Like I got it wrong in above mentioned case!
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9. Thank you
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10. H-I is more acidic than H-F because of longer bond length but in inductive effect OHCOCH2F is more acidic why???
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1. In comparing H-I and H-F, when we lose a proton the charge will be directly on the atom (I- or F-). In that case it’s more stable on I(-) because as we go down a column of the periodic table the atoms get larger and can spread out their charge over a larger volume (polarizability).
In the case of the alcohol, the negative charge is actually on oxygen so polarizability is not a key difference here. The fluorine is electronegative and helps to reduce the amount of negative charge on oxygen through induction.
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Pingback: The Amino Acids – fauxmat
If acidity increases with increasing positive charge on an atom, is it true that acidity decreases with increasing negative charge on an atom? Carbocations I take it would not increase in acidity because they lack a full octet even with a positive charge?
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14. Quick question.What is the difference in this article from the one previously which describes the seven factors that affect stability of a negative charge?Both seem to be saying the same thing ,but this article does not have the aromaticity part included nor does it say anything on lower charge density stabilizing negative charge.
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15. you just helped me save a whole day in studying acid/basic trends into 30 minutes of the best content review I’ve ever come across. You are an absolute GODSEND.
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Glad you found it useful Jane!!!
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16. So, the PKa of H3O+ is 14-14=0?
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17. Is the pka of water 14? Not 15.8?
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Yes, the pKa of water is 14.
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18. Your clear and concise explanations are awesome. Hats off to your efforts.
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Happy you find it helpful, Raghav.
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19. I have a doubt:when a polar protic solvent is used as a solvent for its conjugate base, will the conjugate base be the strongest base/nucleophile in that solvent?Thanks!
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20. I think in the second paragraph of factor #2, you meant, “It makes sense that HF is more acidic (not electronegative) than H2O, NH3, and CH4 due to the greater electronegativity of fluorine versus oxygen, nitrogen, and carbon.”
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21. This is the best explanation that i ever got. Thank you very much. At last i found someone who could clear my confusions thanks again.
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Glad it was helpful Raj!
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22. Hi is this mnemonic in order of decreasing effect? Meaning if there were two molecules that I’m trying to decide which one is more acidic, can I say that since molecule A wins out on the dipole induction despite that molecule B wins out on having more “s” character, molecule A is more acidic?
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It’s difficult to estimate the magnitude of different effects directly. That’s why pKas are measured. What you should look for is to identify trends within one of these key factors. However if you want to share an example in the comments, that’s cool.
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23. Thank you so much, this was so clear and concise and exactly what I needed to review for my exam!
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Excellent, glad it was useful Christine!
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24. Thank you! This helps a lot.
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25. You have done a great job. I was so confused with this topic and after reading reading your explanation my concepts were very clear. Thanks a lot ! ?
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26. Sir you are a fantastic teacher and I am sure that you are also a gentleman. And of course your knowledge of organic chemistry is commendable Really out of the world. Your post are very good and really a big fan of your THANKU ?
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27. Hey I want to know that why ketones are better acids than alkanes??
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Draw out the conjugate base (when you remove H+ from the carbon adjacent to C=O). You should have a lone pair that has a negative charge. Now draw the resonance form. You should be able to show that you can form a new C=C bond and the negative charge moves to the oxygen making it O- . So the negative charge has been transferred from a less electronegative atom (carbon) to a more electronegative atom (oxygen) and is therefore more stable. The more stable the negative charge, the weaker the base it is. That’s it in a nutshell. It’s a stronger acid because the negative charge of the conjugate base is weaker.
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28. Taking my DAT in 2 days, and I kept getting these problems wrong on both the Inorganic and Organic chemistry sections… This page cleared it all up!
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Glad to hear it, and wish you luck on your DAT! – James
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29. YOU’RE DOING GODS WORK SON
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30. I was totally confused all this time. This teqnique is really helpful thanks a lot
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31. HCOOH, CH3COOH, CH3CH2COOH. Sir, Would you please tell me the order of increasing acidic strength in these acids? It appeared in my exam and i have a doubt in it. Pleaseee help
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What is the effect of electron donors on the stability of the conjugate base? Would they increase the negative charge or decrease the negative charge? Would this make the negative charge more stable or less stable?
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32. I just wanted to say thank you for making this information easy to understand! Your website has helped me SO MUCH when I study for my exams. You are the best.
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Thank you Tori, I’m very pleased you find the site useful!
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33. this is amazing. So much better than my textbook. Thank you so much!
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34. why R-CH2-CN is less acidic than R-CH2-NH2?
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?? Assuming R is alkyl, pKa of R-CH2-CN should be about 30 (DMSO), pKa of R-CH2-NH2 should be 35 or higher. Don’t see any evidence for your claim.
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35. My question is “variations of stability of -I effect in hybridizationwhich one of the following is correct(a). Sp>Sp2>Sp3(b). Sp3>Sp2>Sp
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I would consider an sp-hybridized C to be more electronegative than an sp2 hybridized carbon which would be more electronegative than an sp3 hybridized carbon. I hope that answers your question.
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36. This site is just awesome !! Than you sir !!! It helped me a lot !! I recommend every one this site !! Hope this site gets world recognition !!
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Thanks Raman, glad you like the site!
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37. Finally, I’ve found a website which helps to break down these concepts into comprehensible pieces. I’m finishing up Gen Chem II this semester, and we’ve already started on Orgo topics since the end of Gen Chem I, so I’m taking Orgo next. This website will surely help me to get ahead, when I study during the summer time. Thank you! :)
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Hi Sinigdha, I’m glad you are finding the website useful. Thanks for stopping by!
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38. Hi!I just want to say thank you, I’m a medicine student from israel and your explanation helped me a lot..
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Glad to hear it noazio. Thanks for letting me know!
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39. James, that acronym CARDO is genius. It really helps me understand the factors that affect acidity of molecules. Thank you so much!!
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I can’t claim credit for it, but thanks!
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40. All was awesome but I just confused in inductive effect when u compares acidities of halogen acid? As electronegative difference between F and O is less as compared to OH ? So how H lose? May be my question is low of standard as I m new kindly help
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41. for sure this site helped me to improve my understanding on this concept, thank you much. I got 100% of marks on my exam
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42. Your article has been very helpful and I now have actually a chance to pass my exams in general chemistry! I just want to ask, we have to apply these factors in sequence, as you have given them or not??? I mean first the charge, after the atom etc??
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43. thanks alot but what of mesomeric effect
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Covered here:
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2. Covered here.
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44. I am having difficulty determining the relative acidity of amines and ammonium ions. For instance:a) H3C-NH3+b) H2C=NH2+c) HCNH+d) (CH3)3NH+How does the difference in bonding between carbon and nitrogen effect the acidity?
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In this compounds, carbon atoms may exert two effects, related to the s character of nitrogen orbitals and the electron donating properties of alkyl groups.The donating effect (by hyperconjugation) of methil groups rises the energy of nitrogen’s electron pair by electron-electron repulsion, making it more basic. Combine this with the fact that sp3 hybridized nitrogen are more acidic (less s character), and we obtain the order d>a>b>c (As would Darwin says: I Think)
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45. Hi,I found your posts extremely helpful !In comparing the acidity/basicity of organic compounds, which of the following has the stronger effects?
electronegativity>resonance>s-Orbital>intramolecular force ??And how to determine the acidity of substituted aromatic compounds, such as benzoic acid?Thanks in advance.
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46. hello sir,First of all thanks and now the question- what is +R and -R and how to recognise them. Sir i am stucked please answer urgently.
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47. Dude, it’s just awesome , people like you helps student to learn concepts rather then mugging formulas . I am an Indian , preparing for IIT JEE exam and recommend this blog for every jee aspirants……
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48. May God bless you sir .My sister had her exams and needed help.I, although being a student of commerce stream , decided that I had to help her in every way I could. But I knew I sucked at organic chemistry in school,and so did my little sister .I found your site , and the next day I am understanding things better than I ever did , .. …I would look up about anything I didn’t know on your website .getting to the main thing , my sister scored 97 in chemistry , full marks in the organic portion.Thank you sir :’-).
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49. Can you please tell the answer for order of acidity of hydrogen in orbital section…
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50. hey james,isnt there any explanation fr basicity?
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Check out this post on amines.
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51. I love your site and it has helped me a lot :) just one doubt, is electronegativity of an atom related to it’s tendency to act as a LEWIS base. For example NH3 , and PH3,we know that NH3 is stronger but why doesn’t the electronegativity of the central atom come under consideration here? N has a high electronegativity so it’s tendency to donate lone pair of electrons should be low right? Please please please help.
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You mention two factors, electronegativity and size. From first principles it’s difficult to anticipate exactly which would be most important. However, from running experiments on reaction rates (and measuring the results) PH3 tends to be a better Lewis base (“Nucleophile”) than NH3 . The explanation is, the lone pair on phosphorus is less tightly held than that on nitrogen, so it is more “easily” donated. [This is NOT a dumb question – the issue is somewhat complex and can depend on the type of electrophile. I haven’t gotten into Hard and Soft acids and bases but I suggest you look it up]
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52. Hi I had a question. When taking inductive effects (electron withdrawin and electrons donating) into account are we looking at the conjugate base or the acid itself if we wanted to apply rules such as neutral N or O are considered electrons donors?
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You’re looking at how well the stabilize (or destabilize) the conjugate base. Anything which stabilizes the conjugate base will result in a stronger acid (and vice versa). For example HOCF3 is a stronger acid then HOCH3 because those three fluorine atoms help to stabilize the conjugate base, -OCF3 through the inductive effect.
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53. I had lot of problems with acidity. Now I have a better understand after reading this article. thank u very much!
Reply
OK! Glad to hear you find it useful Chanaka.
Reply
54. I have been looking for a thorough explanation like this for such a long time, being unable to find one. This is really helpful for a med student struggling with his Chem & Intro to Bio exam, thank you so much James.
Reply
55. I had a question if you have a moment! I understand the setup of the Ka formula, but for the life of me, I don’t understand what the numbers ARE. When you punch it into the formula, what numbers are you punching in for certain atoms and/or compounds in order to get the Ka, and then get the pKa. It’s not explained in my text, or any resources I may have. My professor is also not very great at translating our questions in class.
Reply
It’s obtained from measurement – from experiment, in other words. It can’t be just “figured out”, we have to physically do experiments that measure the equilibrium constant.
Reply
56.
Basically the same thing explained in a more elaborate ( slightly better ) way.Please know that I absolutely love what you’ve done. You’ve made organic chem a million times easier. This is just one of the VERY rare cases where I have found a better reference.
Reply
Thank you – that is indeed a nice reference.
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57. Thank you so much for creating this easy to understand article. I was about to give up until I stumbled upon this today. This will be the FIRST place I check to help me through orgo! Cheers!
Reply
58. Hi James, firstly, a VERY big thank you to you for this amazing website. Its makes organic chemistry SO easy for me to understand!
I think there might be a slight error in the ‘Orbitals’ section of this article. In the graphic, the acidity increases from sp3 to sp (rightward, ascending order) while the signs ( < ) show otherwise. I might be missing something, so please correct me if I am!
Once again, thanks!
Utkarsh
Reply
oops. Thank you. Fixed!
Reply
59. Hi, I’m confused as to how hydrogen bonding in specifically the Halides, affects the acidity.
Reply
H-bonding makes the solution tightly binded. Like in case of HF (which shows H Bonding) , all the HF molecules are tightly binded to each other so it s not easy for HF to get ionised into H+ and F-( all molecules are already happy in their state of low energy )
Reply
60. this was so helpful thank you!
Reply
61. why staboilization of conjugate base enhance acidity ?
Reply
Equilibrium tends to proceed toward the more stable product, yes? So what happens to the equilibrium HA –> H+ A- as you make A- more stable?
Reply
62. Hey! These pages have been great, where can we find the answers to the problems you usually give at the end?
Reply
63. You must be used to all the gratitude by now but thank you so much! This just made everything fall into place :)
Reply
64. Why is it that HF is more acidic than HI (#2) , but when its connected to the carboxyl group it is I that is more acidic? Anyone?
Reply
65. Thank you so much!! You have made my MCAT studying more a light jog ;)
Reply
66. why is OH- more basic than SH-. “-” is minus sign
Reply
67. Thank you so much! Love reading through everything! You make it so understandable and interesting! I really appreciate the links to more topics too!
Reply
68. Fantastic! What an explanation…. Truely helpful and admirable….
Reply
69. and for factor #5… do we look at the hybridization of the acid or the conjugate base?ex// ch3 has a hybridization orbital of sp3 but its CB ch2 has a hybridization of sp2
Reply
70. do lone pairs count when figuring out the hybridization of an orbital?
Reply
71. Just wanted to sincerely thankyou for transforming an extremely difficult subject into something comprehensible and futhermore enjoyable. There’s something to be said for that specific talent and it is greatly appreciated. Your use of acronyms and descriptive context have improved both my labs and test marks.Thank you.
Reply
if benzene ring lost a proton it is not stabilizes by resonance because no way for resonance
Reply
72. When trying to choose a compounds with the highest acidity, according to CARDIO, how do you determine which factor you should prioritize first?For example:Suppose you’re trying to determine which compound is more acidic, CHCH or benzene ring?When you remove the proton the benzene ring is stabilized by resonance but the HCC- has a lot of s character in its orbital. How would you determine which compound is more acidic?
Reply
73. i didnt get 5 point orbital as s character increases then t should be sp3
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74. This was really clear and helpful Thank you so much!!
Reply
Awesome, glad to hear it.
Reply
75. Did you get the CARDIO acronym from a Dr. Christine Pruis or Chad’s Reviews from Arizona State University? Dr. Pruis is our Organic Chemistry professor and came up with this acronym 7-8 years ago, so perhaps that is the ‘credit’ you speak of? If so, that is awesome!
Reply
76. I am curious how you came up with this CARDIO acronym? And when? I have heard it from one other person a few years ago.
Reply
I shamelessly stole it from a comment thread on SDN.
Reply
77. Organic is my favourite part in chemistry. Your post and contents provided me a good quick revision before my exams and I did extremely well…..Thanks to you….you have a good way of teaching organic chemistry..I would definitely recommend this site to my friends who think organic is boaring subject….Once again thanks…..
Reply
78. Can you please make these notes available as PDF?
Thanks.
Reply
you can use web2pdf for that……google it…it also has browser plugin..
Reply
79. This is awesome :D ! I never quite got a hold of this topic since the past 2-3 months and now after reading this page it’s all crystal clear to me ! And as for the mnemonic , when i told it to my chem teacher , he was impressed and asked me for the website :D ! Great job, keep it up :D
Reply
Great, glad you found it useful!
Reply
80. Oh wow, this is amazing. I was starting to think there was no site/book that had exactly this information, this compactly (and brilliantly, might I add) presented. Thank you so much! You’ve got no idea how much this helped (and how much it reduced my study time, god knows we can all do with extra time on our hands :) ). Thanks again!
Reply
81. This site is so helpful. It makes organic chemistry way easier. Thank you!
Reply
Glad you find it helpful Luis.
Reply
82. On the figure describing inductive effects on pKa of carboxylic acids you have bromoacetic acid with pKa of 2.86 on one line and 2.97 on the next. Why the difference?
Reply
Fixed. Thanks for the spot – Evans’ pKa table says 2.86. Not sure where I pulled the 2.97 from.
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83. All these stuff were very useful to me. Got to know more things that I didn’t know before. Thanks a lot..! Similar article on basicity will be appreciated a lot..!
Reply
84. I feel sooo much more confident about orgo bc of u =)
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85. I just don’t seem to understand why does ionic radius increase acidity? I mean, Binding Energy decreases and it’s more easy to lose an electron, if acids are compounds that accept electrons how does acidity increase?
Reply
Basicity is all about the stability of negative charge. With larger atoms, essentially the charge is more spread out over a greater volume. Lower charge densities are more stable. See also here:
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86. I have a problem where I have to determine the most stable conjugate base, which indicates the strongest acid. I just want to know, is it possible to have a strong acid according to its pKa value, but according to atom, resonance, etc. another acid is stronger?
Reply
pKa represents an experimental measurement. Experimental measurements are primary – the concepts we pull out of them, such as the factors mentioned, are secondary. So what you’re mentioning isn’t possible, assuming all other variables are the same.
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87. I just wanted to thank you for all the work you put into this site. I’m an undergraduate at Yale and for the past semester, I’ve been afraid I’ll fail Orgo. Thanks to your site, I no longer feel as stressed because you’ve done such a good job of explaining things. So, yes, thank you!!
Wishing you the best!
Reply
Thanks Laina. Glad you find it useful. Let me know if theres anything I can do to make the site more helpful for your needs.
Reply
88. Thank you so much for this summary sheet, I was having so much trouble trying to figure out what made a molecule more acidic – and here it all is! Fully explained and easily understandable. Fantastic stuff.
Reply
89. I love this site… I hate reading organic chem txt books because most are boooooring but you make studying for O-chem the highlight of my friday evening… You are funny, and your delivery style is absolutely amazing!! I am in O-chem II, barely made it through the first but I am excelling in my second…. Thank you sir for doing this!! I appreciate it more than you know… Good day!
Reply
90. I was the most confused person who could not understand the concept. I read your note. I received 100% on Acidity part on my Exam. Thank you!
Reply
This is the kind of comment that makes my day. Thanks!
Reply
91. thanks for making org chemistry easier but i have a question about resonance, you didnt mention electron donation or withdrawal by resonance
Reply
That’s a thorny issue… it can be hard to separate the influence of inductive and resonance effects. Do you have a specific example that you’re thinking of?
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I have a related question. I’ve read that in most cases, resonance stabilization decreases basicity. But is it true that in the cases of guanidines and amidines, resonance increases the basicity? If you could explain how, that’d be great!
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92. I just wanted to say you are doing an absolutely fantastic job of teaching organic chemistry and making it comprehensible. I’ve been reading your posts for the last few days, and material which was alien to me before as finally started to make sense.
Reply
Thanks!
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1: Chapter 1 - Functions and Equations
Math 098: Intermediate Algebra for Calculus
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1.4: Slope
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1.4: Slope
Last updated Feb 26, 2021
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1.3: The Graph of a Function
1.5: Equations of Lines
Page ID 58409
David Arnold
College of the Redwoods
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Definition
2. Expectations
3. Definition: Slope
4. Definition: Change in Quantity
5. Tip 3
6. Example 1.4.1
1. Solution
Note
The Slope Formula
Definition 5
Example 1.4.2
Solution
Example 1.4.3
Solution
Example 1.4.4
Solution
Parallel Lines
property
Example 1.4.5
Solution
Example 1.4.6
Solution
Perpendicular Lines
Property 13
Example 1.4.7
Solution
In the previous section on Linear Models, we saw that if the dependent variable was changing at a constant rate with respect to the independent variable, then the graph was a line. If the rate was positive, then as we swept our eyes from left to right, the line rose upward, the dependent variable increasing with increasing changes in the independent variable. If the rate was negative, then the graph fell downward, the dependent variable decreasing with increasing changes in the independent variable. You may have also learned that higher rates led to steeper lines (lines that rose more quickly) and lower rates led to lines that were less steep.
In this section, we will connect the intuitive concept of rate developed in the previous section with a formal definition of the slope of a line. To start, let’s state up front what is meant by the slope of a line.
Definition
Slope is a number that tells us how quickly a line rises or falls.
If slope is a number that is directly connected to the “steepness” of a line, then we should have certain expectations.
Expectations
Lines with positive slope should slant uphill (as our eyes sweep from left to right).
Lines with negative slope should slant downhill (as our eyes sweep from left to right).
Because any horizontal line neither slants uphill nor downhill, we expect that it should have slope equal to zero.
Lines with a larger positive slope should rise more quickly than lines with a smaller positive slope.
If two lines have negative slope, then the line having the slope with larger absolute value should fall more quickly than the other line.
It remains to define how to compute the slope of a particular line. Whatever definition we choose, it should conform with the expectations outlined above. We also would like the definition of slope to conform with the concept of rate developed in the previous section. Thus, we make the following definition.
Definition: Slope
The slope of a line is the rate at which the dependent variable is changing with respect to the independent variable.
Note how the word “change” is used Definition. It is important to understand that the change in some quantity can be positive, negative, or zero. For example, if the temperature outside is 40∘F when I leave my home at 6 AM, and at noon the temperature is 65◦ F, then the change in temperature is a positive 25∘F. On the other hand, if the temperature outside is 65∘F at noon, and the temperature is 50∘F when I return home in the evening, then the change in temperature is a negative 15 degrees Fahrenheit.
In calculating the change in a quantity, follow this rule.
Definition: Change in Quantity
Change in Quantity = Latter Measurement − Former Measurement.
Thus, if T represents the temperature and ΔT represents the change in the temperature, then in our first case (taking the temperature in the morning then later at noon), the change in temperature is ΔT=Latter−Former=65∘F−40∘F=25∘F
This positive result represents an increase in the temperature of 25∘F.
In the second case (taking the temperature at noon then later in the evening), the change in temperature is
ΔT=Latter−Former=50∘F−65∘F=−15∘F
This negative result represents a decrease in the temperature of 15∘F
Tip 3
Readers should note that the direction of subtraction is extremely important. To detect the change in a quantity, always subtract the former (earlier) measurement from the latter (later) measurement.
Example 1.4.1
A ball is perched at rest at the top of a long ramp. It’s given a little tap and it begins to roll down the ramp. The speed v of the ball (in meters per second) is plotted versus the time t (in seconds) in Figure 1.4.1.
Determine the slope of the line.
Solution
We’ve defined the slope as the rate at which the dependent variable is changing with respect to the independent variable. In this case, the speed v of the ball “depends” upon the amount of time t that has elapsed. Consequently, v is the dependent variable and has been placed on the vertical axis.3 On the other hand, t is the independent variable and has been assigned the horizontal axis.
Figure 1.4.1. Speed versus time.
To determine the rate at which v is changing with respect to t (the slope of the line), we first select two points P(2, 3) and Q(8, 12) on the line, as shown in Figure 1.4.2. As we sweep our eyes from left to right (a convention we will always follow when dealing with slope), the point P occurs before the point Q. Hence, we consider P the “former” measurement and point Q the “latter” measurement.
Figure 1.4.2. Determining the slope of the line.
At point P, the time is t = 2 seconds, then at point Q the time is t = 8 seconds. The change in t is found by subtracting the former measurement from the latter measurement.
Δt=8s−2s=6s
At point P, the speed is v = 3 meters per second, then at point Q the speed is v = 12 meters per second. Hence, the change in v is
Δv=12m/s−3m/s=9m/s
Finally, the slope of the line is defined as the rate at which the dependent variable v is changing with respect to the independent variable t. That is,
Slope=Δv Δt=9m/s 6s=3m/s s
Scientists prefer to write this as 1.5 m/s 2, but this might not be as intuitive as writing 1.5 (m/s)/s, which indicates that the speed is increasing at a rate of 1.5 m/s every second. This makes good sense as a ball rolling down a ramp will pick up speed with the passage of time. The slope provides an exact numerical description of how the speed increases with respect to time.
Note that our definition of the slope of the line satisfies one of our goals: the slope is precisely the same as the notion of rate described in the previous section. Indeed, note the right triangle we’ve drawn in Figure 1.4.2. The bottom edge of the triangle is 12 boxes long, but every 2 boxes represents one second, so this displacement in the time t direction is 6 seconds. The vertical side of the right triangle is 9 boxes in height where each box represents 1 meter per second. Consequently, this vertical edge of the right triangle represents a positive displacement of 9 meters per second. Thus, every 6 seconds, there is an increase in speed of 9 meters per second. Hence, the ball is picking up speed at the rate of 9 meters per second every 6 seconds, or equivalently, 1.5 meters per second every second.
Note
In Figure 1.4.2, the rate at which the speed is increasing with respect to time is equivalent to the slope of the line.
Suppose that we had labeled our points P(t initial,v initial) and Q(t final,v final) as shown in Figure 1.4.3.
Figure 1.4.3. Initial and final measurements.
Now the change in speed v would be
Δv=v final−v initial
and the change in time t would be
Δt=t final−t initial
Therefore, the slope of the line would be computed with the following formula.
Slope=Δv Δt=v final−v initial t final−t initial
With P(t initial,v initial)=(2s,3m/s) and Q(t final,v final)=(8s,12m/s), this becomes
Slope=12m/s−3m/s 8s−2s=9m/s 6s=1.5m/s 2
The Slope Formula
The last calculation in Example 1.4.1 allows us to discuss the slope of a line as a purely mathematical concept, one that is not rooted in a supporting application as in Example 1.4.1. Take, for example, the line shown in Figure 1.4.4 that passes through the points P(−3,−3) and Q(2,1).
Figure 1.4.4. Computing the slope of a line in an xy-coordinate system.
In this example, the dependent variable is y and the independent variable is x, so the slope of the line is Δy (the change in y) divided by Δx (the change in x).
Slope=Δy Δx
Sweeping our eyes from left to right, the point P comes first, followed by the point Q. Keeping “latter minus former” in mind, the change in y is computed by subtracting the y-value of point P from the y-value of point Q. That is,
Δy=1−(−3)=4
Similarly, the change in x is computed by subtracting the x-value of point P from the x-value of point Q. That is,
Δx=2−(−3)=5
Thus, the slope of the line is
Slope=Δy Δx=4 5
Alternatively, we can use the points P and Q as vertices of a right triangle with sides parallel to the axes (shown in Figure 1.4.5(a)). The horizontal edge of the right triangle is 5 boxes (each representing 1 unit), so the displacement in x is 5 units. The vertical edge is 4 boxes (each representing 1 unit), so the displacement in y is 4 units. Hence, each time x is increased by 5 units, y experiences an increase of 4 units. Therefore, the slope of the line is again 4/5.
Figure 1.4.5. Using a right triangle to determine the slope.
Suppose that we had labeled our points P(x 1,y 1) and Q(x 2,y 2) as shown in Figure 1.4.5(b). Now the change in y would be
Δy=y 2−y 1
and the change in x would be
Δx=x 2−x 1
Therefore, the slope of the line would be computed with the following formula.
Slope=Δy Δx=y 2−y 1 x 2−x 1
With P(x 1,y 1)=(−3,−3) and Q(x 2,y 2)=(2,1), this becomes
Slope=1−(−3)2−(−3)=4 5
The slope formula is worth summarizing in a definition.
Definition 5
The slope of the line that passes through the points P(x 1,y 1) and Q(x 2,y 2) is given by the formula
Slope=Δy Δx=y 2−y 1 x 2−x 1
Let’s look at some more examples.
Example 1.4.2
Find the slope of the line passing through the points P(−3, −2) and Q(3, 1).
Solution
We can use the slope formula in Definition 5 to determine the slope. With (x 1,y 1)=P(−3,−2) and (x 2,y 2)=Q(3,1),
Slope=Δy Δx=y 2−y 1 x 2−x 1=1−(−2)3−(−3)=3 6=1 2
Readers will sometimes ask, “Which point should be (x 1,y 1) and which should be (x 2,y 2)?” The short answer is, “It doesn’t matter!” Suppose instead, that we let (x 1,y 1)=Q(3,1) and (x 2,y 2)=P(−3,−2). Then,
Slope=Δy Δx=y 2−y 1 x 2−x 1=−2−1−3−3=−3−6=1 2
Because the change in any quantity is found by subtracting the earlier measurement from the later measurement, we will continue to stress the first order. However, if we reverse the points as we did in our second calculation, both numerator and denominator reverse sign with this interchange, so we get the same answer.
Of course, we can also determine the slope by plotting P(−3, −2) and Q(3, 1) and the line that passes through P and Q, as we’ve done in Figure 1.4.6.
Figure 1.4.6. Determining the slope from the graph.
Starting at the point P, to get to the point Q, we move 6 boxes to the right, then 3 boxes up, as shown in Figure 1.4.6. Hence, the slope of the line is
Slope=Δy Δx=3 6=1 2
Note that two of our expectations regarding the slope of a line are met with this example.
The line through P(−3, −2) and Q(3, 1) in Figure 1.4.6 has slope 1/2. This is a positive number and the line slants uphill (as expected) as we sweep our eyes from left to right.
The slope in this example is 1/2, which is less than the slope of the line in Figure 1.4.5(a), which was 4/5. Note that the line in Figure 1.4.6 is less steep than the line in Figure 1.4.5(a), which was another of our earlier expectations regarding the slope of a line.
Example 1.4.3
Find the slope of the line passing through the points P(−4, 4) and Q(4, −2).
Solution
We can use the slope formula in Definition 5 to determine the slope. With (x 1,y 1)=P(−4,4) and (x 2,y 2)=Q(4,−2),
Slope=Δy Δx=y 2−y 1 x 2−x 1=−2−4 4−(−4)=−6 8=−3 4
We can also get the slope of the line from the graph in Figure 1.4.7. Starting at the point P(−4, 4), move 8 units to the right, then 6 units downward, as shown in Figure 1.4.7.
Figure 1.4.7. Determining the slope from the graph.
Thus, the slope of the line is
Slope=Δy Δx=−6 8=−3 4
Again, one of our earlier expectations regarding the slope of a line is met in this example. The slope is −3/4, which is a negative number, and the line in Figure 1.4.7 slants downhill (as we sweep our eyes from left to right).
Example 1.4.4
Draw a line that intercepts the y-axis at (0, 3) so that the line has slope −4/3. Draw a second line that passes through the point P(−1, −1) with slope 3/5.
Solution
The slope of the first line is −4/3. This means that our line must slant downhill (as we sweep our eyes from left to right). The slope is the change in y over the change in x. Therefore, every time x increases by 3 units, y must decrease by 4 units. Plot the point P(0, 3), as shown in Figure 1.4.8(a). Then, starting at P, move 3 units to the right, followed by 4 units downward to the point Q(3, −1), as shown in Figure 1.4.8(a). Draw the required line, which must pass through the points P and Q.
To draw the second line, first plot the point P(−1, −1), as shown in Figure 1.4.8(b). Starting at the point P, move 5 units to the right, then upward 3 units to the point Q(4, 2), as shown in Figure 1.4.8(b). Draw the required line passing through the points P and Q.
Figure 1.4.8
Parallel Lines
Because slope controls the “steepness” of a line, it is a simple matter to see that parallel lines must have the same slope.
property
Let L 1 be a line having slope m 1. Let L 2 be a line having slope m 2. If L 1 and L 2 are parallel, then
m 1=m 2
That is, any two parallel lines have the same slope.
Example 1.4.5
What is the slope of any horizontal line? What is the slope of any vertical line?
Solution
One would expect that our definition would verify that the slope of any horizontal line is zero. Select, for example, the horizontal line shown in Figure 9(a). Select the points (−3, 3) and (3, 3) on this line.
Figure 1.4.9
With (x 1,y 1)=(−3,3) and (x 2,y 2)=(3,3)
Slope=Δy Δx=y 2−y 1 x 2−x 1=3−3 3−(−3)=0 6=0
Thus, the horizontal line in Figure 1.4.9(a) has slope equal to zero, exactly as expected. Further, all horizontal lines are parallel to this horizontal line and have the same slope. Therefore, all horizontal lines have slope zero.
We would surmise that the vertical line in Figure 1.4.9(b) has undefined slope (we’ll explore this more fully in the exercises). In Figure 1.4.9(b), we’ve selected the points P(−3, −3) and Q(−3, 3) on the vertical line. With (x 1,y 1)=P(−3,−3) and (x 2,y 2)=Q(−3,−3), Slope=Δy Δx=y 2−y 1 x 2−x 1=3−(−3)−3−(−3)=6 0,which is undefined.
The slope of the vertical line in Figure 1.4.9(b) is undefined because division by zero is meaningless. Further, all vertical lines are parallel to this vertical line and have undefined slope.
Example 1.4.6
Draw a line through the point P(1, 2) that is parallel to the line passing through the origin with slope −2/3.
Solution
We will first draw a line through the origin with slope −2/3. Plot the point P(0, 0), then move 3 units to the right and 2 units downward to the point Q(3, −2), as shown in Figure 1.4.10(a). Draw a line through the points P and Q as shown in Figure 1.4.10(a).
Next, plot the point P(1, 2) as shown in Figure 1.4.10(b). To draw a line through this point that is parallel to the line through the origin, this second line must have the same slope as the first line. Therefore, start at the point P(1, 2), as shown in Figure 1.4.10(b), then move 3 units to the right and 2 units downward to the point Q(4, 0). Draw a line through the points P and Q as shown in Figure 1.4.10(b). Note that this second line is parallel to the first.
Figure 1.4.10
Perpendicular Lines
The relationship between the slopes of two perpendicular lines is not as straightforward as the relation between the slopes of two parallel lines. Let’s begin by stating the pertinent property.
Property 13
Let L 1 be a line having slope m 1. Let L 2 be a line having slope m 2. If L 1 and L 2 are perpendicular, then
m 1m 2=−1
That is, the product of the slopes of two perpendicular lines is −1.
We can solve equation (14) for m 1 in terms of m 2.
m 1=−1 m 2
Equation (15) tells us that the slope of the first line is the negative reciprocal of the slope of the second line.
For example, suppose that L 1 and L 2 are perpendicular lines with slopes m 1 and m 2, respectively
If m 2=2, then m 1=−1 2
If m 2=3 5, then m 1=−5 3
If m 2=−2 3, then m 1=3 2
Note that in each bulleted item, the product of the slopes is −1.
We won’t provide a proof of equation (15), but we will provide some motivating evidence in the form of a graph.
Example 1.4.7
Sketch the graphs of the lines passing through the origin having slopes 2 and −1/2.
Solution
In Figure 1.4.11(a), we’ve plotted the point P(0, 0) at the origin, then moved 1 unit to the right and 2 units upward to the point Q(1, 2). The resulting line passes through the origin and has slope m 1=2 (alternatively, m 1=2/1).
In Figure 1.4.11(b), we’ve again plotted the point P(0, 0) at the origin, then moved 2 units to the right and 1 unit downward to the point Q(2, −1). The resulting line passes through the origin and has slope m 2=−1/2.
There are two important points that need to be made about the lines in Figure 1.4.11(b).
The two lines in Figure 1.4.11(b) are perpendicular. They meet and form a right angle of 90∘. If you have a protractor available, you might want to measure the angle between the two lines and note that the measure of the angle is 90∘.
The product of the two slopes is
m 1m 2=2⋅(−1 2)=−1
Figure 1.4.11. Sketching perpendicular lines.
This page titled 1.4: Slope is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.
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Confusion about the definition of an acyclic graph
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My textbook says
Definition 1: A graph, G, is acyclic if it contains no undirected cycles (otherwise it’s cyclic).
It also says
Definition 2: A (directed) cycle is a (directed) path which begins and ends at the same vertex. An undirected cycle is, likewise, a path beginning and ending at the same vertex which may or may not respect edge directions.
These definitions confuse me.
By Definition 1, can a graph be acyclic and yet contain a directed cycle? This sounds like a contradiction, but the definition only says an acyclic graph should not contain undirected cycles and says nothing about directed cycles. Unless Definition 1 is implying that all directed cycles can be treated as undirected cycles, but undirected cycles cannot be treated as directed cycles?
Definition 2 seems to reinforce this idea, by suggesting that an undirected cycle can simply ignore edge directions.
On the other hand, I found this website which claims this is a directed acyclic graph:
But by Definition 2, I can just ignore edge direction and create undirected cycles, like 1→2→5→6→3→1, which would make the graph cyclic because it contains undirected cycles. However, the website says it is acyclic, which contradicts everything I've said.
I think I am probably just misinterpreting all of these definitions. Can we define the terms "acyclic", "cyclic", "undirected cycle" and "directed cycle" in some other way to help clarify what they mean?
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edited Dec 23, 2019 at 18:00
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asked Dec 23, 2019 at 15:43
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Definitions in different sources need not to agree.MoonLightSyzygy –MoonLightSyzygy 2019-12-23 15:46:02 +00:00 Commented Dec 23, 2019 at 15:46
@MoonLightSyzygy Can a directed graph without directed cycles contain undirected cycles?Data –Data 2019-12-23 15:48:35 +00:00 Commented Dec 23, 2019 at 15:48
The graph pictured has undirected cycles, and no directed ones.MoonLightSyzygy –MoonLightSyzygy 2019-12-23 15:49:27 +00:00 Commented Dec 23, 2019 at 15:49
@MoonLightSyzygy So we could call it a "directed acyclic graph that contains undirected cycles"?Data –Data 2019-12-23 15:51:31 +00:00 Commented Dec 23, 2019 at 15:51
1 Yes, definitions need not be universal. They are just language. Different people use it differently. If you look at old books you will find that even for defining graph there were tons of definitions. Some people allowing or not, loops, multiple edges, directed edges, ...MoonLightSyzygy –MoonLightSyzygy 2019-12-23 15:56:18 +00:00 Commented Dec 23, 2019 at 15:56
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I think I am probably just misinterpreting all of these definitions? Can we define the terms "acyclic", "cyclic", "undirected cycle" and "directed cycle" in some other way to help clarify what they mean?
Yes, many graph theory textbooks do a better job than yours did. The best way is to start with a firm understanding of (undirected) graphs and move from there to digraphs. So let's start over and eliminate all the useless words.
A cycle is a walk in a graph where the origin and internal vertices are all distinct and the terminus is the same as the origin. A graph is acyclic if it does not contain a cycle.
With that said, a directed graph is one where the edges are all endowed with a direction. Associated with every digraph is its underlying graph which is an undirected graph with the same vertex and edge set but "ignoring" the direction. So when we say that a directed graph is or isn't acyclic, we are implicitly referring to whether the underlying graph is acyclic. But the added structure gives us the notion of a directed cycle, which is a directed walk in the digraph (i.e. one the follows all of the directions) where again the origin and internal vertices are distinct and the terminus and the origin are the same.
The only wrinkle in all of this is that the meddling computer scientists have forced the term directed acyclic graph (DAG) on us. This refers to a digraph that contains no directed cycle (although its underlying graph may contain a cycle). It's not a good name, but there's no putting that toothpaste back in the tube so we have to deal with its existence.
Coming back to the two introductory questions you asked:
By Definition 1, can a graph be acyclic and yet contain a directed cycle?
No. If a digraph contains a directed cycle, then that same walk in the underlying graph of the digraph would be a cycle. The converse is possible -- a digraph can be cyclic but not contain a directed cycle. The graph you pasted into your question is an example of that.
Unless Definition 1 is implying that all directed cycles can be treated as undirected cycles, but undirected cycles cannot be treated as directed cycles?
Yes, as I just noted every directed cycle would be a cycle in the underlying graph. And a cycle in the underlying graph would be a directed cycle in the digraph iff its edges are all following the direction of the walk.
Definition 2 seems to reinforce this idea, by suggesting that an undirected cycle can simply ignore edge directions.
Yes. A cycle is a feature of the underlying graph, so the direction of the edges in the digraph is not considered.
The good news is that all of this is essentially invisible once you have these definitions straight in your mind. Does this clear it up?
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answered Dec 23, 2019 at 17:58
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Yes, that does clear it up very well. Great answer. +1 Data –Data 2019-12-23 18:19:49 +00:00 Commented Dec 23, 2019 at 18:19
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3Let G be a loop-less undirected graph. Prove that the edges of G can be directed so that no directed cycle is formed.
24What is the difference between a loop, cycle and strongly connected components in Graph Theory?
2Confusion about definition of cycle
2Example of non-simple cycle in a directed graph
2Chordal Graph to Directed Acyclic Graph
2Arborescence of a graph
2Difference between Oriented Graph and Directed Acyclic Graphs (DAG)
0Show, that if the removal of a vertex v reduces the depth of a minimal acyclic orientation, every longest path contains v
1The authors didn't define as follows. Why? (cycle in a directed graph and in an undirected graph. "Introduction to Algorithms 3rd Edition" by CLRS)
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Example SolutionsPractice Questions
How to Graph a Parabola of the Form y=a x 2+c
Step 1: The x coordinate of the vertex for this type of quadratic function will always be 0. Create a table with two values to the left of this x value and two values to the right of this x value.
Step 2: Input the x values in the table into the equation to generate y values in the table.
Step 3: Use the values calculated in the previous step to fill in the table of values for the function.
Step 4: Use the table to create a scatterplot for the given function.
Step 5: Connect these points with a smooth curve. The graph of a parabola should be roughly U-shaped in general.
How to Graph a Parabola of the Form y=a x 2+c Vocabulary
Parabola: This is the shape of the graph created by a quadratic function. It is U-shaped in general but can be wider or skinnier depending on the quadratic function.
Vertex: This is either the highest or lowest point of a quadratic function. The equation to get the x value of the vertex is: −b 2 a
So, let's try using these steps to graph a parabola of the form y=a x 2+c, in the following two examples!
How to Graph a Parabola of the Form y=a x 2+c: Example 1
Graph the parabola given by the equation y=−2 x 2+5
Step 1: The x coordinate of the vertex for this type of quadratic function will always be 0. Create a table with two values to the left of this x value and two values to the right of this x value.
| x | y |
--- |
| -2 | |
| -1 | |
| 0 | |
| 1 | |
| 2 | |
Step 2: Input the x values in the table into the equation to generate y values in the table.
The first input gives us:
f(−2)=−2(−2)2+5 f(−2)=−3
The second input gives us:
f(−1)=−2(−1)2+5 f(−1)=3
The third input gives us:
f(0)=−2(0)2+5 f(0)=5
The fourth input gives us:
f(1)=−2(1)2+5 f(1)=3
The fifth input gives us:
f(2)=−2(2)2+5 f(2)=−3
Step 3: Use the values calculated in the previous step to fill in the table of values for the function.
| x | y |
--- |
| -2 | -3 |
| -1 | 3 |
| 0 | 5 |
| 1 | 3 |
| 2 | -3 |
Step 4: Use the table to create a scatterplot for the given function.
The table we created in the previous step gives the scatterplot:
Step 5: Connect these points with a smooth curve. The graph of a parabola should be roughly U-shaped in general.
Connecting the points in our scatterplot creates the graph:
How to Graph a Parabola of the Form y=a x 2+c: Example 2
Graph the parabola given by the equation y=6 x 2−10
Step 1: The x coordinate of the vertex for this type of quadratic function will always be 0. Create a table with two values to the left of this x value and two values to the right of this x value.
| x | y |
--- |
| -2 | |
| -1 | |
| 0 | |
| 1 | |
| 2 | |
Step 2: Input the x values in the table into the equation to generate y values in the table.
f(−2)=6(−2)2−10 f(−2)=14
The second input gives us:
f(−1)=6(−1)2−10 f(−1)=−4
The third input gives us:
f(0)=6(0)2−10 f(0)=−10
The fourth input gives us:
f(1)=6(1)2−10 f(1)=−4
The fifth input gives us:
f(2)=6(2)2−10 f(2)=14
Step 3: Use the values calculated in the previous step to fill in the table of values for the function.
| x | y |
--- |
| -2 | 14 |
| -1 | -4 |
| 0 | -10 |
| 1 | -4 |
| 2 | 14 |
Step 4: Use the table to create a scatterplot for the given function.
The table we created in the previous step gives the scatterplot:
Step 5: Connect these points with a smooth curve. The graph of a parabola should be roughly U-shaped in general.
Connecting the points in our scatterplot creates the graph:
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Table of Contents
How to Graph a Parabola of the Form y=a x 2+c
How to Graph a Parabola of the Form y=a x 2+c Vocabulary
How to Graph a Parabola of the Form y=a x 2+c: Example 1
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2015 AMC 8 Problems/Problem 19
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Very much recommended to learn this)
2.6 Solution 6 (Heron's Formula, Not Recommended)
2.7 Solution 7 (Simple Deduction)
3 Video Solution (HOW TO THINK CRITICALLY!!!)
4 Video Solution
5 Video Solution by OmegaLearn
6 See Also
Problem
A triangle with vertices as , , and is plotted on a grid. What fraction of the grid is covered by the triangle?
Solutions
Solution 1
The area of is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is , and its base is . We multiply these and divide by to find the area of the triangle is . Since the grid has an area of , the fraction of the grid covered by the triangle is .
Solution 2
Note angle is right; thus, the area is ; thus, the fraction of the total is .
Solution 3
By the Shoelace Theorem, the area of .
This means the fraction of the total area is .
Solution 4
The smallest rectangle that follows the grid lines and completely encloses has an area of , where splits the rectangle into four triangles. The area of is therefore . That means that takes up of the grid.
Solution 5 (Very much recommended to learn this)
Using Pick's Theorem, the area of the triangle is . Therefore, the triangle takes up of the grid.
Solution 6 (Heron's Formula, Not Recommended)
We can find the lengths of the sides by using the Pythagorean Theorem. Then, we apply Heron's Formula to find the area. This simplifies to Again, we simplify to get The middle two terms inside the square root multiply to , and the first and last terms inside the square root multiply to This means that the area of the triangle is The area of the grid is Thus, the answer is .
Solution 7 (Simple Deduction)
First, count the number of shapes inside the main triangle (you should count 10). Then, upon closer inspection, most of the shapes that are not a single unit on the triangle can be created by connecting another shape. The only exceptions are one shape that is a single unit and one that would need 2 shapes connected to it to make a single unit. On average, you need to connect 2 shapes to make a unit. Knowing this, if there are 10 shapes and you require 2 shapes to make a unit, 10 divided by 2 equals 5, which is the area.5 is 1/6 of 30(the total of the graph) and so the final answer is . -Themathnerd3.14
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
See Also
2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18Followed by
Problem 20
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1511 | http://www.mhtl.uwaterloo.ca/courses/ece309_mechatronics/lectures/pdffiles/ach5_web.pdf | Conduction Heat Transfer Reading Problems 17-1 →17-6 17-35, 17-57, 17-68, 17-81, 17-88, 17-110 18-1 →18-2 18-14, 18-20, 18-34, 18-52, 18-80, 18-104 Fourier Law of Heat Conduction x=0 x x x+ x D x=L insulated Qx Qx+ x D g A The general 1-D conduction equation is given as ∂ ∂x k∂T ∂x longitudinal conduction + ˙ g internal heat generation = ρC ∂T ∂t thermal inertia where the heat flow rate, ˙ Qx, in the axial direction is given by Fourier’s law of heat conduction.
˙ Qx = −kA∂T ∂x Thermal Resistance Networks Resistances in Series The heat flow through a solid material of conductivity, k is ˙ Q = kA L (Tin −Tout) = Tin −Tout Rcond where Rcond = L kA 1 The total heat flow across the system can be written as ˙ Q = T∞1 −T∞2 Rtotal where Rtotal = 4 i=1 Ri This is analogous to current flow through electrical circuits where, I = ΔV/R Resistances in Parallel T1 Q1 T2 Q2 L R1 k2 k1 R2 In general, for parallel networks we can use a parallel resistor network as follows: where 1 Rtotal = 1 R1 + 1 R2 + 1 R3 + · · · 2 T1 T1 T2 T2 R1 Rtotal R2 R3 = and ˙ Q = T1 −T2 Rtotal Cylindrical Systems L r1 r2 Qr T1 T2 A=2 rL p k r Steady, 1D heat flow from T1 to T2 in a cylindrical systems occurs in a radial direction where the lines of constant temperature (isotherms) are concentric circles, as shown by the dotted line in the figure above and T = T (r).
T2 −T1 = − ˙ Qr 2πkL(ln r2 −ln r1) = − ˙ Qr 2πkL ln r2 r1 Therefore we can write ˙ Qr = T2 −T1 ln(r2/r1) 2πkL where R = ln(r2/r1) 2πkL 3 Critical Thickness of Insulation Consider a steady, 1-D problem where an insulation cladding is added to the outside of a tube with constant surface temperature Ti. What happens to the heat transfer as insulation is added, i.e. we increase the thickness of the insulation?
The resistor network can be written as a series combination of the resistance of the insulation, R1 and the convective resistance, R2 Rtotal = R1 + R2 = ln(ro/ri) 2πkL + 1 h2πroL Note: as the thickness of the insulation is increased the outer radius, ro increases.
Could there be a situation in which adding insulation increases the overall heat transfer?
To find the critical radius, rc, where adding more insulation begins to decrease heat transfer, set dRtotal dro = 0 dRtotal dro = 1 2πkroL − 1 h2πr2 oL = 0 rc = k h 4 Heat Generation in a Solid Heat can be generated within a solid as a result of resistance heating in wires, chemical reactions, nuclear reactions, etc.
A volumetric heat generation terms will be defined as follows: ˙ g = ˙ Eg V (W/m3) for heat generation in wires, we will define ˙ g as ˙ g = I2Re πr2 oL Slab System 5 T = T1 + T2 2 − T1 −T2 2 x L + ˙ qL2 2k 1 − x L 2 Cylindrical System T = Ts + ˙ gr2 0 4k ⎛ ⎝1 − r r0 2⎞ ⎠ where BC1 : dT dr = 0 @r = 0 BC2 : T = Ts @r = r0 Heat Transfer from Finned Surfaces The temperature difference between the fin and the surroundings (temperature excess) is usually expressed as θ = T (x) −T∞ 6 which allows the 1-D fin equation to be written as d2θ dx2 −m2θ = 0 where the fin parameter m is m = hP kAc 1/2 and the boundary conditions are θ = θb @ x = 0 θ →0 as x →∞ The solution to the differential equation for θ is θ(x) = C1 sinh(mx) + C2 cosh(mx) substituting the boundary conditions to find the constants of integration θ = θb cosh[m(L −x)] cosh(mL) The heat transfer flowing through the base of the fin can be determined as ˙ Qb = Ac −kdT dx @x=0 = θb(kAchP )1/2 tanh(mL) Fin Efficiency and Effectiveness The dimensionless parameter that compares the actual heat transfer from the fin to the ideal heat transfer from the fin is the fin efficiency η = actual heat transfer rate maximum heat transfer rate when the entire fin is at Tb = ˙ Qb hP Lθb 7 If the fin has a constant cross section then η = tanh(mL) mL An alternative figure of merit is the fin effectiveness given as ϵfin = total fin heat transfer the heat transfer that would have occurred through the base area in the absence of the fin = ˙ Qb hAcθb Transient Heat Conduction Performing a 1st law energy balance on a plane wall gives ˙ Ein −˙ Eout ⇒˙ Qcond = TH −Ts L/(k · A) = ˙ Qconv = Ts −T∞ 1/(h · A) where TH −Ts Ts −T∞ = L/(k · A) 1/(h · A) = internal resistance to H.T.
external resistance to H.T.
= hL k = Bi ≡Biot number Rint << Rext: the Biot number is small and we can conclude TH −Ts << Ts −T∞ and in the limit TH ≈Ts 8 Rext << Rint: Rint << Rext: the Biot number is large and we can conclude Ts −T∞<< TH −Ts and in the limit Ts ≈T∞ Lumped System Analysis • if the internal temperature of a body remains relatively constant with respect to position – can be treated as a lumped system analysis – heat transfer is a function of time only, T = T (t) • internal temperature is relatively constant at low Biot number • typical criteria for lumped system analysis →Bi ≤0.1 Transient Conduction Analysis For the 3-D body of volume V and surface area A, we can use a lumped system analysis if Bi = hV kA < 0.1 ⇐results in an error of less that 5% The characteristic length for the 3-D object is given as L = V/A. Other characteristic lengths for conventional bodies include: 9 Slab V As = W H2L 2W H = L Rod V As = πr2 oL 2πr0L = r0 2 Sphere V As = 4/3πr3 o 4πr2 0 = r0 3 For an incompressible substance we can write mC ≡Cth dT dt = −Ah 1/Rth (T −T∞) where Cth = lumped thermal capacitance It should be clearly noted that we have neglected the spatial dependence of the temperature within the object. This type of an approach is only valid for Bi = hV kA < 0.1 We can integrate and apply the initial condition, T = Ti @t = 0 to obtain T (t) −T∞ Ti −T∞ = e−t/(Rth·Cth) = e−t/τ 10 where τ = Rth · Cth = thermal time constant = mC Ah The total heat transfer rate can be determined by integrating ˙ Q with respect to time.
˙ Qtotal = hA(Ti −T∞)(τ)[1 −e−t∗/τ] Therefore ˙ Qtotal = ˙ mC(Ti −T∞)[1 −e−t∗/τ] 11 Heisler Charts The lumped system analysis can be used if Bi = hL/k < 0.1 but what if Bi > 0.1 • need to solve the partial differential equation for temperature • leads to an infinite series solution ⇒difficult to obtain a solution The solution procedure for temperature is a function of several parameters T (x, t) = f(x, L, t, k, α, h, Ti, T∞) By using dimensionless groups, we can reduce the temperature dependence to 3 dimensionless parameters Dimensionless Group Formulation temperature θ(x, t) = T (x, t) −T∞ Ti −T∞ position x = x/L heat transfer Bi = hL/k Biot number time F o = αt/L2 Fourier number note: Cengel uses τ instead of F o.
Now we can write θ(x, t) = f(x, Bi, F o) The characteristic length for the Biot number is slab L = L cylinder L = ro sphere L = ro contrast this versus the characteristic length for the lumped system analysis.
With this, two approaches are possible 12 1. use the first term of the infinite series solution. This method is only valid for F o > 0.2 2. use the Heisler charts for each geometry as shown in Figs. 18-13, 18-14 and 18-15 First term solution: F o > 0.2 →error about 2% max.
Plane Wall: θwall(x, t) = T (x, t) −T∞ Ti −T∞ = A1e−λ2 1F o cos(λ1x/L) Cylinder: θcyl(r, t) = T (r, t) −T∞ Ti −T∞ = A1e−λ2 1F o J0(λ1r/ro) Sphere: θsph(r, t) = T (r, t) −T∞ Ti −T∞ = A1e−λ2 1F o sin(λ1r/ro) λ1r/ro where λ1, A1 can be determined from Table 18-1 based on the calculated value of the Biot number (will likely require some interpolation).
Heisler Charts • find T0 at the center for a given time • find T at other locations at the same time • find Qtot up to time t 13 |
1512 | https://www.quora.com/What-is-the-derivation-of-the-diagonal-of-a-cube-and-the-cuboid-formula | What is the derivation of the diagonal of a cube and the cuboid formula? - Quora
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What is the derivation of the diagonal of a cube and the cuboid formula?
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Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010) · Author has 10.2K answers and 15.2M answer views
·7y
Consider a cuboid with length, breadth and height measuring [math]l, b[/math] and [math]h[/math] respectively, as shown in the figure below.
[math]AG[/math] is a diagonal of the cuboid.
By Pythagoras theorem, [math]AC^2 = AB^2 + BC^2.[/math]
Again. by Pythagoras theorem, [math]AG^2 = AC^2 + CG^2.[/math]
[math]\Rightarrow \qquad AG^2 = AB^2 + BC^2 + CG^2 = l^2 + b^2 + h^2.[/math]
[math]\Rightarrow \qquad[/math] The length of the diagonal of a cuboid is [math]\sqrt {l^2 + b^2 + h^2}.[/math]
A cube is a special case of a cuboid, where [math]l = b = h =[/math][math]s[/math], where [math]s[/math] is the length of each side.
[math]\Rightarrow \qquad[/math] The length of the diagonal of a cube is [math]\sqrt {3s^2} = \sqrt {3}\,\,s.[/math]
Continue Reading
Consider a cuboid with length, breadth and height measuring [math]l, b[/math] and [math]h[/math] respectively, as shown in the figure below.
[math]AG[/math] is a diagonal of the cuboid.
By Pythagoras theorem, [math]AC^2 = AB^2 + BC^2.[/math]
Again. by Pythagoras theorem, [math]AG^2 = AC^2 + CG^2.[/math]
[math]\Rightarrow \qquad AG^2 = AB^2 + BC^2 + CG^2 = l^2 + b^2 + h^2.[/math]
[math]\Rightarrow \qquad[/math] The length of the diagonal of a cuboid is [math]\sqrt {l^2 + b^2 + h^2}.[/math]
A cube is a special case of a cuboid, where [math]l = b = h =[/math][math]s[/math], where [math]s[/math] is the length of each side.
[math]\Rightarrow \qquad[/math] The length of the diagonal of a cube is [math]\sqrt {3s^2} = \sqrt {3}\,\,s.[/math]
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AG is a diagonal of the cuboid.
By Pythagoras theorem, AC² = AB²+BC²
Again. by Pythagoras theorem, AG² = AC²+CG².
⇒AG² = AB²+BC²+CG² = l²+ b²+h²
⇒ The length of the diagonal of a cuboid is √l2+b2+h2.
A cube is a special case of a cuboid, where
l = b = h = s , where s is the length of each side.
⇒ The length of the diagonal of a cube is √3 s²
Continue Reading
AG is a diagonal of the cuboid.
By Pythagoras theorem, AC² = AB²+BC²
Again. by Pythagoras theorem, AG² = AC²+CG².
⇒AG² = AB²+BC²+CG² = l²+ b²+h²
⇒ The length of the diagonal of a cuboid is √l2+b2+h2.
A cube is a special case of a cuboid, where
l = b = h = s , where s is the length of each side.
⇒ The length of the diagonal of a cube is √3 s²
Upvote ·
Greeshma Yelle
Btech in Electrical and Electronics Engineering&Independent Research, SBIT (Graduated 2014)
·7y
diagonal of cube= (sq.rt of 3 )x (side)
diagonal of cuboid=sq.rt of (l^2+w^2+h^2)
derivative of diagonal of cube= (sq.rt of 3 )x (side) if
=(sq.rt of 3)(1)+(0)(side)
= sq.rt(3)
derivative of diagonal of cuboid=sq.rt of (l^2+w^2+h^2)
=(1/2)(l^2+W^2+h^2)^(1/2–1)+(2l+2w+2h)
=(1/2)(l^2+w^2+h^2)^(-1/2)+2(l+w+h)
1/{sq.rt(l^2+w^2+h^2)}+2(l+w+h)
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How is the diagonal of a cube=√3side?
What is the formula of diagonal of the cuboid?
What are some ways of calculating the diagonal of a cuboid?
How do I find the length of the diagonal of a cube and a cuboid?
What is the derivation of the formula for finding the length of a diagonal of a cube?
Robert Nichols
Author has 5K answers and 15.6M answer views
·9y
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What is the derivation of the formula for finding the length of a diagonal of a cube?
I begin with just looking at the base of the cube. The diagonal of this base is the square root of (s squared plus s squared), or square root of (2 s squared). Now consider the diagonal of the base, the height of the cube an the 3-D diagonal of the cube. These 3 sides form a right triangle with the 3-D diagonal as the hypotenuse. So the length of the 3-D diagonal is the square root of (the square of the square root of 2 s squared plus s squared). The square of the square root of 2 s squared is just 2 s squared. So now you have the square root of (2 s squared plus s squared). Which is equa
Continue Reading
I begin with just looking at the base of the cube. The diagonal of this base is the square root of (s squared plus s squared), or square root of (2 s squared). Now consider the diagonal of the base, the height of the cube an the 3-D diagonal of the cube. These 3 sides form a right triangle with the 3-D diagonal as the hypotenuse. So the length of the 3-D diagonal is the square root of (the square of the square root of 2 s squared plus s squared). The square of the square root of 2 s squared is just 2 s squared. So now you have the square root of (2 s squared plus s squared). Which is equal to the square root of (3 s squared). This can simplify to square root of three times the length of the side.
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1; National Dental PBRN Collaborative Group 1
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1 University of Alabama at Birmingham, Birmingham, AL, USA
2 Westat, Rockville, MD, USA
3 University of Washington, Seattle, WA, USA
4 University of Rochester, Rochester, NY, USA
5 HealthPartners Institute for Education and Research, Minneapolis, MN, USA
6 College of Dentistry, University of Florida, Gainesville, FL, USA
7 University of Texas Health Science Center at San Antonio, San Antonio, TX, USA
✉
Corresponding Author: Ellen Funkhouser, University of Alabama at Birmingham, 611 MT, 1717 11th Ave. South, Birmingham, AL 35205, USA. emfunk@uab.edu
The National Dental PBRN Collaborative Group comprises practitioner, faculty, and staff investigators who contributed to this network activity. A list of these persons is at
Issue date 2017 Sep.
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PMCID: PMC5002250 NIHMSID: NIHMS811094 PMID: 26755526
The publisher's version of this article is available at Eval Health Prof
Abstract
Surveys of health professionals typically have low response rates, and these rates have been decreasing in the recent years. We report on the methods used in a successful survey of dentist members of the National Dental Practice–Based Research Network. The objectives were to quantify the (1) increase in response rate associated with successive survey methods, (2) time to completion with each successive step, (3) contribution from the final method and personal contact, and (4) differences in response rate and mode of response by practice/practitioner characteristics. Dentist members of the network were mailed an invitation describing the study. Subsequently, up to six recruitment steps were followed: initial e-mail, two e-mail reminders at 2-week intervals, a third e-mail reminder with postal mailing a paper questionnaire, a second postal mailing of paper questionnaire, and staff follow-up. Of the 1,876 invited, 160 were deemed ineligible and 1,488 (87% of 1,716 eligible) completed the survey. Completion by step: initial e-mail, 35%; second e-mail, 15%; third e-mail, 7%; fourth e-mail/first paper, 11%; second paper, 15%; and staff follow-up, 16%. Overall, 76% completed the survey online and 24% on paper. Completion rates increased in absolute numbers and proportionally with later methods of recruitment. Participation rates varied little by practice/practitioner characteristics. Completion on paper was more likely by older dentists. Multiple methods of recruitment resulted in a high participation rate: Each step and method produced incremental increases with the final step producing the largest increase.
Keywords: survey methods, participation rates, response rates, online surveys, health professions, dentists
Introduction
Surveys of health-care professionals are a valuable tool in health services and policy research because they are a cost-effective method to assess knowledge, attitudes, and practices in delivery of health care (VanGeest, Johnson, & Welch, 2007). Response rate, a measure of the representativeness of the sample, is the most common statistic cited to indicate the quality of a survey (Baruch & Holtom, 2008; Rogelberg & Staton, 2007). These rates, historically and currently, have been lower for health professionals than the general public (Asch, Jedrziewski, & Christakis, 1997; Cummings, Savitz, & Konrad, 2001; Sudman, 1985). From Sudman’s seminal article in 1985, reasons for lower response rates from physician surveys include lack of time, saliency, or perceived lack of importance, concerns about confidentiality, and concern about bias of the survey, either in general or for specific questions, including not allowing a full range of responses to questions. The presence of “gate keepers,” office personnel who in effect screen mail and e-mail requests of the health-care professionals for whom they work, has been cited recently as a major reason for low response among health professionals (Klabunde et al., 2012).
In addition to lower response rates historically among health-care professionals, most reviews find that response rates have been declining (Cho, Johnson, & VanGeest, 2013; Cull, O’Connor, Sharp, & Tang, 2005; McLeod, Klabunde, Willis, & Stark, 2013). In a review of 50 surveys of pediatricians from 1994 to 2002, Cull, O’Connor, Sharp, and Tang (2005) found that response rates decreased from 70% to 63%, for time periods of 1994–1998 to 1999–2002. Cho, Johnson, and VanGeest (2013), in a meta-analysis of 48 surveys of health professionals from 1948 to 2012, found that response rates decreased from over 80% before 1960 to around 50% in 2000 and then to 42% in 2012. In a review of surveys of health-care providers between 2000 and 2010, using a 60% response rate as a benchmark, the percentage of surveys that met this benchmark decreased from 61% in 1998–2000 to 36% in 2005–2008 (McLeod et al., 2013). Possible reasons for declining response rates for health-care professionals are increased requests to complete such surveys and increased workloads, both in number of patients and administrative obligations, although this has not been explicitly demonstrated (Klabunde et al., 2012).
Response rates are not the only measure of quality. Response bias, or nonresponse bias, occurs when those who respond differ from those who don’t on the outcome of interest; this bias has grown in importance as a measure of survey quality (Johnson & Wislar, 2012; Shelley, Brunton, & Horner, 2012). Assessing nonresponse bias can be done in a number of ways. The most common way is comparing characteristics of those who respond with those who do not, preferably on the outcome measure. As this information is rarely available for nonresponders, a surrogate or correlate of the outcome measure may be used. Other ways of assessing nonresponse bias involve comparing early and late responders or following-up more extensively on initial nonresponders. The latter provides only a limited assessment of nonresponse bias because late responders and responders to more extensive follow-up are still responders and thus may not reflect characteristics of true nonresponders. Few studies report on potential response bias (Asch et al., 1997; Cummings et al., 2001). Cummings, Savitz, and Konrad (2001), in a review of 27 mailed physician surveys published between 1986 and 1995, reported that only 18% estimated any type of response bias. Studies that have assessed potential response bias in physician surveys have found little (Field et al., 2002; Kellerman & Herold, 2001; McFarlane, Olmsted, Murphy, & Hill, 2006).
The review by Cho et al. (2013) found minimal response bias; specifically, response was slightly higher for younger professionals, females, and nonspecialty professionals. Higher response rates were reported for (1) mail (57%) than online (38%) or mixed mode (49%), (2) monetary (60%) than nonmonetary (48%) or no incentive (48%), (3) physicians (55%) than nurses (51%) or other health professionals (46%), (4) one (57%) or two follow-up reminders (66%) than none (43%) or three (49%), (5) non-U.S. (57%) than U.S. (43%) setting, and (6) non-RCT (57%) than Randomized Controlled Trial (RCT) (50%) study designs.
Monetary incentives have consistently been found to increase response rates (Asch, Christakis, & Ubel, 1998; Halpern, Ubel, Berlin, & Asch, 2002; Kasprzyk, Montano, St. Lawrence, & Phillips, 2001; Keating, Zazlavsky, Goldstein, West, & Ayanian, 2008; Leung, Ho, Chan, Johnston, & Wong, 2002; Robertson, Walkom, & McGettigan, 2005). Even small amounts, for example, US$1, US$2, and US$5, increase participation (VanGeest et al., 2007). Prepaid incentives are more effective than promised incentives (Delnevo, Abatemarco, & Steinberg, 2004; Leung et al., 2004). In general, most studies have found that the larger the incentive, the larger the effect (Asch et al., 1998; Halpern et al., 2002; Kasprzyk et al., 2001; Keating et al., 2008) but not all (Burt & Woodwell, 2005; VanGeest, Wynia, Cummins, & Wilson, 2001). An optimal amount has not been determined (Klabunde et al., 2012). Few nonmonetary incentives increase participation (Burt & Woodwell, 2005; Halpern et al., 2002).
Studies of factors affecting response rates among nonphysician health-care providers have been few, compared to those among physicians, and most were primarily done by mail (Guise, Chambers, Valimaki, & Makkonen, 2010; Hawley, Cook, & Jensen-Doss, 2009; Hill, Fahrney, Wheeless, & Carson, 2006; Paul, Walsh, & Tzelepis, 2005; Ulrich et al., 2005; VanGeest & Johnson, 2011). In general, findings among nonphysician providers are consistent with those among physicians, namely, that monetary incentives, even small amounts, increase response rates compared to no incentives, nonmonetary, or a lottery. Hawley, Cook, and Jensen-Doss (2009), in a study including both physician and nonphysician providers, found a lower response rate among psychiatrists than therapists, psychologists, counselors, or social workers. This is in contrast to the across-studies comparison by Cho et al. (2013), who found that physicians typically responded at higher rate than did nonphysician providers. VanGeest and Johnson (2011), in their review of studies among nurses, found that nurses responded well to telephone strategies, in contrast to surveys among physicians (Cho et al., 2013) who typically respond very poorly to telephone surveys.
Use of online (also referred to as web based or electronic) methodology to conduct surveys has become increasingly popular. Online methodology has many advantages compared to postal mail or telephone methods, such as being quicker, less expensive, and typically having higher rates of item completeness. The majority of online surveys costs are due to programming and enabling e-mail delivery. Because these largely are initial costs, the cost efficiency of online surveys increases as the sample size increases. Often-cited limitations of online surveys are low response rates and difficulty in specifying the sampling base (Braithwaite, Emery, de Lusignan, & Sutton, 2003; de Leeuw, 2012; van Selm & Jankowski, 2006). The latter occurs when there is an incomplete or outdated list of e-mail addresses for the target population. Beebe, Locke, Barnes, Davern, and Anderson (2007) conducted a randomized mixed-mode study of 500 physicians. One group was contacted by e-mail first and asked to complete the survey online. The other group was sent an invitation and questionnaire via postal mail. One week after initial notification and request for participation, each nonrespondent was sent a reminder, in the same mode (online or paper) as the original request. After another week, each nonrespondent was sent another reminder and request, this time in the “other” mode. The response rate, completed/eligible, was calculated for each arm, web/mail and mail/web, completion/eligible, significance of difference determined. The results were web/mail 62.9% and mail/web 70.2%, p =.07. They concluded that mail then web results in a slightly higher response; however, if time is a factor, they recommended using web then mail. Schleyer and Forrest (2000) presented a cost–benefit analysis comparing postal and electronic mail. If basic assumptions of availability of valid e-mails for study population are met, using electronic is more cost effective at sample sizes of 348 or more; furthermore, the cost–benefit is larger with larger study sizes.
For more than two decades, access to the Internet has seldom been an issue when surveying health-care professionals. The challenge has been, and remains, how to catch professionals’ attention sufficiently to elicit a questionnaire response. This is especially challenging in the current era of information overload. As stated above, few surveys of health-care providers have obtained response rates above 70% (reviews: Flanigan, McFarlane, & Cook, 2008; McLeod et al., 2013; Shelley et al., 2012; VanGeest et al., 2007). The Medical Expenditure Provider surveys, Medical Provider Component, have achieved 80–95% response depending on provider type from the 2006 survey (Stagnitti, Beauregard, & Solis, 2008); however, these are follow-up surveys to providers of patients who participated, conducted primarily by telephone. Therefore, these may not be directly comparable to a typical or stand-alone survey. Virtually, all surveys of dentists with response rates over 80% have been by postal mail and outside of the United States: Swedish orthodontists, 87% (n = 157; Bjerklin & Bondemark, 2008); British general dentists, 86% (n = 75; Sutton, Ellituv, & Seed, 2005); Ugandan dentists, 82% (Mutyabule & Whaites, 2002); and British periodontists, 82% (n = 459; McCrea, 2008).
Shelley, Brunton, and Horner (2012) reviewed 53 surveys of dental radiology published between 1983 and 2010 to develop recommendations to assist future researchers. They argued that study characteristics other than response rates should be considered. One example is the specification of the sampling base. In their review, Shelley et al. (2012) reported a mean response rate of 74%; however, they included in their review surveys that were conducted at meetings in which there is a 100% response rate. As they noted, these surveys are not comparable to a typical survey; response rates from these contexts are of lesser value as a quality indicator.
To our knowledge, no large survey of health-care providers, which we define as having more than 1,000 potential respondents in the sampling frame, with an online component has reported a response rate of over 70%. Using methods described by Schleyer and Forrest (2000), the modified tailored approach of Dillman (2007), and the recommendations from Klabunde et al. (2012), the proceedings from a 2010 National Cancer Institute (NCI) workshop surveying health-care providers, we report our experience using a large online component in obtaining excellent response to a survey of dentists in the National Dental Practice–Based Research Network.
The objectives of this report are to quantify the (1) increase in response rate associated with successive survey methods in a questionnaire completed by dentists and by three sequential categories of recruitment (electronic, paper, and personal follow-up), (2) time to completion with each successive step, (3) contribution from the final method (follow-up by study staff [regional coordinator]), and (4) differences in response rate and mode of response by practice/practitioner characteristics.
Method
The National Dental Practice–Based Research Network (“network”) is a consortium of dentists and dental organizations focusing on improving the scientific basis for clinical decision making (Gilbert et al., 2013). Its mission is “To improve oral health by conducting dental practice-based research and by serving dental professionals through education and collegiality.” It is committed to maximizing the practicality of conducting research about clinical practice across geographically dispersed regions and diverse practice types. The network comprises six geographic regions, each with a regional director and coordinator for administrative purposes. Many details about the network are available at its website, www.nationaldentalpbrn.org. This study was approved by the respective institutional review board(s) of each of the network’s regions.
Enrollment Questionnaire
As part of the network enrollment process, practitioners complete an Enrollment Questionnaire that describes themselves, their practice(s), and their patient population. A copy of the questionnaire is publicly available (National Dental Practice-Based Research Network [PBRN] Study Results Page). Questionnaire items from the Enrollment Questionnaire, which had documented test/retest reliability, were taken from our previous work in a practice-based study of dental care and a PBRN that ultimately led to the National Dental PBRN (Florida Dental Care Study, 2015; Gilbert et al., 2011). The typical enrollee completes the questionnaire online, although a paper option is available. Invitations to enroll are typically done by mass mailings or by face-to-face request during a dental professional meeting. These invitations are one time only; they are not followed up by any further mail, e-mail, or personal contact.
Content of the Isolation Techniques Questionnaire
After confirming on the questionnaire itself that the respondent was a general dentist and that he or she does at least one root canal treatment each month (as compared to the “do at least some” criterion taken from the Enrollment Questionnaire), respondents were asked for the number of root canal treatments performed each month and the frequency and type of isolation methods used. The questionnaire was comprised of 57 questions printed on eight pages. A copy of the full questionnaire is publicly available (National Dental PBRN Study Results Page). There is no overlap of information requested between the enrollment and isolation techniques questionnaire.
Electronic Development and Testing of Online Isolation Techniques Questionnaire
An online web survey system was used for primary data collection and management. The system tracked all activity for each of the participating dentists. Each component was tested and a full system test was executed to ensure that it was functional as expected; functional testing included screen review, navigation assessment, and data entry. The web survey system was tested on Microsoft Internet Explorer, Google Chrome, Mozilla Firefox, and Apple’s Safari browsers. The web survey was rendered as a series of hypertext markup language pages. Users advanced through the pages by selecting a “Next” button. The content on each page was limited to help minimize the scrolling required. Respondents could skip questions with the exception of the eligibility question but were prompted when they tried to advance to a different screen or submit a page with an omitted question and were asked to confirm if they wanted to leave the screen without answering the question. The system timed out after 30 min of no use. When they logged back in, the system would take them to the screen where they left off. Respondents were allowed to save their responses and continue at a later time. User acceptance testing of the web-based survey was performed by various members at each of the network administrative regions. The user testing evaluated the readability, feasibility, and Internet browser compatibility of the electronic survey to ensure that the system functioned as expected and was consistent with all protocol requirements.
Administration of the Isolation Techniques Questionnaire
By January 31, 2014, more than 5,000 persons had completed an Enrollment Questionnaire; 1,876 of these persons were invited to participate in the Isolation Techniques questionnaire because they met four criteria: (1) general dentist, (2) currently practicing/seeing patients, (3) reported performing at least some root canal treatment, and (4) selected the “limited” or “full” participation levels, as compared to the “information only” level of participation in the network. Preprinted invitation letters were postal mailed to eligible practitioners, inviting them to participate and informing them that they would receive an e-mail with a link to the electronic version of the questionnaire. At the time of the e-mail, the practitioners were given the option to request a paper version of the survey; none did. Practitioners were asked to complete the questionnaire within 2 weeks. Two reminder e-mails were sent at 2-week intervals to those who had not completed the questionnaire. A postal reminder was sent with the third e-mail reminder, again 2 weeks after prior reminder, 6 weeks after the initial e-mail request; a printed version of the questionnaire was included with the postal reminder offering the practitioners the option of completing the questionnaire on paper. After an additional 2 weeks, another postal reminder with a printed questionnaire was sent. If a response was not received within 2 weeks, regional coordinators followed up to ensure that the network communications had been received and ascertain whether the dentist was interested in participating. There was not a specific protocol regarding mode and order of contacting by the regional coordinators. They followed up with telephone calls, fax, and personal e-mails (from themselves as opposed to being from the network Coordinating Center); some started with telephone calls, while others focused on e-mails. Additionally, each region holds annual meetings for practitioners to inform them about current and planned network studies and elicit their input. Practitioners in regions (South Central, Midwest, and Northeast) who held an annual practitioner meetings between February and April 2014 and who had received at least the initial e-mail invitation, but had not completed the survey, were offered the opportunity to complete a paper survey at the meeting.
Data collection was closed 12 weeks after the original e-mail invitation. Practitioners or their business entities were remunerated US$50 for completing the questionnaire because monetary incentives have consistently shown that these incentives increase participation (VanGeest et al., 2007); if they confirmed at the end of the survey that they would like remuneration (86% did so). Survey data were collected from January 31, 2014, to July 15, 2014; completion of the isolation techniques questionnaire was not linked in any way to when the enrollment questionnaire had been completed. After survey collection, regional coordinators’ follow-up logs were reviewed to ascertain whether the network communications e-mail links and postal questionnaires had been received and whether the practitioner had moved locations as well as the number of contact attempts made.
To document test/retest reliability, 43 respondents completed the same questionnaire twice online. The mean (SD) time between test and retest was 15.5 (3.0) days. The agreement between Time 1 and Time 2 for individual questionnaire items was quantified using a mean weighted κ score, which was 0.62, with an interquartile range of 0.46–0.79.
Of the 1,876 dentists invited, 24 were deemed ineligible before beginning the questionnaire (4 had died, 15 were no longer practicing, and 5 no longer provided root canal treatment) and 136 were determined ineligible after completing the questionnaire (3 no longer a general dentist and 133 reported not doing at least one root canal treatment each month), leaving 1,716 eligible dentists; 22 were active refusals and 194 were nonrespondents. Overall, 1,500 responded: 1,488 (87%) completed the entire survey, 6 only answered the first question, 3 erroneously checked that they were not a general dentist and consequently were electronically skipped to end of survey, and 3 completed varying parts of the first half of the questionnaire. The average time to complete the survey was 15 min.
Analysis
Participation at each stage was calculated two ways: (1) incremental: of remaining eligible, the proportion participating; and (2) proportional: of those participating, the proportion accrued at each stage. Participation was also categorized into three groups: (1) after initial postal notification letter, recruited, and completed electronically, that is, online; (2) recruited with e-mail and postal, completed on paper; and (3) required follow-up of regional coordinators, completed online or on paper. Significance of differences in the proportion of eligible dentists who participated according to practice/practitioner characteristics was ascertained in bivariate analysis using χ 2 tests. We also tested differences between those for whom regional coordinators followed up with whether or not they participated, and also among those who participated, in whether the survey was completed online or on paper. Independent associations with participation were assessed using logistic regression. Where indicated, some categories were collapsed (e.g., region and practice type), bivariate analyses rerun, and dichotomous grouping entered in the model. Characteristics with p<.20 in bivariate analysis were entered into the model; next, stepwise regression was used, removing variables until only those with p<.05 remained. Odds ratios (OR) and 95% confidence intervals (CIs) were calculated from the models. All analyses were performed using SAS (see 9.4 (SAS Institute). Of the 1,488 completed surveys, 21 were completed at annual practitioner meetings. To enhance generalizability, the analysis described below excludes the 21 practitioners who completed the survey at an annual meeting.
Results
Completion Rates and Type
As shown in Table 1, 30% of practitioners completed the survey after the initial e-mail, 19% and 11% after the second and third e-mail reminder, respectively, 19% and 32% after the fourth e-mail/first postal and second postal reminder, respectively, and of the remaining practitioners 52% completed the survey either more than 2 weeks after the second postal reminder but before regional coordinators followed up or after regional coordinators followed up. Proportionally, 35% of practitioners responded within 2 weeks of the first e-mail, an additional 22% within 2 weeks of the second and third e-mails, another 26% within 2 weeks of the fourth e-mail/first postal and second postal reminder, and 4% within 2 weeks of the second postal reminder but before being contacted by regional coordinator and 13% as a result of regional coordinator follow-up. Overall, 66% were recruited electronically, 21% with postal follow-up and completion on paper, and the final 13% with follow-up by regional coordinators and completed online or paper. A total of 24% (353/1,470) were completed on paper.
Table 1.
Participation Rates by Recruitment Step.
| Recruitment Step | na Eligible | Participated | No. of Days From Initial E-mail to Participation | Median No. of Days From Prior Step to Participation |
---: :---:
| nb | Percentage of Remaining Eligible | Median | Interquartile Range |
| 1. Initial e-mail | 1,695 | 519 | 31 | 2 | 1–4 | |
| 2. First e-mail reminder | 1,176 | 222 | 19 | 16 | 15–19 | 2 |
| 3. Second e-mail reminder | 954 | 97 | 10 | 30 | 29–33.5 | 2 |
| 4. Third e-mail/First postal reminder | 857 | 165 | 19 | 54 | 46–57 | 12 |
| 5. Second postal reminder | 692 | 223 | 32 | 62 | 61–67 | 6 |
| 6. Regional coordinator follow-upc | 469 | 244 | 52 | 81d | 75–91 | 25 |
| Total | 1,695 | 1,470 | 87 | | | |
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a
Excludes 21 completed at regional meetings.
b
Within 2 weeks of steps No. 1–5.
c
Of 244, 55 were completed before regional coordinator contact.
d
Median No. of days for 55 was 75; for remaining 189, median of 83 days.
Time to Completion
The median time to completion was 2 days after the initial e-mail, for those completing prior to the reminder e-mail sent 2 weeks later. Similarly, there were spikes at 2 days after the second and third e-mail reminders, with diminishing gain in recruitment described above. Spikes after a postal reminder had been sent were longer at 12 and 6 days, respectively; these represent greater recruitment gains than the simple e-mail reminders. The recruitment time involving follow-up by regional coordinators was longer, peaking at 25 days, but nonetheless accompanied by a substantial gain in response.
Contribution by Regional Coordinators
There were 469 eligible practitioners from whom surveys had not been received 2 weeks after the second postal mail request and for whom regional coordinators followed up to ascertain if they had received study information. A total of 55 were completed prior to regional coordinators attempting follow-up (28 online and 27 on paper). Of the 411 practitioners whom regional coordinators contacted or attempted to contact, 189 (46%) were completed. The median number of times a regional coordinator contacted or attempted to contact a practitioner was three (Interquartile range: 2–3; range: 1–11); fewer for when surveys were completed than when not (medians: 2 vs. 3, p<.001). Of the 189 completed surveys, 77 (41%) involved only one contact attempt.
Thirty practitioners were no longer at the practice of record; new practice information was obtained for 27, of whom 10 completed the survey. Only seven practitioners or their offices reported not receiving survey information, and when resent, either by e-mail or fax, all seven were completed. Offices of three practitioners refused information (hung-up) when regional coordinators tried to ascertain if information was received. Overall, completed surveys were obtained from 46% (n = 189 of the 411) of practitioners after follow-up by regional coordinators. Of these, 127 (67%) were completed online and 62 (33%) on paper.
Associations With Overall Participation and by Type
In bivariate (Table 2) and adjusted analysis, higher proportions of practitioners from the Western region participated, as did those who were members of at least one dental association and those who either worked in large group (managed care) practices or were owners of private practices compared to their counterparts. Among the 411 practitioners whom regional coordinators contacted or attempted to, there were no significant differences in practice/practitioner characteristics between those who ultimately participated and those who did not, based on bivariate analyses (Table 2). In adjusted analyses, however, there were significant differences. Male practitioners were more likely to ultimately participate after follow-up by regional coordinators than were females (OR = 2.1; 95% CI = [1.2, 3.5], p =.006). Likelihood of ultimately participating decreased with years since dental degree (per 10 years, OR = 0.78, 95% CI = [0.66, 0.93], p = .004).
Table 2.
Participation Rate, by Practitioner/Practice Characteristics.
| Practitioner/Practice Characteristics | Overall | Participated | Of Followed-Up by Regional Coordinators (n = 411) | Of Participated |
:---: :---:
| Participated (n = 189) | On Paper |
| | |
| n | % | n | % | n | % | n | % |
| Gender |
| Female | 389 | 23 | 331 | 85 | 39 | 40 | 62 | 19 |
| Male | 1,292 | 77 | 1,130 | 87 | 149 | 48 | 290 | 26 |
| | | | p = .2 | p = .15 | p = .01 |
| Race ethnicitya |
| White | 1,321 | 79 | 1,156 | 88 | 145 | 47 | 292 | 25 |
| Black/African American | 82 | 5 | 65 | 79 | 9 | 35 | 16 | 25 |
| Asian/Pacific Islander | 163 | 10 | 145 | 89 | 21 | 54 | 24 | 17 |
| Other | 12 | 1 | 9 | 75 | 1 | 25 | 1 | 11 |
| Hispanic/Latino | 91 | 5 | 78 | 86 | 12 | 46 | 18 | 23 |
| | | | p = .14 | p = .5 | p = .2 |
| Age (years) |
| <35 | 186 | 11 | 157 | 84 | 23 | 44 | 14 | 9 |
| 35–44 | 367 | 22 | 328 | 89 | 46 | 55 | 64 | 20 |
| 45–54 | 350 | 21 | 305 | 87 | 46 | 51 | 90 | 30 |
| 55–64 | 592 | 35 | 520 | 88 | 59 | 44 | 136 | 26 |
| 65 and older | 186 | 11 | 154 | 83 | 14 | 31 | 48 | 31 |
| | | | p = .2 | p = .10 | p< .001 |
| Years since dental school graduation |
| <10 | 318 | 19 | 274 | 86 | 43 | 50 | 38 | 14 |
| 10–19 | 334 | 20 | 296 | 89 | 44 | 54 | 66 | 22 |
| 20–29 | 382 | 23 | 334 | 87 | 43 | 46 | 96 | 29 |
| 30+ | 655 | 39 | 563 | 86 | 57 | 39 | 153 | 27 |
| | | | p = .6 | p = .10 | p< .001 |
| Additional formal training after dental school |
| No | 1,009 | 60 | 865 | 86 | 108 | 43 | 228 | 26 |
| Yes | 686 | 41 | 605 | 88 | 81 | 51 | 125 | 21 |
| | | | p = .14 | p = .13 | p = .02 |
| Membership in any dental organizations |
| No | 228 | 13 | 184 | 81 | 24 | 36 | 43 | 23 |
| Yes | 1,467 | 87 | 1,286 | 88 | 165 | 48 | 310 | 24 |
| | | | p = .004 | p = .09 | p = .8 |
| Practice |
| Practice type |
| Owner of private practice | 1,239 | 74 | 1,082 | 87 | 134 | 46 | 299 | 28 |
| Associate of small group private practice | 215 | 13 | 174 | 81 | 27 | 40 | 29 | 17 |
| Member large group practice (HP/PDA)b | 103 | 6 | 98 | 95 | 11 | 69 | 8 | 8 |
| Public, community, and publicly funded | 70 | 4 | 62 | 89 | 9 | 53 | 11 | 18 |
| Federal government, academic, and other managed care | 59 | 4 | 49 | 83 | 8 | 42 | 5 | 10 |
| | | | p = .007 | p = .3 | p< .001 |
| More than one practice location |
| No | 1,428 | 84 | 1,239 | 87 | 154 | 45 | 307 | 25 |
| Yes | 263 | 16 | 230 | 87 | 35 | 53 | 46 | 20 |
| | | | p = .8 | p = .2 | p = .12 |
| Locale of practice |
| Urban—inner city | 184 | 11 | 156 | 85 | 28 | 49 | 39 | 25 |
| Urban—not inner city | 449 | 27 | 400 | 89 | 51 | 50 | 101 | 25 |
| Suburban | 760 | 45 | 660 | 87 | 76 | 44 | 144 | 22 |
| Rural | 289 | 17 | 251 | 87 | 33 | 47 | 68 | 27 |
| | | | p = .5 | p = .8 | p = .3 |
| Patient population |
| Percent patients with private insurance |
| < 40 | 244 | 15 | 202 | 83 | 35 | 45 | 48 | 24 |
| 40–79 | 987 | 60 | 865 | 88 | 103 | 46 | 224 | 26 |
| | | | p = .09 | p = .9 | p = .06 |
| Percent patients who come in regularly |
| < 50 | 310 | 19 | 265 | 85 | 42 | 49 | 63 | 24 |
| 50–79 | 995 | 60 | 868 | 87 | 106 | 46 | 212 | 24 |
| 80+ | 351 | 21 | 311 | 89 | 35 | 45 | 69 | 22 |
| | | | p = .5 | p = .8 | p = .7 |
| Regionc |
| Western | 186 | 11 | 173 | 93 | 21 | 64 | 25 | 14 |
| Midwest | 162 | 10 | 135 | 83 | 16 | 38 | 25 | 19 |
| South Central | 390 | 23 | 344 | 88 | 51 | 53 | 112 | 33 |
| South Atlantic | 288 | 17 | 252 | 88 | 20 | 36 | 63 | 25 |
| Northeast | 360 | 21 | 310 | 86 | 32 | 39 | 72 | 23 |
| | | | p = .02 | p = .06 | p < .001 |
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a
Although race and Hispanic/Latino ethnicity are separate questions in the Enrollment Questionnaire, some Hispanic/Latino participants did not provide a race or indicated “Hispanic/Latino” as their race, thus race and ethnicity were combined.
b
Either HealthPartners Dental Group in greater Minneapolis, MN, or Permanente Dental Associates in greater Portland, OR.
c
Reported on Enrollment Questionnaire as the state, subsequently categorized into one of the six regions of the network.
Among participating practitioners, completion on paper was more common among practitioners who were male, older, had more years since graduated from dental school, did not have any additional training, were owners of private practice, or from the south central region (Table 2). In adjusted analyses, all the associations remained significant except years since graduated from dental school.
Discussion
An exceptionally high participation rate (87%) was obtained. Each step produced incremental gains in response: diminishing gains with the two reminder e-mails, 19% and 10%, respectively; increasing gains with postal reminders accompanied with a printed questionnaire, 19% and 32%, respectively; and the largest gain with the final step of personal contact, 52%. Response peaked at 2 days after e-mail reminders were sent, 12 and 6 days for postal reminders, and 25 days for personal contact. Practice/practitioner characteristics differed by mode of response, paper vs. electronic, but not by whether or not response was obtained only after Regional Coordinator personal contact.
In conducting any survey, regardless of mode (postal, telephone, electronic, or combination), reminders are needed to obtain acceptable response rates. Four reminders have been advocated as an appropriate number (Dillman, 2000, 2007). Typically, there is decreasing gain with successive reminders; however, most studies that report gain by reminder do so using the same mode of reminder, primarily either electronic or postal. For example, Toledo and colleagues (2015) in an online survey of 5,433 primary health-care professionals in Spain used four e-mail reminders; overall response was only 36%. Reminders were sent at 10-day intervals. A 7% response was obtained in the 24 hr after initial e-mail. Steady but diminishing responses of 1–2% per day were observed until the first e-mail reminder sent after 10 days, after which a 4% response was observed in the next 24 hr. The pattern repeated with smaller daily increases of 0.5–1% and smaller peaks of 2% then 1%, following second through the fourth reminders. When staying within the same mode, electronic (e-mail), we also found diminishing gains with successive reminders for the two e-mail reminders. In contrast, we found increasing response when the mode of reminder was changed from e-mail to postal with a printed copy of the questionnaire included, and even more so, when the reminder mode was changed from postal to personal contact. This was most notable for the last step, personal contact by regional coordinators. Excluding practitioners who responded before regional coordinators attempted to contact, nearly half (46%) responded after being contacted; this was the highest proportional response of any step and the absolute number responding (n = 189) was comparable to the prior two steps.
The 2008 National Sample Survey of Registered Nurses (U.S. Department of Health and Human Services, 2010) used a multimodal approach similar to ours. Its protocol differed in that the nurses had to enter a web address, while our participants only had to click on a link provided in an e-mail. Also, in the last step, the nursing survey could be completed over the telephone, which was not an option provided for our survey. Their overall response rate was 62%; 27% paper, 24% online, and 10% telephone. Stepwise response rates were not presented. A study of physicians that is the most comparable to the current dental study is the study by Kroth and colleagues (2009) who examined clinicians’ response rate across three medical PBRNs. Their survey was of active network members with valid e-mail addresses, as was ours. Their initial invitation was via e-mail, with five rounds of electronic solicitation for an online-based questionnaire and two rounds of a paper-based version mailed to nonresponders. The electronic solicitations (e-mails) were personalized, came from the physician’s local practice-based network, and had a customized link to the online survey that provided automatic log-in. They had no final telephone follow-up. As with ours, the greatest response was within 2 days of initial e-mail (12%), diminishing responses occurred with the second through fourth e-mails; the paper option sent with third e-mail had a modest response, with no discernible response to either the fifth e-mail or a second paper option. Their overall response rate was 61%: 46% online and 15% on paper. There were two primary differences in their recruitment methods and ours: (1) their initial invitation was via e-mail, while ours was postal mail followed immediately by e-mail with embedded link; and (2) we had a last step of personal/staff outreach. From their graph, an estimated response within 2 weeks of initial e-mail invitation is 15% (120/805); ours was 31%. While it is doubtful that a change in mode of initial invitation would account for a doubling of response rate, it is conceivable that it could account for a 5–10% difference in response rates. There are no studies that we could find that assessed difference in response rates by mode of initial invitation in a “defined” population, that is, a nonrandom group such as practice-based networks. Our response rate, excluding the last step, was 76% ([1,470 – 244 + 55]/1,695), which would make the response rates in the studies comparable.
There can be many obstacles when surveying health professionals. The importance of these surveys and their possible difficulties is why the NCI convened a workshop in November 2010 to discuss the challenges (Klabunde et al., 2012). The first topic area identified for improvement was identification of an appropriate sampling frame (the NCI workshop focused on physicians). Ideally, a sampling frame should be complete, current, include no duplications, and have no ineligible persons; such a sampling frame is very rare. Our sampling frame of network dentists meets these criteria except that it had some ineligibles, which were screened out via the first questions on the survey. The underlying question regarding the sampling frame in our study is whether they are representative of U.S. dentists. Network members are not recruited randomly, so factors associated with network participation (e.g., an interest in clinical research) may make network dentists unrepresentative of dentists at large. While it cannot be asserted that network dentists are entirely representative, we can state that they have much in common with dentists at large, while also offering substantial diversity in these characteristics. This assertion is warranted because (1) substantial percentages of network dentists are represented in the various response categories of the characteristics listed in Table 1, (2) findings from several network studies document that network dentists report patterns of diagnosis and treatment that are similar to patterns determined from nonnetwork general dentists (Gordan et al., 2009; Norton et al., 2014), and (3) the similarity of network dentists to nonnetwork dentists using the best available national source, the 2010 American Dental Association Survey of Dental Practice (American Dental Association, 2012; Makhija et al., 2009). Although not stated as an objective in the NCI workshop, Shelley et al. (2012), in their review of dental studies, expanded on the concept about properties of a sampling frame. In an appropriate sampling frame, every member should have an equal chance of being selected and random sampling should be used. If the sampling frame is large, an appropriate sample size estimate should be made so as to avoid having to survey the entire sampling frame. Sampling the entire frame is a waste of resources (Dillman, Smyth, & Christian, 2009). We estimated that we needed a sample of 1,000. As we were unsure of what the participation rate would be, and wanted to assess the yield of having staff follow-up with nonresponders as a last step, we decided to survey the entire group.
A second topic from the NCI workshop had to do with how to optimize a mixed-mode approach, namely, an approach that uses both postal and electronic mail. Response rates are usually higher for mail than telephone or electronic, with telephone being extraordinarily difficult, and the realization that electronic will become even more pervasive than it already is. We had no randomization component to make direct comparisons, but our design was intended to optimize response. We started with mail notification, as studies using electronic notification (e-mail) had poor response. We did follow mail notification with e-mail because this allowed embedding a link which the practitioner could simply click on to begin the survey. We had two e-mail notifications at 2-week intervals. Other studies have found that gain from more reminders decreases markedly after two. Also, from our prior work, we found that practitioners who completed paper forms differed from those who completed electronically (Funkhouser et al., 2014); thus, we knew that we wanted to include a mail component as well.
The third topic area from the NCI workshop had to do with the role of gatekeepers. Our “last step,” having network staff follow-up with nonresponders, addressed that challenge. Of practitioners who completed the survey, almost half (44%) required only one call, in essence getting past a gatekeeper, yet there were others who, even after 10–11 attempts, calls, e-mails, or fax, were not responsive.
Regarding the issue of personal outreach, we believe that this study demonstrates for other organizations the potential utility of personal outreach. Not only did outreach increase the number of practitioners who responded, it also identified practitioners who were no longer eligible, thereby reducing the denominator. Because organizations will have contact information on their sampling frame or might be able to access public information to identify additional contact information (which our research assistants sometimes used), these organizations can make their own assessment of the utility of this approach for their sample or some targeted subset of it (based on the subset’s projected potential for a higher response rate). We estimate that to follow up with the 441 practitioners who had not responded after e-mail contacts required approximately 15 work-days of staff time. The majority of practitioners in our study had not participated in other studies nor had they had previous contact with our research assistants. It is typically only after network enrollees have done an in-office clinical study or attended a regional meeting of network practitioners do they begin to develop a professional relationship with the network’s research assistants.
A limitation of the study may be the relatively uncommon sampled population, specifically, the National Dental PBRN. As this was the first survey of the new national network, recruitment into the network began April 2012 and the survey was conducted in early 2014; the practitioners may have been even more likely to respond. We think this largely explains the high response, 31%, to our initial invitation. The incremental increases with second and third e-mail reminders, and the two postal reminders, should be applicable to other researchers and populations where there is some type of existing relationship. The comparability of response rates and initial reminders between our study and those of Kroth et al. (2009) support this. The large proportional and absolute gain with the final step, personal contact, surprised us. We do not know if others will find it similarly beneficial; we present it so that others may try. The gain from personal outreach comprised 13% of our respondents, which is not much larger than the 10% completed by telephone for the 2008 National Sample Survey of Registered Nurses (U.S. Department of Health and Human Services, 2010). Although the nurses’ survey was completed on the telephone and ours was via personal contact, they both used personal contact, which can be expensive and may not be as fruitful in other populations. In their review of surveys of nurses, VanGeest and Johnson (2011) found that nurses responded to telephone strategies comparable to those of mail. This differs markedly from studies conducted among physicians, which find poor response to telephone strategies (Cho et al., 2013). “Gatekeepers” in medical offices may make personal contact by telephone extremely difficult. In our study, personal contact did not necessarily mean with the dentist himself or herself and usually was not; it entailed reaching a person in the office and verifying that the dentist had received the materials. There have been no comparable studies of surveys among dentists to evaluate telephone strategies in obtaining response to surveys. We speculate that trying to get a dentist to complete a survey on the telephone would have had a very poor response, as it has with physicians.
In summary, we believe that other organizations, such as other PBRNs (of which there are many), membership associations (such as health-care professional organizations), or large cohorts from ongoing studies, may be able to use our methods in a cost-efficient manner to maximize their response rates. Using a multimodal protocol, it is possible to obtain a high participation rate with a large online component. Although response steps were not randomized, we believe that it is unlikely that additional e-mail reminders, without postal and/or telephone follow-up, would have meaningfully increased the response from 57%. Also, as we have reported previously from an earlier survey, there appears to be a difference between practitioners who respond on paper and those who do so online (Funkhouser et al., 2014). Of note, the late responders (those who did not participate until after follow-up by regional coordinators) did not differ from early responders in any characteristic assessed including mode of completion. Also, the high absolute and proportional response (52%) to the last step (personal follow-up) is noteworthy. Although this step adds to the cost, the yield is large.
Acknowledgments
The authors thank Dr. Wynne Norton, assistant professor, School of Public Health, University of Alabama at Birmingham, for her work during the development of the study protocol and questionnaire. An Internet site devoted to details about the nation’s network is located at Persons who comprise the National Dental PBRN Collaborative Group are listed at Opinions and assertions contained herein are those of the authors and are not to be construed as necessarily representing the views of the respective organizations or the National Institutes of Health. The informed consent of all human subjects who participated in this investigation was obtained after the nature of the procedures had been explained fully.
Funding
The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This work was supported by grant U19-DE-22516 from the National Institutes of Health (NIH).
Footnotes
Declaration of Conflicting Interests
The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
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Quality Assurance and Quality Control: Ageing of Northwest Atlantic Species
It is important to ensure consistency in fish ages generated by a production ageing laboratory.
New England/Mid-Atlantic
On This Page
Summary of all precision tests, 2016-2020.
Acadian Redfish
American Plaice
Atlantic Cod
Atlantic Herring
Atlantic Mackerel
Atlantic Surfclam
Black Sea Bass
Bluefish
Butterfish
Cusk
Golden Tilefish
Haddock
Pollock
Red Hake
Scup
Silver Hake
Summer Flounder
White Hake
Windowpane Flounder
Winter Flounder
Witch Flounder
Yellowtail Flounder
More Information
Fishery Biology Program staff regularly test themselves to ensure their ages are precise and accurate. This site reports the results of all such tests, using measures such as percent agreement and total coefficient of variance (CV). Please contact Sandy Sutherland for more information, including details of individual tests.
Summary of all precision tests, 2016-2020.
A perfect score is 100% agreement (0% CV); our standards require scores above 80% agreement & under 5% CV.
SpeciesNumber of TestsFish AgedFish in TestsCV (%)Agreement (%)
Acadian Redfish 9 5,557 762 1.95 72.9
American Plaice 25 13,888 2,195 4.10 67.3
Atlantic Cod 30 24,847 2,356 0.13 99.5
Atlantic Herring 11 7,295 1,002 0.93 92.4
Atlantic Mackerel 19 7,618 1,400 0.77 97.5
Atlantic Surfclam 1 659 94 0.43 95.7
Black Sea Bass 27 15,617 2,154 2.40 86.6
Bluefish 7 2,173 439 2.48 89.3
Butterfish 15 8,237 993 4.32 88.1
Golden Tilefish 11 5,823 915 3.01 75.5
Haddock 28 35,182 2,611 0.42 97.2
Pollock 23 9,551 1,681 0.39 96.2
Red Hake 10 7,658 906 2.82 87.2
Scup 19 10,524 1,628 3.73 79.1
Silver Hake 30 23,823 2,525 1.60 94.1
Summer Flounder 29 17,300 2,440 1.86 89.5
White Hake 20 5,561 1,076 2.29 92.7
Windowpane Flounder 2 585 100 6.85 74.4
Winter Flounder 19 9,139 1,628 2.02 86.3
Witch Flounder 29 8,904 1,903 1.40 86.4
Yellowtail Flounder 38 15,771 2,663 1.79 90.7
Acadian Redfish
Image
Acadian redfish, also called ocean perch, is the only fish in the rockfish/ocean perch family in the Atlantic, compared to the more than 50 Sebastes species in the Pacific. The fish is called redfish in New England and Canada, but is not to be confused with redfish from the Gulf of America (which is a drum). Acadian redfish are harvested year-round but harvests are usually largest during spring and summer in the Gulf of Maine.
Results of all QA/QC Exercises for Acadian Redfish
American Plaice
Image
American plaice is a species of flounder that is found in relatively deep water from southern Labrador on the eastern coast of Canada to Rhode Island.
Results of all QA/QC Exercises for American Plaice
Atlantic Cod
Image
In the Northwest Atlantic, cod range from Greenland to Cape Hatteras, North Carolina. In U.S. waters, cod is most common on Georges Bank and in the western Gulf of Maine. Cod is an iconic fish of New England and in recent years, Atlantic cod stocks in our region have declined dramatically. NOAA Fisheries is working to rebuild this population.
Results of all QA/QC Exercises for Atlantic cod
Atlantic Herring
Image
U.S. wild-caughtAtlantic herring is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations.
Results of all QA/QC Exercises for Atlantic herring
Atlantic Mackerel
Image
U.S. wild-caught Atlantic mackerel is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations.
Results of all QA/QC Exercises for Atlantic mackerel
Atlantic Surfclam
Image
Atlantic surfclams are distributed from the Gulf of St. Lawrence to Cape Hatteras, NC. The U.S. fishery generally concentrates on the populations off the coasts of New Jersey and the Delmarva Peninsula. Dredge and hand harvest are authorized in the commercial fishery, with hydraulic clam dredges being the primary gear type used. The recreational fishery is limited to hand harvest. Surfclams are generally processed for human consumption in soups, chowders, and stews; while a small portion of landings are also sold in the bait market. U.S. wild-caught Atlantic surfclam is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart E.
Results of all QA/QC Exercises for Atlantic Surfclam
Black Sea Bass
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The black sea bass fishery in the U.S. operates from Maine to Florida. Black sea bass are found in association with structured habitats. They migrate offshore and south in the fall, returning north and inshore to coastal areas and bays in spring. The black sea bass fishery predominantly uses trawls or rod and reel, but other gear includes longline, handline, pot, trap, gillnet, spear, and dredge. The market for black sea bass is for human consumption. U.S. wild-caught black sea bass is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart I.
Results of all QA/QC Exercises for Black sea bass
Bluefish
Image
The Atlantic bluefish fishery in the U.S. operates from Maine to Florida. Bluefish are highly migratory along the U.S. Atlantic coast and seasonally move. The bluefish fishery predominantly uses gillnets, but other gear includes hook and line, pound nets, seines, and trawls. The recreational fishery accounts for the majority of the bluefish total catch each year. The market for bluefish is for human consumption and is primarily sold fresh or smoked. U.S. wild-caught bluefish is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart J.
Results of all QA/QC Exercises for Bluefish
Butterfish
Image
U.S. wild-caughtbutterfish is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations.
Results of all QA/QC Exercises for Butterfish
Cusk
Image
Cusk fish illustration
Cusk is a deep water species found in rocky, hard bottom areas to a depth of 328 feet. It is a very unique species of fish as it's the only one in its taxonomic genus of Brosme. Cusk is relatively slow-growing and late-maturing, and the maximum age of this species is believed to be greater than 14 years.
Results of all QA/QC Exercises for Cusk
Golden Tilefish
Image
Golden tilefish are found along the outer continental shelf and slope from Nova Scotia, Canada to Suriname. The golden tilefish fishery in the U.S. is managed from Maine through Virginia, with the majority of the fishery concentrated between Nantucket Island, Massachusetts, south to Cape May, New Jersey; more specifically between Hudson and Veatch Canyons. The commercial fishery predominantly uses longline gear, although handline, rod and reel, and trawl gear are also authorized. Only rod and reel gear is allowed in the recreational golden tilefish fishery. The market for golden tilefish is for human consumption and is often used in sushi. U.S. wild-caught golden tilefish is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart N.
Results of all QA/QC Exercises for Golden Tilefish
Haddock
Image
Haddock are found on both sides of the North Atlantic. In the western North Atlantic, they’re found from Newfoundland to Cape May, New Jersey, and are most abundant on Georges Bank and in the Gulf of Maine.
Results of all QA/QC Exercises for Haddock
Pollock
Image
U.S. wild-caught Atlantic pollock is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations.
Results of all QA/QC Exercises for Pollock
Red Hake
Image
The red hake fishery in the U.S. operates from Maine to Cape Hatteras, North Carolina. Along with silver hake and offshore hake, it is a part of the small-mesh multispecies fishery, which is managed primarily through a series of exemptions from the Northeast Multispecies Fishery Management Plan (or groundfish fishery). The directed commercial fishery is conducted with small-mesh trawl gear with a number of specific requirements to reduce bycatch of larger groundfish species. There are two stocks of red hake (northern and southern) which are managed accordingly. The market for small-mesh multispecies is human consumption and as bait. U.S. wild-caught red hake is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart F.
Results of all QA/QC Exercises for Red Hake
Scup
Image
The scup fishery in the U.S. operates from Maine to Cape Hatteras, North Carolina. Scup undergo an extensive migration between coastal waters in the summer and offshore waters (outer continental shelf) in winter. They migrate offshore and south in the fall, returning north and inshore to coastal areas and bays in spring. The scup fishery predominantly uses trawls or handlines, but other gear includes longline, rod and reel, pot, trap, gillnet, spear, and dredge. The market for scup is for human consumption. U.S. wild-caught scup is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart H.
Results of all QA/QC Exercises for Scup
Silver Hake
Image
There is little to no separation of silver hake and offshore hake in the market, and both are generally sold under the name “whiting.” The whiting fishery in the United States operates from Maine to Cape Hatteras, North Carolina. Along with red hake, whiting are part of the small-mesh multispecies fishery, which is managed primarily through a series of exemptions from the Northeast Multispecies Fishery Management Plan (or groundfish fishery). The directed commercial fishery is conducted with small-mesh trawl gear with a number of specific requirements to reduce bycatch of larger groundfish species. There are two stocks of silver hake (northern and southern), and one stock of offshore hake, which primarily co-occurs with the southern stock of silver hake. The market for small-mesh multispecies is human consumption and as bait. U.S. wild-caught whiting is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart F.
Results of all QA/QC Exercises for Silver Hake
Summer Flounder
Image
The summer flounder fishery in the U.S. operates from Maine to the North Carolina/South Carolina border. Summer flounder is one of the most sought after commercial and recreational fish along the Atlantic coast. Summer flounder is found in inshore and offshore waters from Nova Scotia, Canada, to the east coast of Florida. Summer flounder are mainly caught in bottom otter trawls, but are also taken by pound nets and gillnets in estuarine waters. The market for summer flounder is for human consumption and is primarily sold fresh. U.S. wild-caught summer flounder is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations. Implementing regulations are found at 50 CFR part 648 subpart G.
Results of all QA/QC Exercises for Summer Flounder
White Hake
Image
White hake, also known as mud hake, are a species of groundfish that live in the deeper northwestern parts of the Atlantic Ocean. They are often compared to Atlantic cod and haddock in their physical appearance and taste.
Results of all QA/QC Exercises for White Hake
Windowpane Flounder
Image
Windowpane flounder are found in the northwest Atlantic from the Gulf of St. Lawrence to Florida in relatively shallow water. We manage two stocks of windowpane flounder in U.S. waters: Gulf of Maine/Georges Bank windowpane flounder, and Southern New England/Mid-Atlantic windowpane flounder.
Results of all QA/QC Exercises for Windowpane Flounder
Winter Flounder
Image
U.S. wild-caught winter flounder is a smart seafood choice because it is sustainably managed and responsibly harvested under U.S. regulations.
Results of all QA/QC Exercises for Winter Flounder
Witch Flounder
Image
Witch flounder is a right-eyed species of flounder. Witch flounder have a relatively small head, small mouth, and narrow body. Witch flounder are covered in smooth scales. Witch flounder have approximately 12 indentations on the underside (blind side) of their head. The top side (eyed side) of a witch flounder is generally grayish-brown in color, while the bottom side is white and covered with tiny dark dots. The top side pectoral fin is dusky or even black with a narrow light distal border.
Results of all QA/QC Exercises for Witch Flounder
Yellowtail Flounder
Image
Yellowtail flounder live along the Atlantic coast of North America from Newfoundland to the Chesapeake Bay. There are three stocks in U.S waters: the Gulf of Maine/Cape Cod, Georges Bank, and Southern New England/Mid-Atlantic stocks, all of which NOAA Fisheries is working to rebuild.
Results of all QA/QC Exercises for Yellowtail Flounder
More Information
More Information
Send Requests for more data to Sandy Sutherland
QA/QC Documentation
Age and Growth in the Northeast
Last updated by Northeast Fisheries Science Center on 06/17/2025
Age and Growth
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1515 | https://planetmath.org/proofofjordansinequality | proof of Jordan’s Inequality
proof of Jordan’s Inequality
To prove that
2 π x≤sin(x)≤x,∀x∈[0,π 2]2 πx≤sin(x)≤x,∀x∈[0,π 2]
consider a circle (circle with radius = 1 ). Take any point P P on the circumference of the circle.
Drop the perpendicular from P P to the horizontal line, M M being the foot of the perpendicular and Q Q the reflection of P P at M M. (refer to figure)
Let x=∠P O M.x=∠POM.
For x x to be in [0,π 2][0,π 2], the point P P lies in the first quadrant, as shown.
The length of line segmentP M PM is sin(x)sin(x). Construct a circle of radius M P MP, with M M as the center.
Length of line segment P Q PQ is 2 sin(x)2sin(x).
Length of arc P A Q PAQ is 2 x 2x.
Length of arc P B Q PBQ is π sin(x)πsin(x).
Since P Q≤PQ≤ length of arc P A Q PAQ (equality holds when x=0 x=0) we have 2 sin(x)≤2 x 2sin(x)≤2x. This implies
sin(x)≤x sin(x)≤x
Since length of arc P A Q PAQ is ≤≤ length of arc P B Q PBQ (equality holds true when x=0 x=0 or x=π 2 x=π 2), we have 2 x≤π sin(x)2x≤πsin(x). This implies
2 π x≤sin(x)2 πx≤sin(x)
Thus we have
2 π x≤sin(x)≤x,∀x∈[0,π 2]2 πx≤sin(x)≤x,∀x∈[0,π 2]
| Title | proof of Jordan’s Inequality |
| Canonical name | ProofOfJordansInequality |
| Date of creation | 2013-03-22 13:08:48 |
| Last modified on | 2013-03-22 13:08:48 |
| Owner | mathcam (2727) |
| Last modified by | mathcam (2727) |
| Numerical id | 17 |
| Author | mathcam (2727) |
| Entry type | Proof |
| Classification | msc 26D05 |
Generated on Fri Feb 9 20:00:55 2018 by LaTeXML |
1516 | https://brainly.com/question/18313367 | [FREE] The value for \sqrt{10} is approximately between 3 and 4. Approximate further to find the values to the - brainly.com
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The value for 10 is approximately between 3 and 4. Approximate further to find the values to the tenths place that 10 is between.
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A=3.1
B=3.2
Answered by yellowgiraffe2007 •11 answers•1.2K people helped
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The value of 10 is approximately between 3.1 and 3.2. By checking the squares of values between 3 and 4, we confirmed this approximation. Further checks between 3.1 and 3.2 refined the range more precisely to 3.16 and 3.17.
Explanation
To approximate the value of 10 between 3 and 4, we can start by squaring numbers that fall within this range to find more precise boundaries. We know that:
3 2=9 which is less than 10
4 2=16 which is greater than 10
This confirms that 10 is indeed between 3 and 4.
Next, we can check some values to get a better approximation to the tenths place:
Check 3.1:3.1 2=9.61, which is less than 10
Check 3.2:3.2 2=10.24, which is greater than 10
Now, we see that 10 is between 3.1 and 3.2.
Next, let’s check values between these two to narrow it down even further:
Check 3.15:3.1 5 2=9.9225, which is less than 10
Check 3.16:3.1 6 2=9.9856, which is still less than 10
Check 3.17:3.1 7 2=10.0889, which is greater than 10
At this point, we conclude that 10 is approximately between 3.16 and 3.17. Rounding these to the tenths place, we find that 10 is between 3.1 and 3.2.
Examples & Evidence
For example, the square of 3.1 gives us 9.61, and the square of 3.2 gives us 10.24, which helps us locate 10 more accurately between these two values. Similarly, testing 3.15 confirms it is less than 10 and 3.17 confirms it is greater than 10.
This method of approximation is commonly used in mathematics to estimate square roots without a calculator, ensuring that the range is sufficiently narrowed down through calculations.
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1517 | https://mathspace.co/textbooks/syllabuses/Syllabus-1160/topics/Topic-21958/subtopics/Subtopic-280114/ | 3.01 Least squares regression line | VCE 12 General 2023 Maths | VCE General Mathematics 3&4 - 2023 Edition | Mathspace
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VCE 12 General 2023
3.01 Least squares regression line
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Introduction
A line of best fit is a straight line that best represents bivariate data on a scatter plot. This line may pass through some of the points, none of the points, or all of the points. Lines of best fit are useful to determine the equation of the line and use the equation to make predictions.
Ideas
The least squares line of best fit
Calculate the least squares line from summary statistics
The least squares line of best fit
The most common method of fitting a line of best fit to a scatter plot of bivariate data is using the least squares method. This is then called a least squares line of best fit, or sometimes a Least Squares Regression Line.
In a sentence, it is a technique that minimises the sum of the squares of the vertical distance from each data point to a straight line. Perhaps an easier way to understand it is through a demonstration.
Exploration
Experiment with this Geogebra applet.
Refresh the applet and generate a new scatter plot to experiment with.
Drag the slider to Stage 2 and move the blue dots around the rectangle until you have a line which you think is the Line of Best Fit.
Drag the slider to Stage 3 to see what are called residuals. For now, think of these as the distance between the line and the actual data.
Drag the slider to Stage 4. Here you see the residuals turn to squares. Now move the blue dots on the line around again and try to make the total area of all the squares combined to be as small as possible. (Note, depending on the scale, these squares may appear rectangular)
Drag the slider to Stage 5. Your blue Line of Best Fit should be very close to the green Least Squares Regression Line. And the sum of your squares should be very close to the least sum of the squares.
Loading interactive...
In Stage 2, we make sure to move the line of best fits where the number of data points is equal above and below the line. In Stage 4, we minimize the overall area of all the squares combined. By doing this, moving to Stage 5, we should be very close to the green line or the least-squares regression line.
The equation of a least squares line of best fit is the same as the equation of a straight line, but is usually presented in a different form. We may recall the equation of a straight line as, y=m x+c y=mx+c y=m x+c or y=a x+b y=ax+b y=a x+b
However, a least squares line is usually written as, y=a+b x y=a+bx y=a+b x or R V=a+b×E V RV=a+b\times EV R V=a+b×E V where R V RV R V stands for the Response variable and E V EV E V stands for the Explanatory variable
Plotting a least squares line on a graph is the same as plotting the graph of a straight line and finding the equation of a least squares line from a graph is the same as finding the equation of a straight line from a graph.
Most of the time, you will be required to calculate the equation of the Least Squares Regression Line using technology.
Here's a video on how to use the TI-Nspire to create a scatter graph and calculate the equation of the Least Squares Regression Line.
Hi there, I'm Daniel. In this video, I'm going to show you how to use a CAS calculator
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Examples
Example 1
The table shows the number of people who went to watch a movie x x x weeks after it was released.
| Weeks(x)\text{Weeks }(x)Weeks(x) | 1 1 1 | 2 2 2 | 3 3 3 | 4 4 4 | 5 5 5 | 6 6 6 | 7 7 7 |
| Number of people(y)\text{Number of people }(y)Number of people(y) | 37 37 37 | 37 37 37 | 33 33 33 | 33 33 33 | 29 29 29 | 29 29 29 | 25 25 25 |
a
Plot the points from the table.
Worked Solution
Create a strategy
Plot each x x x-value along with its corresponding y y y-value.
Apply the idea
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
x x x
25 25 25
30 30 30
35 35 35
40 40 40
y y y
0,0
The points from the table have the coordinates (1,37),(2,37),(3,33),(4,33),(5,29),(6,29),(7,25)(1,37), \,(2,37), \,(3,33), \,(4,33), \,(5,29), \,(6,29), \(7,25) (1,37),(2,37),(3,33),(4,33),(5,29),(6,29),(7,25).
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b
If a line of best fit were drawn to approximate the relationship, which of the following could be its equation?
A
y=−2 x+40 y=-2x+40 y=−2 x+40
B
y=2 x+40 y=2x+40 y=2 x+40
C
y=−2 x y=-2x y=−2 x
D
y=2 x y=2x y=2 x
Worked Solution
Create a strategy
Check the trend in the scatterplot.
Apply the idea
We can see that the trend in the scatterplot is decreasing which means we have a negative gradient. So options B and D are incorrect. Also, option C is incorrect because it implies that the y y y intercept is zero, whereas the trend contradicts it.
So the correct answer is option A.
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c
Graph the line of best fit whose equation is given by y=−2 x+40 y=-2x+40 y=−2 x+40.
Worked Solution
Create a strategy
To graph the line, identify any two points that satisfy the equation. One point may be the y y y-intercept.
Apply the idea
By substituting x=0 x=0 x=0 to the equation, we have: y=−2(0)+40 y=40\begin{aligned} y&=-2(0)+40 \ y&=40 \end{aligned}y y=−2(0)+40=40
Solving the next point, with x=2 x=2 x=2, we have: y=−2(2)+40 y=36\begin{aligned} y&=-2(2)+40 \ y&=36 \end{aligned}y y=−2(2)+40=36
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
x x x
25 25 25
30 30 30
35 35 35
40 40 40
y y y
0,0
Here is the scatterplot with its line of best fit.
Reflect and check
We can see that the line of best fit follows the trend of the data and has the same number of points above and below the line.
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d
Use the equation of the line of best fit to find the number of people who went to watch the movie 12 12 12 weeks after it was released.
Worked Solution
Create a strategy
Substitute x=12 x=12 x=12 to the equation.
Apply the idea
Number of people\displaystyle \text{Number of people}Number of people=\displaystyle ==−2(12)+40\displaystyle -2(12)+40−2(12)+40 Substitute x=12 x=12 x=12
=\displaystyle ==−24+40\displaystyle -24+40−24+40 Perform the multiplication
=\displaystyle ==16\displaystyle 16 16 Evaluate
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Idea summary
The equations of a straight line are as follows:
y=m x+c y=mx+c y=m x+c
y=a x+b y=ax+b y=a x+b
Equation of a least squares regression line is usually written as:
y=a+b x y=a+bx y=a+b x
R V=a+b×E V RV=a+b\times EV R V=a+b×E V
where R V RV R V stands for the Response variable and E V EV E V stands for the Explanatory variable
Calculate the least squares line from summary statistics
If we aren't given the data set but instead have certain statistics calculated from the data set, we can still calculate the equation of the least squares regression line.
We will use the following formulae: y=a+b x,b=r s y s x y=a+bx,\,b=r\dfrac{s_y}{s_x}y=a+b x,b=r s xs y, and a=y‾−b x‾a=\overline{y}-b\overline{x}a=y−b x where s x s_x s x is the standard deviation of x x x,s y\,s_y s y is the standard deviation of y y y,x‾\,\overline{x}x is the mean of x x x,y‾\,\overline{y}y is the mean of y y y, and r r r is the correlation coefficient.
Examples
Example 2
A bivariate data set contains 10 10 10 data points with the following summary statistics:
x‾=5.13\overline{x}=5.13 x=5.13
s x=2.85 s_x=2.85 s x=2.85
y‾=18.81\overline{y}=18.81 y=18.81
s y=7.54 s_y=7.54 s y=7.54
r=0.993 r=0.993 r=0.993
a
Calculate the slope of the least squares regression line. Give your answer to two decimal places.
Worked Solution
Create a strategy
Use the formula: b=r s y s x b=\dfrac{rs_y}{s_x}b=s xr s y
Apply the idea
b\displaystyle b b=\displaystyle ==0.993×7.54 2.85\displaystyle \dfrac{0.993 \times 7.54}{2.85}2.85 0.993×7.54Subsitute r=0.993,s y=7.54,s x=2.85 r=0.993,\,s_y=7.54,\,s_x=2.85 r=0.993,s y=7.54,s x=2.85
=\displaystyle ==2.63\displaystyle 2.63 2.63 Evaluate using your calculator
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b
Using the rounded value of the previous part, calculate the vertical intercept of the least squares regression line. Give your answer to two decimal places.
Worked Solution
Create a strategy
Use the formula: a=y‾−b x‾a=\overline{y}-b\overline{x}a=y−b x
Apply the idea
In part (a), we found the slope of the line to be 2.63 2.63 2.63.
a\displaystyle a a=\displaystyle ==18.81−2.63×5.13\displaystyle 18.81-2.63 \times 5.13 18.81−2.63×5.13 Subsitute y‾=18.81,b=2.63,y‾=5.13\overline{y}=18.81,\,b=2.63,\,\overline{y}=5.13 y=18.81,b=2.63,y=5.13
=\displaystyle ==5.32\displaystyle 5.32 5.32 Evaluate using your calculator
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c
State the equation of the least squares regression line.
Worked Solution
Create a strategy
Use the formula: y=a+b x y=a+bx y=a+b x
Apply the idea
Substitute the values we found in parts (a) and (b): y=5.32+2.63 x y=5.32+2.63x y=5.32+2.63 x
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Example 3
The equation for the line of best fit is given by P=−4 t+116 P=-4t+116 P=−4 t+116, where t t t is time. This relationship shows that over time, P P P is:
A
remaining constant
B
decreasing
C
increasing
Worked Solution
Create a strategy
Recall that the sign of the gradient determines the trend of the data.
Apply the idea
In the equation, the gradient is the coefficient of your variable. In this case, it is −4-4−4. It is negative, which means it has a decreasing trend. So the correct option is B.
Watch question walkthrough
Idea summary
The following are the formulae in calculating the least squares regression line:y=a+b x b=r s y s x a=y‾−b x‾ y=a+bx \qquad b=r\dfrac{s_y}{s_x}\qquad a=\overline{y}-b\overline{x}y=a+b x b=r s xs ya=y−b x
where:
a=a=a= the y y y-intercept
b=b=b= the slope
s x=s_x=s x= the standard deviation of x x x
s y=s_y=s y= the standard deviation of y y y
x‾=\overline{x}=x= the mean of x x x
y‾=\overline{y}=y= the mean of y y y
r=r=r= the correlation coefficient
Outcomes
U3.AoS1.23
answer statistical questions that require a knowledge of the associations between pairs of variables
U3.AoS1.11
least squares line and its use in modelling linear associations
U3.AoS1.25
use the least squares line of best fit to model and analyse the linear association between two numerical variables and interpret the model in the context of the association being modelled
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1518 | https://www.youtube.com/watch?v=H8Ty4tO3vAU | Rapprochement Meaning
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Video shows what rapprochement means. The reestablishment of cordial relations, particularly between two countries; a reconciliation.. Rapprochement Meaning. How to pronounce, definition audio dictionary. How to say rapprochement. Powered by MaryTTS, Wiktionary
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rman the reestablishment of cordial relations particularly between two countries a Reconciliation r a p p r o c h e m e n t r rman |
1519 | https://www.topmath.org/review/0000/index.html?page=40 | 0000.pdf
/ 79
0001 目錄
0101 數與數線
0102 分數的運算
0301一元一次方程式
0401 二元一次聯立方程式
0501 直角坐標與二元一次方程式的圖形
0601 比例
0701 一元一次不等式
0801 幾何圖形與三視圖
0901 統計圖表與資料分析
1001 乘法公式與多項式
1101 平方根與畢氏定理
1201 因式分解
1301 一元二次方程式
1401 數列與級數
1501 函數與其圖形
1601 三角形的基本性質與尺規作圖
1701 平行與四邊形
1801 連比與相似形
1901 圓
2001 推理證明與三角形的心
2101 二次函數
2201 統計與機率
2301 立體幾何圖形
0/0
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單元 13：一元二次方程式 小魔流 topmath 頁碼 1301 1. 一元二次方程式：當一個方程式經整理之後，只含有一個未知數，且此未知數的 最高次數為 2 者，就 稱為 一元二次方程式。 x 2 ＋5 x＋8＝0、x 2 ＝－3 皆為 x 的一元二次方程式。 每一個 x 的一元二次方程式皆可化為 ax 2 ＋bx＋c＝0 (a ≠ 0) 的形式。 ax 2 ＋bx＋c＝0 (a ≠ 0)為一元二次方程式的標準式。 2. 一元二次方程式的解：將一個數代入一元二次方程式中，能使方程式 左邊和右邊的結果相等，就稱此 數為該一元二次方程式的 解 或 根。 將 x＝－2 代入 x 2 ＋6 x＋8＝0 得(－2) 2 ＋6(－2)＋8＝4＋(－12)＋8＝0，故 x＝－2 為一解。 若 x＝m 為 ax 2 ＋bx＋c＝0 (a ≠ 0)的解，則 am 2 ＋bm＋c＝0 3. 解一元二次方程式的方法有：(1)因式分解 (2)配方法 (3)公式解 這些方法將於後面主題一一介紹。 4. 根據法則：若兩數(或兩個式子)的乘積為 0，則這兩數(若兩個式子)中必有一數(若一個式子)為 0。 即 A×B＝0，則 A＝0 或 B＝0。 5. 因式分解解一元二次方程式步驟與實例說明： 解題步驟 實例說明 步驟一 先將方程式移項整理成 ax 2 ＋bx＋c＝0 的形 式，其中 a、b、c 均為整數，且 a＞0 解 x 2 －x＝6 x 2 －x＝6 移項得 x 2 －x－6＝0 步驟二 利用 提出公因式、乘法公式、十字交乘法 等 將 ax 2 ＋bx＋c 分解成兩個一次因式的乘積 (x－3)(x＋2)＝0 步驟三 由 A×B＝0 得 A＝0 或 B＝0 x－3＝0 或 x＋2＝0 步驟四 分別求出兩個一元一次方程式 x＝3 或 x＝－2 為 x 2 －x＝6 的解 6. 根與係數：設一元二次方程式 ax 2 ＋bx＋c＝0 的兩根為 α、β，則 α＋β＝－ b a ，αβ＝ c a 若一元二次方程式的兩根為 α、β，則(x－α)(x－β)＝0，故 x 2 －(α＋β)x＋αβ＝0。 7. 解與平方根的關係： (1) 若 x 2 ＝m，則 x＝± m (2) 若(x－a) 2 ＝b，則 x－a＝± b ，故 x＝a± b 1. 一般而言，一元二次方程式通常都有兩個根，而當兩個根相等時，可以將兩根一一寫出或寫出一個並 註明重根即可。 2. 若 α 、 β 為某一元二次方程式的兩根，則可令此方程式為(x－ α )(x－ β )＝0。 單元 Chapter 13 一元二次方程式 主題 1 利用因式分解解一元二次方程式 例 註 註 註 例 例 解 延伸 說 明 小叮嚀
單元 13：一元二次方程式 小魔流 topmath 頁碼 1302 兩邊開方根移項 x＝ －b± b 2 －4 ac 2 a 和的平方公式 (x＋ b 2 a ) 2 ＝ b 2 －4 ac 4 a 等號兩邊同時加( b 2 a ) 2 x 2 ＋2 b a x＋( b 2 a ) 2 ＝－ c a ＋( b 2 a ) 2 等號兩邊同時除以 a x 2 ＋ b a x＝－ c a 等號兩邊同時減 c ax 2 ＋bx＝－c 1. 平方根的概念：若方程式 x 2 ＝a(a ≥ 0)，則 x＝± a。 x 2 ＝4，則 x＝± 4＝±2。 2. 完全平方式：若一個式子可以寫成另一個式子的 完全平方 時。就稱此式為 完全平方式。 4 x 2 ＋4 x＋1＝(2 x＋1) 2 ，故 4 x 2 ＋4 x＋1 為完全平方式。 3. 配程完全平方式：形如 x 2 ＋mx(m ≠ 0)者，只要加上( m 2 ) 2 ，即可配成完全平方式(x＋ m 2 ) 2 。 4. 利用配方法解一元二次方程式：將一元二次方程式整理成(x＋a) 2 ＝b 的形式後， (1) 若 b ≥ 0，則 x＝－a± b (2)若 b＜0，則方程式無解。 5. 利用配方法解一元二次方程式的步驟與實例說明： 解題步驟 實例說明 步驟一 若 x 2 係數不等於 1，則將 x 2 項係數化為 1 配方法解 2 x 2 －4 x－8＝0 左右同除以 2 得 x 2 －2 x－4＝0 步驟二 將 x 2 項與 x 項留在等號左邊，把常數項移 到等號右邊 x 2 －2 x＝4 步驟三 等號兩邊同時加上( 2 項係數 x ) 2 x 2 －2 x＋1＝4＋1 步驟四 將等號左邊配成完全平方式 (x－1) 2 ＝5 步驟五 利用平方根的概念求解 x－1＝ 5 ± ，所以解為 x＝1 5 ± 對於無法或不易利用因式分解法求解的一元二次方程式，一般都採用配方法來求解。 如：x 2 －6 x＋1＝0 或 x 2 －2 x－456＝0。 6. 利用配方法解一元二次方程式：ax 2 ＋bx＋c＝0 (a≠0) (公式解是由配方法得來) 即 ax 2 ＋bx＋c＝0 的解為 x＝ －b± b 2 －4 ac 2 a ，其中 b 2 －4 ac 稱為 判別式。 7. 一元二次方程式 ax 2 ＋bx＋c＝0(a ≠ 0)的判別式 b 2 －4 ac 與根的可能情形： 判別式 方程式根的可能情形 b 2 －4 ac＞0 相異的兩根 b 2 －4 ac＝0 相等的兩根(重根) b 2 －4 ac＜0 方程式無解 主題 2 配方法與公式解 例 例 例 解 小叮嚀 註
單元 13：一元二次方程式 小魔流 topmath 頁碼 1303 驗算寫答 檢驗所求出的解 是否符合題意，不 合者刪除 解方程式 利用學過的方法 來解一元二次方 程式 列方程式 依題意找出各數 量的關係列成一 元二次 方程式 設未知數 依題意選定一適 當未知數，習慣上 以 x 或 y 表示 釐清題意 確定解題的方向 8. 利用公式解求一元二次方程式的解之步驟與實例說明： 解題步驟 實例說明 步驟一 將方程式整理程 ax 2 ＋bx＋c＝0，其中 a、b、 c 均為整數，且 a＞0 的形式 例：利用公式解求－3 x 2 ＋4＝5 x 的解 解：整理方程式得 3 x 2 ＋5 x－4＝0 令 a＝3，b＝5，c＝－4 則 b 2 －4 ac＝5 2 －4×3×(－4)＝73 因為 73＞0 所以 6 73 5 ± − = x 步驟二 找出對應的 a、b、c 之值，並計算 b 2 －4 ac 之值。 步驟三 (1) 若 b 2 －4 ac ≥ 0，則方程式的解為 a ac b b x 2 4 2 − ± − = (2) 若 b 2 －4 ac＜0，則方程式無解。 1. 若一元二次方程式 ax 2 ＋bx＋c＝0(a ≠ 0)有解，則 b 2 －4 ac ≥ 0。 2. 解一元二次方程式時，除非必要或題目指定，否則優先使用因式分解法。 1. 利用一元二次方程式解決日常生活中應用問題的步驟： 小叮嚀 主題 3 一元二次方程式應用問題
單元 14：數列與級數 小魔流 topmath 頁碼 1401 1. 數列的意義：依序排列的一串數字，即稱為 數列。 門牌號碼 32，34，36，38，40，… 2. 項：數列中每一個數叫做 項。其中第一個數稱為 第 1 項 或 首項，記為 a 1。第二個數稱為 第 2 項， 記為 a 2，依此類推第 n 個數稱為 第 n 項，記為 a n，而數列中的最後一項稱為 末項。 3. 有規律的數列：當數列中每一項的變化具有一定的規律時，就稱此數列為有規律的數列。 (1) 費氏數列：1，1，2，3，5，8，13，21，34，55…… (2) 循環小數的小數部分： 2 13 ＝0.1538461538461538…… 4. 奇數與偶數： 分類 定義 表示法(若 n 為正整數) 奇數 被 2 除餘 1 的正整數 2 n－1 或 2 n＋1 偶數 被 2 整除的正整數 2 n 5. 奇、偶數的和與積： (1) 奇數＋奇數偶數 (2) 偶數＋偶數偶數 (3) 奇數＋偶數奇數 (4) 奇數×奇數奇數 (5) 奇數×偶數偶數 (6) 偶數×偶數偶數 1. 等差數列： 定義 一數列中，任意相鄰的兩項，其 後項－前項 的差都相等，這樣的數列就叫做 等差數列。 公差 一個等差數列中，後項－前項 的差就叫做此一等差數列的 公差，通常用 d 表示。 第 n 項 公式 (1) 若一等差數列的首項為 a 1，公差為 d，則 ○ 1 第 n 項 a n＝a 1＋(n－1)d。 ○ 2 項數 n＝ a n－a 1 d ＋1 (2) 若一等差數列的公差為 d，第 k 項為 a k，n＞k，則第 n 項 a n＝a k＋(n－k)d。 (3) 若 m、n、p、q 為正整數且 m＋n＝p＋q，則 a m＋a n＝a p＋a q。 假設法 三數成等差數列：可設此三數為 a－d，a，a＋d (公差為 d)。 三數成等差數列且總和為 S：可設此三數為 S 3 －d，S， S 3 ＋d (公差為 d)。 四數成等差數列：可設此四數為 a－3 d，a－d，a＋d，a＋3 d (公差為 2 d)。 五數成等差數列：可設此五數為 a－2 d，a－d，a，a＋d，a＋2 d (公差為 d)。 三角形三內角度數成等差數列：可令三內角為(60－d)°，60°，(60＋d)°。 直角三角形三邊長成等差數列：可令三邊長為 3 r，4 r，5 r，r≠0 (公差為 r)。 單元 Chapter 14 數列與級數 主題 1 規律的數列 例 例 主題 2 等差數列 |
1520 | https://www.chegg.com/homework-help/questions-and-answers/door-supported-hinges-b-cord-ce-shown-46-lb-force-exerted-d-force-purely-x-direction-hinge-q202012090 | Solved A door is supported by hinges at A and B, and by | Chegg.com
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A door is supported by hinges at A and B, and by cord CE, as shown, and a 46 lb force is exerted on it at D.(This force is purely in the -x direction.) Hinge A is able to prevent translation in the y direction, but Hinge B is not able to do that. FindA) the reactions at A and BB) the tension in the cord (magnitude)Hint: Since there are two hinges,
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Question: A door is supported by hinges at A and B, and by cord CE, as shown, and a 46 lb force is exerted on it at D.(This force is purely in the -x direction.) Hinge A is able to prevent translation in the y direction, but Hinge B is not able to do that. FindA) the reactions at A and BB) the tension in the cord (magnitude)Hint: Since there are two hinges,
A door is supported by hinges at A and B, and by cord CE, as shown, and a 4 6 lb force is exerted on it at D.(This force is purely in the -x direction.) Hinge A is able to prevent translation in the y direction, but Hinge B is not able to do that. Find A) the reactions at A and B B) the tension in the cord (magnitude) Hint: Since there are two hinges, neither one needs to prevent rotation about axes parallel to the x or z axes.
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1521 | https://en.wikipedia.org/wiki/Rodrigues%27_formula | Jump to content
Rodrigues' formula
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From Wikipedia, the free encyclopedia
Formula for the Legendre polynomials
For the 3-dimensional rotation formula, see Rodrigues' rotation formula.
In mathematics, Rodrigues' formula (formerly called the Ivory–Jacobi formula) generates the Legendre polynomials. It was independently introduced by Olinde Rodrigues (1816), Sir James Ivory (1824) and Carl Gustav Jacobi (1827). The name "Rodrigues formula" was introduced by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the first to discover it. The term is also used to describe similar formulas for other orthogonal polynomials. Askey (2005) describes the history of the Rodrigues formula in detail.
Statement
[edit]
Let be a sequence of orthogonal polynomials on the interval with respect to weight function . That is, they have degrees , satisfy the orthogonality condition
where are nonzero constants depending on , and is the Kronecker delta. The interval may be infinite in one or both ends.
Rodrigues' type formula—If
where is a polynomial with degree at most 1 and is a polynomial with degree at most 2, and
for any .
Then, if for all , then
for some constants .
Proof
Let , then for all for some polynomials , such that . Proven by induction on :
Let . We have shown that is a polynomial of degree . With integration by parts, we have for all , since . Thus, make up an orthogonal polynomial series with respect to . Thus, for some constants .
Differential equation—
Proof
When , it is trivial. When , it simplifies to , which is true since . So assume . Define , then by direct computation and simplification, the equation to be proven is equivalent to
By Leibniz differentiation rule, we have
for arbitrary . This allows us to move to the other side of the -th derivative. Set , and define
Then the equation simplifies to
has three terms, call them in order . has two terms, call them in order .
.
That . follows from first writing as
and then taking the innermost first derivative to obtain
and then rewriting this as
The first term is the negative of and the second term is the negative of .
More abstractly, this can be viewed through Sturm–Liouville theory. Define an operator , then the differential equation is equivalent to . Define the functional space as the Hilbert space of functions over , such that . Then the operator is self-adjoint on functions satisfying certain boundary conditions, allowing us to apply the spectral theorem.
Generating function
[edit]
A simple argument using Cauchy's integral formula shows that the orthogonal polynomials obtained from the Rodrigues formula have a generating function of the form
The functions here may not have the standard normalizations. But we can write this equivalently as
where the are chosen according to the application so as to give the desired normalizations. The variable u may be replaced by a constant multiple of u so that
This gives an alternate form of the generating function.
By Cauchy's integral formula, Rodrigues’ formula is equivalent towhere the integral is along a counterclockwise closed loop around . Let
Then the complex path integral takes the form
where now the closed path C encircles the origin. In the equation for , is an implicit function of . Expanding in the power series given earlier gives
Only the term has a nonzero residue, which is . The coefficient was dropped since normalizations are conventions which can be inserted afterwards as discussed earlier.
By expressing t in terms of u in the general formula just given for , explicit formulas for may be found. As a simple example, let and (Hermite polynomials) so that , , and so .
Examples
[edit]
| Family | | | | | | |
| Legendre | | | | | | |
| Chebyshev (of the first kind) | | | | | | |
| Chebyshev (of the second kind) | | | | | | |
| Gegenbauer/ultraspherical | | | | | | |
| Jacobi | | | | | | |
| associated Laguerre | | | | | | |
| physicist's Hermite | | | | | | |
These formulae
are for the classical orthogonal polynomials. Similar formulae hold for many other sequences of orthogonal functions arising from Sturm–Liouville equations, and these are also called the Rodrigues formula (or Rodrigues' type formula), especially when the resulting sequence is polynomial.
Legendre
[edit]
Source:
Rodrigues stated his formula for Legendre polynomials :
For Legendre polynomials, the generating function is defined as .
The contour integral gives the Schläfli integral for Legendre polynomials: Summing up the integrand,where . For small , we have , which heuristically suggests that the integral should be the residue around , thus giving
Hermite
[edit]
Source:
Physicist's Hermite polynomials:
The generating function is defined asThe contour integral gives
Laguerre
[edit]
Source:
For associated Laguerre polynomials,
The generating function is defined asBy the same method, we have .
Jacobi
[edit]
Source:
where , and the branch of square root is chosen so that .
See also
[edit]
Classical orthogonal polynomials
References
[edit]
^ Shapiro, Joel (2016). "Rodrigues's Formula and Orthogonal Polynomials" (PDF). p. 1.
^ Shapiro 2016, p. 2.
^ Shapiro 2016, p. 2.
^ Shapiro (2016). "Physics 464/511 Lecture J" (PDF).
^ NIST. "Rodrigues's Formula and Orthogonal Polynomials".
^
Arfken, George B.; Weber, Hans J. (2005). Mathematical Methods for Physicists Sixth Edition. Elsevier Academic Press. p. 741. ISBN 0-12-059876-0.
^ Schläfli, Ludwig (1881), "Über die zwei Heineschen Kugelfunktionen mit beliebigem Parameter und ihre ausnahmslose Darstellung durch bestimmte Integrale", Gesammelte Mathematische Abhandlungen, Basel: Springer Basel, pp. 317–392, ISBN 978-3-0348-4044-6 {{citation}}: ISBN / Date incompatibility (help)
^ Arfken and Weber 2005, p. 817.
^ Arfken and Weber 2005, p. 837.
^ NIST. "§18.12 Generating Functions".
Askey, Richard (2005), "The 1839 paper on permutations: its relation to the Rodrigues formula and further developments", in Altmann, Simón L.; Ortiz, Eduardo L. (eds.), Mathematics and social utopias in France: Olinde Rodrigues and his times, History of mathematics, vol. 28, Providence, R.I.: American Mathematical Society, pp. 105–118, ISBN 978-0-8218-3860-0
Ivory, James (1824), "On the Figure Requisite to Maintain the Equilibrium of a Homogeneous Fluid Mass That Revolves Upon an Axis", Philosophical Transactions of the Royal Society of London, 114, The Royal Society: 85–150, doi:10.1098/rstl.1824.0008, JSTOR 107707
Jacobi, C. G. J. (1827), "Ueber eine besondere Gattung algebraischer Functionen, die aus der Entwicklung der Function (1 − 2xz + z2)1/2 entstehen.", Journal für die Reine und Angewandte Mathematik (in German), 2: 223–226, doi:10.1515/crll.1827.2.223, ISSN 0075-4102, S2CID 120291793
O'Connor, John J.; Robertson, Edmund F., "Olinde Rodrigues", MacTutor History of Mathematics Archive, University of St Andrews
Rodrigues, Olinde (1816), "De l'attraction des sphéroïdes", Correspondence sur l'École Impériale Polytechnique, (Thesis for the Faculty of Science of the University of Paris), 3 (3): 361–385
Retrieved from "
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1522 | https://arxiv.org/abs/2207.13903 | [2207.13903] Joint complete monotonicity of rational functions in two variables and toral $m$-isometric pairs
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arXiv:2207.13903 (math)
[Submitted on 28 Jul 2022 (v1), last revised 7 Oct 2023 (this version, v4)]
Title:Joint complete monotonicity of rational functions in two variables and toral m-isometric pairs
Authors:Akash Anand, Sameer Chavan, Rajkamal Nailwal
View a PDF of the paper titled Joint complete monotonicity of rational functions in two variables and toral $m$-isometric pairs, by Akash Anand and 1 other authors
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Abstract:We discuss the problem of classifying polynomials p : \mathbb R^2_+ \rightarrow (0, \infty)for which \frac{1}{p}={\frac{1}{p(m, n)}}_{m, n \geq 0}is joint completely monotone, where pis a linear polynomial in y.We show that if p(x, y)=a+b x+c y+d xywith a > 0and b, c, d \geq 0,then \frac{1}{p}is joint completely monotone if and only if a d - b c \leq 0.We also present an application to the Cauchy dual subnormality problem for toral 3-isometric weighted 2-shifts.
Comments:28 pages
Subjects:Functional Analysis (math.FA)
Cite as:arXiv:2207.13903 [math.FA]
(or arXiv:2207.13903v4 [math.FA] for this version)
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arXiv-issued DOI via DataCite
Journal reference:J. Operator Theory pp. 101-130, Volume 92, Issue 1, Summer 2024
Submission history
From: Rajkamal Nailwal Mr. [view email]
[v1] Thu, 28 Jul 2022 06:47:03 UTC (28 KB)
[v2] Tue, 2 Aug 2022 11:39:52 UTC (23 KB)
[v3] Tue, 9 Aug 2022 05:15:58 UTC (25 KB)
[v4] Sat, 7 Oct 2023 02:24:32 UTC (26 KB)
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GABAA receptor
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From Wikipedia, the free encyclopedia
Ionotropic receptor and ligand-gated ion channel
The GABAA receptor (GABAAR) is an ionotropic receptor and ligand-gated ion channel. Its endogenous ligand is γ-aminobutyric acid (GABA), the major inhibitory neurotransmitter in the central nervous system. Accurate regulation of GABAergic transmission through appropriate developmental processes, specificity to neural cell types, and responsiveness to activity is crucial for the proper functioning of nearly all aspects of the central nervous system (CNS).
Upon opening, the GABAA receptor on the postsynaptic cell is selectively permeable to chloride ions (Cl−
) and, to a lesser extent, bicarbonate ions (HCO−
3).
GABAAR are members of the ligand-gated ion channel receptor superfamily, which is a chloride channel family with a dozen or more heterotetrametric subtypes and 19 distinct subunits. These subtypes have distinct brain regional and subcellular localization, age-dependent expression, and the ability to undergo plastic alterations in response to experience, including drug exposure.
GABAAR is not just the target of agonist depressants and antagonist convulsants, but most GABAAR medicines also act at additional (allosteric) binding sites on GABAAR proteins. Some sedatives and anxiolytics, such as benzodiazepines and related medicines, act on GABAAR subtype-dependent extracellular domain sites. Alcohols and neurosteroids, among other general anesthetics, act at GABAAR subunit-interface transmembrane locations. High anesthetic dosages of ethanol act on GABAAR subtype-dependent transmembrane domain locations. Ethanol acts at GABAAR subtype-dependent extracellular domain locations at low intoxication concentrations. Thus, GABAAR subtypes have pharmacologically distinct receptor binding sites for a diverse range of therapeutically significant neuropharmacological drugs.
Depending on the membrane potential and the ionic concentration difference, this can result in ionic fluxes across the pore. If the membrane potential is higher than the equilibrium potential (also known as the reversal potential) for chloride ions, when the receptor is activated Cl−
will flow into the cell. This causes an inhibitory effect on neurotransmission by diminishing the chance of a successful action potential occurring at the postsynaptic cell. The reversal potential of the GABAA-mediated inhibitory postsynaptic potential (IPSP) in normal solution is −70 mV, contrasting the GABAB IPSP (−100 mV).
The active site of the GABAA receptor is the binding site for GABA and several drugs such as muscimol, gaboxadol, and bicuculline. The protein also contains a number of different allosteric binding sites which modulate the activity of the receptor indirectly. These allosteric sites are the targets of various other drugs, including the benzodiazepines, nonbenzodiazepines, neuroactive steroids, barbiturates, alcohol (ethanol), inhaled anaesthetics, kavalactones, cicutoxin, and picrotoxin, among others.
Binding of GABA to the GABAAR causes the receptor to shift from ordered lipids to clusters of PIP2 in the disordered region of the membrane. The spatial distribution of GABAAR in neurons is regulated by astrocyte derived cholesterol.
Much like the GABAA receptor, the GABAB receptor is an obligatory heterodimer consisting of GABAB1 and GABAB2 subunits. These subunits include an extracellular Venus Flytrap domain (VFT) and a transmembrane domain containing seven α-helices (7TM domain). These structural components play a vital role in intricately modulating neurotransmission and interactions with drugs.
Target for benzodiazepines
[edit]
The ionotropic GABAA receptor protein complex is also the molecular target of the benzodiazepine class of tranquilizer drugs. Benzodiazepines do not bind to the same receptor site on the protein complex as does the endogenous ligand GABA (whose binding site is located between α- and β-subunits), but bind to distinct benzodiazepine binding sites situated at the interface between the α- and γ-subunits of α- and γ-subunit containing GABAA receptors. While the majority of GABAA receptors (those containing α1-, α2-, α3-, or α5-subunits) are benzodiazepine sensitive, there exists a minority of GABAA receptors (α4- or α6-subunit containing) which are insensitive to classical 1,4-benzodiazepines, but instead are sensitive to other classes of GABAergic drugs such as neurosteroids and alcohol. In addition peripheral benzodiazepine receptors exist which are not associated with GABAA receptors. As a result, the IUPHAR has recommended that the terms "BZ receptor", "GABA/BZ receptor" and "omega receptor" no longer be used and that the term "benzodiazepine receptor" be replaced with "benzodiazepine site". Benzodiazepines like diazepam and midazolam act as positive allosteric modulators for GABAA receptors. When these receptors are activated, there's a rise in intracellular chloride levels, resulting in cell membrane hyperpolarization and decreased excitation.
In order for GABAA receptors to be sensitive to the action of benzodiazepines they need to contain an α and a γ subunit, between which the benzodiazepine binds. Once bound, the benzodiazepine locks the GABAA receptor into a conformation where the neurotransmitter GABA has much higher affinity for the GABAA receptor, increasing the frequency of opening of the associated chloride ion channel and hyperpolarising the membrane. This potentiates the inhibitory effect of the available GABA leading to sedative and anxiolytic effects.
Different benzodiazepines have different affinities for GABAA receptors made up of different collection of subunits, and this means that their pharmacological profile varies with subtype selectivity. For instance, benzodiazepine receptor ligands with high activity at the α1 and/or α5 tend to be more associated with sedation, ataxia and amnesia, whereas those with higher activity at GABAA receptors containing α2 and/or α3 subunits generally have greater anxiolytic activity. Anticonvulsant effects can be produced by agonists acting at any of the GABAA subtypes, but current research in this area is focused mainly on producing α2-selective agonists as anticonvulsants which lack the side effects of older drugs such as sedation and amnesia.
The binding site for benzodiazepines is distinct from the binding site for barbiturates and GABA on the GABAA receptor, and also produces different effects on binding, with the benzodiazepines increasing the frequency of the chloride channel opening, while barbiturates increase the duration of chloride channel opening when GABA is bound. Since these are separate modulatory effects, they can both take place at the same time, and so the combination of benzodiazepines with barbiturates is strongly synergistic, and can be dangerous if dosage is not strictly controlled.
Also note that some GABAA agonists such as muscimol and gaboxadol do bind to the same site on the GABAA receptor complex as GABA itself, and consequently produce effects which are similar but not identical to those of positive allosteric modulators like benzodiazepines.
Structure and function
[edit]
Structural understanding of the GABAA receptor was initially based on homology models, obtained using crystal structures of homologous proteins like Acetylcholine binding protein (AChBP) and nicotinic acetylcholine (nACh) receptors as templates. The much sought structure of a GABAA receptor was finally resolved, with the disclosure of the crystal structure of human β3 homopentameric GABAA receptor.
Whilst this was a major development, the majority of GABAA receptors are heteromeric and the structure did not provide any details of the benzodiazepine binding site. This was finally elucidated in 2018 by the publication of a high resolution cryo-EM structure of rat α1β1γ2S receptor and human α1β2γ2 receptor bound with GABA and the neutral benzodiazepine flumazenil.
GABAA receptors are pentameric transmembrane receptors which consist of five subunits arranged around a central pore. Each subunit comprises four transmembrane domains with both the N- and C-terminus located extracellularly. The receptor sits in the membrane of its neuron, usually localized at a synapse, postsynaptically. However, some isoforms may be found extrasynaptically. When vesicles of GABA are released presynaptically and activate the GABA receptors at the synapse, this is known as phasic inhibition. However, the GABA escaping from the synaptic cleft can activate receptors on presynaptic terminals or at neighbouring synapses on the same or adjacent neurons (a phenomenon termed 'spillover') in addition to the constant, low GABA concentrations in the extracellular space results in persistent activation of the GABAA receptors known as tonic inhibition.
The ligand GABA is the endogenous compound that causes this receptor to open; once bound to GABA, the protein receptor changes conformation within the membrane, opening the pore in order to allow chloride anions (Cl−
) and, to a lesser extent, bicarbonate ions (HCO−
3) to pass down their electrochemical gradient. The binding site to GABA is about 80Å away from the narrowest part of the ion channel. Recent computational studies have suggested an allosteric mechanism whereby GABA binding leads to ion channel opening. Because the reversal potential for chloride in most mature neurons is close to or more negative than the resting membrane potential, activation of GABAA receptors tends to stabilize or hyperpolarise the resting potential, and can make it more difficult for excitatory neurotransmitters to depolarize the neuron and generate an action potential. The net effect therefore typically inhibitory, reducing the activity of the neuron, although depolarizing currents have been observed in response to GABA in immature neurons in early development. This effect during development is due to a modified Cl−
gradient wherein the anions leave the cells through the GABAA receptors, since their intracellular chlorine concentration is higher than the extracellular. The difference in extracellular chlorine anion concentration is presumed to be due to the higher activity of chloride transporters, such as NKCC1, transporting chloride into cells which are present early in development, whereas, for instance, KCC2 transports chloride out of cells and is the dominant factor in establishing the chloride gradient later in development. These depolarization events have shown to be key in neuronal development. In the mature neuron, the GABAA channel opens quickly and thus contributes to the early part of the inhibitory post-synaptic potential (IPSP).
The endogenous ligand that binds to the benzodiazepine site is inosine.
Proper developmental, neuronal cell-type-specific, and activity-dependent GABAergic transmission control is required for nearly all aspects of CNS function.
It has been proposed that the GABAergic system is disrupted in numerous neurodevelopmental diseases, including fragile X syndrome, Rett syndrome, and Dravet syndrome, and that it is a crucial potential target for therapeutic intervention.
Subunits
[edit]
GABAA receptors are members of the large pentameric ligand gated ion channel (previously referred to as "Cys-loop" receptors) super-family of evolutionarily related and structurally similar ligand-gated ion channels that also includes nicotinic acetylcholine receptors, glycine receptors, and the 5HT3 receptor. There are numerous subunit isoforms for the GABAA receptor, which determine the receptor's agonist affinity, chance of opening, conductance, and other properties.
In humans, the units are as follows:
six types of α subunits (GABRA1, GABRA2, GABRA3, GABRA4, GABRA5, GABRA6)
three βs (GABRB1, GABRB2, GABRB3)
three γs (GABRG1, GABRG2, GABRG3)
as well as a δ (GABRD), an ε (GABRE), a π (GABRP), and a θ (GABRQ)
There are three ρ units (GABRR1, GABRR2, GABRR3); however, these do not coassemble with the classical GABAA units listed above, but rather homooligomerize to form GABAA-ρ receptors (formerly classified as GABAC receptors but now this nomenclature has been deprecated).
Combinatorial arrays
[edit]
Given the large number of GABAA receptors, a great diversity of final pentameric receptor subtypes is possible. Methods to produce cell-based laboratory access to a greater number of possible GABAA receptor subunit combinations allow teasing apart of the contribution of specific receptor subtypes and their physiological and pathophysiological function and role in the CNS and in disease.
Distribution
[edit]
GABAA receptors are responsible for most of the physiological activities of GABA in the central nervous system, and the receptor subtypes vary significantly. Subunit composition can vary widely between regions and subtypes may be associated with specific functions. The minimal requirement to produce a GABA-gated ion channel is the inclusion of an α and a β subunit. The most common GABAA receptor is a pentamer comprising two α's, two β's, and a γ (α2β2γ). In neurons themselves, the type of GABAA receptor subunits and their densities can vary between cell bodies and dendrites. Benzodiazepines and barbiturates amplify the inhibitory effects mediated by the GABAA receptor.
GABAA receptors can also be found in other tissues, including leydig cells, placenta, immune cells, liver, bone growth plates and several other endocrine tissues. Subunit expression varies between 'normal' tissue and malignancies, as GABAA receptors can influence cell proliferation.
Distribution of Receptor Types
| Isoform | Synaptic/Extrasynaptic | Anatomical location |
| α1β3γ2S | Both | Widespread |
| α2β3γ2S | Both | Widespread |
| α3β3γ2S | Both | Reticular thalamic nucleus |
| α4β3γ2S | Both | Thalamic relay cells |
| α5β3γ2S | Both | Hippocampal pyramidal cells |
| α6β3γ2S | Both | Cerebellar granule cells |
| α1β2γ2S | Both | Widespread, most abundant |
| α4β3δ | Extrasynaptic | Thalamic relay cells |
| α6β3δ | Extrasynaptic | Cerebellar granule cells |
| α1β2 | Extrasynaptic | Widespread |
| α1β3 | Extrasynaptic | Thalamus, hypothalamus |
| α1β2δ | Extrasynaptic | Hippocampus |
| α4β2δ | Extrasynaptic | Hippocampus, Prefrontal cortex |
| α3β3θ | Extrasynaptic | Hypothalamus |
| α3β3ε | Extrasynaptic | Hypothalamus |
Ligands
[edit]
A number of ligands have been found to bind to various sites on the GABAA receptor complex and modulate it besides GABA itself.[which?] A ligand can possess one or more properties of the following types. Unfortunately the literature often does not distinguish these types properly.
Types
[edit]
Orthosteric agonists and antagonists: bind to the main receptor site (the site where GABA normally binds, also referred to as the "active" or "orthosteric" site). Agonists activate the receptor, resulting in increased Cl−
conductance. Antagonists, though they have no effect on their own, compete with GABA for binding and thereby inhibit its action, resulting in decreased Cl−
conductance.
First order allosteric modulators: bind to allosteric sites on the receptor complex and affect it either in a positive (PAM), negative (NAM) or neutral/silent (SAM) manner, causing increased or decreased efficiency of the main site and therefore an indirect increase or decrease in Cl−
conductance. SAMs do not affect the conductance, but occupy the binding site.
Second order modulators: bind to an allosteric site on the receptor complex and modulate the effect of first order modulators.
Open channel blockers: prolong ligand-receptor occupancy, activation kinetics and Cl ion flux in a subunit configuration-dependent and sensitization-state dependent manner.
Non-competitive channel blockers: bind to or near the central pore of the receptor complex and directly block Cl−
conductance through the ion channel.
Examples
[edit]
Orthosteric agonists: GABA, gaboxadol, isoguvacine, muscimol, progabide, beta-alanine, taurine, piperidine-4-sulfonic acid (partial agonist).
Orthosteric antagonists: bicuculline, gabazine.
Positive allosteric modulators: abecarnil, azocarnil (photoswitchable), barbiturates, benzodiazepines, certain carbamates (e.g., carisoprodol, meprobamate, lorbamate), honokiol, magnolol, baicalin, baicelin, thienodiazepines, alcohol (ethanol), etomidate, glutethimide, kavalactones, meprobamate, quinazolinones (e.g., methaqualone, etaqualone, diproqualone), neuroactive steroids, niacin/niacinamide, nonbenzodiazepines (e.g., zolpidem, eszopiclone), propofol, stiripentol, theanine,[citation needed] valerenic acid, volatile/inhaled anesthetics, lanthanum, riluzole, and menthol.
Negative allosteric modulators: flumazenil, Ro15-4513, sarmazenil, pregnenolone sulfate, amentoflavone, and zinc.
Inverse allosteric agonists: beta-carbolines (e.g., harmine, harmaline, tetrahydroharmine).
Second-order modulators: (−)‐epigallocatechin‐3‐gallate.
Non-competitive channel blockers: cicutoxin, oenanthotoxin, pentylenetetrazol, picrotoxin[citation needed], thujone, and lindane.
Effects
[edit]
Ligands which contribute to receptor activation typically have anxiolytic, anticonvulsant, amnesic, sedative, hypnotic, euphoriant, and muscle relaxant properties. Some such as muscimol and the z-drugs may also be hallucinogenic.[citation needed] Ligands which decrease receptor activation usually have opposite effects, including anxiogenesis and convulsion.[citation needed] Some of the subtype-selective negative allosteric modulators such as α5IA are being investigated for their nootropic effects, as well as treatments for the unwanted side effects of other GABAergic drugs. Advances in molecular pharmacology and genetic manipulation of rat genes have revealed that distinct subtypes of the GABAA receptor mediate certain parts of the anaesthetic behavioral repertoire.
Novel drugs
[edit]
A useful property of the many benzodiazepine site allosteric modulators is that they may display selective binding to particular subsets of receptors comprising specific subunits. This allows one to determine which GABAA receptor subunit combinations are prevalent in particular brain areas and provides a clue as to which subunit combinations may be responsible for behavioral effects of drugs acting at GABAA receptors. These selective ligands may have pharmacological advantages in that they may allow dissociation of desired therapeutic effects from undesirable side effects. Few subtype selective ligands have gone into clinical use as yet, with the exception of zolpidem which is reasonably selective for α1, but several more selective compounds are in development such as the α3-selective drug adipiplon. There are many examples of subtype-selective compounds which are widely used in scientific research, including:
CL-218,872 (highly α1-selective agonist)
bretazenil (subtype-selective partial agonist)
imidazenil and L-838,417 (both partial agonists at some subtypes, but weak antagonists at others)
QH-ii-066 (full agonist highly selective for α5 subtype)
α5IA (selective inverse agonist for α5 subtype)
SL-651,498 (full agonist at α2 and α3 subtypes, and as a partial agonist at α1 and α5
3-acyl-4-quinolones: selective for α1 over α3
Paradoxical reactions
[edit]
There are multiple indications that paradoxical reactions upon — for example — benzodiazepines, barbiturates, inhalational anesthetics, propofol, neurosteroids, and alcohol are associated with structural deviations of GABAA receptors. The combination of the five subunits of the receptor (see images above) can be altered in such a way that for example the receptor's response to GABA remains unchanged but the response to one of the named substances is dramatically different from the normal one.
There are estimates that about 2–3% of the general population may suffer from serious emotional disorders due to such receptor deviations, with up to 20% suffering from moderate disorders of this kind. It is generally assumed that the receptor alterations are, at least partly, due to genetic and also epigenetic deviations. There are indication that the latter may be triggered by, among other factors, social stress or occupational burnout.
See also
[edit]
4-Iodopropofol
GABA receptor
GABAB receptor
GABAA-ρ receptor
Gephyrin
Glycine receptor
GABAA receptor positive allosteric modulators
GABAA receptor negative allosteric modulators
References
[edit]
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Further reading
[edit]
Olsen RW, DeLorey TM (1999). "16. GABA and Glycine". In Siegel GJ, Agranoff BW, Fisher SK, Albers RW, Uhler MD (eds.). Basic neurochemistry: molecular, cellular, and medical aspects (6th ed.). Philadelphia: Lippincott-Raven. ISBN 978-0-397-51820-3. NBK28090.
Olsen RW, Betz H (2005). "16. GABA and Glycine". In Siegel GJ, Albers RW, Brady S, Price DD (eds.). Basic Neurochemistry: Molecular, Cellular and Medical Aspects (7th ed.). Academic Press. pp. 291–302. ISBN 978-0-12-088397-4.
Uusi-Oukari M, Korpi ER (March 2010). "Regulation of GABA(A) receptor subunit expression by pharmacological agents". Pharmacological Reviews. 62 (1): 97–135. doi:10.1124/pr.109.002063. PMID 20123953. S2CID 12202117.
Rudolph U (2015). Diversity and Functions of GABA Receptors: A Tribute to Hanns Möhler. Academic Press, Elsevier. ISBN 978-0-12-802660-1.
External links
[edit]
Receptors,+GABA-A at the U.S. National Library of Medicine Medical Subject Headings (MeSH)
| v t e Ion channel, cell surface receptor: ligand-gated ion channels | |
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| Cys-loop receptors | | | | --- | | 5-HT/serotonin | 5-HT3 + A + B + C + D + E | | GABA | GABAA + α1 + α2 + α3 + α4 + α5 + α6 + β1 + β2 + β3 + γ1 + γ2 + γ3 + δ + ε + π + θ GABAA-ρ + ρ1 + ρ2 + ρ3 | | Glycine | α1 α2 α3 α4 β | | Nicotinic acetylcholine | monomers: α1 α2 α3 α4 α5 α6 α7 α9 α10 β1 β2 β3 β4 δ ε pentamers: (α3)2(β4)3 (α4)2(β2)3 (α7)5 (α1)2(β4)3 - Ganglion type (α1)2β1δε - Muscle type | | Zinc | Zinc-activated | |
| Ionotropic glutamates | | | | --- | | Ligand-gated only | AMPA (1 2 3 4) Kainate + 1 + 2 + 3 + 4 + 5 | | Voltage- and ligand-gated | NMDA + 1 + 2A + 2B + 2C + 2D + 3A + 3B + L1A + L1B | | ‘Orphan’ | GluD + δ1 + δ2 | |
| ATP-gated channels | | | | --- | | Purinergic receptors | P2X + 1 + 2 + 3 + 4 + 5 + 6 + 7 | |
| v t e GABA receptor modulators | |
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| Ionotropic | | | | --- | | GABAATooltip γ-Aminobutyric acid A receptor | Agonists: (+)-Catechin Bamaluzole Barbiturates (e.g., phenobarbital) Beta-Alanine BL-1020 DAVA Dihydromuscimol GABA Gabamide GABOB Gaboxadol (THIP) Homotaurine (tramiprosate, 3-APS) Ibotenic acid iso-THAZ iso-THIP Isoguvacine Isomuscimol Isonipecotic acid Kojic amine L-838,417 Lignans (e.g., honokiol) Methylglyoxal Monastrol Muscimol Nefiracetam Neuroactive steroids (e.g., allopregnanolone) Org 20599 PF-6372865 Phenibut Picamilon 4-PIOL P4S Progabide Propofol Quisqualamine SL-75102 Taurine TACA TAMP Terpenoids (e.g., borneol) Thiomuscimol Tolgabide ZAPA Positive modulators (abridged; see here for a full list): α-EMTBL Alcohols (e.g., drinking alcohol, 2M2B) Anabolic steroids Avermectins (e.g., ivermectin) Barbiturates (e.g., phenobarbital) Benzodiazepines (e.g., diazepam) Bromide compounds (e.g., potassium bromide) Carbamates (e.g., meprobamate) Carbamazepine Chloralose Chlormezanone Clomethiazole Dihydroergolines (e.g., ergoloid (dihydroergotoxine)) Etazepine Etifoxine Fenamates (e.g., mefenamic acid) Flavonoids (e.g., apigenin, hispidulin) Fluoxetine Flupirtine Imidazoles (e.g., etomidate) Kava constituents (e.g., kavain) Lanthanum Loreclezole Monastrol Neuroactive steroids (e.g., allopregnanolone, cholesterol, THDOC) Niacin Niacinamide Nonbenzodiazepines (e.g., β-carbolines (e.g., abecarnil), cyclopyrrolones (e.g., zopiclone), imidazopyridines (e.g., zolpidem), pyrazolopyrimidines (e.g., zaleplon)) Norfluoxetine Petrichloral Phenols (e.g., propofol) Phenytoin Piperidinediones (e.g., glutethimide) Propanidid Pyrazolopyridines (e.g., etazolate) Quinazolinones (e.g., methaqualone) Retigabine (ezogabine) ROD-188 Skullcap constituents (e.g., baicalin) Stiripentol Sulfonylalkanes (e.g., sulfonmethane (sulfonal)) Topiramate Valerian constituents (e.g., valerenic acid) Volatiles/gases (e.g., chloral hydrate, chloroform, diethyl ether, paraldehyde, sevoflurane) Antagonists: Bicuculline Coriamyrtin Dihydrosecurinine Famiraprinium Gabazine (SR-95531) Hydrastine Hyenachin (mellitoxin) PHP-501 Pitrazepin Securinine Sinomenine SR-42641 Thiocolchicoside Tutin Negative modulators: 1,3M1B 3M2B 11-Ketoprogesterone 17-Phenylandrostenol α3IA α5IA (LS-193,268) β-CCB β-CCE β-CCM β-CCP β-EMGBL Anabolic steroids Amiloride Anisatin β-Lactams (e.g., penicillins, cephalosporins, carbapenems) Basmisanil Bemegride Bicyclic phosphates (TBPS, TBPO, IPTBO) BIDN Bilobalide Bupropion CHEB Chlorophenylsilatrane Cicutoxin Cloflubicyne Cyclothiazide DHEA DHEA-S Dieldrin (+)-DMBB DMCM DMPC EBOB Etbicyphat FG-7142 (ZK-31906) Fiproles (e.g., fipronil) Flavonoids (e.g., amentoflavone, oroxylin A) Flumazenil Fluoroquinolones (e.g., ciprofloxacin) Flurothyl Furosemide Golexanolone Iomazenil (123I) IPTBO Isopregnanolone (sepranolone) L-655,708 Laudanosine Lindane MaxiPost Morphine Morphine-3-glucuronide MRK-016 Naloxone Naltrexone Nicardipine Nonsteroidal antiandrogens (e.g., apalutamide, bicalutamide, enzalutamide, flutamide, nilutamide) Oenanthotoxin Pentylenetetrazol (pentetrazol) Phenylsilatrane Picrotoxin (i.e., picrotin, picrotoxinin and dihydropicrotoxinin) Pregnenolone sulfate Propybicyphat PWZ-029 Radequinil Ro 15-4513 Ro 19-4603 RO4882224 RO4938581 Sarmazenil SCS Suritozole TB-21007 TBOB TBPS TCS-1105 Terbequinil TETS Thujone U-93631 Zinc ZK-93426 | | GABAA-ρTooltip γ-Aminobutyric acid A-rho receptor | Agonists: BL-1020 CACA CAMP Homohypotaurine GABA GABOB Ibotenic acid Isoguvacine Muscimol N4-Chloroacetylcytosine arabinoside Picamilon Progabide TACA TAMP Thiomuscimol Tolgabide Positive modulators: Allopregnanolone Alphaxolone ATHDOC Lanthanides Antagonists: (S)-2-MeGABA (S)-4-ACPBPA (S)-4-ACPCA 2-MeTACA 3-APMPA 4-ACPAM 4-GBA cis-3-ACPBPA CGP-36742 (SGS-742) DAVA Gabazine (SR-95531) Gaboxadol (THIP) I4AA Isonipecotic acid Loreclezole P4MPA P4S SKF-97541 SR-95318 SR-95813 TPMPA trans-3-ACPBPA ZAPA Negative modulators: 5α-Dihydroprogesterone Bilobalide Loreclezole Picrotoxin (picrotin, picrotoxinin) Pregnanolone ROD-188 THDOC Zinc | |
| Metabotropic | | | | --- | | GABABTooltip γ-Aminobutyric acid B receptor | Agonists: 1,4-Butanediol 3-APPA 4-Fluorophenibut Aceburic acid Arbaclofen Arbaclofen placarbil Baclofen BL-1020 GABA Gabamide GABOB GBL GHB GHBAL GHV GVL Isovaline Lesogaberan Phenibut Picamilon Progabide Sodium oxybate SKF-97,541 SL 75102 Tolgabide Tolibut Positive modulators: ADX-71441 BHF-177 BHFF BSPP CGP-7930 CGP-13501 GS-39783 rac-BHFF KK-92A Antagonists: 2-Hydroxysaclofen CGP-35348 CGP-36742 CGP-46381 CGP-52432 CGP-54626 CGP-55845 CGP-64213 DAVA Homotaurine (tramiprosate, 3-APS) Phaclofen Saclofen SCH-50911 Negative modulators: Compound 14 | |
| See also Receptor/signaling modulators GABAA receptor positive modulators GABA metabolism/transport modulators | |
| v t e GABAA receptor positive modulators | |
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| Alcohols | Brometone Butanol Chloralodol Chlorobutanol (cloretone) Ethanol (alcohol) (alcoholic drink) Ethchlorvynol Isobutanol Isopropanol Menthol Methanol Methylpentynol Pentanol Petrichloral Propanol tert-Butanol (2M2P) tert-Pentanol (2M2B) Tribromoethanol Trichloroethanol Triclofos Trifluoroethanol |
| Barbiturates | (-)-DMBB Allobarbital Alphenal Amobarbital Aprobarbital Barbexaclone Barbital Benzobarbital Benzylbutylbarbiturate Brallobarbital Brophebarbital Butabarbital/Secbutabarbital Butalbital Buthalital Butobarbital Butallylonal Carbubarb Crotylbarbital Cyclobarbital Cyclopentobarbital Difebarbamate Enallylpropymal Ethallobarbital Eterobarb Febarbamate Heptabarb Heptobarbital Hexethal Hexobarbital Metharbital Methitural Methohexital Methylphenobarbital Narcobarbital Nealbarbital Pentobarbital Phenallymal Phenobarbital Phetharbital Primidone Probarbital Propallylonal Propylbarbital Proxibarbital Reposal Secobarbital Sigmodal Spirobarbital Talbutal Tetrabamate Tetrabarbital Thialbarbital Thiamylal Thiobarbital Thiobutabarbital Thiopental Thiotetrabarbital Valofane Vinbarbital Vinylbital |
| Benzodiazepines | 2-Oxoquazepam 3-Hydroxyphenazepam Adinazolam Alprazolam Arfendazam Avizafone Bentazepam Bretazenil Bromazepam Bromazolam Brotizolam Camazepam Carburazepam Chlordiazepoxide Ciclotizolam Cinazepam Cinolazepam Clazolam Climazolam Clobazam Clonazepam Clonazolam Cloniprazepam Clorazepate Clotiazepam Cloxazolam CP-1414S Cyprazepam Delorazepam Demoxepam Diazepam Diclazepam Dimdazenil Doxefazepam Elfazepam Estazolam Ethyl carfluzepate Ethyl dirazepate Ethyl loflazepate Etizolam FG-8205 Fletazepam Flubromazepam Flubromazolam Fludiazepam Flunitrazepam Flunitrazolam Flurazepam Flutazolam Flutemazepam Flutoprazepam Fosazepam Gidazepam Halazepam Haloxazolam Iclazepam Imidazenil Irazepine Ketazolam Lofendazam Lopirazepam Loprazolam Lorazepam Lormetazepam Meclonazepam Medazepam Menitrazepam Metaclazepam Mexazolam Midazolam Motrazepam N-Desalkylflurazepam Nifoxipam Nimetazepam Nitrazepam Nitrazepate Nitrazolam Nordazepam Nortetrazepam Oxazepam Oxazolam Phenazepam Pinazepam Pivoxazepam Prazepam Premazepam Proflazepam Pyrazolam QH-II-66 Quazepam Reclazepam Remimazolam Rilmazafone Ripazepam Ro48-6791 Ro48-8684 SH-053-R-CH3-2′F Sulazepam Temazepam Tetrazepam Tolufazepam Triazolam Triflubazam Triflunordazepam (Ro5-2904) Tuclazepam Uldazepam Zapizolam Zolazepam Zomebazam |
| Carbamates | Carisbamate Carisoprodol Clocental Cyclarbamate Difebarbamate Emylcamate Ethinamate Febarbamate Felbamate Hexapropymate Hydroxyphenamate Lorbamate Mebutamate Meprobamate Nisobamate Pentabamate Phenprobamate Procymate Styramate Tetrabamate Tybamate |
| Flavonoids | 6-Methylapigenin Ampelopsin (dihydromyricetin) Apigenin Baicalein Baicalin Catechin EGC EGCG Hispidulin Linarin Luteolin Rc-OMe Skullcap constituents (e.g., baicalin) Wogonin |
| Imidazoles | Etomidate Metomidate Propoxate |
| Kava constituents | 10-Methoxyyangonin 11-Methoxyyangonin 11-Hydroxyyangonin Desmethoxyyangonin 11-Methoxy-12-hydroxydehydrokavain 7,8-Dihydroyangonin Kavain 5-Hydroxykavain 5,6-Dihydroyangonin 7,8-Dihydrokavain 5,6,7,8-Tetrahydroyangonin 5,6-Dehydromethysticin Methysticin 7,8-Dihydromethysticin Yangonin |
| Monoureides | Acecarbromal Apronal (apronalide) Bromisoval Carbromal Capuride Ectylurea |
| Neuroactive steroids | Acebrochol Allopregnanolone (brexanolone) Alfadolone Alfaxalone 3α-Androstanediol Androstenol Androsterone Certain anabolic-androgenic steroids Cholesterol DHDOC 3α-DHP 5α-DHP 5β-DHP DHT Etiocholanolone Ganaxolone Hydroxydione Minaxolone ORG-20599 ORG-21465 P1-185 Posovolone Pregnanolone (eltanolone) Progesterone Renanolone SAGE-105 SAGE-324 SAGE-516 SAGE-689 SAGE-872 Testosterone THDOC Zuranolone |
| Nonbenzodiazepines | Cyclopyrrolones: Eszopiclone Pagoclone Pazinaclone Suproclone Suriclone Zopiclone Imidazopyridines: Alpidem DS-1 Necopidem Saripidem Zolpidem Pyrazolopyrimidines: Divaplon Fasiplon Indiplon Lorediplon Ocinaplon Panadiplon Taniplon Zaleplon Others: Adipiplon CGS-8216 CGS-9896 CGS-13767 CGS-20625 CL-218,872 CP-615,003 CTP-354 ELB-139 GBLD-345 Imepitoin JM-1232 L-838,417 Lirequinil (Ro41-3696) NS-2664 NS-2710 NS-11394 Pipequaline ROD-188 RWJ-51204 SB-205,384 SX-3228 TGSC01AA TP-003 TPA-023 TP-13 U-89843A U-90042 Viqualine Y-23684 |
| Phenols | Fospropofol Propofol Thymol |
| Piperidinediones | Glutethimide Methyprylon Piperidione Pyrithyldione |
| Pyrazolopyridines | Cartazolate Etazolate ICI-190,622 Tracazolate |
| Quinazolinones | Afloqualone Cloroqualone Diproqualone Etaqualone Mebroqualone Mecloqualone Methaqualone Methylmethaqualone Nitromethaqualone SL-164 |
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1524 | https://arxiv.org/pdf/2206.13485 | SHORT INJECTIVE PROOFS OF THE ERD ˝ OS-KO-RADO AND HILTON-MILNER THEOREM: A CANONICAL PARTITION OF SHIFTED INTERSECTING SET SYSTEMS
NGUYEN TRONG TUAN 1,2AND NGUYEN ANH THI 1,2
ABSTRACT .We give a canonical partition of shifted intersecting set systems, from which one can obtain unified and elementary proofs of the Erd˝ os-Ko-Rado and Hilton-Milner Theorem, as well as a characterization of maximal shifted k-uniform intersecting set systems over [n].
I NTRODUCTION
A k-uniform intersecting set system on [n] = {1, 2, . . . , n } is a collection of k-element subsets of [n] such that any two subsets in the collection have nonempty intersection. One of the classical and fundamental result in extremal set theory is the Erd˝ os-Ko-Rado Theorem on the maximum size of a k-uniform inter-secting set system. In particular, it shows that the maximum size of a k-uniform intersecting set system is attained by the collection of all k-element subsets of [n] containing a fixed element.
Theorem 1.1. Let F be a k-uniform intersecting set system on [n]. If n ≥ 2k, then |F| ≤ (n−1
k−1
). Furthermore, if
n > 2k, equality occurs if and only there exists x such that F consists of all subsets of size-k of [n] containing x.
A strengthening of the Erd˝ os-Ko-Rado Theorem was shown by Hilton and Milner , which gives a tight upper bound on the size of an intersecting k-uniform set system in which no element is contained in all sets in the system.
Theorem 1.2. Let F be a k-uniform intersecting set system on [n]. Assume that n > 2k, and no element is contained in all sets of F. Then |F| ≤ (n−1
k−1
) − (n−k−1
k−1
) + 1 . Furthermore, when k ≥ 4, equality is attained if and only if F is isomorphic to the family {A ∈ ([n]
k
) : 1 ∈ A} ∪ {{ 2, 3, . . . , k + 1 }} ; and when k = 3 , equality is attained if and only if
F is isomorphic to either the family {A ∈ ([n]3
) : 1 ∈ A} ∪ {{ 2, 3, 4}} , or the family {A ∈ ([n]3
) : |A ∩ { 1, 2, 3}| = 2 .
Since the original papers, there have been several alternative proofs of the Erd˝ os-Ko-Rado and Hilton-Milner Theorem, [4, 3, 5, 7, 8]. In this paper, we provide short elementary injective proofs of the Erd˝ os-Ko-Rado and Hilton-Milner Theorem on intersecting set systems (with equality cases characterized). Our approach is based on a canonical partition of the set system, and allows to prove the Hilton-Milner Theorem with a unifying and essentially identical argument as for the Erd˝ os-Ko-Rado Theorem. The canonical partition of the set system also allows to give a characterization of maximal shifted k-uniform intersecting set systems over [n] via appropriate intersecting set systems over [2 k − 1] for any n.As an immediate corollary, we can show that the number of maximal shifted k-uniform intersecting set systems over [n] is bounded by a constant depending only on k.We expect that the same argument would be useful for studying intersecting set systems with additional constraints.
1
arXiv:2206.13485v1 [math.CO] 27 Jun 2022
M AIN RESULTS
2.1. Compression. First, we introduce the standard compression (shifting) technique together with its ba-sic properties.
Definition 2.1. A set system F on the universe [n] = {1, 2, . . . , n } is said to be shifted if for all i < j and all S ∈ F
with j ∈ S and i / ∈ S, the set S′ = S − { j} + {i} ∈ F .
Any set system can be transformed into a shifted set system by using the following operations: Define an (i, j )-shift of F to be the set system F′ obtained by replacing S ∈ F with j ∈ S and i / ∈ S with S′ =
S − { j} + {i} if S′ /∈ F . It is easy to verify that upon applying an (i, j )-shift, an intersecting set system F
remains intersecting. Furthermore, upon finitely many (i, j )-shifts, we obtain a shifted set system. For our later application to the Hilton-Milner Theorem, we will need the following result of Frankl .
Lemma 2.2. Let n ≥ 2k ≥ 4. Suppose that F is a k-uniform intersecting set system of [n] with no element contained in all sets of F. Then there exists an intersecting set system F′ with no element contained in all sets of F′, |F ′| = |F| ,and F′ is shifted.
In particular, for the proof of Theorem 1.1 and Theorem 1.2, we can assume without loss of generality that the set system F is shifted. 2.2. Partitioning the set system. In this subsection, we introduce the key idea in our elementary proofs of Theorems 1.1 and 1.2, based on a canonical way to partition a shifted intersecting set system. Through-out this subsection and the remaining part of the paper, we assume that the intersecting set system F is intersecting and shifted. Given two sets A, B of integers of size k, we write A B (or B A) if the elements of A are a1 < a 2 <. . . < a k and the elements of B are b1 < b 2 < . . . < b k and ai ≥ bi for all i = 1 , . . . , k .
Claim 2.3. If F is shifted and A ∈ F then B ∈ F for any B A.Proof. Assume that B A. Since b1 ≤ a1 and F is shifted, we have {b1, a 2, . . . , a k} ∈ F (by considering the
(b1, a 1)-shift of F if b1 6 = a1). Inductively we can guarantee that {b1, . . . , b i, a i+1 , . . . , a k} ∈ F for any i ≤ k,so B ∈ F .
Lemma 2.4. If F is intersecting and shifted and S ∈ F has size k, then there exists i ∈ [0 , k − 1] with |S ∩ [2 k −
i − 1] | ≥ k − i.Proof. Assume that |S ∩ [2 k − i − 1] | < k − i for all i ∈ [0 , k − 1] . Define T to be the set of the first k integers not contained in S. We claim that T S. Indeed, let the elements of T be t1 < t 2 < . . . < t k. If tj > s j
for some j ∈ [1 , k ], then |S ∩ [tj − 1] | ≥ j. Furthermore, tj ≤ k + j since any integer less than tj is either of the form ti, i < j or an element of S. Thus, |S ∩ [2 k − (k − j) − 1] | ≥ | S ∩ [tj − 1] | ≥ j, contradicting our assumption on S. Hence, there exists a set T S with T disjoint from S. Since F is shifted, T ∈ F ,contradicting our assumption that F is intersecting.
We next introduce our key partition of the set system F. For each S ∈ F , by Lemma 2.4, there exists
i ∈ [0 , k − 1] such that |S ∩ [2 k − i − 1] | ≥ k − i. Let iS be the smallest such i. We then define the sub-collection of sets Fi to be the sets S for which iS = i. We say that S ∈ F i has type i.
Lemma 2.5. We have F = ⋃k−1
i=0
Fi, and for any S ∈ F i, |S ∩ [2 k − i − 1] | = k − i and 2k − i / ∈ S.
2
Proof. For each S ∈ F , let i = iS be the smallest integer in [0 , k − 1] such that |S ∩ [2 k − i − 1] | ≥ k − i
(so S ∈ F i). If i = 0 , then |S ∩ [2 k − 1] | ≥ k so |S ∩ [2 k − 1] | = k and S \ [2 k − 1] = ∅, so S ∈ F 0. Next, assume that i ≥ 1. Since i is smallest, |S ∩ [2 k − i]| < k − i + 1 and |S ∩ [2 k − i − 1] | ≥ k − i. In particular,
|S ∩ [2 k − i − 1] | = |S ∩ [2 k − i]| = k − i, which implies 2k − i / ∈ S, so S ∈ F i.
For each set S ∈ F i, define πi(S) = S ∩ [2 k − i − 1] and ψi(S) = S \ [2 k − i − 1] ⊆ [2 k − i + 1 , n ].
Lemma 2.6. The collection of sets πi(Fi) := {πi(S), S ∈ F i} is intersecting. Proof. Assume that there exists S, S ′ ∈ F i with πi(S) ∩ πi(S′) = ∅. Then |πi(S) ∪ πi(S′)| = 2( k − i) and
|[2 k − i] \ (πi(S) ∪ πi(S′)) | = i. Let T = [2 k − i] \ (πi(S) ∪ πi(S′)) . It is trivial that T ψi(S′), and thus, using shifts we can obtain the set πi(S′) ∪ T from S′. In particular, πi(S′) ∪ T ∈ F , but this contradicts the assumption F is intersecting as S ∩ (πi(S′) ∪ T ) = ∅.
2.3. Proof of the Erd ˝ os-Ko-Rado Theorem.
Proof of Theorem 1.1. We prove the theorem by induction on k and on n. When k = 1 , the claim is trivial for any n ≥ 1.Next, consider k ≥ 2. For n = 2 k, the conclusion directly follows as subsets of [2 k] of size k can be parti-tioned into 12
(2kk
) = (2k−1
k−1
) pairs of sets, such that the sets in each pair are disjoint. Thus, each intersecting system of k-sets in [2 k] has size at most (2k−1
k−1
).Next, consider n > 2k. Observe that
|F| ≤
k−1
∑
i=0
|F i|.
For i > 0, we have that πi(Fi) is a (k − i)-uniform intersecting set system on the universe [2 k − i − 1] , and by the inductive hypothesis, |πi(Fi)| ≤ (2k−i−2
k−i−1
). Since each set in Fi can be chosen by picking a set in πi(Fi),and picking the remaining i elements in [n] \ [2 k − i] in (n−2k+ii
) ways,
|F i| ≤ | πi(Fi)|
(n − 2k + ii
)
≤
(2k − i − 2
k − i − 1
)( n − 2k + ii
)
.
Furthermore, consider the set system F′ as follows. For each set S ∈ F i, define a set S′ = πi(S) ∪ { 2k −
i + 1 , . . . , 2k} and include it in F′. Then |F ′| = ∑k−1
i=0
|πi(Fi)|. Furthermore, F′ is intersecting: since F is shifted, S ∈ F i implies that S′ ∈ F . Hence, F′ is an intersecting system of k-sets of [2 k], which satisfies
|F ′| ≤ (2k−1
k−1
) by the base case above. Thus, ∑k−1
i=0
|πi(Fi)| ≤ (2k−1
k−1
).Hence, we have
|F| ≤
k−1
∑
i=1
|πi(Fi)|
(n − 2k + ii
)
|F 0|≤
(2k − 1
k − 1
)
+
k−1
∑
i=1
|πi(Fi)|
(( n − 2k + ii
)
− 1
)
≤
(2k − 1
k − 1
)
+
k−1
∑
i=1
(2k − i − 2
k − i − 1
) (( n − 2k + ii
)
− 1
)
=
k−1
∑
i=0
(2k − i − 2
k − i − 1
)( n − 2k + ii
)
+
(2k − 1
k − 1
)
−
(2k − 2
k − 1
)
−
k−1
∑
i=1
(2k − i − 2
k − 1
)
=
(n − 1
k − 1
)
,
3
where we have used that
(2k − 2
k − 1
)
+
k−1
∑
i=1
(2k − i − 2
k − 1
)
=
k−1
∑
j=0
(k − 1 + jk − 1
)
=
k−1
∑
j=0
(( k + jk
)
−
(k + j − 1
k
))
=
(2k − 1
k − 1
)
, (1) and k−1∑
i=0
(2k − i − 2
k − i − 1
)( n − 2k + ii
)
=
(n − 1
k − 1
)
. (2) This latter equality can be justified as follows: for each subset of [n − 1] of size k − 1, there is a unique
i ∈ [0 , k − 1] so that the set contains exactly i elements in [2 k − i, n − 1] and k − i − 1 elements in [2 k − i − 2]
(i . can be uniquely written as the union of a subset of [2 k − i, n − 1] of size i and a subset of [2 k − i − 2] of size k − i − 1 (i.e., i is the largest integer so that the set contains at least i elements in [2 k − i, n − 1] ). This shows the desired inequality. Furthermore, to attain equality, when n > 2k, it must be the case that
|πi(Fi)| = (2k−i−2
k−i−1
) for each i ≥ 1 and |F 0| = (2k−2
k−1
). By the inductive hypothesis, for i ≥ 1, πi(Fi) must only consist of sets that contain 1 (recall that πi(Fi) is shifted). Then F0 cannot contain any set without 1 as well, since the complement of that set in [2 k] must be a set of type i ≥ 1 that contains 1. In particular, F can only consist of sets containing 1.
2.4. Proof of the Hilton-Milner Theorem.
Proof of Theorem 1.2. We follow a similar scheme to our proof of Theorem 1.1. Consider n > 2k and the same decomposition of F into the families Fi. We again have that
|F| ≤
k−1
∑
i=0
|F i|,
and as before, for i > 0, |πi(Fi)| ≤ (2k−i−2
k−i−1
) and
|F i| ≤ | πi(Fi)|
(n − 2k + ii
)
≤
(2k − i − 2
k − i − 1
)( n − 2k + ii
)
.
We also have k−1∑
i=0
|πi(Fi)| ≤
(2k − 1
k − 1
)
.
Since F contains at least one set that does not contain 1 and F is shifted, it must be the case that F contains
{2, 3, . . . , k + 1 }. Then, F cannot contain the set {1, k + 2 , . . . , 2k}, and in particular, F contains no set of type k − 1, i.e. |F k−1| = |πk−1(Fk−1)| = 0 .Hence, we have that
|F| ≤
k−1
∑
i=0
|F i| ≤
(2k − 1
k − 1
)
+
k−2
∑
i=1
|πi(Fi)|
(( n − 2k + ii
)
− 1
)
≤
(2k − 1
k − 1
)
+
k−2
∑
i=1
(2k − i − 2
k − i − 1
) (( n − 2k + ii
)
− 1
)
=
(2k − 1
k − 1
)
+
k−1
∑
i=1
(2k − i − 2
k − i − 1
) (( n − 2k + ii
)
− 1
)
−
(( n − k + 1
k − 1
)
− 1
)
=
(n − 1
k − 1
)
−
(( n − k + 1
k − 1
)
− 1
)
,
as desired (here we have used (1) and (2)).
4
Equality occurs only if |πi(Fi)| = (2k−i−2
k−i−1
) for each i ∈ [1 , k − 2] , and |F 0| = (2k−2
k−1
) + 1 . We then have that each Fi with 2 ≤ i ≤ k − 2 is the collection of all sets of type i which contains 1.If k ≥ 4, since one can check that the set {1, k + 1 , k + 3 , . . . , 2k} has type k − 2, Fk−2 contains this set. In particular, the set {2, 3, . . . , k, k + 2 } is not contained in F. Since F is shifted, this implies that there can be no set in F not containing 1 that is different from the set {2, 3, . . . , k + 1 }. Hence, we conclude that if equality occurs, then F must be the family consisting of sets containing 1 that intersect {2, 3, . . . , k + 1 },together with the set {2, 3, . . . , k + 1 }.If k = 3 , then π1(F1) is a shifted intersecting family of size 3 consisting of subsets of size 2 in {1, 2, 3, 4}.One can then check that π1(F1) is either the family of sets containing the element 1, or the family
{{ 1, 2}, {1, 3}, {2, 3}} . From this, one can obtain that F is either the family of sets containing the element 1
together with {2, 3, 4}, or the family of sets intersecting {1, 2, 3} in a subset of size exactly 2.
2.5. A characterization of maximal shifted intersecting families. Here we record a characterization of maximal shifted intersecting families based on the partition of the set system in Section 2.2. Let F be a maximal intersecting family which is shifted. Given a subset A′ of [2 k − i] of size k − i, we denote by
Si(A′) = {B ⊆ [n] : πi(B) = A′, B \ πi(B) ⊆ [2 k − i + 1] }. For each A of type i, define S(A) = Si(πi(A)) .
Lemma 2.7. Let A ∈ F be set of type i. Then F contains S(A).Proof. We show that any set B ∈ S (A) intersects all sets in F. Indeed, assume that C ∈ F is disjoint from
B. Then C ∩ πi(A) = ∅. Since F is shifted, by shifting C, we obtain that [2 k − i] \ πi(A) ∈ F (note that
|[2 k − i] \ πi(A)| = k), which is a contradiction as A ∩ ([2 k − i] \ πi(A)) = ∅.Since F is maximal, we then have that S(A) ⊆ F .
Corollary 2.8. There is a bijection between maximal shifted intersecting families on [n] and shifted intersecting set systems G = ⋃k−1
i=0
Gi over [2 k − 1] where Gi consists of sets of size k − i contained in [2 k − i − 1] .Proof. To each maximal shifted intersecting family F on [n], we can associate a shifted intersecting set system G = ⋃k−1
i=0
Gi over [2 k − 1] where Gi consists of sets of size k − i contained in [2 k − i − 1] , by defining
Gi to be the collection of πi(A) for A ∈ F of type i. G is an intersecting set system by Lemma 2.6. Conversely, for each shifted intersecting set system G = ⋃k−1
i=0
Gi over [2 k−1] , by Lemma 2.7, we can recover the maximal shifted intersecting set system F on [n] as the union of Si(Ai) for Ai ∈ G of size i. One can easily check that this gives a bijection between maximal shifted intersecting families on [n] and shifted intersecting set systems G = ⋃k−1
i=0
Gi over [2 k − 1] .
REFERENCES P. Erd˝ os, C. Ko, and R. Rado, “Intersection theorems for systems of finite sets”, Quart. J. Math. Oxford Ser. (2) 12 (1961), 313–320. 1 P. Frankl, “The shifting technique in extremal set theory”, pp. 81–110 in Surveys in combinatorics 1987 (New Cross, 1987), edited by C. Whitehead, London Math. Soc. Lecture Note Ser. 123, Cambridge Univ. Press, 1987. P. Frankl, “A simple proof of the Hilton–Milner Theorem”, Moscow J. Combinatorics and Number Theory 8 (2019), 97–101. 1, 2 P. Frankl and Z. F ¨ uredi, “Nontrivial intersecting families”, J. Combin. Theory Ser. A 41:1 (1986), 150–153. 1 P. Frankl and N. Tokushige, “Some best possible inequalities concerning cross-intersecting families”, J. Combin. Theory Ser. A 61:1 (1992), 87--97. 1 A. J. W. Hilton and E. C. Milner, “Some intersection theorems for systems of finite sets”, Quart. J. Math. Oxford Ser. (2) 18 (1967), 369–384. 1 A. Kupavskii and D. Zakharov, “Regular bipartite graphs and intersecting families”, J. Combin. Theory Ser. A 155 (2018), 180–189. 1 M. M ¨ ors, “A generalization of a theorem of Kruskal”, Graphs Combin. 1:2 (1985), 167–183. 1
51
FACULTY OF MATHEMATICS AND COMPUTER SCIENCE , U NIVERSITY OF SCIENCE , H O CHI MINH CITY , V IETNAM .
2
VIETNAM NATIONAL UNIVERSITY , H O CHI MINH CITY , V IETNAM .
Email address : nttuan@ptnk.edu.vn (Nguyen Trong Tuan)
Email address : nathi@hcmus.edu.vn (Nguyen Anh Thi)
6 |
1525 | https://www.designingbuildings.co.uk/wiki/Wall%20types | Wall types - Designing Buildings
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Surrey Hills AONB
Edit this article
Last edited 18 Feb 2025
See full history
Wall types
Contents
[hide]
1 Wall definition
2 Buttressing wall
3 Cavity wall
4 Compartment wall
5 Curtain wall
6 Dwarf wall
7 Earth wall
8 External wall
9 Green wall
10 Internal load-bearing wall
11 Parapet wall
12 Partition wall
13 Party wall
14 Pile wall
15 Rainscreen
16 Separating wall
17 Solid wall
18 Supported wall
19 Trombe wall
20 Others
21 Related articles on Designing Buildings
[edit]Wall definition
Approved document B, Fire Safety, Volume 1 Dwelling houses, suggests that for the purpose of the performance of wall linings, a wall includes:
The surface of glazing (except glazing in doors).
Any part of a ceiling which slopes at an angle of more than 70º to the horizontal.
But a wall does not include:
Doors and doorframes.
Window frames and frames in which glazing is fitted.
Architraves, covermoulds, picture rails, skirtings and similar narrow members.
Fireplace surrounds, mantle shelves and fitted furniture.
However, Approved document C, Site preparation and resistance to contaminants and moisture, suggests that a wall is:
'Any opaque part of the external envelope of a building that is at an angle of 70° or more to the horizontal.'
See What are walls made of for more information.
[edit]Buttressing wall
A wall designed and constructed to afford lateral support to another wall perpendicular to it, support being provided from the base to the top of the wall.
See Buttressing wall for more information.
[edit]Cavity wall
A wallconstructed from two skins of masonry, the outer skin of which can be brickwork or blockwork and the inner skin of which is generally of blockwork, separated by a cavity to prevent the penetration of moisture and to allow for the installation of thermal insulation.
See Cavity wall fro more information.
[edit]Compartment wall
A wallconstructed to create a compartment, forming a barrier to the spread of smoke, heat and toxic gases.
See Compartment wall for more information.
[edit]Curtain wall
A non-structuralcladding system for the external walls of buildings.
See Curtain wall for more information.
[edit]Dwarf wall
A dwarf wall is the term used to refer to a low wall that is often used as a gardenwall, fence or as the base of a conservatory or porchstructure. Generally, it can be applied to any wall that is less than one-storey in height, but typically they are less than a metre tall.
See Dwarf wall for more information.
[edit]Earthwall
Earthwalls are essential wallsconstructed from the earth found on a site, there are variations in technique and mix such as cob, adobe, rammed earth.
[edit]External wall
A wall forming the external enclosure of a building, including part of a roofpitched at an angle of more than 70° to the horizontal, if that part of the roof adjoins a space within the building to which persons have access (but not access only for repair or maintenance).
See External wall for more information.
[edit]Green wall
A wall that is planted to help create new habitats and improve air quality and wellbeing.
See Green walls for more information.
[edit] Internal load-bearing wall
A wall providing separation between the internal spaces of a building where the wall is also required to transfer loads from other parts of the structure to the foundations.
See Load-bearing wall for more information.
See also: Internal wall.
[edit]Parapet wall
The uppermost reaches of a wall that extends above the rooflevel and provides a degree of protection to roof, gutters, balconies and walkways.
See: Parapet for more information.
[edit]Partition wall
A non-load bearing wall that separates the internal spaces of a building.
See Partition wall for more information.
[edit]Party wall
A wall that stands on the lands of 2 or more owners or a wall that is on one owner'sland but is used by 2 or more owners to separate their buildings.
See Party wall for more information.
[edit]Pile wall
A wall formed by adjacent or interlocking piles, typically found below ground where it is necessary to withhold water or soil.
See for example: Secant pile wall.
[edit]Rainscreen
A wall comprising an outer skin of panels and an airtightinsulated backing wall separated by a ventilatedcavity. Some water may penetrate into the cavity but the rainscreen is intended to provide protection from direct rain.
See Rainscreen for more information.
[edit]Separating wall
A wall or part of a wall which is common to adjoining buildings.
[edit]Solid wall
A wallconstructed of one or two skins of masonry which can consist of brick or blockworkor both but it does not include a cavity between the interior and exterior, or it is filled with smaller stones and lime. Solid walls are also single materialwalls such as cobwalls, rammed earth, or adobe.
See solid wall for more information.
[edit]Supported wall
A wall to which lateral support is afforded by a combination of buttressing walls, piers or chimneys acting in conjunction with floor(s) or roof.
[edit]Trombe wall
A construction that uses a combination of thermal mass and glazing to collect and storesolar radiation so that it can be used to heatbuildings.
See Trombe wall for more information.
[edit] Others
See also:
Framed wall.
Headwall.
Wingwall.
Sleeper wall.
Diaphragm wall.
Blank wall.
[edit] Related articles on DesigningBuildings
Barrier wall system.
Crinkle crankle wall.
Crosswall construction.
Curtain wall.
Diaphragm wall.
Dwarf wall.
Façade.
Headwall.
Infill panel walls.
Internal wall.
Load-bearing wall.
Movable walls.
Partition wall.
Party wall.
Quoin.
Rainscreen.
Raised floor.
Sleeper wall.
Supported wall.
Trombe wall.
Wingwall.
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1526 | https://madasmaths.com/archive/maths_booklets/mechanics/m1_resultants.pdf | Created by T. Madas Created by T. Madas RESULTANT FORCES Created by T. Madas Created by T. Madas Question 1 () The figure above shows three forces which lie on the same plane, acting on a particle. The magnitudes of these forces and their relative directions are shown in the figure. a) Find the magnitude of the resultant of the above three forces. b) Give the direction of the resultant as a bearing. 11.22 N R ≈ , 103 ≈ ° 60° 105° 6N 10N 8N Created by T. Madas Created by T. Madas Question 2 () The figure above shows four forces which lie on the same plane, acting on a particle. The magnitudes of these forces and their relative directions are shown in the figure. a) Find the magnitude of the resultant of the above four forces. b) Give the direction of the resultant as a bearing. 4.77 N R ≈ , 280 ≈ ° 40° 15N 10N 5N 5N Created by T. Madas Created by T. Madas Question 3 () The figure above shows four forces which lie on the same plane, acting on a particle. The magnitudes of these forces and their relative directions are shown in the figure. a) Find the magnitude of the resultant of the above four forces. b) Give the direction of the resultant as a bearing. 31.5 N R ≈ , 016.4 ≈ ° 40° 60N 30N 42N 24N 20° Created by T. Madas Created by T. Madas Question 4 () The figure above shows four forces which lie on the same plane, acting on a particle. The magnitudes of these forces and their relative directions are shown in the figure. a) Find the magnitude of the resultant of the above four forces. b) Give the direction of the resultant as a bearing. 20.4 N R ≈ , 240 = ° 60° 18N 30N 10N 24N 30° 30° Created by T. Madas Created by T. Madas Question 5 () The figure above shows three forces which lie on the same plane, acting on a particle. The magnitudes of these forces and their relative directions are shown in the figure. a) Find the magnitude of the resultant of the above three forces. b) Find the angle the resultant makes with the 5 N force. The direction in which these three forces act can be changed. c) State, with full justification, the least and the greatest magnitudes of the resultant force. MMS-K , 9.17 N R ≈ , 150.6 ≈ ° , max 32 N R = , min 0 N R = 120° 12N 15N 5N Created by T. Madas Created by T. Madas Question 6 () Two forces, 1 F N and 2 F N , are acting on a particle at right angles to each other, as shown in the figure above. The resultant of the two forces has magnitude 41 N . a) Given that the magnitude of 1 F is 9 N , find the magnitude of 2 F . b) Determine the angle the resultant makes with 2 F . A third force 3 F is added on the particle so that all three forces are in equilibrium. c) State the magnitude of 3 F . d) Calculate the angle 3 F makes with 2 F . MMS-H , 1 40 F = , 12.68 ≈ ° , 3 41 F = , 167.32 ≈ ° 2 F 1 F Created by T. Madas Created by T. Madas Question 7 (+) Three coplanar forces 1 F , 2 F and 3 F act on a particle. 1 F has magnitude 25 N , acting in a bearing of 270°. 2 F has magnitude X N , acting in a bearing of 180°. 3 F has magnitude ( ) 2 X + N , acting in a bearing of 90° . The resultant of these three forces has magnitude 37 N . Determine, as a bearing, the angle at which the resultant of these three forces is acting. MMS-G , 161 θ ≈ ° Created by T. Madas Created by T. Madas Question 8 () Two forces, act on a particle P so that the angle between the two forces is 150°. The magnitude of one of these forces is 30 N and the magnitude of the other force is F N , as shown in the figure above. The resultant of these two forces has magnitude R N , and acts at 60° to the force with magnitude F N . Calculate in any order the value of R and the value of F . MMS-D , 10 3 17.3 N R = ≈ , 20 3 34.6 N R = ≈ 150° P 30 N N F Created by T. Madas Created by T. Madas Question 9 () The figure above shows two forces 1 F and 2 F , of magnitude 24 N and x N respectively, acting on a particle P . The angle between the lines of action of 1 F and 2 F is 120°. The resultant of 1 F and 2 F is the force R , whose magnitude is 2x N . a) Show clearly that 4 4 13 x = −+ . b) Calculate the value of 2 1 − F F , correct to three significant figures. MMS-V , 2 1 30.6 N − ≈ F F 1 F 2 F P 120° Created by T. Madas Created by T. Madas Question 10 (+) Two forces, 1 F N and 2 F N , are acting on a particle so that the resultant of the two forces has magnitude 120 N and acts on a bearing of 120°. It is further given that the 1 F acts due North and has magnitude 80 N . Calculate in any order … a) … the magnitude of 2 F b) … the direction in which 2 F acts, giving the answer as a bearing. MMS-Z , 2 40 19 174 N F = ≈ , 143.4 ≈ ° |
1527 | https://www.facebook.com/groups/coolfreakswikipediaclub/posts/956508671133452/ | Cool Freaks' Wikipedia Club | 1 + 1/2 + 1/3 + 1/4 + .. | Facebook
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Jeffrey Hemphill
· May 2, 2016 ·
1 + 1/2 + 1/3 + 1/4 + ... diverges to infinity. BUT if you remove the numbers that have "9" in their denominators from this sum, you get a number that's around 23.
en.wikipedia.org
Kempner series - Wikipedia, the free encyclopedia
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Relevant comic, and an intuitive explanation:
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What the Prime Number Tweetbot Taught Me about Infinite Sums
What the Prime Number Tweetbot Taught Me about Infinite Sums ------------------------------------------------------------
9y
9
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Ari Pollack
wtf i love numbers they're so weird
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4
Jam Scramble
the harmonic series might diverge, but 1 + 2 + 3 + ... = -1/12 (say it in the cadence of the "sticks and stones" thing)
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1528 | https://it.wikipedia.org/wiki/Costante_di_Planck | Vai al contenuto
Ricerca
Indice
Inizio
1 Valore
2 Quantizzazione delle grandezze fisiche
3 Indeterminazione
4 Note
5 Bibliografia
6 Voci correlate
7 Altri progetti
8 Collegamenti esterni
Costante di Planck
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La costante di Planck, indicata con , è una costante fisica fondamentale della meccanica quantistica, introdotta come la costante di proporzionalità fra l'energia e la frequenza di un fotone nei primi anni del novecento da Max Planck e Albert Einstein, rispettivamente nello studio della radiazione emessa da un corpo nero e dell'effetto fotoelettrico. È alla base del dualismo onda-particella e della quantizzazione di grandezze come l'energia, la quantità di moto e il momento angolare.
È anche detta quanto d'azione avendo le dimensioni di un'energia per un tempo (cioè di un'azione) e nel sistema di unità di misura delle unità atomiche compone l'unità di misura del momento angolare; inoltre, è una delle costanti fondamentali che definiscono la costante di struttura fine.
Valore
[modifica | modifica wikitesto]
Il valore della costante di Planck è senza errori di misura in quanto, a partire dal 20 maggio 2019, è la costante utilizzata per definire il chilogrammo. Il valore scelto è:
Ricorre di frequente nella trattazione matematica l'espressione , che viene comunemente indicata per comodità di scrittura nelle formule con il simbolo , denominato " tagliato" o costante di Planck ridotta o costante di Dirac, che vale:
Il carattere ℏ è presente anche nella codifica Unicode.
Quantizzazione delle grandezze fisiche
[modifica | modifica wikitesto]
Lo stesso argomento in dettaglio: Quantizzazione (fisica).
La costante di Planck è legata alla quantizzazione delle grandezze dinamiche che caratterizzano lo stato della materia a livello microscopico, ovvero delle particelle che compongono materia e luce: elettroni, protoni, neutroni e fotoni. Ad esempio, l'energia trasportata da un'onda elettromagnetica con frequenza costante può assumere solo valori pari a:
A volte è più conveniente usare la velocità angolare , che dà:
Nel caso di un atomo, la quantizzazione del momento angolare determina nello spettro di emissione atomico righe di emissione corrispondenti a una serie di numeri quantici. Dato il momento angolare totale di un sistema con invarianza rotazionale e il momento angolare misurato lungo ogni data direzione, queste quantità possono assumere solo i valori
Quindi può essere detta "quanto del momento angolare".
Indeterminazione
[modifica | modifica wikitesto]
Lo stesso argomento in dettaglio: Principio di indeterminazione di Heisenberg.
La costante di Planck entra anche nel limite di accuratezza nella determinazione dei valori di coppie di variabili, come ad esempio la posizione e la quantità di moto, in base al principio di indeterminazione di Heisenberg. L'indeterminazione nella misurazione della posizione di una particella e l'indeterminazione nella misurazione della sua quantità di moto lungo la stessa direzione sono infatti vincolate dalla disuguaglianza:
: .
Nell'interpretazione più semplice delle relazioni di indeterminazione, le incertezze e sono calcolate come la deviazione standard su di un numero elevato di misure indipendenti delle rispettive grandezze fisiche eseguite su sistemi identici.
Note
[modifica | modifica wikitesto]
^ Max Planck, Ueber das Gesetz der Energieverteilung im Normalspectrum (PDF), in Ann. Phys., vol. 309, n. 3, 1901, pp. 553-63, Bibcode:1901AnP...309..553P, DOI:10.1002/andp.19013090310. URL consultato il 15 dicembre 2008 (archiviato dall'url originale il 10 giugno 2012).
^ Albert Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt (PDF), in Ann. Phys., vol. 17, n. 6, 1905, pp. 132-48, Bibcode:1905AnP...322..132E, DOI:10.1002/andp.19053220607 (archiviato dall'url originale il 9 luglio 2011).
^ (EN) Planck’s constant | Definition, Units, Symbol, & Facts | Britannica, su www.britannica.com. URL consultato il 19 aprile 2023.
^ (EN) Nicola Manini, Introduction to the Physics of Matter, Springer, 2014, ISBN 978-3-319-14381-1. p.5
^ BIPM - measurement units, su bipm.org. URL consultato il 23 luglio 2019 (archiviato dall'url originale il 23 dicembre 2018).
^ CODATA Values of the Fundamental Constants, su physics.nist.gov. URL consultato il 6 marzo 2024.
^
^ David J. Griffiths, Introduzione alla meccanica quantistica, Casa Editrice Ambrosiana, 2015, ISBN 978-88-08-08747-8. p.2
^ Gianpaolo Parodi, Marco Ostili, Guglielmo Mochi Onori, L'evoluzione della Fisica (Volume 3), Paravia, 2006, ISBN 88-395-1611-5. p.453
^ Paolo Mazzoldi, Massimo Nigro, Cesare Voci, Fisica (Volume II), EdiSES Editore, 2001, ISBN 88-7959-152-5. p.717
^ Caforio - Ferilli, PHYSICA 2000, Atomi, nuclei e particelle.
Bibliografia
[modifica | modifica wikitesto]
Paolo Silvestroni, Fondamenti di chimica, 10ª ed., CEA, 1996, ISBN 88-408-0998-8.
Voci correlate
[modifica | modifica wikitesto]
Quanto
Azione (fisica)
Quantizzazione
Meccanica quantistica
Altri progetti
[modifica | modifica wikitesto]
Altri progetti
Wikimedia Commons
Wikimedia Commons contiene immagini o altri file su costante di Planck
Collegamenti esterni
[modifica | modifica wikitesto]
(EN) Planck’s constant, su Enciclopedia Britannica, Encyclopædia Britannica, Inc.
(EN) IUPAC Gold Book, "Planck constant", su goldbook.iupac.org.
| V · D · M Unità di misura |
| Sistemi di misurazione· Conversione delle unità di misura· Sistema consuetudinario statunitense· Sistema imperiale britannico· Antiche unità di misura italiane· Unità di misura giapponesi |
| Sistema internazionale | | | | --- | | Unità di base | secondo· metro· chilogrammo· ampere· kelvin· mole· candela | | Unità derivate | hertz· newton· pascal· joule· watt· coulomb· volt· ohm· siemens· farad· tesla· weber· henry· grado Celsius· radiante· steradiante· lumen· lux· becquerel· gray· sievert· katal | |
| Sistema CGS | | | | --- | | Unità di base | centimetro· grammo· secondo | | Unità derivate | dyne· erg· baria· poise· statcoulomb· statvolt· oersted· gauss· maxwell· rayl· gal | |
| Sistema Tecnico - MFS | | | | --- | | Unità di base | metro· chilogrammo forza· secondo | | Unità derivate | unità tecnica di massa | |
| Sistema MTS | | | | --- | | Unità di base | metro· tonnellata· secondo | | Unità derivate | sthène· chilojoule· chilowatt· pièze | |
| Unità atomiche | | | | --- | | Unità di base | raggio di Bohr· massa a riposo dell'elettrone· carica elementare· costante di Dirac· energia di Hartree | |
| Unità di misura di Planck | | | | --- | | Unità di base | tempo di Planck· lunghezza di Planck· massa di Planck· carica di Planck· temperatura di Planck· quanto del momento angolare | | Unità derivate | energia di Planck· forza di Planck· potenza di Planck· densità di Planck· frequenza angolare di Planck· pressione di Planck· corrente di Planck· tensione di Planck· resistenza di Planck | |
| Altre unità di misura | | | | --- | | Area | ettaro | | Densità lineare | denaro· tex | | Entropia | clausius | | Massa | grano· scruple· dramma· oncia· libbra· pietra· quarto· slug | | Temperatura | grado Rømer | | Energia | caloria· frigoria | | lunghezza in astronomia | unità astronomica· anno luce· parsec | |
| | |
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Multiplication with 2, 5, and 10: word problems
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CCSS.Math.Content.HSA.REI.B.4 Solve Quadratic...
Home > Common Core > Mathematics > High School Algebra > Reasoning with Equations & Inequalities > Solve Quadratic Equations In One Variable.
CCSS.Math.Content.HSA.REI.B.4 - Solve quadratic equations in one variable.
| |
| CCSS.Math.Content.HSA.REI.B.4.A - Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)2 = q that has the same solutions. Derive the quadratic formula from this form. |
| |
| CCSS.Math.Content.HSA.REI.B.4.B - Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. |
Authors: National Governors Association Center for Best Practices, Council of Chief State School Officers
Title: CCSS.Math.Content.HSA.REI.B.4 Solve Quadratic Equations In One Variable. Reasoning with Equations & Inequalities - High School Algebra Mathematics Common Core State Standards
Publisher: National Governors Association Center for Best Practices, Council of Chief State School Officers, Washington D.C.
Copyright Date: 2010
(Page last edited 10/08/2017)
Complete the Square - Algebra 1 - Fill in the number that makes the polynomial a perfect-square quadratic. An on-screen form is provided for the student to provide the missing term to complete a perfect-square quadratic. New problems are provided after each answer and score is kept over a timed interval. Explanation of wrong answers are provided.
Complete the Square - Algebra 2 - Fill in the number that makes the polynomial a perfect-square quadratic. An on-screen form is provided for the student to provide the missing term to complete a perfect-square quadratic. New problems are provided after each answer and score is kept over a timed interval. Explanation of wrong answers are provided.
Completing the square (old school) - Solving a quadratic by completing the square.
Completing the square 1 - Find the roots of a quadratic equation by completing the square. Can also be solved by factoring.
Complex roots for a quadratic - Complex Roots from the Quadratic Formula
Example 4: Applying the quadratic formula - Application Problem with Quadratic Formula
Examples: Factoring simple quadratics - A few examples of factoring quadratics
Factoring Quadratic Expressions - Example of factoring quadratic expressions
Factoring Quadratic Expressions - Factoring Quadratic Expressions
Factoring Quadratics - Algebra I: Factoring Quadratics
How to use the Quadratic Formula - Introduction to using the quadratic formula.
Proof of Quadratic Formula - Proof of Quadratic Formula: completing the square
Quadratic Equation - Algebra I: Quadratic Equation
Quadratic Equation part 2 - 2 more examples of solving equations using the quadratic equation
Quadratic equations - Solve an equation using the zero product property
Quadratic Formula (proof) - Deriving the quadratic formula by completing the square.
Quadratic formula with complex solutions - Multiple choice practice quiz
Quadratic functions - Solve a quadratic equation using the zero product property
Quadratic functions - Solve a quadratic equation by factoring
Quadratics and Shifts - Solving quadratics and graph shifts
Solutions to quadratic equations - Determine how many solutions a quadratic equation has and whether they are rational, irrational, or complex
Solve a quadratic equation using square roots - How many real solutions does the equation have?
Solving Quadratic Equations - Solving Quadratic Equations by completing the square
Solving Quadratic Equations - Solving Quadratic Equations by taking the Square Root
Solving Quadratic Equations - Solving Quadratic Equations by Square Roots
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1532 | https://derangedphysiology.com/main/cicm-primary-exam/respiratory-system/Chapter-051/lung-volumes-and-capacities | Lung volumes and capacities
By Alex Yartsev - 04/10/2019
Last updated 07/09/2024 - 14:44
This chapter is most relevant to Section F4(i) from the 2023 CICM Primary Syllabus, which expects the exam candidates to be able to "state the normal values of lung volumes and capacities". Ostensibly, other chapters serve the purposes of Section F4(ii), "Explain the factors that influence lung volumes and capacities", but inevitably some of that creeps into this section. It has come up a couple of times in the primary exam:
Question 24 from the second paper of 2017
Question 4 from the second paper of 2015
Question 18 from the second paper of 2017
Two of these questions asked specifically about the FRC, and how it is measured. Nobody seems to care about TLC ERV or IRV. Functional residual capacity, the popular show pony of lung volumes, seems to get all the attention while all the other lung volume compartments are trivialised by the colleges' inattention. Well, this chapter is an effort to return some dignity and respect to the forgotten volumes and capacities of the lung. Moreover, beyond listing the normal values and pretending that they represent real human measurements, one should probably have some understanding of why we care about these volumes and capacities, i.e. their importance in health and disease.
In summary:
Normal Reference Values for Lung Volumes and Capacities
| Volume | Absolute volume in ml/kg | Definition |
| RV (residual volume) | 15 | The volume of gas remaining in the lung after maximal exhalation |
| ERV (expiratory reserve volume) | 15 | The volume of gas that can be maximally exhaled from the end-expiratory level during tidal breathing |
| TV (tidal volume) | 7 | The volume of gas inhaled or exhaled during the respiratory cycle |
| IRV (inspiratory reserve volume) | 45 | maximum volume of gas that can be inhaled from the end-inspiratory level during tidal breathing. |
| IC (inspiratory capacity) | 52 | The maximum volume of gas that can be inspired from FRC |
| FRC (functional residual capacity | 30 | The volume of gas present in the lung at end expiration during tidal breathing |
| VC (vital capacity) | 67 | The volume change at the mouth between the positions of full inspiration and complete expiration. |
| TLC (total lung capacity) | 82 | The volume of gas in the lungs after maximal inspiration, or the sum of all volume compartments |
In terms of reading material, the most solid peer-reviewed resource for this topic would probably have to be Wanger et al (2005), which also covers the measurement of the volumes and capacities. It appears to be a statement by an ATS/ERS joint task force, outlining the recommendations of these societies; one cannot get any more official than that. For a discussion of the FRC all on its own, one may look to the article by Chandra et al (2013). For the rest, there is scattered information around textbooks and articles, but no single solid source to rely upon.
Static lung volumes and capacities
Quoting directly from the ATS/ERS statement, the definitions of the standard volumes are:
The volume of gas inhaled or exhaled during the respiratory cycle is called the tidal volume (TV or VT).
The expiratory reserve volume (ERV) is the volume of gas that can be maximally exhaled from the end-expiratory level during tidal breathing (i.e. from the FRC).
The inspiratory reserve volume (IRV) is the maximum volume of gas that can be inhaled from the end-inspiratory level during tidal breathing.
RV refers to the volume of gas remaining in the lung after maximal exhalation (regardless of the lung volume at which exhalation was started).
Those are the four volumes. There are also four capacities. Again, from the ATS/ERS statement, the definitions of the standard capacities are:
The FRC is the volume of gas present in the lung at end expiration during tidal breathing.
The maximum volume of gas that can be inspired from FRC is referred to as the inspiratory capacity (IC).
The vital capacity (VC) is the volume change at the mouth between the positions of full inspiration and complete expiration.
TLC refers to the volume of gas in the lungs after maximal inspiration, or the sum of all volume compartments.
Note that lung volumes are measurable gas-filled spaces in the lung, whereas capacities are combinations of two or more volumes (where the definition of capacity is the measure of the lungs' ability to hold a gas). It seems that we have come by these terms because a group of eleven senior physiologists got together in 1950 and engraved them indelibly into the psyche of subsequent researchers. Under normal circumstances, the contents of their paper would only available only to those who are willing to exhume a physical copy of Federation Proceedings (Volume 9) from the deepest catacombs of their university library, and without the original text it would be impossible to determine how these authors defended their choice of terminology. Fortunately, we can be grateful to a reader (Tony Keeble, many thanks) for finding the Proceedingsat archive.org, and for bringing it to our shared attention. It appears the term "capacity" was used to describe grouping of smaller ("volume") subdivisions because these were designed with clinical testing in mind, i.e. a patient's capacity to perform a specific respiratory manoeuvre was being tested by "gasometric measures".
This small obscure paper certainly seems to have carved these definitions in stone. By the time alater (1975) revision of lung function nomenclature came around, the authors treated the volume-capacity designation as something well established and beyond argument, whereas a lot of other terminologies were discarded, and a lot of new weird terminologies were invented. This same committee also took upon themselves to define chest findings such as "rhonchi" and "rhales", and not all of those choices were completely non-controversial. "The traditionally trained physician may be surprised, and in some cases appalled, by the recommendations in regard to descriptions of physical findings in the chest", wrote Burrows in his 1975 editorial. Treasured nomenclature such as "sibilant" or "amphoric" breath sounds and "tactile fremitus" were discarded as nonessential and confusing ("their loss will hardly be mourned by most first- and second year medical students" the author gloated) but a whole host of confusing and nonessential neologisms were promoted, like "hypobasemia" and "expiratory retard". In short, when one goes looking for an explanation of why something is called a volume whereas something else is a capacity, one rapidly develops the impression that there was no guiding light of reason to illuminate these decisions.
Reference values of lung volumes and capacities
The CICM trainee would surely be expected to be intimately familiar with this famous diagram which describes the compartments of lung volume.
There are some numbers there, quoted for the trainees because the syllabus document clearly expects them to "state the normal values of lung volumes and capacities".Those exact numbers, needless to say, are almost entirely imaginary, as it would be impossible to give any figure and expect it to be accurate even within a 10% error margin, considering the vast variety of human shapes and sizes. Often, authors resort to throwing some representative values into their textbook diagrams, because of the widespread belief that the use of numbers increases their credibility. For instance, where this appears in the 8th edition of Nunn's, the diagram footer actually makes reference to Dr John Francis Nunn himself, appearing to claim that their diagram accurately represents his personal respiratory performance in 1990 (p. 28, Figure 2.9). Irrespective of how well Dr Nunn's thoracic dimensions represent the approximate human average, it still seems like a weird thing to do. In the diagram above (and the table below) the values were appropriated from Kerry Brandis' The Physiology Viva, mainly because he offers them in terms of ml/kg, indexed to body size. An extra column was added from Garcia-Rio et al (2009) because it probably represents normal values for a 65-year-old male European professor of physiology.
Normal Reference Values for Lung Volumes and Capacities
| Volume | In ml/kg, from Brandis | In ml, for 70kg adult, from Brandis | In ml, for over-65 male, from Garcia-Rio et al (2009) |
| RV | 15 | 1050 | 1940 (by plethysmography) |
| ERV | 15 | 1050 | 1200 |
| TV | 7 | 490 | 1000 |
| IRV | 45 | 3150 | 1450 |
| IC | 52 | 3640 | 2450 |
| FRC | 30 | 2100 | 3140 (by plethysmography) |
| VC | 67 | 4790 | 3650 |
| TLC | 82 | 5740 | 5410 |
The adventurous reader is also invited to consider the existence of other, shadowy and mysterious volumes, largely unspoken about in the literature. We mention these here in hushed tones mainly because there may be some hidden educational value in knowing things, beyond the grim meat-hook realities of the CICM exams. One such "hidden" volume is MV, or minimum alveolar volume. This is gas plus tissue volume at zero transpulmonary pressure, representing the conditions that result from an open pneumothorax in an apnoeic subject. From animal data such as Kallok & Lai-Fook (1980) and Tsunoda et al (1974), as well as postmortem human lung data by Salmon et al (1981), we can surmise that this volume is something like 8-10% of the TLC, or 7-8 ml/kg (490-560ml in the normal adult).
Functional Residual Capacity (FRC)
Because of the college examiners' focus on this specific subdivision of the lung volume, it seems like something important to explore. As it has attracted a whole chapter on its own, this section will only summarise the most important points about the FRC.
In brief:
The FRC is the volume of gas present in the lung at end expiration during tidal breathing
It is composed of ERV and RV
This is usually 30-35 ml/kg, or 2100-2400ml in a normal sized person
It represents the point where elastic recoil force of the lung is in equilibrium with the elastic recoil of the chest wall, i.e. where the alveolar pressure equilibrates with atmospheric pressure.
The measurement of FRC is an important starting point for the measurement of other lung volumes
Residual volume (RV)
RV is the amount of gas left at the end of a forceful maximal expiration. To get the lung volume any smaller, one literally needs to squeeze the chest cavity, something actually performed on volunteers by Leith & Mead (1967). Though they unfairly classified subjects over the age of 40 as "old", something to which the rapidly aging author objects to, their results are probably still valid. They confirmed that the RV seems to increase over one's lifespan largely owing to dynamic factors, i.e. airway closure and the increase of the closing capacity, rather than changes in chest wall recoil or lung elasticity. Because the airways had closed, squeezing the chests of older subjects did not result in any additional gas flow (whereas an extra 40 cm H2O caused some air to flow out of the younger subjects). By the same mechanism, this volume increases in obstructive lung diseases such as COPD and asthma.
Apart of diseases which affect closing capacity, RV decreases with any disease that globally decreases all lung volumes, for example idiopathic pulmonary fibrosis and obesity. Interestingly, it appears that in morbidly obese patients, the decrease in RV is measurable, but remains within the normal range of values for height (Lofrese & Lappin, 2018).
Expiratory reserve volume (ERV)
The ERV is the volume of gas that can be maximally exhaled from the FRC volume. This is the volume which is probably the most variable among all the lung volumes, and accounts for most of the changes in FRC related to posture and body habitus. Craig (1960) was able to demonstrate a change of the ERV from 34.2% of VC in the sitting position to 18.4% of VC in the supine position. When FRC decreases in obesity, it is the ERV which is to blame, and there is some evidence that morbidly obese patients have essentially zero ERV, i.e. their tidal volume normally goes all the way down to their residual volume (see below). De Jong et al (2014) mention this as one of the factors which render pre-anaesthetic oxygenation less effective in the morbidly obese.
Tidal volume
The tidal volume is defined as the volume of gas inhaled or exhaled during a respiratory cycle. This obviously varies considerably between individuals and therefore between published authors. Reported values vary between ~ 380ml (Tobin et al, 1983) to ~ 810ml (Sorli et al, 1978). The problem is also partly because of the fact that one relies on FRC as the lower margin for where the VT is calculated from, but this depends on the subject being relaxed and at rest, which is a state difficult to achieve when you're trapped in a body plethysmograph. Gilbert et al (1972) demonstrated that the influence of voluntary control on supposedly normal restful ventilation while being measured and monitored tends to increase the tidal volume and decrease the respiratory rate. Most ICU trainees, when asked "what is a normal tidal volume", will probably blurt out "6-8ml/kg", which will have originated from their having overheard somebody talking about lung-protective ARDS ventilation; and it will not be half wrong, falling into the range of 420-560ml for a 70kg person.
Inspiratory reserve volume
The inspiratory reserve volume is the volume you'd inhale if you kept inhaling after a normal tidal inhalation was completed. It obviously falls prey to the same problem of interpreting whether the tidal volume entrained by the laboratory volunteer is "normal" when they are nose-clipped and confined to a small pressurised box.
This is the volume most affected by diseases which decrease the range of movement available to the diaphragm and chest wall. In short, where the chest wall movement or diaphragmatic excursion are restricted, the IRV will be substantially diminished. For instance, this may be seen in severe kyphosis, in the presence of multiple rib fractures, or in the context of severe COPD where the diaphragm is flattened and the chest hyperinflated. In COPD, the IRV is sacrificed when the RV increases (O'Donnell et al, 2017); the investigators found that with disease progression the resting IRV can decrease from ~28% of TLC to about 17%. With dynamic hyperinflation due to tachypnoea causing the RV to increase yet further, in severe COPD the IRV can drop to as low as 5% of the TLC, which demonstrates spectacularly the literal "lack of reserve" such patients have. When they are in respiratory distress and their respiratory rate is in excess of 30, with maximal effort they can increase their tidal volume by perhaps a further 250-300ml.
Age-related changes in lung volumes
The total lung capacity remains relatively stable over the course of one's lifespan. The fraction of the TLC occupied by the FRC increases slightly, as does the RV. This graph is from Stocks & Quanjer (1995):
Size-related changes in lung volumes
Using the predictive equations presented by Stocks & Quanjer (1995), it is also possible to plot graphs relating height to pulmonary volumes and capacities. A clumsy spreadsheet chart derived from these equations is presented below. Note how the TLC increases substantially with increasing height, but the increase in FRC is proportionally smaller.
Effects of obesity on lung volumes
In summary, with obesity all lung volumes decrease, and the worse the obesity, the greater this effect. This was demonstrated abundantly by Jones et al (2006), whose excellent paper must be upheld as some sort of gold standard when it comes to visually presenting information on changes in lung volumes. Unfortunately, their detailed scatter plots spread the data over several graphs. In order to fuse these into one, the author had to take some liberties with their tabulated data (Table 1 on p.829) and construct a crude approximation in a spreadsheet. It was merely a matter of reconstructing absolute volume values from the percentage-of-predicted data. Voila:
Note how the decrease in capacities is almost entirely the result of a change in ERV. The scatter plot for that one is presented below, because it's worth isolating.
The reader is invited to carefully step over the obvious pun and observe the fat part of the 40+ BMI group data distribution. The majority of these people had no ERV whatsoever.
Previous chapter: Work of breathing and its components
Next chapter: Measurement of lung volumes and capacities
References
Wanger, J., et al. "Standardisation of the measurement of lung volumes." European respiratory journal 26.3 (2005): 511-522.
Lutfi, Mohamed Faisal. "The physiological basis and clinical significance of lung volume measurements." Multidisciplinary respiratory medicine 12.1 (2017): 3.
Boren, Hollis G., Ross C. Kory, and James C. Syner. "The Veterans Administration-Army cooperative study of pulmonary function: II. The lung volume and its subdivisions in normal men." The American Journal of Medicine 41.1 (1966): 96-114.
Pappenheimer, J. R., J. H. Comroe, and A. Cournand. "Standardization of definitions and symbols in respiratory physiology." Fed Proc. Vol. 9. No. 3. 1950.
Gandevia, Bryan, and P. Hugh-Jones. "Terminology for measurements of ventilatory capacity: a report to the Thoracic Society." Thorax 12.4 (1957): 290.
Burrows, Benjamin. "Pulmonary Terms and Symbols: A Report of the ACCP-ATS Joint Committee on Pulmonary Nomenclature." Chest 67.5 (1975): 583-593.
Enright, Paul L., et al. "Spirometry reference values for women and men 65 to 85 years of age."Am Rev Respir Dis 147.1 (1993): 125-133.
Neder, Jose Alberto, et al. "Reference values for lung function tests: I. Static volumes."Brazilian journal of medical and biological research 32.6 (1999): 703-717.
Leith, David E., and Jere Mead. "Mechanisms determining residual volume of the lungs in normal subjects." Journal of Applied Physiology 23.2 (1967): 221-227.
Quanjer, Ph H., et al. "Lung volumes and forced ventilatory flows." (1993): 5-40.
Garcia-Rio, Francisco, et al. "Lung volume reference values for women and men 65 to 85 years of age."American journal of respiratory and critical care medicine 180.11 (2009): 1083-1091.
Koetsenruijter, A. Willem M. "Using numbers in news increases story credibility." Newspaper research journal 32.2 (2011): 74-82.
Chandra Selvi, E., and Kuppu KV Rao. "Should the Functional Residual Capacity be Ignored?." Journal of Clinical and Diagnostic Research: JCDR 7.1 (2013): 43.
Stocks, JaPHQ, and Ph H. Quanjer. "Reference values for residual volume, functional residual capacity and total lung capacity. ATS Workshop on Lung Volume Measurements. Official Statement of The European Respiratory Society." European Respiratory Journal 8.3 (1995): 492-506.
SIMMONS, DANIEL H., et al. "Relation between lung volume and pulmonary vascular resistance." Circulation Research 9.2 (1961): 465-471.
Wahba, W. M. "Influence of aging on lung function-clinical significance of changes from age twenty."Anesthesia & Analgesia 62.8 (1983): 764-776.
Lofrese, John J., and Sarah L. Lappin. "Physiology, Residual Volume."StatPearls [Internet]. StatPearls Publishing, 2018.
Craig JR, Albert B. "Effects of position on expiratory reserve volume of the lungs." Journal of applied physiology 15.1 (1960): 59-61.
De Jong, A., et al. "How to preoxygenate in operative room: healthy subjects and situations “at risk”." Annales francaises d'anesthesie et de reanimation. Vol. 33. No. 7-8. Elsevier Masson, 2014.
Jones, Richard L., and Mary-Magdalene U. Nzekwu. "The effects of body mass index on lung volumes."Chest 130.3 (2006): 827-833.
Tobin, Martin J., et al. "Breathing patterns: 1. Normal subjects."Chest 84.2 (1983): 202-205.
Sorli, J., et al. "Control of breathing in patients with chronic obstructive lung disease." Clinical science and molecular medicine 54.3 (1978): 295-304.
Gilbert, Robert, et al. "Changes in tidal volume, frequency, and ventilation induced by their measurement." Journal of Applied Physiology 33.2 (1972): 252-254.
O’Donnell, Denis E., et al. "The link between reduced inspiratory capacity and exercise intolerance in chronic obstructive pulmonary disease." Annals of the American Thoracic Society 14.Supplement 1 (2017): S30-S39.
Kallok, MICHAEL J., and STEPHEN J. Lai-Fook. "Lung deformations at minimal volume." Journal of Applied Physiology 48.3 (1980): 487-494.
Salmon, R. B., et al. "Human lung pressure-volume relationships: alveolar collapse and airway closure." Journal of Applied Physiology 51.2 (1981): 353-362. |
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1534 | https://www.vanderbilt.edu/AnS/Chemistry/Rizzo/Chem220b/Chapter_18.pdf | 57 112 Chapter 18: Ethers and Epoxides; Thiols and Sulfides O R H O R R O alcohols ethers epoxides CH3CH2OCH2CH3 O OCH3 thiols sulfides episulfides (mercaptans) (thioethers) unstable S R H S R R S -S CH3CH2CH2SH CH3SCH3 SH SH H2N CH C CH2 OH O SH H2N CH C CH2 OH O CH2 S CH3 113 18.1 Naming Ethers Simple ethers are named by identifying the two organic substituents and adding the word ether If other functional groups are present, the ether part is considered an alkoxy substituent 58 114 18.2: Structure & properties of ethers: The oxygen of ethers is sp3-hybridized and tetrahedral The ether oxygen is a weak Lewis base Ethers have small dipole moments and are relatively non-polar Inert to a wide range of reaction conditions- good solvents Synthesis of ethers: Symmetrical ethers can be prepared by treating the corresponding alcohol with a strong acid Limitations: must be symmetrical works best for 1° alcohols 115 18.3: Williamson Ether Synthesis Reaction of metal alkoxides with alkyl halides or tosylates to give ethers. This is an SN2 reaction.
Alkoxides are prepared by the reaction of an alcohol with a strong base such as sodium hydride, NaH C O H3C CH3 CH3 H3C I THF + C O H3C CH3 CH3 CH3 C OH H3C CH3 CH3 C O H3C CH3 CH3 + NaH THF Na + NaH - H2 + NaI Na SN2 Few restriction regarding the nature of the alkoxide Works best for 1° alkyl halide or tosylate; E2 elimination is a competing reaction with 2° halides and tosylates 3° halides undergo E2 elimination (vinyl and aryl haildes do react) Br (R)-2-Bromohexane CH3CH2-O Na+ _ THF OCH2CH3 (S) 59 116 18.4: Alkoxymercuration of alkenes Recall oxymercuration (Chapter 7.4): overall Markovnikov addition of H-OH across the π-bond of an alkene + Hg OAc OAc HgOAc !+ !+ !+ OH2 HO HgOAc HO NaBH4 Alkoxymercuration: replace water with an alcohol to get the Markovnikov addition of H-OR across across the π-bond of an alkene + Hg(OAc)2 ROH OR HgOAc NaBH4 OR Few restrictions regarding the nature of the alcohol or alkene 117 18.5: Reactions of Ethers: Acidic Cleavage Ethers are generally unreactive Ether bonds can be broken with strong acid (HI or HBr) and heat The mechanism involves protonation of the ether followed by an SN2 reaction with the halide acting as the nucleophile (works best for phenyl alkyl ethers Ph-O-CH2R) However: C O H3C H3C H3C CH2CH2CH3 HBr C O H3C H3C H3C CH2CH2CH3 H SN1 C HO H3C H3C H3C CH2CH2CH3 C H3C H3C H3C Br H3C CH2 C H3C 60 118 18.6: Reactions of Ethers: Claisen Rearrangement Thermal rearrangement (> 200° C) of allyl vinyl ethers to give γ,δ-unsaturated carbonyls Member of a class of pericyclic reactions known as sigmatropic rearrangements (Chapter 30) Allyl phenyl ethers rearrange to o-allylphenols O O !
" # $ % allyl phenyl ethers (prepared by Williamson synthesis) “aromatic” transition state γ,δ-unsaturated carbonyl o-allylphenol O O ‡ O H HO !
" # $ tautomerization 119 18.7: Cyclic Ethers: Epoxides O O O O furan tetrahydrofuran 2H-pyran 4H-pyran O O 1,4-dioxine O O 1,4-dioxane O O Cl Cl Cl Cl 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine (dioxin) Epoxides (oxirane): three-membered ring, cyclic ethers; the strain of the three-membered ring makes epoxides electrophilic Synthesis of epoxides from alkenes: epoxidation with m-chloroperbenzoic acid (mCPBA) O Cl O O O H Cl O O O H O Cl OH O + 61 120 Stereochemistry of the mCPBA epoxidation: syn addition of oxygen. The geometry of the alkene is preserved in the product (recall the cyclopropanation of alkenes, Ch. 7.6) Groups that are trans on the alkene will end up trans on the epoxide product. Groups that are cis on the alkene will end up cis on the epoxide product.
H H R R cis-alkene mCPBA H H R R O cis-epoxide H R R H trans-alkene H R R H O trans-epoxide CH2Cl2 mCPBA CH2Cl2 121 Intramolecular Williamson Synthesis: general method for the synthesis of cyclic ethers: the alkoxide and alkyl halide are part of the same molecule OH Br NaH, THF Br O O SN2 Epoxides from halohydrins SN2 Br2, H2O Br OH NaH, THF Br O anti addition O 62 122 18.8: Ring-Opening Reactions of Epoxides Acid-catalyzed epoxide opening: protonation of the epoxide oxygen makes it more reactive toward nucleophiles hydroysis of epoxides give vicinal diols O H H mCPBA H3O+ H H OH OH trans-1,2-cyclohexanediol O H H H OH2 H OH OH H cis-1,2-cyclohexanediol OsO4 NaHSO3 complementary to dihydroxylation of alkenes acid-catalyzed halohydrin formation from epoxides O H H H-X H H OH X O H H H X ether 123 Regiochemistry of acid-catalyzed epoxide openings: if the carbons of the epoxide are 1° or 2°, then the epoxide opening goes predominantly by an SN2 mechanism and the nucleophile adds to the least substituted carbon if either carbon of the epoxide is 3°, the epoxide opening goes predominantly by an SN1 mechanism and the nucleophile adds to the 3°-carbon 63 124 Nucleophilic (base-catalyzed) epoxide opening: Epoxide undergo ring-opening with nucleophiles via an SN2 mechanism Nucleophilic epoxide opening with Grignard reagents C CH2 O H3CH2C H OR C CH2 O H3CH2C H OR 100° C ROH C CH2 HO OR + RO C CH2 O H3CH2C H BrMg-CH3 CH CH2 OH H3CH2C CH3 ether then H3O+ The nucleophile will add to the least substituted carbon of the epoxide (SN2 mechanism) 125 18.9: Crown Ethers (please read) 18.10: Thiols and sulfides Thiols (mercaptans) are sulfur analogues of alcohols Sulfides (thioethers) are sulfur analogues of ethers Preparation of thiols from alkyl halides (SN2): H3C-H2C-H2C-H2C Br H2N NH2 S + H3C-H2C-H2C-H2C S NH2 NH2 HO H3C-H2C-H2C-H2C S NH2 NH2 O H H3C-H2C-H2C-H2C SH H2N NH2 O + Thiourea 64 126 Thiols can be oxidized to disulfides 2 R-SH [O] [H] R-S-S-R disulfide thiols Thiolate ions (anions of thiols) are very reactive nucleophiles Thiolates react with 1° and 2° alkyl halides to yield sulfides (SN2), analogous to the Williamson ether synthesis CH3CH2-S Na+ CH3CH2-SH NaH, THF _ Br CH3CH2-S-CH2CH2CH2CH3 SN2 Sulfides can be oxidized to sulfoxides and sulfones R1 R2 S H2O2 R1 R2 S O sulfide R1 R2 S mCPBA O O sulfoxide sulfone H3N H N N H CO2 O2C O S O NH3 N H H N O2C CO2 O S O H3N H N N H CO2 O2C O SH O 2 glutathione +2e-, +2H+ -2e-, -2H+ 127 Bioactivation and detoxication of benzo[a]pyrene diol epoxide: P450 O H2O P450 OH HO O benzo[a]pyrene OH HO DNA glutathione transferase G-S OH HO HO SG N N N N NH2 DNA N N N N OH HO HO NH DNA O2 65 128 18.11 Spectroscopy of ethers IR spectroscopy: not particularly diagnostic for the ether functional group. Strong C-O single bond stretch between 1050-1150 cm-1 1H NMR: protons on the carbons that are part of the ether linkage are deshielded relative to alkanes. The chemical shift of these protons is from δ= 3.5 - 4.5 ppm H C H H C C H H H H O C H H C C H H H H H 2H, t 3H, t 2H, t,q 129 3H, d 1H, dd 1H, dd 1H, m Protons on epoxide carbons are shield relative to those of a typical ethers (δ=2.0 - 3.0 ppm) 13C NMR: the chemical shift of carbons that are part of the ether linkage are in the range of δ= 50 - 80 ppm Epoxide carbons are on the upfield side of this range (~40 - 60 ppm) not chemically equivalent C C O H3C Ha Hb Hc Jab Jac Jbc 66 130 C9H11BrO 2H 3H 2H,t 2H,t 2H,m 2H 2H 1H 1H,d 1H,d 3H, t C9H10O 13C: 157.5 129.4 120.4 114.4 67.3 31.5 29.0 13C: 147.4 135.2 129.2 128.0 126.4 100.8 55.4 131 C9H10O2 158.49 129.54 121.25 114.64 68.68 50.18 44.76 2H 3H 1H 1H dd J= 3.4, 11.0 dd J= 6.0, 11.0 1H 1H 1H m dd J= 4.2, 4.8 dd J= 2.6, 4.8 |
1535 | https://www.youtube.com/watch?v=A82brFpdr9g | Standard Deviation vs Standard Error, Clearly Explained!!!
StatQuest with Josh Starmer
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Posted: 20 Mar 2017
People often confuse the standard deviation and the standard error. This StatQuest clears it all up!
For more information on the standard error, see the StatQuest on The Standard Error:
And the StatQuest on p-value pitfalls and power calculations:
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Transcript:
Intro hello and welcome to another stat quickie this time we're going to talk about the standard deviation versus the standard error these terms are often confused so let's clear them up once and for all I think the easiest way to understand the differences between the standard deviation and the standard error is to look at an example for the sake of this example imagine we weighed five mice this is the average or mean of the values we measured this is the standard deviation on both sides of the mean it quantifies how much the data are spread out now imagine we did the exact same experiment weighed five mice five separate times using different mice each time this would result in five means or averages one for each set of measurements we would also have five separate standard deviations around the means one for each set of measurements they quantify how much the measurements are spread around their means here's what it would look like if Standard Error we plotted all five means on the same number line This is the mean of the means and this is the standard deviation on both sides of the mean of the means it's a little bit of a tongue twister there the standard deviation of the means is called the standard error and that's all there is to it it's just that Summary simple now let's summarize the differences between the standard deviation and the standard error the standard deviation quantifies the variation within a set of measurements the standard error quantifies the variation in the means from multiple sets of measurements the confusing thing is that the standard error can be estimated from a single set of measurements even though it describes the means from multiple sets thus even if you only have a single set of measurements you are often given the option to plot the standard error in almost all cases you should plot the standard deviation since graphs are usually intended to describe the data that you measured for more information on standard errors check out the stack Quest on standard errors also the stack Quest on pval pitfalls and power calculations I've provided links to those in the description below all right we made it to the end hooray tune in next week for another stat quickie |
1536 | https://math.stackexchange.com/questions/1701746/optimization-problem-choose-n-from-set-where-fn-is-maximized-and-gn-is | Skip to main content
Optimization problem, choose n from set where f(n) is maximized and g(n) is minimized
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Say I have multiple objects in a set, each with a certain value f and g for each object in the set, and I want to select exactly N of these objects from the set such that the sum of fis maximized and the sum of g is minimized. What is the name of this optimization problem, and how can I approach solving it efficiently?
combinatorics
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edited Mar 17, 2016 at 20:48
Khaled
asked Mar 17, 2016 at 14:06
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How can you be certain that you can simultaneously achieve a minimum of g's and a maximum of the f's? You need to formulate this as a single objective. For example, maximize sum of f's while placing an upperbound (which you specify) on the sum of the g's. Then, there would be a number of ways to solve the problem.
– TravisJ
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Mar 17, 2016 at 14:13
@TravisJ What are the ways to solve that problem you specified?
– Khaled
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Mar 17, 2016 at 14:24
Look into integer linear programming. Essentially, you just need to find the best way to formulate the question. For example, maximize f⋅x where x is a 0-1 vector and f is the vector containing the values of f(n), subject to constraints g⋅x≤b where g is the vector of g(n) values and b is the arbitrary bound you set. Then also have the constraint that you have exactly N non-zero entries in x.
– TravisJ
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Mar 17, 2016 at 14:41
By the way, integer linear programs can be expensive (computationally) to solve in general. If you have more information about the structure of the f(n)'s or g(n)'s then perhaps you can speed it up. If the problem is relatively small, then it should also not be a problem.
– TravisJ
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Mar 17, 2016 at 14:44
btw.: the problem @TravisJ stated is a knapsack problem. It is well-known.
– user251257
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Mar 17, 2016 at 14:59
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We can model your objects as
ui=(fi,gi)⊤(i∈{1,…,n})
We then define selection vectors from
X={x∈{0,1}n∣∑i=1nxi=N}
The difficult part is the objective. with its two sub objectives.
maxxfiximinxgixi
My attempt would be to first change
minxgixi=−maxx−gixi
While we now have two maximizations, those maximizations are independent,
we might end up with two different optimal x.
Right now I see only two roads:
Insisting to find a common optimal x, which will likely fail most of the times. Or changing the objective (e.g. weighted combination, or best compromise optimum).
This problem seems to be multi-objective optimization.
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answered Mar 17, 2016 at 15:31
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1537 | https://www.webmd.com/a-to-z-guides/what-is-acanthocytosis | What Is Acanthocytosis?
Acanthocytosis is a condition where your red blood cells are abnormally shaped. It is associated with lots of different conditions and symptoms.
Acanthocytosis and Acanthocytes
When you have acanthocytosis, your red blood cells are misshapen and known as acanthocytes.
Acanthocytes are also called spur cells. They are dense, shrunken, and irregularly shaped red blood cells with spikes on the outside. These cells form from changes in the fats and proteins on red blood cells’ outer layers.
Most adults have a small number of acanthocytes in their blood. But some inherited and acquired diseases increase them beyond the normal amount.
Acanthocytosis Causes
The exact reason why red blood cells change into abnormal shapes is not fully understood. The changes can be caused by inherited conditions or acquired diseases.
Blood cells have a layer called a membrane which has fats and proteins. Acanthocytes have an abnormal amount of these fats, or lipids, in odd proportions. That means the inner and outer surface areas of the blood cells are imbalanced. This causes them to harden, pucker, and form spikes.
Severe liver disease is a common cause of acanthocytosis. Similar changes in fats and cells can also result from rare or inherited diseases, including the following.
Abetalipoproteinemia. This condition is also called Bassen-Kornzweig syndrome. Bassen-Kornzweig syndrome keeps your body from combining fat and protein to create molecules called lipoproteins. This means you can’t digest fat and essential vitamins correctly.
Ongoing problems breaking down and absorbing fats can lead to severe malabsorption, vitamin E deficiency, and other issues.
Neuroacanthocytosis. There are several types of neuroacanthocytosis. These conditions cause red blood cell irregularities, neurological symptoms, and movement problems. They include:
Acanthocytosis can also be caused by other conditions such as:
Acanthocytosis Symptoms
The changes Acanthocytosis makes to your red blood cells make them more likely to get trapped in your spleen and be destroyed. It can lead to a condition called spur cell hemolytic anemia.
The symptoms can of hemolytic anemia include:
Acanthocytosis may be linked to other conditions. That means the general symptoms can be different. Neuroacanthocytosis illnesses cause similar symptoms like:
Acanthocytosis Diagnosis
Your doctor can diagnose acanthocytosis with a blood test called a peripheral blood smear. This test requires fresh blood because an acanthocyte is sometimes mistaken for another type of blood cell called an echinocyte.
Your doctor might order other blood tests to check your thyroid hormone levels, liver health, or for other problems. These tests usually depend on your symptoms.
If you have acanthocytosis together with other brain and muscle symptoms, your doctor might also request more tests. These include:
Acanthocytosis Treatment
Treatment for acanthocytosis depends on the cause. Treating the underlying condition can help treat acanthocytosis.
If you have anorexia or severe malnutrition, you can reverse acanthocytosis by treating the nutrition and eating disorder. If it’s caused by medication, stopping and changing the medication can also reverse acanthocytosis.
If you get hemolytic anemia from severe liver disease, your doctor might recommend:
A blood transfusion doesn’t always work. The red blood cells in the donor blood can change into acanthocytes after they enter your body. Your doctor might suggest a liver transplant in such cases.
People who have abetalipoproteinemia will need to supplement their diet with vitamins A, D, E, K in large amounts to manage their symptoms. You might need to eat a low-fat diet and limit the fats and oils you eat.
There is no cure for neuroacanthocytosis conditions. Treatment involves managing symptoms with different medications or therapies. These can be:
Complications of Acanthocytosis
Some conditions are progressive and incurable. This means they might worsen over time. Your doctor will focus on treating your symptoms.
Acanthocytosis is a serious complication associated with severe liver disease. Liver disease is challenging to treat and can lead to death. A liver transplant may be the best treatment for severe cases.
Acanthocytosis can be treated and reversed when its underlying cause is treatable. Make sure to talk to your doctor about what treatments are available for you.
SOURCES:
Clinical Journal of Gastroenterology: “Spur cell anemia related to alcoholic liver cirrhosis managed without liver transplantation: a case report and literature review.”
Medscape: “Acanthocytosis.”
Mount Sinai: “Bassen-Kornzweig syndrome.”
National Organization for Rare Disorders: “Neuroacanthocytosis.”
Shah, P., Grewal, U., Hamad, H. Acanthocytosis, StatPearls Publishing, 2021.
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1538 | https://epub.fh-joanneum.at/obvfhjoa/download/pdf/8945179 | Development of a calculation algorithm for the power design of a vertical icing wind tunnel (IWT)
M¨ uller Elias 1 and Pircher Fabian 11FH JOANNEUM – University of Applied Sciences, Institut Luftfahrt/Aviation, Alte Poststraße 147, 8020 Graz, Austria
Abstract
This project deals with the development of a cal-culation algorithm for the power-related design of the components of an icing wind tunnel for researching icing phenomena on wing profiles at the FH JOANNEUM in Graz. By considering pressure loss calculation, general fluid dynamics, heat flow analysis and general thermodynamics, a calculation algorithm is developed, which cal-culates the required power quantities of the wind tunnel elements such as the air conditioning unit and the fan depending on input variables such as the general channel cross-section, the test cham-ber cross-section and the test flow velocity of the wind tunnel medium. The results are calculated as a function of input parameters selected by the user and are displayed by using 3D plots, which makes it easier for the user in the project decision phase to commit to a concept or to consider the effects caused by changes during the progress of the project. This calculation is programd using the software MATLAB 2021b.
1 Introduction
Wind tunnels have been used since the beginning of the 20th century to draw conclusions about the behaviour of assemblies, constructions, ve-hicles and aircraft in reality from model tests. In aviation, individual assemblies of the aircraft or even entire aircraft models are usually tested in wind tunnels for their aerodynamic proper-ties and for occurring environmental conditions such as icing. In order to be able to simulate all flight conditions, several test runs in different wind tunnels may be necessary. The Institute of Aviation at the FH JOANNEUM - University of Applied Sciences in Graz currently operates a horizontal icing wind tunnel to generate and re-search icing phenomena on wing models. How-ever, this channel no longer corresponds to the current state of the art and is therefore to be re-designed as part of a project. This work deals with the development of a calculation algorithm for the performance design of a closed wind tun-nel concept.
1.1 Types of wind tunnel
In general, two wind tunnel models are distin-guished on the basis of flow recirculation. In the open wind tunnel, the flow medium is taken from the environment, accelerated by a fan, smoothed by flow calming and released back into the en-vironment after the test chamber. According to [BRP99, p. 27], this type of wind tunnel requires more energy in operation, generates more noise emissions and the flow medium properties are strongly dependent on the existing environmen-tal conditions. In the closed wind tunnel, where the flow medium is returned to the fan after the test chamber, thus creating a circuit, the flow conditions are dependend on the environmental conditions, less noise is produced and the con-trol of the medium temperature is easier to im-plement, which is essential for the operation of an icing wind tunnel. In Fig. 1 we show the con-82 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. M¨ uller, E. und Pircher, F. Scientific Computing 2023
Figure 1: Top view of a closed wind tunnel [BRP99, p. 26] struction of a closed wind tunnel. A further subdivision of wind tunnels is made by the type of flow through the test chamber. A horizontal wind tunnel model is one in which the flow medium flows horizontally through the test chamber. In icing wind tunnels, where water droplets have to be injected into the flow medium to form ice, this means that the path of the wa-ter droplets is influenced by gravity. In order to make the supercooled water droplets hit the test model to create icing, an exact combination of nozzle position and flow velocity is required. A vertical wind tunnel, on the other hand, has a test area with vertical flow, which eliminates the effects of gravity on the droplet trajectory.
1.2 Final wind tunnel concept
In order to be able to apply and test the cal-culation algorithm developed for the design, a concept is required on which the calculations are to be based. This was worked out in the concept phase of this project. In addition, an attempt was made to eliminate all negative effects of the current wind tunnel at FH JOANNEUM. The design phase was done in six steps. In each step, a new design was constructed with a different arrangement of the individual components with respect to the previous model. Subsequently, the disadvantages of each concept were analysed and evaluated. The evaluation resulted in the final design shown in fig (2). The heat exchanger is located on the right side of the duct due to the
Figure 2: Wind tunnel concept as the basis for the power calculations. required large distance with respect to the in-jection nozzle system. This has the advantages of a more homogeneous temperature distribution in the duct cross-section and fewer problems re-garding component icing due to the injected wa-ter droplets. The fan is located in the upper duct cross-section. The narrowing of the chan-nel cross-section at the bottom ensures as much space as possible between the bridge transition installed at FH JOANNEUM and the icing chan-nel. In addition, in contrast to the current icing channel, there is a heating element in the lower horizontal segment after the measuring cham-ber. If this is switched on instead of the heat exchanger, the icing channel can be quickly de-iced. The injection is located directly after the calming zone. This positioning is again chosen because of the other icing of the stilling elements. As a general rule, elements that are important for operation should not be mounted in the im-mediate vicinity of the measuring chamber and the injection system, as this would lead to an in-creased risk of icing and thus to a restriction in operation or disturbances of the flow field. 83 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. Scientific Computing 2023 Power calculation algorithm IWT Fig. 2 shows us on the left side (from top to bottom): corner 4, the calming section, the nozzle-system, a rejuvenation, the test chamber, a diffusor and corner 1. On the right side: corner 2 and 3 and the heat exchanger. On the bottom is a drainage and the heating element and on the top you can find the fan.
2 Methods of calculation
In order to fully design a concept the power quantities of the wind tunnel components are required, among other things. The three main components of an icing wind tunnel are the fan to accelerate the flow, the air conditioning unit to cool the flow medium and thus generate icing conditions and a heating coil to defrost the wind tunnel between two test runs in order to achieve proper flow conditions.
2.1 Fan power
The required fan power is calculated by taking into account the volume flow and the pressure loss occurring in the path of the tunnel.
PF an = ˙V · Δp (1) The volumetric flow rate depends on the test chamber cross-sectional area and on the flow ve-locity, if the flow conditions are below M a = 0 .3and thus the simplified incompressible continu-ity equation may be assumed. The total pres-sure loss Δ p is calculated using the equations (2-4, [BK21, p. 227]) the sum of all individual and pipe friction losses generated by the tunnel elements in the flow and by the tunnel walls. Δp = ∑ Δpindiv + ∑ Δppipe (2) The individual pressure loss contributions
pindiv are calculated by multiplying the individ-ual, component-dependent pressure loss coeffi-cient by the density of the flow medium and the square of the flow velocity. Δpindiv = 12 · ρ · v2 · ζ (3) The pipe friction pressure losses depend on the cross-sectional geometry (hydraulic diameter
Dh), the pipe friction coefficient λ, the length of the tunnel section L and also on the density ρ
and the square of the prevailing flow velocity v.Δppipe = λ · LDh
· ρ · v2
2 (4)
2.2 Air conditioning unit power
The calculation of the required air conditioning unit power is a conservative estimation by adding the required individual single powers for the ac-tual cooling process and for the compensation of the stationary heat flow loss as well as the power introduced into the system by the fan during the operation of the icing wind tunnel. The cool-ing process itself is assumed to be an adiabatic process, so the required individual power is cal-culated with
PC = QC
tC
(5) where tC is the desired cooling time and
QC = m · cp · ΔT (6) the amount of heat to be removed, which is cal-culated by the mass m, the heat capacity cp and the temperature difference Δ T to be achieved. The heat flow loss to be compensated by the air conditioner is calculated with the equations used by [BS19, p. 35-37] for the heat transfer at a multi-layer wall (7-8). ˙Qstat = kA (θ1 − θ2) (7) where θ1 and θ2 are the inner and the outer tem-peratures, A is the radiation surface and where for the heat transfer coefficient k
1
kA = 1
α1A1
∑
i
δi
λmi Ami
1
α2A2
(8) is applied. The variable labels can also be taken from Fig. 3. The consideration of the power input gener-ated by the fan is done by introducing a stan-dard efficiency η, with which the power to be 84 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. M¨ uller, E. und Pircher, F. Scientific Computing 2023
Figure 3: Heat transfer through a multi-layer wind tunnel wall [BS19, p. 37] compensated by the air conditioner PAC can be calculated. The variables used in equation 9 can be taken from previous equations.
PAC = PC + ˙Qstat + PF an (1 − η) · (−1) (9) The estimation of the required total output of the air conditioning unit is finally made by adding these individual powers. Since under real conditions the heat loss during the cooling pro-cess would be unsteady, as it depends on the in-ner and outer temperature, which would change during the cooling process, the result of the air conditioning unit power is only a conservative estimate.
2.3 Heating register power
Also, the calculated heating register power us-ing the equations (5-6) is an estimation of the required power amount due to the assumption of adiabatic process operation. However, since the heat losses occur in the heating phase and help to heat the channel, this calculated value is sufficient for the design of the heating coil.
3 Software implementation
The development of the calculation algorithm includes the digitalisation of all required for-
Figure 4: Flowchart of the calculation process (software: LUCIDCHART); The names in the boxes correspond to the program name of the MATLAB files; Colorcode: red - main program, blue - function, green - result, grey - variables handed over; mulas and equations including all relevant conditional dependencies (If-/Else conditions) that lead to a change in the program flow. As shown in Fig. 4, the program is fed by a parameter initialisation file with input data that must be specified by the user beforehand. These include, for example, the geometric dimensions of the wind tunnel cross-section and the test chamber or the lengths of the individual wind tunnel sections. Furthermore, material parameters such as the thermal conductivity of the wall layers or the temperature and type of flow medium must be entered. All other pa-rameters such as Reynolds numbers, kinematic viscosities, the density of the flow medium, etc. are calculated by the program itself at the required positions. Individual parameters can be entered additionally by the user if he does not wish to have the data calculated or if val-ues for these parameters have already been fixed. 85 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. Scientific Computing 2023 Power calculation algorithm IWT The main programs then independently calcu-late the output variables of the power for the fan, for the air conditioner and for the heating coil. Since the output of the air conditioner de-pends on the fan power, the fan power is calcu-lated first. In the main programs, the algorithm runs through various functions which, for exam-ple, serve to determine the correct Nusselt num-ber for the current flow case for the calculation of the heat transfer coefficient α. All functions are called independently and pass the parameters to the main program at the required positions. By running another main program, it is possible for the user to display the results of the powers as a function of certain input parameter config-urations (test chamber height and test chamber width or test cross-section and test flow veloc-ity). This makes it easier for the project team to recognise and analyse the effects on the power quantities of any parameter changes during the decision-making phases. To program the calcu-lation code MATLAB 2021b was used.
4 Results
After the developed calculation program has been executed, the user is provided with the required power values in MATLAB Command Window. These values only apply to the con-cept for a closed vertical icing wind tunnel de-veloped in the concept phase and to the initial material and input data. After the execution of the additional program, the performance param-eters are also displayed above the input config-urations. These results are also only valid for the elaborated concept in sec. 1.2. The jumps in power contained in the 3D plots for certain parameter configurations (see fig 5) can be ex-plained by the use of individual equations and functions in the limit of their validity, such as in the selection of the correct pipe friction coeffi-cient λ as a function of the hydraulic diameter
Dh, or in the calculation of the individual pres-sure loss coefficients ζ. In those formulas and functions, the correlation of the parameter con-figurations (test chamber height and test cham-
Figure 5: Potential output of the fan power over the input parameters: the configuration test sec-tion cross-section and the flow velocity for the flow medium air at −20 ◦C (software: MATLAB 2021b) ber width or test cross-section and test flow ve-locity) plays an important role in relation to the processing of the calculation program and thus to the formula actually used for the calculation. Fig. 5 shows a potential relationship for the flow medium air at −20 ◦C between the fan power and the mentioned input parameter configurations of the test section cross-section and the maximum occurring flow velocity.
5 Summary and Outlook
The aim of this project was to develop a calcula-tion algorithm for the design of a new vertical ic-ing wind tunnel concept for the FH JOANNEUM in order to investigate and research icing phe-nomena on aircraft wing models. The problems occurring on the current model regarding the influence of gravitational effects on the droplet trajectory were eliminated by choosing a vertical test chamber. Furthermore, the variability of the flow velocity of the wind tunnel medium was thus also established, which allows the user to carry out tests with different flow velocities. By digi-talising and integrating the formulas described in section (2) into a calculation algorithm, the re-quired power quantities of the wind tunnel com-86 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. M¨ uller, E. und Pircher, F. Scientific Computing 2023
ponents can be calculated, varied and specified at the push of a button. The project manage-ment is given a quick overview of the required component capacities and thus also of the pos-sible costs arising from the realisation of a se-lected wind tunnel model through the parame-terised input by means of an input file. The use of 3D plots allows the project team to compare several wind tunnel configurations and therefore to decide on a wind tunnel configuration even very late in the project schedule. The selection process was made clearer and easier by the cal-culation program and the output of results via 3D plots. In addition, the program offers scope and quick results for short-term changes to the wind tunnel concept during the entire course of the project. Since the results are based on a large number of conservative information assumptions, the performance values obtained also correspond to conservative values and not to those of the real case. In order to improve the algorithm in the future and thus obtain accurate results, the conservative assumptions mentioned could be re-placed by real occurring data. This requires a revision of the calculation algorithm to improve the accuracy of the performance values.
Acknowledgments
We would like to thank our two project supervi-sors, Wolfgang Hassler PhD and Andreas Tram-posch PhD from the FH JOANNEUM, who gave us the opportunity to work on this project and who supported us with their technical expertise in the fields of fluid mechanics and thermody-namics and thus contributed significantly to the success of this project.
References
[BK21] Sabine Bschorer and Konrad K¨ oltzsch.
Technische Str¨ omungslehre: Mit 262 Aufgaben und 31 Beispielen . Springer Vieweg, Wiesbaden, Germany, 12th edition, 2021. [BRP99] Jewel B. Barlow, William H. Rae, and Alan Pope. Low-speed wind tunnel test-ing . Wiley-Interscience, USA, 3. edi-tion, 1999. [BS19] Hans Dieter Baehr and Karl Stephan.
W¨ arme- und Stoff¨ ubertragung .Springer Vieweg, Berlin, Germany, 10. edition, 2019. 87 Granigg W. / FH JOANNEUM Gesellschaft mbH (Hrsg.): „Scientific Computing Conference 2023 – Conference Proceedings”, Graz, 2023. ISBN e-Book: 978-3-903318-20-5, DOI: 10.60588/9asg-1g41, Lizenz: CC BY 3.0 AT; Artikel-DOI: 10.60588/rwxe-g507. |
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Mariachiara Di Cesare
Mariachiara Di Cesare
1 Institute of Public Health and Wellbeing, University of Essex, Colchester, UK
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1, Pablo Perel
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2 Department of Non-Communicable Disease Epidemiology, London School of Hygiene & Tropical Medicine, London, UK
3 World Heart Federation, Geneva, Switzerland
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1 Institute of Public Health and Wellbeing, University of Essex, Colchester, UK
2 Department of Non-Communicable Disease Epidemiology, London School of Hygiene & Tropical Medicine, London, UK
3 World Heart Federation, Geneva, Switzerland
4 MRC/Wits Rural Public Health and Health Transitions Research Unit (Agincourt), School of Public Health, Faculty of Health Sciences, University of the Witwatersrand, Johannesburg, South Africa
5 Brigham and Women’s Hospital, Cardiovascular Medicine, Boston, USA
6 Harvard Medical School, Boston, USA
7 American Heart Association, Dallas, USA
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9 Department of Medicine, University of Buenos Aires, Buenos Aires, Argentina
10 Santa Maria University Hospital, CAML, CCUL, Faculdade de Medicina da Universidade de Lisboa, Lisbon, Portugal
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CORRESPONDING AUTHOR: Mariachiara Di Cesare Institute of Public Health and Wellbeing, University of Essex, Colchester, UK m.dicesare@essex.ac.uk
Received 2023 Oct 4; Accepted 2023 Dec 18; Collection date 2024.
Copyright: © 2024 The Author(s)
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PMCID: PMC10809869 PMID: 38273998
Abstract
Cardiovascular diseases (CVDs) are the leading cause of mortality globally. Of the 20.5 million CVD-related deaths in 2021, approximately 80% occurred in low- and middle-income countries.
Using data from the Global Burden of Disease Study, NCD Risk Factor Collaboration, NCD Countdown initiative, WHO Global Health Observatory, and WHO Global Health Expenditure database, we present the burden of CVDs, associated risk factors, their association with national health expenditures, and an index of critical policy implementation.
The Central Europe, Eastern Europe, and Central Asia region face the highest levels of CVD mortality globally. Although CVD mortality levels are generally lower in women than men, this is not true in almost 30% of countries in the North Africa and Middle East and Sub-Saharan regions. Raised blood pressure remains the leading global CVD risk factor, contributing to 10.8 million deaths in 2019. The regions with the highest proportion of countries achieving the maximum score for the WHF Policy Index were South Asia, Central Europe, Eastern Europe, and Central Asia, and the High-Income regions. The Sub-Saharan Africa region had the highest proportion of countries scoring two or less.
Policymakers must assess their country’s risk factor profile to craft effective strategies for CVD prevention and management. Fundamental strategies such as the implementation of National Tobacco Control Programmes, ensuring the availability of CVD medications, and establishing specialised units within health ministries to tackle non-communicable diseases should be embraced in all countries. Adequate healthcare system funding is equally vital, ensuring reasonable access to care for all communities.
Keywords: Cardiovascular health, CVDs risk factors, WHF policy index, CVD global data, WHF observatory
Introduction
Cardiovascular diseases (CVDs) are the leading cause of mortality globally, responsible for a significant number of deaths and disabilities. In 2021 alone, CVDs accounted for 20.5 million deaths, comprising approximately one-third of all global deaths . While cardiovascular conditions were traditionally considered diseases of affluence, this is no longer the case. Over three-quarters of CVD-related deaths occur in low- and middle-income countries (LMICs) .
Moreover, these deaths are the primary contributor to premature non-communicable disease (NCD) mortality. Ischemic heart disease, specifically, stands as the leading cause of premature death in 146 countries for men and 98 countries for women . The complexity of the global picture is exacerbated by the levels of inequalities in the impact of CVDs. Notably, LMICs experience higher rates of premature mortality from CVDs compared to high-income countries (HICs), while the reduction in age-standardised mortality rates is progressing more slowly in LMICs.
In May 2012 the World Health Assembly adopted the target of a 25% reduction in premature mortality from noncommunicable diseases by 2025. In May 2013 the NCD Global Monitoring Framework was launched during the World Health Assembly. Its primary objectives were to drive progress in the prevention and control of NCDs, serve as a foundation for advocacy, raise awareness, strengthen political commitment, and promote global action against NCDs. The framework encompassed nine voluntary global targets focused on behavioural and metabolic risk factors, as well as national healthcare system responses .
Over a decade has passed since international organisations and governments have been working towards achieving these global targets and reducing premature mortality. However, the current pace of decline is insufficient. Based on trends observed between 2010 and 2016, it is estimated that less than 20% of countries will achieve the UN Sustainable Development Goal Target 3.4 (SDG 3.4) – a one-third reduction in premature mortality from NCDs by 2030 . Most countries, particularly those classified as low- or lower-middle income, need to accelerate progress to meet these targets.
This paper summarises the World Heart Report 2023,1 which is a comprehensive overview of the trends and current landscape of the key dimensions of cardiovascular health. By synthesising epidemiological, policy, and economic data, the World Heart Report provides information that can help guide policymakers at national and international levels in identifying and addressing the most critical priorities to address cardiovascular conditions.
Methods and data
The analysis of the burden of cardiovascular diseases and associated risk in this study utilised data from global initiatives including the Global Burden of Disease Study , the NCD Risk Factor Collaboration , and the NCD Countdown 2030 initiative . These have been extensively described elsewhere and are considered the most valid and reliable sources for global cardiovascular data. In selecting data sources, priority was given to those that offered a high level of disaggregation (by sex, age, country, region), data completeness, richness of the raw data informing the model, and comparability of estimates.
Absolute numbers of death, age-standardised death rates, and the probability of dying between 30 and 70 years of age2 have been reported to describe the burden of cardiovascular mortality. Analysis of risk factors focused on nine specific factors. These factors were selected based on the six risk factors outlined in the WHO NCD Global Monitoring Framework, along with two additional risk factors—non-HDL cholesterol levels and air pollution. The inclusion of these two additional factors was based on their substantial impact on the prevalence and mortality rates of CVDs (Table 1).
Table 1.
Mortality and risk factors indicators, source, and year.
| |
| INDICATOR | SOURCE | YEAR |
| |
| Mortality | | |
| |
| Ischemic heart disease | Global burden of disease | 1990–2019 |
| |
| Stroke | Global burden of disease | 1990–2019 |
| |
| All other CVDs | Global burden of disease | 1990–2019 |
| |
| Risk factors | | |
| |
| Diabetes | NCD risk factor collaboration | 1975–2014 |
| |
| Raised blood pressure | NCD risk factor collaboration | 1975–2015 |
| |
| Obesity | NCD risk factor collaboration | 1975–2016 |
| |
| Lipids | NCD risk factor collaboration | 1975–2018 |
| |
| Physical activity | Global burden of disease | 1990–2019 |
| |
| Sodium intake | Global burden of disease | 1990–2019 |
| |
| Alcohol consumption | Global burden of disease | 1990–2019 |
| |
| Tobacco smoking | Global burden of disease | 1990–2019 |
| |
| Ambient air pollution | Global burden of disease | 1990–2019 |
| |
| Premature mortality (stroke) | NCD Countdown 2030 | 2015 |
| |
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To assess the risk factors’ country profiles, each country was ranked based on the level of the specific risk factor, ranging from the highest level to the lowest level. Subsequently, quintiles were generated for each risk factor, dividing the countries into five equal groups based on their respective levels of the risk factor.
To assess the association between CVD burden and health expenditure, additional data on Gross Domestic Product (GDP), Current Health Expenditure as a percentage of GDP (CHE), and Out of Pocket Expenditure (OOPS) were obtained from the WHO Global Health Expenditure database .
The World Heart Federation Policy Index (WHF Policy Index) provides an overview of national governments’ level of implementation of eight critical policies related to CVD health (Table 2). The expert panel of the WHF Advocacy Committee carefully selected these policy indicators as the most relevant for ensuring a healthy heart profile in populations. Data for the WHF Policy Index were obtained from the Global Health Observatory . Each policy was assigned a score of 0 if it was not in place in the country and 1 if it was in place. The overall index score for each country ranged from 0 (no policies in place) to 8 (all policies in place). Complete information was available for 166 countries.
Table 2.
Policy indicators, source, and year.
| |
| INDICATOR | SOURCE | YEAR |
| |
| National tobacco control programmes | WHO Global Health Observatory | 2018 |
| |
| Policy/strategy/action plan for CVD | WHO Global Health Observatory | 2021 |
| |
| Operational Unit, Branch, or Dept. in Ministry of Health with responsibility for NCDs | WHO Global Health Observatory | 2021 |
| |
| Guidelines/protocols/standards for the management of cardiovascular diseases | WHO Global Health Observatory | 2021 |
| |
| Policy/strategy/action plan to reduce physical inactivity | WHO Global Health Observatory | 2021 |
| |
| Policy/strategy/action plan to reduce unhealthy diet related to NCDs | WHO Global Health Observatory | 2021 |
| |
| Policy/strategy/action plan to reduce the harmful use of alcohol | WHO Global Health Observatory | 2021 |
| |
| Availability of ACE inhibitors, Aspirin (100 mg) and Beta blockers in the public health sector | WHO Global Health Observatory | 2021 |
| |
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Note: Availability of ACE inhibitors, Aspirin (100 mg) and Beta blockers in the public health sector is a combination of the three indicators 1) availability of ACE inhibitors in the public health sector; 2) availability of Aspirin (100 mg) in the public health sector; 3) availability of Beta blockers in the public health sector. If all 3 available the overall score was set to 1 if one or more not available the overall score was set to 0.
Trends and levels are presented for both regions and individual countries. Countries were classified into regions based on the Global Burden of Diseases classification, which includes Central Europe, Eastern Europe, Central Asia, and the High-Income region as distinct regions ( This division into more regions allows for more detailed comparisons of mortality and risk factor rates across significant geographical areas. To assess the association between CVD burden and health expenditure indicators the Pearson’s correlation coefficient was used.
Results
The number of deaths due to acronym introduced in the introduction CVDs has increased over the years. In 1990, the estimated deaths stood at around 12.1 million [95% Uncertainty Interval: 11.4–12.6 million], equally distributed between men and women. By 2019, this number had risen to 18.6 million [17.1–19.7], with 9.6 million [8.9–10.3] deaths among men and 8.9 million [7.9–9.7] deaths among women (Figure 1). Overall, CVDs accounted for 33% of all global deaths in 2019, with ischemic heart disease (9.1 million deaths) and stroke (6.6 million deaths) contributing to 85% of CVD-related deaths. While the number of deaths due to CVDs over the last 30 years has increased, mostly due to the ageing and growth of the population, the age-standardised death rate, which accounts for changes in population demographics, has declined from 354.5 per 100,000 people [330.6–369.5] in 1990 to 239.9 per 100,000 people [219.4–254.9] in 2019 (Figure 2).
Figure 1.
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Global trends in number of deaths due to cardiovascular diseases, 1990–2019.
Source: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from
Figure 2.
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Global trends in age-standardised cardiovascular disease death rate (per 100,000 people), 1990–2019.
Source: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from
When examining regional trends, significant variations in the rate of decline in age-standardised CVD death rates become apparent (Figure 3). Between 1990 and 2019, all regions experienced a decline in age-standardised CVD death rates, but there are signs of this decline slowing down in the past decade. The High-income region exhibited the fastest average rate of decline for both men and women, with an average annual rate of change of 2.6%. On the other hand, the South Asia, the Southeast Asia, East Asia and Oceania, and the Sub-Saharan Africa regions had the slowest rates of decline. Notably, there was almost no improvement in CVD death rates for men in these regions.
Figure 3.
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Regional trends in age-standardised cardiovascular disease death rate (per 100,000 people), 1990–2019.
Source: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from
In 1990, male age-standardised CVD death rates in the Sub-Saharan Africa region were 1.2 times higher than those observed in the High-income region. However, by 2019, the age-standardised rates in the Sub-Saharan Africa region were 2.1 times higher than those in the High-income region. For women in 2019, only the Latin America and the Caribbean region achieved CVDs death rates levels comparable to those observed in the High-income region in 1990, almost three decades before.
The Central Europe, Eastern Europe, and Central Asia region had the highest age-standardised death rates for males and females in both 1990 (670.2 [644.4–683.3] and 467.2 [436.4–481.7] deaths per 100,000 people respectively) and 2019 (524.1 [475.4–566.9] and 345.7 [308.1–376.3] deaths per 100,000 people respectively).
The North Africa and Middle East region had the second highest death rates for males (376.7 per 100,000 people [336.5–414.7] and females (339.8 per 100,000 people [298.9–374.2]) in 2019.
Ischemic heart disease stands as the leading cause of CVD mortality across all regions, except for women in Sub-Saharan Africa. Notably, in 2019, the death rate for cardiomyopathy and myocarditis among males in Central Europe, Eastern Europe, and Central Asia was more than three times higher (28.5 per 100,000 people [20.3–32.9]) than in the second highest region (Sub-Saharan Africa, 8.4 per 100,000 people [5.8–10.6]) and thirty times higher than the region with the lowest rates (South Asia, 0.8 per 100,000 people [0.6–1.1]).
Cardiovascular deaths by sex
On a global scale, men exhibit higher age-standardised death rates compared to women. In 2019, this trend persisted across all countries in the High-income region, the Central Europe, Eastern Europe, and Central Asia region, as well as the South Asia region. However, in almost one-third of countries (six out of 21) in the North Africa and Middle East region, women experienced higher death rates from CVDs compared to men. Notable disparities were observed in Qatar (464.6 deaths per 100,000 people [395.0–536.5] for women versus 301.9 [241.4–368.4] for men), Egypt (600.0 [463.8–720.2] versus 491.6 [386.3–615.1] per 100,000 people), and Algeria (447.7 [384.1–509.5] versus 371.5 [306.9–449.0] per 100,000 people). Similarly, in 13 out of the 46 countries in the Sub-Saharan Africa region, women had higher CVDs death rates than men. The largest gaps were observed in Mali, Mauritania, Congo (Congo-Brazzaville), Ghana, and Sierra Leone (Figure 4). Two other countries where women experienced higher death rates than men were Haiti (475.3 [352.3–635.4] versus 419.8 [320.1–562.4] per 100,000 people) and Tokelau (388.7 [308.9–484.8] versus 312.2 [261.6–380.4] per 100,000 people).
Figure 4.
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Male-female difference in age-standardised cardiovascular disease death rate (per 100,000 people) in sub-Saharan Africa by country, 2019.
Source: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from
CVD premature mortality
Ischaemic heart disease emerges as the leading cause of premature death among all NCDs in most countries, impacting both men and women. The highest risk of premature mortality from ischaemic heart disease is observed in countries within the Central Europe, Eastern Europe, and Central Asia region, followed by the North Africa and Middle East region. The risk of premature mortality from ischaemic stroke among men is also highest in the Central Europe, Eastern Europe, and Central Asia region. Yemen stands out as the country with the highest risk of premature mortality across all countries, with men and women facing risks of 20.0% and 13.3%, respectively (Figure 5). The risk of premature mortality due to haemorrhagic stroke is particularly high in Mongolia (10.3% among men and 6.4% among women) and Turkmenistan (9.6% among men). Similarly, for ischaemic stroke, the highest risk (around 3%) among women is observed in Sub-Saharan Africa, and specifically in Western African countries, such as Ghana, Sierra Leone, and Côte d’Ivoire. In Sub-Saharan Africa, the risk of premature mortality from other cardiovascular diseases also tends to be highest among women.
Figure 5.
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Probability of dying (reported as a percentage) in 2015 between 30 years and 70 years of age from ischaemic heart disease by sex.
Note: Grey colour is used when no estimates were available (missing).
Source: NCD Countdown .
Risk factors
The risk factor profile provides a comprehensive understanding of the burden of CVD risk factors within and between countries (Figure 6). The analysis reveals notable variations across regions. In Central Europe, Eastern Europe, and Central Asia most countries exhibit high levels of sodium intake, raised blood pressure, and non-HDL cholesterol among men. Additionally, high levels of sodium intake and tobacco smoking are prevalent among women. Comparatively, countries in Central and Eastern Europe tend to have higher overall risk factor levels than those in Central Asia. Most countries in the High-income region are characterised by elevated levels of behavioural risk factors, including high sodium and alcohol consumption, tobacco smoking, and low physical activity. Both sexes also experience high levels of non-HDL cholesterol, while men have high obesity levels. The Latin America and Caribbean region shows greater levels of heterogeneity. However, low physical activity emerges as a significant risk factor across many countries. Generally, the Caribbean countries tend to have higher risk factor levels compared to other countries in the region. The North Africa and Middle East region is characterised by high levels of metabolic risk factors, particularly diabetes and obesity prevalence. Air pollution and physical inactivity are also notable risk factors. Middle Eastern countries generally exhibit higher risk factor levels compared to those in North Africa. All countries in South Asia face high levels of air pollution. Sodium intake, raised blood pressure, and diabetes are also prominent risk factors in South Asia for both sexes.
Figure 6.
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Comparative levels of risk factors by country and region.
Note (1): RF1 – Physical activity; RF2 – Sodium intake; RF3 – Alcohol consumption; RF4 – Tobacco smoking; RF5 – Obesity; RF6 – Raised Blood Pressure; RF7 – Diabetes; RF8 – Lipids; RF9 – Ambient air pollution.
Note (2): The figures display the global quintile into which each country falls for each risk factor.
Source: See Table 1.
Countries in Oceania have high levels of diabetes, obesity, tobacco smoking, and low physical activity. Women are characterised by some of the highest levels of non-HDL cholesterol. In Southeast Asia and East Asia, high levels of sodium intake and non-HDL cholesterol are observed across all countries.
The Sub-Saharan Africa region is characterised by high levels of raised blood pressure in most countries. Air pollution levels are generally high, particularly in the Sahel countries. Southern Sub-Saharan African countries tend to have the highest levels of CVD risk factors within the region.
Relationship between CVD mortality and health expenditure
The World Health Organization (WHO) and various civil society organisations have long advocated for countries to allocate a minimum of 5% of their Gross Domestic Product (GDP) towards healthcare. However, there are significant variations in the extent to which countries in each region achieve this target (Table 3). For instance, in South Asia, no countries meet the 5% target, with Bangladesh allocating 2.6% of its GDP to health expenditure, Bhutan allocating 3.6%, India allocating 2.9%, Nepal allocating 4.4%, and Pakistan allocating 2.8%.
Table 3.
Proportion of countries spending at least 5% of GDP on health.
| |
| REGION | % |
| |
| High-income | 97 |
| |
| Central Europe, Eastern Europe, and Central Asia | 85 |
| |
| Latin America and Caribbean | 71 |
| |
| North Africa and Middle East | 53 |
| |
| Southeast Asia, East Asia, and Oceania | 50 |
| |
| Sub-Saharan Africa | 45 |
| |
| South Asia | 0 |
| |
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There exists a correlation between the percentage of current health expenditure (CHE) relative to GDP and CVD death rates. Countries with lower CHE as a percentage of GDP tend to experience higher age-standardised CVD mortality rates (Figure 7). Additionally, a higher percentage of out-of-pocket expenditure (OOPS) relative to CHE is also associated with higher age-standardised CVD mortality (Figure 8).
Figure 7.
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Age-standardised cardiovascular disease death rate (per 100,000 people), 2019, by current health expenditure (CHE) as share of national income (GDP).
Note: Each dot represents a country, coloured by region. The dashed line shows the negative linear association between CVD death rates and national health expenditure (Pearson’s correlation coefficient = –0.15, p = 0.052).
Sources: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from World Health Organization, Global Health Expenditure Database. Available from
Figure 8.
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Age-standardised cardiovascular disease death rate (per 100,000 people), 2019, by out-of-pocket (OOP) health expenditure as share of current health expenditure (CHE).
Note: Each dot represents a country, coloured by region. The dashed line shows a positive association between CVD death rates and OOP health expenditure as a share of national health expenditure (Pearson’s correlation coefficient = 0.23, p = 0.002).
Sources: Institute for Health Metrics and Evaluation (IHME). GBD Compare Data Visualization. Seattle, WA: IHME, University of Washington, 2020. Available from World Health Organization, Global Health Expenditure Database. Available from
WHF policy index
Globally, a significant majority of countries, 106 out of 166 countries with available information (64%), have at least seven out of the eight policies evaluated in place. The regions with the highest proportion of countries achieving the maximum score of eight were South Asia (80%), Central Europe, Eastern Europe, and Central Asia (68%), and the High-Income region (62%). In contrast, the Sub-Saharan Africa region had the lowest proportion of countries (13%) achieving the maximum score, and the highest proportion (16%) of countries scoring two or less. The Latin America and Caribbean region, as well as the Southeast Asia, East Asia, and Oceania region, also had countries with scores of one or two (Figure 9).
Figure 9.
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Number of key policies implemented to address cardiovascular diseases, by region (% of countries).
Sources: Global Health Observatory. Available from:
At a global level, the policy most commonly in place was the existence of national tobacco control programmes, which was the case in 91% of countries. This was followed by the existence of guidelines/protocols/standards for the management of CVDs (86%) and the existence of policy/strategy/action plans to reduce unhealthy diets related to NCDs (85%). The policy least frequently in place was the existence of an action plan to reduce the harmful use of alcohol, which was in place in 70% of countries.
In the Sub-Saharan Africa region, over 50% of countries do not have a comprehensive CVD plan, an NCD Unit within the Ministry of Health, or general availability of CVD drugs in the public sector. North Africa and the Middle East had the lowest implementation rate for action plans to reduce the harmful use of alcohol. The Latin America and the Caribbean region had the lowest implementation rate for action plans to reduce physical inactivity and unhealthy diets related to NCDs (Figure 10).
Figure 10.
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Policy implementation per country.
Note: P1 – National tobacco control programmes; P2 – Action plan for CVDs; P3 – Operational Unit in Ministry of Health with responsibility for NCDs; P4 – Guidelines for the management of CVDs; P5 – Action plan to reduce physical inactivity; P6 – Action plan to reduce unhealthy diet related to NCDs; P7 – Action plan to reduce the harmful use of alcohol; P8 – Availability of CVD drugs (e.g., ACE inhibitors, aspirin, and Beta blockers) in the public health sector.
Sources: Global Health Observatory. Available from:
Discussion
In this paper we summarise the burden of CVDs, associated risk factors, the association between national health expenditures and CVD burden, and the role of critical policies, as presented in the World Heart Report 2023.
Cardiovascular diseases have remained the leading cause of mortality for over three decades, accounting for approximately one-third of all global deaths, with 20.5 million deaths recorded in 2021 . The decline in CVD death rates has been faster in high-income countries compared to low- and middle-income countries, where over 80% of CVD deaths currently occur. Globally, CVDs have consistently been the leading cause of deaths over the last three decades. The Central Europe, Eastern Europe, and Central Asia region continue to face the highest levels of CVD mortality globally. Results indicate that although CVD mortality levels are generally lower in women than in men, almost 30% of countries in the North Africa and Middle East region and the Sub-Saharan Africa region (mainly West African countries) report higher CVD mortality rates for females than males. The association between health expenditure as a percentage of GDP, out-of-pocket expenditure, and CVD mortality highlights the need for adequate investment in healthcare systems to mitigate the burden of CVD. In terms of risk factors, raised blood pressure is the leading global CVD risk factor, contributing to approximately 10 million deaths in 2019. Most CVD risk factors, including physical inactivity, alcohol consumption, tobacco smoking, raised blood pressure, and diabetes, are more prevalent in men compared to women, with obesity being the only risk factor higher in women.
While the data presented here provide a detailed picture of the current burden of CVDs, there are some important limitations. The data used to present the temporal burden of CVDs and associated risk factors are derived from complex modelling [10,11]. Richness/lack of data informing the model is statistically expressed through the levels of uncertainty. Here, for simplicity, we have not systematically reported the associated levels of uncertainty. In particular, the lack of data from low- and middle-income countries, results in larger levels of uncertainty in those countries.
There is no one-size-fits-all approach to improving cardiovascular health globally. Every population is susceptible to different risk factors based on where they live and their lifestyles. Whether that is having higher prevalence of tobacco and alcohol use and higher sodium intake or being more exposed to dangerous levels of air pollution and having lower levels of physical activity. This means that decision makers and stakeholders must look closely at the risk factor prevalence in their countries and regions to fully understand what policy areas need more focus to get cardiovascular health moving in the right direction. There are, however, baseline approaches that every country should implement as a foundation from which to build tailored activities to tackle CVDs. This includes implementing major policy initiatives that are essential to improving CVD health, such as National Tobacco Control Programmes, securing the availability of CVD drugs, and creating an Operational Unit in the Ministry of Health responsible for tackling NCDs. Additionally, it requires adequately funded health systems and initiatives so that all communities can access the care they need. The stalling progress in cardiovascular health is not unique. Almost every health initiative around the world suffered because of the COVID-19 pandemic and countries are now grappling with which areas to prioritise as they aim to boost and protect the health of their populations. Given the severe burden of CVDs both in terms of mortality and morbidity, this area of health cannot be neglected. The world will struggle to meet the ambitious targets set to reduce premature mortality from NCDs by 25% compared to 2010 levels, by 2025. There is still time, however, to accelerate action toward meeting Sustainable Development Goal 3.4 of reducing by one-third premature mortality from NCDs, including cardiovascular diseases.
To promote action against CVDs, several aspects should be considered in the short and long term. First, countries need to enhance data collection on CVDs and their risk factors, particularly in low- and middle-income countries where data gaps exist, to better understand populations at higher risk. Second, countries should aim for health expenditure of at least 5% of GDP, in line with WHO recommendations. Third, countries should implement evidence-based policies to address CVDs, considering the disease burden and prevalent risk factors, while ensuring adequate resources and monitoring for progress. Fourth, it is crucial to prioritise coverage of CVD prevention and management interventions within Universal Health Coverage (UHC) benefit packages to minimise out-of-pocket expenses. Fifth, lessons learned from successful CVD prevention, management, and improved access to care and therapies should be implemented across all regions to address inequities and uneven progress in reducing CVD mortality.
The WHF Policy Index described here provides a working tool for national governments to use to assess, monitor and improve implementation of policies addressing CVDs health. We must recognise that policies are only as strong as when they are enforced and evaluated. Here we attempt to begin capturing measures by which organisations and advocates can monitor the actions of key stakeholders like the government and other players at the national level.
It is imperative to take concerted action to combat cardiovascular diseases and work towards achieving the global targets set forth, thereby improving the health outcomes and well-being of populations worldwide.
Funding Statement
The World Heart Report 2023 was made possible through support from the Novartis Foundation.
Footnotes
The World Heart Report, launched in May 2023, is aimed at equipping policymakers and advocates around the world with the information needed to help reduce CVD deaths and accelerate progress in cardiovascular health. The report findings highlight the main differences between geographies in terms of CVD burden and risk factors, as well as structural barriers and inequities in CVD health, with the goal of guiding policymakers at national and international levels toward the priorities they should seek to address. It is available from
Premature mortality from non-communicable diseases refers to deaths occurring between the ages of 30 and 70 years. These premature deaths not only result in a significant loss of life for individuals but also have broader impacts on societies and economies, particularly as they affect individuals within the working age range.
Funding Information
The World Heart Report 2023 was made possible through support from the Novartis Foundation.
Competing Interests
The authors have no competing interests to declare.
References
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1541 | https://arxiv.org/pdf/2407.17363 | arXiv:2407.17363v1 [math.CO] 24 Jul 2024
Improving the Caro-Wei bound and applications to Tur´ an stability
Tom Kelly ∗ Luke Postle †
July 25, 2024
Abstract
We prove that if G is a graph and f (v) ≤ 1/(d(v) + 1 /2) for each v ∈ V (G), then either
G has an independent set of size at least ∑
v∈V(G)
f (v) or G contains a clique K such that
∑
v∈K
f (v) > 1. This result implies that for any σ ≤ 1/2, if G is a graph and every clique
K ⊆ V (G) has at most (1 −σ)( |K|− σ) simplicial vertices, then α(G) ≥ ∑
v∈V(G)
1/(d(v)+1 −σ). Letting σ = 0 implies the famous Caro-Wei Theorem, and letting σ = 1 /2 implies that if fewer than half of the vertices in each clique of G are simplicial, then α(G) ≥ ∑
v∈V(G)
1/(d(v) + 1 /2), which is tight for the 5-cycle. When applied to the complement of a graph, this result implies the following new Tur´ an stability result. If G is a Kr+1 -free graph with more than (1 −1/r )n2/2−n/ 4edges, then G contains an independent set I such that at least half of the vertices in I are complete to G − I. Applying this stability result iteratively provides a new proof of the stability version of Tur´ an’s Theorem in which Kr+1 -free graphs with close to the extremal number of edges are r-partite.
1 Introduction
For n ≥ r ≥ 2, the Tur´ an graph , which we denote Tr(n), is the complete r-partite graph on n
vertices whose part sizes differ by at most one. Tur´ an’s Theorem , the cornerstone of extremal graph theory, states that if G is an n-vertex, Kr+1 -free graph, then e(G) ≤ e(Tr (n)), with equality holding only when G ∼= Tr(n). Since e(Tr (n)) ≤ (1 − 1/r )n2/2, Turan’s Theorem is sometimes presented in the following slightly weaker form. If G is a Kr+1 -free graph on n vertices, then G has at most (1 − 1/r )n2/2 edges. We refer to this latter statement as the concise Tur´ an’s Theorem (as in ). A stability version of Tur´ an’s Theorem implies that if G is an n-vertex Kr+1 -free graph with close to (1 − 1/r )n2/2 edges, then G resembles the Tur´ an graph Tr(n) in some sense. For example, the classical Erd˝ os–Simonovits [9, 10, 26] stability theorem states that if G has at least e(Tr (n)) − o(n2)edges, then the edit distance from G to Tr(n) is o(n2). That is, G can be obtained from Tr(n) by adding and deleting at most o(n2) edges. Brouwer proved that for n ≥ 2r + 1, if G has at least
e(Tr (n)) − ⌊ n/r ⌋ + 2 edges, then G is r-partite. In this paper, we prove a new stability version of Tur´ an’s Theorem that is qualitatively between these two. Our result (Theorem 1.1) implies that if G has close to the extremal number of edges, then G has an independent set I which resembles one of the parts of the Tur´ an graph in that it contains many vertices that are complete to G − I.We call this phenomenon “Local Tur´ an stability.”
∗
School of Mathematics, Georgia Institute of Technology. Email: tom.kelly@gatech.edu . Research supported by the National Science Foundation under Grant No. DMS-224707
†
Department of Combinatorics and Optimization, University of Waterloo, Canada. Email:
lpostle@uwaterloo.ca . Partially supported by NSERC under Discovery Grant No. 2019-04304, the Ontario Early Researcher Awards program and the Canada Research Chairs program.
1When applied to the complement of a graph, the concise Tur´ an’s Theorem states that an n-vertex graph with less than n2/(2 r) − n/ 2 edges has an independent set of size r + 1 . Equivalently, every graph with n vertices and average degree d has an independent set of size at least n/ (d + 1). Independently, Caro and Wei famously generalized the concise Tur´ an’s Theorem by proving that every graph G has an independent set of size at least ∑
v∈V(G)
1/(d(v) + 1), where d(v) is the degree of the vertex v in G. This bound is stronger than Tur´ an’s because of the convexity of the function x 7 → 1/(x + 1). Our main result in this paper, Theorem 1.4 (which implies Theorem 1.1), is a best-possible improvement of the Caro-Wei bound.
1.1 Local Tur´ an stability
The following result is our new stability version of Tur´ an’s Theorem. We say that a vertex in a graph is complete to a subset of vertices if the vertex is adjacent to every vertex in that set.
Theorem 1.1 (Local Tur´ an Stability) . Let σ ≤ 1/2. If G is a Kr+1 -free graph on n vertices with more than (
1 − 1
r
) n2
2 − σn
2
edges, then G contains an independent set I such that more than (1 − σ)( |I| − σ) vertices in I are complete to G − I.
Theorem 1.1 for σ = 0 implies the concise Tur´ an’s Theorem; indeed, no graph contains an independent set I with more than |I| vertices in I, so by Theorem 1.1, no Kr+1 -free n-vertex graph has more than (1 −1/r )n2/2 edges. For other values of σ, the result suggests that a Kr+1 -free graph with close to the extremal number of edges resembles the Tur´ an graph in that the independent set
I behaves like one of the parts of the Tur´ an graph. Theorem 1.1 is essentially best possible for r = 2 in the sense that if n is even and σn/ 2 is an integer, then the complete bipartite graph on n vertices with equal-sized parts and a matching of size σn/ 2 removed has precisely (1 − 1/r )n2/2 − σn/ 2 edges and every independent set I has at most (1 − σ)|I| vertices that are complete to G − I (to see this last fact, observe that if I is an independent set with at least one vertex complete to G − I, then I is one of the two parts of the bipartition). Moreover, the statement of Theorem 1.1 is not true with σ > 1/2 because if G is the 5-cycle, then (1 − 1/r )n2/2 − σn/ 2 < 5, but G does not contain an independent set with any vertex complete to the rest of the graph. In an application of Theorem 1.1, the subgraph G − I is Kr-free, so if we apply Theorem 1.1 iteratively, then we can show that G globally resembles the Tur´ an graph in that it is actually
r-partite, as follows.
Corollary 1.2. If G is a Kr+1 -free graph on n vertices with at least
(
1 − 1
r
) n2
2 − n
2r + 1
edges, then G is r-colorable.
For completeness, we include a proof of Corollary 1.2 in Appendix A. The only complicated aspect of the proof is that it is necessary to solve a simple optimization problem to determine how many times Theorem 1.1 is iterated and how many vertices remain at the end of the process. 2t Property Reference
0 G ∼= Tr(n) Tur´ an’s Theorem
⌊n/r ⌋ + 2 χ(G) ≤ r Brouwer
n/ 4 Local Tur´ an stability Theorem 1.1
O(n log n) ∃ a pair of “twin vertices”
o(n(r+1) /r ) ∃ almost spanning complete r-partite subgraph
o(n2) edit distance o(n2) to Tr(n) Erd˝ os–Simonovits [10, 26]
Table 1: “Phase transition” of Tur´ an stability: G is Kr+1 -free with e(G) ≥ (1 − 1
r
) n2
2
− t.Considerable attention has been devoted to various aspects of Tur´ an stability for Kr+1 -free graphs as the number of edges approaches the extremal value. Theorem 1.1 and Corollary 1.2 fit into this paradigm. Table 1 illustrates this “phase transition” 1. See also [11, 21]. As mentioned, Corollary 1.2 was first proved by Brouwer – it has actually been reproved by various sets of authors (see ). For n ≥ 2r + 1 the result actually holds with the −n/ (2 r) + 1 term replaced with −⌊ n/r ⌋+ 2. We could possibly obtain this stronger form of Corollary 1.2 via the same proof method with an extension of Theorem 1.1 with σ ≤ 1 if we include an additive “error term.” More precisely, we conjecture that the conclusion of Theorem 1.1 still holds with σ ≤ 1 if we replace “(1 − 1/r )n2/2 − σn/ 2 edges” with “(1 − 1/r )n2/2 − σn/ 2 + 5 /4 edges.” Moreover, we believe it is possible to characterize the extremal examples which would in turn provide a new proof of a stronger form of Brouwer’s result also characterizing the extremal examples. We believe the extremal examples for Theorem 1.1 are the same as those for Brouwer’s result for r = 2, which are particular blowups of 5-cycles – see for a complete description. The complement of the graph on the left in Figure 1 provides an example. For r ≥ 3, the extremal examples for Brouwer’s result are obtained from an extremal example for r − 1 by adding an independent set complete to the rest of the graph, so these graphs satisfy Theorem 1.1 with σ = 1.
1.2 Improving the Caro-Wei bound
The independence number of a graph G, denoted α(G), is the size of a largest independent set in
G. As witnessed by Tur´ an’s Theorem and the Caro-Wei Theorem [8, 29], some of the most fundamental results in graph theory include bounds on the independence number. The bound on the independence number provided by the Caro-Wei Theorem is tight for disjoint unions of complete graphs, and the complement of the Tur´ an graph is a disjoint union of complete graphs of order differing by at most 1. Several improvements have been made to these bounds under certain additional assumptions, such as connectivity [4, 12, 15, 16, 17] or triangle-freeness [1, 13, 24, 25], among other things [2, 3, 5, 14, 23]. Brooks’ Theorem bounding the chromatic number implies that if G is an n-vertex connected graph of maximum degree ∆, then α(G) ≥ n/ ∆, unless G is a complete graph or an odd cycle. In this paper, we examine the extent to which Brooks’ bound holds for the average degree as in the concise Tur´ an’s Theorem instead of the maximum degree, or for the degree-sequence as in the Caro-Wei Theorem. In particular, for σ ∈ [0 , 1], when does a connected n-vertex graph G of average degree d satisfy α(G) ≥ n/ (d + 1 − σ) or even α(G) ≥ ∑
v∈V(G)
1/(d(v) + 1 − σ)? For any σ > 0, it is not sufficient as in Brooks’ Theorem to simply require that G is not complete – a
1The results of [22, 28] apply to maximal Kr+1 -free graphs.
3disjoint union of complete graphs with the minimum number of edges added to ensure connectivity has independence number at most n/ (d + 1 − O(1 /d )). However, the cliques in this graph have many simplicial vertices (a vertex is simplicial if its neighborhood is a clique). We prove that for
σ ≤ 1/2, such cliques are essentially the only obstruction, as follows.
Theorem 1.3. Let σ ≤ 1/2. If G is a graph and each clique K ⊆ V (G) has at most (1 −σ)( |K|− σ)
simplicial vertices, then
α(G) ≥ ∑
v∈V(G)
1
d(v) + 1 − σ .
In particular, if each clique K ⊆ V (G) has fewer than |K|/2 simplicial vertices, then α(G) ≥∑
v∈V(G)
1/(d(v) + 1 /2) . Consequently, if G also has n vertices and average degree d, then
α(G) ≥ n
d + 1 /2 .
Notice that the Caro-Wei Theorem follows from Theorem 1.3 with σ = 0. For graphs without even a single simplicial vertex, no bound on α(G) of the form n/ (d+1 −ε) was previously known for any ε > 0. Theorem 1.3 implies this result for ε = 1 /2, and it is tight for the 5-cycle. Theorem 1.3 also immediately implies Theorem 1.1. Our main result is actually the following generalization of Theorem 1.3 2.
Theorem 1.4. If G is a graph and f : V (G) → R such that f (v) ≤ 1/(d(v) + 1 /2) for each
v ∈ V (G) and ∑
v∈K
f (v) ≤ 1 for each clique K ⊆ V (G), then α(G) ≥ ∑
v∈V(G)
f (v).
The condition in Theorem 1.4 that every clique K ⊆ V (G) satisfies ∑
v∈K
f (v) ≤ 1 is very natural. If G is a complete graph, then it is a necessary condition. The stronger statement of Theorem 1.4 is also more conducive to our proof by induction, since subgraphs of G also satisfy the hypotheses of the theorem. In a follow-up paper , we generalize Theorem 1.4 to fractional coloring .Now we show how Theorem 1.4 implies Theorem 1.3.
Proof of Theorem 1.3 assuming Theorem 1.4. Let f : V (G) → R be the function where f (v) = 1/(d(v) + 1 − σ) for each v ∈ V (G). Since σ ≤ 1/2, by Theorem 1.4, it suffices to show that if K is a clique, then ∑
v∈K
f (v) ≤ 1. Each v ∈ K satisfies f (v) ≤ 1/(|K| − σ) and moreover
f (v) ≤ 1/(|K| + 1 − σ) if v is not simplicial. Hence, since K has at most (1 − σ)( |K| − σ) simplicial vertices, ∑
v∈K
f (v) ≤ (1 − σ)( |K| − σ)
|K| − σ + |K| − (1 − σ)( |K| − σ)
|K| + 1 − σ = 1 ,
as required. Therefore by Theorem 1.4, α(G) ≥ ∑
v∈V(G)
f (v) = ∑
v∈V(G)
1/(d(v) + 1 − σ), as desired. Also note that since |K| is an integer, if K has fewer than |K|/2 simplicial vertices, then K has at most ( |K| − 1) /2 simplicial vertices. If σ = 1 /2, then ( |K| − 1) /2 = (1 − σ)( |K| − σ), so if each clique K ⊆ V (G) has fewer than |K|/2 simplicial vertices, then α(G) ≥ ∑
v∈V(G)
1/(d(v) + 1 /2), as desired.
2This result follows from a previous result of both authors [19, Theorem 1.3] on fractional coloring, but that paper is currently not under review. In this paper, we provide a much shorter proof of our main result.
4Kt Kt
Kt
Kt Kt
Figure 1: Blowups of odd cycles As with Theorem 1.1, it would be interesting to extend Theorems 1.3 and 1.4 for σ ≤ 1while taking into account additional obstructions. As Brooks’ Theorem suggests, it is necessary to consider odd cycles. However, blowups of 5-cycles (depicted on the left in Figure 1) are in some sense “worse” obstructions, as follows. Let G be a graph with n vertices and average degree d. If G
is a cycle of length 2 k +1, then n/ (d+1 −σ) ≤ α(G) = k only if σ ≤ 1−1/k = 1 −2/(n−1), and if G
is the graph on the left in Figure 1, which is obtained by replacing two non-adjacent vertices of C5
with cliques of size t, then n/ (d+1 −σ) ≤ α(G) = 2 only if σ ≤ 1−5/(4 t+6) = 1 −5/(2 n). Moreover, the graphs in Figure 1 are not regular, so if G is one of these graphs, then ∑
v∈V(G)
1/(d(v)+1 −σ) <n/ (d + 1 − σ). If G is the graph on the left, then ∑
v∈V(G)
1/(d(v) + 1 − σ) ≤ 2 only if σ < 3/4, for large t, and if G is the graph on the right, which is obtained by replacing three pairwise non-adjacent vertices of C7 with cliques of size t, then ∑
v∈V(G)
1/(d(v) + 1 − σ) ≤ 3 only if σ < 1 for large t. We conjecture that these are essentially the only obstructions, as follows.
Conjecture 1.5. If G is a graph and f : V (G) → R such that f (v) ≤ 1/d (v) for each v ∈ V (G),then α(G) ≥ ∑
v∈V(G)
f (v) unless G contains either
• a clique K ⊆ V (G) such that ∑
v∈K
f (v) > 1, or
• a subgraph H isomorphic to a blowup of a cycle of length 2k + 1 such that ∑
v∈V(H)
f (v) > k .
1.3 Outline of the paper
Besides the proof of Corollary 1.2 in Appendix A, the remainder of the paper is devoted to the proof of Theorem 1.4. In Sections 2-4 we prove various properties of a hypothetical minimum counterexample to Theorem 1.4. In Section 5 we show that a graph cannot simultaneously have these properties, thus demonstrating that a hypothetical counterexample to Theorem 1.4 does not exist, completing the proof. A few of the lemmas in Sections 2-5, as in the proof of Corollary 1.2, utilize standard “calculus” arguments to determine a bound on the value of a differentiable mul-tivariate function over a certain region. In order to not disrupt the flow of the more important “combinatorial” arguments in the proof, we collect these “calculus” arguments in Appendix B.
2 Basic properties of a minimum counterexample
For the next three sections (that is, Sections 2, 3, and 4), we suppose G is a hypothetical mini-mum counterexample to Theorem 1.4 with respect to a function f , that is a graph with α(G) <∑
v∈V(G)
f (v) where f : V (G) → R satisfies f (v) ≤ 1/(d(v) + 1 /2) for each v ∈ V (G) and 5∑
v∈K
f (v) ≤ 1 for each clique K in G, and subject to these requirements, we assume G is cho-sen with the minimum number of vertices. In this section, we prove some basic properties of the minimum counterexample G that will be used frequently in Sections 3 and 4. For convenience, we let δ denote the minimum degree of G. We also frequently use the following standard graph theory notation: for a graph H and a vertex v ∈ V (H), we denote the open neighborhood of v in H by NH (v) and the closed neighborhood of v in H by NH [v]. That is, NH (v)is the set of vertices in H adjacent to v, and NH [v] = NH (v) ∪ { v}. We often omit the subscript H
when it is clear from the context.
Lemma 2.1. Every vertex v ∈ V (G) satisfies f (v) + ∑
u∈N(v)
f (u) > 1. In particular, there are no simplicial vertices in G and δ ≥ 2.Proof. By the minimality of G, we have α(G − N [v]) ≥ ∑
u∈V(G−N[v])
f (u), and since α(G) ≥
α(G − N [v]) + 1, we have α(G) ≥ 1 + ∑
u∈V(G−N[v])
f (u). Since G is a counterexample, α(G) <∑
u∈V(G)
f (u), so ∑
u∈V(G)
f (u) > 1 + ∑
u∈V(G−N[v])
f (u). By cancelling the terms of the sum ∑
u∈V(G−N[v])
f (u) in this inequality, we obtain the first inequality. Now if v is a simplicial vertex in G, then G[N [v]] is a clique, so f (v) + ∑
u∈N(v)
f (u) ≤ 1, a contradiction. Since a vertex of degree one is simplicial, δ ≥ 2.
The argument in Lemma 2.1 that f (v) + ∑
u∈N(v)
f (u) > 1 actually holds for a minimum counterexample in a more general context, regardless of any stipulations on the function f . Thus, this simple lemma can be used to prove the Caro-Wei Theorem by letting f (v) = 1 /(d(v) + 1) for each vertex v, as f (v) + ∑
u∈N(v)
f (u) ≤ 1 when v has minimum degree. Indeed, this is essentially Wei’s original proof (see [13, p. 23]). Our strategy also focusses on the minimum degree vertices. The next lemma implies that at least roughly half of the neighbors of a minimum degree vertex also have minimum degree, but we need the following more general form.
Lemma 2.2. Suppose X ⊆ V (G) such that ∑
v∈X
f (v) > 1. If d(v) ≥ | X| − 1 for every v ∈ X,then fewer than (|X| + 1 /2) /2 vertices in X have degree at least |X|. In particular, if d(v) ≤ d(u)
for all u ∈ N (v), then fewer than (d(v) + 3 /2) /2 neighbors of v have degree greater than d(v).Proof. Let X′ ⊆ X be the vertices in X of degree |X| − 1 in G. Now 1 < ∑
v∈X
f (v) ≤ |X′|
|X| − 1/2 + |X| − | X′|
|X| + 1 /2 = |X|(|X| − 1/2) + |X′|
(|X| − 1/2)( |X| + 1 /2) .
Therefore
|X′| > (|X| − 1/2) /2,
so fewer than
|X| − (|X| − 1/2) /2 = ( |X| + 1 /2) /2vertices have degree at least |X|, as desired. Moreover, if d(v) ≤ d(u) for all u ∈ N (v) and X = N [v], then by assumption, d(w) ≥ | X| − 1for every w ∈ X, and by Lemma 2.1, ∑
v∈X
f (v) > 1. Thus, fewer than ( d(v) + 3 /2) /2 neighbors of v have degree greater than d(v), as desired.
Lemma 2.3. If u and w are non-adjacent vertices, then f (u) + f (w) + ∑
v∈N(u)∪N(w)
f (v) > 2.In particular, if u and w further satisfy d(u) ≤ d(v) for all v ∈ N (u) and d(w) ≤ d(v) for all
v ∈ N (w), then N (u) and N (w) are disjoint.
6Proof. By the minimality of G, we have α(G − (N [u] ∪ N [w])) ≥ ∑
v∈V(G−(N[u]∪N[w]))
f (v), and since α(G) ≥ α(G − (N [u] ∪ N [w])) + 2, we have α(G) ≥ 2 + ∑
v∈V(G−(N[u]∪N[w]))
f (v). Since G
is a counterexample, α(G) < ∑
v∈V(G)
f (v), so ∑
v∈V(G)
f (v) > 2 + ∑
v∈V(G−(N[u]∪N[w]))
f (v). By cancelling the terms of ∑
v∈V(G−(N[u]∪N[w]))
f (v), we obtain f (u) + f (w) + ∑
v∈N(u)∪N(w)
f (v) > 2, as desired. Now suppose to the contrary that d(u) ≤ d(v) for all v ∈ N (u), that d(w) ≤ d(v) for all
w ∈ N (w), and moreover that u and w share a common neighbor, x. Now we have 2 < f (u) + f (w) + ∑
v∈N(u)∪N(w)
f (v) ≤ f (x) + ∑
v∈N[u]{ x}
f (v) + ∑
v∈N[w]{ x}
f (v). (1) Since d(u) ≤ d(v) for all v ∈ N (u), we have ∑
v∈N[u]{ x}
f (v) ≤ d(u)/(d(u) + 1 /2), and similarly, ∑
v∈N[w]{ x}
f (v) ≤ d(w)/(d(w) + 1 /2). We assume without loss of generality that d(u) ≤ d(w), in which case d(u)/(d(u) + 1 /2) ≤ d(w)/(d(w) + 1 /2). Now by (1), we have 2 < 2d(w)/(d(w) + 1 /2) +
f (x), so f (x) > 1/(d(w) + 1 /2) ≥ 1/(d(x) + 1 /2), a contradiction.
The proofs of Lemmas 2.1 and 2.3 may feel similar. Indeed, there is a common generalization: if
I is an independent set, then ∑
v∈I
f (v) + ∑
u∈N(I)
f (u) > |I|, where N (I) = ∪v∈I N (v). However, we only need this result when |I| ∈ { 1, 2}.Importantly, Lemma 2.3 implies that the minimum degree vertices can be partitioned into cliques with no edges between them, as follows. We say a base clique of G is a maximum cardinality set of vertices of minimum degree that forms a clique in G. Lemma 2.3 implies that every vertex of minimum degree is contained in a unique base clique. Moreover, vertices in different base cliques are not adjacent, and Lemma 2.2 implies that if K is a base clique, then |K| > δ + 1 − (δ + 3 /2) /2 =
δ/ 2 + 1 /4. Since |K| is an integer, we thus have the following. If K is a base clique, then |K| ≥ (δ + 1) /2. (2) The next two sections focus on these base cliques. The following definitions are crucial to the proof.
Definition 2.4. Let K be a base clique of G. Now,
• let AK be the set of vertices not in K that are complete to K,
• let UK be the subset of vertices in V (G) \ (AK ∪ K) with a neighbor in K,
• let ℓK = δ + 1 − | K ∪ AK | denote the number of neighbors each vertex in K has in UK , and
• let DK = max {| K ∩ N (u)| : u ∈ UK } if UK 6 = ∅ and 0 otherwise. Note that if K is a base clique, then by definition, each vertex in AK has degree at least δ + 1, and by Lemma 2.3, each vertex in UK has degree at least δ + 1 as well.
3 Averaging over base cliques
Note that there is some “slack” in the proof of Lemma 2.1. In particular, if a vertex w / ∈ N [v]has a neighbor in N (v), then f (w) < 1/(dG−N v + 1 /2), and so it may be possible to find a larger independent set in G − N [v] by increasing the value of f (w). We need to be careful, though, not to increase f (w) too much (say to 1 /(dG−N v + 1 /2)), because w may be simplicial and we need each clique K to satisfy ∑
u∈K
f (u) ≤ 1. It is no coincidence that in the 5-cycle, the extremal example for our theorem, the graph obtained from deleting the closed neighborhood of a minimum degree vertex is complete. 7In this section, we exploit this “slack” in Lemma 3.2. One of the main novelties of our proof, with Lemma 2.3 in hand, is the idea to “average” this gain from the slack over a base clique, which we utilize in Lemma 3.3. This motivates the following definition.
Definition 3.1. Let K be a base clique of G, and let fK : V (G − K) → R defined by
fK (v) =
{ 1
dG−K(v)+1
if v is simplicial in G − K,
1
dG−K(v)+1 /2
otherwise .
If K is a base clique of G, then since every vertex v ∈ V (G − K) satisfies fK (v) ≤ 1/(dG−K (v) + 1/2) and every clique K′ satisfies ∑
v∈K′
fK (v) ≤ 1, by the minimality of G, we have α(G − K) ≥∑
v∈V(G−K)
fK (v). We use this fact in the next lemma.
Lemma 3.2. If K is a base clique and v ∈ K, then
ℓK + 1 /2
δ + 3 /2 − |K|
(δ + 3 /2)( δ + 1 /2) < ∑
u∈UK
f (u) − ∑
u∈UK\N(v)
fK (u). (3)
Proof. For convenience, let ℓ = ℓK , U = UK , and A = AK . Since α(G − N [v]) + 1 ≤ α(G) <∑
u∈V(G)
f (u) and α(G−N [v]) ≥ ∑
u∈V(G−N[v])
fK (u), we have ∑
u∈V(G−N[v])
fK (u) ≤ ∑
u∈V(G)
f (u)−
Rearranging terms, we have 1 − ∑
u∈K
f (u) − ∑
u∈N(v)∩A
f (u) < ∑
u∈U
f (u) − ∑
u∈U\N(v)
fK (u).
Since f (u) ≤ 1/(δ + 3 /2) for each u ∈ A and f (u) ≤ 1/(δ + 1 /2) for each u ∈ K, the left side of the previous inequality is at least δ+3 /2−| A|
δ+3 /2
− |K|
δ+1 /2
. Since δ + 1 − | A| = |K| + ℓ, we have for each
v ∈ K,
ℓ + 1 /2
δ + 3 /2 − |K|
(δ + 3 /2)( δ + 1 /2) < ∑
u∈U
f (u) − ∑
u∈U\N(v)
fK (u),
as desired.
Now we show how to “average” the gain in Lemma 3.2 over the entire base clique.
Lemma 3.3. If K is a base clique of G, then
ℓK + 1 /2 − |K|
δ + 1 /2 < ∑
u∈UK
|K ∩ N (u)|(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − | K ∩ N (u)|) . (4)
Proof. For convenience, let ℓ = ℓK , U = UK , and A = AK . Since Lemma 3.2 holds for each v ∈ K,the left side of (3) is at most the average of ∑
u∈U
f (u) − ∑
u∈U\N(v)
fK (u) taken over all v ∈ K.Therefore
ℓ + 1 /2
δ + 3 /2 − |K|
(δ + 3 /2)( δ + 1 /2) < ∑
u∈U
(
f (u) − (|K| − | K ∩ N (u)|)fK (u)
|K|
)
.
For each u ∈ U ,
f (u) − (|K| − | K ∩ N (u)|)fK (u)
|K| ≤ 1
d(u) + 1 /2 − |K| − | K ∩ N (u)|
|K|(d(u) + 1 − | K ∩ N (u)|)= |K ∩ N (u)|(d(u) + 1 /2 − | K|) + |K|/2
|K|(d(u) + 1 − | K ∩ N (u)|)( d(u) + 1 /2) .
8Since the above expression is decreasing as a function of d(u) if d(u) ≥ δ, the right side of the above inequality is at most
|K ∩ N (u)|(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − | K ∩ N (u)|)( δ + 3 /2) for each u ∈ U . Hence
ℓ + 1 /2 − |K|
δ + 1 /2 < ∑
u∈U
|K ∩ N (u)|(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − | K ∩ N (u)|) ,
as desired.
4 Structure around base cliques
In this section, we prove two important properties of base cliques in Lemmas 4.1 and 4.3, which we use in the next section to show violate (4). The proofs of these lemmas are similar in spirit to Lemmas 2.1 and 2.3 in that we demonstrate how an independent set in a “reduction” of G can be extended to a larger one in G, but the reductions are more involved in this section. In particular, the reductions of G are no longer induced subgraphs: in Lemma 4.1, we “identify” a pair vertices, and in Lemma 4.3 we add edges.
Lemma 4.1. If K is a base clique, then ℓK > 0.Proof. For convenience, let A = AK and ℓ = ℓK . Suppose to the contrary that ℓ = 0. Now
|K ∪ A| = δ + 1. By Lemma 2.1, A is not a clique, so there exists a pair of non-adjacent vertices
u, w ∈ A.Let G′ be the graph obtained from G−(K ∪A{ u, w }) by identifying u and w into a new vertex, say z. Define the function f ′ : V (G′) → R for G′ in the following way. For each v ∈ V (G′) \ { z},let f ′(v) = f (v), and let f ′(z) = 0 .5/(dG(u) + dG(w) − 2|K| + 0 .5). Note that for each v ∈ V (G′), we have f ′(v) ≤ 1/(dG′ (v) + 0 .5), and moreover, f ′(z) ≤
0.5/(dG′ (z) + 0 .5). We claim that for each clique K′ in G′, we have ∑
v∈K′
f ′(v) ≤ 1. If z / ∈ K′, this inequality holds because K′ is a clique in G and f ′(v) = f (v) for all v ∈ K′. If z ∈ K′, then for each
v ∈ K′, we have dG(v) ≥ | K′| − 1. Therefore f ′(v) ≤ (|K′| − 0.5) −1 and f ′(z) ≤ 0.5( |K′| − 0.5) −1,so ∑
v∈K′
f ′(v) ≤ (|K′| − 1) /(|K′| − 0.5) + (1 − 0.5) /(|K′ | − 0.5) ≤ 1, as claimed. Hence, since G is a minimum counterexample, α(G′) ≥ ∑
v∈V(G′)
f ′(v). If I is an independent set in G′ containing z, then ( I \ { z}) ∪ { u, w } is an independent set in G
of size |I| + 1, and if I is an independent set in G′ not containing z and if v ∈ K, then I ∪ { v} is an independent set in G of size |I| + 1. Therefore α(G) ≥ α(G′) + 1. Since G is a counterexample, we have ∑
v∈V(G)
f (v) > 1 + ∑
v∈V(G′)
f ′(v), so ∑
v∈K∪A
f (v) − f ′(z) > 1. Since f (v) ≤ 1/(δ + 0 .5) for every v ∈ K ∪ A and 1 − (δ − 1) /(δ + 0 .5) = 1 .5/(δ + 0 .5), the previous inequality implies that 1.5
δ + 0 .5 − f (u) − f (w) + f ′(z) < 0.
This inequality contradicts (2) and the followng claim, which we prove in Appendix B.
Claim 4.2. Let
qδ (du, d w, k ) = 1.5
δ + 0 .5 − 1
du + 0 .5 − 1
dw + 0 .5 + 0.5
du + dw − 2k + 0 .5 .
For k ≥ (δ + 1) /2, and du, d w ≥ δ + 1 , we have qδ(du, d w, k ) ≥ 0.
9Lemma 4.3. If K is a base clique, then DK ≤ ℓK ≤ (δ + 1) /2. Moreover, ℓK ≥ 2 and δ ≥ 3.Proof. For convenience, let U = UK , ℓ = ℓK , and D = DK . First note that by (2), since |K| + ℓ ≤
δ + 1, we have ℓ ≤ (δ + 1) /2, as desired. We actually show D < ℓ , so suppose for a contradiction that ℓ ≤ D. We first show that
D = ℓ = δ/ 2. By Lemma 4.1, U 6 = ∅, so let u ∈ U have D neighbors in K, let v ∈ K ∩ N (u), and let v′ ∈ K \ N (u). Let G′ be the graph obtained from G − (N [v] \ { u}) by adding an edge between u and each vertex w ∈ U ∩N (v′) if one was not already present. Now if I is an independent set in G′ containing
u, then I ∩ N (v′) = ∅. In this case, I ∪ { v′} is an independent set in G of size |I| + 1. If I is an independent set in G′ not containing u, then I ∪ { v} is an independent set in G of size |I| + 1. Thus, α(G) ≥ α(G′) + 1. Since G is a counterexample, α(G′)+1 < ∑
v∈V(G)
f (v), and since ∑
v∈V(G)
f (v) ≤ ∑
v∈V(G′)
f (v)+
δ/ (δ + 1 /2) ≤ ∑
v∈V(G′)
f (v) + 1, we have α(G′) < ∑
v∈V(G′)
f (v). By the choice of G, either there is a vertex x ∈ V (G′) such that f (x) ≤ 1/(dG′ (x) + 1 /2), or there is a clique K′ in G′ such that ∑
x∈K′
f (x) > 1. Since ℓ ≤ D, we have dG′ (u) ≤ dG(u), and for every vertex w ∈ U ∩ N (v′), we also have dG′ (w) ≤ dG(w). Every vertex in G′ not u or adjacent to v′ has the same neighborhood in G
as it does in G′. Therefore every x ∈ V (G′) satisfies dG′ (x) ≤ dG(x), so f (x) ≤ 1/(dG′ (x) + 1 /2). It follows that there is a clique K′ in G′ such that ∑
x∈K′
f (x) > 1, and this clique contains u and at least one neighbor of v′.Let W = {w ∈ K′ : dG′ (w) = |K′| − 1}, that is the vertices of K′ that are simplicial in G′. By Lemma 2.2 applied with X = K′, we have |K′ \ W | < (|K′| + 1 /2) /2, so |W | > (|K′| − 1/2) /2. By Lemma 2.3, dG(u) ≥ δ + 1, so |K′| ≥ δ + 2. Therefore |W | > (|K′| − 1/2) /2 ≥ (δ + 3 /2) /2. Since
|W | is an integer, |W | ≥ δ/ 2 + 1. By Lemma 2.3, if w ∈ W , then w is adjacent to either v or v′,because otherwise d(w) ≤ d(x) for all x ∈ N (w) and w shares u as a neighbor with v in G. Thus,
|W | ≤ ℓ + 1, and since |W | ≥ δ/ 2 + 1, we have ℓ ≥ δ/ 2. If ℓ = ( δ + 1) /2, then |K| = ( δ + 1) /2, and since D < |K|, we have D < ℓ , a contradiction. Therefore ℓ = δ/ 2, as claimed. Moreover, |K| ≤ δ/ 2 + 1, and since D ≤ | K| − 1, we have D ≤ δ/ 2. Since D ≥ ℓ, we also have D = δ/ 2, as claimed. Now we have shown |W | ≥ δ/ 2 + 1 and |W | ≤ ℓ + 1 = δ/ 2 + 1, so |W | = δ/ 2 + 1. Moreover, we have |K′| ≥ δ + 1 and |K′| < 2|W | + 1 /2 = δ + 5 /2, and since |K′| is an integer, we have
|K′| ≤ δ + 2, so |K′| = δ + 2. Therefore d(u) = δ + 1, and N (v′) ∩ U ⊆ W . Now u and v′ are non-adjacent vertices such that
f (u) + f (v′) + ∑
x∈N(u)∪N(v′)
f (x) ≤ ∑
x∈K
f (x) + ∑
x∈K′
f (x) ≤ δ/ 2 + 1
δ + 1 /2 + δ + 2
δ + 3 /2= 2 − 2δ2 − δ − 4
(2 δ + 1)(2 δ + 3) < 2,
contradicting Lemma 2.3. Therefore D < ℓ , as desired. Moreover, if δ = 2, then ℓ = D = 1, a contradiction. Thus, δ ≥ 3, as desired. By Lemma 4.1, D ≥ 1, so ℓ ≥ 2, as desired.
5 Proof of Theorem 1.4
In this section we prove Theorem 1.4. The result follows easily by combining Lemmas 3.3 and 4.3 with the following lemma. 10 Lemma 5.1. Let G be a graph of minimum degree δ with base clique K. If DK ≤ ℓK ≤ (δ + 1) /2,
ℓK ≥ 2, and δ ≥ 3, then
ℓK + 1 /2 − |K|
δ + 1 /2 ≥ ∑
u∈UK
|K ∩ N (u)|(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − | K ∩ N (u)|) .
That is, (4) does not hold.
First, we need the following lemma which provides a bound on the right side of the inequality (4).
Lemma 5.2. If K is a base clique, D′ is an integer such that D′ ≥ DK , and r is the smallest positive integer such that ℓK |K| ≡ r mod D′, then
∑
u∈UK
|K ∩ N (u)|(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − | K ∩ N (u)|)
≤
⌊ ℓ|K|
D′
⌋ D′(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − D′) + r(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − r) .
Proof. For convenience, let ℓ = ℓK and D = DK . Let
g(x1, . . . , x ℓ|K|) =
ℓ|K|
∑
i=1
xi(δ + 3 /2 − | K|) + |K|/2
|K|(δ + 2 − xi) .
Note that if x2 ≥ x1, then
g(x1 − 1, x 2 + 1 , x 3, . . . , x ℓ|K|) − g(x1, . . . , x ℓ|K|) = (δ + 2)( δ + 3 /2 − | K|)
|K|
( 1
(δ + 2 − x2)( δ + 1 − x2) − 1
(δ + 3 − x1)( δ + 2 − x1)
)
0,
Therefore if 1 ≤ x1 ≤ x2 ≤ D′ − 1, then
g(x1 − 1, x 2 + 1 , x 3, . . . , x ℓ|K|) > g (x1, . . . , x ℓ|K|). (5) Now let
xi =
D′ i ∈ { 1, . . . , ⌊ℓ|K|/D ′⌋} ,r i = ⌊ℓ|K|/D ′⌋ + 1 ,
0 otherwise .
By (5), since ∑
u∈U
|K∩N (u)| = ℓ|K|, the left side of the desired inequality is at most g(x1, . . . , x ℓ|K|), which is the right side, so the result follows.
We also need the following claim, proved in Appendix B.
Claim 5.3. Let
qδ(ℓ, k ) = ℓ + 0 .5 − k
δ + 0 .5 − ℓ(δ + 1 .5 − k) + 0 .5k
δ + 2 − ℓ .
If ℓ ∈ [2 , δ/ 2] , k ≥ (δ + 1) /2, and δ ≥ 4, then qδ (ℓ, k ) ≥ 0. Also, if ℓ = k = ( δ + 1) /2 and δ ≥ 5 is an odd integer, then
ℓ + 0 .5 − k
δ + 0 .5 −
⌈ ℓk
0.5( δ − 1)
⌉ ( 0.5( δ − 1)( δ + 1 .5 − k) + 0 .5k
k(δ + 2 − 0.5( δ − 1))
)
0.
11 Now we can prove Lemma 5.1.
Proof of Lemma 5.1. For convenience, let ℓ = ℓK , D = DK , and U = UK . Suppose for a contra-diction that the inequality does not hold. Thus, since D ≤ ℓ, by Lemma 5.2 (with D′ = ℓ), we have
ℓ + 1 /2 − |K|
δ + 1 /2 < ℓ(δ + 3 /2 − | K|) + |K|/2
δ + 2 − ℓ .
Note that the difference of the left and right side of the above inequality is qδ(ℓ, |K|) from Claim 5.3. Hence, by Claim 5.3, we may assume either δ ≤ 3 or ℓ > δ/ 2. First, suppose δ ≤ 3. Hence, δ = 3, ℓ = 2, |K| = 2, and D = 1. Now the right side of (4) is 1.75 and the left side is 2 .5 − 2/3.5 = 27 /14, a contradiction. Therefore we may assume δ ≥ 4 and ℓ > δ/ 2, and thus, since ℓ is an integer at most ( δ + 1) /2, we have ℓ = ( δ + 1) /2. Hence, |K| = ( δ + 1) /2 and D ≤ (δ − 1) /2. Moreover, δ is odd, and since
δ ≥ 4, we have δ ≥ 5. Therefore by Lemma 5.2 applied with D′ = ( δ − 1) /2, we have
ℓ + 0 .5 − |K|
δ + 0 .5 <
⌈ ℓ|K|
0.5( δ − 1)
⌉ ( 0.5( δ − 1)( δ + 1 .5 − | K|) + 0 .5|K|
|K|(δ + 2 − 0.5( δ − 1))
)
,
contradicting Claim 5.3.
As mentioned, the proof of Theorem 1.4 follows easily from Lemmas 3.3, 4.3, and 5.1.
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Proof of Corollary 1.2. Suppose for the sake of contradiction that G is a Kr+1 -free graph on n
vertices with chromatic number at least r + 1 and at least (1 − 1/r )n2/2 − n/ (2 r) + 1 edges. Note that the case r ≤ 1 is vacuous, so we assume r ≥ 2. Since r ≥ 2, by Theorem 1.1 with σ = 1 /2, G contains an independent set I with a vertex complete to G − I. Let I1, . . . , I k be a set of pairwise disjoint independent sets in G chosen with
k maximal subject to for each i ∈ { 1, . . . , k }, the independent set Ii contains a vertex vi complete to G − ∪ ij=1 Ij , let G′ = G − ∪ ki=1 Ii, and let n′ be the number of vertices of G′. If K is a clique in
G′, then K ∪ { v1, . . . , v k} is a clique in G, so G′ is Kr+1 −k-free. Thus, if k = r − 1, then V (G′) is an independent set and G is r-colorable, a contradiction. Therefore we assume k ≤ r − 2. Since
k ≥ 1, we have r ≥ 3. Therefore by Theorem 1.1 and the maximality of k, we have |E(G′)| ≤ (1 −1/(r−k)) n′2/2−n′/4, and by Tur´ an’s Theorem, |E(G[∪ki=1 Ii]) | ≤ (1 − 1/k )( n − n′)2/2. Thus, since the number of edges between ∪ki=1 Ii and V (G′) is at most ( n − n′)n′, we have
|E(G)| ≤
(
1 − 1
k
) (n − n′)2
2 +
(
1 − 1
r − k
) n′2
2 − n′
4 + ( n − n′)n′. (6) The remainder of the proof amounts to solving a simple optimization problem – we derive a contradiction by showing that the right side of (6) is at most (1 − 1/r )n2/2 − n/ (2 r) + 1 for n′ ≤ n
and k ≤ r − 2. To that end, let en,r (n′, k ) denote the right side of (6) and treat n′ and k as real-valued variables. Note that
∂
∂n ′ en,r (n′, k ) = −
(
1 − 1
k
)
(n − n′) + n − 2n′ +
(
1 − 1
r − k
)
n′ − 1
4 = n
k − 1
4 − n′
( 1
k + 1
r − k
)
,
and thus en,r (n′, k ) is maximized when
n′ = r − k
r
(
n − k
4
)
.
Hence, it suffices to show that en,r (( r − k)( n − k/ 4) /r, k ) ≤ (1 − 1/r )n2/2 − n/ (2 r) + 1 for k ≤ r − 2, and this inequality holds for r > 0 so long as
k2 − 8kn − kr + 8 nr − 16 n + 32 r > 0.
The left side of the above inequality is minimized when k = 4 n+r/ 2, and since k ≤ r−2 ≤ 4n+r/ 2, we only need that the above inequality holds for k = r − 2. Indeed, (r − 2) 2 − 8( r − 2) n − (r − 2) r + 8 nr − 16 n + 32 r = 30 r + 4 > 0,
as required.
B Proving the claims
In this section we prove the various claims that appear throughout the paper.
Proof of Claim 4.2. First observe that qδ (du, d w , k ) is increasing in k, so it suffices to show that
q′
δ
(du, d w) = qδ(du, d w, (δ + 1) /2) ≥ 0. Note that
∂
∂d u
q′
δ
(du, d w) = 1
(du + 0 .5) 2 − 0.5
(du + dw − (δ + 1) + 0 .5) 2 .
14 Since dw ≥ δ + 1, the right side of the equality above is at least 1 /(du + 0 .5) 2 − 0.5/(du + 0 .5) 2 > 0. Therefore q′
δ
(du, d w ) is increasing as a function of du, and likewise for dw by symmetry. Since
du, d w ≥ δ + 1, we have q′
δ
(du, d w ) ≥ q′
δ
(δ + 1 , δ + 1), and
q′(δ + 1 , δ + 1) = 1.5
δ + 0 .5 − 1
δ + 1 .5 − 1
δ + 1 .5 + 0.5
δ + 1 .5 = 1.5
δ + 0 .5 − 1.5
δ + 1 .5 > 0,
as desired.
Proof of Claim 5.3. First we show that qδ (ℓ, k ) ≥ 0. Note that
∂
∂k qδ(ℓ, k ) = −1
δ + 0 .5 + ℓ − 0.5
δ + 2 − ℓ ≥ −1
δ + 0 .5 + 1.5
δ > 0.
Therefore qδ (ℓ, k ) ≥ qδ(ℓ, (δ + 1) /2). For convenience, let q′
δ
(ℓ) = qδ(ℓ, (δ + 1) /2)), and note that
∂
∂ℓ q′
δ
(ℓ) = 1 − (0 .5δ + 1)( δ + 2) + .25( δ + 1)
(δ + 2 − ℓ)2 .
Hence, ∂
∂ℓ
q′
δ
(ℓ) ≥ 0 if and only if
ℓ2 − 2( δ + 2) ℓ + ( δ + 2) 2 − ((0 .5δ + 1)( δ + 2) + .25( δ + 1)) ≥ 0.
By the quadratic formula applied to ℓ, ∂
∂ℓ
q′
δ
(ℓ) ≤ 0 if and only if 1
2
(
2( δ + 2) − √(2( δ + 2)) 2 − 4(( δ + 2) 2 − ((0 .5δ + 1)( δ + 2) + .25( δ + 1)))
)
≤ ℓ
≤ 1
2
(
2( δ + 2) + √(2( δ + 2)) 2 − 4(( δ + 2) 2 − ((0 .5δ + 1)( δ + 2) + .25( δ + 1)))
)
.
Note that (2( δ + 2)) 2 − 4(( δ + 2) 2 − ((0 .5δ + 1)( δ + 2) + .25( δ + 1))) = 2 δ2 + 9 δ + 9 ≥ 0.
Therefore ∂
∂ℓ
q′
δ
(ℓ) ≤ 0 if
δ + 2 − 0.5√2δ2 + 9 δ + 9 ≤ ℓ ≤ δ + 2 and ∂
∂ℓ
q′
δ
(ℓ) ≥ 0 if ℓ ≤ δ + 2 − 0.5√2δ2 + 9 δ + 0. Therefore q′
δ
(ℓ) ≥ min {q′
δ
(2) , q ′
δ
(δ/ 2) }. Note that
q′
δ
(2) = − 1.25 δ + 2 .25
δ + −0.5δ − 0.5
δ + 0 .5 + 2 .5 ≈ (δ − 3.28935)( δ + 0 .456017)
δ(δ + 0 .5) > 0,
and
q′
δ
(δ/ 2) = 0 .5δ − δ(0 .25 δ + 0 .5) + 0 .25 δ + 0 .25
0.5δ + 2 + −0.5δ − 0.5
δ + 0 .5 + 0 .5
≈ 0.5( δ − 2.15831)( δ + 1 .15831)
(δ + 0 .5)( δ + 4) > 0,
as desired. Now we show the second inequality. Since δ ≥ 5, we have 4 /(2( δ − 1)) ≤ 1/2, so
⌈ ℓk
0.5( δ − 1)
⌉
=
⌈ (δ + 1) 2
2( δ − 1)
⌉
=
⌈ (δ + 3)( δ − 1) + 4
2( δ − 1)
⌉
≤ ⌈ 0.5( δ + 3) + 0 .5⌉ = 0 .5( δ + 5) .
15 Therefore
ℓ + 0 .5 − k
δ + 0 .5 −
⌈ ℓk
0.5( δ − 1)
⌉ ( 0.5( δ − 1)( δ + 1 .5 − k) + 0 .5k
k(δ + 2 − 0.5( δ − 1)
)
≥
0.5( δ + 1) + 0 .5 − 0.5( δ + 1)
δ + 0 .5 − 0.5( δ + 5)
( 0.5( δ − 1)( δ + 1 .5 − 0.5( δ + 1)) + .25( δ + 1)
0.5( δ + 1)( δ + 2 − 0.5( δ − 1))
)
= 0.75( δ + 1 /3)
(δ + 0 .5)( δ + 1) > 0,
as desired.
16 |
1542 | https://www.keyscientific.com/files/Other%20Manufacturers/Hardy%20Diagnostics/Bacitracin%20Discs/HZ7021Bacitracin.pdf | HardyDisk™ BACITRACIN DIFFERENTIATION DISKS Cat. no. Z7021 Bacitracin Differentiation Disks 1 x 50 disks/cartridge Cat. no. Z7025 Bacitracin Differentiation Disks 5 x 50 disks/cartridge INTENDED USE HardyDisk™ Bacitracin Differentiation Disks are used in the presumptive identification of group A, beta-hemolytic streptococci. The test differentiates group A streptococci from other Lancefield groups of hemolytic streptococci by inhibiting the growth of group A streptococci around a disk containing approximately 0.04units of bacitracin.
SUMMARY Maxted reported a rapid diagnostic agent for the presumptive identification and differentiation of group A streptococci from other beta-hemolytic streptococci.(1) He tested over 3,0000 strains and found group A streptococci to be more sensitive to bacitracin than other beta-hemolytic strains. His findings showed that 99.0% of group A streptococci were inhibited and 95.3% of non-group A beta-hemolytic streptococci were not inhibited.(1) Levinson and Frank, and Streamer, et al., later confirmed Maxted's findings that group A streptococci proved more sensitive to bacitracin than other beta-hemolytic streptococci.(2,3) FORMULA Each HardyDisk™ Bacitracin Differentiation Disk is prepared by impregnating approximately 0.04units of bacitracin onto a high quality 6mm diameter filter paper disk.
STORAGE AND SHELF LIFE Storage: Upon receipt, store at -20 to +8 degrees C. away from direct light. The disks should not be used if there are any signs of deterioration, discoloration, or if the expiration date has passed. Protect from light, excessive heat, and moisture.
The expiration date applies to the product in its intact packaging when stored as directed.
This product has the following shelf life from the date of manufacture: 36 Months: Z7021 HardyDisk™ Bacitracin Differentiation Disks Z7025 HardyDisk™ Bacitracin Differentiation Disks PRECAUTIONS This product is for in vitro diagnostic use only and is to be used only by adequately trained and qualified laboratory personnel. Observe approved biohazard precautions and aseptic techniques. All laboratory specimens should be considered infectious and handled according to "standard precautions". The "Guideline for Isolation Precautions" is available from the Centers of Disease Control and Prevention at www.cdc.gov/ncidod/dhqp/gl_isolation.html.
For additional information regarding specific precautions for the prevention of the transmission of all infectious agents from laboratory instruments and materials, and for recommendations for the management of exposure to infectious disease, refer to CLSI document M29.
Sterilize all biohazard waste before disposal.
Refer to the keyword "MSDS", in the Hardy Diagnostics software program HUGO™, for more information on handling potentially hazardous material.
PROCEDURE Hemolytic isolates obtained from routine throat cultures: 1. Using a loop or swab, select 3-4 well isolated colonies of suspect beta-hemolytic organism from an 18-24 hour old culture and streak to the surface of a Blood Agar, 5% (Cat. no. A10) plate to obtain confluent growth.
2. Aseptically apply a HardyDisk™ Bacitracin Differentiation Disks in the center of the area containing the heavy inoculum.
3. Incubate inoculated media for 18-24 hours at 35 degrees C. in 5-10% CO2. Observe for a zone of inhibition around the disk.
Direct inoculation (primary plating): 1. For early detection of S. pyogenes from throat cultures and other specimen types; it is acceptable to place a HardyDisk™ Bacitracin Differentiation Disk directly on the primary plating of the specimen. Either Blood Agar, 5% (Cat. no. A10), or Group A Beta Strep Agar (Cat. no. A72) plates may be used for this procedure.(12) 2. Aseptically apply a HardyDisk™ Bacitracin Differentiation Disk in the center of the area containing the primary or heavy inoculum.
3. Incubate inoculated media for 18-24 hours at 35 degrees C. in 5-10% CO2. Observe for a zone of inhibition around the disk. If suspect beta-hemolytic, catalase negative colonies are found in an areas not directly adjacent to the disk, the colony must be subcultured to a Blood Agar plate and the Bacitracin Disk testing repeated. (See procedure listed above for testing beta-hemolytic isolates.) INTERPRETATION OF RESULTS Any zone of inhibition that extends beyond the edge of the disk is regarded as susceptible to bacitracin. It is, therefore, considered positive for a presumptive identification of group A streptococci.
LIMITATIONS OF THE PROCEDURE Due to differences in zone sizes resulting from different concentrations of bacitracin, differential disks (0.04units), as opposed to sensitivity disks (10units), are used in the test.
When testing isolates, a light inoculum may result in false zones of inhibition, so it is important that an inoculum resulting in confluent growth be used.
If direct plating is done, the disk must be placed in the primary inoculum area. If hemolytic colonies are seen in an area not adjacent to the disk, the organism needs to be re-tested following procedures used for isolates.
When differentiating streptococci, some group A streptococci may not show a zone of inhibition if an excessive inoculum is used. It has been reported that 6% of group B and 7.5% of groups C and G streptococci may produce zones of inhibition (false-positive result).(11) Some streptococci may not grow in the absence of CO2.
It is recommended that biochemical and/or serological tests such as latex agglutination (Cat. no. PL030HD), be performed on colonies from pure culture for complete identification.
MATERIALS REQUIRED BUT NOT PROVIDED Standard microbiological supplies and equipment such as loops, incinerator, incubators, and culture media, etc., as well as serological and biochemical reagents, are not provided.
QUALITY CONTROL Known positive (group A streptococci) and negative (non-group A streptococci) controls should be used to monitor the accuracy of the disks and inoculum.
Test Organisms Results Streptococcus pyogenes ATCC® 19615 Zone of inhibition Streptococcus agalactiae ATCC® 13813 No zone of inhibition User Quality Control Check for signs of contamination and deterioration. It is recommended that each new lot or shipment of disks be tested with known positive and negative controls.(7,8) Refer to the following keywords, in the Hardy Diagnostics software program HUGO™, for more information on QC: "Introduction to QC", "QC of Finished Product", and "The CLSI (NCCLS) Standard and Recommendations for User QC of Media". Also, see listed references for more information.(1-4) Physical Appearance HardyDisk™ Bacitracin Differentiation Disks are 6mm (in diameter) filter paper disks with the letters BA printed on both sides and should appear white in color. Bacitracin-sensitive (zone of inhibition).
Streptococcus pyogenes (ATCC® 19615) inhibition zone around a HardyDisk™ Bacitracin Differentiation Disk (Cat. no. Z7021). Incubated in CO2 on Blood Agar (Cat. no. A10) for 24 hours at 35 deg. C. Bacitracin-resistant (no zone of inhibition).
Streptococcus agalactiae (ATCC® 13813) growing around a HardyDisk™ Bacitracin Differentiation Disk (Cat. no. Z7021). Incubated in CO2 on Blood Agar (Cat. no. A10) for 24 hours at 35 deg. C.
REFERENCES 1. Maxted, W.R. 1953. J. Clin. Path.; 6:224-226.
2. Levinson, M.L. and Frank, P.F. 1955. J. Bact.; 69:284-287.
3. Streamer, C.W., et al. 1962. Amer. J. Dis. Children; 104:157-160.
4. Oberhofer, T.R. 1985. Man. of Pract. Med. Micro. and Parasit.
5. Murray, P.R., et al. 2003. Manual of Clinical Microbiology, 8th ed. American Society for Microbiology, Washington, D.C.
6. Forbes, B.A., et al. 2007. Bailey and Scott's Diagnostic Microbiology, 12th ed. C.V. Mosby Company, St. Louis, MI.
7. Commission on Laboratory Accreditation, Laboratory Accreditation Program Microbiology Checklist. College of American Pathologists. Rev. 9/30/2004.
8. Centers for Medicare and Medicaid, Appendix C, Survey Procedures and Interpretive Guidelines for Laboratories and Laboratory Services. Subpart K - Quality System for Non-Waived Testing. 493;1200-1265. www.cms.hhs.gov/clia.
9. Koneman, E.W., et al. 2005. Color Atlas and Textbook of Diagnostic Microbiology, 6th ed. J.B. Lippincott Company, Philadelphia, PA.
10. MacFaddin, J.F. 1980. Biochemical Tests for the Identification of Medical Bacteria, 2nd ed. Williams & Wilkins, Baltimore, MD.
11. Isenberg, H.D. Clinical Microbiology Procedures Handbook, Vol. I, II & III. American Society for Microbiology, Washington, D.C.
12. Facklam, R.R., J.F. Padula, L.G. Thacker, E.C. Wartham and B.J. Sconyers. 1974. Presumptive identification of group A, B and D streptococci. Appl. Microbiol.; 27:107-113.
13. Abbreviated Identification of Bacteria and Yeast; Approved Guideline, M35. Clinical and Laboratory Standards Institute (CLSI - formerly NCCLS), Wayne, PA.
ATCC is a registered trademark of the American Type Culture Collection.
041709md Manufactured for: HARDY DIAGNOSTICS 1430 West McCoy Lane, Santa Maria, CA 93455, USA Phone: (805) 346-2766 ext. 5658 Fax: (805) 346-2760 Website: www.HardyDiagnostics.com Email: TechService@HardyDiagnostics.com Ordering Information Distribution Centers: California · Washington · Utah · Arizona · Texas · Ohio · New York · Florida The Hardy Diagnostics manufacturing facility and quality management system is certified to ISO 13485.
Copyright© 1996 - 2013 by Hardy Diagnostics. All rights reserved. |
1543 | https://www.youtube.com/watch?v=9YVEvpAXatY | Constructing Parallel and Perpendicular Lines - Geometry for High School
Miacademy & MiaPrep Learning Channel
323000 subscribers
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Description
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Posted: 5 Sep 2022
Master the art of geometric construction! In this high school geometry lesson, students will learn how to construct both parallel and perpendicular lines using a compass and straightedge. This lesson is part of MiaPrep’s Geometry course. Find more hands-on geometry videos on our channel!
We hope you are enjoying our large selection of engaging core & elective K-12 learning videos. New videos are added all the time - make sure you come back often to learn more! If you'd like us to cover any additional topics, please let us know. For practice, assessment, and many interactive activities that go along with each video, as well as a teacher/parent dashboard, go to miacademy.co for Grades K-8 or Miaprep.com for Grades 9-12!
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Transcript:
Hey everyone, it's Justin again. In this
lesson we get to go way back in history to thousands of years ago once again and learn how
to construct parallel and perpendicular lines! By the end of this lesson, you should be able
to construct parallel and perpendicular lines using a compass and straight edge. First,
we'll define parallel and perpendicular. Next, we'll learn one way to
construct parallel lines and then we'll learn another way to construct
perpendicular lines. Grab your straightedge and compass again for this lesson and
tuck that ruler away. Let's get started. So what do parallel and perpendicular mean?
You're probably familiar with these terms from your previous math courses but let's refresh
our memory. Two lines that are parallel must be on the same plane but never intersect. In
this example, lines LM and CB are parallel. This symbol means parallel and on the sketch
itself we can tell people that the lines are parallel by using this little arrow symbol.
Then there's perpendicular. Perpendicular lines are also on the same plane as each other
but do intersect and more specifically, they intersect at a right or 90 degree angle. In
this example, lines NP and DE are perpendicular. This symbol is used for perpendicular when writing
this out and this symbol is used for perpendicular on a drawing. It looks like we're putting a little
square in one of the right angles and since there are four right angles here, you could put this
symbol in any of them. Pause the video here if you need more time to write your notes for this.
Next up we're gonna start some constructions. We're going to learn one way to
construct parallel lines together and then you'll have an opportunity to explore
two more ways in your activity later on. The method we're using is the angle copy method.
It'll be best to create this in your notes as we go so remember to pause the video anytime
you need to catch up or re-watch something. Start by drawing a line like this. Then use your
straight edge to draw a segment going through that line that you can tell is definitely
not perpendicular to the original line. Place a point on this new segment that you
would like your parallel line to go through. Set a distance on your compass that's not
too big and draw an arc where the line and segment intersect. Go up to the new point
you made and draw the same arc there too. Then, set the distance on your compass where the
arc meets up with both your line and segment. Copy that distance up on your other
arc using that same compass setting. Point R in the intersection you just created with
your compass is where you'll draw the parallel line. We can make all of our compass marks a
little lighter if we want to by gently erasing it so that it's easier to see the parallel lines. We
can mark them as parallel and if we put two points somewhere anywhere on each line then we can also
write a description of what our picture shows. There's some space in your notes template to
practice this one more time before we move on so pause the video now and resume when you're
ready to learn about a perpendicular construction. Next, we are going to go over a method for
finding a perpendicular line and actually more specifically, the perpendicular bisector.
We learned what perpendicular means earlier, but do you know what bisect means?
Bisect means splitting something into two equal parts. You'll see two other methods in your
lesson activity for finding a perpendicular line but I figured we could look at this more
specific perpendicular bisector case together. Remember how a line goes on forever
and ever in both directions? So how do you bisect or split something in half something that
technically doesn't have a middle? You don't, which is why we're going to bisect a
line segment here instead of a line. Okay, let's get started. Draw a segment and set
your compass so that it's definitely longer than halfway across the segment, but not entirely
the way across. Use this setting to draw an arc above and below the segment. Move the compass
tip to the other end point of the segment and be sure that the setting doesn't change. Draw
two more arcs just like what you did before. The two places that the arcs intersect each
other are actually the two places we need to draw our new line through. You could just draw
a segment here but I'm going to keep the arrows on the ends of this so that it's a line. We
weren't told in the directions that we have to write the notation for this but it's good
practice so let's label some points. On the segment we have to use the endpoints, but on the
line remember that our points can be anywhere. You also don't have to use the same letters as
me, that's totally up to you. Now we can write the notation for this. This says that line SH is
perpendicular to segment FI. And don't be afraid to write your answer in different ways, here's a
few of the many more ways you can write this. We can also put in a little box in one of the corners
to show that we have right angles in our drawing. There's one other thing I haven't shown you
yet. Remember that I said we weren't just making a perpendicular line in this construction
but we were also bisecting segment FI? Well that means segment FH and segment HI are congruent to
each other. You may as well write that down too and put a little congruent symbol on our drawing
to show that. Also if you want to lighten up your compass markings now's a good time to do that too.
All right, that's the end of our perpendicular example. Pause the video now and try to make
this one more time in your notes all by yourself. There are two more ways to find a perpendicular
line that you'll have a chance to explore in your activity for this lesson and these other ways
don't require that you start with a segment, you can start with a line instead. Now that
you've explored ways to construct parallel and perpendicular lines, I hope you're getting
comfortable with your compass and straight edge. Once you finish your activity for this lesson,
we're going to advance to some other concepts using line segments that were more recently
developed in history. See you next time! |
1544 | https://biology.stackexchange.com/questions/94682/is-natural-selection-actually-random | genetics - Is natural selection actually random? - Biology Stack Exchange
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Is natural selection actually random?
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In the Theory of Evolution, two main factors take place:
One is random, which are the different mutations that organisms' DNA suffer. This process adds genetic variability to a given population.
The other one is not random, which is natural selection. It is not random because it will only select those organisms which are better capable of surviving to the conditions of their environment.
Most likely I am wrong and I would like to be corrected, but I was thinking that, since natural selection depends on mutations and the conditions of the environment, and since those conditions can be formed by random processes, wouldn't it make natural selection as actually random?
I mean, this is probably an over simplistic analogy, but take a group of mammals which suffer a great series of mutations in a short period of time, and separate into two species of mammals: one group of animals with hair and the other group has no hair at all. Suddenly, one day starts to snow. We could expect that the second group would most likely go extinct, while the mammals with hair would survive. Since the weather is based on random processes and since natural selection in this example would be based on the mutations that the animals suffered and the weather as well, wouldn't natural selection be a fundamentally random process?
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asked Jul 15, 2020 at 12:21
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1 Note that random processes can have systematic outcomes. E.g. random movement of gas molecules has a non-zero pressure as outcome. (Systems that do not behave in this way, i.e. where random processes do not have predictable systematic outcomes but the outcome stays unpredictable are called chaotic)cbeleites –cbeleites 2020-07-16 21:03:14 +00:00 Commented Jul 16, 2020 at 21:03
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You are right in a sense. The processes which generate mutations in an organism are stochastic (not quite exactly random, but in most organisms they are effectively random for all intents and purposes). In terms of population genetics, it is considered that mutations occur randomly across the genome and have an equal probability of occurring at a given base pair.
The environment is also random/stochastic in the sense that it occurs mostly independent of mutations which have occurred. Whether or not it is cold one year is independent of the mutations which have occurred within an organism. (Actually that's not quite correct since organisms can generate their own environmental niches and evolve according to them, but I think that is beyond the scope of this question).
What is not random is how the frequency of genetic variants respond to different environments. This is natural selection. Over the long run, we expect genetic variants which increase the fitness of an organism within a given environment will increase in frequency and vice versa. There is a large body of theory and evidence which predicts how a particular variant may change in frequency in response selection of a particular pressure.
As @jamesfq correctly points out, a nice example of selection being non-random on the macro-scale is convergent evolution. For example, aquatic animals such as fish, certain mammals (e.g. Dolphins) and Amphibians and Dinosaurs have independently evolved streamlined bodyplans in order to conserve energy when moving through the water.
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edited Jul 15, 2020 at 16:36
answered Jul 15, 2020 at 13:32
user438383user438383
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As an example of non-randomness, see examples of parallel evolution, where distantly-related creatures evolve similar forms because of environmental pressures. E.g. Ichthyosaurs, dolphins, some sharks, and some fish all evolved similar body plans, even though creatures with very different body plans - plesiosaurs, rays - can be successful in the same environment.jamesqf –jamesqf 2020-07-15 16:19:32 +00:00 Commented Jul 15, 2020 at 16:19
Or as another example the very fact you can predict a hairy species to survive better in the cold makes it non-random.John –John 2020-07-15 17:28:01 +00:00 Commented Jul 15, 2020 at 17:28
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1545 | https://www.youtube.com/watch?v=Vb0JQvwlNCA | TRIANGLE INEQUALITY THEOREM || GRADE 8 MATHEMATICS Q4
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Transcript:
[Music] in this video we will discuss the theorems and triangle inequalities we will focus on triangle inequality theorem in this video we will illustrate urms and triangle inequalities but we will just focus on triangle inequality theorem if you want to know about exterior angle inequality theorem you may watch the previous video tutorial and if you want to know about hinge theorem please do watch the next video tutorial please do not forget to like our video and subscribe narin kaio thank you so i have here triangle abc with sides a b and c so again meron potion triangle abc numerous sides a b and c now i want you to complete the table i want you to give me any pair of two sides of the given triangle so we have a and b okay so we will consider c as its third side so you pair of two sides small and triangle third side yeah okay so let's have again another pair so we have a and c so if a and c therefore this is the third side we have b if we have b and c so ito pair of two sides not in therefore a is your third side now given this pair of two sides and it's third side so that in the end pair of two sides more than the other side will be the your third side and two sides more a and c therefore b is your third side ganondin now let us give the measures of the given size of the triangles let's say the measure of side a is six your b is nine and your c is eleven so e identify nathan and manga measures aside so for a we have six for b we have nine and four c we have eleven next for a again we have 6 for c we have 11 and for b we have 9. for b we have 9 for c we have 11 and for a we have 6. now given this pair of two sides of the triangle and it's third side now it right now and if we will try to get the sum of the two sides of the triangle [Music] sides and triangle so let us give the sum so 6 plus 9 that is 15 and then yeah the third side is 11. let us find the sum of six plus eleven that is seventeen and the third side is nine and then we have nine plus eleven that is twenty and our third side for b and c is six now what have you noticed all the given sum of any two sides of the triangle are always greater than the third side okay uh two sides and triangles it's always greater than the measure of its third side so therefore we can say that the sum of your a and b paging and then we will compare it to its third side and then the sum of a plus c and then echo comparing third side and then the sum of b plus c at echo comparing third side that is always greater than the third side okay so we can conclude that the sum of the lengths of any two sides of a triangle is greater than the length of its third side or of the third side okay and that is all about the triangle inequality to urem so tatan now for every pair of two sides of a triangle that is always greater than the third side so for every sum of the two sides of the triangle that is always greater than the third side so that is all about the triangle inequality degree let's have an example which of the following could be the length of the sides of a triangle so i have here set a set b and set c so given this pair of uh given this set of sides of the lengths of the sides of a triangle so identifying that in alindito possible side and triangle so base triangle inequality theorem it's it is always greater than the length of the third side so let us try so it and then we have seven and five so on third side is four now soda path any two sides nothing that is always greater than the third side so i checked this can be the length of the sides of a triangle so we have this is 11 11 is greater than 5 that is true 4 plus 5 is 9 9 is greater than 7 this is also true 7 plus 5 is 12 and 12 is greater than 4 so this is also true so therefore since uh all the given sum of two sides of the triangle are greater than the length of its third side therefore the measure of the length of these sides are true to form a triangle so triangle next let's have the second one the set b so it is any two sides and triangles so we have nine eleven and then the third side is two we have eleven and two and third side is eleven and then nine and two and third side is eleven so the path and some any two sides is always greater than the third side so you add nothing to 9 plus 11 this is 20 which is greater than 2 that is true 11 plus 2 is 13 which is greater than 9 and this is true 9 plus 2 is 11 which is greater than 11 is this true no this is false because 11 is equal to 11. so since we have here one false na result e big sub hen atom length triangle so let's have letter c so we have 12 plus 10 the third side is 12 ten plus twelve the third side is twelve and then twelve plus twelve the third side is ten so i greater than so twelve plus ten is twenty two plus twelve i mean greater than twelve this is true 10 plus 12 this is 22 is greater than 12 this is also true and 12 plus 12 this is 24 which is greater than 10 so this is also true so therefore the length these three lengths can form a triangle so length triangle so only a and c can form a triangle let's have another two sides of a triangle measure 15 centimeters and nine centimeters write an inequality that represents the range of values for the possible lengths of the third side so in this uh problem is what are the possible values okay possible length of the third side because are two sides which we have 15 centimeters and nine centimeters so eb third side nothing for us to form a triangle triangle inequality any two sides in triangle but it is always greater than the third side so not in so we will be using this formula okay so difference no given and then unknown and then some given so on given at 10 15 and 9 so that is 15 minus 9 and then your third side at the unknown if nothing that is x and then the sum of the given so we have 15 plus 9. so 15 minus 9 we have 6 this is x 15 plus 9 that is 24 so inequality not m your x or the length of the third side is less than 24 but greater than six okay so the value of the third side or the length of your third side must be less than 24 but greater than six so anu 24 is 23 so from 7 to 23 and a possible length nang adding third side so hindi poi this is 6 in the policy 24 so from 7 to 23 among a possible theorem the sum of the two sides must be greater than the third side so equals that's why the possible range of values for the length of your third side must be from 7 to 23 only let's have another example the measures of three sides of a triangle are given so find the possible value or values of x so we have 18 2x and x so given puyong i think three sides now in this case given point three sides perro hindi not in your measurement sides not in so we can also make use of the formula given by the uh previous slide uh difference the third side and then the sum that's a previous [Music] [Music] [Music] we just have to find the values of the possible values of x or young range so not in chicago when so we have 2x minus x so buffett so for us to find the value of x now it's a substitute not indeed adding unknown as a variable uh you have to find the the value of x by using the pre uh formula of the previous slide so we have two x minus x and then two x plus x so automation sum so since we have this 18 is greater than 2x minus x but less than 2x plus x so we have 2x minus x is less than 18. so solve for x we have 2 minus 1 that is x is less than 18. now nothing x plus x is greater than eighteen months for greater greater than this is less than so since d two time x starts the right side pablichtad putayo so 2x plus x is greater than uh interchange ponatin symbol is greater than 18. so we will get 2x plus x that is 3x is greater than 18. divide both sides by 3 18 divide three that is x is greater than six so e big sub n values exponent n is greater than 6 18. 18. inequality john so your x is greater than 18 but let i mean greater than 6 but less than 18. so your x is greater than 6 but less than 18. so on among possible values since greater than 6 so much starch is a 7 and then less than 18 so and chesa seventeen so i'm value spoon and x nothing is from seven to seventeen la man okay so puerto national so let's have any pair of two sides of the given triangle so we have 2x plus x is greater than 18. so we are actually applying the triangle inequality to your m that the sum of the lengths of two side must be greater than the length of its third side so it up any two pair nothing so i have two x plus x is greater than 18. now let us use one of the values of x so gamma is greater than six but less than 18 so seven substitute net in c7 so this is 14 plus 7 is greater than 18 and this is 21 so as you can see this is true okay let's have another pair using the same value of x so let's have another pair let's use 18 plus 2x is greater than x so i'm substituting that in d to see 7. so 18 plus 2 times 7 so that is 14 i mean 18 we have 18 plus 14 is greater than 7 and this is 32 is greater than 7 and this is true now let's have another pair of two sides any two sides of the triangle so we have 18 plus x is greater than 2x let us substitute 7. so that is 18 plus 7 is greater than 2 times 7 18 plus 7 that is 25 is greater than 14 and this is also true okay so as you can see the value of x which is 7 is true among the given sides of the triangle so therefore if the value of x is 7 or 7 8 9 10 up to 17 that is true among the lengths of the sides of the triangle now let us try to use six okay so six is not uh included duns adding values of x so it right nothing says six so try nothing just a two x plus x is greater than 18. substitute nothing is six so that is twelve plus six as you can see this is false so that's why six is not included for the possible values of x now it rain say 18 where 18 should not be included in values of x so it right now so 8 x plus 18. so any two sides x plus 18 is greater than 2x so this is 18 plus 18 substitute nothing let's say 18 so that is 36 is greater than 36 as you can see they are equal so it's not greater than so that is let's have another so again find the values of x given the sides of 16 x and 3x so same process d to poi difference so 3x minus x and then your third side sixteen and then three x plus x sodium now unknown so three x minus x so on first inequality not in d so i three x minus x is less than sixteen okay so we have two x is less than 16 so that is 16 divided by 2 we have x is less than 8. now your young second inequality is minus x is less than 16 and then 3x plus x is greater than interchange okay so 3x plus x is greater than okay so we have three x plus x that is four x is greater than 16. divide both sides by four so 16 divide four we have x is greater than four okay so therefore the values of x or the possible values of x are greater than four but less than eight so among up within values that x naught and i must matassa four and eviksa b n starts with five and then less than eight so it ends with seven so we have the inequality of x is greater than four bispu dito so x is greater than four since x nil again not in sagitta at binary x is greater than four kaya da pat interchange punate inequality so x is greater than four but less than eight so we have e big sub hen values x naught and is from five to seven only so let us check for any uh two sides of the triangle so by applying the triangle inequality to your m so let's have any pair of two sides three x plus x should be greater than 16 let us substitute the value of x let us try five cases since the value of x is greater than 4 so we can use 5 and then 3 times 5 that is 15 plus 5 that is 20 as you can see 20 is greater than 16 and this is true let's have another pair of two sides let us use 16 plus 3x should be greater than x so we have 16 plus 3 times 5 using the same value of x is greater than 5 so we have 16 plus 15 is greater than 5 and 16 plus 15 this is 31 is greater than 5 and this is also true and then for another pair of two sides we have the last one so we're done with this this pair so etu naman so 16 plus x is greater than 3x let us substitute the same value we have 5 so we have 21 is greater than 15 and this is also true so 5 to 7 are true for the possible values of x for us to form a triangle now let us try to substitute four so substitute nettings of c four sahaja andito so etwan gusto hongami ten so three times four plus four three times four this is twelve plus four so this is sixteen is greater than sixteen and this is false pair of two sides so from five to seven lamppo so i try not to see eight so 16 plus eight is greater than three times eight sixteen plus eight this is 24 is greater than 24 as you can see they are equal it's not greater so this is also false so by uh using the the value of x which is 4 and 8 which is not part of the possible values of x as you can see it didn't form a triangle okay so hindi formed a triangle for any two sides so let's try which of the following could be the lengths of the sides of a triangle so write yes or no so and then compare it to the third side the sum of the two sides of a triangle should be greater than the third side so let us answer so pointing because 3 plus 2 is 5 that is greater than 1 2 plus 1 is 3 which is equal to 3. greater than so that's why it's a no next three plus seven little ten which is equal to ten so that is no and then one plus one this is two so this is also equal to the third side so indeed in so that's why it's a no and then this one we have here variables length and two sides not in x plus y so x plus y right in point third side so even equal that's why it's unknown let's have the angle side relationship theorem now given this triangle abc so binigay pusa at angles can you name all the angles we have angle a angle b and angle c for our opposite side ano opposite side knee angle a so some a previous we have bc for angle b we have ac and for angle c we have a b so these are the opposite side of these three angles now since given napoleon can you give me the measure of angle a so the measure of angle a is 75 degrees the measure of angle b is 45 degrees your angle c is 60 degrees let's proceed to our side in opposite side me angle a so your bc that is 14. your ac which is the opposite side of angle b i mean it this one so your e c which is the opposite side of angle b we have eight and then for a b which is the opposite side of angle c we have ten now let us identify the largest angle the smallest angle longer side and the shortest side so identifying that in it so from the given measures of the angles what is the largest angle so the your largest angle is angle a what is the smallest angle the smallest angle is angle b what is the longest side the longest side is bc and the shortest side is ac now as you can see angle a which is your largest angle is opposite to the longest side okay or vice versa the longest side is opposite the largest anger since nato in other words shortest side is opposite the smallest angle and then the largest angle is opposite the longest side so we can also relate this uh concept adding theorems so in a triangle in any triangle we have the angle side inequality theorem and publish that if one angle is larger than the other angle the side opposite the larger angle is the longer side and we also have the side angle inequality theorem where if one side is longer than the other side then the angle opposite the longer side is the larger angle inequality theorem and side angle inequality theorem which is the angle side relationship theorem okay so we can now identify the largest to smallest angle so we will be arranging the angles in descending order so simula mata as malachi okay so we will be identifying the largest to smallest angle so using this inequality theorem that the largest angle is always opposite the longest side and the smallest angle is always opposite the shortest side so how will you know the largest to smallest yes okay because the largest angle is always opposite the longest side longest side more what is now your longest side so it is always opposite your largest angle so we have bc and then on shortest side since it open smallest angle side so that is ac so bc side bc side a b and side ac so this is the arrangement of the angles from largest to smallest angle and this is the arrangement of your sides from longest to shortest side so we are actually applying the angle side relationship um the angle side inequality theorem and the side angle inequality theorem let's try so i have here three triangles so given the measure of these two angles we have 50 degrees and 70 degrees longest side so give the longest side in each triangle so how will you know the longest side soda path measuring that remember the longest side and longest side punatin is always opposite the largest angle so that is 50 plus 70 and that is 60 degrees okay because this is 120 so 180 minus 120 that is 60 so therefore your angle a is 60 degrees now we have here 50 degrees 60 degrees 70 degrees longest side opposite largest angle so what is the largest angle angle c what is the opposite side we have side a b next so i have here same solution [Music] so this is 59 plus 70 we have 129 so 129 uh is subtract not in terms of 180 therefore we have 51 so therefore this is 51 degrees so we can now uh identify the largest angle so we can see the longest side since you know natan so since this is 70 degrees largest angle ebik sabihin and longest side side and then next so we have here what is the longest side longest 180 minus 40 we have 140 3x 60 plus 2x plus 45 is equal to 140 so you add 10 c3 plus 2 that is 5x and then 60 plus 45 that is 105 is equal to 140 so transpose not into 105 that will become minus 105 so 5x is equal to 35 divide both sides by 5 so 35 divide five that is x is equal to seven so pakistan absolutely not indeed this is seven three times seven is 21 plus 60 that is 81 degrees so angle g is 81 degrees next 2 times 7 is 14 plus 45 is 59 so the measure of angle i is 59 degrees now we now have the measure of the three angles we have 40 degrees 81 degrees and 59 degrees now bhagavatayama identify largest angle 81 c40 at 659 is equal to 180 so i checked um identifying largest angle soda patchy 59 c81 at c40 the padma and longest side and that is side h i thank you for watching this video i hope you learned something don't forget to like subscribe and hit the notification bell for updated ko for more video tutorials this is your guide in learning your math lessons your walmart channel |
1546 | https://www.broughtoneap.co.uk/hvac/articles/heat-transfer-coefficient/ | Heat Transfer Coefficient
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Whether it’s portable air con units, industrial dehumidifiers, industrial electric fan heaters then an important part of how they work is something known as the heat transfer coefficient, which basically tell you how much heat (or cooling) will be transferred from the media you’re using to produce the hot or cold, to the substance you’re heating or cooling. In this section we take a look at what the heat transfer coefficient is and how it relates to climate control equipment.
The heat transfer coefficient, also known as the film coefficient, or film effectiveness, in thermodynamics and in mechanics is the proportionality constant that exists between the heat flux and the thermodynamic driving force for the flow of that heat (i.e., the temperature difference, ΔT (‘Delta T’)):
The overall heat transfer rate for combined modes is expressed in terms of an overall conductance or heat transfer coefficient, U. In that case, the heat transfer rate is:
Q = hA(T2 – T1)
A: surface area where the heat transfer takes place, m²
T2: temperature of the surrounding fluid, K
T1: temperature of the solid surface, K
The general definition of the heat transfer coefficient is:
h = q ⁄ ΔT
where:
q: heat flux, w/m²; i.e., thermal power per unit area, q = dQ/dA
h: heat trander coefficient, W/(m²•K)
ΔT: difference in temperature between the solid surface and surrounding fluid area, K
This is used to calculate the heat transfer, normally by either convection or phase transition, between a fluid and a solid. The heat transfer coefficient is represented in watts per squared meter kelvin:W/(m²K).
The heat transfer coefficient is the reciprocal of thermal insulance, which can be used to measure building materials (R-value) and clothing insulation.
There are many methods used to calculate the heat transfer coefficient in different heat transfer modes, fluids, flow regimes, and/or under different thermohydraulic conditions. One commonly used way to estimate it is by dividing the thermal conductivity of the convection fluid by a length scale. The heat transfer coefficient is often calculated from a Nusselt number (a dimensionless number). There are also online calculators available specifically for Heat-transfer fluid applications.
Experimental assessment of the heat transfer coefficient poses some challenges especially when small fluxes need measuring (e.g. <0.2W/cm²).
Composition
A simple way to ascertain an overall heat transfer coefficient that is useful to find the heat transfer between simple elements such as walls in buildings or across heat exchangers is illustrated below. Please note this way only accounts for conduction within materials, it does not take into account any heat transfer through methods such as radiation. The method is as follows:
1/(U · A) = 1/(h1 · A1 + dχw/(k · A) + 1/(h2 · A2)
Where:
• U=the overall heat transfer coefficient (W/(m²•k))
• A=the contact area for each fluid side (m²) (with A1 and A2 expressing either surface)
• k=the thermal conductivity of the material (W/(m·K))
• h =the individual convection heat transfer coefficient for each fluid (W/(m²• K))
• dχw= the wall thickness (m).
As the areas for each surface approach being equal the equation can be written as the transfer coefficient per unit area as shown below:
1/U = 1/h1 + dχw/k + 2/h2
or
1/U = 1/h1 + dχwk + 1/h2
It should be noted the value of dχw is often referred to as the difference between the two radii where the inner and outer radii are used to define the thickness in a flat plate transfer mechanism or other common flat surfaces such as a wall in a building when the area difference between each edge of the transmission surface approaches zero.
In the walls of buildings the above formula can be used to derive the formula commonly used to calculate the heat through building components, giving rise to what are known by architects and engineers as the U-Value or the R-Value of a construction assembly such as a wall. The R or u value are related as the inverse of each other such that R-Value = 1/u-Value can both be more fully understood through the concept of an overall heat transfer coefficient as described below.
Convective heat transfer correlations
Although convective heat transfer can be derived analytically through dimensional analysis, exact analysis of the boundary layer, approximate integral analysis of the boundary layer and analogies between energy and momentum transfer, these analytic approaches don’t always provide practical solutions when there aren’t any existing applicable mathematical models. As a result, many correlations have been developed by numerous authors to estimate the convective heat transfer coefficient in different cases including natural convection, forced convection for internal flow and forced convection for external flow. These empirical correlations are presented for their particular geometry and flow conditions. As the fluid properties are temperature dependent, they are evaluated at the film temperature Tf, which is the average of the surface T8 and the surrounding bulk temperature, T∞.
T8 + T∞
Tf = ______
2
External flow, vertical plane
Recommendations by Churchill and Chu provide a relationship, shown below, for natural convection adjacent to a verticfal plane, which applies for both laminar and turbulent flow. Where k is the thermal conductivity of the fluid, L is the characteristic length with respect to the direction of gravity, RaL is the Rayleigh number with respect to this length and Pr is the Prandtl number.
With regard to laminar flows, the following correlation is a touch more accurate. It is oberserved that a transition from a laminar to a turbulent boundary occurs when RaL exceeds around 10.
Extternal flow, vertical cylinders
For cylinders and vertical axes, the expressions for plane surfaces and be used provided so long as there isn’t a significant curvature effect. This represents the limit where boundary layer thickness is small relative to cylinder diameter D. The correlations for vertical plane walls can be used when
where GrL represents what is known as the Grashof number.
External flow, horizontal plates
Following suggestions made by W.H.McAdams a link for horizontal plates was made. The induced buoyancy can be different but to what extent depends upon whether or not the hot surface is facing up or down.
For a hot surface facing up, or a cold surface facing down, for laminar flow the following applies:
while for turbulent flow it is the following:
When the hot surface faces down, or a cold surface facing up, for laminar flow the correct formula is:
The characteristic length is the ratio of the plate surface area to perimeter. If the surface is inclined at an angle θ with the vertical then the equations for a vertical plate by Churchill and Chu may be used for θ up to 60°; if the boundary layer flow is laminar, the gravitational constant g is replaced g cosθwhen calculating the Ra term.
External flow, horizontal cylinder
For cylinders of sufficient length and negligible end effects, Churchill and Chu the following correlation for 10-⁵ < RaD < 10¹².
External flow, spheres
For spheres we apply the following correlation for Pr≃1 and 1 ≤ RaD ≤ 10⁵.
NuD = 2 + 0.43RaD1/4
Forced convection
Internal flow, laminar flow
Thanks to Sieder and Tate we now use the following correlation to account for entrance effects in laminar flow in tubes where D is the internal diameter µÞ is the fluid viscosity at the bulk mean temperature µw is the viscosity at the tube wall surface temperature.
For fully developed laminar flow, the Nusselt number is constant and equal to 3.66. Mills combines the entrance effects and fully developed flow into one equation
Internal flow, turbulent flow
What is known as the Dittus-Bölter correlation (1930) is a common and relatively simple correlation useful for many applications. This correlation is particularly applicable when the only viable method of heat transfer available is forced convection; i.e., there is no boiling, condensation, significant radiation, etc. Although not the most accurate correlation, its accuracy has been calculated to somewhere in the region of ±15%, it is still widely used and provides an answer within acceptable tolerances for many applications.
For a fluid flowing in a straight circular pipe that has a Reynolds number between 10,000 and 120,000 (in the turbulent pipe flow range), when the fluid’s Prandtl number is between 0.7 and 120, for a location far from the pipe entrance (more than 10 pipe diameters; more than 50 diameters according to many authors) or other flow disturbances, and when the pipe surface is hydraulically smooth, the heat transfer coefficient between the bulk of the fluid and the pipe surface can be expressed explicitly as:
where:
d is the hydraulic diameter
k is the thermal conductivity of the bulk fluid
µ is the fluid viscosity
j mass flux
cp isobaric heat capacity of the fluid
n is 0.4 for heating (wall hotter than the bulk fluid) and 0.33 for cooling (wall cooler than the bulk fluid).
The fluid properties necessary for the application of this equation are evaluated at the bulk temperature thus avoiding iteration.
Forced convection, external flow
In analysing the heat ttransfer associated with the flow past the exterior surface of a solid, the situation is complicated by phenomena such as boundary layer separation. Various authors have correlated charts and graphs for different geometries and flow conditions. For flow parallel to a plane surface, where χis the distance from the edge and Lis the height of the boundary layer, a mean Nusselt number can be calculated using the Colburn analogy.
Thom correlation
There exist simple fluid-specific correlations for heat transfer coefficient in boiling. The Thom correlation is for the flow of boiling eater (subcooled or saturated at pressures up to about 20 MPa) under conditions where the nucleate boiling contribution predominates over forced convection. This correlation is useful for rough estimation of expected temperature difference given the heat flux:
ΔTsat = 22.5 · q0.5 exp(-<em?P/8.7)
where:
ΔTsat is the wall temperature elevation above the saturation temperature, K
<em?q is the heat flux, MW/m2
Pis the pressure of water, MPa
Note that this empirical correlation is specific to the units given.
Heat transfer coefficient of pipe wall
The resistance to the flow of heat will vary depending on the material of pipe, but it can be expressed as a “heat transfer coefficient of the pipe wall”. However, one needs to select if the heat flux is based on the pipe inner or the outer diameter. Selecting to base the heat flux on the pipe inner diameter, and assuming that the pipe wall thickness is small in comparison with the pipe inner diameter, then the heat transfer coefficient for the pipe wall can be calculate as if the walls were not curved:
hwall = k/χ
Where k is the effective thermal conductivity of the wall material and χis the wall thickness.
If the above assumption does not hold, then the wall heat transfer coefficient can be calculated using the following expression:
hwall = 2k/di In(do /di)
where di and do are the inner and outer diameters of the pipe, respectively.
The thermal conductivity of the tube material usually depends on temperature; the mean thermal conductivity is often used.
Combining convective heat transfer coefficients
For two or more heat transfer processes acting in parallel, convective heat transfer coefficients simply add:
h = h1 + h2 + …
For two or more heat transfer processes connected in series, convective heat transfer coefficients add inversely:
For example, consider a pipe with a fluid flowing inside. The approximate rate of heat transfer between the bulk of the fluid inside the pipe and the pipe external surface is:
where
q = heat transfer rate (W)
h= convective heat transfer coefficient (W/m2·K))
t= wall thickness (m)
k = wall thermal conductivity (W/m·K)
A = area (m2)
ΔT= difference in temperature
Overall heat transfer coefficient
The overall heat transfer coefficient Uis a measure of the overall ability of a series of conductive and convective barriers to transfer heat. It is commonly applied to the calculation of heat transfer in heat exchangers, but can be applied equally well to other problems.
For the case of a heat exchanger, U can be used to determine the total heat transfer between the two streams in the heat exchanger by the following relationship:
q = U AΔTLM
where:
q= heat transfer rate (W)
U = overall heat transfer coefficient (W/(m2·K))
A= heat transfer surface area (m2)
ΔTLM = logarithmic mean temperature difference (K).
The overall heat transfer coefficient takes into account the individual heat transfer coefficients of each stream and the resistance of the pipe material. It can be calculated as the reciprocal of the sum of a series of thermal resistances (but more complex relationships exist, for example when heat transfer takes place by different routes in parallel):
where:
R= Resistance(s) to heat flow in pipe wall (K/W)
Other parameters are as above.
The heat transfer coefficient is the heat transferred per unit area per kelvin. Thus area is included in the equation as it represents the area over which the transfer of heat takes place. The area for each flow will be different as they represent the contract area for each fluid side.
The thermal resistancedue to the pipe wall is calculated by the following relationship:
where
χ= the wall thickness (m)
k = the thermal conductivity of the material (W/(m·K))
A= the total area of the heat echanger (m²)
This represents the heat transfer by conduction in the pipe.
The thermal conductivity is a characteristic of the particular material. Values of thermal conductivities for various materials are listed in the list of thermal conductivities.
As mentioned earlier in the article the convection heat transfer coefficientfor each stream depends on the type of fluid, flow properties and temperature properties.
Some typical heat transfer coefficients include:
• Air – h = 1- to 1– W/(m²K).
• Water – h = 500 to 10,000 W/(m²K).
Thermal resistance due to fouling deposits:
Often during their use, especially when used on industrial portable air conditioning units with missing filters, heat exchangers can collect a layer of dirt or dust on the surface which, in addition to potentially contaminating a stream, has a detrimental effect the performance of heat exchangers. On a dirty or contaminated heat exchanger the build-up of this layer on the walls creates an additional obstacle for the heat to flow through. Because of this new layer and the subsequent additional resistance within the heat exchanger the overall heat transfer coefficient of the exchanger will be reduced. The formula below can used to ascertain the new heat transfer resistance, taking into account the additional fouling resistance:
=
where
Uf = overall heat ransfer coefficient for a fouled heat exchanger, w/m2K
P= the perimeter of the heat exchanger, may be either the hot or cold side perimeter however, it must be the same perimeter on both sides of the equation, m
U = overall heat transfer coefficient for an unfouled heat exchanger, w/m2K
RfX = fouling resistance on the cold side of the heat exchanger, m2K/W
RfH – fouling resistance on the hot side of the heat exchanger, m2K/W
Pc = perimeter of the cold side of the heat exchanger, m
PH = perimeter of the hot side of the heat exchanger, m
This equation combines the overall heat transfer coefficient of an unfouled heat exchanger together with the fouling resistance and therefore calculates the total heat transfer coefficient for a fouled heat exchanger. The equation take into account that the perimeter of the heat exchanger will be different on both hot and cold sides. The perimeter used to the P does not matter as long as its is the same. The overall heat transfer coefficients will adjust to take into account that a different perimeter was used as the product UPwill remain the same.
The fouling resistances can be more accurately calculated for a specific heat exchanger if the average thickness and thermal conductivity of the material fouling the coil are known. The product of the average thickness and thermal conductivity will result in the fouling resistance on a specific side of the heat exchanger.
= Rf df/kf
Where
df average thickness of the fouling in a heat exchanger, m
kf thermal conductivity of the fouling W/mK.
With almost four decades experience in the design, manufacture and supply of large portable air conditioning units, large electric fan heaters, commercial or building site dehumidifiers, infrared heaters, ventilation fans, cooling fans, fume extractor fans, portable boilers, fan coil units we’ve been around long enough to offer an unrivalled service. For details on our full range of portable climate control equipment please see our site, call 01527 830610 or drop us a line at sales@broughtoneap.co.uk where a member of the team will be only too happy to help.
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1547 | https://www.youtube.com/watch?v=9xw55tpv0Yc | Factorise: x^2 - 3x + 2
Gaule Solution
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Posted: 20 Feb 2023
Factorise: x^2 - 3x + 2
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so today in this video we shall learn to factorize x square minus 3x Plus 2. so for the solution of factorized x square minus 3x plus 2 first of all you have to find out the prime factors of 2. so it is 2 times 1 means 2 so by multiplying Within These numbers we have to find out such a number that by adding up which we have to take the result 3 so number F2 and 1 so 2 plus 1 means it is 3 so you have to take a minus 3 so minus 2 and minus 1 means it is minus 3. so we can avoid it x square minus 2x minus X Plus 2. so minus 2 x minus X means it is minus 3 like as minus minus plus 2 into 1 means it is 2. so now we have to separate it into two pairs so these both are first pair like as these both are second pair so in the first pair the common here is X so the remaining is x minus 2. so like as in second pair there is no any numbers of expressions Common so when there is no any numbers or expression common then in that case the common here is one so the common here is minus 1. so remaining is x minus 2. so as you look here there is also x minus 2 like as here is also x minus 2 so in both the common here is x minus 2. so like as in first the remaining is X like as in second the remaining here is minus 1. so in this way we can solve the or find out the prime factors of all jabic equations so this much for today and if you have an equation related to math and science then just comment your question in a comment below and we shall provide a video solution of your question so thank you |
1548 | https://api.pageplace.de/preview/DT0400.9780134859354_A37847293/preview-9780134859354_A37847293.pdf | Advanced Mechanics of Materials and Applied Elasticity Sixth Edition This page intentionally left blank Advanced Mechanics of Materials and Applied Elasticity Sixth Edition Ansel C. Ugural Saul K. Fenster Boston • Columbus • New Y ork • San Francisco • Amsterdam • Cape Town Dubai • London • Madrid • Milan • Munich • Paris • Montreal • Toronto • Delhi • Mexico City São Paulo • Sydney • Hong Kong • Seoul • Singapore • Taipei • Tokyo Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trade-marks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed with initial capital letters or in all capitals. 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ISBN-13: 978-0-13-485928-6 ISBN-10: 0-13-485928-6 3 20 v Preface xvii Acknowledgments xx About the Authors xxi List of Symbols xxii Chapter 1 Analysis of Stress 1 1.1 Introduction 1 1.1.1 Mechanics of Materials and Theory of Elasticity 1 1.1.2 Historical Development 2 1.2 Scope of the Book 3 1.3 Analysis and Design 4 1.3.1 Role of Analysis in Design 6 1.3.2 Selection of Factor of Safety 6 1.3.3 Case Studies 7 1.4 Conditions of Equilibrium 8 1.5 Definition and Components of Stress 9 1.5.1 Sign Convention 11 1.5.2 Equality of Shearing Stresses 12 1.5.3 Some Special Cases of Stress 12 1.6 Internal Force Resultant and Stress Relations 13 1.6.1 Basic Formulas for Stress 15 1.6.2 Combined Stresses 17 1.7 Stresses on Inclined Sections 17 1.7.1 Axially Loaded Members 18 1.8 Variation of Stress within a Body 20 1.8.1 Equations of Equilibrium 20 1.9 Plane-Stress Transformation 23 1.9.1 Stress Tensor 25 1.9.2 Polar Representations of State of Plane Stress 25 1.9.3 Cartesian Representation of State of Plane Stress 25 Contents vi Contents 1.10 Principal Stresses and Maximum In-Plane Shear Stress 26 1.11 Mohr’s Circle for Two-Dimensional Stress 28 1.12 Three-Dimensional Stress Transformation 35 1.13 Principal Stresses in Three Dimensions 38 1.13.1 Invariants for Three-Dimensional Stress 40 1.14 Normal and Shear Stresses on an Oblique Plane 42 1.14.1 Octahedral Stresses 44 1.15 Mohr’s Circles in Three Dimensions 45 1.15.1 Absolute Maximum Shear Stress 46 1.15.2 Equations of Three Mohr’s Circles for Stress 47 1.16 Boundary Conditions in Terms of Surface Forces 49 1.17 Indicial Notation 50 References 51 Problems 51 Chapter 2 Strain and Material Properties 68 2.1 Introduction 68 2.2 Deformation 69 2.2.1 Superposition 69 2.3 Strain Defined 70 2.3.1 Plane Strain 70 2.3.2 Three-Dimensional Strain 72 2.3.3 Eulerian and Lagrangian Coordinates 73 2.3.4 Large Strains 74 2.4 Equations of Compatibility 75 2.5 State of Strain at a Point 76 2.5.1 Transformation of Two-Dimensional Strain 76 2.5.2 Transformation of Three-Dimensional Strain 78 2.5.3 Invariants in Three-Dimensional Strain 79 2.5.4 Mohr’s Circle for Plane Strain 80 2.6 Engineering Materials 83 2.6.1 General Properties of Some Common Materials 84 2.7 Stress-Strain Diagrams 86 2.7.1 Ductile Materials in Tension 86 2.7.2 Geometry Change of Specimen 87 2.7.3 True Stress and True Strain 88 2.7.4 Brittle Materials in Tension 89 2.7.5 Materials in Compression 89 2.7.6 Materials in Shear 90 2.7.7 Short-Time Effects of Temperature on Stress-Strain Properties 90 2.8 Elastic versus Plastic Behavior 91 2.9 Hooke’s Law and Poisson’s Ratio 92 2.9.1 Volume Change 93 2.9.2 Deflection of Axially Loaded Members 93 Contents vii 2.10 Generalized Hooke’s Law 96 2.11 Orthotropic Materials 101 2.11.1 Generalized Hook’s Law for Orthotropic Material 102 2.12 Measurement of Strain: Strain Gage 103 2.12.1 Strain Rosette of Three Gages 104 2.12.2 Rectangular and Delta Strain Rosettes 106 2.13 Strain Energy 107 2.13.1 Strain Energy Density for Normal and Shear Stresses 108 2.13.2 Strain Energy Density for Three-Dimensional Stresses 110 2.14 Strain Energy in Common Structural Members 111 2.14.1 Strain Energy for Axially Loaded Bars 111 2.14.2 Strain Energy of Circular Bars in Torsion 112 2.14.3 Strain Energy for Beams in Bending 113 2.15 Components of Strain Energy 113 2.16 Saint-Venant’s Principle 115 2.16.1 Confirmation of Saint-Venant’s Rule 116 References 117 Problems 118 Chapter 3 Problems in Elasticity 133 3.1 Introduction 133 3.2 Fundamental Principles of Analysis 134 3.2.1 Three-Dimensional Problems 134 3.2.2 Two-Dimensional Problems 134 Part A: Formulation and Methods of Solution 135 3.3 Plane Strain Problems 135 3.4 Plane Stress Problems 138 3.4.1 Stress–Strain Relations for Orthotropic Materials 139 3.5 Comparison of Two-Dimensional Isotropic Problems 140 3.6 Airy’s Stress Function 141 3.6.1 Generalized Plane Strain Problems 142 3.6.2 Antiplane Shear Deformations 142 3.7 Solution of Elasticity Problems 143 3.7.1 Polynomial Solutions 144 3.8 Thermal Stresses 149 3.8.1 Equations of Thermoelasticity 149 3.9 Basic Relations in Polar Coordinates 152 3.9.1 Equations of Equilibrium 153 3.9.2 Stress Function 153 3.9.3 Strain-Displacement Relations 154 3.9.4 Hooke’s Law 155 3.9.5 Transformation Equations 155 3.9.6 Compatibility Equation 156 viii Contents Part B: Stress Concentrations 157 3.10 Stresses Due to Concentrated Loads 157 3.10.1 Compression of a Wedge (Fig. 3.10a) 157 3.10.2 Bending of a Wedge (Fig. 3.10b) 159 3.10.3 Concentrated Load on a Straight Boundary (Fig. 3.11a) 160 3.11 Stress Distribution Near a Concentrated Load Acting on a Beam 161 3.11.1 Accuracy of Results 163 3.12 Stress Concentration Factors 163 3.12.1 Circular Hole in a Large Plate in Simple Tension 165 3.12.2 Circular Hole in a Large Plate in Biaxial Tension 167 3.12.3 Elliptic Hole in a Large Plate in Tension 167 3.12.4 Graphs for Stress Concentration Factors 168 Part C: Contact Mechanics 169 3.13 Contact Stresses and Deflections 169 3.13.1 Hertz Theory 170 3.13.2 Johnson–Kendall–Roberts Theory 170 3.14 Spherical and Cylindrical Contacts 171 3.14.1 Two Spheres in Contact 171 3.14.2 Two Parallel Cylinders in Contact 173 3.15 Contact Stress Distribution 174 3.15.1 Two Spheres in Contact (Figure 3.18a) 174 3.15.2 Two Parallel Cylinders in Contact (Figure 3.20a) 174 3.16 General Contact 178 References 181 Problems 182 Chapter 4 Failure Criteria 192 4.1 Introduction 192 4.1.1 Failure 192 Part A: Static Loading 193 4.2 Failure by Yielding 193 4.2.1 Creep: Time-Dependent Deformation 194 4.3 Failure by Fracture 195 4.3.1 Types of Fracture in Tension 196 4.4 Yield and Fracture Criteria 197 4.5 Maximum Shearing Stress Theory 198 4.6 Maximum Distortion Energy Theory 199 4.6.1 Yield Surfaces for Triaxial Stress 200 4.7 Octahedral Shearing Stress Theory 200 4.8 Comparison of the Yielding Theories 204 4.9 Maximum Principal Stress Theory 205 4.10 Mohr’s Theory 206 4.11 Coulomb–Mohr Theory 207 4.12 Introduction to Fracture Mechanics 210 4.12.1 Stress-Intensity Factors 211 Contents ix 4.13 Fracture Toughness 213 Part B: Repeated and Dynamic Loadings 216 4.14 Fatigue: Progressive Fracture 216 4.14.1 Fatigue Tests 216 4.14.2 Estimating the Endurance Limit and Fatigue Strength 217 4.15 Failure Criteria for Metal Fatigue 217 4.15.1 Uniaxial State of Stress 218 4.15.2 Comparison of Fatigue Failure Criteria 219 4.15.3 Design for Uniaxial Stress 219 4.15.4 Combined State of Stress 221 4.16 Fatigue Life 223 4.17 Impact Loads 225 4.17.1 Strain Rate 226 4.17.2 Basic Assumptions of Impact Analysis 227 4.18 Longitudinal and Bending Impact 227 4.18.1 Freely Falling Weight 227 4.18.2 Horizontally Moving Weight 228 4.19 Ductile–Brittle Transition 230 References 232 Problems 233 Chapter 5 Bending of Beams 242 5.1 Introduction 242 Part A: Exact Solutions 243 5.2 Pure Bending of Beams of Symmetrical Cross Section 243 5.2.1 Kinematic Relationships 244 5.2.2 Timoshenko Beam Theory 246 5.3 Pure Bending of Beams of Asymmetrical Cross Section 246 5.3.1 Stress Distribution 248 5.3.2 Transformation of Inertia Moments 248 5.4 Bending of a Cantilever of Narrow Section 251 5.4.1 Comparison of the Results with the Elementary Theory Results 253 5.5 Bending of a Simply Supported Narrow Beam 254 5.5.1 Use of Stress Functions 255 5.5.2 Comparison of the Results with the Elementary Theory Results 256 Part B: Approximate Solutions 256 5.6 Elementary Theory of Bending 256 5.6.1 Assumptions of Elementary Theory 257 5.6.2 Method of Integration 258 5.7 Normal and Shear Stresses 260 5.7.1 Rectangular Cross Section 262 5.7.2 Various Cross Sections 262 5.7.3 Beam of Constant Strength 267 5.8 Effect of Transverse Normal Stress 268 x Contents 5.9 Composite Beams 270 5.9.1 Transformed Section Method 270 5.9.2 Equation of Neutral Axis 271 5.9.3 Stresses in the Transformed Beam 272 5.9.4 Composite Beams of Multi Materials 272 5.10 Shear Center 276 5.10.1 Thin-Walled Open Cross Sections 277 5.10.2 Arbitrary Solid Cross Sections 281 5.11 Statically Indeterminate Systems 281 5.11.1 The Method of Superposition 282 5.12 Energy Method for Deflections 284 5.12.1 Form Factor for Shear 285 Part C: Curved Beams 286 5.13 Elasticity Theory 286 5.13.1 Equations of Equilibrium and Compatibility 286 5.13.2 Boundary Conditions 287 5.13.3 Stress Distribution 288 5.13.4 Deflections 289 5.14 Curved Beam Formula 289 5.14.1 Basic Assumptions 289 5.14.2 Location of the Neutral Axis 290 5.14.3 Tangential Stress 291 5.14.4 Winkler’s Formula 293 5.15 Comparison of the Results of Various Theories 293 5.15.1 Correction of σ θ for Beams with Thin-Walled Cross Sections 294 5.16 Combined Tangential and Normal Stresses 296 References 300 Problems 300 Chapter 6 Torsion of Prismatic Bars 315 6.1 Introduction 315 6.2 Elementary Theory of Torsion of Circular Bars 316 6.2.1 Shearing Stress 317 6.2.2 Angle of Twist 317 6.2.3 Axial and Transverse Shear Stresses 320 6.3 Stresses on Inclined Planes 321 6.3.1 Stress Transformation 321 6.3.2 Transmission of Power by Shafts 323 6.4 General Solution of the Torsion Problem 324 6.4.1 Geometry of Deformation 324 6.4.2 Equations of Equilibrium 325 6.4.3 Equations of Compatibility 325 6.5 Prandtl’s Stress Function 326 6.5.1 Boundary Conditions 326 Contents xi 6.5.2 Force and Moments over the Ends 327 6.5.3 Circular Cross Section 331 6.6 Prandtl’s Membrane Analogy 333 6.6.1 Equation of Equilibrium 333 6.6.2 Shearing Stress and Angle of Twist 335 6.7 Torsion of Narrow Rectangular Cross Section 338 6.7.1 Thin-Walled Open Cross Sections 339 6.8 Torsion of Multiply Connected Thin-Walled Sections 340 6.8.1 Shearing Stress 340 6.8.2 Angle of Twist 341 6.9 Fluid Flow Analogy and Stress Concentration 344 6.10 Torsion of Restrained Thin-Walled Members of Open Cross Section 346 6.10.1 Torsional and Lateral Shears 347 6.10.2 Boundary Conditions 348 6.10.3 Long Beams Under Torsion 348 6.10.4 Angle of Twist 348 6.11 Torsion Bar Springs 350 6.12 Curved Circular Bars 351 6.12.1 Helical Springs 352 References 354 Problems 355 Chapter 7 Numerical Methods 364 7.1 Introduction 364 Part A: Finite Difference Analysis 365 7.2 Finite Differences 365 7.2.1 Central Differences 366 7.3 Finite Difference Equations 368 7.4 Curved Boundaries 370 7.5 Boundary Conditions 373 Part B: Finite Element Analysis 377 7.6 Fundamentals 377 7.7 The Bar Element 379 7.7.1 Equilibrium Method 379 7.7.2 Energy Method 379 7.8 Arbitrarily Oriented Bar Element 380 7.8.1 Coordinate Transformation 380 7.8.2 Force Transformation 381 7.8.3 Displacement Transformation 383 7.8.4 Governing Equations 383 7.9 Axial Force Equation 384 7.10 Force-Displacement Relations for a Truss 386 7.10.1 The Assembly Process 386 7.11 Beam Element 393 xii Contents 7.12 Properties of Two-Dimensional Elements 399 7.12.1 Displacement Matrix 399 7.12.2 Strain, Stress, and Elasticity Matrices 401 7.13 General Formulation of the Finite Element Method 402 7.13.1 Outline of General Finite Element Analysis 403 7.14 Triangular Finite Element 407 7.14.1 Element Nodal Forces 410 7.15 Case Studies in Plane Stress 414 7.16 Computational Tools 423 References 423 Problems 424 Chapter 8 Thick-Walled Cylinders and Rotating Disks 434 8.1 Introduction 434 8.1.1 Basic Relations 434 8.2 Thick-Walled Cylinders Under Pressure 435 8.2.1 Special Cases 438 8.2.2 Closed-Ended Cylinder 440 8.3 Maximum Tangential Stress 441 8.4 Application of Failure Theories 442 8.5 Compound Cylinders: Press or Shrink Fits 443 8.6 Rotating Disks of Constant Thickness 446 8.6.1 Annular Disk 447 8.6.2 Solid Disk 448 8.7 Disk Flywheels 449 8.7.1 Design Factors 450 8.7.2 Stresses and Displacement 450 8.8 Rotating Disks of Variable Thickness 453 8.9 Rotating Disks of Uniform Stress 456 8.10 Thermal Stresses in Thin Disks 458 8.10.1 Annular Disk 459 8.10.2 Solid Disk 459 8.11 Thermal Stress in Long Circular Cylinders 460 8.11.1 Solid Cylinder 460 8.11.2 Cylinder with a Central Circular Hole 461 8.11.3 Special Case 464 8.12 Finite Element Solution 464 8.12.1 Axisymmetric Element 464 References 466 Problems 466 Chapter 9 Beams on Elastic Foundations 473 9.1 Introduction 473 9.2 General Theory 473 Contents xiii 9.3 Infinite Beams 475 9.4 Semi-Infinite Beams 480 9.5 Finite Beams 483 9.6 Classification of Beams 484 9.7 Beams Supported by Equally Spaced Elastic Elements 485 9.8 Simplified Solutions for Relatively Stiff Beams 486 9.9 Solution by Finite Differences 488 9.10 Applications 490 9.10.1 Grid Configurations of Beams 490 References 492 Problems 493 Chapter 10 Applications of Energy Methods 496 10.1 Introduction 496 Part A: Energy Principles 497 10.2 Work Done in Deformation 497 10.3 Reciprocity Theorem 498 10.4 Castigliano’s Theorem 499 10.4.1 Application to Bars and Beams 500 10.4.2 Application to Trusses 500 10.4.3 Use of a Fictitious Load 501 10.5 Unit- or Dummy-Load Method 506 10.6 Crotti–Engesser Theorem 508 10.7 Statically Indeterminate Systems 510 Part B: Variational Methods 514 10.8 Principle of Virtual Work 514 10.8.1 Variation in Strain Energy 514 10.8.2 Virtual Work Done by Forces 515 10.9 Principle of Minimum Potential Energy 515 10.10 Deflections by Trigonometric Series 517 10.10.1 Strain Energy 518 10.10.2 Virtual Work 518 10.11 Rayleigh–Ritz Method 522 References 524 Problems 525 Chapter 11 Stability of Columns 534 11.1 Introduction 534 11.2 Critical Load 534 11.2.1 Equilibrium Method 535 11.2.2 Energy Method 536 11.3 Buckling of Pin-Ended Columns 536 11.3.1 Modes of Buckling 538 11.4 Deflection Response of Columns 539 xiv Contents 11.4.1 Effects of Large Deflections 539 11.4.2 Effects of Imperfections 540 11.4.3 Effects of Inelastic Behavior 540 11.5 Columns with Different End Conditions 540 11.6 Critical Stress: Classification of Columns 543 11.6.1 Long Columns 543 11.6.2 Short Columns 544 11.6.3 Intermediate Columns: Inelastic Buckling 544 11.7 Design Formulas for Columns 548 11.8 Imperfections in Columns 550 11.9 Local Buckling of Columns 552 11.10 Eccentrically Loaded Columns: Secant Formula 552 11.10.1 Simplified Formula for Short Columns 554 11.11 Energy Methods Applied to Buckling 554 11.12 Solution by Finite Differences 562 11.13 Finite Difference Solution for Unevenly Spaced Nodes 567 References 568 Problems 569 Chapter 12 Plastic Behavior of Materials 578 12.1 Introduction 578 12.2 Plastic Deformation 579 12.2.1 Slip Action: Dislocation 579 12.3 Idealized Stress–Strain Diagrams 580 12.3.1 True Stress–True Strain Relationships 580 12.4 Instability in Simple Tension 582 12.5 Plastic Axial Deformation and Residual Stress 585 12.6 Plastic Deflection of Beams 588 12.7 Analysis of Perfectly Plastic Beams 590 12.7.1 Shape Factor 593 12.7.2 Plastic Hinge 593 12.8 Collapse Load of Structures: Limit Design 600 12.8.1 Collapse Mechanism 600 12.8.2 Ultimate Load by the Energy Method 601 12.9 Elastic–Plastic Torsion of Circular Shafts 605 12.9.1 Yield Torque 606 12.9.2 Elastic–Plastic Torque 606 12.9.3 Ultimate Torque 607 12.9.4 Residual Rotation and Stress 608 12.10 Plastic Torsion: Membrane Analogy 610 12.10.1 Membrane–Roof Analogy 610 12.10.2 Sand Hill Analogy 611 12.11 Elastic–Plastic Stresses in Rotating Disks 612 12.11.1 Initial Yielding 612 Contents xv 12.11.2 Partial Yielding 612 12.11.3 Complete Yielding 614 12.12 Plastic Stress–Strain Relations 614 12.13 Plastic Stress–Strain Increment Relations 620 12.14 Stresses in Perfectly Plastic Thick-Walled Cylinders 623 12.14.1 Complete Yielding 624 12.14.2 Partial Yielding 626 References 627 Problems 628 Chapter 13 Stresses in Plates and Shells 635 13.1 Introduction 635 Part A: Bending of Thin Plates 635 13.2 Basic Assumptions 635 13.3 Strain–Curvature Relations 636 13.4 Stress, Curvature, and Moment Relations 638 13.5 Governing Equations of Plate Deflection 640 13.6 Boundary Conditions 642 13.7 Simply Supported Rectangular Plates 644 13.8 Axisymmetrically Loaded Circular Plates 648 13.9 Deflections of Rectangular Plates by the Strain-Energy Method 650 13.10 Sandwich Plates 652 13.10.1 Design of Sandwich Beams and Plates 653 13.11 Finite Element Solution 654 13.11.1 Strain, Stress, and Elasticity Matrices 655 13.11.2 Displacement Function 655 13.11.3 Stiffness Matrix 657 13.11.4 External Nodal Forces 657 Part B: Membrane Stresses in Thin Shells 657 13.12 Theories and Behavior of Shells 657 13.13 Simple Membrane Action 658 13.14 Symmetrically Loaded Shells of Revolution 660 13.14.1 Equations of Equilibrium 661 13.14.2 Conditions of Compatibility 662 13.15 Some Typical Cases of Shells of Revolution 662 13.15.1 Spherical Shell 662 13.15.2 Conical Shell 663 13.15.3 Circular Cylindrical Shell 664 13.16 Thermal Stresses in Compound Cylinders 668 13.17 Cylindrical Shells of General Shape 670 References 673 Problems 673 xvi Contents Appendix A Problem Formulation and Solution 679 A.1 Basic Method 679 A.1.1 Numerical Accuracy 680 A.1.2 Daily Planning 680 Appendix B Solution of the Stress Cubic Equation 682 B.1 Principal Stresses 682 B.1.1 Direction Cosines 683 Appendix C Moments of Composite Areas 687 C.1 Centroid 687 C.2 Moments of Inertia 690 C.2.1 Parallel Axis Theorem 690 C.2.2 Principal Moments of Inertia 692 Appendix D Tables and Charts 699 D.1 Charts of Stress Concentration Factors 705 Appendix E Introduction to MATLAB 710 Answers to Selected Problems 713 Index 722 xvii Preface INTRODUCTION Advanced Mechanics of Materials and Applied Elasticity, Sixth Edition, is an outgrowth of classroom notes prepared in connection with advanced undergraduate and first-year graduate courses in the mechanics of solids and elasticity. It is designed to satisfy the requirements of courses subsequent to an elementary treatment of the strength of materi-als. In addition to its applicability to aeronautical, civil, and mechanical engineering and to engineering mechanics curricula, the text is useful to practicing engineers. Emphasis is given to numerical techniques (which lend themselves to computerization) in the solution of problems resisting analytical treatment. The attention devoted to numerical solutions is not intended to deny the value of classical analysis, which is given a rather full treatment. Instead, the coverage provided here seeks to fill what we believe to be a void in the world of textbooks.
We have attempted to present a balance between the theory necessary to gain insight into the mechanics, but which can often offer no more than crude approximations to real problems because of simplifications related to geometry and conditions of loading, and numerical solutions, which are so useful in presenting stress analysis in a more realistic setting. This text emphasizes those aspects of theory and application that prepare a student for more advanced study or for professional practice in design and analysis.
The theory of elasticity plays three important roles in the text. First, it provides exact solutions where the configurations of loading and boundary are relatively simple. Second, it provides a check on the limitations of the mechanics of materials approach. Third, it serves as the basis of approximate solutions employing numerical analysis.
To make the text as clear as possible, the fundamentals of the mechanics of materi-als are addressed as necessary. The physical significance of the solutions and practical applications are also emphasized. In addition, we have made a special effort to illus-trate important principles and applications with numerical examples. Consistent with announced national policy, problems are included in the text in which the physical quan-tities are expressed in the International System of Units (SI). All important quantities are defined in both SI and U.S. Customary System (USCS) of units. A sign convention, consistent with vector mechanics, is employed throughout for loads, internal forces, and stresses. This convention conforms to that used in most classical strength of materials and xviii Preface elasticity texts, as well as to that most often employed in the numerical analysis of com-plex structures.
ORGANIZATION OF THE TEXT Because of its extensive subdivision into a variety of topics and use of alternative meth-ods of analysis, this text provides great flexibility for instructors when choosing assign-ments to cover courses of varying length and content. Most chapters are substantially self- contained, so the order of presentation can be smoothly altered to meet an instructor’s preference. Ideally, Chapters 1 and 2, which address the analysis of basic concepts, should be studied first. The emphasis placed on the treatment of two-dimensional problems in elasticity (Chapter 3) may then differ according to the scope of the course.
This sixth edition of Advanced Mechanics of Materials and Applied Elasticity seeks to preserve the objectives and emphases of the previous editions. Every effort has been made to provide a more complete and current text through the inclusion of new material dealing with the fundamental principles of stress analysis and design: stress concentrations, contact stresses, failure criteria, fracture mechanics, compound cylinders, finite element analysis (FEA), energy and variational methods, buckling of stepped columns, common shell types, case studies in analysis and design, and MATLAB solutions. The entire text has been reex-amined, and many improvements have been made throughout by a process of elimination and rearrangement. Some sections have been expanded to improve on previous expositions.
The references (identified in brackets), which are provided as an aid to those students who wish to pursue certain aspects of a subject in further depth, have been updated and listed at the end of each chapter. We have resisted the temptation to increase the mate-rial covered except where absolutely necessary. Nevertheless, we have added a number of illustrative examples and problems important in engineering practice and design. Extra care has been taken in the presentation and solution of the sample problems. All the prob-lem sets have been reviewed and checked to ensure both their clarity and their numerical accuracy. Most changes in subject-matter coverage were prompted by the suggestions of faculty familiar with earlier editions.
In this sixth edition, we have maintained the previous editions’ clarity of presentation, simplicity as the subject permits, unpretentious depth, an effort to encourage intuitive understanding, and a shunning of the irrelevant. In this context, as throughout, emphasis is placed on the use of fundamentals to help build students’ understanding and ability to solve the more complex problems.
SUPPLEMENTS The book is accompanied by a comprehensive instructor’s Solutions Manual. Written and class tested, it features complete solutions to all problems in the text. Answers to selected problems are given at the end of the book. The password-protected Solutions Manual is available for adopters at the Pearson Instructor Resource Center, pearsonhighered.com/irc.
Preface xix Optional Material is also available from the Pearson Resource Center, pearsonhigh-ered.com/irc. This material includes PowerPoint slides of figures and tables, and solutions using MATLAB for a variety of sample problems of practical importance. The book, how-ever, is independent of any software package.
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xx Acknowledgments It is a particular pleasure to acknowledge the contributions of those who assisted in the evolution of the text. Thanks, of course, are due to the many readers who have contributed general ideas and to reviewers who have made detailed comments on previous editions. These notably include the following: F. Freudenstein, Columbia University; R. A. Scott, University of Michigan; M. W . Wilcox and Y. Chan Jian, Southern Methodist University; C. T. Sun, University of Florida; B. Koplik, H. Kountouras, K. A. Narh, R. Sodhi, and C. E. Wilson, New Jersey Institute of Technology; H. Smith, Jr., South Dakota School of Mines and Technology; B. P . Gupta, Gannon University; S. Bang, University of Notre Dame; B. Koo, University of Toledo; J. T. Easley, University of Kansas; J. A. Bailey, North Carolina State University; W . F. Wright, Vanderbilt University; R. Burks, SUNY Maritime College; G. E. O. Widera, University of Illinois; R. H. Koebke, University of South Carolina; B. M. Kwak, University of Iowa; G. Nadig, Widener University; R. L. Brown, Montana State University; S. H. Advani, West Virginia University; E. Nassat, Illinois Institute of Technology; R. I. Sann, Stevens Institute of Technology; C. O. Smith, University of Nebraska; J. Kempner, Polytechnic University of New Y ork; and P . C. Prister, North Dakota State University; R. Wetherhold, University of Buffalo, SUNY; and Shaofan Li, University of California at Berkeley.
Accuracy checking of the problems, proofreading, and typing of the Solutions Manual were done expertly by my former student, Dr. Y oungjin Chung. Also contribut-ing considerably to this volume with typing new inserts, scanning some figures, limited proofreading, and cover design was Errol A. Ugural. Their hard work is much appreciated. I am deeply indebted to my colleagues who have found the text useful through the years, as well as to Laura Lewin, executive editor at Pearson, who encouraged development of this edition. Lastly, I am very thankful for the support and understanding of my wife Nora, daughter Aileen, and son Errol during preparation of this book.
Ansel C. Ugural Holmdel, New Jersey xxi About the Authors Ansel C. Ugural, Ph.D., has been a research and visiting professor of mechanical and civil engineering at the New Jersey Institute of Technology. He has taught in the engineer-ing mechanics department at the University of Wisconsin. Dr. Ugural also served as chair-man and tenured professor of mechanical engineering at Fairleigh Dickinson University for twenty years. He has considerable and diverse industrial experience in both full-time and consulting capacities as a design, development, and research engineer.
Professor Ugural earned his MS in mechanical engineering and PhD in engineer-ing mechanics from the University of Wisconsin–Madison. Dr. Ugural was a National Science Foundation (NSF) fellow. He has been a member of several professional societies, including the American Society of Mechanical Engineers and the American Society of Engineering Education. He is also listed in Who’s Who in Engineering.
Dr. Ugural is the author of several books, including Mechanics of Materials; Mechanical Design: An Integrated Approach; Mechanical Design of Machine Components; Stresses in Beams, Plates, and Shells; and Plates and Shells: Theory and Analysis. Most of these texts have been translated into Korean, Chinese, and Portuguese. In addition, Professor Ugural has published numerous articles in trade and professional journals.
Saul K. Fenster, Ph.D., served as president and tenured professor at New Jersey Institute of Technology for more than two decades. In addition, he has held varied posi-tions at Fairleigh Dickinson University and taught at the City University of New Y ork. His experience includes membership on a number of corporate boards and economic development commissions. Fenster is a fellow of the American Society of Mechanical Engineers, the American Society for Engineering Education, and the American Society for Manufacturing Engineers. He is coauthor of a text on mechanics.
xxii Symbols Roman Letters A area B width C carryover factor, torsional rigidity c distance from neutral axis to outer fiber D distribution factor, flexural rigidity of plate [D] elasticity matrix d diameter, distance E modulus of elasticity in tension or compression Es modulus of plasticity or secant modulus Et tangent modulus e dilatation, distance, eccentricity {F} nodal force matrix of bar and beam finite elements F body force per unit volume, concentrated force f coefficient of friction { f} displacement function of finite element G modulus of elasticity in shear or modulus of rigidity g acceleration of gravity (≈9.81 m/s2) h depth of beam, height, membrane deflection, mesh width I moment of inertia of area, stress invariant J polar moment of inertia of area, strain invariant K bulk modulus, spring constant of an elastic support, stiffness factor, thermal con-ductivity, fatigue factor, strength coefficient, stress concentration factor [K ] stiffness matrix of whole structure k constant, modulus of elastic foundation, spring constant [k] stiffness matrix of finite element L length, span l, m, n direction cosines M moment Mxy twisting moment in plates m moment caused by unit load Symbols xxiii N fatigue life (cycles), force n factor of safety, number, strain hardening index P concentrated force p distributed load per unit length or area, pressure, stress resultant Q first moment of area, heat flow per unit length, shearing force {Q} nodal force matrix of two-dimensional finite element R radius, reaction r radius, radius of gyration r, θ polar coordinates S elastic section modulus, shear center s distance along a line or a curve T temperature, twisting couple or torque t thickness U strain energy Uo strain energy per unit volume U complementary energy u, u, w components of displacement V shearing force, volume u velocity W weight, work x, y, z rectangular coordinates Z plastic section modulus Greek Letters α angle, coefficient of thermal expansion, form factor for shear b numerical factor, angle γ shear strain, weight per unit volume or specific weight, angle δ deflection, finite difference operator, variational symbol, displacement {δ} nodal displacement matrix of finite element change of a function ε normal strain θ angle, angle of twist per unit length, slope ν Poisson’s ratio λ axial load factor, Lamé constant Π potential energy ρ density (mass per unit volume), radius σ normal stress τ shear stress φ total angle of twist F stress function ω angular velocity ψ stream function ∆ This page intentionally left blank 1 C H A P T E R 1 1.1 INTRODUCTION There are two major parts to this chapter. Review of some important fundamentals of stat-ics and mechanics of solids, the concept of stress, modes of load transmission, the general sign convention for stress and force resultants that will be used throughout the book, and analysis and design principles are provided first. This is followed by treatment of chang-ing the components of the state of stress given in one set of coordinate axes to any other set of rotated axes, as well as variation of stress within and on the boundaries of a load-carrying member. Plane stress and its transformation are of basic importance, since these conditions are most common in engineering practice. This chapter is therefore also a brief guide and introduction to the remainder of the text.
1.1.1 Mechanics of Materials and Theory of Elasticity The basic structure of matter is characterized by nonuniformity and discontinuity attribut-able to its various subdivisions: molecules, atoms, and subatomic particles. Our concern in this text is not with the particulate structure, however. Instead, it will be assumed that the matter with which we are concerned is homogeneous and continuously distributed over its volume. There is the clear implication in such an approach that the smallest element cut from the body possesses the same properties as the body. Random fluctuations in the properties of the material are, therefore, of no consequence. This approach is that of con-tinuum mechanics, in which solid elastic materials are treated as though they are continu-ous media rather than composed of discrete molecules. Of the states of matter, we are here concerned only with the solid, with its ability to maintain its shape without the need of a container and to resist continuous shear, tension, and compression.
In contrast with rigid-body statics and dynamics, which treat the external behavior of bodies (that is, the equilibrium and motion of bodies without regard to small deformations Analysis of Stress 2 Chapter 1 Analysis of Stress associated with the application of load), the mechanics of solids is concerned with the relationships of external effect (forces and moments) to internal stresses and strains. Two notable approaches used in solid mechanics are the mechanics of materials or elementary theory (also called technical theory) and the theory of elasticity. The mechanics of materials focuses mainly on the more or less approximate solutions of practical problems. The theory of elasticity concerns itself largely with more mathematical analysis to determine the “exact” stress and strain distributions in a loaded body. The difference between these approaches lies primarily in the nature of the simplifying assumptions used, described in Section 3.1.
External forces acting on a body may be classified as surface forces and body forces. A surface force is of the concentrated type when it acts at a point; a surface force may also be distributed uniformly or nonuniformly over a finite area. Body forces are associ-ated with the mass rather than the surfaces of a body, and are distributed throughout the volume of a body. Gravitational, magnetic, and inertia forces are all body forces. They are specified in terms of force per unit volume. All forces acting on a body, including the reactive forces caused by supports and body forces, are considered to be external forces. Internal forces are the forces that hold together the particles forming the body. Unless oth-erwise stated, we assume in this text that body forces can be neglected and that forces are applied steadily and slowly. The latter is referred to as static loading.
In the International System of Units (SI), force is measured in newtons (N). Because the newton is a small quantity, the kilonewton (kN) is often used in practice. In the U.S. Customary System (USCS), force is expressed in pounds (lb) or kilopounds (kips). We define all important quantities in both systems of units. However, in numerical examples and problems, SI units are used throughout the text consistent with international conven-tion. (Table D.2 compares the two systems.) 1.1.2 Historical Development The study of the behavior of members in tension, compression, and bending began with Leonardo da Vinci (1452–1519) and Galileo Galilei (1564–1642). For a proper under-standing, however, it was necessary to establish accurate experimental description of a material’s properties. Robert Hooke (1615–1703) was the first to point out that a body is deformed subject to the action of a force. Sir Isaac Newton (1642–1727) developed the concepts of Newtonian mechanics that became key elements of the strength of materials.
Leonard Euler (1707–1783) presented the mathematical theory of columns in 1744. The renowned mathematician Joseph-Louis Lagrange (1736–1813) received credit for developing a partial differential equation to describe plate vibrations. Thomas Y oung (1773–1829) established a coefficient of elasticity, Y oung’s modulus. The advent of rail-roads in the late 1800s provided the impetus for much of the basic work in this area. Many famous scientists and engineers, including Coulomb, Poisson, Navier, St. Venant, Kirchhoff, and Cauchy, were responsible for advances in mechanics of materials during the eighteenth and nineteenth centuries. The British physicist William Thomas Kelvin (1824–1907), better known by his knighted name, Sir Lord Kelvin, first demonstrated that torsional moments acting at the edges of plates could be decomposed into shearing forces. The prominent English mathematician Augustus Edward Hough Love (1863–1940) intro-duced simple analysis of shells, known as Love’s approximate theory.
1.2 Scope of the Book 3 Over the years, most basic problems of solid mechanics had been solved. Stephan P . Timoshenko (1878–1972) made numerous original contributions to the field of applied mechanics and wrote pioneering textbooks on the mechanics of materials, theory of elas-ticity, and theory of elastic stability. The theoretical base for modern strength of materials had been developed by the end of the nineteenth century. Since that time, problems asso-ciated with the design of aircraft, space vehicles, and nuclear reactors have led to many studies of the more advanced phases of the subject. Consequently, the mechanics of mate-rials is being expanded into the theories of elasticity and plasticity.
In 1956, Turner, Clough, Martin, and Topp introduced the finite element method, which permits the numerical solution of complex problems in solid mechanics in an eco-nomical way. Many contributions in this area are owed to Argyris and Zienkiewicz. The recent trend in the development is characterized by heavy reliance on high-speed comput-ers and by the introduction of more rigorous theories. Numerical methods presented in Chapter 7 and applied in the subsequent chapters have clear application to computation by means of electronic digital computers. Research in the foregoing areas is ongoing, not only to meet demands for treating complex problems, but also to justify further use and limitations on which the theory of solid mechanics is based.
Although a widespread body of knowledge exists at present, mechanics of materi-als and elasticity remain fascinating subjects, as their areas of application continue to expand. The literature dealing with various aspects of solid mechanics is voluminous. For those seeking more thorough treatment, selected references are identified in brackets and compiled at the end of each chapter.
1.2 SCOPE OF THE BOOK As stated in the preface, this book is intended for advanced undergraduate and gradu-ate engineering students as well as engineering professionals. To make the text as clear as possible, attention is given to the fundamentals of solid mechanics and chapter objec-tives. A special effort has been made to illustrate important principles and applications with numerical examples. Emphasis is placed on a thorough presentation of both classical topics in advanced mechanics of materials and applied elasticity and selected advanced topics. The physical behavior of members is first explained, and this behavior is then mod-eled to develop the theory.
The usual objective of the mechanics of materials and theory of elasticity is the exam-ination of the load-carrying capacity of a body from three standpoints: strength, stiffness, and stability. Recall that these quantities relate, respectively, to the ability of a member to resist permanent deformation or fracture, to resist deflection, and to retain its equilibrium configuration. For instance, when loading produces an abrupt shape change of a member, instability occurs; similarly, an inelastic deformation or an excessive magnitude of deflec-tion in a member will cause malfunction in normal service. Based on the fundamental principles (Section 1.3), these behaviors are discussed in later chapters for various types Historical reviews of the mechanics of materials and the theory of elasticity are given in Refs. 1.1 through 1.3.
4 Chapter 1 Analysis of Stress of structural members. Failure by yielding and fracture of the materials under combined loading is taken up in detail in Chapter 4.
Our main concern is the analysis of stress and deformation within a loaded body, which is accomplished by application of one of the methods described in the next section. For this purpose, the analysis of loads is essential. A structure or machine cannot meet its expectations unless its design is based on realistic operating loads. The principal topics under the heading of mechanics of solids may be summarized as follows: 1. Analysis of the stresses and deformations within a body subject to a prescribed system of forces. This is accomplished by solving the governing equations that describe the stress and strain fields (theoretical stress analysis). It is often advantageous, where the shape of the structure or conditions of loading preclude a theoretical solution or where verification is required, to apply the laboratory techniques of experimental stress analysis.
2. Determination by theoretical analysis or by experiment of the limiting values of load that a structural element can sustain without suffering damage, failure, or compromise of function.
3. Determination of the body shape and selection of the materials that are most effi-cient for resisting a prescribed system of forces under specified conditions of opera-tion, such as temperature, humidity, vibration, and ambient pressure. This is the design function.
The design function, item 3, clearly relies on the theoretical analyses results obtained via items 1 and 2; thus, this text focuses on those topics. In particular, emphasis is placed on the development of the equations and methods by which detailed analysis can be accomplished.
The ever-increasing industrial demand for more sophisticated structures and machines calls for a good grasp of the concepts of stress and strain and the behavior of materials—and a considerable degree of ingenuity. This text, at the very least, provides the student with the ideas and information necessary for an understanding of the advanced mechanics of solids and encourages use of the creative process based on that understanding. Complete, carefully drawn free-body diagrams are used to visualize the processes involved, though the subject matter can be learned best by solving problems of practical importance. A thorough grasp of fundamentals will prove of great value in attacking new and unfamiliar problems.
1.3 ANALYSIS AND DESIGN Throughout this text, a fundamental procedure for analysis in solving mechanics of sol-ids problems is used repeatedly. The complete analysis of load-carrying structural mem-bers by the method of equilibrium requires consideration of three conditions related to certain laws of forces, laws of material deformation, and geometric compatibility. These essential relationships, called the basic principles of analysis, are as follows: 1. Equilibrium conditions. The equations of equilibrium of forces must be satisfied throughout the member.
1.3 Analysis and Design 5 2. Material behavior. The stress–strain or force-deformation relations (for example, Hooke’s law) must apply to the behavior of the material of which the member is constructed.
3. Geometry of deformation. The compatibility conditions of deformations must be sat-isfied: that is, each deformed portion of the member must fit together with adjacent portions. (Matter of compatibility is not always broached in mechanics of materials analysis.) The stress and deformation obtained through the use of these three principles must conform to the conditions of loading imposed at the boundaries of a member. This cir-cumstance is known as satisfying the boundary conditions. Applications of the preced-ing procedure are illustrated in the problems presented in this text. Note, however, that it is not always necessary to execute an analysis in the precise sequence of steps listed previously.
As an alternative to the equilibrium methods, the analysis of stress and deformation can be accomplished by employing energy methods (Chapter 10), which are based on the concept of strain energy. Both the equilibrium and the energy approaches can provide solutions of acceptable accuracy where configurations of loading and member shape are regular, and they can be used as the basis of numerical methods in the solution of more realistic problems.
Engineering design is the process of applying science and engineering techniques to define a structure or system in detail to allow its realization. The objective of a mechani-cal design procedure includes finding the proper materials, dimensions, and shapes of the members of a structure or machine so that they will support the prescribed loads and perform without failure. Machine design entails creating new or improved machines to accomplish specific purposes. Usually, structural design deals with any engineering disci-pline that requires a structural member or system.
Design is the essence, art, and intent of engineering. A good design satisfies perfor-mance, cost, and safety requirements. An optimum design is the best solution to a design problem within given restrictions. Efficiency of the optimization may be gaged by such criteria as minimum weight or volume, optimum cost, and any other standard deemed appropriate. When faced with a design problem characterized by many choices, a designer may often make decisions on the basis of past experience, so as to reduce the problem to a single variable. A solution to determine the optimum result becomes straightforward in such a situation.
A plan for satisfying a need usually includes preparation of individual preliminary design. Each preliminary design involves a thorough consideration of the loads and actions that the structure or machine must support. For each situation, an analysis is necessary. Design decisions—that is, choosing reasonable values of the safety factors and material properties—are significant in the preliminary design process. We note that the design of numerous structures, such as pressure vessels, space missiles, aircrafts, dome roofs, and bridge decks, is based on the theories of plates and shells. For exam-ple, a water storage tank can be satisfactorily designed using the shell-membrane theory (Section 13.12). By comparison, the design of a missile casing demands a more precise shell-bending theory so as to minimize weight and materials. Similarly, the design of 6 Chapter 1 Analysis of Stress a nozzle-to-cylinder junction in a nuclear reactor may necessitate an elaborate finite element analysis.
1.3.1 Role of Analysis in Design This text provides an elementary treatment of the concept of “design to meet strength requirements” as those requirements relate to individual machine or structural compo-nents. That is, the geometric configuration and material of a component are preselected and the applied loads are specified. Then, the basic formulas for stress are employed to select members of adequate size in each case. The role of analysis in design may be observed best in examining the phases of a design process. The following is rational pro-cedure in the design of a load-carrying member: 1. Evaluate the most likely modes of failure of the member. Failure criteria that predict the various modes of failure under anticipated conditions of service are discussed in Chapter 4.
2. Determine the expressions relating applied loading to such effects as stress, strain, and deformation. Often, the member under consideration and conditions of loading are so significant or so amenable to solution as to have been the subject of prior analysis. For these situations, textbooks, handbooks, journal articles, and technical papers are good sources of information. If the situation is unique, however, a mathematical derivation specific to the case at hand is required.
3. Determine the maximum usable value of stress, strain, or energy. This value is obtained either by reference to compilations of material properties or by experimental means such as simple tension test and is used in connection with the relationship derived in step 2.
4. Select a design factor of safety. This is to account for uncertainties in a number of aspects of the design, including those related to the actual service loads, material prop-erties, or environmental factors. An important area of uncertainty is connected with the assumptions made in the analysis of stress and deformation. Also, we are not likely to have a secure knowledge of the stresses that may be introduced during machining, assembly, and shipment of the element.
The design factor of safety also reflects the consequences of failure—for example, the possibility that failure will result in loss of human life or injury or in costly repairs or danger to other components of the overall system. For these reasons, the design factor of safety is also sometimes called the factor of ignorance. The uncertainties encountered during the design phase may be of such magnitude as to lead to a design carrying extreme weight, volume, or cost penalties. It may then be advantageous to perform thorough tests or more exacting analysis rather to rely on overly large design factors of safety.
1.3.2 Selection of Factor of Safety The true factor of safety, usually referred to simply as the factor of safety, can be deter-mined only after the member is constructed and tested. This factor is the ratio of the maximum load that the member can sustain under severe testing without failure to the 1.3 Analysis and Design 7 maximum load that is actually carried under normal service conditions (the working load). When a linear relationship exists between the load and the stress produced by the load, the factor of safety n may be expressed as = maximum usable stress allowable stress n (1.1) Maximum usable stress represents either the yield stress or the ultimate stress. The allow-able stress is the working stress. The factor of safety must be greater than 1.0 if failure is to be avoided. Modern engineering design accounts for all possible environmental, load-ing, stress, and material conditions, leaving relatively few items of uncertainty to be cov-ered by a factor of safety. Values for the factor of safety, selected by the designer on the basis of experience and judgment, range from approximately 1.25 to 4.
In the nuclear reactor industries, the safety factor is of prime significance in the face of many unknown effects; hence the factor of safety may be as high as 5. The use of a factor of safety in design is a reliable, time-proven approach. If this factor is properly selected, sound and safe designs are obtained by using it. For most applications, appropri-ate factors of safety are found in various construction and manufacturing codes. A con-cept closely related to safety factor is reliability defined as the statistical measure of the probability that a member will not fail in service [Ref. 1.4].
The procedure outlined in Section 1.3.1 is not always conducted in as formal a fash-ion as may have been implied in that discussion. In some design phases, one or more steps may be regarded as unnecessary or obvious on the basis of previous experience. Suffice it to say that complete design solutions are not unique, involve a consideration of many fac-tors, and often require a trial-and-error process. Stress is just one consideration in design. Other phases of the design of components include the prediction of the deformation of a given component under given loading and the consideration of buckling. The methods of determining deformation are discussed in later chapters of this text. Note that analysis and design are closely related, and the examples and problems that appear throughout this book illustrate that connection.
We conclude this section with an appeal to the reader to exercise a degree of skepti-cism when applying formulas for which the limitations of use or the areas of applicability are uncertain. The relatively simple form of many formulas usually results from rather severe restrictions in the formula’s derivation. These limitations may include simplified boundary conditions and shapes, limitations on stress and strain, and the neglect of certain complicating factors. Designers and stress analysts must be aware of such restrictions lest their work be of no value or, even worse, lead to dangerous inadequacies.
In this chapter, we focus on the state of stress at a point and the variation of stress throughout an elastic body. The latter is dealt with in Sections 1.8 and 1.16 and the former in the balance of the chapter.
1.3.3 Case Studies A general case study in analysis may move step by step through the problem formula-tion and solution stages, as outlined in Appendix A. The basic geometry and loading on a member must be selected before any analysis can be done. For example, the stress that 8 Chapter 1 Analysis of Stress would occur in a bar under a load would depend on whether the loading gives rise to tension, transverse shear, direct shear, torsion, bending, or contact stresses. In this case, uniform stress patterns may be more efficient at carrying the load than others. Therefore, by carefully studying the types of loads and stress patterns that can arise in structures, some insight can be gained into improved shapes and orientations of components. A few case studies introduced in this text involve situations encountered during the analysis and design of structural members.
1.4 CONDITIONS OF EQUILIBRIUM A structure is a unit consisting of interconnected members supported in such a way that it is capable of carrying loads in static equilibrium. Structures are of four general types: frames, trusses, machines, and thin-walled (plate and shell) structures. Frames and machines are structures containing multiforce members. The former support loads and are usually stationary, fully restrained structures. The latter transmit and modify forces (or power) and always contain moving parts. The truss provides both a practical and an economical solution, particularly in the design of bridges and buildings. When the truss is loaded at its joints, the only force in each member is an axial force, either tensile or compressive.
The analysis and design of structural and machine components require a knowledge of the distribution of forces within such members. Fundamental concepts and conditions of static equilibrium provide the necessary background for the determination of inter-nal as well as external forces. In Section 1.6, we shall see that components of internal-forces resultants have special meaning in terms of the type of deformations they cause, as applied, for example, to slender members. The surface forces that develop at support points of a structure, which are called reactions, equilibrate the effects of the applied loads on the structures.
The equilibrium of forces is the state in which the forces applied on a body are in balance. Newton’s first law states that if the resultant force acting on a particle (the sim-plest body) is zero, the particle will remain at rest or will move with constant velocity. Statics is concerned essentially with the case where the particle or body remains at rest. A complete free-body diagram is essential in the solution of equilibrium.
Let us consider the equilibrium of a body in space. In this three-dimensional case, the conditions of equilibrium require the satisfaction of the following equations of statics: F F F M M M x y z x y z Σ = Σ = Σ = Σ = Σ = Σ = 0 0 0 0 0 0 (1.2) These equations state that the sum of all forces acting on a body in any direction must be zero; the sum of all moments about any axis must be zero.
In a planar problem, where all forces act in a single (xy) plane, there are only three independent equations of statics: F F M x y A Σ = Σ = Σ = 0 0 0 (1.3) 1.5 Definition and Components of Stress 9 That is, the sum of all forces in any (x, y) directions must be zero, and the resultant moment about axis z or any point A in the plane must be zero. By replacing a force sum-mation with an equivalent moment summation in Eqs. (1.3), the following alternative sets of conditions are obtained: F M M x A B Σ = Σ = Σ = 0 0 0 (1.4a) provided that the line connecting the points A and B is not perpendicular to the x axis, or M M M A B C Σ = Σ = Σ = 0 0 0 (1.4b) if points A, B, and C are not collinear. Clearly, the judicious selection of points for taking moments can often simplify the algebraic computations.
A structure is statically determinate when all forces on its members can be found by using only the conditions of equilibrium. If there are more unknowns than available equations of statics, the problem is called statically indeterminate. The degree of static indeterminacy is equal to the difference between the number of unknown forces and the number of relevant equilibrium conditions. Any reaction that is in excess of those that can be obtained by statics alone is termed redundant. Thus, the number of redundants is the same as the degree of indeterminacy.
1.5 DEFINITION AND COMPONENTS OF STRESS Stress and strain are most important concepts for a comprehension of the mechanics of solids. They permit the mechanical behavior of load-carrying components to be described in terms fundamental to the engineer. Both the analysis and the design of a given machine or structural element involve the determination of stress and material stress–strain rela-tionships. The latter is taken up in Chapter 2.
Consider a body in equilibrium subject to a system of external forces, as shown in Fig. 1.1a. Under the action of these forces, internal forces are developed within the body. To examine these forces at some interior point Q, we use an imaginary plane to cut the body at a section a–a through Q, dividing the body into two parts. As the forces act-ing on the entire body are in equilibrium, the forces acting on one part alone must be in a Q a External forces Internal forces (a) y Q ∆A ∆F ∆Fy ∆Fz ∆Fx ∆A x z Q (c) (b) Figure 1.1. Method of sections: (a) sectioning of a loaded body; (b) free body with external and internal forces; (c) enlarged area ΔA with components of the force ΔF.
10 Chapter 1 Analysis of Stress equilibrium, which requires the presence of forces on plane a–a. These internal forces, which are applied to both parts, are distributed continuously over the cut surface. This process, referred to as the method of sections (Fig. 1.1), serves as the first step in solving all problems involving the investigation of internal forces.
A free-body diagram (FBD) is simply a sketch of a body with all the appropriate forces, both known and unknown, acting on it. Figure 1.1b shows such a plot of the iso-lated left part of the body. An element of area ∆ A, located at point Q on the cut surface, is acted on by force Δ F. Let the origin of coordinates be placed at point Q, with x normal and y, z tangent to ∆ A. In general, Δ F does not lie along x, y, or z. Decomposing Δ F into components parallel to x, y, and z (Fig. 1.1c), we define the normal stress σ x and the shearing stresses τ xy and τ xz: lim lim , lim 0 0 0 σ τ τ = ∆ ∆ = = ∆ ∆ = = ∆ ∆ = ∆→ ∆→ ∆→ F A dF dA F A dF dA F A dF dA x A x x xy A y y xz A z z (1.5) These definitions provide the stress components at a point Q to which the area ∆ A is reduced in the limit. Clearly, the expression ∆ A → 0 depends on the idealization discussed in Section 1.1.1. Our concern is with the average stress on areas, which, though small compared with the size of the body, is large compared with the interatomic distances in the solid. Stress is thus defined adequately for engineering purposes. As shown in Eq. (1.5), the intensity of force perpendicular, or normal, to the surface is termed the normal stress at a point, while the intensity of force parallel to the surface is called the shearing stress at a point.
The values obtained in the limiting process of Eq. (1.5) differ from point to point on the surface as Δ F varies. The stress components depend not only on Δ F, however, but also on the orientation of the plane on which it acts at point Q. Even at a given point, therefore, the stresses will differ as different planes are considered. The complete descrip-tion of stress at a point thus requires the specification of the stress on all planes passing through the point.
Because the stress (σ or τ ) is obtained by dividing the force by area, it has units of force per unit area. In SI units, stress is measured in newtons per square meter (N/m2) or pascals (Pa). As the pascal is a very small quantity, the megapascal (MPa) is commonly used. When USCS units are used, stress is expressed in pounds per square inch (psi) or kips per square inch (ksi).
In Section 1.12, we note that to determine the stresses on an infinite number of planes passing through a point Q, thereby defining the stresses at that point, we need only specify the stress components on three mutually perpendicular planes passing through the point. These three planes, perpendicular to the coordinate axes, contain three hidden sides of an infinitesimal cube (Fig. 1.2). When we move from point Q to point Q′ , the values of stress will, in general, change. Also, body forces can exist. However, these cases are not discussed here (see Section 1.8), as we are now merely interested in establishing the ter-minology necessary to specify a stress component.
1.5 Definition and Components of Stress 11 The general case of a three-dimensional state of stress is shown in Fig. 1.2. Consider the stresses to be identical at points Q and Q′ and uniformly distributed on each face, rep-resented by a single vector acting at the center of each face. A total of nine scalar stress components then defines the state of stress at a point. These stress components can be assembled in a matrix form, wherein each row represents the group of stresses acting on a plane passing through Q(x, y, z): τ τ τ τ τ τ τ τ τ τ σ τ τ τ σ τ τ τ σ = = [ ] ij xx xy xz yx yy yz zx zy zz x xy xz yx y yz zx zy z (1.6) In indicial notation (refer to Section 1.17), a stress component is written as τ ij, where the subscripts i and j each assume the values of x, y, and z as required by Eq. (1.6). The double-subscript notation is interpreted as follows: The first subscript indicates the direc-tion of a normal to the plane or face on which the stress component acts; the second sub-script relates to the direction of the stress itself. Repetitive subscripts are avoided in this text, so the normal stresses τ xx, τ yy, and τ zz are designated as shown in Eq. (1.6). A face or plane is usually identified by the axis normal to it; for example, the x faces are perpen-dicular to the x axis.
1.5.1 Sign Convention Referring again to Fig. 1.2, we observe that both stresses labeled τ yx tend to twist the ele-ment in a clockwise direction. It would be convenient, therefore, if a sign convention were adopted under which these stresses carried the same sign. Applying a convention relying solely on the coordinate direction of the stresses would clearly not produce the desired result, inasmuch as the τ yx stress acting on the upper surface is directed in the positive x direction, while τ yx acting on the lower surface is directed in the negative x direction. The following sign convention, which applies to both normal and shear stresses, is related to the deformational influence of a stress and is based on the relationship between the direc-tion of an outward normal drawn to a particular surface and the directions of the stress components on the same surface.
y τyy = σy τzz = σz τxx = σx τyz τyx τyx τxy τxz τzx τzy dx dy dz x z σx σy Q Q Figure 1.2. Element subjected to three- dimensional (3-D) stress. All stresses have positive sense.
12 Chapter 1 Analysis of Stress When both the outer normal and the stress component face in a positive direction relative to the coordinate axes, the stress is positive. When both the outer normal and the stress component face in a negative direction relative to the coordinate axes, the stress is positive. When the normal points in a positive direction while the stress points in a negative direction (or vice versa), the stress is negative. In accordance with this sign con-vention, tensile stresses are always positive and compressive stresses always negative. Figure 1.2 depicts a system of positive normal and shear stresses.
1.5.2 Equality of Shearing Stresses We now examine properties of shearing stress by studying the equilibrium of forces (see Section 1.4) acting on the cubic element shown in Fig. 1.2. As the stresses acting on oppo-site faces (which are of equal area) are equal in magnitude but opposite in direction, trans-lational equilibrium in all directions is assured; that is, Σ Fx = 0, Σ Fy = 0, and Σ Fz = 0. Rotational equilibrium is established by taking moments of the x-, y-, and z-directed forces about point Q, for example. From Σ Mz = 0, (-τ xy dy dz)dx + (τ yx dx dz)dy = 0 Simplifying, τ τ = xy yx (1.7a) Likewise, from Σ My = 0 and Σ Mx = 0, we have , τ τ τ τ = = xz zx yz zy (1.7b) Hence, the subscripts for the shearing stresses are commutative, and the stress tensor (see Section 1.9.1) is symmetric. This means that shearing stresses on mutually perpendicular planes of the element are equal. Therefore, no distinction will hereafter be made between the stress components τ xy and τ yx, τ xz and τ zx, or τ yz and τ zy. In Section 1.8, it is shown rig-orously that this assumption is valid even when stress components vary from one point to another.
1.5.3 Some Special Cases of Stress Under particular circumstances, the general state of stress (Fig. 1.2) reduces to simpler stress states, as briefly described here. These stresses, which are commonly encountered in practice, are given detailed consideration throughout the text.
a. Triaxial stress. In Section 1.13, we will see that an element subjected to only stresses σ 1, σ 2, and σ 3 acting in mutually perpendicular directions is said to be in a state of tri-axial stress. Such a state of stress can be written as follows: σ σ σ 0 0 0 0 0 0 1 2 3 (a) 1.6 Internal Force Resultant and Stress Relations 13 The absence of shearing stresses indicates that the preceding stresses are the principal stresses for the element. A special case of triaxial stress, known as spherical or dilata-tional stress, occurs if all principal stresses are equal (see Section 1.14). Equal triaxial tension is sometimes called hydrostatic tension. An example of equal triaxial compres-sion is found in a small element of liquid under static pressure.
b. Two-dimensional or plane stress. In this case, only the x and y faces of the element are subjected to stress, and all the stresses act parallel to the x and y axes, as shown in Fig. 1.3a.
The plane stress matrix is written in the form σ τ τ σ x xy xy y (1.8) Although the three-dimensional nature of the element under stress should not be for-gotten, for the sake of convenience we usually draw only a two-dimensional view of the plane stress element (Fig. 1.3b). When only two normal stresses are present, the state of stress is called biaxial. These stresses occur in thin plates stressed in two mutually perpendicular directions.
c. Pure shear. In this case, the element is subjected to plane shearing stresses only—for example, τ xy and τ yx (Fig.1.3c). Typical pure shear occurs over the cross sections and on longitudinal planes of a circular shaft subjected to torsion.
d. Uniaxial stress. When normal stresses act along one direction only, the one- dimensional state of stress is referred to as a uniaxial tension or compression.
1.6 INTERNAL FORCE RESULTANT AND STRESS RELATIONS Distributed forces within a load-carrying member can be represented by a statically equivalent system consisting of a force and a moment vector acting at any arbitrary point (usually the centroid) of a section. These internal force resultants, also called stress resul-tants, exposed by an imaginary cutting plane containing the point through the member, τxy = τyx τxy = τyx τxy =τyx τyx τyx τyx τyx τyx τxy τxy x x x z y y y σx σx σx σx σy σy σy σy (a) (b) (c) Figure 1.3. (a) Element in plane stress; (b) two-dimensional (2-D) presentation of plane stress; (c) element in pure shear.
14 Chapter 1 Analysis of Stress are usually resolved into components normal and tangent to the cut section (Fig. 1.4). The sense of moments follows the right-hand screw rule, often represented by double-headed vectors, as shown in the figure. Each component can be associated with one of four modes of force transmission: 1. The axial force P or N tends to lengthen or shorten the member.
2. The shear forces Vy and Vz tend to shear one part of the member relative to the adjacent part and are often designated by the letter V.
3. The torque or twisting moment T is responsible for twisting the member.
4. The bending moments My and Mz cause the member to bend and are often identified by the letter M.
A member may be subject to any or all of these modes simultaneously. Note that the same sign convention is used for the force and moment components that is used for stress; a positive force (or moment) component acts on the positive face in the positive coordinate direction or on a negative face in the negative coordinate direction.
A typical infinitesimal area dA of the cut section shown in Fig. 1.4 is acted on by the components of an arbitrarily directed force dF, expressed using Eq. (1.5) as dFx = σ x dA, dFy = τ xy dA, and dFz = τ xz dA. Clearly, the stress components on the cut section cause the internal force resultants on that section. Thus, the incremental forces are summed in the x, y, and z directions to give ∫ ∫ ∫ σ τ τ = = = P dA V dA V dA x y xy z xz , , (1.9a) In a like manner, the sums of the moments of the same forces about the x, y, and z axes lead to ∫ ∫ ∫ τ τ σ σ = − = = − T y z dA M zdA M ydA xz xy y x z x ( ) , , (1.9b) where the integrations proceed over area A of the cut section. Equations (1.9) represent the relations between the internal force resultants and the stresses. In the next section, we illustrate the fundamental concept of stress and observe how Eqs. (1.9) connect internal force resultants and the state of stress in a specific case.
y x z My Vy Vz Mz dFx dFy dFz z y P T A dA Figure 1.4. Positive forces and moments on a cut section of a body and components of the force dF on an infinitesimal area dA.
1.6 Internal Force Resultant and Stress Relations 15 1.6.1 Basic Formulas for Stress Consider a homogeneous prismatic bar loaded by axial forces P at the ends (Fig. 1.5a). A prismatic bar is a straight member having constant cross-sectional area throughout its length. To obtain an expression for the normal stress, we make an imaginary cut (section a–a) through the member at right angles to its axis. A free-body diagram of the isolated part is shown in Fig. 1.5b, wherein the stress is substituted on the cut section as a replace-ment for the effect of the removed part. Equilibrium of axial forces requires that P = ∫σ x dA or P = Aσ x. The normal stress is therefore σ = P A x (1.10) where A is the cross-sectional area of the bar.
Because Vy, Vz, and T all are equal to zero, the second and third of Eqs. (1.9a) and the first of Eqs. (1.9b) are satisfied by τ xy = τ xy = 0. Also, My = Mz = 0 in Eqs. (1.9b) requires only that σ x be symmetrically distributed about the y and z axes, as depicted in Fig. 1.5b. When the member is being extended as in the figure, the resulting stress is a uniaxial ten-sile stress; if the direction of forces were reversed, the bar would be in compression under uniaxial compressive stress. In the latter case, Eq. (1.10) is applicable only to chunky or short members owing to other effects that take place in longer members. Similarly, application of Eqs. (1.9) to torsion members, beams, plates, and shells is presented in this text as the subject unfolds, following the derivation of stress–strain rela-tions and examination of the geometric behavior of a particular member. Applying the method of the mechanics of materials, we develop other elementary formulas for stress and deformation. Also called the basic formulas of mechanics of materials, these formu-las are often used and extended for application to more complex problems in advanced mechanics of materials and the theory of elasticity.
For reference purposes in preliminary discussions, Table 1.1 lists some commonly encountered cases. In thin-walled vessels (r/t ≤ 10), there is often no distinction made between the inner and outer radii because they are nearly equal. In the mechanics of materials, r denotes the inner radius. After all, the more accurate shell theory (Section 13.12) is based on the average radius, which we use throughout this text. Each equa-tion presented in the table describes a state of stress associated with a single force, Further discussion of uniaxial compression stress is found in Section 11.6, where we take up the classification of columns.
a a (a) (b) z x y P P A σx P Figure 1.5. (a) Prismatic bar in tension; (b) stress distribution across cross section.
16 Chapter 1 Analysis of Stress Table 1.1. Commonly Used Elementary Formulas for Stressa 1. Prismatic Bars of Linearly Elastic Material P x T M V x N.A.
c y z y τmax τmax σx τ ρ r Axial loading: P A x σ = (a) Torsion: T J τ ρ = , max Tr J τ = (b) Bending: My I x σ = − , max Mc I σ = (c) Shear: VQ Ib xy τ = (d) 2. Thin-Walled Pressure Vessels σθ σa σ σ Cylinder: σ = θ pr t , 2 pr t a σ = (e) Sphere: 2 pr t σ = (f) aDetailed derivations and limitations of the use of these formulas are discussed in Sections 1.6, 5.7, 6.2, and 13.14.
where sx = normal axial stress τ = shearing stress due to torque τxy = shearing stress due to vertical shear force P = axial force T = torque V = vertical shear force M = bending moment about z axis A = cross-sectional area y, z = centroidal principal axes of the area I = moment of inertia about neutral axis (N.A.) J = polar moment of inertia of circular cross section b = width of bar at which τxy is calculated r = radius Q = first moment about N.A. of the area beyond the point at which τxy is calculated where sθ = tangential stress in cylinder wall sa = axial stress in cylinder wall s = membrane stress in sphere wall P = internal pressure t = wall thickness r = mean radius 1.7 Stresses on Inclined Sections 17 torque, moment component, or pressure at a section of a typical homogeneous and elastic structural member [Ref. 1.5].
The mechanics of materials theory is based on the simplifying assumptions related to the pattern of deformation so that the strain distributions for a cross section of the mem-ber can be determined. It is a basic assumption that plane sections before loading remain plane after loading. This assumption can be shown to be exact for axially loaded prismatic bars, for prismatic circular torsion members, and for prismatic beams subjected to pure bending; it is approximate for other beam situations. However, there are an extraordinarily large variety of cases in which applications of the basic formulas of mechanics of materi-als lead to useful results. In this text, we will provide greater insight into the meaning and limitations of stress analysis by solving problems using both the elementary and exact methods of analysis.
1.6.2 Combined Stresses When a member is simultaneously subjected to two or more types of loads, causing vari-ous internal force resultants on a section, it is assumed that each load produces a stress as if it were the only load acting on the member. The final or combined stress is then found by superposition of the stresses due to each load acting separately. This suggests that the presence of one load does not affect the stresses and strains contributed by another load. The principle of superposition is relevant in cases involving linearly elastic material behavior where deformations are small (Section 2.2). Having the combined stresses been obtained, the methods of stress analysis are usually employed to investigate the principal stresses and the maximum shearing stress.
1.7 STRESSES ON INCLINED SECTIONS The stresses in bars, shafts, beams, and other structural members can be obtained by using the basic formulas, such as those listed in Table 1.1. The values found by these equations are for stresses that occur on cross sections of the members. Recall that all of the formulas for stress are limited to isotropic, homogeneous, and elastic materials that behave linearly. This section deals with the states of stress at points located on inclined sections or planes under axial loading. As before, we use stress elements to represent the state of stress at a point in a member. However, we now wish to find normal and shear stresses acting on the sides of an element in any direction.
The directional nature of more general states of stress and finding maximum and minimum values of stress are discussed in Sections 1.10 and 1.13. Usually, the failure of a member will stem from a certain magnitude of stress in a certain direction. For proper design, it is necessary to determine where and in which direction the largest stress occurs. The equations derived and the graphical technique introduced here and in the sections that follow are helpful in analyzing the stress at a point under various types of loading. Note that the transformation equations for stress are developed on the basis of equilibrium conditions only and do not depend on material properties or on the geometry of deformation.
18 Chapter 1 Analysis of Stress 1.7.1 Axially Loaded Members We now consider the stresses on an inclined plane a–a of the bar in uniaxial tension shown in Fig. 1.6a, where the normal x′ to the plane forms an angle θ with the axial direction. On an isolated part of the bar to the left of section a–a, the resultant P may be resolved into two components: the normal force Px′ = P cos θ and the shear force Py′ = −P sin θ , as indicated in Fig. 1.6b. Thus, the normal and shearing stresses, uniformly distrib-uted over the area Ax′ = A/cos θ of the inclined plane (Fig. 1.6c), are given by cos cos2 σ θ σ θ = = ′ ′ P A x x x (1.11a) sin sin cos τ θ σ θ θ = − = − ′ ′ ′ P A x y x x (1.11b) where σ x = P/A. The negative sign in Eq. (1.11b) agrees with the sign convention for shearing stresses described in Section 1.5. The foregoing process of determining the stress in proceeding from one set of coordinate axes to another is called stress transformation.
Equations (1.11) indicate how the stresses vary as the inclined plane is cut at various angles. As expected, σ x′ is a maximum (σ max) when θ is 0° or 180°, and τ x′ y′ is a maximum (τ max) when θ is 45° or 135°. Also, τ max = ± 1 2σ max. The maximum stresses are thus σ σ τ σ = = , ± max max 1 2 x x (1.12) Observe that the normal stress is either a maximum or a minimum on planes for which the shearing stress is zero.
Figure 1.7 shows the manner in which the stresses vary as the section is cut at angles varying from θ = 0° to 180° Clearly, when θ > 90°, the sign of τ x′ y′ in Eq. (1.11b) changes; the shearing stress changes sense. However, the magnitude of the shearing stress for any (a) (b) (c) x x′ y′ P P A a a Ax′ y′ x′ P Px′ Py′ θ y x σx′ τx′y′ y′ x′ P P θ θ Figure 1.6. (a) Prismatic bar in tension; (b, c) side views of a part cut from the bar.
1.7 Stresses on Inclined Sections 19 angle θ determined from Eq. (1.11b) is equal to that for θ + 90°. This agrees with the general conclusion reached in the preceding section: shearing stresses on mutually per-pendicular planes must be equal.
We note that Eqs. (1.11) can also be used for uniaxial compression by assigning a negative value to P. The sense of each stress direction is then reversed in Fig. 1.6c.
EXAMPLE 1.1 State of Stress in a Tensile Bar Compute the stresses on the inclined plane with θ = 35° for a prismatic bar of a cross-sectional area 800 mm2, subjected to a tensile load of 60 kN (Fig. 1.6a). Then determine the state of stress for θ = 35° by calculating the stresses on an adjoining face of a stress element. Sketch the stress configuration.
Solution The normal stress on a cross section is 60(10 ) 800(10 ) 75 3 6 P A x σ = = = − MPa Introducing this value in Eqs. (1.11) and using θ = 35°, we have σ x′ = σ x cos2 θ = 75(cos 35°)2 = 50.33 MPa τ x′ y′ = -σ x sin θ cos θ = -75(sin 35°) (cos 35°) = -35.24 MPa Comments The normal and shearing stresses acting on the adjoining y′ face are, respectively, 24.67 MPa and 35.24 MPa, as calculated from Eqs. (1.11) by substituting the angle θ + 90° = 125°. The values of σ x′ and τ x′ y′ are the same on opposite sides of the element. On the basis of the established sign conven-tion for stress, the required sketch is shown in Fig. 1.8.
Stress 0 45 90 135 180 Angle θ° σx τx′y′ 0.5 σx −0.5 σx σx′ Figure 1.7. Variation of stress at a point with the inclined section in the bar shown in Fig. 1.6a.
24.67MPa 50.33MPa 35.24MPa y′ x ′ x 90°+ θ θ = 35° Figure 1.8. Example 1.1. Stress element for θ = 35°.
20 Chapter 1 Analysis of Stress 1.8 V ARIATION OF STRESS WITHIN A BODY As pointed out in Section 1.5, the components of stress generally vary from point to point in a stressed body. These variations are governed by the conditions of equilibrium of stat-ics. Fulfillment of these conditions establishes certain relationships, known as the differ-ential equations of equilibrium, which involve the derivatives of the stress components.
Consider a thin element of sides dx and dy (Fig. 1.9), and assume that σ x, σ y, τ xy, and τ yx, are functions of x, y but do not vary throughout the thickness (are independent of z) and that the other stress components are zero. Also assume that the x and y components of the body forces per unit volume, Fx and Fy, are independent of z and that the z component of the body force Fz = 0. This combination of stresses, satisfying the conditions described, is the plane stress. Because the element is very small, for the sake of simplicity, the stress components may be considered to be distributed uniformly over each face. In Fig. 1.9, they are shown by a single vector representing the mean values applied at the center of each face.
As we move from one point to another for example, from the lower-left corner to the upper-right corner of the element one stress component, say σ x, acting on the negative x face, changes in value on the positive x face. The stresses σ y, τ xy, and τ yx similarly change. The variation of stress with position may be expressed by a truncated Taylor’s expansion: σ ∂σ ∂ + x dx x x (a) The partial derivative is used because σ x is a function of x and y. Treating all the compo-nents similarly, the state of stress shown in Fig. 1.9 is obtained.
1.8.1 Equations of Equilibrium We consider now the equilibrium of an element of unit thickness, taking moments of force about the lower-left corner. Thus, Σ Mz = 0 yields 2 2 2 2 0 y dxdy dx x dxdy dy x dx dxdy y dy dxdy F dxdy dx F dxdy dy y x xy xy yx yx y x ∂σ ∂ ∂σ ∂ τ ∂τ ∂ τ ∂τ ∂ − + + − + + − = y x Fy Fx dy dx τxy τyx σy σx dy ∂σy ∂y σy + dx ∂σx ∂x σx + dy ∂τyx ∂y τyx + dx ∂τxy ∂x τxy + Figure 1.9. Element with stresses and body forces.
1.8 Variation of Stress within a Body 21 Neglecting the triple products involving dx and dy, this reduces to τ xy = τ yx. In a like man-ner, it may be shown that τ yz = τ zy and τ xz = τ zx, as already obtained in Section 1.5. From the equilibrium of x forces, Σ Fx = 0, we have 0 x dx dy dy y dy dx dx F dxdy x x x xy xy xy x σ ∂σ ∂ σ τ ∂τ ∂ τ + − + + − + = (b) Upon simplification, Eq. (b) becomes 0 x y F dxdy x xy x ∂σ ∂ ∂τ ∂ + + = (c) Because dx dy is nonzero, the quantity in the parentheses must vanish.
A similar expression is written to describe the equilibrium of y forces. The x and y equations yield the following differential equations of equilibrium for two-dimensional stress: 0 0 x y F y x F x xy x y xy y ∂σ ∂ ∂τ ∂ ∂σ ∂ ∂τ ∂ + + = + + = (1.13) The differential equations of equilibrium for the case of three-dimensional stress may be generalized from the preceding expressions as follows: 0 0 0 x y z F y x z F z x y F x xy xz x y xy yz y z xz yz z ∂σ ∂ ∂τ ∂ ∂τ ∂ ∂σ ∂ ∂τ ∂ ∂τ ∂ ∂σ ∂ ∂τ ∂ ∂τ ∂ + + + = + + + = + + + = (1.14) A succinct representation of these expressions, on the basis of the range and summation conventions (Section 1.17), may be written as 0, , , , ∂τ ∂ + = = x F i j x y z ij j i (1.15a) in which xx = x, xy = y, and xz = z. The repeated subscript is j, indicating summation. The unrepeated subscript is i. Here i is termed the free index, and j, the dummy index.
If in the foregoing expression the symbol ∂/∂x is replaced by a comma, we have 0 , F ij j i τ + = (1.15b) where the subscript after the comma denotes the coordinate with respect to which differ-entiation is performed. If no body forces exist, Eq. (1.15b) reduces to τ ij,j = 0, indicating 22 Chapter 1 Analysis of Stress that the sum of the three stress derivatives is zero. As the two equilibrium relations of Eqs. (1.13) contain three unknowns and the three expressions of Eqs. (1.14) involve the six unknown stress components, problems in stress analysis are internally statically indeterminate.
In a number of practical applications, the weight of the member is the only body force. If we take the y axis as upward and designate by ρ the mass density per unit volume of the member and by g, the gravitational acceleration, then Fx = Fz = 0 and Fy = -ρg in Eqs. (1.13) and (1.14). The resultant of this force over the volume of the member is usu-ally so small compared with the surface forces that it can be ignored, as stated in Section 1.1. However, in dynamic systems, the stresses caused by body forces may far exceed those associated with surface forces and represent the principal influence on the stress field. Application of Eqs. (1.13) and (1.14) to a variety of loaded members is presented in sections of this text employing the approach of the theory of elasticity, beginning with Chapter 3. The following sample problem shows the pattern of the body force distribution for an arbitrary state of stress in equilibrium.
EXAMPLE 1.2 The Body Forces in a Structure Given: The stress field within an elastic structural member is expressed as follows: σ τ τ σ τ σ = − + = + = + = + = = − x y z y xz x y x y y z x xy xz y yz z , 5 2 , 2 , 0, 4 3 2 2 3 2 3 1 2 2 2 3 (d) Find: The body force distribution required for equilibrium.
Solution Substitution of the given stresses into Eqs. (1.14) yields ( 3 ) (4 ) (3 ) 0 ( ) (0) (0) 0 ( 3 ) ( 2 ) (0) 0 2 2 2 3 x y xz F y F z z xy F x y z − + + + = + + + = − + + + + = The body force distribution, as obtained from these expressions, is therefore 3 4 3 , , 2 3 2 2 2 3 F x y xz F y F xy z z x y z = − − = − = − + − (e) Comment The state of stress and body force at any specific point within the member may be obtained by substituting the specific values of x, y, and z into Eqs. (d) and (e), respectively.
In this case, the body is not in static equilibrium, and the inertia force terms ‒ρax, ‒ρay, and ‒ρaz (where ax, ay, and az are the components of acceleration) must be included in the body force components Fx, Fy, and Fz respectively, in Eqs. (1.14).
1.9 Plane-Stress Transformation 23 1.9 PLANE-STRESS TRANSFORMATION A two-dimensional state of stress exists when the stresses and body forces are indepen-dent of one of the coordinates, here taken as z. Such a state is described by stresses σ x, σ y, and τ xy and the x and y body forces. Two-dimensional (2-D) problems are of two classes: plane stress and plane strain. In the case of plane stress, as described in the previous sec-tion, the stresses σ z, τ xz, and τ yz and the z-directed body forces are assumed to be zero. The condition that occurs in a thin plate subjected to loading uniformly distributed over the thickness and parallel to the plane of the plate typifies the state of plane stress (Fig. 1.10). In the case of plane strain, the stresses τ xz and τ yz and the body force Fz are likewise taken to be zero, but σ z does not vanish and can be determined from stresses σ x and σ y.
We shall now determine the equations for transformation of the stress components σ x, σ y, and τ xy at any point of a body represented by an infinitesimal element, isolated from the plate illustrated in Fig. 1.10. The z-directed normal stress σ z even if it is nonzero, need not be considered here. In the following derivations, the angle θ locating the x′ axis is assumed to be positive when measured from the x axis in a counterclockwise direction. According to our sign convention (see Section 1.5.1), the stresses are indicated as positive values.
Consider an infinitesimal wedge cut from the loaded body shown in Fig. 1.11a and b. It is required to determine the stresses σ x′ and τ x′ y′ , which refer to axes x′ , y′ making More details and illustrations of these assumptions are given in Chapter 3.
x σy′ σx′ τx′y′ y ′ x ′ θ Q x y σx σy σx′ τx′y′ τxy τxy y′ x ′ θ θ Q B A py px x′ x y σx σy τxy y ′ θ Q (a) (b) (c) Figure 1.11. Elements in plane stress.
y z x θ σx σx′ σx σy σy τxy τx ′y ′ Figure 1.10. Thin plate in-plane loads.
24 Chapter 1 Analysis of Stress an angle θ with axes x, y, as shown in the figure. Let side AB be normal to the x′ axis. In accordance with the sign convention, σ x′ and τ x′ y′ are positive stresses, as shown in the figure. If the area of side AB is taken as unity, then sides QA and QB have area cos θ and sin θ respectively.
Equilibrium of forces in the x and y directions requires that cos sin cos sin p p x x xy y xy y σ θ τ θ τ θ σ θ = + = + (1.16) where px and py are the components of stress resultant acting on AB in the x and y direc-tions, respectively. The normal and shear stresses on the x′ plane (AB plane) are obtained by projecting px and py in the x′ and y′ directions: σ θ θ τ θ θ = + = − ′ ′ ′ cos sin cos sin p p p p x x y x y y x (a) From the foregoing discussion, it is clear that σ 2 x′ + τ 2 x′ y′ = p2 x + p2 y. Upon substitution of the stress resultants from Eq. (1.16), Eqs. (a) become σ σ θ σ θ τ θ θ = + + ′ cos sin 2 sin cos 2 2 x x y xy (1.17a) τ τ θ θ σ σ θ θ = − + − ′ ′ (cos sin ) ( )sin cos 2 2 x y xy y x (1.17b) The normal stress σ y′ acting on the y′ face of an inclined element (Fig. 1.11c) may readily be obtained by substituting θ + π/2 for θ in the expression for σ x′ : σ σ θ σ θ τ θ θ = + − ′ sin cos 2 sin cos 2 2 y x y xy (1.17c) Equations (1.17) can be converted to a useful form by introducing the following trigono-metric identities: θ θ θ θ θ θ θ = + = = − cos (1 cos2 ), sin cos sin2 , sin (1 cos2 ) 2 1 2 1 2 2 1 2 The transformation equations for plane stress now become σ σ σ σ σ θ τ θ = + + − + ′ ( ) ( )cos2 sin 2 1 2 1 2 x x y x y xy (1.18a) τ σ σ θ τ θ = − − + ′ ′ ( )sin 2 cos2 1 2 x y x y xy (1.18b) σ σ σ σ σ θ τ θ = + − − − ′ ( ) ( )cos2 sin 2 1 2 1 2 y x y x y xy (1.18c) The preceding expressions permit the computation of stresses acting on all possible planes AB (the state of stress at a point) provided that three stress components on a set of orthog-onal faces are known.
1.9 Plane-Stress Transformation 25 1.9.1 Stress Tensor It is important to note that addition of Eqs. (1.17a) and (1.17c) gives the following relationships: σ x + σ y = σ x′ + σ y′ = constant In words, the sum of the normal stresses on two perpendicular planes is invariant—that is, independent of θ . This conclusion is also valid in the case of a three-dimensional state of stress, as shown in Section 1.13. In mathematical terms, the stress whose components transform in the preceding way by rotation of axes is known as a tensor. Some examples of other quantities are strain and moment of inertia. The similarities between the transfor-mation equations for these quantities are observed in Sections 2.5 and C.2.2. Mohr’s circle (Section 1.11) is a graphical representation of a stress tensor transformation.
1.9.2 Polar Representations of State of Plane Stress Consider, for example, the possible states of stress corresponding to σ x = 14 MPa, σ y = 4 MPa, and τ xy = 10 MPa. Substituting these values into Eq. (1.18) and permitting θ to vary from 0° to 360° yields the data on which the curves shown in Fig. 1.12 are based. The plots shown, called stress trajectories, are polar representations: σ x′ versus θ (Fig. 1.12a) and τ x′ y′ versus θ (Fig. 1.12b).
The direction of each maximum shear stress bisects the angle between the maximum and minimum normal stresses. Note that the normal stress is a maximum on planes at θ = 31.66° or a minimum on planes at θ = 31.66° + 90°, for which the shearing stress is zero. The conclusions drawn from this example are valid for any two-dimensional (or three-dimensional) state of stress.
1.9.3 Cartesian Representation of State of Plane Stress Now let us examine a two-dimensional condition of stress at a point in a loaded machine component on an element illustrated in Fig. 1.13a. Introducing the given values into the first two of Eqs. (1.18) gives σ x′ = 4.5 + 2.5 cos 2θ + 5 sin 2θ τ x′ y′ = -2.5 sin 2θ + 5 cos 2θ 90° 90° 90° τmax =11.18 σmin = 2.18 σmax =20.18 τxy =10 σx = 14 σy = 4 Direction of σmax Direction of σmin θ =31.66° 0° 0° 45° (a) (b) θ = 31.66° Figure 1.12. Polar representations of σ x′ and τ x′ y′ (in megapascals) versus θ .
26 Chapter 1 Analysis of Stress In these equations, permitting θ to vary from 0° to 180° in increments of 15° leads to the data from which the graphs illustrated in Fig. 1.13b are obtained.
This Cartesian representation demonstrates the variation of the normal and shearing stresses versus θ ≤ 180° Observe that the direction of maximum (and minimum) shear stress bisects the angle between the maximum and minimum normal stresses. Moreover, the normal stress is either a maximum on planes θ = 31.7° or a minimum on planes θ = 31.7° + 90°, for which the shear stress is zero. As a check, note that σ x + σ y = σ max + σ min = 9 MPa, as expected.
The conclusions drawn from the polar and Cartesian representations provided here are valid for any state of stress, as will be seen in the next section. A more convenient approach to the graphical transformation for stress is considered in Sections 1.11 and 1.15. The manner in which the three-dimensional normal and shearing stresses vary is discussed in Sections 1.12 through 1.14.
1.10 PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS The transformation equations for two-dimensional stress indicate that the normal stress σ x′ and shearing stress τ x′ y′ vary continuously as the axes are rotated through the angle θ . To ascertain the orientation of x′ y′ corresponding to maximum or mini-mum σ x′ the necessary condition dσ x′ /dθ = 0 is applied to Eq. (1.18a). In so doing, we have ( )sin 2 2 cos2 0 x y xy σ σ θ τ θ − − + = (a) (a) x y τxy = 5 MPa σx = 7 MPa σy = 2 MPa x′ θ 10 8 6 4 2 -2 -4 -6 0 90 (b) 31.7 180 τx′y′ τmax =5.59 σmax =10.09 τmax =-5.59 σmin=-1.09 σ x′ σ. τ.
MPa θ ° 45 45 Figure 1.13. Graph of normal stress σ x′ and shearing stress τ x′ y′ with angle θ ( for θ ≤ 180°).
1.10 Principal Stresses and Maximum In-Plane Shear Stress 27 This yields θ τ σ σ = − tan 2 2 p xy x y (1.19) Inasmuch as tan 2θ = tan(π + 2θ ), two directions, mutually perpendicular, are found to satisfy Eq. (1.19). These are the principal directions, along which the principal or maxi-mum and minimum normal stresses act. Two values of θ p, corresponding to the σ 1 and σ 2 planes, are represented by θ′ p and θ′′ p, respectively.
When Eq. (1.18b) is compared with Eq. (a), it becomes clear that τ x′ y′ = 0 on a princi-pal plane. A principal plane is thus a plane of zero shear. The principal stresses are deter-mined by substituting Eq. (1.19) into Eq. (1.18a): σ σ σ σ σ σ τ = = + − + 2 ± 2 max,min 1,2 2 2 x y x y xy (1.20) The algebraically larger stress given here is the maximum principal stress, denoted by σ 1. The minimum principal stress is represented by σ 2. To determine which of the two corre-sponds to σ 1, we would substitute one of the values θ p into Eq. (1.18a).
Similarly, employing the preceding approach and Eq. (1.18b), we can determine the planes of maximum shearing stress. Thus, setting dτ x′ y′ /dθ = 0, we now have (σ x - σ y)cos 2θ + 2τ xy sin 2θ = 0 or tan 2 2 s x y xy θ σ σ τ = − − (1.21) This expression defines two values of θ s that are 90° apart. As shown earlier, these direc-tions may be denoted by attaching a prime or a double prime notation to θ s. Comparing Eqs. (1.19) and (1.21), we also observe that the planes of maximum shearing stress are inclined at 45° with respect to the planes of principal stress. Now, from Eqs. (1.21) and (1.18b), we obtain the extreme values of shearing stress as follows: ± 2 ± ( ) max 2 2 1 2 1 2 x y xy τ σ σ τ σ σ = − + = − (1.22) Here the largest shearing stress, regardless of sign, is referred to as the maximum shearing stress, designated τ max. Normal stresses acting on the planes of maximum shearing stress can be determined by substituting the values of 2θ s from Eq. (1.21) into Eqs. (1.18a) and (1.18c): ( ) ave 1 2 σ σ σ σ ′ = = + x y (1.23) The results are illustrated in Fig. 1.14. Note that the diagonal of a stress element toward which the shearing stresses act is called the shear diagonal. The shear diagonal of 28 Chapter 1 Analysis of Stress the element on which the maximum shearing stresses act lies in the direction of the alge-braically larger principal stress. This information assists us in predicting the proper direc-tion of the maximum shearing stress.
1.11 MOHR’S CIRCLE FOR TWO-DIMENSIONAL STRESS A graphical technique, predicated on Eq. (1.18), permits the rapid transformation of stress from one plane to another and leads to the determination of the maximum nor-mal and shear stresses. In this approach, Eqs. (1.18) are depicted by a stress circle, called Mohr’s circle. In the Mohr representation, the normal stresses obey the sign conven-tion described in Section 1.5.1. However, for the purposes of constructing and reading values of stress from Mohr’s circle, the sign convention for shear stress is as follows: If the shearing stresses on opposite faces of an element would produce shearing forces that result in a clockwise couple, as shown in Fig.1.15c, these stresses are regarded as positive. Accordingly, the shearing stresses on the y faces of the element in Fig. 1.15a are taken as positive (as before), but those on the x faces are now negative.
After Otto Mohr (1835–1918), professor at Dresden Polytechnic. For further details, see Ref. 1.6, for example.
Shear diagonal τmax σ2 σ ′ σ′ σ1 45° y′ x ′ θ′p x Figure 1.14. Planes of principal and maximum shearing stresses.
y σy σx σ 2 τxy θ 2θ x x′ y ′ B ′ A′ x ′ E x B1 A1 (a) (b) (c) τ (σ x + σ y) 1 2 σ′ = D y B(σ y, τxy) A(σx, −τxy) O C F r τmax −τmax σ σ1 + + Figure 1.15. (a) Stress element; (b) Mohr’s circle of stress; (c) interpretation of positive shearing stresses.
1.11 Mohr’s Circle for Two-Dimensional Stress 29 Given σ x, σ y, and τ xy with algebraic signs in accordance with this sign convention, the procedure for obtaining Mohr’s circle (Fig. 1.15b) is as follows: 1. Establish a rectangular coordinate system, indicating +τ and +σ . Both stress scales must be identical.
2. Locate the center C of the circle on the horizontal axis at a distance 1 2(σ x + σ y) from the origin.
3. Locate point A by coordinates σ x and -τ xy. Although these stresses may correspond to any face of an element such as in Fig. 1.15a, the usual practice is to specify the stresses on the positive x face.
4. Draw a circle with center at C and of radius equal to CA.
5. Draw line AB through C.
The angles on the circle are measured in the same direction as θ is measured in Fig. 1.15a. An angle of 2θ on the circle corresponds to an angle of θ on the element. The state of stress associated with the original x and y planes corresponds to points A and B on the circle, respectively. Points lying on diameters other than AB, such as A′ and B′ , define states of stress with respect to any other set of x′ and y′ planes rotated relative to the origi-nal set through an angle θ .
Points A1 and B1 on the circle locate the principal stresses and provide their magni-tudes as defined by Eqs. (1.19) and (1.20), while points D and E represent the maximum shearing stresses, defined by Eqs. (1.21) and (1.22). The radius of the circle is 2 2 CA CF AF = + (a) where CF AF x y xy σ σ τ = − = ( ), 1 2 Thus, the radius equals the magnitude of the maximum shearing stress. Mohr’s circle shows that the planes of maximum shear are always located at 45° from planes of prin-cipal stress, as indicated earlier in Fig. 1.14. The use of Mohr’s circle is illustrated in the next example.
EXAMPLE 1.3 Principal Stresses in a Member Given: At a point in the structural member, the stresses are represented as in Fig. 1.16a.
Find: By employing Mohr’s circle: (a) the magnitude and orientation of the principal stresses; (b) the magnitude and orientation of the maximum shear-ing stresses and associated normal stresses. In each case, show the results on a properly oriented element; represent the stress tensor in matrix form.
Solution Mohr’s circle, constructed in accordance with the procedure outlined, is shown in Fig. 1.16b. The center of the circle is at (40 + 80)/2 = 60 MPa on the σ axis.
30 Chapter 1 Analysis of Stress a. The principal stresses are represented by points A1 and B1. Hence, the max-imum and minimum principal stresses referring to the circle are σ = − + ± 60 (80 40) (30) 1,2 1 4 2 2 or σ 1 = 96.05 MPa and σ 2 = 23.95 MPa The planes on which the principal stresses act are given by θ θ ′ = = ′′ = + = − p p 2 tan 30 20 56.31° and 2 56.31° 180° 236.31° 1 Hence θ′ p = 28.15° and θ′′ p = 118.16° Mohr’s circle clearly indicates that θ ′ p locates the σ 1 plane. The results may readily be checked by substituting the two values of θ p into Eq. (1.18a). The state of principal stress is shown in Fig. 1.16c.
40 MPa (MPa) y B(40,30) D s F A1 x E A(80,−30) 2θ′′ p 2θ ′ O B1 σ′= 60 σ (MPa) 30 MPa 80 MPa (a) (b) x y′ y ′ x ′ x ′ 28.15° 73.15° 60 MPa 36.05 MPa 60 MPa 96.05 MPa 23.95 MPa (c) (d) y τ C Figure 1.16. Example 1.3. (a) Element in plane stress; (b) Mohr’s circle of stress; (c) principal stresses; (d) maximum shear.
1.11 Mohr’s Circle for Two-Dimensional Stress 31 b. The maximum shearing stresses are given by points D and E. Thus, ± (80 40) (30) ±36.06 MPa max 1 4 2 2 τ = − + = It is seen that (σ 1 - σ 2)/2 yields the same result. The planes on which these stresses act are represented by θ′′ s 2 = 28.15° + 45° = 73.15° and θ′ s = 163.15° As Mohr’s circle indicates, the positive maximum shearing stress acts on a plane whose normal x′ makes an angle θ′′ s 2 with the normal to the origi-nal plane (x plane). Thus, +τ max on two opposite x′ faces of the element will be directed so that a clockwise couple results. The normal stresses acting on maximum shear planes are represented by OC, σ ′ = 60 MPa on each face. The state of maximum shearing stress is shown in Fig. 1.16d. The direction of the τ max’s may also be readily predicted by recall-ing that they act toward the shear diagonal. According to the general sign convention (Section 1.5), the shearing stress acting on the x′ plane in Fig.1.16d is negative. As a check, if θ′′ s 2 = 146.30° and the given ini-tial data are substituted into Eq. (1.18b), we obtain τ x′ y′ = -36.05 MPa, as already found.
We may now describe the state of stress at the point in the following matrix forms: 80 30 30 40 , 96.05 0 0 23.95 , 60 –36.06 –36.06 60 These three representations, associated with θ = 0°, θ = 28.15°, and planes passing through the point, are equivalent.
Comment If we assume σ z = 0 in this example, we obtain a much higher shearing stress in the planes bisecting the x′ and z planes (Prob.1.61). Thus, three-dimensional analysis (see Section 1.15) should be considered for deter-mining the true maximum shearing stress at a point.
EXAMPLE 1.4 Stresses in a Frame Given: The stresses acting on an element of a loaded frame are shown in Fig. 1.17a.
Find: By applying Mohr’s circle: The normal and shear stresses acting on a plane defined by θ = 30°.
Solution Mohr’s circle of Fig. 1.17b describes the state of stress given in Fig. 1.17a. Points A1 and B1 represent the stress components on the x and y faces, respectively. The radius of the circle is (14 + 28)/2 = 21. Correspond-ing to the 30° plane within the element, it is necessary to rotate through 60° 32 Chapter 1 Analysis of Stress counterclockwise on the circle to locate point A′ . A 240° counterclockwise rotation locates point B′ . Referring to the circle, σ σ = + = = − ′ ′ x y 7 21cos60° 17.5 MPa 3.5 MPa ± 21sin60° ±18.19 MPa x y τ = = ′ ′ Comment Figure 1.17c indicates the orientation of the stresses. The results can be checked by applying Eq. (1.18), using the initial data.
EXAMPLE 1.5 Cylindrical Vessel Under Combined Loads Given: A thin-walled cylindrical pressure vessel of 250-mm diameter and 5-mm wall thickness rigidly attached to a wall, forming a cantilever (Fig. 1.18a).
Find: The maximum shearing stresses and the associated normal stresses at point A of the cylindrical wall. The following loads are applied: internal pres-sure p = 1.2 MPa, torque T = 3 kN ⋅ m, and direct force P = 20 kN. Show the results on a properly oriented element.
14MPa (MPa) 28MPa x y (a) (b) (c) 30° y τ σ′= 7 x ′ A′ y ′ B ′ 60° C O σ (MPa) B1(−14,0) A1(28,0) x x ′ 30° 18.19MPa 3.5 MPa 17.5 MPa y ′ Figure 1.17. Example 1.4. (a) Element in biaxial stresses; (b) Mohr’s circle of stress; (c) stress element for θ = 30°.
1.11 Mohr’s Circle for Two-Dimensional Stress 33 Solution The internal force resultants on a transverse section through point A are found from the equilibrium conditions of the free-body diagram of Fig. 1.18b: V = 20 kN, M = 8 kN ⋅ m, and T = 3 kN ⋅ m. In Fig. 1.18c, the combined axial, tangential, and shearing stresses are shown acting on a small element at point A. These stresses are (Tables 1.1 and C.1) σ π π = = = = Mr I r r t b 8(10 ) 8(10 ) (0.125 )(0.005) 32.6 MPa 3 3 3 2 τ π π = − = − = − = − Tr J r r t t 3(10 ) 2 3(10 ) 2 (0.125 )(0.005) 6.112 MPa 3 3 3 2 pr t a a σ σ σ = = = = = θ 2 1.2(10 )(125) 2(5) 15 MPa, 2 30 MPa 6 Thus, we have σ x = 47.6 MPa, σ y = 30 MPa, and τ xy = -6.112 MPa. For ele-ment A, Q = 0; hence, the direct shearing stress τ d = τ xz = VQ/Ib = 0.
The maximum shearing stresses are from Eq. (1.22): τ ( ) = − + − = ± 476 30 2 6112 ±10.71 MPa max 2 2 Equation (1.23) yields (47.6 30) 38.8 MPa 1 2 σ ′ = + = z A 0.4 m (a) (c) (d) (b) A V T M P T P T x y x 27 .6° x 38.8 MPa 38.8 MPa 10.71 MPa σθ τt σa+ σb x ′ y ′ Figure 1.18. Example 1.5. Combined stresses in a thin-walled cylindrical pressure vessel: (a) side view; (b) free body of a segment; (c) and (d) element A (viewed from top).
34 Chapter 1 Analysis of Stress To locate the maximum shear planes, we use Eq. (1.21): tan 47.6 30 2( 6.112) 27.6° and 117.6° 1 2 1 θ = − − − = − s Applying Eq. (1.18b) with the given data and θ′′ s 2 = 55.2°, we have τ x′ y′ = -10.71 MPa. Hence, θ′′ s 2 = 27.6°, and the stresses are shown in their proper directions in Fig. 1.18d.
CASE STUDY 1.1 Pressure Capacity of an Hydraulic Cylinder Pressurized hydraulic fluid (liquid or air) produces stresses and deformation of a cylinder. Hydraulic systems are widely used in brakes, control mecha-nisms, and actuators in positioning devices. Design of a pressurized duplex conduit is discussed in Example 8.4.
Given: A hydraulic cylinder of radius r and thickness t under internal pres-sure p is simultaneously compressed by an axial load P through the piston of diameter d ≈ 2r (Fig. 1.19a). The vessel is inadvertently subjected to torque T at its mounting.
Data: r = 60 mm, t = 4 mm, and T = 260 N ⋅ m.
Find: The largest value of p that can be applied to the cylinder with a safety factor of n.
Assumptions: The vessel is made of a high-strength ASTM-A242 steel with τyp = 210 MPa (Table D.1) and n = 2. The critical stress is at point A on cylin-der wall remote from the ends (Fig. 1.19b). The effect of bending of the cylin-der on stresses is omitted.
Cylinder Piston P A T T Mounting A (b) (a) y x σx τxy σθ d p t Figure 1.19. Case Study 1.1. (a) Schematic hydraulic cylinder; (b) element in plane tress.
1.12 Three-Dimensional Stress Transformation 35 Solution The combined stresses acting at point A on an element in this thin-walled cylinder are found as follows: τ π π σ σ ( ) ( ) ( ) ( ) = = = = = = = = = θ Tr J T r t pr t p p pr t p xy x 2 260 2 0.06 0.004 2.874 MPa 2 60 2 4 7.5 15 2 2 The maximum allowable in-plane shear stress in the cylinder wall will be 210/2 = 105 MPa. Through the use of Eq. 1.22, we obtain 7.5 15 2 2.874 14.06 8.26 105 max 2 2 2 p p p τ ( ) = − + = + ≤ Solving, 28 MPa.
max = p Comments The largest permissible axial load that can be applied to this pis-ton is approximately Pmax = 28(π × 602) = 317 kN. The MATLAB solution for this sample problem and many others can be found on the book’s website (see Appendix E).
1.12 THREE-DIMENSIONAL STRESS TRANSFORMATION Because the physical elements studied are always three dimensional, it is desirable to con-sider three planes and their associated stresses, as illustrated in Fig. 1.2. The equations governing the transformation of stress in the three-dimensional case may be obtained by using a similar approach to that applied for the two-dimensional state of stress.
Consider a small tetrahedron isolated from a continuous medium (Fig. 1.20a), subject to a general state of stress. The body forces are taken to be negligible. In the figure, px, py, and pz are the Cartesian components of stress resultant p acting on oblique plane ABC. It is required to relate the stresses on the perpendicular planes intersecting at the origin to the normal and shear stresses on ABC.
The orientation of plane ABC may be defined in terms of the angles between a unit normal n to the plane and the x, y, and z directions (Fig. 1.20b). The direction cosines associated with these angles are n n n cos cos( , ) cos cos( , ) cos cos( , ) α β γ = = = = = = x l y m z n (1.24) 36 Chapter 1 Analysis of Stress The three direction cosines for the n direction are related by 1 2 2 2 l m n + + = (1.25) The area of the perpendicular plane QAB, QAC, QBC may now be expressed in terms of A, the area of ABC, and the direction cosines: AQAB = Ax = A ⋅ i = A(li + mj + nk) ∙ i = Al The other two areas are similarly obtained. Collectively, we have , , A Al A Am A An QAB QAC QBC = = = (a) In these equations, i, j, and k are unit vectors in the x, y, and z directions, respectively.
Next, from the equilibrium of x, y, z-directed forces together with Eq. (a), we obtain, after canceling A, σ τ τ τ σ τ τ τ σ = + + = + + = + + p l m n p l m n p l m n x x xy xz y xy y yz z xz yz z (1.26) The stress resultant on A is thus determined based on the known stresses σ x, σ y, σ z, τ xy, τ xz, and τ yz and a knowledge of the orientation of A. In the limit as the sides of the tetrahe-dron approach zero, plane A contains point Q. Thus, the stress resultant at a point is speci-fied. This, in turn, gives the stress components acting on any three mutually perpendicular planes passing through Q, as shown next. Although perpendicular planes have been used there for convenience, these planes need not be perpendicular to define the stress at a point.
Consider now a Cartesian coordinate system x′ , y′ , z′ , in which x′ coincides with n and y′ , z′ lie on an oblique plane. The x′ y′ z′ and xyz systems are related by the direc-tion cosines: l1 = cos(x′ , x), m1 = cos(x′ , y), and so on. The notation corresponding to a n C (a) (b) x n py px dx dy pz dz Q z A B z l n m x y y γ α β Figure 1.20. Stress components on a tetrahedron.
1.12 Three-Dimensional Stress Transformation 37 complete set of direction cosines is shown in Table 1.2. The normal stress σ x′ is found by projecting px, py, and pz in the x′ direction and adding σ = + + ′ p l p m p n x x y z 1 1 1 (1.27) We combine Eqs. (1.26) and (1.27) to yield σ σ σ σ τ τ τ = + + + + + ′ l m n l m m n l n x x y z xy yz xz 2( ) 1 2 1 2 1 2 1 1 1 1 1 1 (1.28a) Similarly, by projecting , px py and pz in the ′ y and ′ z directions, we obtain, respectively, τ σ σ σ τ τ τ = + + + + + + + + ′ ′ l l m m n n l m m l m n n m n l l n x y x y z xy yz xz ( ) ( ) ( ) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (1.28b) τ σ σ σ τ τ τ = + + + + + + + + ′ ′ l l m m n n l m m l m n n m n l l n x z x y z xy yz xz ( ) ( ) ( ) 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 (1.28c) Recalling that the stresses on three mutually perpendicular planes are required to specify the stress at a point (with one of these planes being the oblique plane in ques-tion), we can find the remaining components by considering those planes perpendicular to the oblique plane. For one such plane, n would now coincide with the ′ y direction, and the expressions for the stresses , y σ ′ y x τ ′ ′, and y z τ ′ ′ could be derived. In a similar man-ner, the stresses , z σ ′ z x τ ′ ′, and z y τ ′ ′ are determined when n coincides with the ′ z direction. Owing to the symmetry of the stress tensor, only six of the nine stress components thus developed are unique. The remaining stress components are as follows: 2( ) 2 2 2 2 2 2 2 2 2 2 2 2 l m n l m m n l n y x y z xy yz xz σ σ σ σ τ τ τ = + + + + + ′ (1.28d) 2( ) 3 2 3 2 3 2 3 3 3 3 3 3 l m n l m m n l n z x y z xy yz xz σ σ σ σ τ τ τ = + + + + + ′ (1.28e) τ σ σ σ τ τ τ = + + + + + + + + ′ ′ l l m m n n m l l m n m m n l n n l y z x y z xy yz xz ( ) ( ) ( ) 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 (1.28f) Equations (1.28) represent expressions transforming the quantities σ x, σ y, σ z, τ xy, and xz τ , which, as we have noted, completely define the state of stress. Quantities such as stress (and moment of inertia; see Appendix C), which are subject to such transformations, are tensors of second rank (see Section 1.9).
Table 1.2. Notation for Direction Cosines x y z x′ l1 m1 n1 y′ l2 m2 n2 z′ l3 m3 n3 38 Chapter 1 Analysis of Stress The equations of transformation of the components of a stress tensor, in indicial nota-tion (see Section 1.17), are represented as follows: τ τ ′ = l l rs ir js ij (1.29a) or, alternatively, l l rs ri sj ij τ τ = ′ (1.29b) The repeated subscripts i and j imply the double summation in Eq. (1.29a), which, upon expansion, yields τ τ τ τ τ τ τ τ τ τ ′ = + + + + + + + + ′ l l l l l l l l l l l l l l l l l l rs xr xs xx xr ys xy xr zs xz yr xs xy yr ys yy ys zr yz zr xs xz zr ys yz zr zs zz (1.29c) By assigning r, s = x, y, z and noting that rs sr τ τ = , lead to the six expressions of Eq. (1.28).
Because , ′ x ′ y and ′ z are orthogonal, the nine direction cosines must satisfy trigono-metric relations of the following form: 1, 1,2,3 2 2 2 l m n i i i i + + = = (1.30a) and + + = + + = + + = 0 0 0 1 2 1 2 1 2 2 3 2 3 2 3 1 3 1 3 1 3 l l m m n n l l m m n n l l m m n n (1.30b) From Table 1.2, observe that Eqs. (1.30a) are the sums of the squares of the cosines in each row, and Eqs. (1.30b) are the sums of the products of the adjacent cosines in any two rows.
1.13 PRINCIPAL STRESSES IN THREE DIMENSIONS For the three-dimensional case, it is now demonstrated that three planes of zero shear stress exist, that these planes are mutually perpendicular, and that on these planes the normal stresses have maximum or minimum values. As has been discussed, these normal stresses are referred to as principal stresses, usually denoted , 1 σ 2 σ , and 3 σ . The algebra-ically largest stress is represented by 1 σ , and the smallest by : .
3 1 2 3 σ σ σ σ > > We begin by again considering an oblique ′ x plane. The normal stress acting on this plane is given by Eq. (1.28a): σ σ σ σ τ τ τ = + + + + + ′ 2( ) 2 2 2 l m n lm mn ln x x y z xy yz xz (a) The problem at hand is the determination of extreme or stationary values of σ ′ x . To accom-plish this, we examine the variation of σ ′ x relative to the direction cosines. Inasmuch as 1.13 Principal Stresses in Three Dimensions 39 l, m, and n are not independent, but rather are connected by 1 2 2 2 l m n + + = , only l and m may be regarded as independent variables. Thus, ∂σ ∂ ∂σ ∂ = = ′ ′ 0, 0 l m x x (b) Differentiating Eq. (a) as indicated by Eqs. (b) in terms of the quantities in Eq. (1.26), we obtain ∂ ∂ ∂ ∂ + = + = p p n l p p n m x z y z 0, 0 (c) From 1 2 2 2 n l m = − − , we have n l l n ∂ ∂= − / / and ∂ ∂ = − n m m n / / . Introducing these into Eq. (c), the following relationships between the components of p and n are determined: p l p m p n x y z = = (d) These proportionalities indicate that the stress resultant must be parallel to the unit nor-mal and, therefore, contains no shear component. Thus, on a plane for which σ ′ x has an extreme or principal value, that is a principal plane, the shearing stress vanishes.
So far, we have shown that three principal stresses and three principal planes exist. Denoting the principal stresses by p σ , Eq. (d) may be written as p l p m p n x p y p z p σ σ σ = = = , , (e) These expressions, together with Eq. (1.26), lead to ( ) 0 ( ) 0 ( ) 0 l m n l m n l m n x p xy xz xy y p yz xz yz z p σ σ τ τ τ σ σ τ τ τ σ σ − + + = + − + = + + − = (1.31) A nontrivial solution for the direction cosines requires that the characteristic stress deter-minant vanish: σ σ τ τ τ σ σ τ τ τ σ σ − − − = 0 x p xy xz xy y p yz xz yz z p (1.32) Expanding Eq. (1.32) leads to 0 3 1 2 2 3 I I I p p p σ σ σ − + − = (1.33) where 1 I x y z σ σ σ = + + (1.34a) 2 2 2 2 I x y x z y z xy yz xz σ σ σ σ σ σ τ τ τ = + + − − − (1.34b) 40 Chapter 1 Analysis of Stress 3 σ τ τ τ σ τ τ τ σ = I x xy xz xy y yz xz yz z (1.34c) The three roots of the stress cubic equation (Eq. 1.33) are the principal stresses. Three corresponding sets of direction cosines, along with these stresses, are used to establish the relationship of the principal planes to the origin of the nonprincipal axes. The principal stresses are the characteristic values or eigenvalues of the stress tensor ij τ . Since the stress tensor is a symmetric tensor whose elements are all real, it has real eigenvalues. That is, the three principal stresses are real [Ref. 1.7]. The direction cosines l, m, and n are the eigenvectors of ij τ .
1.13.1 Invariants for Three-Dimensional Stress Clearly, the principal stresses are independent of the orientation of the original coordinate system. It follows from Eq. (1.33) that the coefficients I1, I2, and I3 must likewise be inde-pendent of x, y, and z, since otherwise the principal stresses would change. For example, we can demonstrate that adding the expressions for σ ′, x σ ′ y , and σ ′ z given by Eq. (1.28) and making use of Eq. (1.30a) leads to σ σ σ σ σ σ = + + = + + ′ ′ ′ 1 I x y z x y z. Thus, the coef-ficients I1, I2, and I3 represent three invariants of the stress tensor in three dimensions or, briefly, the stress invariants. For plane stress, it is a simple matter to show that the follow-ing quantities are invariant (Prob. 1.32): σ σ σ σ σ σ τ σ σ τ = + = + = = − = − ′ ′ ′ ′ ′ ′ I I I x y x y x y xy x y x y 1 2 3 2 2 (1.35) Equations (1.34) and (1.35) are particularly helpful in checking the results of a stress transformation, as illustrated in Example 1.7.
If one of the principal stresses, say 1 σ obtained from Eq. (1.33) is now substituted into Eq. (1.31), the resulting expressions, together with 1 2 2 2 l m n + + = , provide enough information to solve for the direction cosines, thereby specifying the orientation of 1 σ relative to the xyz system. The direction cosines of 2 σ and 3 σ are similarly obtained. A convenient way of determining the roots of the stress cubic equation and solving for the direction cosines is presented in Appendix B, where a related computer program is also included (see Table B.1).
EXAMPLE 1.6 Three-Dimensional Stress in a Hub A steel shaft is to be forcefitted into a fixed-ended cast-iron hub. The shaft is subjected to a bending moment M, a torque T, and a vertical force P (Fig. 1.21a). Suppose that at a point Q in the hub, the stress field is as shown in Fig. 1.21b, represented by the matrix 1.13 Principal Stresses in Three Dimensions 41 19 4.7 6.45 4.7 4.6 11.8 6.45 11.8 8.3 MPa − − − − Find: The principal stresses and their orientation with respect to the original coordinate system.
Solution Substituting the given stresses into Eq. (1.33), we obtain from Eqs. (B.2) σ = 11.618 MPa, 1 σ = −9.001 MPa, 2 σ = −25.316 MPa 3 Successive introduction of these values into Eq. (1.31), together with Eq. (1.30a), or application of Eqs. (B.6) yields the direction cosines that define the orienta-tion of the planes on which , 1 σ 2 σ , and 3 σ act: l l l m m m n n n = = − = = − = = = − = − = − 0.0266, 0.6209, 0.7834 0.8638, 0.3802, 0.3306 0.5031, 0.6855, 0.5262 1 2 3 1 2 3 1 2 3 Note that the directions of the principal stresses are seldom required for pur-poses of predicting the behavior of structural members.
EXAMPLE 1.7 Three-Dimensional Stress in a Machine Component Given: The stress tensor at a point in a machine element with respect to a Cartesian coordinate system is given by the following array: τ = ii [ ] 50 10 0 10 20 40 0 40 30 MPa (f) Q P T M (a) 11.8 y 8.3 z Q (b) 4.6 4.7 6.45 19 x Figure 1.21. Example 1.6. (a) Hub-shaft assembly. (b) Element in three-dimensional stress.
42 Chapter 1 Analysis of Stress Find: The state of stress and I1, I2, and I3 for an x′ , y′ , z′ coordinate system defined by rotating x, y through an angle of 45 θ = ° counterclockwise about the z axis (Fig.1.22a).
Solution The direction cosines corresponding to the prescribed rotation of axes are given in Fig.1.22b. Thus, through the use of Eq. (1.28), we obtain τ = − − ′ ′ i j [ ] 45 15 28.28 15 25 28.28 28.28 28.28 30 MPa (g) Comments When the arrays (f) and (g) are substituted into Eq. (1.34), both yield I1 = 100 MPa, I2 = 1400 (MPa)2, and I3 = -53,000 (MPa)3, and the invari-ance of I1, I2, and I3 under the orthogonal transformation is confirmed.
1.14 NORMAL AND SHEAR STRESSES ON AN OBLIQUE PLANE A cubic element subjected to principal stresses , 1 σ 2 σ , and 3 σ acting on mutually perpendicular principal planes is described as being in a state of triaxial stress (Fig. 1.23a). In this figure, the x, y, and z axes are parallel to the principal axes. Clearly, this stress condition is not the general case of three-dimensional stress, which was taken up in the last two sections. Nevertheless, we sometimes need to determine the shearing and normal stresses acting on an arbitrary oblique plane of a tetrahedron, as in Fig. 1.23b, given the principal stresses or triaxial stresses acting on perpendicular planes. In the figure, the x, y, and z axes are parallel to the principal axes. Denoting the direction cosines of plane ABC by l, m, and n, Eqs. (1.26) with , 1 x σ σ = 0 xy xz τ τ = = and so on, reduce to , , 1 2 3 p l p m p n x y z σ σ σ = = = (a) Referring to Fig. 1.23a and definitions (a), the stress resultant p is related to the prin-cipal stresses and the stress components on the oblique plane by the expression 2 1 2 2 2 2 2 3 2 2 2 2 p l m n σ σ σ σ τ = + + = + (1.36) y x′ x′ x 1 2 1 2 0 −1 2 0 (a) (b) 0 1 1 2 0 y z y′ z′ y′ z x θ Figure 1.22. Example 1.7. Direction cosines for θ = 45° 1.14 Normal and Shear Stresses on an Oblique Plane 43 The normal stress σ on this plane, from Eq. (1.28a), is found as 1 2 2 2 3 2 l m n σ σ σ σ = + + (1.37) Substituting this expression into Eq. (1.36) leads to 2 1 2 2 2 2 2 3 2 2 2 l m n τ σ σ σ σ = + + − (1.38a) or ( ) 2 1 2 2 2 2 2 3 2 2 1 2 2 2 3 2 2 l m n l m n τ σ σ σ σ σ σ = + + − + + (1.38b) Expanding and using the expressions 1 ,1 , 2 2 2 2 2 2 l m n n l m − = + − = + and so on, the fol-lowing result is obtained for the shearing stress τ on the oblique plane: ( ) ( ) ( ) 1 2 2 2 2 2 3 2 2 2 3 1 2 2 2 1/2 l m m n n l τ σ σ σ σ σ σ = − + − + − (1.39) This clearly indicates that if all the principal stresses are equal, the shear stress vanishes, regardless of the choices made for the direction cosines.
For situations in which both shear and normal stresses act on perpendicular planes (Fig. 1.23b), we have , px py, and pz defined by Eqs. (1.26). Then, Eq. (1.37) becomes 2( ) 2 2 2 l m n lm mn ln x y z xy yz xz σ σ σ σ τ τ τ = + + + + + (1.40) Hence, τ σ τ τ τ σ τ τ τ σ σ = + + + + + + + + − l m n l m n l m n x xy xz xy y yz xz yz z ( ) ( ) ( ) 2 2 2 2 1/2 (1.41) where σ is given by Eq. (1.40). Formulas (1.37) through (1.41) represent the simplified transformation expressions for the three-dimensional stress.
(b) Q z x y n A B C σ1 σ σ3 σ2 τ py pz px y z (a) x σ1 σ3 σ2 Q Figure 1.23. Elements in triaxial stress.
44 Chapter 1 Analysis of Stress Substitution of the direction cosines from Eqs. (a) into Eq. (1.25) leads to 1 1 2 2 2 3 2 p p p x y z σ σ σ + + = (1.42) which is a stress ellipsoid having its three semiaxes as the principal stresses (Fig.1.24). This geometric interpretation helps explain the earlier conclusion that the principal stresses are the extreme values of the normal stress. If 1 2 3 σ σ σ = = , a state of hydrostatic stress exists, and the stress ellipsoid becomes a sphere. In this case, any three mutually perpendicular axes can be taken as the principal axes.
1.14.1 Octahedral Stresses The stresses acting on an octahedral plane are represented by face ABC in Fig. 1.23b with QA = QB = QC. The normal to this oblique face, in turn, has equal direction cosines rela-tive to the principal axes. Since 1 2 2 2 l m n + + = we have 1 3 l m n = = = (b) Plane ABC is clearly one of eight such faces of a regular octahedron (Fig. 1.25). Equations (1.39) and (b) are now applied to provide an expression for the octahedral shearing stress, which may be rearranged as follows: [( ) ( ) ( ) ] oct 1 3 1 2 2 2 3 2 3 1 2 1/2 τ σ σ σ σ σ σ = − + − + − (1.43) py px pz σ1 σ3 σ2 Figure 1.24. Stress ellipsoid.
Octahedral plane C B A σoct τoct σ2 σ3 σ1 Q Figure 1.25. Stresses on an octahedron.
1.15 Mohr’s Circles in Three Dimensions 45 Through the use of Eqs. (1.37) and (b), we obtain the octahedral normal stress: ( ) oct 1 3 1 2 3 σ σ σ σ = + + (1.44) Thus, the normal stress acting on an octahedral plane is the average of the principal stresses, the mean stress. The orientations of oct σ and oct τ are indicated in Fig. 1.25. The fact that the normal and shear stresses are the same for the eight planes is a powerful tool that can be used in failure analysis of ductile materials (see Section 4.7). Another useful form of Eq. (1.43) is developed in Section 2.15.
1.15 MOHR’S CIRCLES IN THREE DIMENSIONS Consider the wedge shown in Fig. 1.26a, which is cut from a cubic element subjected to triaxial stresses (Fig. 1.23a). The only stresses on the inclined ′ x face (parallel to the z axis) are the normal stress σ ′ x and the shear stress τ ′ ′ x y acting in the ′ ′ x y plane. These stresses are determined from force equilibrium equations the ′ ′ x y plane, so they are inde-pendent of the stress. Thus, the transformation equations of plane stress (Section 1.9) and Mohr’s circle can be employed to obtain the stresses x σ ′ and x y τ ′ ′. The same is true for normal and shear stresses acting on inclined faces cut through the element parallel to the x and y axes.
The stresses acting on elements oriented at various angles to the principal axes can be visualized with the aid of Mohr’s circle. The cubic element (Fig. 1.23a) viewed from three different directions is sketched in Figs. 1.27a to c. A Mohr’s circle is drawn corresponding to each projection of an element. The cluster of three circles represents Mohr’s circles for triaxial stress (Fig. 1.27d). The radii of the circles are equal to the maximum shear stresses, as indicated in the figure. The normal stresses acting on the planes of maximum shear stresses have the magnitudes given by the abscissa as of the centers of the circles.
45° σ3 σ2 σ1 (b) σ2 σ3 σ1 σx ′ τx′y ′ (a) y′ x′ x θ Figure 1.26. Triaxial state of stress: (a) wedge; (b) planes of maximum shear stress.
46 Chapter 1 Analysis of Stress 1.15.1 Absolute Maximum Shear Stress The largest shear stresses occur on planes oriented at 45° to the principal planes. The shear stress is a maximum located at the highest point on the outer circle. The value of the absolute maximum shearing stress is therefore ( ) ( ) ( ) max 13 1 2 1 3 a a τ τ σ σ = = − (1.45) acting on the planes that bisect the planes of the maximum and minimum principal stresses, as shown in Fig. 1.26b. Alternatively, the planes of maximum shear stress may be ascertained by substituting 1 2 2 2 n l m = − − into Eq. (1.38b), differentiating with respect to l and m, and equating the resulting expressions to zero (Prob. 1.86).
Determining the absolute value of maximum shear stress is a significant step when designing members made of ductile materials, as the strength of the material depends on its ability to resist shear stress (Section 4.5). Obviously, in terms of the stress mag-nitudes, the largest circle is the most significant one. However, all stresses in their var-ious transformations may play a role in causing failure, and it is usually instructive to plot all three principal circles of stress. One such example arises with thin-walled pres-surized cylinders, where , 1 σ σ = θ 2 a σ σ = , and 0 3 r σ σ = = at the outer surface (Table 1.1). In the special case where two or all principal stresses are equal, a Mohr’s circle becomes a point.
σ2 σ1 σ3 x y (a) σ3 σ1 σ2 x z (b) z y σ2 σ3 σ1 (c) (d) τ σ1 σ σ3 O In-plane τ max Absolute (τ max)a (τ 13)max (τ 23)max (τ 12)max C2 C3 C1 σ2 Figure 1.27. (a–c) Views of elements in triaxial stresses on different principal axes; (d) Mohr’s circles for three-dimensional stress.
1.15 Mohr’s Circles in Three Dimensions 47 1.15.2 Equations of Three Mohr’s Circles for Stress So far, we have shown that, given the values of the principal stresses and of the direction cosines for any oblique plane (Fig. 1.23b), we can determine the normal and shear stresses on the plane by applying Eqs. (1.37) and (1.38). This goal may also be accomplished by means of a graphical technique due to Mohr [Ref. 1.8]. Mohr’s procedure was used in the early history of stress analysis, but today it is employed only as a heuristic device.
In the following discussion, we demonstrate that the previously given equations together with the relation 1 2 2 2 l m n + + = are represented by three circles of stress, and the coordinates ( , ) σ τ locate a point in the shaded area of Fig. 1.27d. These simultaneous equations are 1 2 2 2 1 2 2 2 3 2 2 1 2 2 2 2 2 3 2 2 2 l m n l m n l m n σ σ σ σ τ σ σ σ σ = + + = + + = + + − (a) where l ≥0, 2 m ≥0 2 , and n ≥0.
2 Solving for the direction cosines, results in σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ = + − − − − ≥ = + − − − − ≥ = + − − − − ≥ l m n ( )( ) ( )( ) 0 ( )( ) ( )( ) 0 ( )( ) ( )( ) 0 2 2 2 3 1 2 1 3 2 2 3 1 2 3 2 1 2 2 1 2 3 1 3 2 (1.46) Given that σ σ σ > > 1 2 3, the numerators of Eqs. (1.46) satisfy σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ + − − ≥ + − − ≤ + − − ≥ ( )( ) 0 ( )( ) 0 ( )( ) 0 2 2 3 2 3 1 2 1 2 (b) the denominators of Eqs. (1.46) are σ σ − > ( ) 0 1 2 and σ σ − > ( ) 0, 1 3 σ σ − > ( ) 0 2 3 and σ σ − < ( ) 0, 2 1 and σ σ − < ( ) 0 3 1 and < σ σ − ( ) 0 3 2 , respectively.
Finally, the preceding inequalities may be expressed as follows: [ ( )] ( ) ( ) [ ( )] ( ) ( ) [ ( )] ( ) ( ) 2 1 2 2 3 2 1 4 2 3 2 23 max 2 2 1 2 1 3 2 1 4 1 3 2 13 max 2 2 1 2 1 2 2 1 4 1 2 2 12 max 2 σ σ σ σ σ σ τ σ σ σ σ σ σ τ σ σ σ σ σ σ τ + − + ≥ − = + − + ≤ − = + − + ≥ − = (1.47) Equations (1.47) represent the formulas of the three Mohr’s circles for stress, shown in Fig. 1.26d. Stress points (σ , τ ) satisfying the equations for circles centered at C1 and C2 48 Chapter 1 Analysis of Stress lie on or outside the circles, but for the circle centered at C3 lie on or inside the circle. We conclude that an admissible state of stress must lie on Mohr’s circles or within the shaded area enclosed by these circles.
EXAMPLE 1.8 Analysis of Three-Dimensional Stresses in a Member Given: The state of stress on an element of a structure illustrated in Fig.1.28a.
Find: Using Mohr’s circle: (a) the principal stresses; (b) the maximum shear-ing stresses. Show the results on a properly oriented element. (c) Apply the equations developed in Section 1.14 to calculate the octahedral stresses.
Solution a. First, Mohr’s circle for the transformation of stress in the xy plane is sketched in the usual manner as shown, centered at C2 with diameter A2A3 (Fig. 1.28b). Next, we complete the three-dimensional Mohr’s cir-cle by drawing two additional circles of diameters A1A2 and A1A3 in the figure. Referring to the circle, the principal stresses are 100 MPa, 1 σ = σ σ = = − 40 MPa, and 60 MPa.
2 3 Angle 26.56°, as tan 2 4/3 p θ θ ′′′= = . The results are sketched on a properly oriented element in Fig. 1.28c.
b. The absolute maximum shearing stress, point B3, equals the radius of the circle centered at C3 of diameter A1 A3. Thus, ( ) ( ) [100 ( 60)] 80 MPa 13 max max 1 2 a τ τ = = −− = (b) (a) 40MPa 40MPa 100 MPa 20MPa x z y τ (MPa) B3(20, 80) σ (MPa) A1 100 60 A3 (−40, −40) y C2 C3 A2 C1 −10 (c) 40 MPa 60 MPa 100 MPa z x ′ y ′ y x θp ′″ 2θp ′″ 40 Figure 1.28. Example 1.8. (a) Element in three-dimensional stress; (b) Mohr’s circles of stress; (c) stress element for 26.56° p θ′′′= .
1.16 Boundary Conditions in Terms of Surface Forces 49 The maximum shearing stress occurs on the planes 45° from the y′ and z faces of the element of Fig. 1.28c.
c. The octahedral normal stress, from Eq. (1.44), is (100 40 60) 26.7 MPa oct 1 3 σ = + − = The octahedral shearing stress, using Eq. (1.43), is [(100 40) (40 60) ( 60 100) ] 66 MPa oct 1 3 2 2 2 1 2 τ = − + + + − − = Comment Comparing the results (see Fig. 1.28b) reveals that σ σ τ τ < < a and ( ) oct 1 oct max That is, the maximum principal stress and absolute maximum shear stress are greater than their octahedral counterparts.
1.16 BOUNDARY CONDITIONS IN TERMS OF SURFACE FORCES We now consider the relationship between the stress components and the given surface forces acting on the boundary of a body. The equations of equilibrium that must be satis-fied within a body were derived in Section 1.8. The distribution of stress in a body must also be able to accommodate the conditions of equilibrium with respect to externally applied forces. Thus, these external forces may be regarded as a continuation of the inter-nal stress distribution.
Consider the equilibrium of the forces acting on the tetrahedron shown in Fig.1.20a, and assume that the oblique face ABC is coincident with the surface of the body. The com-ponents of the stress resultant p now become the surface forces per unit area, or tractions, , px , py and pz. The equations of equilibrium for this element, representing boundary con-ditions, are, from Eqs. (1.26), p l m n p l m n p l m n x x xy xz y xy y yz z xz yz z σ τ τ τ σ τ τ τ σ = + + = + + = + + (1.48) For example, if the boundary is a plane with an x-directed surface normal, Eqs. (1.48) give , px x σ = py xy τ = , and pz xz τ = ; under these circumstances, the applied surface force components , p p x y, and pz are balanced by , x σ xy τ , and , xz τ respectively.
Sometimes, instead of prescribing the distribution of surface forces on the bound-ary, the boundary conditions of a body may be given in terms of displacement compo-nents. Furthermore, we may be given boundary conditions that prescribe surface forces 50 Chapter 1 Analysis of Stress on one part of the boundary and displacements on another. When displacement bound-ary conditions are given, the equations of equilibrium express the situation in terms of strain, through the use of Hooke’s law and subsequently in terms of the displacements by means of strain-displacement relations (Section 2.3). In engineering problems, how-ever, the usual practice is to specify the boundary conditions in terms of surface forces, as in Eq. (1.48), rather than surface displacements. We will adhere to this practice in this text.
1.17 INDICIAL NOTATION This text uses a system of symbols called indicial notation, index notation, or tensor nota-tion to represent components of force, stress, displacement, and strain. Within this system, a particular class of tensor, a vector, requires only a single subscript to describe each of its components. Other tensors, however, may require more than a single subscript for defini-tion. For example, second-order or second-rank tensors, such as those of stress or inertia, require double subscripts: , ij τ Iij. Quantities such as temperature and mass are scalars, classified as tensors of zero rank.
Tensor or indicial notation offers the advantage of succinct representation of lengthy equations through the minimization of symbols. In addition, physical laws expressed in tensor form are independent of the choice of coordinate system, such that similarities in seemingly different physical systems may become obvious. That is, indi-cial notation generally provides insight and understanding not readily apparent to the relative newcomer to the field. It results in a saving of space and serves as an aid in non-numerical computation.
The displacement components u, υ, and w, for instance, are written u1, u2, u3 (or ux, uy, uz) and collectively as ui, as with the understanding that the subscript i can be 1, 2, and 3 (or x, y, z). Similarly, the coordinates themselves are represented by x1, x2, x3, or simply ( 1,2,3) x i i = , and , , x x x x y z, or ( , , ) x i x y z i = . Many equations of elasticity become unwieldy when written in full, unabbreviated term; see, for example, Eqs. (1.28). As the complexity of the situation described increases, so does that of the formulations, tending to obscure the fundamentals in a mass of symbols. For this reason, the more compact indi-cial notation is sometimes found in publications.
Two simple conventions enable us to write most equations developed in this text in indicial notation. These conventions, relative to range and summation, are as follows: Range convention: When a lowercase alphabetic subscript is unrepeated, it takes on all values indicated.
Summation convention: When a lowercase alphabetic subscript is repeated in a term, then summation over the range of that subscript is indicated, making unnecessary the use of the summation symbol.
The introduction of the summation convention is attributed to Albert Einstein (1879– 1955). This notation, in conjunction with the tensor concept, has far-reaching conse-quences not restricted to its notational convenience [Refs. 1.9 and 1.10].
Problems 51 REFERENCES 1.1. Timoshenko, S. P . History of Strength of Materials. New Y ork: Dover, 2011.
1.2. Todhunter, L., and Pearson, K. History of the Theory of Elasticity and the Strength of Materials, Vols. I and II. New Y ork: Dover, 1960.
1.3. Love, A. E. H. A Treatise on the Mathematical Theory of Elasticity, 4th ed. New Y ork: Dover, 2011.
1.4. Ugural, A. C. Mechanical Design of Machine Components, 2nd ed. Boca Raton, FL: CRC Press, Taylor & Francis Group, 2016.
1.5. Ugural, A. C. Mechanics of Materials, Hoboken, NJ: Wiley, 2008.
1.6. Boresi, A. P ., and Chong, K. P . Elasticity in Engineering Mechanics, 2nd ed. Hoboken, NJ: Wiley, 2000.
1.7. Sokolnikoff, I. S. Mathematical Theory of Elasticity, 2nd ed. Melbourne, FL: Krieger, 1986.
1.8. Ugural, A. C., and Fenster, S. K. Advanced Strength and Applied Elasticity, 4th ed. Englewood Cliffs, NJ: Prentice Hall, 2003.
1.9. Reismann, H., and Pawlik, P . S. Elasticity: Theory and Applications. Hoboken, NJ: Wiley, 1980.
1.10. Chou, P . C., and Pagano, N. J. Elasticity. New Y ork: Dover, 1992.
PROBLEMS Sections 1.1 through 1.8 1.1. Two prismatic bars of a by b rectangular cross section are glued as shown in Fig. P1.1. The allowable normal and shearing stresses for the glued joint are 700 and 560 kPa, respectively. Assuming that the strength of the joint controls the design, what is the largest axial load P that may be applied? Use 40°, φ = a = 50 mm, and b = 75 mm.
1.2. A prismatic steel bar of a = b = 50-mm2 cross section is subjected to an axial ten-sile load 125 kN = P (Fig. P1.1). Calculate the normal and shearing stresses on all faces of an element oriented at (a) 70° φ = and (b) 45°.
φ = 1.3. A prismatic bar is under an axial load, producing a compressive stress of 75 MPa on a plane at an angle 30° θ = (Fig. P1.3). Determine the normal and shearing stresses on all faces of an element at an angle of 50° θ = .
75 MPa θ = 30° Figure P1.3.
P a φ P Glue b Figure P1.1. |
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2024 AMC 10A Problems/Problem 14
Contents
1 Problem
2 Diagram
3 Solution 1
4 Note
5 Solution 2 (Quick Guess)
6 Solution 3
7 Solution 4 (oh no)
8 Solution 5 (Pure Geometry Approach — Region DCE minus Circular Segment)
9 Solution 6
10 Video Solution(Fast! 30-60-90 Triangle solution)
11 Video Solution by Number Craft
12 Video Solution by Pi Academy
13 Video Solution 1 by Power Solve
14 Video Solution by SpreadTheMathLove
15 Video Solution by Just Math⚡
16 Video Solution by Dr. David
17 Video solution by TheNeuralMathAcademy
18 See Also
Problem
One side of an equilateral triangle of height lies on line . A circle of radius is tangent to line and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line can be written as , where , , and are positive integers and is not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM
Solution 1
Call the bottom vertices of the triangle and (the one closer to the circle is ) and the top vertex . The tangency point between the circle and the side of the triangle is , and the tangency point on line , and the center of the circle is .
Draw radii to the tangency points, the arc is degrees because is , and since is supplementary, it's .(Using Angle of Intersecting Secants Theorem) The sum of the angles in a quadrilateral is , which means is
Triangle ODC is -- triangle so CD is . Since we have congruent triangles ( and ), the combined area of both is . The area of the arc is which is , so the answer is
is which is
~ASPALAPATI75 ~andy_liu766 (latex)
edits by KR
Note
There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line and the base of the equilateral triangle. However, since the area in this configuration is simply we can infer that the problem is talking about the configuration in Solution 1.
~dbnl
Solution 2 (Quick Guess)
Since this problem involves equilateral triangles, the only possible number under the square root is . Now subtracting all of the answer choices by , we get:
Due to the even parity of the problem, we can safely assume that the answer is either or , but as is a multiple of and , we get the answer of .
~megaboy6679
Solution 3
(pardon the diagrams:D)
say the area we want to find is x.
since the equilateral triangle has an internal angle of 60, the exterior angle formed by the triangle and the line is 120. simplifying the diagram you will get:
\ #####
\ ########
########
#####_____
make three of these that each circle is tangent to the other 2 circles
\
\ #####
\ ########
#############
#_####____
#/
#####/########
/ ########
/ ######
Since they are 3 congruent triangles, you can make an equilateral triangle using their radius(12), with each vertex at the center of each circle. This will make an equilateral triangle of side length 24. if you look now, the area within the equilateral triangle consists of 3 of a circle, and 3 of x.
we first find the area of the triangle, which is , we then find the area of of a circle, which is , we subtract by , and divide by 3, yielding the area of x.
-
___________ =
3
is which is
~Yiguo Zhang
Solution 4 (oh no)
Setting up the problem graphically:
Define the parametric function of the circle r(θ), centered at (0,12) with radius 12, as:
The line ℓ is taken to be the x-axis, so y=0. The circle is tangent to ℓ at (0,0).
We only care about the side of the triangle that is also tangent to the circle; call this line f(x). Since the triangle is equilateral of height 24, one vertex is at (0,24) (directly above the tangent point), but for now we focus on the particular side tangent to the circle.
Because we expect that line to make a 330∘ angle with the positive x-axis (or a slope of −3), let
We also know the perpendicular distance from the circle’s center (0,12) to f(x) must be 12, because the circle of radius 12 is externally tangent to that line. The distance from a point (x 0,y 0) to a line A x+B y+C=0 is:
Here, f(x)=−3 x+c can be rewritten as 3 x+y−c=0. Plugging in (x 0,y 0)=(0,12) and setting the distance to 12 yields:
Solving this and noting that f(x) has a negative y-intercept gives:
Next, we find the point at which r(θ) intersects f(x). Substituting r(θ)=(12 cos(θ),12 sin(θ)+12) into f(x)=−3 x−12, we set:
One helpful approach is to note that a line perpendicular to f(x) and passing through (0,12) has slope 1/3, so that line is g(x)=x 3+12. Intersecting g(x) with f(x) gives:
which solves to x=−6 3. Then y=6. So the intersection point of the circle and the line is (−6 3,6).
To find the corresponding θ, use:
From the first, cos(θ)=−3/2. This corresponds to θ=5 π/6 or 7 π/6. Checking the y-coordinate shows θ=7 π/6 indeed yields sin(7 π/6)=−1/2, so 12 sin(7 π/6)+12=6.
The circle also meets ℓ (the x-axis) at (0,0). In polar terms, that is θ=3 π/2. We want the area under the circle’s parametric curve from θ=7 π/6 to θ=3 π/2.
Parametrically, the area under y(θ) from θ 1 to θ 2 can be found via:
Here x(θ)=12 cos(θ), so x′(θ)=−12 sin(θ). Hence the area under the circle from θ=7 π/6 to θ=3 π/2 is:
This simplifies (pulling out constants) to:
Evaluating these integrals yields:
Next, find where f(x)=0:
However, we also need the relevant intersection interval for the region. The line and the circle intersect at x=−6 3. So the area under f(x) from x=−6 3 to x=−4 3 is:
Carrying out that integral (or carefully checking the geometry) gives 6 3.
Subtracting this triangular “cap” (6 3) from the circle’s sector area (54 3−24 π) gives the final region of interest:
Hence the area can be written as
so a=48, b=3, and c=24. Therefore,
D)75
~meihk_neiht
P.S.: Please don’t whip out calculus like this on an AMC 10. Yes,it’s“doable”without a calculator(I speak from painful experience). Yes,I did it,but it took forever,and trust me,nobody at the AMC office expects calculus.
Solution 5 (Pure Geometry Approach — Region DCE minus Circular Segment)
Step 1: Area of triangle
Given: (radius of the circle), and the corresponding height from to is . - The area formula for a triangle is:
Explanation: This step finds the area of triangle using its base and height.
Step 2: Area of sector
The radius is , and the central angle is . - The formula for the area of a sector is:
Explanation: The sector's area is a fraction () of the full circle's area.
Step 3: Area of triangle
, and the included angle . - The formula for an equilateral triangle's area:
Explanation: Triangle is equilateral, so we use the standard area formula for such a triangle.
Step 4: Area of the circular segment (arc minus chord )
Explanation: The segment is the region between the arc and the chord .
Step 5: Final required region
Explanation: The required area is the triangle minus the circular segment. This gives the area in the desired region bounded by the triangle, the circle, and the line.
Step 6: Answer in the form
Final Answer:
, and
-clicksong
Solution 6
Step 1: Area of triangle (using Heron's formula)
By geometry, is a triangle with sides , , . - Let , , . - The semi-perimeter is:
The area by Heron's formula:
However, the actual required region is only of this triangle (since the shaded region is less than the whole triangle). - So,
But based on precise geometry (as in AMC standard), the correct area to use for this region is (check the triangle's effective area).
Step 2: Area of sector
The radius is and the central angle is . - The formula for the area of a sector is:
Step 3: Area of triangle
, . - The formula for an equilateral triangle's area:
Step 4: Area of the circular segment (arc minus chord )
Step 5: Final required region
Explanation: The required area is the triangle minus the circular segment.
Step 6: Expressing the answer as
Final Answer:
, and
also by clicksong
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A level Physics (12.2) Potential difference and power (Electric current)
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A level Physics (12.2) Potential difference and power (Electric current)
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1551 | https://www.quora.com/If-R-is-the-radius-of-the-earth-and-g-is-the-acceleration-due-to-gravity-on-earths-surface-what-is-the-mean-density-of-the-earth | If R is the radius of the earth and g is the acceleration due to gravity on earth's surface, what is the mean density of the earth? - Quora
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If R is the radius of the earth and g is the acceleration due to gravity on earth's surface, what is the mean density of the earth?
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Victor Mazmanian
Former Associate Prof. Of Physics (Retired) at United States Air Force Academy · Author has 1.5K answers and 4.7M answer views
·7y
Density is defined as Mass / Unit Volume.
Taking the globe as a sphere (it’s actually an oblong spheroid) the volume is calculated by: 4/3 pi R^3.
The acceleration due to gravity at the surface of the Earth “g” = G M(e) / R^2
This gives us the mass of the Earth: M(e) = g R^2 / G
Using definition of Density: g R^2 / G / 4/3 pi R^3 = (3 g) / (4 G pi R)
Plugging in book values, we obtain D = 5,500 kg / m^3 which matches most commonly accepted value for density of Earth.
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What is the relationship between acceleration due to gravity and mean density in terms of the gravitational constant and the radius of the earth?
What is the value of acceleration due to gravity (g) at a distance of 2R from the surface of earth, where R is the radius of the earth?
What is the value of g (the acceleration due to gravity) at Earth's surface? What is the radius of Earth?
What is the formula for calculating acceleration due to gravity on Earth's surface using mass and radius values of Earth?
The mass of a planet is twice that of Earth and its density is twice of Earth. What is the ratio of the accelerations due to gravity of the planet and Earth?
Kaiser Tarafdar
Physician at Covenant Medical Group · Author has 2.6K answers and 5.2M answer views
·Updated 6y
Let's take an object, say it's a ball that you threw upwards then its weight mg which is the gravitational force acting on it will be equal to Newton's equation for universal gravitation which is F = GMm/d^2, where G is Newton's universal gravitational constant, M is earth's mass, m is the mass of object in free fall, and d distance from the center of the earth, being the radius R here. You can ignore the height it reaches as it is such a tiny increase in the R, and also all things on earth's surface have weights which are given by mg, with the same g, and the distance between any object on ea
Continue Reading
Let's take an object, say it's a ball that you threw upwards then its weight mg which is the gravitational force acting on it will be equal to Newton's equation for universal gravitation which is F = GMm/d^2, where G is Newton's universal gravitational constant, M is earth's mass, m is the mass of object in free fall, and d distance from the center of the earth, being the radius R here. You can ignore the height it reaches as it is such a tiny increase in the R, and also all things on earth's surface have weights which are given by mg, with the same g, and the distance between any object on earth's surface and the center of earth's mass is the radius of the earth, R with little variation with latitudes. So we can write it mathematically as follows : GMm/R^2 = mg, which gives you g = GM/R^2 which when solved for M gives you M = gR^2/G. (So you can see also that g varies directly with the mass of any planet or moon, and inversely to the second power of the radius, and M varies with g, and the second power of the radius directly). Now we have a mathematical expression for the mass of the earth M = gR^2/G as derived above ; just beautiful mathematics ! As you can see the smaller mass m cancels out and is never a determinant for the g or the M. The earth's volume is given by the volume of a sphere which is 4/3pR^3 where p is pi. To calculate density you have to divide mass by volume. So let's do that here and get an equation for the density of the earth as follows : D = M/V, hence D = (gR^2/G)/4/3pR^3 which gives you the mean density equation : D = (3/4)g/pGR or D = 3g/4pGR. The only variable here is R in the above equation for the earth, with 3g/4pG being a constant for earth, and both g and R for other planets, and moons will vary with different values of g and R. The acceleration due to gravity on earth, g, varies slightly only based on latitudes, as earth is not a perfect sphere, but an oblate spheroid as stated by Newton 331 years ago !!! As you can see if you keep g constant and vary the R or the radius, then greater the radius the lower the density, and smaller the radius the greater the density(with mass remaining the same inherently) showing an inverse relationship of mean density and the radius R. If you keep R constant and vary g, then greater the g, greater is the density, and smaller the g, the lower is the density(showing variable mass within the same volume) in a direct proportionality. It is to be noted that Newton was able to estimate the mass of the earth working from his own creation, the universal law of gravitation and his equation, with an ingenious approach(before G would be experimentally calculated by Cavendish hundred years later) and was able to estimate the mass of the earth off by a very tiny percentage, way less than 0.1 % ; Just sheer genius ! Kaiser T, MD(Life long physics, mathematics, and science enthusiast).
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To find the mean density of the Earth, we can use the relationship between gravitational acceleration at the surface, the radius of the Earth, and the Earth's density.
The formula for the acceleration due to gravity [math]g[/math] at the surface of a spherical body is given by:
[math]g = \frac{G M}{R^2}[/math]
where:
[math]G[/math] is the gravitational constant, approximately [math]6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}[/math]
[math]M[/math] is the mass of the Earth
[math]R[/math] is the radius of the Earth
The mean density [math]\rho[/math] of the Earth can be expressed in terms of its mass and volume. The volume [math]V[/math] of a sphere is given by:
[math]V = \frac{4}{3}[/math]
Continue Reading
To find the mean density of the Earth, we can use the relationship between gravitational acceleration at the surface, the radius of the Earth, and the Earth's density.
The formula for the acceleration due to gravity [math]g[/math] at the surface of a spherical body is given by:
[math]g = \frac{G M}{R^2}[/math]
where:
[math]G[/math] is the gravitational constant, approximately [math]6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}[/math]
[math]M[/math] is the mass of the Earth
[math]R[/math] is the radius of the Earth
The mean density [math]\rho[/math] of the Earth can be expressed in terms of its mass and volume. The volume [math]V[/math] of a sphere is given by:
[math]V = \frac{4}{3} \pi R^3[/math]
Thus, the mean density [math]\rho[/math] is:
[math]\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}[/math]
Now, substituting [math]M[/math] from the gravitational equation into the density formula:
From the equation for [math]g[/math]:
[math]M = \frac{g R^2}{G}[/math]
Substituting [math]M[/math] into the density equation:
[math]\rho = \frac{3}{4 \pi R^3} \cdot \frac{g R^2}{G}[/math]
Simplifying this gives:
[math]\rho = \frac{3g}{4 \pi G R}[/math]
Constants:
[math]g \approx 9.81 \, \text{m/s}^2[/math]
[math]G \approx 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}[/math]
[math]R \approx 6.371 \times 10^6 \, \text{m}[/math]
Calculation:
Now we can plug in the values:
[math]\rho \approx \frac{3 \times 9.81 \, \text{m/s}^2}{4 \pi (6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(6.371 \times 10^6 \, \text{m})}[/math]
Calculating the denominator:
[math]4 \pi G R \approx 4 \pi (6.674 \times 10^{-11})(6.371 \times 10^6) \approx 5.303 \times 10^6 \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}[/math]
Now substituting back:
[math]\rho \approx \frac{3 \times 9.81}{5.303 \times 10^6} \approx \frac{29.43}{5.303 \times 10^6} \approx 5.55 \times 10^3 \, \text{kg/m}^3[/math]
Conclusion:
The mean density of the Earth is approximately:
[math]\rho \approx 5.55 \, \text{g/cm}^3[/math]
This value is consistent with the generally accepted mean density of the Earth, which is around [math]5.5 \, \text{g/cm}^3[/math].
Upvote ·
Nrusingha Charan Behera
B.Tech(Agril. Engg ) in Mathematics&Physics, Orrisa University of Agriculture & Technology, Bhubaneswar (Graduated 1993) · Author has 5.9K answers and 5.1M answer views
·7y
It is 3g/4πGRe.Ans…the explanation follows below.
Continue Reading
It is 3g/4πGRe.Ans…the explanation follows below.
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Related questions
What is the relationship between acceleration due to gravity and mean density in terms of the gravitational constant and the radius of the earth?
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What is the value of g (the acceleration due to gravity) at Earth's surface? What is the radius of Earth?
What is the formula for calculating acceleration due to gravity on Earth's surface using mass and radius values of Earth?
The mass of a planet is twice that of Earth and its density is twice of Earth. What is the ratio of the accelerations due to gravity of the planet and Earth?
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If the diameter of Earth were to double, while maintaining the same mass and density, what would be the new acceleration due to gravity on Earth's surface?
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Related questions
What is the relationship between acceleration due to gravity and mean density in terms of the gravitational constant and the radius of the earth?
What is the value of acceleration due to gravity (g) at a distance of 2R from the surface of earth, where R is the radius of the earth?
What is the value of g (the acceleration due to gravity) at Earth's surface? What is the radius of Earth?
What is the formula for calculating acceleration due to gravity on Earth's surface using mass and radius values of Earth?
The mass of a planet is twice that of Earth and its density is twice of Earth. What is the ratio of the accelerations due to gravity of the planet and Earth?
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1554 | https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/12.9/primary/lesson/induction-proofs-pcalc/ | Induction Proofs | CK-12 Foundation
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12.9
Induction Proofs
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Induction Proofs
Written by:CK-12 |Mark Spong
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Last Modified: Aug 01, 2025
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Lesson
Induction is one of many methods for proving mathematical statements about numbers. The basic idea is that you prove a statement is true for a small number like 1. This is called the base case. Then, you show that if the statement is true for some random number k, then it must also be true for k+1.
An induction proof is like dominoes set up in a line, where the base case starts the falling cascade of truth. Once you have shown that in general if the statement is true for k then it must also be true for k+1, it means that once you show the statement is true for 1, then it must also be true for 2, and then it must also be true for 3, and then it must also be true for 4 and so on.
What happens when you forget the base case?
Proof by Induction
Induction is a method of proof usually used to prove statements about positive whole numbers (the natural numbers). Induction has three steps:
The base case is where the statement is shown to be true for a specific number. Usually this is a small number like 1. This is the first domino to fall, creating a cascade and thus proving the statement true for every number greater than the base case.
The inductive hypothesis is where the statement is assumed to be true for k.
The inductive step/proof is where you show that then the statement must be true for k+1.
These three logical pieces will show that the statement is true for every number greater than the base case.
Suppose you wanted to use induction to prove: n≥1,2+2 2+2 3+⋯+2 n=2 n−1−2.
Start with the Base Case. Show that the statement works when n=1:
2 1=2 and 2 1+1−2=4−2=2. Therefore, 2 1=2 1+1−2. (Both sides are equal to 2)
Next, state your Inductive Hypothesis. Assume that the statement works for some random number k:
2+2 2+⋯+2 k=2 k+1−2 (You are assuming that this is a true statement)
Next, you will want to use algebra to manipulate the previous statement to prove that the statement is also true for k+1. So, you will be trying to show that 2+2 2+⋯+2 k+1=2 k+1+1−2. Start with the inductive hypothesis and multiply both sides of the equation by 2. Then, do some algebra to get the equation looking like you want.
Inductive Hypothesis (starting equation): 2+2 2+⋯2 k=2 k+1−2
Multiply by 2: 2(2+2 2+⋯+2 k)=2(2 k+1−2)
Rewrite: 2 2+2 3+⋯+2 k+1=2 k+1+1−4
Add 2 to both sides: 2+2 2+2 3+⋯+2 k+1=2 k+1+1−4+2
Simplify: 2+2 2+⋯+2 k+1=2 k+1+1−2
This is exactly what you were trying to prove! So, first you showed that the statement worked for n=1. Then, you showed that the if the statement works for one number than it must work for the next number. This means, the statement must be true for all numbers greater than or equal to 1.
The idea of induction can be hard to understand at first and it definitely takes practice. One thing that makes induction tricky is that there is not a clear procedure for the “proof” part. With practice, you will start to see some common algebra techniques for manipulating equations to prove what you are trying to prove.
Examples
Example 1
Earlier, you were asked what happens if you forget the base case in induction. If you forget the base case in an induction proof, then you haven’t really proved anything. You can get silly results like this “proof” of the statement: “1=3”
Base Case: Missing
Inductive Hypothesis:k=k+1 where k is a counting number.
Proof: Start with the assumption step and add one to both sides.
k=k+1 k+1=k+2
Thus by transitivity of equality:
k=k+1=k+2 k=k+2
Since k is a counting number,k could equal 1. Therefore:
1=3
Example 2
There is something wrong with this proof. Can you explain what the mistake is?
F o r n≥1:1 2+2 2+3 2+⋯+n 2=n(n+1)(2 n+1)6
Base Case:1=1 2=1(1+1)(2⋅1+1)6=1⋅2⋅3 6=6 6=1
Inductive Hypothesis: Assume the following statement is true:
1 2+2 2+3 2+⋯+k 2=k(k+1)(2 k+1)6
Proof: You want to show the statement is true for k+1.
“Since the statement is assumed true for k, which is any number, then it must be true for k+1. You can just substitute k+1 in.”
1 2+2 2+3 2+⋯+(k+1)2=(k+1)((k+1)+1)(2(k+1)+1)6
This is the most common fallacy when doing induction proofs. The fact that the statement is assumed to be true for k does not immediately imply that it is true for k+1 and you cannot just substitute in k+1 to produce what you are trying to show. his is equivalent to assuming true for all numbers and then concluding true for all numbers which is circular and illogical.
Example 3
Write the base case, inductive hypothesis and what you are trying to show for the following statement. Do not actually prove it.
1 3+2 3+3 3+⋯+n 3=n 2(n+1)2 4
Base Case:1 3=1 2(1+1)2 4(Both sides are equal to 1)
Inductive Hypothesis:Assume the following statement is true:
1 3+2 3+3 3+⋯+k 3=k 2(k+1)2 4
Next, you would want to prove that the following is true:
1 3+2 3+3 3+⋯+k 3+(k+1)3=(k+1)2((k+1)+1)2 4
Example 4
Prove the following statement: F o r n≥1,1 3+2 3+3 3+⋯n 3=(1+2+3+⋯n)2.
Base Case(s):Two base cases are shown however only one is actually necessary.
1 3=1 2 1 3+2 3=1+8=9=3 2=(1+2)2
Inductive Hypothesis: Assume the statement is true for some number k . In other words, assume the following is true:
1 3+2 3+3 3+⋯k 3=(1+2+3+⋯k)2
You want to show the statement is true for k+1. It is a good idea to restate what your goal is at this point. Your goal is to show that:
1 3+2 3+3 3+⋯k 3+(k+1)3=(1+2+3+⋯k+(k+1))2
You need to start with the assumed case and do algebraic manipulations until you have created what you are trying to show (the equation above):
1 3+2 3+3 3+⋯k 3=(1+2+3+⋯k)2
From the work you have done with arithmetic series you should notice:
1+2+3+4+⋯+k=k 2(2+(k−1))=k(k+1)2
Substitute into the right side of the equation and add (k+1)3 to both sides:
1 3+2 3+3 3+⋯k 3+(k+1)3=(k(k+1)2)2+(k+1)3
When you combine the right hand side algebraically you get the result of another arithmetic series.
1 3+2 3+3 3+⋯k 3+(k+1)3=((k+1)(k+2)2)2=(1+2+3+⋯k+(k+1))2
∴
The symbol ∴is one of many indicators like QED that follow a proof to tell the reader that the proof is complete.
Example 5
Prove the following statement using induction:
For n≥1,1+2+3+4+⋯+n=n(n+1)2
Base Case:1=1(1+1)2=1⋅2 2=1
Inductive Hypothesis:1+2+3+4+⋯+k=k(k+1)2
Proof: Start with what you know and work to showing it true for k+1.
Inductive Hypothesis: 1+2+3+4+⋯+k=k(k+1)2
Add k+1 to both sides: 1+2+3+4+⋯+k+(k+1)=k(k+1)2+(k+1)
Find a common denominator for the right side: 1+2+3+4+⋯+k+(k+1)=k 2+k 2+2 k+2 2
Simplify the right side: 1+2+3+4+⋯+k+(k+1)=k 2+3 k+2 2
Factor the numerator of the right side: 1+2+3+4+⋯+k+(k+1)=(k+1)(k+2)2
Rewrite the right side: 1+2+3+4+⋯+k+(k+1)=(k+1)((k+1)+1)2
∴
Summary
Induction is a method for proving mathematical statements about numbers. There are three steps:
The base case shows the statement is true for a specific number, usually a small number like 1.
The inductive hypothesis assumes the statement is true for some random number k.
The inductive step/proof shows that if the statement is true for k,it must also be true for k+1,proving the statement for all numbers greater than or equal to the base case.
Review
For each of the following statements: a) show the base case is true; b) state the inductive hypothesis; c) state what you are trying to prove in the inductive step/proof. Do not prove yet.
For n≥5,4 n<2 n.
For n≥1,8 n−3 n is divisible by 5.
For n≥1,7 n−1 is divisible by 6.
For n≥2,n 2≥2 n.
For n≥1,4 n+5 is divisible by 3.
For n≥1,0 2+1 2+⋯+n 2=n(n+1)(2 n+1)6
Now, prove each of the following statements. Use your answers to problems 1-6 to help you get started.
For n≥5,4 n<2 n.
For n≥1,8 n−3 n is divisible by 5.
For n≥1,7 n−1 is divisible by 6.
For n≥2,n 2≥2 n.
For n≥1,4 n+5 is divisible by 3.
For n≥1,0 2+1 2+⋯+n 2=n(n+1)(2 n+1)6
You should believe that the following statement is clearly false. What happens when you try to prove it true by induction?
For n≥2,n 2<n
Explain why the base case is necessary for proving by induction.
The principles of inductive proof can be used for other proofs besides proofs about numbers. Can you prove the following statement from geometry using induction?
The sum of the interior angles of any n−g o n is 180(n−2) for n≥3.
Review (Answers)
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Asked by Students
Here are the top questions that students are asking Flexi for this concept:
What is the Induction Hypothesis?
Induction is a method of proof usually used to prove statements about positive whole numbers (the natural numbers). Induction has three steps:
The base case is where the statement is shown to be true for a specific number. Usually this is a small number like 1. This is the first domino to fall, creating a cascade and thus proving the statement true for every number greater than the base case.
The inductive hypothesis is where the statement is assumed to be true for k.
The inductive step/proof is where you show that then the statement must be true for k+1.
These three logical pieces will show that the statement is true for every number greater than the base case.
There are 3 consecutive integers that add up to 0. What is the value of the greatest integer?
Let's denote the three consecutive integers as n, n+1, and n+2.
According to the problem, their sum is 0, so we can write the equation as follows:
n+(n+1)+(n+2)=0
Solving this equation gives us the value of n:
3 n+3=0
3 n=−3
n=−1
The greatest integer is n+2, so we substitute n=−1 into n+2:
n+2=−1+2
n+2 = 1
What is the inductive hypothesis?
In an induction proof, the inductive hypothesis is the step where you assume the statement is true for k.
How does induction work?
Induction is a powerful mathematical proof technique used to show that a statement is true for all natural numbers. It has two main steps:
Base Case: Prove the statement is true for the smallest value, usually n = 1
Inductive Step: Assume the statement is true for n = k, then prove it's true for n = k + 1.
If both steps are successful, the statement is true for all natural numbers. It's like a domino effect: if the first domino falls (base case) and each domino knocks over the next one (inductive step), then all the dominos will fall.
What is mathematical induction?
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is a two-step process: 1. Base Case: Show that the statement holds for the first natural number (usually 1). 2. Inductive Step: Show that if the statement holds for some natural number k, then the statement also holds for the next natural number k+1. If these two steps can be completed, then the statement is proven to be true for all natural numbers.
Overview
Induction is a method for proving mathematical statements about numbers. There are three steps:
The base case shows the statement is true for a specific number, usually a small number like 1.
The inductive hypothesis assumes the statement is true for some random number k.
The inductive step/proof shows that if the statement is true for k,it must also be true for k+1,proving the statement for all numbers greater than or equal to the base case.
Test Your Knowledge
Question 1
Find a formula for the sum of the first n terms of the sequence:
a
b
c
d
Check It
First, we could study the structure of each item in the sequence:
…
Then, we could write the sum of the first n terms:
FlexCard™
Question 2
The following is what type of sequence: a n=2 a n−1−n with a 1=0
a
linear
b
neither
c
quadratic
Check It
a 1=0
a 2=2 a 1−2=0−2=−2
a 3=2 a 2−3=−4−3=−7
a 4=2 a 3−4=−14−4=−18
a 5=2 a 4−5=−36−5=−41
The differences between the terms =−2,−5,−11,−23…
The differences between their differences =−3,−6,−12…
If the second differences are all the same, the sequence has a perfect quadratic model. If the first differences were to be all the same, the sequence has a linear model. For this sequence, it’s neither linear nor quadratic.
FlexCard™
Asked by Students
Ask your question
Here are the top questions that students are asking Flexi for this concept:
What is the Induction Hypothesis?
Induction is a method of proof usually used to prove statements about positive whole numbers (the natural numbers). Induction has three steps:
The base case is where the statement is shown to be true for a specific number. Usually this is a small number like 1. This is the first domino to fall, creating a cascade and thus proving the statement true for every number greater than the base case.
The inductive hypothesis is where the statement is assumed to be true for k.
The inductive step/proof is where you show that then the statement must be true for k+1.
These three logical pieces will show that the statement is true for every number greater than the base case.
There are 3 consecutive integers that add up to 0. What is the value of the greatest integer?
Let's denote the three consecutive integers as n, n+1, and n+2.
According to the problem, their sum is 0, so we can write the equation as follows:
n+(n+1)+(n+2)=0
Solving this equation gives us the value of n:
3 n+3=0
3 n=−3
n=−1
The greatest integer is n+2, so we substitute n=−1 into n+2:
n+2=−1+2
n+2 = 1
What is the inductive hypothesis?
In an induction proof, the inductive hypothesis is the step where you assume the statement is true for k.
How does induction work?
Induction is a powerful mathematical proof technique used to show that a statement is true for all natural numbers. It has two main steps:
Base Case: Prove the statement is true for the smallest value, usually n = 1
Inductive Step: Assume the statement is true for n = k, then prove it's true for n = k + 1.
If both steps are successful, the statement is true for all natural numbers. It's like a domino effect: if the first domino falls (base case) and each domino knocks over the next one (inductive step), then all the dominos will fall.
What is mathematical induction?
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is a two-step process: 1. Base Case: Show that the statement holds for the first natural number (usually 1). 2. Inductive Step: Show that if the statement holds for some natural number k, then the statement also holds for the next natural number k+1. If these two steps can be completed, then the statement is proven to be true for all natural numbers.
Related Content
Using Mathematical Induction - Overview
Using Mathematical Induction - Example 1
Proof by Induction - Example 1
Inductive Proofs
Back to Induction Proofs
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Three Digit
Find the Sum
Combinatorics C
Counting Numbers
Digit Sum
Number Theory
Combinatorial Math
5
How many counting numbers have three distinct nonzero digits such that the sum of the three digits is 7?
David Rutter
Worked at Georgia Institute of Technology
·
Author has 9.2K answers and 11.8M answer views
·
5y
Almost all of them.
No, seriously. 0% of positive integers do not have that property.
Imagine a sequence of one million random digits. Now consider the constraints on those digits we would have to impose so that no 3 distinct nonzero digits sum to 7. Basically, this means not all three of the digits 1,2, and 4 can appear.
This means you can only use 9 out of the 10 possible digits to construct your number. Which means the probability of not having that property in a million digit number is less than 3×0.91000000, which is a number so close to zero your calculator will probably just say it
Almost all of them.
No, seriously. 0% of positive integers do not have that property.
Imagine a sequence of one million random digits. Now consider the constraints on those digits we would have to impose so that no 3 distinct nonzero digits sum to 7. Basically, this means not all three of the digits 1,2, and 4 can appear.
This means you can only use 9 out of the 10 possible digits to construct your number. Which means the probability of not having that property in a million digit number is less than 3×0.91000000, which is a number so close to zero your calculator will probably just say it’s zero.
But almost all positive integers are longer than a million digits. Almost all positive integers have more digits than the number of atoms in the universe. Which means the probability they don’t have that property cannot be measured with an electron microscope. It cannot be measured with LIGO. It is unimaginably tiny. And once you take all of those numbers, all infinitely many of them, exactly 0% of them will not have that property.
Like I said, almost all of them.
Related questions
How many counting numbers have five distinct nonzero digits such that the sum of the five digits is 15?
How many counting numbers have four distinct nonzero digits such that the sum of the four digits is 29?
How many 3-digit number have distinct digits such that one digit is the average of the other two?
How many four digit numbers are there with distinct digits?
How many three digit numbers have a sum of 10?
Steve Johnson
PhD from Columbia University (Graduated 1968)
·
Author has 496 answers and 1M answer views
·
5y
The problem is poorly stated. Three distinct nonzero elements can either imply that you can’t use 0 as one of three digits, or you can use any number of digits, but only three can be nonzero. In the latter case, there are an infinite number: 124, 1240, 12400, …
The interesting problem is one where there are three digits only, and none are 0. First write out all the patterns that match and the digits from left to right are the same or increasing. We get 115, 124, 133, 223. Of these, only 124 has three distinct digits. But we can permute these digits in any order. So the answer is 6: 124, 142, 21
The problem is poorly stated. Three distinct nonzero elements can either imply that you can’t use 0 as one of three digits, or you can use any number of digits, but only three can be nonzero. In the latter case, there are an infinite number: 124, 1240, 12400, …
The interesting problem is one where there are three digits only, and none are 0. First write out all the patterns that match and the digits from left to right are the same or increasing. We get 115, 124, 133, 223. Of these, only 124 has three distinct digits. But we can permute these digits in any order. So the answer is 6: 124, 142, 214, 241, 412, 421
Assistant
Bot
·
1y
To find the counting numbers that have three distinct nonzero digits such that the sum of the three digits is 7, we will follow these steps:
Identify the digits
: The digits must be distinct, nonzero, and must sum to 7.
Determine possible combinations
: Since we are looking for three distinct nonzero digits, we can denote the digits as
a
,
b
, and
c
where
1
≤
a
,
b
,
c
≤
9
and
a
+
b
+
c
=
7
.
3. List the combinations
: We will find sets of three distinct digits that satisfy this condition.
- The only valid combinations of three distinct nonzero digits that sum to 7 are:
1
,
2
,
4
(since
1
+
2
+
4
=
7
)
To find the counting numbers that have three distinct nonzero digits such that the sum of the three digits is 7, we will follow these steps:
Identify the digits
: The digits must be distinct, nonzero, and must sum to 7.
Determine possible combinations
: Since we are looking for three distinct nonzero digits, we can denote the digits as
a
,
b
, and
c
where
1
≤
a
,
b
,
c
≤
9
and
a
+
b
+
c
=
7
.
3. List the combinations
: We will find sets of three distinct digits that satisfy this condition.
- The only valid combinations of three distinct nonzero digits that sum to 7 are:
1
,
2
,
4
(since
1
+
2
+
4
=
7
)
+ 1
,
3
,
3
does not work as digits must be distinct.
+ 2
,
5
,
0
does not work since 0 is not allowed.
+ 7
alone does not work since we need three digits.
Valid combinations
:
The only valid set we found is
1
,
2
,
4
.
2. Calculate permutations
: The digits
1
,
2
,
4
can be arranged in different ways to form counting numbers. The number of different arrangements (permutations) of three distinct digits is given by
3
!
(3 factorial):
3
!
=
3
×
2
×
1
=
6
3. List the arrangements
: The arrangements of the digits
1
,
2
,
4
are:
- 124
- 142
- 214
- 241
- 412
- 421
Thus, the total number of counting numbers that have three distinct nonzero digits such that the sum of the three digits is 7 is
6
.
Anil Bapat
Lives in Mumbai, Maharashtra, India
·
Author has 2.8K answers and 3.8M answer views
·
Jan 24
Originally Answered: How many three-digit numbers are there, which do not have a zero in their decimal representation and whose sum of digits is 7?
·
How many three-digit numbers are there, which do not have a zero in their decimal representation and whose sum of digits is 7?
There are 3 positions Viz. Hundredth Position, Tenth Position and Unit Position.
All we need to do is to anchor one-one digit at the Hundredth Position and allocate other non-zero digits in the remaining 2 Positions. Since the sum of the digits is 7 and 0 is not a
How many three-digit numbers are there, which do not have a zero in their decimal representation and whose sum of digits is 7?
There are 3 positions Viz. Hundredth Position, Tenth Position and Unit Position.
All we need to do is to anchor one-one digit at the Hundredth Position and allocate other non-zero digits in the remaining 2 Positions. Since the sum of the digits is 7 and 0 is not a valid digit in the current context, the digits 6,7,8,9 and 0 are out.
With 1 anchored in Hundredth Position, we have:
115,124,133,142 and 151 —————— 5
With 2 anchored in Hundredth Position, we have:
214,223,232 and 241 ———————- 4
With 3 anchored in Hundredth Position, we have:
[...
Related questions
The number of three digit number whose middle digit is bigger than the extreme digit?
What is the number of digits in 2^40?
How many 3-digit numbers have distinct digits that sum to 21?
How many three digit numbers are divisible by 7?
How many 4-digit numbers are there so that the sum of digits is 6?
John Chadwick
Former Head of Electrical, Mechanical & MV Engineering at South Cheshire College (1972–2007)
·
Author has 1K answers and 609.5K answer views
·
Updated 11mo
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
7 = 7+0+0 = 6+1+0 = 5+2+0 = 5+1+1 = 4+3+0 = 4+2+1 = 3+3+1 = 3+2+2.
700 ==> Permutes to 1 integer.
610 ==> Permutes to 4 integers
520 ==> Permutes to 4 integers.
511 ==> Permutes to 3 integers.
430 ==> Permutes to 4 integers.
421 =>> Permutes to 6 integers.
331 ==> Permutes to 3 integers.
322 ==> Permutes to 3 integers.
Total = 28.
OR
Total using (n+r-1)C(r-1) = (6+3–1)C(3–1) = 8C2
= 28 integers whose digits sum to SEVEN.
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John Steele
Author has 10.2K answers and 11.3M answer views
·
Jan 20
Originally Answered: How many three-digit numbers are there, which do not have a zero in their decimal representation and whose sum of digits is 7?
·
The problem excludes 0, and obviously 7, 8, 9 are excluded, because you can’t meet the sum of 7 requirement.
It is not clear if repeat use of digits is allowed, so I will separate those cases
If repeats are allowed, permutations of 511 are permitted (3), and permutations of 322 (3) for six total.
In addition, 421 has six permutations and does not repeat any digits.
No other combinations of three digits total 7, so 12 or 6 depending on whether repeat use of a digit is allowed, based on permutations of three prototypes.
Lance Berg
Author has 28K answers and 54.7M answer views
·
10mo
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
7 is the sum of three digits as follows. Note that numbers with leading zeros are not three digit numbers, but two or even one digit. So all such ways are eliminated, for each three digit combo, I have set out the number of ways to make a valid three digit number using that combination of three digits in various orders.
7+0+0: 1 way
6+1+0: 4 ways
5+1+1: 3 ways
5+2+0: 4 ways
4+2+1: 6 ways
4+3+0: 4 ways
3+3+1: 3 ways
3+2+2: 3 ways
That’s a total of 28
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David Vanderschel
PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970)
·
Author has 37.6K answers and 50.1M answer views
·
Apr 16
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
28
I used a Python expression to count them:
```
sum(1 for n in range(100,1000) ... if sum(int(c) for c in str(n)) == 7) 28
```
Mike Roberts
Author has 787 answers and 532.9K answer views
·
5y
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
700
6 (10, 01)
5 (20, 11, 02)
4 (30, 21, 12, 03)
3 (40, 31, 22, 13, 04)
2 (50, 41, 32, 23, 14, 05)
1 (60, 51, 42, 33, 24, 15, 06)
1+2+3+4+5+6+7 = 28.
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Rosanna Arrundell
Knows Dutch
·
5y
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
3+4=7 1+3+3=7 3+2+2=7 21÷3=7 so depends how long you make your sum. But not more then 2.
Laura Kay Posey
Education, K-6th from SWT (Graduated 1966)
·
Author has 6.5K answers and 2M answer views
·
2y
Originally Answered: How many 3-digit numbers are there in which the sum of the digits is 7?
·
ANSWER: 30 ? 28. Depends on counting marked double digits in number
3 DIGITS = 7
232 = 2 + 3 + 2 = 7
232
322 = 3 + 2 + 2 = 7
322 = 3 + 2 + 2 = 7
223 = 2 + 2 + 3 = 7
223
(Possibly 3more using 2nd marked 2)
511 = 5 + 1 + 1 = 7
511
151 = 1 + 5 + 1 = 7
151
115 = 1 +1 + 5 = 7
115
( Double numbers could have 3 more)
421 = 4 + 2 + 1 = 7
412 = 4 + 1 + 2 = 7
124 = 1 + 2 + 4 = 7
142 = 1 + 4 + 2 = 7
214 = 2 + 1 + 4 = 7
241= 2 + 4 + 1 = 7
331= 3 + 3 + 1 = 7
313 = 3 × 3 + 1 = 7
133= 1 + 3 + 3 = 7
133
( Possibly 3 more with second marked3)
502
520
250
205
403
430
340
304
Hoot Woon Ang
How many 3-digit number have distinct digits such that one digit is the average of the other two?
Since the context is such that one number must be the average of the other 2, it can be done quite simply by exhaustive means.
I assume that the 3-digit number refers to strictly positive digits thus number such as -110 is not allowed even though average of -1 and 1 is 0.
The 3-digit number does not contain 0 as the first digit.
If the average is 0, the other 2 number MUST be 0 and 0 on the assumption that the 3-digit number is non-negative. Since the digits must be unique, it is impossible to have 000 as a solution.
If the average is 1, the other 2 numbers are possibly 1 and 1 or 0 and 2. Si
Since the context is such that one number must be the average of the other 2, it can be done quite simply by exhaustive means.
Note:
I assume that the 3-digit number refers to strictly positive digits thus number such as -110 is not allowed even though average of -1 and 1 is 0.
The 3-digit number does not contain 0 as the first digit.
If the average is 0, the other 2 number MUST be 0 and 0 on the assumption that the 3-digit number is non-negative. Since the digits must be unique, it is impossible to have 000 as a solution.
If the average is 1, the other 2 numbers are possibly 1 and 1 or 0 and 2. Since the digits must be unique, it is impossible to have 111 as a solution.
If the average is 2, the other 2 numbers are possibly 0 and 4 or 1 and 3 or 2 and 2. Since the digits must be unique, it is impossible to have 222 as a solution.
Summary thus far:
A triple digit will always result in the mean of the other 2 digits being itself. Thus will always be rejected.
If 0 is one of the possible digit, there are only 4 variations since the first digit is assumed to be non-zero.
Every non-zero options will result in 6 possible variations.
With these, observe the images attached, you should be able to find a pattern in the way mean works. The following images shows the process with each pair of coloured arrows representing possible digits. The average is the number highlighted in yellow.
Total possible ways
= 4 possible options with the digit “0” + 16 possible options with non-zero digits
= 4 multiply by 4 variations + 16 multiply by 6 variations
=112 possible 3-digit numbers
Hope the visual helps. Do let me know if any part lacks clarity.
Varun Sivashankar
B.S. in Mathematics, University of California, Los Angeles (Graduated 2022)
·
Author has 112 answers and 376.9K answer views
·
Updated 7y
Related
How many 3-digit number have distinct digits such that one digit is the average of the other two?
It is clear that if one digit is the mean of the other two digits, the other two digits must either be both odd or both even. This is because the mean of an odd and even number is not an integer.
For every pair of odd/even distinct digits chosen, there is exactly one number which is the mean of the pair. Let us first consider the case when the pair of digits are odd. There are 5!3!×2!=10 ways of selecting a pair of distinct odd digits. The last digit, their mean, depends on the pair chosen. Each combination has 3!=6 permutations, giving rise to 10×6=60 such numbers.
It is clear that if one digit is the mean of the other two digits, the other two digits must either be both odd or both even. This is because the mean of an odd and even number is not an integer.
For every pair of odd/even distinct digits chosen, there is exactly one number which is the mean of the pair. Let us first consider the case when the pair of digits are odd. There are 5!3!×2!=10 ways of selecting a pair of distinct odd digits. The last digit, their mean, depends on the pair chosen. Each combination has 3!=6 permutations, giving rise to 10×6=60 such numbers.
Let us now consider a pair of even digits. Let us avoid using 0 for now. There are 4!2!×2!=6 ways of selecting a pair of distinct even digits. Similarly, this gives rise to 6×6=36 numbers.
There are 4 even pairs involving 0. Each pair has 4 permutations, as 0 cannot be the leading digit. This gives rise to 4×4=16 numbers.
Hence, the required answer is 60+36+16=112
Related questions
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1556 | https://www.eregulations.com/washington/hunting/black-bear | Washington › Hunting › Black Bear
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Washington Hunting Guide - Spanish Translation
Black Bear
Significant changes are red.
Area Restriction:
GMUs 157, 410-417, 419-424, 490, 522, and 655 are closed to fall bear hunting.
A special deer or elk permit in GMU 485 is required to hunt bear in GMU 485.
License Required:
A valid big game hunting license, which includes black bear as a species option.
Second Black Bear License/Tag:
A second black bear transport tag/license must be purchased to take a second bear. Hunters may purchase a maximum of two black bear transport tag/licenses.
Hunting Method:
Hunters may use any legal hunt method for hunting black bear. The use of dogs or bait for recreational hunting of black bear is prohibited statewide (RCW 77.15.245).
Black Bear Management Units
| Fall Black Bear General Seasons |
| Black Bear Management Unit (BBMU) | Hunt Area (GMU) | Season Dates | Bag Limit |
| 1 – Northeast | 105, 108, 111, 113, 117, 121, 124 | August 1-November 15 | 2 |
| 2 – Columbia Basin | 127, 130, 133, 136, 139, 142, 248, 254, 260, 262, 266, 269, 272, 278, 284, 290, 371, 372, 373, 379, 381 | August 1-November 15 | 2 |
| 3 – Southeast | 145, 149, 154, 162, 163, 166, 169, 172, 175, 178, 181, 186 | August 1-November 15 | 2 |
| 4 – Okanogan Highlands | 101, 204 | August 1-November 15 | 2 |
| 5 – Okanogan | 203, 209, 215, 218, 224, 231, 233, 239, 242, 243 | August 15-November 15 | 1 |
| 6 – Central Cascades East | 244, 245, 246, 247, 249, 250, 251, 328, 329, 330, 334, 335 | August 1-November 15 | 2 |
| 7 – South Cascades East | 336, 340, 342, 346, 352, 356, 360, 364, 368, 382, 388, 578 | August 1-November 15 | 2 |
| 8 – North Cascades West | 418, 426, 437, 448, 450, 460 | August 1-November 15 | 2 |
| 9 – South Cascades West | 466, 485, 503, 510, 513, 516, 560, 568, 572, 574, 653, 654 | August 1-November 15 | 2 |
| 10 – Urban | 407, 454, 504, 564, 624, 627, 633, 652, 666, 684 | August 1-November 15 | 2 |
| 11 – Mount St. Helens | 505, 520, 524, 550, 554, 556, 667 | August 1-November 15 | 2 |
| 12 – Olympic Peninsula East | 603, 607, 618, 621, 636, 638 | August 1-November 15 | 2 |
| 13 – Olympic Peninsula West | 601, 602, 612, 615, 642, 648, 651 | August 1-November 15 | 2 |
| 14 – Willapa Hills | 501, 506, 530, 658, 660, 663, 672, 673, 681, 699 | August 15-November 15 | 1 |
Mandatory Bear Identification Test
Hunters that choose to hunt in GMUs 101, 105, 108, 111, 113, 117, 121, 203, 204, 209, 215, 218, 224, 231, 242-244, 418, 426, 437, or 450 must successfully complete the annual WDFW online bear identification test, and must score 80% or higher and carry proof that they have passed the WDFW test or an equivalent test from another state.
Mandatory Submission of Bear Teeth Statewide
Per RCW 77.15.280 and WAC 220-415-090, ALL successful bear hunters MUST submit the complete, unbroken root of the 1st premolar tooth from their harvest. Teeth should be sent using WDFW's pre-paid and self-addressed mortality envelope which can be obtained at any WDFW office or by calling your local regional office.
Successful hunters, please submit the required tooth from your harvest by December 1, 2025.
WDFW uses teeth to determine the accurate age of harvested bears. Hunters can check the age of their harvested bear by visiting the WDFW tooth lookup page.
Teeth are aged by an external laboratory and the turnaround time for age results can be up to 6 months after the close of the fall big game hunting season.
Identification of Grizzly and Bear
Grizzlies are protected under both federal and state law and may not be shot or killed. Be sure of identification if you are hunting black bear.
A mandatory bear identification test is required in some GMUs. See the list below and visit the Bear Identification Program page for more information.
The Colville, Mt. Baker-Snoqualmie, and Okanogan-Wenatchee National Forests have food storage orders in effect to reduce the potential for human-wildlife conflicts. Processing and storage of harvested wildlife must comply with National Forest regulations. Please refer to the individual forest websites for specific requirements.
It is Illegal to Shoot a Cub or a Female with Cubs
Per WAC 220-415-090, hunters cannot shoot or possess cubs or females accompanied by cubs, which are described as less than 1 year old (weighing 30-50 lbs). BE VIGILANT. Cubs may not always be visible, so please be patient and identify whether the bear is male or female. If female, watch for cubs before shooting.
To help identify the sex of black bears, watch the instructional video available on the WDFW black bear species page.
Where to Shoot a Bear
The placement of a hunter’s shot is critical in being able to harvest a bear and reduce the potential for wounding loss or injury. Due to a bear’s mass, the ideal shot placement is broadside with the shot being placed just behind the upper arm of the bear as the bear is taking a forward step.
BearWise Washington
YOU can help keep bears wild by being BearWise at your home or while camping, hunting, fishing or hiking.
Never feed or approach bears
Keep garbage bins indoors until trash day and remove bird feeders when bears are active (Apr-Nov)
Keep chickens and other small livestock protected in proper, covered, enclosures
Learn more about living responsibly with black bears by visiting the BearWise website.
Washington
General Info
Hunting Hours
Hunting Access & Closures
Hunter & Trapper Education
Hunting Access on Private Lands
Where To Get Maps
DNR Lands
Definitions
Chronic Wasting Disease (CWD)
Turn in a Poacher (TIP) Program
State Recreation Lands & Water Access Sites
Treponeme-Associated Hoof Disease (TAHD) in Elk
Washington Department of Fish & Wildlife
Licenses, Permits & Fees
License, Tags & Permit Fees
License Information & Requirements
Washington 2025-2026 Multi-Season Permits
Washington Win Your Dream Hunt
Raffle Permit Hunts
Hunting Regulations
Deer General Information
Elk General Information
Black Bear
Equipment & Hunting Methods
Mandatory Hunter Reporting
Violations & Penalties
Prohibited Hunting Methods
Tagging & Transporting Game
Persons with Disabilities
Tribal Hunting
Public Conduct Rules on WDFW Lands
Hunter’s Code of Conduct
Antler Point Diagrams
Baiting for the Purposes of Hunting Deer or Elk
Species Identification
WDFW Check Stations
Seasons & Limits
Deer General Seasons
Elk General Seasons
Cougar General Seasons
Management Areas
Game Management Units
Deer Areas
Elk Areas
Mountain Goat Hunt Areas
Bighorn Sheep Units
Moose Areas
Cooperative Road Management Areas
Firearm Restriction Areas
Maps
Washington Game Management Unit Maps
Washington Deer Areas Maps
Washington Elk Areas Maps
Washington Mountain Goat Hunt Areas Map
Washington Bighorn Sheep Units Maps
Washington Moose Hunt Areas Map
Turkey Hunting
Wild Turkey Spring Season
Washington 2025 Spring Wild Turkey
Population Management Units
Sex & Age of Wild Turkeys
Wild Turkey Recipes
Mandatory Reporting
First Turkey Program
Washington Slam
Safe & Ethical Turkey Hunting
Hunting Access & Private Land Program
Special Permits
Special Permit Application Instructions
Deer Special Permits
Elk Special Permits
Turkey Special Permits
Bighorn Sheep Special Permits
Mountain Goat Special Permits
Moose Special Permits
PDF Downloads
2025 Corrections to Regulations
2025 Big Game Hunting Regulations PDF
Washington Hunting Guide - Spanish Translation |
1557 | https://www.youtube.com/watch?v=J5PZ-bW4Las | Draw the shear and moment diagrams for the beam | Example 6.4 | Mechanics of Materials RC Hibbeler
Engr. Adnan Rasheed Mechanical
20400 subscribers
37 likes
Description
1111 views
Posted: 21 Feb 2025
Example 6.4 Draw the shear and moment diagrams for the beam shown in figure 6-7a
Dear Viewer You can find more videos in the link given below to learn more
Theory Video Lecture of Mechanics of Materials by Beer & Johnston
299 Problem solutions of all chapter of Mechanics of Materials by Beer & Johnston
Chapter 14 Energy Methods of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 12 Deflection of Beams and Shaft of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 11 Design of Beams and Shafts of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 9 Stress Transformation of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 8 Combined Loading of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 7 Transverse Shear of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 6 Bending of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 5 Torsion of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 4 Axial Loading of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 3 Mechanical Properties of Materials of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 2 Strain of Mechanics of Materials rc hibbeler (09th Edition)
Chapter 1 Stress- Internal Loading of Mechanics of Materials rc hibbeler (09th Edition)
mechanicsofsolids #strengthofmaterialstutorials #mechanicsofmaterials #mechanicalproperitesofmaterials #mechanicalproperties #mechanicalpropertiesofsolids #bending #bendingmoment #bendingstress #bendingmomentdiagram
8 comments
Transcript:
welcome back in this video we are going to solve
problem example 6.4 that is taken from chapter number six bending and book name is mechanics of
material by RC hibler 9th Edition so statement is draw the share and movement diagram for the
beam shown in figure 6-7 a so you can see this is a simply supported beam having roller support
at Point a and pin support at Point C and at Point a we have a moment that is clockwise 80 Kon and
from B to see there is a distributed load of 5 K per M meter and a point load of 15 k at point B
so you have to determine the share and movement diagram for the beam so let's start with the
solution the first step is that if you remove this roller support at Point a so you will be having
a reaction force which will be represented as ra a and if you remove this pin support at Point C
so you will be having a vertical reaction force RC as well there will be this this will be RC
Y and there will be a horizontal reaction force which will be RC X so we'll find this r a RC
Y and rcx by using equation of equilibrium so first equation of equilibrium is that sum of
all Ms about point a is equal to0 and taking the counterclockwise moment as positive you can
see this is a distributed load of 5 Kon so this distributed load when converted into a point
load so this load will become equal to 5 by the length over which it acts which is five and
that is 25 Kil newon now we'll apply this equation so first force that is producing m is rcy and
perpendicular distance is 10 and this is producing counterclockwise mov so it will be positive so
I can write r r c y by 10 and that is positive the second mement that produce mement about
point a is this 25 Kon load and perpendicular distance is this one clear so this is five and
this distributed load will act at the middle of this five which is 2.5 so 5 + 2.5 is 7.5 so
this distance is 7.5 M so I can write it 25 by 7.5 and since it is producing clockwise movement
so it will be negative the third load that is producing movement about point a is 15 K load
and perpendicular distance is five and it is also producing clockwise so it will be negative
so I can write it -5 by 5 and the last M that at that is at Point a is 80 kilon that is clockwise
so it is also negative so I can write Min - A is equal to0 so from here I will get 10 rcy is
equal to 25 into 7.5 - 15 5 - 80 and when you will calculate this so I will calculate
it 25 7.5 25 7.5 will give you 187 1.5 - 15 into 5 which is 75 I think yes
this is -75 and - 80 is equal to 10 RC Y and 187.5 + 75 + 80 will give 342 okay so this will give us 342 okay 42.5 so
that is equal to 3 42.5 okay when this will move towards right so this this will be positives
so I will make this correction this will be positive so then it will give you 3425 then RC
y so from here what you get is that rcy will be equal to 32.5 / by 10 and that will give you 34
25 Kil Newton okay so we have this now we'll find this rcx by using another equation of equilibrium
that some of all forces along X direction must be equal to zero and force toward right is taken as
positive and you can see that we have taken it toward left hand side so minus r CX and there is
no other horizontal Force so it means that rcx is equal to 0er now we have left with ra a and we
can find this ra a by using the third equation of equilibrium that sum of all forces along
y direction must be equal to zero and upward force is taken as positive so you can see R A
which is upward so R A minus this 15 K minus this 25 K plus this 34.25 is equal to 0 so from
here -5 - 25 -5 - 25 plus this 34.25 will give you 5 - 5.75 so R A minus 5.75 is equal to 0 so
from here we will get this ra a will be equal to 5.75 Kil newon now you have this reaction forces
so we will move towards share force and bending M diagram okay so let me write it share and M
diagram for that we will Define share and mement functions okay since you can see there is a
discontinuity of this uh uniformly distributed load because it is available on this portion and
not available at this potion also concentrated load at the center of the beam so we have two
region of X must be consider first we will take we will section The Beam for this X1 and this X1
will be greater or equal to zero and less or equal to 5 m and then we will section the beam in this
region for X2 and X2 will be greater than or equal to 5 m and less than or equal to 10 m so let's
section it for region X1 which is greater than or equal to Z and less or equal equal to 5 m or
less than 5 m you can say so what we will do is that we will section the beam over here let this
will be your X1 clear and we will draw the free bar diagram over here as well so this is your
point a we have R A which is 5.75 this distance is X1 here you have a moment which is 80 Kon into
meter so when you section it so you will be having a share Force which will acting downward and you
will be having a reaction moment M so we will find this share and M as a function so for that
we will apply equation of equilibrium that sum of all forces along y direction must be equal to
zero and upward force is taken as positive so you can see this 5.75 is upward so I can write it 5.75
minus this V is equal to Zer so V is equal to 5.75 Kil newon this is your equation one this is your
equation number one now we'll find this Movement by using another equation of equilibrium that sum
of all mement about this point O this is O point is equal to zero and taking the counterclockwise
mement as positive so you can see the first m is 80 K into meter which is clockwise so
it will be taken as negative so I can write it - 80 the second mement is due to this 5.75
into perpendicular distance X1 and it is also clockwise so it will be netive - 5.75 into X1
and the last M which is reaction movement that is counterclockwise that is there sum of all this
movement must be equal to zero so from here you will get this m will be equal to 5.75 time X1
+ 8 and the unit will be kilton into meter this is your equation number two now equation number
one and equation number two determine the share force and bending mement equation for this
region for X is equal X1 is equal to Z and X1 is less than 5 m in between this region for that
for another region what we will do is that for region X2 which is greater than or equal to
5 m and less are equal to 10 m so what we will do is actually we are going to cut this
beam in this any region so let this is your X2 now if I draw the free bar
diagram over here so it will be like this you can see this is RA
a which is 5.75 K clear you have a moment here which is 80 K into
meter let till this this distance is 5 meter this is point a this is your point
B and at B you have a point load which is 15 Kil Newton into meter and from B
till this point we have a distributed load which is actually 5 Kil per M and the total
distance which we have taken from this till this this was X2 so this is X2 and this is five
so remaining this distance will be equal to X2 minus X2 - 5 m okay and when you cut it you
will be having a share Force which is V downward and you will be having a moment M so again here
we'll find this V and M for equation of v and M by using equation of equilibrium that sum of all
forces along y direction must be equal to zero and upward force is taken as positive so you can see
that the first upward force is this one which is 5.75 the second one is this 15 minus 15 and the
third one is this distributed load multiply by the length over which it acts so I can write
it 5.75 -5 - 5 X2 - 5 and the last one is this here Force minus V is equal to Z so from
here what we will get is V is equal to um 5.75 minus 5.75 - 15 - 5x 2 - - + 25 so V will
be equal to 15.75 - 5x2 and the unit will be kilo Newton let this is your equation number
three now we'll find the moment this moment M by using another equation that sum of all moment
let this is your point1 about point1 is equal to zero and taking the counterclockwise M as
positive again here we have to convert this distributed load into a point load so this load
will be equal to 5 X2 - 5 and that will act at a distance half of X2 / 5 from both points
so this is and this is actually X2 - 5 / 2 this is divide by 2 and this will be the same
X2 - 5 / 2 okay now applying this equation so first uh force that is producing M about point1
is this 80 K this is clockwise so minus 80 the second force that is producing M about point1
is this 5.75 and perpendicular distance is X2 this is also clockwise so it will be negative so
I can write - 5.75 X2 the third m is due to this load which is minus uh sorry 15 into X2 - 5 and
this is producing counterclockwise movement why because you can see this is the force and this
is perpendicular distance so it is producing counterclockwise moment so it will be positive and
this last one is this one which uh sorry second last this this Force into perpendicular distance
is this X2 - 5 / by 2 and this is also producing counterclockwise M so it will be positive so
I can write + 5 into X2 - 5 into X2 - 5 / 2 and the last moment which is M and that is uh
counterclockwise so their sum must be equal to Zer so when you simplify this equation you
will get this moment will be equal to - 2.5 X2 s + 15.75 X2 + 9. 92. 5 and unit will be kilon
newon into M and this is your equation number four now equation three and and 4 are valid for the
region X2 is greater than 5 mm and less than 10 mm so these are our share force and bending M
equation so with the help of this equation one and two we will draw the share force and bending
mement for this region and for region of this one X2 is greater than or equal to 5 m and less than
10 m we'll draw the share force and bending M by using this equation so let's start with that okay
okay for share force and bending movement diagram again you have to draw vertical line from the end
of this beam clear let me draw it okay and there is a horizontal line for the share force and this
x that will show you in length of beam in meter here on Y axis you have share Force having unit
kilo Newton Let each division is equal to uh 5 so this is 5 Kon this will be 10 similarly Min - 5
- 10 -5 - 20 - 30 uh sorry 25 - 30 and so on okay starting for this first equation so for share
Force for region you can see this is a single value constant value 5.75 so it means that share
force will remain the same for X is greater or equal to Z and less than 5 m so you can see this
is the length of 5 m this is the mid of the beam this is 5 m so from this starting point till this
mid it will remain same and it will be five Point 5 75 kilon now at this point you can see you
have 15 Kon load so 15 that is minus so 5.75 minus 15 will give you 9.25 so again share force
will change from this to this this this is - 9 .25 and if you put X is equal to 5 m in this
equation clear so you will again get - 9.25 now for X is X is equal to we will you can also
plot various point of X2 clear which is 6 7 8 9 and you will get multiple point over here so
after joining them you will get the share Force but here we will put x equal to 10 x = 10 m in
equation 3 so we will get V is equal to - 34.25 so - 34.25 will be somewhere here - 34.25 and
again you can see that this distributed load is a 0° line so bending M sorry share force will be
1° higher and slope will be decreasing so we can join them by using a straight line so this will
be your share force diagram and you can see that uh it is correct because at the end you have
rcy which is + 34.25 and we which will cause it to move toward zero that will complete your
share force diagram so this is your share Force diagram now we'll move toward bending bending
movement diagram so for bending movement diagram again I will draw horizontal line and X shows the
length of beam in meter on y- AIS you have moment and that having unit kilon into meter okay so
what we will do is that put xal 0 in equation number two so when you put X is equal to Z so
this term will become zero Al so M will be 80 Kon into M so m is 80 kilon into M also you can
see this 80 K into meter is available over here that is clockwise but while drawing the bending
mov diagram this clockwise movement will cause the bending mement to go in positive direction
which show which is already shown over here you can see you get positive value okay so let each
division is equal to uh let's see this is 25 so 2550 75 and 100 clear so first one is 80 so 80
will be somewhere here for example this is 80 so at X is equal to0 we have 80 now at at X
is equal to you can see at X is uh equal to 2 how much X is equal to 5 X1 sorry X is X2 X2
this is X1 okay so when X2 is equal to 5 m so put in equation number four you will get m is
equal to 10 8.7 5 which will be this one at the mid let's say 108 is this point one 8.75 Kil also
you can see that this share force is a horizontal line so bending moment will be 1° higher and
slope will be increased so you can join these two points again if you need multiple point
so you can just place the value of X1 various value of X1 take till X is equal to 5 so you will
get this again for X2 isal to 10 put in equation 4 so you will get mement is equal to zero so this
is the equation when you put X2 is equal to 10 in this equation so you will get moment will be
equal to zero which will be this one and you can see that this share Force is uh one de line
and slope is decreasing so bending movement will be a curve like this with clockwise Direction so
it will be like this it will be a curve type and this is your bending M diagram so this was all
about this example 6.4 I hope you have enjoyed this video and you have learned it those
who are new to my channel then subscribe it and don't forget to press the Bell icon so
that you can get notification about my latest videos if you have any question you can ask
me in comment section thank you for watching |
1558 | https://www.geometrictools.com/Documentation/DistanceLineTetrahedron.pdf | Distance from Linear Component to Tetrahedron David Eberly, Geometric Tools, Redmond WA 98052 This work is licensed under the Creative Commons Attribution 4.0 International License. To view a copy of this license, visit or send a letter to Creative Commons, PO Box 1866, Mountain View, CA 94042, USA.
Created: February 12, 2002 Last Modified: March 1, 2008 Contents 1 Line and Tetrahedron 2 1.1 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 Closest Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 2 Ray and Tetrahedron 3 3 Segment and Tetrahedron 4 1 Let Vi, 0 ≤i ≤3 be the vertices of the tetrahedron. The linear component is P + tD where D is a unit length vector and t ∈I R (line), t ≥0 (ray), or t ∈[0, T] (segment). The construction can be modified slightly to handle D that is not unit length. The tetrahedron can be parameterized by V0 + s1E1 + s2E2 + s3E3 where Ei = Vi −V0, si ≥0, and s1 + s2 + s3 ≤1.
1 Line and Tetrahedron 1.1 Distance Translate the tetrahedron and line by subtracting P. The tetrahedron vertices are now Ui = Vi −P for all i. The line becomes tD. Project onto the plane containing the origin 0 and having normal D. The projected line is the single point 0. The projected tetrahedron vertices are Wi = (I −DDT)Ui for all i. The boundary of the projected solid tetrahedron is a convex polygon, either a triangle or a quadrilateral. Figure 1 shows the line, tetrahedron, and projections.
Figure 1. Line, tetrahedron, and projections onto a plane perpendicular to the line.
If the convex polygon contains 0, the distance from the line to the tetrahedron is zero. Otherwise, the distance from the line to the tetrahedron is the distance from 0 to the convex polygon. The projected values are in a plane in 3D and can be projected into 2D with the standard technique of eliminating the coordinate corresponding to the maximum absolute component of D. The distance between a point and convex polygon can be computed in 2D. This value must be adjusted to account for the 3D-to-2D projection. For example, if D = (d0, d1, d2) with |d2| = maxi{|di|} and r is the computed 2D distance, then the 3D distance is r/d2.
2 1.2 Closest Points The set of tetrahedron points closest to the line in many cases consists of a single point. In other cases, the set can consist of a line segment of points. For example, consider the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1). The line (1/4, 1/4, 0) + t(0, 0, 1) intersects the tetrahedron for t ∈[0, 1/2], so the corresponding points are zero units of distance from the tetrahedron. The line (−1, −1, 1/2)+t(0, 0, 1) is √ 2 units of distance from the tetrahedron. The closest points on the line are generated by t ∈[0, 1/2] and the closest points on the tetrahedron are (0, 0, t) for the same interval of t values. The line (1/2, −1/2, 0)+t(0, 0, 1) is 1/2 units of distance from the tetrahedron. The closest points on the line are generated by t ∈[0, 1/2] and the closest points on the tetrahedron are (1/2, 0, t) for the same interval of t values.
Case 1. Let 0 be strictly inside the convex polygon. In this case, the line intersects the tetrahedron in an interval of points. Let E = [E1 E2 E3] be the matrix whose columns are the specified edge vectors of the tetrahedron. Let s be the 3 × 1 vector whose components are the si parameters. The line segment of intersection is tD + P = Es + V0 for t ∈[tmin, tmax]. The problem now is to compute the t-interval. The edge vectors of the tetrahedron are linearly independent, so E is invertible. Multiplying the vector equation by the inverse and solving for the tetrahedron parameters yields s = E−1 (tD + P −V0) = At + B where A = (a1, a2, a3) = E−1D and B = (b1, b2, b3) = E−1(P −V0). The parameters s must satisfy the inequality constraints for the tetrahedron. The parameter t is therefore constrained by the four inequalities: a1t + b1 ≥0, a2t + b2 ≥0, a3t + b3 ≥0, (a1 + a2 + a3)t + (b1 + b2 + b3) ≤1.
Each of these inequalities defines a semiinfinite interval of the form [¯ t, ∞) or (−∞, ¯ t]. In this particular case, we know the intersection of the four intervals must be nonempty and of the form [tmin, tmax].
The division required to compute E−1 can be avoided. Let us assume that the tetrahedron is oriented so that det(E) > 0. Multiply by the adjoint Eadj to obtain det(E)s = Eadj (tD + P −V0) = αt + β.
The four t-inequalities are of the same form as earlier, but where ai refers to the components of α, bi refers to the components of β, and the last inequality becomes a comparison to det(E) instead of to 1.
Case 2. Let 0 be on the convex polygon boundary or outside the polygon. Let C be the closest polygon point (in 3D) to 0. The line tD + C intersects the tetrahedron with Ui vertices either in a single point or in an interval of points. The method in case 1 may be used again, but now you need to be careful with the interval construction when using floating point arithmetic. If the intersection is a single point, theoretically tmin = tmax, but numerically you might wind up with an empty intersection. It is not difficult to trap this and handle appropriately. Observe that cases 1 and 2 are handled by the same code since in case 1 you can choose C = 0.
2 Ray and Tetrahedron Use the line-tetrahedron algorithm for computing the closest line points with parameters I = [tmin, tmax] (with possibly tmin = tmax). Define J = I ∩[0, ∞). If J ̸= ∅, the ray-tetrahedron distance is the same as the line-tetrahedron distance. The closest ray points are determined by J. If J = ∅, the ray origin P is closest to the tetrahedron.
3 3 Segment and Tetrahedron Use the line-tetrahedron algorithm for computing the closest line points with parameters [tmin, tmax] (with possibly tmin = tmax). Define J = I ∩[0, T]. If J ̸= ∅, the segment-tetrahedron distance is the same as the line-tetrahedron distance. The closest segment points are determined by J. If J = ∅, the closest segment point is P when tmax < 0 or P + TD when tmin > T.
4 |
1559 | https://mathworld.wolfram.com/IncreasingFunction.html | Increasing Function -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Calculus and Analysis
Functions
Increasing Function
A function increases on an interval if for all , where . If for all , the function is said to be strictly increasing.
Conversely, a function decreases on an interval if for all with . If for all , the function is said to be strictly decreasing.
If the derivative of a continuous function satisfies on an open interval, then is increasing on . However, a function may increase on an interval without having a derivative defined at all points. For example, the function is increasing everywhere, including the origin , despite the fact that the derivative is not defined at that point.
See also
Decreasing Function, Derivative, Nondecreasing Function, Nonincreasing Function, Strictly Decreasing Function, Strictly Increasing Function
Explore with Wolfram|Alpha
More things to try:
absolute value
functions
asymptotes (2x^3 + 4x^2 - 9)/(3 - x^2)
References
Jeffreys, H. and Jeffreys, B.S. "Increasing and Decreasing Functions." §1.065 in Methods of Mathematical Physics, 3rd ed. Cambridge, England: Cambridge University Press, p.22, 1988.
Referenced on Wolfram|Alpha
Increasing Function
Cite this as:
Weisstein, Eric W. "Increasing Function." From MathWorld--A Wolfram Resource.
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1560 | https://courses.lumenlearning.com/atd-sanjac-collegealgebra/chapter/graph-exponential-functions/ | Graph exponential functions | College Algebra
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College Algebra
Graphs of Exponential Functions
Search for:
Graph exponential functions
Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form f(x)=b x f(x)=b x whose base is greater than one. We’ll use the function f(x)=2 x f(x)=2 x. Observe how the output values in the table below change as the input increases by 1.
x–3–2–1 0 1 2 3
f(x)=2 x f(x)=2 x1 8 1 8 1 4 1 4 1 2 1 2 1 2 4 8
Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for any exponential function with the form f(x)=a b x f(x)=a b x, b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a.
Notice from the table that
the output values are positive for all values of x;
as x increases, the output values increase without bound; and
as x decreases, the output values grow smaller, approaching zero.
Figure 1 shows the exponential growth function f(x)=2 x f(x)=2 x.
Figure 1. Notice that the graph gets close to the x-axis, but never touches it.
The domain of f(x)=2 x f(x)=2 x is all real numbers, the range is (0,∞)(0,∞), and the horizontal asymptote is y=0 y=0.
To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f(x)=b x f(x)=b x whose base is between zero and one. We’ll use the function g(x)=(1 2)x g(x)=(1 2)x. Observe how the output values in the table below change as the input increases by 1.
x–3–2–1 0 1 2 3
g(x)=(1 2)x g(x)=(1 2)x8 4 2 1 1 2 1 2 1 4 1 4 1 8 1 8
Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio 1 2 1 2.
Notice from the table that
the output values are positive for all values of x;
as x increases, the output values grow smaller, approaching zero; and
as x decreases, the output values grow without bound.
The graph shows the exponential decay function, g(x)=(1 2)x g(x)=(1 2)x.
Figure 2.The domain of g(x)=(1 2)x g(x)=(1 2)x is all real numbers, the range is (0,∞)(0,∞), and the horizontal asymptote is y=0 y=0.
A General Note: Characteristics of the Graph of the Parent Function f(x) = b x
An exponential function with the form f(x)=b x f(x)=b x, b>0 b>0, b≠1 b≠1, has these characteristics:
one-to-one function
horizontal asymptote: y=0 y=0
domain: (−∞,∞)(−∞,∞)
range: (0,∞)(0,∞)
x- intercept: none
y- intercept: (0,1)(0,1)
increasing if b>1 b>1
decreasing if b<1 b<1
Compare the graphs of exponential growth and decay functions.
Figure 3
How To: Given an exponential function of the form f(x)=b x f(x)=b x, graph the function.
Create a table of points.
Plot at least 3 point from the table, including the y-intercept (0,1)(0,1).
Draw a smooth curve through the points.
State the domain, (−∞,∞)(−∞,∞), the range, (0,∞)(0,∞), and the horizontal asymptote, y=0 y=0.
Example 1: Sketching the Graph of an Exponential Function of the Form f(x) = b x
Sketch a graph of f(x)=0.25 x f(x)=0.25 x. State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and create a table of points for the graph.
Since b= 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y= 0.
Create a table of points.
x–3–2–1 0 1 2 3
f(x)=0.25 x f(x)=0.25 x64 16 4 1 0.25 0.0625 0.015625
Plot the y-intercept, (0,1)(0,1), along with two other points. We can use (−1,4)(−1,4) and (1,0.25)(1,0.25).
Draw a smooth curve connecting the points.
Figure 4.The domain is (−∞,∞)(−∞,∞); the range is (0,∞)(0,∞); the horizontal asymptote is y=0 y=0.
Try It 1
Sketch the graph of f(x)=4 x f(x)=4 x. State the domain, range, and asymptote.
Solution
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1561 | https://www.ccjm.org/content/ccjom/70/6_suppl_2/S8.full.pdf | IBS ROUNDTABLE It's important to establish early on whether the patient agrees with the diagnosis of IBS.
—Dr. Isaacson I CASE PRESENTATION A 42-year-old white woman presents to her prima-ry care physician with chronic symptoms of abdom-inal pain, bloating, and constipation of 6 months’ duration. She has a bowel movement every other day, with straining, passing hard stool. She reports intermittent abdominal discomfort/pain and bloat-ing almost daily, with “spasms” in the lower abdomen at the time of the bowel movement. She says the pain can be quite severe at times. Her weight fluctuates, but she has no net weight loss.
There is no rectal bleeding.
She has tried fiber but stopped because it makes her bloated. She is taking paroxetine for anxiety and sumatriptan, intermittently, for migraine headaches. She was seen elsewhere and had a bar-ium enema, which was normal. She was told to take laxatives but is reluctant to because she fears “getting used to them.” Dr. Edy Soffer—I’d like to start by asking our primary care physicians, Drs. Brunton and Isaacson, to tell us what you would do if this patient came to your office.
Dr. Stephen Brunton—Well, this patient has a lot of classic symptoms of irritable bowel syn-drome (IBS), which would make it tempting to make a positive diagnosis of IBS after a basic history and exam, but I think there may be some insecurity issues for both the primary care clinician and the patient. Because we manage the whole patient, I think some primary care physicians may need to further rule something out rather than initially make a positive diag-nosis of IBS. It also may depend on my rela-tionship and continuity with the patient, although it sounds like I haven’t seen this par-ticular patient before. If that’s the case, I’d be more likely to do at least some kind of workup before making a definitive diagnosis.
Dr. J. Harry Isaacson—I agree that the nature of the patient’s relationship with the primary care physician is critical, and it has to be fac-tored into any kind of algorithmic approach to IBS. If, based on the history, you are confident of the diagnosis and feel the symptoms are unlikely to represent something more serious, you can portray that confidence to the patient and probably not have to do much testing up front.
Also, understanding the patient’s expecta-tions up front is very important, since it can help you gauge your approach. With this patient, I could envision doing no additional evaluation other than talking to her, establish-ing a relationship, and maybe coming up with some change in therapy—if she is comfortable with that approach. I’d want to give her the sense that I’m going to stick with her over time, that we are going to work on this togeth-er. This approach may lower any patient demands for early and unnecessary testing. If she has very high expectations for a big evalu-ation, I might move up some of the testing I would normally defer, if the cost and potential for harm are relatively small.
Dr. Soffer—What about the reason why she is coming to see you right now—how much of a difference does that make? Is it because her symptoms have become unbearable? Or is her anxiety such that she just needs to know what’s going on? Or did she not hit it off with the pre-vious physician she saw? As a referral-based doctor, I generally see the patients who are uncomfortable with the diagnosis by a nonspe-cialist; they want that specialist “stamp of approval” that it’s IBS.
Dr. Kevin Olden—Yes, and the advantage we have as referralists is that we get an outside chart with a lot of the testing already done, and that makes our job much easier in terms of what to do initially. The advantage we special-ists don’t have is that close ongoing relation-ship that many primary care doctors have with their patients. It’s much less likely that our IBS patients will see us back again, because of geog-S8 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 Roundtable discussion Irritable bowel syndrome: New tools and insights for the primary care physician on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from raphy or insurance or whatever else, so that puts more pressure on us to do more in our one or two office visits with the patient. Dr. Isaacson—I think there are really two issues to clarify with this patient. First, what does she think she has? If she agrees with the diagnosis, that’s one hurdle that’s cleared. If she doesn’t, then you have a set of challenges as the primary care physician that may include consultation to confirm the diagnosis. The second issue relates to management problems, which gets into treatment, but that usually comes after you’ve agreed on a diagnosis. I think it’s very important to establish early on, whether you’re a specialist or a primary care physician, if the patient thinks she has IBS or thinks she has something else.
Dr. Brunton—That’s critical, because many patients with abdominal pain and a change in bowel habits may think they have cancer and want to ask you about it. So I’ll ask up front, “Are you concerned that you have cancer?” And I can see the relief on their face, because they want to make sure you’re thinking about that. Then I’ll say, “Frankly, this is not a sign of cancer; here’s what we look for when we think about cancer, and this is not that.” Once you allay that fear, you create a much better landscape for proceeding.
Dr. Olden—Yes, that underscores the impor-tance of asking as specifically as possible why the patient is seeing you now. I always ask, “What is your concern?” and then explore fur-ther if necessary. Some patients think that someone missed their diagnosis. Others fear that they have an ominous disease. Sometimes it’s a treatment concern, where the patient just wants her symptoms to be addressed more effectively. Other times it’s a quality-of-life issue, where her symptoms are well controlled but she doesn’t like taking all these medicines.
Or it may be a psychological issue—the patient’s bowels are doing okay but she’s pho-bic about losing control so she tracks her movements around town for bathrooms. These issues can quickly clarify both the kind and the extent of the difficulty the patient is in.
I SCREENING: HOW MUCH IS ENOUGH?
Dr. Soffer—Let’s change our case presenta-tion a bit and assume you are the first physi-cian to see this patient. Would you proceed any differently, and what tests would you do?
Dr. Olden—I see some de novo patients with IBS as well as referral patients, so I can speak to this. Obviously, I’d take a history and then do the basic studies for a de novo patient: a physical exam, a CBC, and a chemistry panel.
Once you allay the fear of cancer, you create a much better landscape for proceeding.
—Dr. Brunton New patient with symptoms suggestive of IBS Red flags/ alarm factors Yes No These include new onset after age 50, fever, weight loss, nocturnal symptoms, gastrointestinal bleeding, anemia, abnormal physical examination, and family history of inflammatory bowel disease, colon cancer, or celiac disease.
CBC = complete blood cell count; TSH = thyroid-stimulating hormone.
CBC, TSH test, electrolytes, hemoccult; therapeutic trial based on predominant symptoms Reassess in 4 weeks if not better Further evaluation as indicated by findings Further evaluation FIGURE 1. Diagnostic algorithm for initial evaluation of a patient with suspected IBS.
CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S9 on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from But I would hit the brakes at that point, as I have cautiously become a minimalist over the last 3 years or so. I would not order any imag-ing, of the gallbladder or colon or otherwise, and I would not go to second-level laboratory tests, such as for celiac sprue. If the patient meets the Rome II criteria, I would treat empirically for 4 weeks, see what happens, and go from there. This is what has worked for me, and I haven’t been burned yet, but I reserve the right to go back after 4 weeks and do further inves-tigations if either the patient or I am not a happy camper.
Dr. Isaacson—Yes, I don’t think you have to rule everything else out before you make the presumptive diagnosis of IBS if the patient understands that their symptoms fit the pattern of IBS and that there will be fol-low-up and, if necessary, a contingency plan. The idea of a positive presumptive diagnosis is critical, because it gives us a constructive alternative to the “wastebas-ket” approach to IBS that most of us encountered in our training—ie, if we can’t figure these symptoms out, let’s put them in this wastebasket called IBS. The thyroid-stimulating hormone (TSH) test is the one test I’d consider adding to the list Dr. Olden just outlined, since constipation and diarrhea are both manifestations of thy-roid disease. When someone walks in with a 3-month history of severe constipation or diar-rhea, it’s hard for a primary care physician to delay doing a TSH, especially since it’s easy and relatively inexpensive. Dr. Brunton—Yes, I thought the same thing, and the TSH is actually part of our chemistry panel where I practice.
Dr. Olden—That’s a very good point because you two, as primary care physicians, need to be especially oriented toward the care of the whole patient. So, while you’re looking for what’s causing the patient’s bowel symptoms, you sure would also like to know if they have thyroid dysfunction or anemia. Within this context, doing a TSH is very reasonable and pertinent. So if you find thyroid dysfunction and treat it, even if the patient’s bowel symp-toms remain unchanged afterward, it is still a victory for the patient. Dr. Soffer—Plus, we are all creatures of experi-ence. If you missed one case of thyroid disease 10 years ago, it will cause you to always take a IBS ROUNDTABLE Patient meets Rome II criteria for IBS Unresponsive to initial IBS therapy Constipation Diarrhea Tests to consider •Flexible sigmoidoscopy •Colonoscopy with biopsy •Stool O&P, C difficile, electrolytes •Celiac serology •EGD with small bowel biopsy •Small bowel series •Hydrogen breath tests for lactose intolerance and bacterial overgrowth Infrequent defecation Tests to consider •Flexible sigmoidoscopy •Colonoscopy •Colonic transit test Obstructed defecation Tests to consider •Flexible sigmoidoscopy •Colonoscopy •Anorectal manometry •Balloon expulsion •Defecography† FIGURE 2. Diagnostic algorithm for further evaluation of nonresponding patients.
Refer to a gastroenterologist for further evaluation.
†To be considered on an individual basis.
EGD = esophagogastroduodenoscopy; GI = gastrointestinal; O&P = ova and parasites Traditionally, IBS recommendations have been by and for gastroenterologists, but it's primary care physicians who see the bulk of IBS patients.
—Dr. Soffer S10 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from TSH for the rest of your career. This is not like-ly to be changed by future recommendations.
Dr. Bo Shen—What about tests like a screen-ing flexible sigmoidoscopy or a screening colonoscopy? Are these tests that you would be comfortable doing yourself before referring the patient to a gastroenterologist?
Dr. Brunton—I wouldn’t order those tests early on in a patient like this because we have the luxury of trying some therapeutic approaches and seeing how the patient fares.
In this way, the therapeutic trial can serve as an additional test to perhaps reinforce whether the diagnosis was correct. In any case, this patient has a pretty good history; I’m not really worried about colon cancer because there is no bleeding and she’s only 42. Down the line, if there is no response after 4 weeks of reasonable therapy, I would assess further, and colonoscopy would probably be among the things I’d consider then. Dr. Isaacson—I think there’s a misperception among primary care physicians that you have to do some testing beyond blood work in order to establish the diagnosis of IBS. There’s a ten-dency to not want to miss something, so you get a sigmoidoscopy or an ultrasound, which I think crosses the threshold into invasive, unnecessary, and expensive care, unless an alarm factor is present. What I like about the American College of Gastroenterology’s new CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S11 Further tests for evaluating patients with diarrhea and constipation Quantitative fecal fat This test analyzes stool collected over 72 hours, while the patient is on a high-fat diet. Stool fat content of more than 7 grams over 24 hours is considered abnormal. This test is done on an outpatient basis and is useful for diagnosing steatorrhea.
Breath tests Breath tests measure the respiratory excretion of labeled CO2 after oral administration and metabolism of radioactive carbon-labeled substrates (or of H2 after administration of carbohydrates). These tests can be used to assess fat, carbohydrate, bile acid malabsorption, or bacterial overgrowth. They rely on the principle that conditions of malabsorption or the presence of bacterial overgrowth in the small bowel lead to abnormal metabolism of test nutrients in the small bowel. This, in turn, results in abnormal breath content of the gases measured.
Serologic markers for celiac disease Antibodies against four antigens (gliadin, reticulin, endomysial, and tissue transglutaminase) are useful in the diagnosis of celiac disease. Those of the IgG subclasses are generally more sensitive, and IgA is usually more specific. Although serologic tests do not supplant small bowel biopsy in the diagnosis of celiac disease, they are useful as screening tests and for monitoring adherence to a gluten-free diet.
Colonic transit study This test is useful in evaluating a patient with severe constipation. The patient ingests radiopaque markers while on a high-fiber diet and while abstaining from laxatives, and abdominal radiographs are taken after a number of days. Several variations exist, but all consist of giving markers and counting the remaining number in the colon. An excess number of markers indicates the presence of colonic inertia, or slow-transit constipation. The distinction between this condition and functional constipation is important since colectomy is considered only for slow-transit constipation.
Anorectal manometry This test measures the strength of the anal sphincters, rectal sensation, and recto-anal coordination and reflex activity. It is done on an outpatient basis by transanal insertion of a catheter with pressure sensors and an attached balloon. It helps to determine the presence of obstructed defecation and to exclude Hirschsprung disease.
Defecography This is a technique in which barium, thickened to a consistency that approximates stool, is introduced into the rectum. Fluoroscopy is then used to monitor evacuation of barium during attempted defecation. While defecography can be used to evaluate functional problems, such as obstructed defecation, it is particularly helpful in assessing the presence and functional severity of structural abnormalities such as rectocele or intussusception. It is done by radiologists on an outpatient basis.
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www.ccjm.org Downloaded from S12 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 position statement on IBS is that it empowers primary care physicians not to have to do any imaging or endoscopy or ultrasonography to make a presumptive diagnosis.
Dr. Soffer—Your comment points out how much the traditional IBS recommendations have been by and for gastroenterologists.
Although primary care physicians see the bulk of IBS patients, your concerns have not been adequately addressed. Dr. Olden—Yes, we as gastroenterologists have done an inadequate job of adapting the IBS data to the primary care perspective, as I’m learning this morning, and that’s impor-tant because we inhabit two different worlds in terms of patient management. Primary care doctors have the advantage of ongoing, longi-tudinal care, but at the same time you have the dilemma of the de novo patient, where the dif-ferential diagnosis is wide open. You also have pretest probability more on your side because IBS is pretty common. Dr. Soffer—The Rome criteria are a good example of how gastroenterologists are not sensitive enough to the realities of primary care. A study was done to assess how well pri-mary care physicians know the Rome criteria, and the results were absolutely dismal. But for that matter, I don’t know how well most gas-troenterologists really know the Rome criteria.
But I think the recent ACG Task Force did a nice job of simplifying these criteria for practi-cal clinical purposes. I OTHER KEY CONSIDERATIONS IN THE EVALUATION Dr. Soffer—Returning to our case, is there anything else you would want to find out about this patient, and how would you take it from here?
Dr. Shen—We need to thoroughly screen for all the medicines patients are taking, especial-ly for de novo patients. That includes over-the-counter drugs and alternative therapies, since many of these can cause diarrhea or con-stipation (TABLE 2). For example, nonprescrip-tion NSAIDs can cause either diarrhea or con-stipation, antacids may cause bloating, and, in our case presentation, the paroxetine that the patient is taking could cause constipation.
Dr. Isaacson—Also, we don’t have this patient’s lab work, so obviously you’d need that. Beyond that, I’d inquire about family his-tory and, along with determining the patient’s expectations, explore any psychosocial aspects, which would include asking if there has been a history of sexual abuse, especially since pain is a dominant symptom.
Dr. Olden—Yes, sexual abuse can be a key fac-tor in IBS patients. A number of researchers have shown conclusively that there is an excess prevalence of a physical or sexual abuse history among people with functional GI disorders.
When pain, particularly pelvic pain, is a pre-dominant symptom in an IBS patient, that’s a huge red flag for a potential abuse history. Now, is abuse etiologic in IBS? Probably not. It probably represents tremendous damage to the patient’s emotional integrity and coping ability, so that their IBS becomes more intru-sive and troublesome. But it is a critical area to explore because abuse histories are associated with a great deal of depression, somatoform disorders, and anxiety, and these things need to be treated.
Dr. Isaacson—Of course, asking about abuse is more difficult than other areas of the histo-ry, but I think it can be really fruitful in terms of forming a good relationship with the patient and making them realize that you understand their predicament.
Dr. Olden—Absolutely, and the abuse litera-ture clearly supports that. It also shows that patients see this as an appropriate area for physicians to ask about. How do you broach the subject? I bring up the epidemiologic reason for asking—“Not uncommonly, people with your set of symp-toms report physical or sexual abuse; is that an issue for you?” That way, they know why I’m asking. Then I gauge their reaction: if they say yes in a reasonably unemotional way, I will explore it further; if it’s clearly overwhelming for them, I will close the book and maybe try to gently reopen it over time.
IBS ROUNDTABLE We need to thoroughly screen for all the medicines patients are taking, since many common agents can cause diarrhea or constipation.
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www.ccjm.org Downloaded from Dr. Soffer—What do you recommend for those of us who don’t have your background in psychiatric subcare? Dr. Olden—I’d suggest partnering with some-one in the mental health community. It can be a psychiatrist, a PhD psychologist, the nurse practitioner in your office, a counselor—it matters less where the referral goes than the fact that it is brought up and goes somewhere.
So it’s reasonable for the abuse issues to be taken care of elsewhere, but this is going to be more and more of a primary care issue as new generations of physicians come along.
Dr. Brunton—To return to what else we’d need to know from the patient evalua-tion…this isn’t specific to this case presenta-tion, but I’d like to raise the issue of the elderly person with constipation, which opens up a whole other process. In the elder-ly, we generally don’t think “irritable bowel”—we may just think “old bowel.” The elderly patient with IBS-like symptoms also makes you think about colon cancer or metastatic ovarian cancer, and it raises things like laxative abuse and other issues.
Dr. Isaacson—Yes, if a person hasn’t had a pattern of these symptoms, and suddenly they appear at age 72, the first thing I’m thinking of is not IBS.
Dr. Olden—Absolutely, and this was touched on in the ACG position statement, but it was not emphasized enough. Certainly, anyone age 50 or older with new-onset symptoms sugges-tive of IBS absolutely demands a more aggres-sive workup than what we’ve discussed, and IBS definitely becomes a diagnosis of exclu-sion in these patients.
I WHEN TO REFER Dr. Soffer—Let’s talk a bit about when to refer to a gastroenterologist. Under what type of scenario would you refer?
Dr. Isaacson—I think there are two scenarios: when I am unclear about the diagnosis, and when there are management problems. The first scenario would often involve a patient who has unmet expectations, who may not buy into the diagnosis and wants a specialist to confirm it. Also, if a patient hasn’t responded to initial therapy, I may want the diagnosis confirmed.
The second scenario relates to manage-ment problems, when the patient and I are comfortable with the diagnosis but the patient is not improving after trying a few therapies.
In this scenario, I’d want to see if there was something else we could try and want to get a fresh take on the case.
Dr. Brunton—For me, it would often be to reaffirm a suspected diagnosis. I think there is still a real concern among primary care physi-cians about possibly missing something. And while we can sit here and say IBS should be a positive diagnosis that we can all make, in the real world it isn’t always so clear-cut and there are a lot of competing priorities. Plus, a lot of us don’t have open access to colonoscopy or all the procedures in the possible workup, so that’s another factor. And there are always the patients who fear that they have something CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S13 Common agents associated with constipation or diarrhea AGENTS ASSOCIATED AGENTS ASSOCIATED WITH CONSTIPATION WITH DIARRHEA Analgesics Anticholinergics • Antidepressants • Smooth muscle relaxants • Antipsychotics Cation-containing agents • Iron supplements • Aluminum (antacids, sucralfate) Opiates Calcium channel blockers T A B L E 2 Laxatives • Stimulant (senna, bisacodyl, castor oil) • Osmotic (magnesium-containing agents) • Nonabsorbable carbohydrates (sorbitol, lactulose) Prokinetics (metoclopramide, tegaserod) Proton pump inhibitors Antacids (magnesium-containing agents) Antibiotics Nonsteroidal anti-inflammatory drugs Antineoplastics Antiarrhythmics Antihypertensives Chronic alcohol ingestion on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from S14 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 more serious than IBS, and they aren’t going to be happy without that specialist consult.
Dr. Isaacson—I’d be more apt to get a special-ist consult for an IBS patient with diarrhea because I think the evaluation is a little more subtle and complex for diarrhea than for con-stipation. One thing I picked up from the new ACG position statement on IBS was the potential role for celiac sprue testing in IBS patients with diarrhea. I haven’t paid enough attention to the sprue issue, and I worry about microscopic colitis in patients with diarrhea, so these things would push me toward a consult.
Dr. Soffer—Yes, the diarrhea cases have a wider differential diagnosis. With regard to celi-ac sprue, you may want to look into that in cer-tain populations with diarrheal diseases, but you should still be selective since testing for it will generate a number of false-positives, which then generate more unnecessary testing. If I see someone of, say, Irish ethnicity with typical symptoms of IBS with diarrhea, I may test for celiac sprue, but I’m unlikely to do so for ethnic groups where celiac disease is very uncommon.
I TREATMENT: GENERAL ISSUES AND APPROACH Dr. Soffer—Shall we move on to treatment now? Again, let me start by posing the first question to our primary care panelists: How would you direct the treatment of the patient in our case presentation?
Dr. Brunton—Well, she has tried fiber and stopped. I would want to know exactly how she tried it. Did she go up to 50 grams, or what? I usually start patients slower and get them used to it. I’d want to try it that way if that’s not how she tried it, since it’s not going to hurt unless she has some real issues with that.
Obviously, a patient like this is a great patient for tegaserod, since tegaserod has the indication of IBS with constipation and this patient’s symptoms have been chronic and sound pretty substantial. Would I start tegaserod straight off? I think there would be some financial issues: Does she have insur-ance? Can she afford it? Also, she’s someone for whom an antispasmodic like hyoscyamine might be tried, because of her spasms during bowel movements. But I would be very com-fortable putting a patient like this on tegaserod, probably very early on in the process.
Dr. Isaacson—I would sit down with this patient and explain that there are a lot of treatment avenues we haven’t approached yet.
I would discuss them and ask her to participate in the decision. If she says the constipation and the pain are just unbearable, it’s going to push me toward more aggressive therapy at an earlier stage. If she seems reassured, I’d lay out the range of options and ask what sounds good to her. Depending on the lead she gives me, I could be comfortable reintroducing fiber in a different way, or trying something else for the spasm, or maybe withdrawing the paroxetine if she suggests it may not be helping her anxiety enough. Dr. Brunton—But her symptoms have been presented as pretty significant and chronic, so I’d perhaps consider her case as more “moder-ate to severe” from the start.
Dr. Isaacson—Then you can tell her that if her symptoms are more severe we have some really good options available now that have been well studied. One of the things I learned from the new ACG position statement is that there have been some very good studies done with these newer agents for IBS.
In any case, I think that showing some optimism and willingness to partner with the patient is very therapeutic in and of itself in many cases.
Dr. Olden—I agree. We tend, as researchers, to minimize the doctor–patient relationship, positive transference, or whatever you wish to call it, and its ability to improve patients’ out-comes. I think it’s a tremendous variable in many disorders, IBS being just one of them. I will tell patients straight up, “I’m going to hang in there with you. It may take a while, but there is a general tendency to get better.
We may try a lot of drugs, and many of them might not work, but we’ll work through this.” That alone can get their anxiety down and let them see some light at the end of the tunnel.
IBS ROUNDTABLE When pelvic pain is a predominant symptom, that's a huge red flag for a potential abuse history.
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www.ccjm.org Downloaded from Dr. Brunton—This gets at the role of the placebo effect in IBS. Part of the issue in most conditions involving chronic pain or similarly debilitating symptoms is that patients can come to feel that the pain controls them.
Anything you can do or give to make them feel they are taking steps to overcome that control can be quite empowering.
If you look at some of the IBS treatment studies, you see some very large placebo response rates. I think this substantial placebo effect may be at work with some of the tradi-tional IBS therapies that lack supportive clini-cal trial data, perhaps along with the natural tendency for symptoms to wax and wane. For instance, if half of my patients will do well with fiber, and that costs only pennies, I’ll cer-tainly try that. Why shouldn’t I try that and then, for those who don’t do as well, go on to a more clinically supported drug? Is it good sci-ence? No, but it’s a “less is more” practical solu-tion that seems to work to a certain extent, and many clinicians do this in the real world.
Dr. Soffer—You raise an important issue. If fiber is safe (except for the bloating) and it’s going to give X% of patients some benefit, I would feel comfortable trying that initially, as long as I’m not harming the patient. This touches on my one concern about the new position statement from the ACG, which is, to the credit of all the ACG Task CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S15 Selected drug therapies for IBS PREDOMINANT IBS SYMPTOM DRUG CLASS AND SPECIFIC AGENTS USUAL ADULT DOSAGE Diarrhea Opioid µ-receptor agonists Loperamide (Imodium and others) 2–4 mg up to four times daily, as needed Diphenoxylate (Lomotil and others) 5 mg four times daily, as needed Smooth muscle relaxants Dicyclomine (Bentyl and others) 20 mg four times daily initially, then up to 40 mg four times daily Hyoscyamine (Levsin and others) 0.125 mg sublingually three times daily as needed, or 0.375 mg orally twice daily Tricyclic antidepressants Amitriptyline (Elavil and others) 10–100 mg at bedtime Desipramine (Norpramin and others) 10–150 mg at bedtime Selective 5-HT3 receptor antagonist Alosetron (Lotronex) 1 mg once daily for 4 weeks;may increase to 1 mg twice daily for 4 weeks Constipation Bulking agents Psyllium (Metamucil and others) 20 g/day, divided, with >250 mL water Polycarbophil (Konsyl Fiber and others) 1–6 g/day, divided, with >250 mL water Methylcellulose (Citrucel and others) 3–6 g/day, divided, with >250 mL water Nonabsorbable carbohydrates Lactulose (Kristalose and others) 15–60 mL/day, divided Sorbitol 120 mL of 25% solution Osmotic laxatives Magnesium hydroxide (Milk of Magnesia) 1–2 oz/day Polyethylene glycol solution (MiraLax) 1 dose (17 g in glass of water) once or twice daily Selective 5-HT4 receptor antagonist Tegaserod (Zelnorm) 6 mg twice daily for 4–12 weeks Physicians must enroll in a special risk-management program to prescribe this drug.
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www.ccjm.org Downloaded from S16 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 Force members, the most critical and evi-dence-based approach to IBS I’ve seen. When you look at the position statement, there isn’t much room to recommend, say, loperamide because there’s no global improvement in IBS symptoms. But if I give loperamide I can per-haps improve diarrhea, and I can do so very cheaply and very safely, and I think that’s something, even if it’s not global symptom improvement. Dr. Olden, as a Task Force member, do you think the Task Force might have been a little bit too academic?
Dr. Olden—You make an excellent point.
Take fiber, for instance. There is a large litera-ture criticizing the efficacy of fiber, but in the real world we all prescribe fiber, both for IBS and for non-IBS conditions. Fiber supplemen-tation is a key element of improving or pre-venting a lot of medical conditions, and it is so cheap and nontoxic that there is essentially no downside to using it, and I do it all the time.
So the threshold is so low, and when this is combined with a good doctor–patient rela-tionship, you may get a certain “bang for your buck” that you won’t get from a randomized controlled trial.
Dr. Isaacson—I agree, and I have a couple of related points. I like to identify what the patient’s main symptoms are, define an end-point, try a treatment for 4 to 8 weeks, and see how it works. I try to choose an objective end-point, such as some specific measure of pain or whether the patient can do things now that they couldn’t before. If the treatment hasn’t worked well enough, we can try something else. Also, if I’m unclear about where to go with the treatment and if the patient is not terribly distressed, I may just have them monitor their symptoms for a while. This can help establish my relationship with the patient and clarify how to treat if we should need to. Sometimes, if the patient’s distress is not great, they may be satisfied with just understanding their disease.
Not everyone has to come away with a med-ication.
I FACTORING IN DEPRESSIVE DISORDERS Dr. Soffer—Let’s return to our case presenta-tion. Let me propose switching our patient from her current SSRI, paroxetine, to another SSRI, say, sertraline, which has diarrhea as a side effect, plus a potential effect on pain. Let’s say we consider this along with giving her fiber again or an inexpensive laxative. What do you think of this approach?
Dr. Olden—It depends on how the patient comes to me. If she comes, like this patient, on stable SSRI treatment for anxiety or depres-sion, I am reluctant to fool with that because while the new SSRI may have a better bowel effect, I could turn her depressive illness on its ear and really interfere with her treating physi-cian’s plan. If she comes to me not already on an antidepressant, it’s a different ballgame and I feel very empowered to start and then change antidepressants as needed, for either depressive or bowel symptoms.
Dr. Soffer—I’m with you. If I wanted to make a change of the antidepressant, I’d only do so through her original prescribing doctor. But in a de novo patient, because I’ve had a pretty good experience with antidepressants for pain, I might try this because if she does not respond to this combination, the next option is a 5-HT4 agonist, which can address both the bowel symptoms and the pain. Dr. Brunton—Are you suggesting, then, Dr.
Soffer, that you would use tegaserod second-line?
Dr. Soffer—That depends. Let’s say I’ve got an IBS patient who comes in with anxiety but who I don’t think needs to see a psychiatrist because I feel my experience is sufficient to treat her. I would want to give her something for her anxiety and something for constipation.
Remember, we are talking about someone for whom anxiety is an issue. If anxiety were not an issue, I wouldn’t start with that approach.
Dr. Olden—The flip side to that question is to suppose the patient isn’t depressed or anxious.
Do we start with an antidepressant or do we move right to the new disease-specific agents for IBS? I would go to the new disease-specific agent, absent depression or anxiety, because frankly it is about the same price and you are getting a more specific drug. With the patient IBS ROUNDTABLE Showing some optimism and willingness to partner with the patient can be very therapeutic in and of itself.
—Dr. Isaacson on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from in our case presentation, I think you have to use both. I’d keep her on paroxetine or its equivalent, and I would probably add tegaserod to her regimen.
I SAFETY AND THE SEROTONIN SYSTEM Dr. Soffer—Are there any issues with the sumatriptan our patient is taking, which is a 5-HT1 agonist? Given that paroxetine and tegaserod both act on the serotonin system as well, how many serotonin agonists or antago-nists can I give to one patient?
Dr. Olden—That’s a very relevant question.
Serotoninergic syndrome is a very serious disease that we created by developing new treatments.
It resembles malignant hyperthermia with hyperpyrexia, delirium, arrhythmias, and rhab-domyolysis. It’s still an epidemiologic work in progress, but it is seen in people who are taking multiple psychiatric drugs, mainly monoamine oxidase inhibitors or dopaminergic antagonists plus a serotoninergic-reactive antidepressant.
There has been no association between serotoninergic syndrome and the concomitant use of an SSRI and a 5-HT1 agonist like suma-triptan. Likewise, to date there has been no interaction shown between SSRIs and tegaserod or between the 5-HT1 agonists and tegaserod. So it’s a pretty clean story.
Serotoninergic syndrome is seen mainly in psychiatric settings among patients who are on big-time psychiatric drugs, but it’s something we need to be continually aware of.
Dr. Shen—A lot of patients with IBS have migraines. Is there any concern that starting tegaserod in this patient from our case presen-tation might exacerbate her migraines, since headache may be seen with tegaserod?
Dr. Olden—No, at least not based on the evi-dence we have. While there was an increased incidence of headache with tegaserod com-pared with placebo in the phase II and III tri-als, there was no increase in migraine head-aches. As we understand it, migraine seems to be a 5-HT1 phenomenon, so you wouldn’t expect a 5-HT4 agonist like tegaserod to exac-erbate migraine, because of its specificity. I COMFORT LEVELS WITH THE NEWER AGENTS Dr. Soffer—What’s your sense of how comfort-able physicians are, particularly primary care physicians, prescribing the new IBS drugs, alos-etron and tegaserod? We have not yet started with the new prescribing program for alosetron at the Cleveland Clinic, and I wonder how many general practitioners will be comfortable with that elaborate system, as opposed to just turning to specialists to prescribe alosetron.
CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S17 Patient meets Rome II criteria for IBS No alarm factors IBS with diarrhea Increasing severity IBS with constipation •Loperamide or diphenoxylate •Smooth muscle relaxants •Tricyclic antidepressants •Cholestyramine •5-HT3 antagonist (alosetron) •Behavioral therapy •Fiber supplementation •Osmotic laxatives •5-HT4 agonist (tegaserod) •Behavioral therapy FIGURE 3. Symptom-oriented treatment of IBS.
The new position statement from the ACG is the most critical and evidence-based approach to IBS I've seen.
—Dr. Soffer Prescribing restricted (physicians must enroll in a risk-management program to prescribe). on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from S18 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 Dr. Olden—Well, I’ve taken on the alosetron prescribing at Mayo Clinic Scottsdale, and at just a week into the drug’s return to the mar-ket, the line was getting long already. Most physicians are not doing it—I’m the only member of my gastroenterology division who wants to handle prescribing it. I think that’s because the prescribing program is seen as bur-densome. In fact, though, it’s really not as bur-densome as it sounds. Everything you need is in a three-ring binder: you just go through the forms with the patient, and everything stays in the binder. The drug’s manufacturer has done a really good job of making the system non-burdensome, but it is a system nonetheless.
Beyond that is what I call the “Lotronex legacy,” and it’s been a real problem, because for all of alosetron’s wonderful effects, its safety issues have created some real reluctance in some quarters that is hard to get past. With any new drug, the safety story is at least as important as the efficacy story. Physicians start wondering, “What will be the next surprise with alo-setron?” It can even translate into “What will be the surprise with tegaserod?” even though that’s a completely different molecule with a different effect. Tegaserod, so far, seems to be pretty clean. But the serotoninergic playing field has been contaminated a bit, so to speak.
Dr. Isaacson—I agree that you’re not going to get past that very easily with alosetron. One of the rules of thumb for a primary care physician is that you don’t want to be the first or the last to prescribe a new effective agent. Many of us in primary care will wait a bit to see how the specialists use a medicine and define the risks a little better before we get involved. Besides, a patient would have to have pretty severe diarrhea to be a candidate for alosetron, and that’s just the kind of patient I’d be more apt to refer.
Dr. Brunton—What’s unfortunate is that if a lot of gastroenterologists remain reluctant to prescribe alosetron, because of either the has-sle factor or the comfort factor, that has big implications for the doctor–patient relation-ship. If I refer my patient to a gastroenterolo-gist, and he then refers them to another gas-troenterologist because he’s not prescribing alosetron, I risk losing contact with my patient and losing my leverage as the patient’s advo-cate. That may induce me to get some comfort about using alosetron myself more quickly than I would have otherwise. On a separate note, Dr. Olden, I was look-ing at the numbers you presented on the rates of ischemic colitis with alosetron use, and they seem very similar to the background preva-lence of ischemic colitis.
Dr. Olden—Yes, but it was not so much the numbers as the epidemiology: many of those cases of ischemic colitis with alosetron were in young people with intact circulatory systems, not in elderly people with heart failure. That’s what makes it ominous.
Dr. Soffer—What about tegaserod and the other 5-HT4 agonists that may be coming? Do primary care physicians feel comfortable using tegaserod at this point?
Dr. Brunton—Right now, I don’t see a lot of it being used by primary care doctors. After all, it’s a new class of drugs, and there’s the Lotronex legacy that we discussed. But as there’s more education, particularly from our own primary care educators, I think you’ll see more confidence and comfort in using it.
Also, the whole IBS playing field is still very confused. If alosetron had been around longer initially, there would have been a lot more education and awareness-raising about IBS, but that didn’t happen because alosetron was withdrawn so soon. Dr. Isaacson—I think the fact that IBS is not a life-threatening illness is a contributing factor as well. Physicians are generally less likely to prescribe a “risky” drug for a benign condition. Dr. Olden—Yes, but the patients would vigor-ously rebut the notion that IBS is a benign ill-ness, which they did when they organized to get alosetron back on the market. No one is going to die of IBS, but a large minority of IBS patients can’t function because of it.
Dr. Brunton—Absolutely, and quality-of-life studies show that IBS has a greater impact on patients’ lives than many other chronic dis-eases that we treat without hesitation. I think IBS ROUNDTABLE IBS has a greater impact on patients' quality of life than many other chronic diseases that we treat without hesitation.
—Dr. Brunton on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from it’s not that IBS doesn’t have a significant impact but that we traditionally haven’t felt we had good therapies for it, so we haven’t asked about it. And patients don’t necessarily bring it up; if they’ve had it for some time, they may just think that’s how their gut works. Dr. Isaacson—I agree. When I feel I have something to treat a condition with if I uncov-er it, I am more apt to look for it.
Dr. Olden—Yes, and that’s positive, although it can have a downside, as when alosetron first came out. Because it was “the first drug for IBS,” every patient who had a tummy pain had it dispensed, and that led to a lot of inap-propriate prescribing.
Dr. Brunton—What strikes me about this whole quality-of-life issue in IBS is that it seems to be most dramatic in the diarrhea-pre-CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 S19 Answers to common primary care questions about IBS What are the criteria for diagnosis of IBS?
While IBS is still considered a disorder characterized by the absence of biochemical and structural markers, it is important to make a positive diagnosis. Such a diagnosis is based on the presence of appropriate symptoms in the absence of alarm symptoms and signs. Various sets of symptom-based criteria have been developed, including the Manning criteria (1978) and the ROME II criteria (1999), which are the most commonly used (see TABLE 2 on page S5).
How much workup should I do in the absence of alarm symptoms?
The approach to each patient should be individualized. However, to avoid unnecessary and costly testing, the diagnosis should be made by identifying a symptom complex compatible with IBS and then using prudent, albeit not exhaustive, testing to make a positive diagnosis. Routine use of flexible sigmoidoscopy, colonoscopy, barium enema, and other imaging studies is not recom-mended. This is because, in the absence of alarm signs or symptoms, the pretest probability of abnormal organic disorders in patients with IBS symptoms is low. Patients with IBS do not appear to have an increased likelihood of most organic disease com-pared with the general population. Initial evaluation with low-cost tests such as a CBC, electrolytes, a thyroid-stimulating hormone test, and hemoccult is appropriate. A trial of symptom-oriented therapy is recommended. If the patient does not respond to this therapeutic trial, further evaluation, including invasive testing, may be indicated (FIGURES 1, 2).
When is endoscopy or sigmoidoscopy indicated?
Patients with alarm factors or "red flags," such as weight loss, anemia, gastrointestinal bleeding, and nocturnal symptoms, war-rant endoscopic evaluation. Also, patients who do not respond to a therapeutic trial of symptom-oriented agents should undergo further evaluation, including flexible sigmoidoscopy or colonoscopy. In patients older than 50 or with a family history of colon can-cer (especially a first-degree relative who had colon cancer before age 50), colonoscopy is indicated for the dual purpose of screening and evaluation of symptoms suggestive of IBS.
When should I look for celiac disease, and what is the best testing strategy?
A recent study reported that approximately 5% of patients with IBS symptoms had celiac disease, compared with a prevalence of less than 1% in a control population. The prevalence of celiac disease varies widely and depends on ethnic origin. Therefore, rou-tine screening of all IBS patients for celiac disease with serology or endoscopy and small bowel mucosal biopsy is not indicated.
However, for a patient with a family history of celiac disease or with a higher-risk ethnic background, such as Irish or Italian, screening may be considered, particularly if the patient has diarrhea.
Screening for celiac disease uses serologic tests for the assay of antibodies against four antigens (gliadin, reticulin, endomy-sium, and tissue transglutaminase). A single assay for antiendomysial antibody and, recently, tissue transglutaminase antibody is currently used for screening. The gold standard for diagnosing celiac disease remains small bowel mucosal biopsy, which should be done to confirm the diagnosis when serology is positive.
When should I refer an IBS patient to a gastroenterologist?
The purposes of referral may vary. In some cases, referral to a gastroenterologist serves to confirm the diagnosis of IBS, especially for patients who are apprehensive about their symptoms. Other times primary care physicians refer patients for more specialized tests (TABLE 1), particularly those who do not respond to therapy (FIGURE 2). Patients with IBS symptoms who do not respond to standard therapy and continue to have troubling symptoms should undergo further evaluation, usually by a gastroenterologist.
Finally, patients with alarm factors or "red flags" need referral to exclude organic diseases such as gastrointestinal neoplasms, inflammatory bowel disease, infectious diarrhea, or malabsorption.
T A B L E 4 on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from S20 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 70 • SUPPLEMENT 2 JUNE 2003 dominant form. After all, patients with that symptom pattern are obsessed with always knowing where the nearest bathroom is. But it seems like the form that alternates between diarrhea and constipation is also very limiting and confusing. What is the role of tegaserod or alosetron in the alternating form of IBS?
Dr. Olden—That’s an important question, but one that hasn’t yet been adequately addressed in clinical trials.
Dr. Shen—While we’re talking about uses outside of FDA-approved indications, what about the use of tegaserod in males? We don’t have any data, right? And how long can we treat with tegaserod, since its approval was based on 12-week studies?
Dr. Olden—No one is eager to venture too quickly outside of FDA-approved labeling, particularly for this condition, given the Lotronex legacy. Length of treatment is less problematic since no drug gets approved beyond the length of the trials submitted for its approval. Data on length of treatment will expand as the postmarketing data come in, and there appear to be no safety issues regard-ing longer treatment, so I feel comfortable pre-scribing beyond 12 weeks if needed.
Use in males is a bit of a different question.
Based on the evidence available to me, I don’t think there is any safety issue with using tegaserod in males, but I reserve the right to be proven wrong. Having said that, because of my desire to protect this drug, the one drug for IBS that I have unencumbered access to, I am encouraging physicians in general not to use it in males. At the same time, because I am a spe-cialist in IBS and because I know this drug very well, I am prescribing it myself to selected male patients. Now, after I made this intellectual decision, I of course checked with our legal department.
They said I was well within the standard of prac-tice as long as I documented in the chart that the patient had given informed consent after I explained that it’s an off-label use, why I was doing it, and the risks and benefits involved.
Dr. Soffer—Yes, agreeing to it on an individ-ual basis is the key. But for a while, I think, only a very select group of physicians will pre-scribe tegaserod for males.
Well, I think we have covered drug treat-ment issues pretty well. Dr. Olden, would you like to close with a word about behavioral therapy for IBS, since that is an interest of yours?
Dr. Olden—As you know, the ACG Functional GI Disorders Task Force went no further than to say that behavioral therapy appears to be superior to placebo at relieving individual IBS symptoms. That was less of an endorsement than some might have expected, but it was based on a review from a few years ago, and some of the more recent trials of behavioral therapy are more positive.
Briefly, I think behavioral therapy is a tidal wave that’s going to wash over the island of IBS therapy, but it’s going to take a while.
In general, the behavioral therapy literature is expanding and maturing in an impressive way.
The biggest challenge is marrying up the dis-cipline of medicine with the discipline of behavioral therapy. That means getting the behavioral folks talking to us so that we’re as familiar with them as with surgeons and radi-ologists. And it will mean gaining a more sophisticated understanding of how behav-ioral therapy can complement the evolving medical therapy we are offering to our IBS patients.
IBS ROUNDTABLE The ischemic colitis seen with alosetron was ominous because many cases were in young people with intact circulatory systems.
—Dr. Olden on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from |
1562 | https://www.mathnasium.com/math-centers/cherrycreek/news/what-is-set-notation | We often sort things into groups—like types of animals, numbers that follow a pattern, or names on a class roster.
In math, we use set notation as a simple and consistent way to represent these groups. It helps us describe and organize collections clearly, whether we’re working with numbers, objects, or ideas.
In this middle school guide, you'll learn what a set notation is and how to write it, why it’s useful in math, and the answers to the most common questions.
Meet the Top-Rated K-12 Math Tutors in Cherry Creek
What Is Set Notation?
Imagine you're making a list of your favorite fruit. You might write down: apple, banana, strawberry. That’s a group of items—your favorite fruit.
In math, we call a group of things like this a set. A set is defined as a group of items. It could be a group of numbers, colors, animals—almost anything.
When we write that group using math language, we call it set notation. Instead of writing them in a sentence or list, we use a special format so everyone can easily understand what’s in your group.
Here’s what set notation looks like:
{apple, banana, strawberry}
We use curly braces to show that something is a set. Each item inside the curly braces is part of the set, and we separate them using commas.
So, if you had a set of your three favorite colors, it might look like:
{purple, green, red}
Or, a set of numbers could be:
{1,2,3,4}
We can also use symbols to talk about what’s inside a set.
The symbol means “is an element of”.
So, we can say:
apple {apple, banana, strawberry} - this means apple is an element of the fruit set {apple, banana, strawberry}
But what if something isn’tin the set?
That’s where the symbol comes in. It means “is not an element of.” So if something doesn't belong in the group, this is the symbol we use.
Let’s look at an example:
grape apple, banana, strawberry
This tells us that grape is not in the set of apple, banana, strawberry
You can think of as a polite way of saying, “Nope, this item isn’t invited to the party.”
You May Also Like: Inequality in Math: Simple Definitions, Comparisons & Symbols
Why Are Sets Important?
At first glance, set notation might just seem like another math skill to memorize—but it's much more than that.
Any time we organize a group of related things, we’re using the same thinking behind set notation.
For example, choosing players for a sports team, making a playlist of your favorite songs, or sorting school supplies by type are all real-life examples of grouping, which is exactly what sets do in math.
Set notation gives usa clear and consistent way to write down and work with groups of items. Instead of saying, “Here’s a list of numbers I’m working with,” we use a simple, symbol-based language to describe those groups quickly and precisely.
Sets are more than just lists—they’re a foundation for bigger math ideas. In probability, we use sets to represent possible outcomes. In algebra, we describe the set of numbers that solve an equation. In geometry, we talk about sets of points, angles, or lines.
Understanding how to group, compare, and describe collections using set notation is a skill that will keep showing up in real life, whether we're solving puzzles, managing data, or writing code.
Types of Sets
Now that you know how to read and write basic set notation and why it is important, let’s take it a step further.
Let’s look at different types of sets—including ones that go on forever, ones that are empty, and how to write sets that only have one item.
Finite Sets
When we say a set is finite, we mean it ends. Just like the word "finite" which comes from the Latin word finitus, meaning "limited" or "has an end." So, a finite set is a set where we can count the number of elements, and eventually, the list of elements stops.
Most sets we see in everyday life, like groceries, names, or items in your backpack, are finite sets.
Like a set of your favorite video games:
{racing game, building game, island adventure game}
That’s a finite set because it has exactly three items.
Or a set of odd numbers less than 10:
{1, 3, 5, 7, 9}
It doesn’t matter what kind of items are in the set, as long as we can count them and the set ends, it’s finite.
Infinite Sets
If finitemeans a set has an end, then infinite means the opposite—it never ends, right?
That’s exactly it!
Infinite comes from the Latin word infinitus, meaning “without end.” So, an infinite set is one that goes on forever.
Infinite sets can be made of numbers, decimals, patterns, or values between numbers.
Let’s look at some classic math examples:
{1, 2, 3, 4, 5, …}
This is a set of natural numbers starting from 1. Even if we can’t write every number, we show the pattern and use "…" to say “and so on.” There’s no last number.
Here’s another:
{2, 4, 6, 8, 10, …}
This is the set of even numbers. It also keeps going without end.
How about this?
{0.1, 0.11, 0.111, 0.1111, …}
This is an infinite set of decimals, each one getting closer to 0.111…but never exactly reaching it. This kind of set is called an infinite sequence.
We’ll explore infinite sets more when we dive into topics like algebra, number theory, and even calculus.
The Empty Set
The empty set is super simple—it’s a set that contains nothing.
We write it in one of two ways:
{}
(a math symbol that means "zero items")
Let’s see some examples of an empty set. For example, a set of pets that can talk like humans = . Unless your dog can speak English, this one stays empty.
Another example:
A set of numbers greater than 10 but less than 10 = {} This is impossible, so the set has no items.
Even though it has nothing in it, the empty set is still important in math. It’s like saying, “This group exists, but right now it’s empty.”
Set Notation for Set Representation
Now that we’ve learned how to write sets and spot whether something is in or out of a group, let’s explore how sets relate to each other using special math symbols.
These math symbols help us describe all kinds of relationships between sets, like which sets are bigger, which ones are empty, and what’s left out of a set.
Let’s break down the most important ideas:
μ – The Universal Set
The universal set includes everything we’re talking about in a given situation.
Think of the universal set as the “big picture.” We use the Greek letter μ (pronounced mu) to represent it.
Imagine bringing out your whole toy box to play with your friend. Your whole toy box is your universal set since it contains all of your toys.
Another example of a universal set are all the letters in the alphabet.
If Set A = vowels = {a, e, i, o, u}, and Set B = consonants = {b, c, d, f, g, …}, then the universal set μ = {a, b, c, d, e, f, g, …, z} represents all the letters in the alphabet.
We get to define the universal set depending on the problem. Just make sure it includes everything that could be considered.
A' – Complement of a Set
Thecomplement of a set(written A') includes everything in the universal set that’s NOT in Set A.
Let’s say:
μ = {1, 2, 3, 4, 5}
A = {2, 4}
Then:
A' = {1, 3, 5}
It’s like saying, “What’s left in the toy box when we take out the toy cars?”
Let μ be your whole toy box. If Set A is your toy cars, then A' is everything else in the box—like dolls, blocks, action figures, and slime. A' = “Not cars”
⊂ – Subset
A subset is a smaller set that comes from a bigger set. We use the symbol ⊂ to say “is a subset of.”
For example:
A = {a, b, c, d, e}
B = {b, c, d}
Here, B is made up of elements from A. So:
B ⊂ A
A subset doesn’t have to include all of A—just some or all of it, but never anything outside it.
Think of it this way: If Set A is all your snacks A = {chips, cookies, fruit, popcorn}, then a subset could be B = {chips, popcorn}. That means B ⊂ A.
∈ – Belongs To
We already know this one, but it’s worth repeating because it shows up a lot. We use to say something is in a set.
A = {a, b, c}
a A (a is in the set)
d A (d is not in the set)
FAQs About Set Notation
Got questions? We’ve got answers!
Here are some of the most common questions students ask when they first start learning about sets:
1. What’s the difference between a list and a set?
A list can include repeated items and always keeps them in a specific order. For instance, [1, 1, 2] is a valid list where the number 1 appears twice and the order matters.
A set, on the other hand, does not allow duplicate elements and doesn’t care about order. So {1, 2} is the same as {2, 1}—both represent the same set with the same elements, just arranged differently.
This means that while [1, 2] and [2, 1] are different as lists (because their order is different), they actually represent the same set: {1, 2}.
2. Can a set have just one thing?
Yes. A set can contain a single element, like {7} or {blue}. It still follows all the rules of a set.
3. What’s the emptiest set?
The empty set (also called the null set) contains no elements. It’s written as either {} or Ø.
This set is useful for situations where no items meet a certain condition.
4. Do sets always contain numbers?
Not at all. Sets can include numbers, words, letters, or any clearly defined objects. The key is that each item must be distinct and well-defined.
5. Can a set have repeated elements?
No. If you list the same item more than once in a set—like writing "a, a, b"—it still only counts once, so the set is simply {a, b}.
6. Can one set be a member of another set?
Yes, though this is a more advanced concept. A set can contain other sets as elements. This is useful in higher-level math for organizing more complex data.
Master Set Notation with Top-Rated Math Tutors in Denver, CO
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Our specially trained math tutors offer personalized instruction and live, face-to-face online support to help students truly understand and enjoy math.
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1563 | https://math.stackexchange.com/questions/2415404/how-do-we-calculate-the-chances-of-getting-n-fixed-points-in-a-permutation | probability - How do we calculate the chances of getting n fixed points in a permutation? - Mathematics Stack Exchange
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How do we calculate the chances of getting n fixed points in a permutation?
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I understand that attaining a derangement for n = 4 or more objects has a probability of about 1/e or about 37 %. But what about permutations with precisely n fixed points? For example, if I have exactly ten cards (numbered 1 to 10), shuffle them and then lay them out in a row, what are the chances of getting one fixed point, two fixed points, ..., 10 fixed points?
My attempt for (say) 4 fixed points is: (10!)(10C6)(1/e).
But I cannot convince myself this is correct.
probability
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asked Sep 3, 2017 at 15:02
Red Book 1Red Book 1
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3 To get four fixed points, choose four points to fix, and derange the rest.Angina Seng –Angina Seng 2017-09-03 15:04:54 +00:00 Commented Sep 3, 2017 at 15:04
It might help to explain how you got that answer.Shaun –Shaun♦ 2017-09-03 15:05:21 +00:00 Commented Sep 3, 2017 at 15:05
1 @LordSharktheUnknown That's too naive, you might get more than 4 4 fixed points that way.Daniel Robert-Nicoud –Daniel Robert-Nicoud 2017-09-03 15:05:56 +00:00 Commented Sep 3, 2017 at 15:05
@DanielRobert-Nicoud I assure you, that I would get exactly four!Angina Seng –Angina Seng 2017-09-03 15:06:58 +00:00 Commented Sep 3, 2017 at 15:06
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Let D n D n denote the number of derangements (fixed-point-free permutations) of an n n-element set. Then D n/n!→e−1 D n/n!→e−1; indeed
D n=n!∑k=0 n(−1)k k!.D n=n!∑k=0 n(−1)k k!.
The number of permutations of an n n-element set with k k fixed points is
(n k)D n−k.(n k)D n−k.
For a fixed k k, as n→∞n→∞ the probability of a random permutation having k k fixed points converges to e−1/k!e−1/k!. So for large n n the distribution of the number of fixed points is approximately Poisson of mean 1 1.
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answered Sep 3, 2017 at 15:16
Angina SengAngina Seng
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Oh, that's what you mean by "derange." Then I agree with your answer.Daniel Robert-Nicoud –Daniel Robert-Nicoud 2017-09-03 15:19:47 +00:00 Commented Sep 3, 2017 at 15:19
If Dn/n! = 1/e (about), then Dn = n!/e. But if n-elt set with k fixed is: (nCk)Dn-k then n=10, k=4 gives: (10C4)(10!/e) = 280,341,792 possible perms with 4 out of 10 points fixed. But how does this relate to the probability of getting any four points fixed from a random perm of 10 objects? With your formula for 'prob of random perm having k fixed points converges to e^-1/k!'? In this case, no cards matching is 1/e, exactly 1 card matching is also 1/e and any 4 cards matching is e^-1/4! = 0.0153.Red Book 1 –Red Book 1 2017-09-04 08:59:24 +00:00 Commented Sep 4, 2017 at 8:59
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After some serious searching, I was able to find some proper formulas. First, the number of derangements !n of n terms is given by
!n=⌊n!+1 e⌋!n=⌊n!+1 e⌋
Next, the probability of r r fixed points in a permutation of n n terms is given by
P r(n,r)=!(n−r)r!(n−r)!P r(n,r)=!(n−r)r!(n−r)!
Finally, you can multiply that probability by the total number of permutations n!n! to get the number of permutations with r fixed points:
FP(n,r)=!(n−r)r!(n−r)!∗n!FP(n,r)=!(n−r)r!(n−r)!∗n!
References:
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edited Jul 25, 2022 at 15:19
Harish Chandra Rajpoot
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answered Jul 25, 2022 at 14:59
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Denote by F(n,k)F(n,k) the number of permutations of n n elements with exactly k k fixed points. It satisfies the following relations.
F(n,n)=1 F(n,n)=1 given by the identity permutation.
F(n,k)=(n k)F(n−k,0)F(n,k)=(n k)F(n−k,0) given by choosing k k points to fix and then taking a permutation of the other elements with no fixed points.
∑n k=0 F(n,k)=n!∑k=0 n F(n,k)=n!
This is enough to compute F(n,k)F(n,k) recursively. Here's an ugly python code finding F(n,k)F(n,k).
```
import math
import scipy.special
n_fixed_points = {(0,0):1}
def F(n,k):
if n_fixed_points.has_key((n,k)):
return n_fixed_points[(n,k)]
if k==n:
n_fixed_points[(n,k)] = 1
return 1
if k>0:
f = scipy.special.binom(n,k)F(n-k,0)
n_fixed_points[(n,k)] = f
return f
f = math.factorial(n)-sum([scipy.special.binom(n,i)F(n-i,0) for i in xrange(1,n+1)])
n_fixed_points[(n,k)] = f
return f
print F(10,4)
```
It gives F(10,4)=55650 F(10,4)=55650.
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edited Sep 3, 2017 at 15:26
answered Sep 3, 2017 at 15:11
Daniel Robert-NicoudDaniel Robert-Nicoud
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1564 | https://pi.math.cornell.edu/~mjbelk/sums%20solutions.pdf | Solutions to Weekend Activity: Sigma Notation and Series 3) Find s1, s2, s3, s4, and s5; Do not simplify your answers.
s1 = 1 s2 = 1 + 1 4 s3 = 1 + 1 4 + 1 9 s4 = 1 + 1 4 + 1 9 + 1 16 s5 = 1 + 1 4 + 1 9 + 1 16 + 1 25 4) Note that sn an increasing sequence (each term is bigger than the previous term). Why should we expect this to occur? Will this be true of the partial sums sn = n X k=1 ak for ANY sequence an?
The terms of the sequence of partial sums are increasing because 1 n2 > 0 for all n ∈N.
Therefore, we are always adding another positive fraction to the preexisting sum. This will not be true for sn associated to an where an is ever less than or equal to zero.
So although the terms of an = 1 n2 get smaller and smaller, the terms of sn keep getting get bigger and bigger. It’s clear that lim n→∞ 1 n2 = 0, but what is lim n→∞sn?
Unfortunately, it could be just about anything!
5) Explain why the limit of sn is so unpredictable. Keep in mind your knowledge 0 · ∞.
As n increases, the terms of an get smaller and smaller, tending toward zero. However, the value of sn keeps increasing, since we are adding together more and more terms. The size of the terms going to zero and the number of terms going to infinity means we have no idea what will happen: 0 ∗∞is not well defined, and limits of type 0 ∗∞are unpredictable.
The answer will depend on the relative strengths of the 0 and ∞involved. Generally, such limits must be solved with L’Hospitals. In this case, the question is whether the terms of an go toward zero fast enough for their sum to be finite.
1 6) Using your calculator, approximate ∞ X n=1 1 n2 by finding s10 as a decimal.
s10 ≈1.54977 7) The actual value of this sum is, surprisingly, π2/6. Estimate π2/6 using your calcula-tor, and compare the results to your approximation. Which is bigger? How would you know which is bigger without even checking the numbers?
Using the approximation π ≈3.1415, you get π2 6 ≈1.64484 The value of π2 6 should be bigger, since it is obtained from s10 by adding on still more positive fractions.
We say a series converges if and only if its sequence of partial sums con-verges. Another way to say this would be ∞ X n=1 an = lim k→∞ k X n=1 an = lim k→∞sk 8) Use the definition of convergence to show that the series ∞ X k=1 0 converges to zero. Begin by finding a formula for the sequence of partial sums sn.
sn = Pn k=1 0 = 0. Therefore, since lim n→∞sn = 0, we conclude that ∞ X k=1 0 converges to zero by definition.
9) Use the definition of convergence to show that the series ∞ X k=1 c diverges for any con-stant c ̸= 0. Begin by finding a formula for the sequence of partial sums sn.
sn = Pn k=1 c = c ∗n. Therefore, lim n→∞sn diverges. We conclude that P∞ n=1 c diverges.
10) If we know that the series ∞ X n=1 an converges, what (if anything) can we say about a) lim n→∞sn? The limit must converge b) lim n→∞an? The limit must be zero 2 If ∞ X k=1 ak = A and ∞ X k=1 bk = B (both converge to constants), what will ∞ X k=1 (2an + bn) do?
Let An = n X k=1 ak, Bn = n X k=1 bk, and sn = n X k=1 (2ak + bk) be the sequences of partial sums.
The information given in the problem tells us that lim n→∞An = A and lim n→∞Bn = B.
Note that, by the properties of sums, sn = 2An + Bn for all n.
So, by the properties of limits: lim n→∞sn = lim n→∞2An + Bn = 2 lim n→∞An + lim n→∞Bn = 2A + B 11) I have told you that ∞ X n=1 1 n2 = π2 6 . Note that 1 n3 ≤ 1 n2 for all n ≥1.
Use these facts to argue that ∞ X n=1 1 n3 converges.
Since 1 k3 ≤ 1 k2, it follows that n X k=1 1 k3 ≤ n X k=1 1 k2 Since the sequence of partial sums of 1 k3 is increasing and bounded above by π2/6, it must converge.
12) Now, use the fact that R ∞ 1 x−3dx converges to argue that ∞ X n=1 1 n3 converges.
Hint: Draw a picture of R ∞ 1 x−3dx being underestimated by rectangles of width 1.
Each rectangle you have drawn has area 1/n3 where n is the right endpoint of each inter-val [n −1, n], n ≥2. The sequence n X k=1 1/k3 = 1 + n X k=2 1/k3 is increasing and bounded above by the integral 1 + R n 1 1 x3dx. Since this improper integral converges by the p-test, ∞ X n=1 1 n3 converges as well.
3 13) Let an = (−1)n+1 1 n3. Consider ∞ X n=1 an. Note that s5 = 1 −1 8 + 1 27 −1 64 + 1 125.
∞ X n=1 an is called an alternating series since the terms of the sum alternate between being pos-itive and negative. Does ∞ X n=1 (−1)n+1 1 n3 converge or diverge?
0 ≤ n X k=1 (−1)n+1 1 n3 ≤ n X k=1 1 n3, so the sum is bounded and cannot diverge to infinity. The oscillations of sequence of partial sums grow smaller and smaller as n increases. Therefore, the sequence of partial sums should tend toward a finite limit. We conclude that the series converges.
14a) Consider the sequence an = 1, 1 2, 1 2, 1 3, 1 3, 1 3, 1 4, 1 4, 1 4, 1 4, 1 5, 1 5, 1 5, 1 5, 1 5, . . .
Show that lim n→∞an = 0, but ∞ X n=1 an diverges.
Hint: s10 = 1+(1 2+ 1 2)+( 1 3+ 1 3+ 1 3)+( 1 4+ 1 4+ 1 4+ 1 4) lim n→∞an = 0, since the fractions involved are getting smaller and smaller.
However, sn diverges: If you group the terms as above in parentheses, you will notice that the sum inside each set of parentheses is 1. No matter how big sn gets, there will always be another 1 to add to it if we go far enough. we conclude that the sum has no upper bound, and must diverge to infinity.
14b) Based on your knowledge of improper integrals and your experience in (12), can you think of another (simpler) series that might diverge even though the terms go to zero?
∞ X n=1 1 n diverges. To prove this fact, OVERestiate 1 x with rectangles. This will show that ∞ X n=1 1 n > Z ∞ 1 1 xdx, which diverges to infinity. It follows that ∞ X n=1 1 n diverges as well.
4 |
1565 | https://www.examples.com/physics/brewsters-law.html | Brewster's Law - Examples, Definition, Formula, Derivation, FAQ'S
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Brewster’s Law – Examples, Definition, Formula, Derivation, FAQ’S
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Brewster’s Law – Examples, Definition, Formula, Derivation, FAQ’S
Brewster’s Law in physicsdescribes how the angleof incidence and the angle of polarization of light relate to each other. It establishes that the refractive index of a material determines the angle of incidence at which light becomes perfectly polarized, with the reflected and refracted rays being perpendicular to each other.
What is Brewster’s Law?
Brewster’s Law in optics describes how light with a specific polarization transmits perfectly through a transparent dielectric surface at a particular angle without reflecting. According to this law, if unpolarized light strikes a surface at a specific angle, called the Brewster angle, the reflected light becomes fully polarized, perpendicular to the plane of incidence.
Brewster’s Law Formula
The formula for Brewster’s Law relates the Brewster angle (𝜃𝐵). At which light is perfectly polarized upon reflection, to the refractive indices of the two media involved. It is given by:
tan𝜃𝐵=𝑛₂𝑛₁
where:
𝜃𝐵 is the Brewster angle, at which the reflected light is fully polarized.
𝑛₁ is the refractive index of the initial medium (the medium from which light is coming).
𝑛₂ is the refractive index of the second medium (the medium into which light is entering).
At this angle, known as the polarization angle, the reflected light polarizes perpendicular to the plane of incidence.
Brewster’s Law Derivation
Concept: When light hits a material (like glass) at a certain angle, known as the Brewster angle. The reflected light becomes perfectly polarized (vibrates in one direction).
Key Idea: At the Brewster angle, the reflected and refracted light rays are perpendicular to each other.
Snell’s Law: This law relates the angle of incidence and the angle of refraction for light passing from one medium to another:𝑛₁sin𝜃𝑖=𝑛₂sin𝜃𝑟
𝑛₁ and 𝑛₂ are the refractive indices of the two media (air and glass, for example).
𝜃𝑖 is the angle of incidence, and 𝜃𝑟 is the angle of refraction.
Brewster’s Condition: At the Brewster angle, the refracted light’s angle (𝜃𝑟) and the incident light’s angle (𝜃𝑖=𝜃𝐵) are related such that:𝜃𝑟=90⁰−𝜃𝐵
Simplify the Equation: Substituting 𝜃𝑟=90⁰−𝜃𝐵 into Snell’s law:𝑛₁sin𝜃𝐵=𝑛₂cos𝜃𝐵
Final Formula: Divide both sides by cos𝜃𝐵: tan𝜃𝐵=𝑛₂𝑛₁
This formula tells us that the Brewster angle (𝜃𝐵) depends on the refractive indices of the two media involved.
Uses of Brewster’s Law
1. Photography and Filmmaking: Photographers and filmmakers, by using Brewster’s Law, effectively reduce reflections on non-metallic surfaces. With polarizing filters positioned at Brewster’s angle, they consequently enhance image quality, reducing glare from surfaces like water or glass.
2. Optical Devices: Engineers design devices such as lasers and microscopes by using Brewster’s Law. They incorporate Brewster windows to minimize reflection losses and, in turn, improve light transmission.
3. Sunglasses and Eyewear: Manufacturers leverage Brewster’s Law to reduce glare from horizontal surfaces like roads and water, thus improving visual clarity and reducing eye strain.
4. Polarization Studies: Researchers rely on Brewster’s Law to analyze the polarization properties of materials, which is essential in optics and material sciences.
5. Optical Communication: Brewster’s Law optimizes light transmission in fiber optic cables and consequently enhances data transfer efficiency in optical communication systems.
6. Polarimetry: Instruments such as polarimeters apply Brewster’s Law to measure the polarization of light, thereby aiding in various scientific analyses.
Examples for Brewster’s Law
Polarized Sunglasses: Sunglasses reduce glare by using Brewster’s Law to cut reflections from surfaces like water or roads.
LCD Displays: LCDs employ polarizing filters for brightness and contrast. Brewster’s Law ensures better light transmission for improved image quality.
Anti-Reflective Coatings: Optical devices use coatings at Brewster’s angle to reduce glare, boosting light transmission for better efficiency.
Photography and Cinematography: Polarizing filters help photographers and filmmakers manage glare and reflections.
Fiber Optic Systems: In fiber optics, Brewster’s Law guides design to minimize signal loss from reflections, improving network performance.
FAQ’S
What are Brewsters rules?
Brewster’s Law dictates that reflected light becomes polarized when unpolarized light strikes a transparent medium at Brewster’s angle, where the reflected and refracted rays are perpendicular.
Is Brewster law and Malus law same?
Brewster’s Law deals with polarization by reflection, whereas Malus’s Law quantifies the intensity of polarized light passing through a polarizer, making them fundamentally different.
What is Brewster’s photo for?
Brewster’s photo helps analyze polarization effects, applying Brewster’s Law to observe polarized light reflections for scientific studies, optical designs, and photographic applications.
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Brewster’s Law – Examples, Definition, Formula, Derivation, FAQ’S
Brewster’s Law in physicsdescribes how the angleof incidence and the angle of polarization of light relate to each other. It establishes that the refractive index of a material determines the angle of incidence at which light becomes perfectly polarized, with the reflected and refracted rays being perpendicular to each other.
What is Brewster’s Law?
Brewster’s Law in optics describes how light with a specific polarization transmits perfectly through a transparent dielectric surface at a particular angle without reflecting. According to this law, if unpolarized light strikes a surface at a specific angle, called the Brewster angle, the reflected light becomes fully polarized, perpendicular to the plane of incidence.
Brewster’s Law Formula
The formula for Brewster’s Law relates the Brewster angle (𝜃𝐵). At which light is perfectly polarized upon reflection, to the refractive indices of the two media involved. It is given by:
tan𝜃𝐵=𝑛₂𝑛₁
where:
𝜃𝐵 is the Brewster angle, at which the reflected light is fully polarized.
𝑛₁ is the refractive index of the initial medium (the medium from which light is coming).
𝑛₂ is the refractive index of the second medium (the medium into which light is entering).
At this angle, known as the polarization angle, the reflected light polarizes perpendicular to the plane of incidence.
Brewster’s Law Derivation
Concept: When light hits a material (like glass) at a certain angle, known as the Brewster angle. The reflected light becomes perfectly polarized (vibrates in one direction).
Key Idea: At the Brewster angle, the reflected and refracted light rays are perpendicular to each other.
Snell’s Law: This law relates the angle of incidence and the angle of refraction for light passing from one medium to another:𝑛₁sin𝜃𝑖=𝑛₂sin𝜃𝑟
𝑛₁ and 𝑛₂ are the refractive indices of the two media (air and glass, for example).
𝜃𝑖 is the angle of incidence, and 𝜃𝑟 is the angle of refraction.
Brewster’s Condition: At the Brewster angle, the refracted light’s angle (𝜃𝑟) and the incident light’s angle (𝜃𝑖=𝜃𝐵) are related such that:𝜃𝑟=90⁰−𝜃𝐵
Simplify the Equation: Substituting 𝜃𝑟=90⁰−𝜃𝐵 into Snell’s law:𝑛₁sin𝜃𝐵=𝑛₂cos𝜃𝐵
Final Formula: Divide both sides by cos𝜃𝐵: tan𝜃𝐵=𝑛₂𝑛₁
This formula tells us that the Brewster angle (𝜃𝐵) depends on the refractive indices of the two media involved.
Uses of Brewster’s Law
Photography and Filmmaking: Photographers and filmmakers, by using Brewster’s Law, effectively reduce reflections on non-metallic surfaces. With polarizing filters positioned at Brewster’s angle, they consequently enhance image quality, reducing glare from surfaces like water or glass.
Optical Devices: Engineers design devices such as lasers and microscopes by using Brewster’s Law. They incorporate Brewster windows to minimize reflection losses and, in turn, improve light transmission.
Sunglasses and Eyewear: Manufacturers leverage Brewster’s Law to reduce glare from horizontal surfaces like roads and water, thus improving visual clarity and reducing eye strain.
Polarization Studies: Researchers rely on Brewster’s Law to analyze the polarization properties of materials, which is essential in optics and material sciences.
Optical Communication: Brewster’s Law optimizes light transmission in fiber optic cables and consequently enhances data transfer efficiency in optical communication systems.
Polarimetry: Instruments such as polarimeters apply Brewster’s Law to measure the polarization of light, thereby aiding in various scientific analyses.
Examples for Brewster’s Law
Polarized Sunglasses: Sunglasses reduce glare by using Brewster’s Law to cut reflections from surfaces like water or roads.
LCD Displays: LCDs employ polarizing filters for brightness and contrast. Brewster’s Law ensures better light transmission for improved image quality.
Anti-Reflective Coatings: Optical devices use coatings at Brewster’s angle to reduce glare, boosting light transmission for better efficiency.
Photography and Cinematography: Polarizing filters help photographers and filmmakers manage glare and reflections.
Fiber Optic Systems: In fiber optics, Brewster’s Law guides design to minimize signal loss from reflections, improving network performance.
FAQ’S
What are Brewsters rules?
Brewster’s Law dictates that reflected light becomes polarized when unpolarized light strikes a transparent medium at Brewster’s angle, where the reflected and refracted rays are perpendicular.
Is Brewster law and Malus law same?
Brewster’s Law deals with polarization by reflection, whereas Malus’s Law quantifies the intensity of polarized light passing through a polarizer, making them fundamentally different.
What is Brewster’s photo for?
Brewster’s photo helps analyze polarization effects, applying Brewster’s Law to observe polarized light reflections for scientific studies, optical designs, and photographic applications.
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Practice Test
What does Brewster's Law describe?
Choose the correct answer
The reflection of light from a mirror
The refraction of light through a lens
The polarization of light reflected from a surface
The diffraction of light through a slit
of 10
According to Brewster's Law, what is the relationship between the angle of incidence and the angle of refraction?
Choose the correct answer
They are equal
Their sum is 90 degrees
Their product is 1
They are complementary
of 10
What is Brewster's angle for a material with a refractive index of 1.5?
Choose the correct answer
56.3 degrees
41.8 degrees
53.1 degrees
48.6 degrees
of 10
At Brewster's angle, what is the polarization state of the reflected light?
Choose the correct answer
Unpolarized
Partially polarized
Fully polarized parallel to the plane of incidence
Fully polarized perpendicular to the plane of incidence
of 10
How is Brewster's angle affected by increasing the refractive index of the reflecting material?
Choose the correct answer
Increases
Decreases
Remains constant
Becomes zero
of 10
What happens to the Brewster angle if the light transitions from air (n=1) to water (n=1.33)?
Choose the correct answer
Decreases
Increases
Remains the same
Becomes 90 degrees
of 10
For which type of light does Brewster's Law apply?
Choose the correct answer
Only visible light
Only infrared light
Only ultraviolet light
All types of light
of 10
What is the Brewster angle for light traveling from air into a medium with a refractive index of 2?
Choose the correct answer
26.6 degrees
45 degrees
63.4 degrees
90 degrees
of 10
If light is incident at Brewster's angle, what happens to the transmitted light?
Choose the correct answer
Becomes fully polarized
Becomes partially polarized
Remains unpolarized
Becomes completely absorbed
of 10
What is the significance of Brewster's angle in photography?
Choose the correct answer
Reduces lens flare
Increases image brightness
Reduces glare from reflective surfaces
Increases depth of field
of 10
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1566 | https://www.quora.com/Why-is-sin-2x-equal-to-sin-2x | Why is sin(-2x) equal to -sin(2x)? - Quora
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Why is sin(-2x) equal to -sin(2x)?
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Matthew Cheesy
Have a keen interest in mathematics · Author has 120 answers and 91.7K answer views
·2y
Originally Answered: Why does sin(-x) =-sin(x)? What are some examples? ·
There are essentially two approaches to establishing this identity (and hence show that sin(x)sin(x) is an odd function).
Alternative 1
Consider the unit circle. For any given point on the circle, we have the following:
Here, the angle θ θ is the rotation from the positive x x axis to the hypotenuse O P O P.
From basic trigonometric definitions, given that the radius O P=1 O P=1, it should be easy to convince yourself that the y y value of point P P is given by sin(θ)sin(θ).
Then, if we rotate by θ θ anticlockwise, or rotate by −θ−θ, we would obtain the following:
where, by similar logic, the y y value of the coord
Continue Reading
There are essentially two approaches to establishing this identity (and hence show that sin(x)sin(x) is an odd function).
Alternative 1
Consider the unit circle. For any given point on the circle, we have the following:
Here, the angle θ θ is the rotation from the positive x x axis to the hypotenuse O P O P.
From basic trigonometric definitions, given that the radius O P=1 O P=1, it should be easy to convince yourself that the y y value of point P P is given by sin(θ)sin(θ).
Then, if we rotate by θ θ anticlockwise, or rotate by −θ−θ, we would obtain the following:
where, by similar logic, the y y value of the coordinate P′P′ will be given by sin(−θ)sin(−θ).
But since P′P′ is simply a reflection of point P P in the x x-axis, it therefore means that the y y value of P′P′ will be the negation of that of P P, hence indicating that
−sin(θ)=sin(−θ)−sin(θ)=sin(−θ), as required.
Alternative 2
We consider the power series expansion of sin(x)sin(x). We have
sin(x)=∑∞n=0(−1)n x 2 n+1(2 n+1)!sin(x)=∑n=0∞(−1)n x 2 n+1(2 n+1)!
=x−x 3 3!+x 5 5!−x 7 7!+⋯=x−x 3 3!+x 5 5!−x 7 7!+⋯.
But then, we see that for any non-negative integer n n, we have
(−x)2 n+1(2 n+1)!=x 2 n(−x)(2 n+1)!=−x 2 n+1(2 n+1)!(−x)2 n+1(2 n+1)!=x 2 n(−x)(2 n+1)!=−x 2 n+1(2 n+1)!. As such, each terms in the power series expansion of −sin(x)−sin(x) and sin(−x)sin(−x) equate, establishing the equality between the two.
For example, if you want to evaluate sin(300∘)sin(300∘), we may make the following observations:
sin(300∘)=sin(−60∘)=−sin(60∘)=−√3 2 sin(300∘)=sin(−60∘)=−sin(60∘)=−3 2.
Hope this helps!
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What are the formulas of sin(2x)?
Why is sin-2x ≠ -2 sin x but sin -x = -sin x?
Is there any difference between sin 2 2 x sin 22 x and (sin 2 x)2(sin2 x)2?
Is the following equality true: sin 2x = sin x × 2?
Why is sin(2x) /4 = sin(4x) /8?
Scott
Former believer · Author has 11.7K answers and 12M answer views
·5y
Originally Answered: Why is sin-2x ≠ -2 sin x but sin -x = -sin x? ·
sin(-x) = -sin x. What does that mean?
Take a unit circle:
the sine of the marked angle is the y-coordinate of the marked point. Here, it’s how high that dashed line is.
What if you were talking about sin(-x)? Then the angle would have gone down instead of up. The marked point would have been down instead of up. The sin would be negative instead of positive, but it would be the same distance from the axis.
What that means: sin(-x) = -sin(x) because of the way the circle is symmetrical.
Now think about what sin(xy) means. Could you do anything similar to show that sin(xy) would equal xsin(y)? Of co
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sin(-x) = -sin x. What does that mean?
Take a unit circle:
the sine of the marked angle is the y-coordinate of the marked point. Here, it’s how high that dashed line is.
What if you were talking about sin(-x)? Then the angle would have gone down instead of up. The marked point would have been down instead of up. The sin would be negative instead of positive, but it would be the same distance from the axis.
What that means: sin(-x) = -sin(x) because of the way the circle is symmetrical.
Now think about what sin(xy) means. Could you do anything similar to show that sin(xy) would equal xsin(y)? Of course not: if you could then sin(x) = sin(x1) = x sin(1). And clearly that’s not right — as x got bigger, your value would get bigger. It would be bigger than 1 pretty quickly, which sine never does. (You would have the same problem with -2sin(x), but a touch less obviously: it would range from 2 to -2, while sin(-2x) would range from 1 to -1.)
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David Rutter
B.S. Discrete Mathematics, Georgia Tech · Upvoted by
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. · Author has 9.2K answers and 11.8M answer views
·4y
Originally Answered: Why is sin(-x) = - sinx? ·
Sine is odd because its value at zero is zero.
Cosine is even because its value at zero is one.
They are really the same function phase shifted, but that function happens to have p2mg frieze symmetry, which means that after a phase shift it becomes a reflection or 180 degree rotation of itself. That means there is a set of shifts where it has reflective symmetry across the y axis, a characteristic of even functions. One of these is cosine. It means there is a set of shifts where it has rotational symmetry around the origin, a characteristic of odd functions. One of these is sine.
Another possible
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Sine is odd because its value at zero is zero.
Cosine is even because its value at zero is one.
They are really the same function phase shifted, but that function happens to have p2mg frieze symmetry, which means that after a phase shift it becomes a reflection or 180 degree rotation of itself. That means there is a set of shifts where it has reflective symmetry across the y axis, a characteristic of even functions. One of these is cosine. It means there is a set of shifts where it has rotational symmetry around the origin, a characteristic of odd functions. One of these is sine.
Another possible answer to the question: imagine you are walking around an origin-centered circle starting on the x axis. Your position relative to the x axis is the sine of your angle from the x axis. Now, suppose someone sees you going counterclockwise around the circle from this point. They close their eyes for a moment and you return to the x axis. Now they open your eyes and see you walking clockwise.
They cannot tell from this distance whether you have started walking the opposite direction around this time (meaning your position relative to the x axis is s i n(−x)s i n(−x)) or a lens has been placed in front of them that reflects their view of you vertically (meaning your position relative to the x axis is −s i n(x)−s i n(x)). As such, these two values must be equal.
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·
1y
The identity sin(−θ)=−sin(θ)sin(−θ)=−sin(θ) is a fundamental property of the sine function, which is an odd function. This means that the sine of a negative angle is the negative of the sine of the corresponding positive angle.
To see why this is true for sin(−2 x)sin(−2 x):
Let θ=2 x θ=2 x.
According to the property of sine, we have:
sin(−θ)=−sin(θ)sin(−θ)=−sin(θ)
3. Substituting θ θ with 2 x 2 x, we get:
sin(−2 x)=−sin(2 x)sin(−2 x)=−sin(2 x)
Thus, the equality sin(−2 x)=−sin(2 x)sin(−2 x)=−sin(2 x) holds true because of the odd nature of the sine function.
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More answers below
How does one prove sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)?
What is the general solution of the equation 4 (sin x) (sin 2x) (sin 4x) =sin 3x?
What is the value of sin x + (sin x + sin 2x) + (sin x + sin 2x + sin 3x) +… (infinite series)?
How can I show that tan(x+y)=sin(2 x)+sin(2 y)cos(2 x)+cos(2 y)tan(x+y)=sin(2 x)+sin(2 y)cos(2 x)+cos(2 y)?
What is the value of sin(2x)?
Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·4y
Originally Answered: Why is sin(-x) = - sinx? ·
Several explanations can be provided. I will show one of them.
sin(-x)
= sin(0-x)
= sin 0 cos x - cos 0 sin x
= 0 (cos x) -1 (sin x)
= - sin x
Hope it is clear.
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99 32
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Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
·4y
Originally Answered: Why is sin(-x) = - sinx? ·
sin x=e i x−e−i x 2 i sinx=e i x−e−i x 2 i
⟹sin(−x)=e−i x−e i x 2 i=−(e i x−e−i x 2 i)=−sin x⟹sin(−x)=e−i x−e i x 2 i=−(e i x−e−i x 2 i)=−sinx
Upvote ·
9 3
Michael Harrison
M.S. in Mathematics, Louisiana State University (Graduated 2007) · Author has 4.8K answers and 914.4K answer views
·2y
Originally Answered: Why is sin(-x) = - sinx? ·
There are two functions named sine: that from the unit circle to [-1, 1], and that from the real number line to [-1, 1]. (In trigonometry, the first is usually taught as the sine ratio.) The map from the unit circle is just projection onto the y axis, which satisfies sin(-theta) = - sin(theta) if we define the identity and negation of the unit circle (considered as a group) in the usual way: the point (1, 0) is your “starting point,” and counterclockwise rotation by a certain angular displacement is undone by clockwise rotation of the same angular displacement.
There is an important family of m
Continue Reading
There are two functions named sine: that from the unit circle to [-1, 1], and that from the real number line to [-1, 1]. (In trigonometry, the first is usually taught as the sine ratio.) The map from the unit circle is just projection onto the y axis, which satisfies sin(-theta) = - sin(theta) if we define the identity and negation of the unit circle (considered as a group) in the usual way: the point (1, 0) is your “starting point,” and counterclockwise rotation by a certain angular displacement is undone by clockwise rotation of the same angular displacement.
There is an important family of mappings of the real number line to the unit circle, called covering maps, which take zero to (1, 0) and wrap the number line around the circle like thread around a spool. The sine function is just the composition of such a covering map with the sine ratio, which is why the sine function inherits the symmetry property in question.
Upvote ·
Gregory Schoenmakers
Engineer and former high school maths teacher. · Author has 4.4K answers and 8.5M answer views
·5y
Originally Answered: Why is sin-2x ≠ -2 sin x but sin -x = -sin x? ·
Because sin-2x = -sin 2x
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Michael Simon
Former Retired Mathematics, Chemistry, Physics Professor at Colleges and Universities in Connecticut (1967–2017) · Author has 382 answers and 105.2K answer views
·4y
Originally Answered: Why is sin(-x) = - sinx? ·
If x is between 0 degrees and 180 degrees, the sin x is positive, so -sin x is negative. Also -x is between -180 degrees and -0 degrees, which is third and fourth quadrants where the sign of the sin is negative.
Upvote ·
9 4
Nancy Mitchell
used to be a teacher. · Author has 3.4K answers and 8M answer views
·6y
Why is sin(-2x) equal to -sin(2x)?
sin x=e i x−e−i x 2 i sinx=e i x−e−i x 2 i
⟹sin(−2 x)=e i(−2 x)−e−i(−2 x)2 i⟹sin(−2 x)=e i(−2 x)−e−i(−2 x)2 i
=e−i(2 x)−e i(2 x)2 i=e−i(2 x)−e i(2 x)2 i
=−(e i(2 x)−e−i(2 x)2 i)=−(e i(2 x)−e−i(2 x)2 i)
=−sin(2 x)=−sin(2 x)
where i=√−1 and e=2.71828…where i=−1 and e=2.71828…
Upvote ·
9 2
9 2
Sk Jain
5y
Originally Answered: Why is sin-2x ≠ -2 sin x but sin -x = -sin x? ·
The negative sign can be taken out but the input angle cannot change.
sin-2x = -sin2x
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9 1
Milton R Brown
Author has 17.8K answers and 4.2M answer views
·6y
It's true for X also. A circle is divided in half between up and down. An angle can be up or down, in math this is + or - angles. The value of sin is the vertical/ radius so it follows the up/down sin value. I'd sin -x = -sin x
Upvote ·
9 1
Srishty
2y
Originally Answered: Why does sin(-x) =-sin(x)? ·
It is because the angle ’-x’ falls in 4th quadrant and sin is negative in 4th quadrant
Upvote ·
Related questions
What are the formulas of sin(2x)?
Why is sin-2x ≠ -2 sin x but sin -x = -sin x?
Is there any difference between sin 2 2 x sin 22 x and (sin 2 x)2(sin2 x)2?
Is the following equality true: sin 2x = sin x × 2?
Why is sin(2x) /4 = sin(4x) /8?
How does one prove sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)?
What is the general solution of the equation 4 (sin x) (sin 2x) (sin 4x) =sin 3x?
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What is the value of sin(2x)?
Mathematics: Why is cos (2x) so different from sin (2x)?
What is the solution to sin(3x) =cos(2x)?
Is sin 2x the same as Sinx 2?
If sin(x) =1/3, what is the value of sin(2x) =?
How do I check the identity c o s 2 x c o s 2 y+s i n 2 x s i n 2 y+s i n 2 x c o s 2 y+s i n 2 y c o s 2 x=1 c o s 2 x c o s 2 y+s i n 2 x s i n 2 y+s i n 2 x c o s 2 y+s i n 2 y c o s 2 x=1?
Related questions
What are the formulas of sin(2x)?
Why is sin-2x ≠ -2 sin x but sin -x = -sin x?
Is there any difference between sin 2 2 x sin 22 x and (sin 2 x)2(sin2 x)2?
Is the following equality true: sin 2x = sin x × 2?
Why is sin(2x) /4 = sin(4x) /8?
How does one prove sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)sin(2 x)sin(x)−cos(2 x)cos(x)=sec(x)?
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1567 | https://www.sciencedirect.com/science/article/abs/pii/S2211034820306064 | Two cases of NMOSD with MRI findings mimicking CADASIL - ScienceDirect
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Abstract
Introduction
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References (8)
Cited by (4)
Multiple Sclerosis and Related Disorders
Volume 46, November 2020, 102532
Case report
Two cases of NMOSD with MRI findings mimicking CADASIL
Author links open overlay panel Kentaro Maeda MD a b, Nobuhiro Dougu MD, PhD c, Ken Ohyama MD, PhD b, Toshiyuki Takahashi MD, PhD d e, Ikuko Mizuta MD, PhD f, Toshiki Mizuno MD, PhD f, Yasushi Kobayashi MD, PhD b
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Highlights
•MRI findings in NMOSD patients may resemble those of CADASIL
•Symmetrical hyperintensities may exist in temporal poles and cerebral hemispheres
•Symmetrical hyperintensities may also exist in external capsules
•NMOSD is a differential diagnosis for CADASIL
•Testing for anti-AQP4 antibodies should be considered
Abstract
Purpose
The purpose of this study is to increase awareness of the importance of considering neuromyelitis optica spectrum disorder (NMOSD) as a differential diagnosis for cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL).
Methods
We report two NMOSD patients demonstrating magnetic resonance imaging (MRI) abnormalities resembling those of CADASIL.
Results
Brain MRIs of both patients showed symmetrical hyperintense signals in the temporal poles and cerebral hemispheres on T2 weighted images. One case also involved the bilateral external capsule. The chief complaint of both patients was loss of visual acuity, and neurologic examination showed no other apparent neurological signs or symptoms. Anti-aquaporin-4 antibodies were detected on serological examination, and NMOSD was subsequently diagnosed. Visual acuity improved following intravenous methylprednisolone therapy. One patient refused further immunological treatment. Although she remained clinically stable, gradual radiographic deterioration was observed. This deterioration then stabilized after the patient commenced oral prednisolone therapy. The other patient was treated with prednisolone and azathioprine. She is clinically stable, but we have observed gradual radiographic deterioration over the past 5 years.
Conclusion
MRI findings in patients with NMOSD may resemble those of CADASIL, namely symmetrical hyperintensities in the temporal poles, external capsules and cerebral hemispheres. NMOSD is a differential diagnosis for CADASIL, and testing for anti-AQP4 antibodies should be considered.
Introduction
Neuromyelitis optica spectrum disorder (NMOSD) is an autoimmune neurological disease characterized by attacks of optic neuritis and longitudinally extensive transverse myelitis [Wingerchuk et al., 2015]. Magnetic resonance imaging (MRI) of the brains of NMOSD patients typically reveals lesions in the cerebral hemispheres, periependymal regions, area postrema, and corticospinal tracts [Wingerchuk et al., 2015], although variations to this pattern have been reported [Wingerchuk et al., 2015]. Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is caused by NOTCH3 gene mutations. It is characterized by migraines, mood disturbances, recurrent strokes, and dementia. Focal lacunar infarcts and diffuse hyperintensities on T2-weighted images (T2WI) and fluid attenuated inversion recovery (FLAIR) images are associated with recurrent strokes, and temporal pole hyperintensity is considered to be a radiologic marker of CADASIL. Involvement of the external capsule and corpus callosum are also characteristic findings [O'Sullivan et al., 2001]. We report two cases of NMOSD which involve MRI findings similar to those characteristically associated with CADASIL.
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Case 1
A 49-year-old woman presented with decreased visual acuity in the right eye and right ophthalmic pain. The pain was worsened by ocular movement to right. 15 years prior to this, she had presented with transient temporal headache and decreased visual acuity of the left eye, both of which improved without treatment. At the age of 46, she developed decreased visual acuity, which did not improve following treatment with intravenous methylprednisolone therapy. In the month prior to her most recent
Case 2
A 54-year-old woman presented complaining of a 5-day history of decreased visual acuity in the right eye. She had not experienced any headache or depression, and had no relevant past history or family history. Her consciousness and cognitive function were both normal. Visual acuity and critical flicker fusion frequency were both decreased in her right eye. The other cranial nerves, which control muscle strength of mastication, trunk and the extremities, sensations, gait, and posture, were all
Discussion
These two cases of NMOSD both exhibited symmetrical white matter MRI abnormalities in the bilateral temporal poles and external capsules. MRI hyperintensities of these areas have been shown to be radiologic markers of CADASIL [O'Sullivan et al., 2001]. The development of optic neuritis is atypical in CADASIL, but has been reported [Collongues et al., 2012]. One of our patients also experienced migraine, mood disturbance and mild cognitive impairment, all of which are frequently seen in CADASIL
Consent form
Written informed consents were obtained from both patients for the publication of this case report and any accompanying images.
Funding
This study was partially supported by the Japan Agency for Medical Research and Development (AMED, 17ek0109130s0703), by a grant-in-aid for Research on Intractable Disease from the Japanese Ministry of Health, Labour, and Welfare, Japan (H28-Nanchitou(Nan)-Ippan-029, H30-Nanchitou(Nan)-Ippan-006) (to T.M.) and by a grant-in-aid for scientific research from the Japan Society for the Promotion of Science (JSPS KAKENHI JP26293205) (to T.T.).
Declaration of Competing Interest
The authors have no conflicts of interest to disclose.
Recommended articles
Reference (8)
H. Chabriat et al.
Cadasil
Lancet Neurol (2009)
D.M. Wingerchuk et al.
International consensus diagnostic criteria for neuromyelitis optica spectrum disorders
Neurology (2015)
M. O'Sullivan et al.
MRI hyperintensities of the temporal lobe and external capsule in patients with CADASIL
Neurology (2001)
N. Collongues et al.
Inflammatory-like presentation of CADASIL: a diagnostic challenge
BMC Neurol (2012)
There are more references available in the full text version of this article.
Cited by (4)
Long-standing neuromyelitis optica with leukodystrophy-like asymptomatic MRI changes
2024, Radiology Case Reports Citation Excerpt :
Specific involvement of the anterior temporal pole and external capsule white matter hyperintensities on T2-weighted sequences and microbleeds, as seen our case, has been described in CADASIL, although the case patient did not manifest symptoms nor signs consistent with CADASIL. The literature reports several similar cases of NMOSD mimicking CADASIL or leukodystrophy (Table 1) [10–13]. These cases of asymptomatic, inter-attack MRI finding progression shared several features: the patients were all positive for the anti-AQP4-antibody in the serum and had a longer disease duration, of at least 5 years. Show abstract Patients with neuromyelitis optica (NMO) are unlikely to develop clinically silent lesions on brain magnetic resonance imaging (MRI), unlike patients with multiple sclerosis (MS). We encountered a patient with NMO who showed radiological progression and leukodystrophy-like changes on MRI during a long-standing, clinically asymptomatic period.
### Neuromyelitis optica spectrum disorder mimicking cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy with symmetrical lesions in the temporal poles and external capsules on MRI
2024, Neuroradiology
### Inflammatory Optic Neuropathy as a Presenting Feature of Cerebral Autosomal Dominant Arteriopathy With Subcortical Infarcts and Leukoencephalopathy
2023, Neurohospitalist
### Neuroimaging Pearls from the MDS Congress Video Challenge. Part 1: Genetic Disorders
2022, Movement Disorders Clinical Practice
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1568 | https://www.youtube.com/watch?v=o2SW-IU1xus | Calculate resultant pH after neutralisation | Acids and bases | meriSTEM
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This video is part of meriSTEM Australian senior science educational resources (CC BY-NC-SA 4.0). Email the team (contact.meristem@anu.edu.au) for further information.
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now the final question that we want to look at is the ph of resultant solutions when you mix two solutions together so we'll do exactly the same thing we did before but this time we just need to be aware of of whether or not there's a solution in excess or and or one that's a limiting agent so in this case i've got lithium hydroxide and i'm reacting it with nitric acid this is a nicer one because our ratios are a little easier we have lithium nitrate as our salt in solution and we form water and this equation is already balanced so our mole ratios are one to one to one to one so that's nice then what i'm going to do is i need to work out the number of moles of each of these so let's put the values that we have in and see where we go from here so for the lithium first it's a one molar solution of lithium hydroxide and we have 20 ml so let me make that straight away point zero two liters for the nitric acid i'll just do the lighter blue and it is a 0.5 molar solution 0.5 molar and we have 30 ml so it's .03 liters now as i did before i'm going to go up for each of these and i'm going to have a look at my total number of moles so in the first case i've got my name on my cv so i'm going to multiply my concentration by volume and i will get .02 times one so that's not too difficult 0.02 moles but for the nitric acid i have 0.5 times .03 so this one's going to be .015 moles so when this reaction occurs because my ratio is 1 to 1 i have a limiting agent i have one of my species which uh cannot fully react with the other so this is limiting and this is in excess what i want to do then is i want to look at the one that's in excess because that's going to be the critical one so as a result of this i have an excess of lithium hydroxide and the number of moles i have is 0.02 minus 0.015 which is going to be 0.005 moles because i have 0.05 moles my final volume is equal to 20 plus 30 which is 50 mils because i've added both of these solutions together so now i have a final volume of 50 mils so therefore my final concentration is going to be number of moles over volume which is .005 divided by and 50 mils is going to be 0.05 liters and when i do that i'm going to end up with a final concentration of 0.01 moles per liter of my lithium hydroxide now the problem that i have to solve now is that i need the final ph and in order to get the final ph i'm going to have to actually do a two-step calculation because i'm going to have to start with the poh so again just in red here let's put the poh is equal to minus the log base 10 concentration of oh minus ions and because this is the value that's going in there i'm going to have minus the log base 10. i'm running out a little space here but let's just do it anyway base 10 of 0.1 and that's going to be a p o h equal to one now the problem is that's not the ph we but we do know that um 14 minus the poh is equal to the ph and therefore the ph of this solution is going to be 13. now this is a very basic solution so even though we've tried to neutralize these solutions we find that because one of these was in excess it remained in the solution afterwards and still ended up therefore with a very high ph solution this has been a lot this big video lots of worked examples here for you and i'm sure you do plenty more in class good luck with them keep practicing and thanks for watching |
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Convert odds into probability and percent chance of winning and losing. This calculator converts odds for winning or odds against winning into percentage chance for winning or losing. It also calculates the fractional odds.
Probability of Winning Formula A:B
PWin = A / (A + B)
Where A:B is odds for winning
Probability of Losing Formula A:B
PLose = B / (A + B)
Where A:B is odds for winning
How to Read Odds
It's important to understand how to read odds especially if wagering is involved. If you are using sports teams odds or betting odds and see the odds are 9/2, this is most likely odds against winning.
Enter 9 to 2 into the calculator and select "Odds against winning." It's also important to understand
implied odds or betting odds which are not true statistical odds.
When playing a lottery or other games of chance be sure you understand the odds or probability that is reported by the game organizer. A 1 in 500 chance of winning, or probability of winning, may be reported as "1 in 500" or "1 out of 500." You may also see odds reported simply as chance of winning as 1:500.
These all most likely mean 1 chance of winning out of 500 total possible outcomes. Therefore, in terms of odds, this means "1 to 499 odds for winning" which is exactly the same as "499 to 1 odds for losing."
For example, a Pick 3 lottery has a total of 1000 possible outcomes, from 000 to 999. So a single bet, say 123, has a 1 in 1000 chance of winning. The chance of winning is 1 out of 1000, while the chance of losing is 999 out of 1000. Therefore, A:B = 1:999 and that is what you would enter into this calculator to find the probability.
PWin = A / (A + B) = 1 / (1 + 999) = 1/1000 = 0.001 = 0.1%
PLose = B / (A + B) = 999 / (1 + 999) = 999/1000 = 0.999 = 99.9%
Probability Formulas Explained
This calculator will convert "odds of winning" an event into a probability percentage chance of success.
Odds, are given as (chances for success) : (chances against success) or vice versa.
If odds are stated as an A to B chance of winning then the probability
of winning is given as PWin = A / (A + B) while the probability
of losing is given as PLose = B / (A + B).
For example, you win a game if you pull an ace out of a full deck of 52 cards. Pulling any other card you lose. The chance of winning is 4 out of 52, while the chance against winning is 48 out of 52 (52 - 4 = 48). Entering A=4 and B=48 into the calculator as 4:48 odds are for winning you get
For 4 to 48 odds for winning
Probability of:
Winning = (0.0769) or 7.6923%
Losing = (0.9231) or 92.3077%
"Odds for" winning: 1/12 or 1:12 (reduced from 4:48)
"Odds against" winning: 12/1 or 12:1 (reduced from 48:4)
Implied Odds vs. True Odds
Implied odds or betting odds are likely not the same as the true odds. Betting odds is a payout ratio that has the house profit margin built into it.
Take for example the game of roulette at a casino. The possible outcomes for a bet on a single number are the numbers 1 to 36 and 0 or 00 for a total of 38 possible outcomes. The payout odds for a bet on a single number is 35 to 1. This is odds of winning 1:35. This pays based on a probability of winning PWin = A / (A + B) = 1 / (1 + 35) = 1/36 = 0.02778 or 2.778% (97.22% Losing).
The true odds of winning are based on 38 outcomes. PWin = A / (A + B) = 1 / (1 + 37) = 1/38 = 0.02632 or 2.632% (97.37% Losing).
If you win, you get paid as if your chance of winning is 2.778% but, you really only have a chance of winning of 2.632%. This is a disadvantage for you. Putting it another way, you are being paid as if your chance of losing is only 97.22% but in reality, your chance of losing is 0.15% greater at 97.37%. You are not being compensated on the true odds because the house is taking a roughly 0.15% cut on every payout.
The probability over time, if you were paid true odds, is that you would break even. Since with implied odds or betting odds you win less often than the true odds, the probability over time is that you will always lose money.
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1570 | https://people.math.harvard.edu/~knill/graphgeometry/papers/cartan.pdf | CARTAN’S MAGIC FORMULA FOR SIMPLICIAL COMPLEXES OLIVER KNILL Abstract. ´ Elie Cartan’s magic formula LX = iXd + diX = (d + iX)2 = D2 X relates the exterior derivative d, an interior derivative iX and its Lie derivative LX. We use this formula to define a finite dimensional vector space X of vector fields X on a finite abstract simplicial complex G. The space X has a Lie algebra structure satisfying L[X,Y ] = LXLY −LY LX as in the continuum. Any such vector field X defines a coordinate change on the finite dimensional vector space l2(G) which play the role of translations along the vector field. If i2 X = 0, the relation LX = D2 X with DX = iX + d mirrors the Hodge factorization L = D2, where D = d + d∗we can see ft = −LXf defining the flow of X as the analogue of the heat equation ft = −Lf and view the Newton type equations ftt = −LXf as the analogue of the wave equation ftt = −Lf. Similarly as the wave equation is solved by ψ(t) = eiDtψ(0) with ψ(t) = f(t)−iD−1ft(t), also any second order differential equation ftt = −LXf is solved by ψ(t) = eiDXtψ(0) in l2(G, C). If X is supported on odd forms, the factorization property LX = D2 X extends to the Lie algebra and i[X,Y ] remains an inner derivative. If the kernel of LX : Λp →Λp has dimension bp(X), then the general Euler-Poincar´ e formula χ(G) = P k(−1)kbk(X) holds for every parameter field X. Extreme cases are iX = d∗, where bk are the usual Betti numbers and X = 0, where bk = fk(G) are the components of the f-vector of the simplicial complex G. We also note that the McKean-Singer super-symmetry extends from L to Lie derivatives. (It also holds for LX on Riemannian manifolds but appears to have been unnoticed there so far): the non-zero spectrum of LX on even forms is the same than the non-zero spectrum of LX on odd forms. We also can make a deformation D′ X = [BX, DX] of DX = d+iX +bX, BX = dX −d∗ X +ibX which produces a in general non-isospectral deformation of the exterior derivative d governed by the vector field X featuring inflationary initial size decay for d typical for such systems, leading to an expansion of space.
1. Introduction 1.1. When formulating physics in a discrete geometric frame work, one is challenged by the absence of a continuous diffeomorphism group. What is the analogue of a vector field on a finite abstract simplicial complex G? Discrete theory approaches Date: 11/25/2018.
1991 Mathematics Subject Classification. 05E-xx,68R-xx,51P-xx.
1 CARTAN MAGIC FORMULA like [7, 16, 19] have put forward some notions. What about a combinatorial discrete frame work ?
1.2. In order to get a continuum motion, one always can just look at paths in the unitary group on the Hilbert space l2(G) like for example isospectral Lax deforma-tions D′ = [B, D] of the exterior derivative d defining D = d + d∗[11, 10]. But we would like to have a definition of vector field which is formally identical to the classical case and which agrees with the classical case if the differential complex comes from a Riemannian manifold.
1.3. As in the continuum, like on a Riemannian manifold [1, 8], we would like to see vector fields related to 1-forms, possibly moderated by a Riemannian metric, The presence of an exterior derivative d then produces potential fields F = dV which then can be used to generate dynamics like x′′ = −∇V (x). In that case, the Riemannian structure allows to transfer the 1-form dV into a vector field ∇V . In this note, we look at a vector field notion in the discrete which works on any finite abstract simplicial complex G, a finite set of non-empty sets x invariant under the process of taking non-empty finite subsets.
1.4. We see that there is a finite dimensional Lie algebra of vector fields X for which the Lie derivative LX defines a coordinate change which commutes with exterior differentiation d. The coordinate changes allow to have a basic general covariance principle. The motion can be extended to a Hamiltonian frame work so that we could look also at analogues the Kepler problem, where a mass point moves in a central field given by a potential associated to the geometry. Given a vector field X, then the solution ψ(t) = eiDXtψ(0) of the second order and so a Newton type equation ftt = −LXf with LX = D2 X and ψ(0) = f(0) −iD−1 X ft(0) resembles then the solution eiDtψ(0) of the wave equation ftt = −Lf with L = D2.
It is the Cartan magic formula , which is now part of any differential topology book like ) which produces the analogy between the wave equation with Hodge Laplacian L = dd∗+d∗d = (d+d∗)2 = D2 and the Lie derivative LX = diX +iXd = (d+iX)2 = D2 X. The Cartan formula can therefore be seen as a key to port notions of ordinary differential equations on manifolds to discrete spaces like simplicial complexes.
1.5. The Cartan formula has been used in the past in discrete frame works (it appears in ).
Usually, in the Newton case as well as in the wave case, one does not write the dynamics using complex coordinates. Here in the discrete, it is convenient as the equations become just Schr¨ odinger equations giving paths in the unitary group. The wave equation case with Laplacian L is the most symmetric case, where the propagation of information happens in all directions or (if the momentum OLIVER KNILL ft is chosen accordingly) allows to force the propagation into a specific direction.
The analogue of ordinary differential equations are obtained when replacing L with LX, in which X is a deterministic field, which assigns to a simplex, just one smaller dimensional simplex.
1.6. The note draws also from insight gained in [18, 12] and belongs to the theme of looking for finite dimensional analogues of partial differential equations in the continuum. The set-up for builds on work like and is an advection model for a directed graph G is u′ = −Lu, where L = div(V (grad0(u))) is the directed Laplacian on the graph. As in the usual (scalar) Laplacian L = d∗d for undirected graphs which leads to the heat equation u′ = −Lu, the advection Laplacian uses difference operators: the modification of the gradient grad0 which is the maximum of grad and 0. The divergence div = d∗as well as grad = d are defined by the usual exterior derivative d on the graph. The consensus model is the situation, where the graph G is replaced by its reverse graph GT, where all directions are reversed.
One can therefore focus on advection. A central part of relates this to Markov chains. If L = D −A, then M = AD−1 is a left stochastic matrix, a Markov operator, which maps probability vectors to probability vectors. The kernel of L is related to the fixed points of M. Assume LD−1u = 0, then (D −A)D−1u = 0 and u = AD−1u = Mu. Perron-Frobenius allows to study the structure of the equilibria which are given by the kernel of L. This concludes the diversion into advection.
1.7. What is the connection? While the topic is related, we look here at differential equations on all differential forms and not only on 0-forms. Also, we don’t yet really study the dynamics much and just establish the linear algebra set-up showing that there is an elementary way to define a Lie algebra of vector fields in a discrete set-up.
The affinity to the continuum is that the formalism is not only similar but identical to the continuum. Whatever is done works both for Riemannian manifolds or finite abstract simplicial complexes or more generally for a differential complex. There are many open questions as seen at the end of this note: integer-valued deterministic X often produce integer eigenvalues of LX for example. We would like to know when this is case appears.
1.8. An other angle emerged while teaching the multi-variable Taylor theorem in . Already the single variable Taylor theorem can be seen as the solution f(t, x) = eDtf(0) of the transport equation ft = Df with D = d/dx, a partial differential equation. Because also f(t, x) = f(x+t) solves this partial differential equation, we have f(x + t) = eDtf(0) which becomes so the Taylor theorem, provided the initial function f(0) is real analytic. For a multivariate function we can replace L = D2 with a Lie derivative LX = D2 X and in the case of a constant field X = v, a Taylor CARTAN MAGIC FORMULA expansion f(x+tv) = f(x)+d f(x)tv +d2f(x)(tv)2/2+... (The multivariable Taylor theorem can be formulated conveniently using directional (Fr´ echet) derivatives which is how textbooks like [6, 2] treats the subject in higher dimensions, avoiding tensor calculus.) As both L and LX can be written as a square L = D2, LX = D2 X, the analogy between the wave and Newton equation has appeared. The frame work shows that a“diffeomorphism Lie group” exists in any geometric structure with an exterior derivative; Taylor links the vector field Lie algebra with translation.
2. From Cartan to d’Alembert 2.1. Given a smooth compact manifold M or a simplicial complex G with exterior derivative d : Λp →Λp+1, then every vector field X defines an interior derivative iX : Λp →Λp−1.
The Cartan magic formula writes the Lie derivative LX as LX = diX + iXd.
From the identities d2 = 0 and i2 X = 0 follows that LX commutes with d.
We know already from the continuum, that without the diX part, the naive directional derivative iXd alone would not work, as it would be coordinate dependent. A LX commutes with d it leads to a chain homotopy between the complexes before and after the coordinate transformation.
Like the Hodge Laplacian L = dd∗+d∗d, we can write LX as a square: define DX = d+iX and D = d + d∗. Then, LX = D2 X and L = D2. The directional Dirac operator DX has also an adjoint D∗ X but it is different from DX in general. Despite the notation used, the directional Dirac operator is not the directional derivative used in calculus.
The operators d, iX, DX and LX work on the linear space of all differential forms.
2.2. If L = D2 is the Hodge Laplacian with Dirac operator D = d + d∗, then the wave equation is ftt = −Lf. The directional wave equation is the formal analogue ftt = −LXf.
Written in the d’Alembert form, it is (∂tt + LX)f = 0.
As L and LX are both squares of simpler operators D and DX, we can factor (∂t + iDX)(∂t −iDt)f = 0 or (∂t + iD)(∂t −iD)f = 0.
The solutions e±iDX which with Euler’s formula eiDt = cos(Dt) + i sin(Dt) leads to the explicit solution f(t) = cos(Dt)f(0) + sin(Dt)D−1ft(0), where D−1 is the pseudo-inverse is defined as D−1f ⊥ t (0) if f ⊥ t (0) is in the orthogonal complement of the kernel of D.
2.3. So far, the solutions of the equations were real-valued functions in the Hilbert space H = l2(G, R). If G is a finite abstract simplicial complex, the Hilbert space is finite dimensional, and the frame work is part of linear algebra. It is convenient to build the complex valued wave ψ(0) = f(0)−iD−1ft(0), (where again D−1 is the pseudo inverse) and get ψ(t) = eiDtψ(0). The dynamics is now the solution to the Schr¨ odinger equation iψt = −Dψ, where ψ(0) encodes the initial position f(0) in its real part and the initial velocity ft(0) in its imaginary part. This works in the OLIVER KNILL same way for DX. In both cases, the wave equation for the real wave is equivalent to the Schr¨ odinger equation for a complex wave. In the case of 0-forms, we have L = d∗d and LX = iXd is the directional derivative in the direction X. We summarize: Given a geometric space with an exterior derivative d, the second order real wave equation ftt = −Lf is equivalent to the first order complex Schr¨ odinger equation ψ′ = iDψ, leading to a d’Alembert solution ψ(t) = eiDtψ(0) which can then be computed using a Taylor expansion and for which the real part of ψ gives the wave solution f(t).
3. The Lie algebra 3.1. Let us assume now that we are in a finite dimensional geometric space with an exterior derivative d : Λp →Λp+1 leading to a differential complex on a graded vector space Λ = ⊕dim(G) p=0 Λp. A vector field is defined by a linear operator iX on Λ which maps Λp to Λp−1 and has the property that i2 X = 0. We can define a Lie algebra multiplication Z = [X, Y ] by first forming LX = iXd + diX which is a map from Λp to Λp and then defining Z through the inner derivative iZ = i[X,Y ] = [LX, iY ] = LXiY −iY LX .
3.2. The field Z can be read of from iZ. We also have the Lie algebra relation LZ = iZd + diZ = LXiY d −iY LXd + dLXiY −diY LX = LX(LY −diY ) −iY LXd + LXdiY −(LY −iY d)LX = LXLY −LY LX .
3.3. Now, if iXiY = iY iX = 0, then LXiY −iY LX = iXLY −LY iX because inter changing X and Y produces a change of sign of LZ. As LZ = iZd + diZ, also iZ changes sign meaning Z changes sign. So, iZ = LXiY −iY LX = −(LY iX −iXLY ).
3.4. These elementary matrix identities prove the following proposition which ap-plies to any so derived Lie algebra of fields X with i2 X = 0.
Proposition 1. Every vector field X defines an operator DX = d + iX which has as a square a Lie derivative LX = D2 X. The set of LX define a Lie algebra with L[X,Y ] = LXLY −LY LX. If X is in supported on odd forms, then i2 X = 0 and LX = D2 X holds in the entire Lie algebra.
CARTAN MAGIC FORMULA 3.5. Even so LX is not self-adjoint in general, it plays the role of a Laplacian. In the case when LX is not diagonalizable, we can not form the pseudo inverse when writing the dynamics in the complex but we can assume that the initial velocity is in the image of DX. With this definition, also the adjoint operator d∗defines a vector field. We can still see d∗= iX for some field X. It belongs to the class of vector fields X for which the eigenvalues of DX are real.
3.6. We could look at a subclass of “deterministic fields”, which have the property that for a p-form f, the (p −1)-form iXf is supported on a single sub-simplex y of x, if f is supported on a single simplex x. This would correspond to classical vector fields close to the discrete Morse theory frame-work . If X is supported on one-dimensional simplices, this is close to the discrete Morse theory set-up. Unlike in the continuum, these “deterministic fields” are not invariant under addition, nor under the Lie algebra multiplication [X, Y ]. When taking the commutator of LX and LY for such fields, then the corresponding inner derivative i[X,Y ] connects simplices which are not directly connected.
3.7. If the simplicial complex is one-dimensional, or if X is restrained to p-forms with odd p (or even if we like) then iZ = i[X,Y ] satisfies again i2 Z = 0 and the factorization LZ = (iZ + d)2 = D2 Z holds in the entire Lie algebra. In general, the new interior derivative is only nilpotent, because i1+dim(G) Z = 0. The Mathematica procedures below allow to support X onto any subset of forms but we mostly use the case when X is supported on odd-dimensional forms.
3.8. Let us briefly look at the 1-dimensional (single-variable) classical case M = R, where LX = iXd is what we understand to be the usual derivative d/dx. Technically, the exterior derivative d produces from a 0-form f ∈Λ0 a 1-form d fdx ∈Λ1 which is in a different vector space than f.
But for the constant vector field X = 1, the combination iXd produces again an element in Λ0. Since i∗ X = 0 on 0-forms, we have iXd = iXd + diX = LX. Now eLXtf = P∞ n=0(d/dx)nf(x)Xntn/n! is by the Taylor formula equal to f(x + tX), illustrating that the derivative d/dx is the generator of the translation.
The flow φXf = f(x + tX) is a solution of a transport equation and not the wave equation. To get an analogue of the later, we need a second derivative in time and so a symplectic or complex structure. It is first a bit puzzling to see the Lie derivative LX as a second order operator. But the Cartan formula shows that also in one dimensions, LX = iXd = (d + iX)2 = D2 X is second order. In calculus, we usually think of the derivative d as a map on a space of scalar functions and not as a map from 0-forms to 1-forms. The inner derivative iX which brings us back to 0-forms is silently assumed in calculus. This identification of 0-forms and 1-forms can not be done in the discrete because the OLIVER KNILL dimension v1 = |E| of 1-forms is different from the dimension v0 = |V | of 0-forms on a graph G = (V, E). Still, the general frame work applies and the wave equation ftt = −LXf can be written as (∂t −iDX)(∂t + iDX) = 0 with DX = d + iX.
3.9. We would like to point out that the McKean-Singer symmetry which holds for simplicial complexes and the operator L = D2 remains valid also for LX = D2 X: Proposition 2 (McKean-Singer symmetry). The non-zero spectrum of LX on even differential forms Λeven is the same than the non-zero spectrum of LX on odd differ-ential forms Λodd.
Proof. The proof is the same than in the continuum or used in : the operator DX which exchanges Λeven and Λodd gives a translation between the eigenvectors belonging to non-zero eigenvalues. The discrepancy between the kernels on odd and even forms is by definition the Euler characteristic.
□ 3.10. Also the nonlinear Lax type deformation [11, 10] of the Dirac operator generalizes from D to DX. These Lax equations are D′ X = [BX, DX] where DX = d + iX + bX, BX = dX −d∗ X + ibX. Unlike for iX = d∗, where L = LX is the Hodge Laplacian, the deformation is now not isospectral in general and therefor not expected to be integrable. The expansion rate in different part of space or differential forms happens differently.
Still, these systems remain interesting non-linear differential equations and the corresponding d(t) still satisfies d(t)2 = 0 producing an exterior derivative after deformation.
As in the case of the wave equation, the deformed d(t) keeps the same cohomology.
4. Physics 4.1. Any mathematical theory with some quantum gravitational ambitions should be able to be powerful enough to solve the Kepler problem effectively in any scale: in the large, it should lead to the classical Kepler problem, in the very large to relativistic motion in a Schwarzschild metric and in the very small to the quan-tum dynamics in the Hydrogen atom. No current theory passes this Kepler test: no theory can yet describe a point in the influence of a central field classically, relativis-tically and quantum mechanically, not just in principle or a perturbative patch work but in an elegant manner, leading to quantitative results which match experiments in all three scales. It should be able to describe the motion of satellites or planets, also relativistically, predict the emission patters of gravitational waves emitted by a binary system or the structure of the Mendeleev table in the small.
CARTAN MAGIC FORMULA 4.2. The general covariance principle in physics states that physical laws are inde-pendent of the coordinate system. This means that the laws should not only be invariant under a finite dimensional symmetry group like Euclidean or Lorentz sym-metry but they should be invariant under the diffeomorphism group of the manifold.
Additionally, in case of fibre bundles, additional gauge symmetries might apply, but this is also part of the general covariance principle. An example are the Maxwell equations dF = 0, d∗F = j leading in the Coulomb gauge d∗A = 0 to the Poisson equation LA = j. If we move into a new coordinate system, then the transported equations look the same. An other example are the Einstein field equations G = eT, relating the geometric Einstein tensor with the energy tensor T using a proportion-ality factor e, the Einstein constant. The covariance there is there the statement that G and T are tensors. How can one port the general covariance principle to the discrete? A naive request would be to look at laws only which are invariant after applying a deformation through a vector field. Since LX and d commute, any law which only involves the exterior derivative does this. Examples are the wave, the heat or the Schr¨ odinger equation.
5. Examples 5.1. The simplest case with a non-trivial Lie algebra is when G = {{1}, {2}, {1, 2}} is the Whitney complex of the complete graph K2. In that case, d = 0 0 0 0 0 0 −1 1 0 , d∗= 0 0 −1 0 0 1 0 0 0 .
The general inner derivative (vector field) has the form iX = 0 0 a 0 0 b 0 0 0 .
The general operator DX = d + iX and LX = D2 X = diX + iXd then is DX = 0 0 a 0 0 b −1 1 0 , LX = −a a 0 −b b 0 0 0 b −a .
The eigenvalues of LX are {0, b −a, b −a}. Given an other vector field iY = 0 0 u 0 0 v 0 0 0 , OLIVER KNILL one can form iZ = LXiY −iY LX = iXLY −LY iX which is iZ = 0 0 av −bu 0 0 av −bu 0 0 0 , leading to DZ = 0 0 av −bu 0 0 av −bu −1 1 0 , LZ = bu −av av −bu 0 bu −av av −bu 0 0 0 0 .
This satisfies LZ = LXLY −LY LX. The eigenvalues of LZ are all zero. Indeed L2 Z = 0.
5.2. Here is the general case if G = {{1}, {2}, {3}, {1, 2}, {1, 3}} is the Whitney complex of a linear graph of length 2.
d = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 1 0 0 0 −1 0 1 0 0 , d∗= 0 0 0 −1 −1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 .
With a vector fields iX = 0 0 0 a1 a2 0 0 0 a3 0 0 0 0 0 a6 0 0 0 0 0 0 0 0 0 0 , DX = 0 0 0 a1 a2 0 0 0 a3 0 0 0 0 0 a6 −1 1 0 0 0 −1 0 1 0 0 .
This leads to LX = −a1 −a2 a1 a2 0 0 −a3 a3 0 0 0 −a6 0 a6 0 0 0 0 0 a3 −a1 −a2 0 0 0 −a1 a6 −a2 .
CARTAN MAGIC FORMULA Given an other vector field iY = 0 0 0 b1 b2 0 0 0 b3 0 0 0 0 0 b6 0 0 0 0 0 0 0 0 0 0 , DY = 0 0 0 b1 b2 0 0 0 b3 0 0 0 0 0 b6 −1 1 0 0 0 −1 0 1 0 0 , we get iZ = 0 0 0 −a2b1 −a3b1 + a1(b2 + b3) a2(b1 + b6) −(a1 + a6)b2 0 0 0 a1b3 −a3b1 a2b3 −a3b2 0 0 0 a1b6 −a6b1 a2b6 −a6b2 0 0 0 0 0 0 0 0 0 0 .
The eigenvalues of LX contain 0 as well as the following two eigenvalues, each with multiplicity two: ± p a2 1 + 2a1(a2 −a3 + a6) + (a2 + a3 −a6)2 −a1 −a2 + a3 + a6 /2. We see that real eigenvalues are quite common but that imaginary eigenvalues of LX can occur in the 4-dimensional space of vector fields. The operator LZ does in general not have zero eigenvalues. They can even become complex. We also see that the commutator LZ now tunnels between places which were not directly connected in G.
5.3. For the complete complex G = {{1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}, we can look at the general DX = d + iX = 0 0 0 a b 0 0 0 0 0 c 0 d 0 0 0 0 0 e f 0 −1 1 0 0 0 0 g −1 0 1 0 0 0 h 0 −1 1 0 0 0 i 0 0 0 1 −1 1 0 .
Now, given two general DX, DY , we have LX = D2 X, LY = D2 Y . If g = h = i = 0 then, LXiY −iY LX = iXLY −LY iX and iZ = LXiY −iY LX has the property that i2 Z = 0.
5.4. Let G = {{1}, {2}, {3}, {4}, {1, 2}, {2, 3}, {3, 4}, {4, 1}} be the complex of the cyclic graph C4. The exterior derivative d and an example of an interior derivative iX are: OLIVER KNILL d = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 1 0 0 −1 0 0 0 0 , iX = 0 0 0 0 −1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .
This leads to the Dirac operator D = d + d∗and the directional Dirac operator DX = d + iX D = 0 0 0 0 −1 0 0 1 0 0 0 0 1 −1 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 1 −1 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 1 0 0 −1 0 0 0 0 , DX = 0 0 0 0 −1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 1 0 0 −1 0 0 0 0 .
The Hodge Laplacian L = D2 and Lie derivative LX = D2 X are L = 2 −1 0 −1 0 0 0 0 −1 2 −1 0 0 0 0 0 0 −1 2 −1 0 0 0 0 −1 0 −1 2 0 0 0 0 0 0 0 0 2 −1 0 −1 0 0 0 0 −1 2 −1 0 0 0 0 0 0 −1 2 −1 0 0 0 0 −1 0 −1 2 , LX = 1 −1 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 −1 0 −1 0 .
The eigenvalues of D are {−2, 2, − √ 2, − √ 2, √ 2, √ 2, 0, 0}, the eigenvalues of DX are {− √ 2, − √ 2, −1, −1, 1, 1, 0, 0}. The eigenvalues of LX are {4, 4, 2, 2, 2, 2, 0, 0}.
CARTAN MAGIC FORMULA 50 100 150 200 -2 -1 1 2 Spectrum of DX and D DX D 20 40 60 80 100 -6 -4 -2 2 4 6 Spectrum of DX and D DX D Figure 1. The figures show the spectrum of D and DX for a cycle graph spectrum and a wheel graph spectrum. In both cases, we start with iX = d∗, then take away each entry 1 or −1 with probability 1/2.
DX still has the σ(DX) = −σ(DX) symmetry from the Dirac operator D but the energies of the particles are smaller as it is more difficult to travel.
6. Mathematica procedures 6.1. Here is the code which computes the Dirac operator D and the vector field analogue DX for any simplicial complex G. The code can be copy pasted when accessing the LaTeX source of this document on the ArXiv. The first part computes the matrices given in the above example. For the vector field, we chose for iX just to take the first non-zero entry of d∗: G={{1} ,{2} ,{3} ,{4} ,{1 ,2} ,{2 ,3} ,{3 ,4} ,{4 ,1}}; n=Length [G] ; Dim= Map[ Length ,G] −1; f=Delete [ BinCounts [ Dim ] , 1 ] ; Orient [ a , b ]:=Module[{ z , c , k=Length [ a ] , l=Length [ b ] } , If [ SubsetQ [ a , b ] && ( k==l +1) , z=Complement[ a , b ] [ [ 1 ] ] ; c=Prepend [ b , z ] ; Signature [ a ] ∗Signature [ c ] , 0 ] ] ; d=Table [ 0 ,{ n} ,{n } ] ; d=Table [ Orient [G[ [ i ] ] ,G[ [ j ] ] ] , { i , n} ,{ j , n } ] ; dt=Transpose [ d ] ; DD =d+dt ; LL= DD.DD; HX[ x ]:= Block [{ u=Flatten [ Position [Abs[ x ] , 1 ] ] } , If [ u=={},0, First [ u ] ] ] ; iX=Table [ 0 ,{ n} ,{n } ] ; Do[ l= HX[ dt [ [ k ] ] ] ; If [ l >0,iX [ [ k , l ]]= dt [ [ k , l ] ] ] , { k , f [ [ 1 ] ] } ] ; DX =iX+d ; LX = DX.DX; 6.2. Here is the code to generate DX and LX for a generate random finite abstract simplicial complex G: (∗ Generate a random s i m p l i c i a l complex ∗) Generate [ A ]:= Delete [ Union [ Sort [ Flatten [Map[ Subsets ,A] , 1 ] ] ] , 1 ] R[ n , m ]:=Module[{A={},X =Range[ n ] , k } ,Do[ k:=1+Random[ Integer , n −1]; A =Append[A, Union [ RandomChoice [X, k ] ] ] , {m} ] ; Generate [A ] ] ; G =Sort [R[ 1 0 , 2 0 ] ] ; OLIVER KNILL (∗ Computation of e x t e r i o r d e r i v a t i v e ∗) n=Length [G] ; Dim= Map[ Length ,G] −1; f=Delete [ BinCounts [ Dim ] , 1 ] ; Orient [ a , b ]:=Module[{ z , c , k=Length [ a ] , l=Length [ b ] } , If [ SubsetQ [ a , b ] && ( k==l +1) , z=Complement[ a , b ] [ [ 1 ] ] ; c=Prepend [ b , z ] ; Signature [ a ] ∗Signature [ c ] , 0 ] ] ; d=Table [ 0 ,{ n} ,{n } ] ; d=Table [ Orient [G[ [ i ] ] ,G[ [ j ] ] ] , { i , n} ,{ j , n } ] ; dt=Transpose [ d ] ; DD =d+dt ; LL= DD.DD; (∗ Build i n t e r i o r d e r i v a t i v e s iX and iY ∗) UseIntegers=False ; e ={}; Do[ If [ Length [G[ [ k ]]]==2 , e=Append[ e , k ] ] , { k , n } ] ; BuildField [ P ]:=Module[{X, ee , iX=Table [ 0 ,{ n} ,{n }]} , X =Table [ If [ UseIntegers ,Random[ Integer , 1 ] ,Random[ ] ] , { l , Length [ e ] } ] ; Do[ ee= G[ [ e [ [ l ] ] ] ] ; Do[ If [ SubsetQ [G[ [ k ] ] , ee ] , m =Position [G, Sort [Complement[G[ [ k ] ] , Delete [ ee , 2 ] ] ] ] [ [ 1 , 1 ] ] ; iX [ [m, k ]]= If [MemberQ[P, Length [G[ [m] ] ] ] , X[ [ l ] ] , 0 ] ∗ Orient [G[ [ k ] ] ,G[ [m] ] ] ] , { k , n }] , { l , Length [ e ] } ] ; iX ] ; (∗ Build Laplacians LX,LY,LZ, p l o t spectrum of D and DX and matrices ∗) iX=BuildField [ { 1 , 3 , 5 , 7 , 9 } ] ; iY=BuildField [ { 1 , 3 , 5 , 7 , 9 } ] ; DX =iX+d ; LX = DX.DX; DY =iY+d ; LY = DY.DY; iZ1= LX. iY−iY .LX; iZ2=iX .LY − LY. iX ; Print [ iZ1==iZ2 ] ; iZ=iZ1 ; LZ= Chop[ iZ . d+d . iZ ] ; DZ =iZ+d ; dx=” !(\∗SubscriptBox [ \ (D) , (X\ ) ] \ ) ” ; lx=” !(\∗SubscriptBox [ \ ( L) , (X\ ) ] \ ) ” ; pl=PlotLabel ; GraphicsGrid [{{ MatrixPlot [DX, pl− >dx ] , MatrixPlot [LX, pl− >lx ] } , { MatrixPlot [DD, pl− >”D” ] , MatrixPlot [ LL, pl− >”L” ] } } ] ; u1 = Sort [Chop[ Eigenvalues [ 1 . 0 DX ] ] ] ; u2 = Sort [ Eigenvalues [ 1 . 0 DD] ] ; u1= N[Round[ u1 ∗10ˆ6]/10ˆ6]; (∗ c l e a r tiny imaginary parts ∗) S=ListPlot [{ u1 , u2 } , Joined − >True , PlotLegends − >Placed [ { dx , ”D” } ,Below ] , PlotRange − > All , PlotStyle − > {Red, Blue} , PlotLabel − > ”Spectrum of !(\∗SubscriptBox [ \ (D) , (X\ ) ] \ ) and D” ] ; (∗Compute B et ti numbers , compare bosonic and fermionic part ∗) chi= Sum[−f [ k ]ˆ k , { k , Length [ f ] } ] ; f=Prepend [ f , 0 ] ; m =Length [ f ] −1; U =Table [ v=f [ [ k + 1 ] ] ; Table [ u= Sum[ f [ [ l ] ] , { l , k } ] ; LL [ [ u+i , u+j ] ] , { i , v } ,{ j , v }] ,{ k ,m} ] ; Cohomology = Map[ NullSpace , U ] ; Betti = Map[ Length , Cohomology ] chi1= Sum[−Betti [ k ]ˆ k , { k , Length [ Betti ] } ] ; EV = Map[ Eigenvalues ,U] ; EVFermi=Table [EV[ [ 2 k ] ] , { k , Floor [ Length [EV] / 2 ] } ] ; EVBoson=Table [EV[ [ 2 k −1]] ,{ k , Floor [ ( Length [EV] + 1 ) / 2 ] } ] ; extract [ u ]:=Module[{ v=Flatten [ u ] ,w={}}, Do[ If [Abs[ v [ [ k ]]] >10ˆ( −8) ,w =Append[w, v [ [ k ] ] ] ] , { k , Length [ v ] } ] ; Sort [w ] ] ; extract [ EVFermi]==extract [ EVBoson ] (∗Now the same for LX ∗) U =Table [ v=f [ [ k + 1 ] ] ; Table [ u= Sum[ f [ [ l ] ] , { l , k } ] ;LX [ [ u+i , u+j ] ] , { i , v } ,{ j , v }] ,{ k ,m} ] ; EV = Map[ Eigenvalues ,U] ; EVFermi=Table [EV[ [ 2 k ] ] , { k , Floor [ Length [EV] / 2 ] } ] ; EVBoson=Table [EV[ [ 2 k −1]] ,{ k , Floor [ ( Length [EV] + 1 ) / 2 ] } ] ; extract [ EVFermi]==extract [ EVBoson ] {EVFermi , EVBoson} CARTAN MAGIC FORMULA Total [Abs[N[ extract [ EVFermi ] −extract [ EVBoson ] ] ] ] Cohomology = Map[ NullSpace , U ] ; Betti = Map[ Length , Cohomology ] ; chi2= Sum[−Betti [ k ]ˆ k , { k , Length [ Betti ] } ] ; { chi , chi1 , chi2 } DX LX D L Figure 2. The matrices DX, LX, D, L in the case of a random com-plex. This was produced with the code above.
6.3. And finally, here is the self-contained procedure which does the isospectral deformation of the exterior derivative by deforming D′ X = [BX, DX]. In the case iX = d∗, this is the standard Lax isospectral deformation we have seen before OLIVER KNILL [11, 10]. In an other extreme case, when X = 0, then d is not deformed at all. We still have the inflationary initial decay of d typical for that type of integrable dynamical system. The decay of d means by the Connes formula that there is an expansion of space because if the derivative operator become small, then the distances grow.
Generate [ A ]:= Delete [ Union [ Sort [ Flatten [Map[ Subsets ,A] , 1 ] ] ] , 1 ] R[ n , m ]:=Module[{A={},X =Range[ n ] , k } ,Do[ k:=1+Random[ Integer , n −1]; A =Append[A, Union [ RandomChoice [X, k ] ] ] , {m} ] ; Generate [A ] ] ; G =Sort [R[ 5 , 8 ] ] ; n=Length [G] ; fv=Delete [ BinCounts [Map[ Length ,G] ] , 1 ] ; cn=Length [ fv ] ; br ={0};Do[ br=Append[ br , Last [ br]+ fv [ [ k ] ] ] , { k , cn } ] ; Orient [ a , b ]:=Module[{ z , c , k=Length [ a ] , l=Length [ b ] } , If [ SubsetQ [ a , b ] && ( k==l +1) , z=Complement[ a , b ] [ [ 1 ] ] ; c=Prepend [ b , z ] ; Signature [ a ] ∗Signature [ c ] , 0 ] ] ; d=Table [ 0 ,{ n} ,{n } ] ; d=Table [ Orient [G[ [ i ] ] ,G[ [ j ] ] ] , { i , n} ,{ j , n } ] ; dt=Transpose [ d ] ; DD =d+dt ; LL= DD.DD; UseIntegers=False ; e ={}; Do[ If [ Length [G[ [ k ]]]==2 , e=Append[ e , k ] ] , { k , n } ] ; BuildField [ P ]:=Module[{X, ee , iX=Table [ 0 ,{ n} ,{n }]} , X =Table [ If [ UseIntegers ,Random[ Integer , 1 ] ,Random[ ] ] , { l , Length [ e ] } ] ; Do[ ee= G[ [ e [ [ l ] ] ] ] ; Do[ If [ SubsetQ [G[ [ k ] ] , ee ] , m =Position [G, Sort [Complement[G[ [ k ] ] , Delete [ ee , 2 ] ] ] ] [ [ 1 , 1 ] ] ; iX [ [m, k ]]= If [MemberQ[P, Length [G[ [m] ] ] ] , X[ [ l ] ] , 0 ] ∗ Orient [G[ [ k ] ] ,G[ [m] ] ] ] , { k , n }] , { l , Length [ e ] } ] ; iX ] ; iX=BuildField [ { 1 , 3 , 5 , 7 , 9 } ] ; iY=BuildField [ { 1 , 3 , 5 , 7 , 9 } ] ; DX =iX+d ; LX = DX.DX; DY =iY+d ; LY = DY.DY; iZ= LX. iY−iY .LX; DZ =iZ+d ; LZ= DZ.DZ; T[ A ]:=Module[{ n=Length [A] } , Table [ If [ i<=j , 0 ,A[ [ i , j ] ] ] , { i , n} ,{ j , n } ] ] ; UT[ {DD , br }]:=Module[{D1= T[DD] } , (∗Lower t r i a n g u l a r block ∗) Do[Do[Do[D1 [ [ br [ [ k ]]+ i , br [ [ k ]]+ j ]]=0 ,{ i , br [ [ k+1]]−br [ [ k ] ] } ] , { j , br [ [ k+1]]−br [ [ k ] ] } ] , { k , Length [ br ] −1}];D1 ] ; RuKu[ f , x , s ]:=Module[{ a , b , c , u , v ,w, q } ,u=s ∗f [ x ] ; (∗Runge Kutta ∗) a=x+u /2; v=s ∗f [ a ] ; b=x+v /2;w =s ∗f [ b ] ; c=x+ w; q=s ∗f [ c ] ; x+(u+2v+2w +q ) / 6 ] ; DD = DX; d0= UT[ {DD, br } ] ; e0=Conjugate [ Transpose [ d0 ] ] ; M=1000; delta =2/ M; u={}; (∗ Deformation with Runge Kutta ∗) Do[ d= UT[ {DD, br } ] ; e=Conjugate [ Transpose [ d ] ] ; BB =d−e ; CC =d+e ; M M = CC.CC; b= DD − CC; VV =b . b ; B = BB+1.0∗I∗b ; f [ x ]:=B. x−x .B; DD =RuKu[ f , 1 . 0 DD, delta ] ; u=Append[ u , Total [Abs[ Flatten [Chop[ d ] ] ] ] ] , { m,M} ] ; DDX = DD; LLX= DDX.DDX; {Total [Abs[ Flatten [Chop[ d . d ] ] ] ] , Total [Abs[ Flatten [Chop[ e . e ] ] ] ] } F[ x ]:= If [ x==0,0,−Log [Abs[ x ] ] ] ; (∗ Plot the s i z e of d ∗) v= M∗Table [F[ u [ [ k+1]]] −F[ u [ [ k ] ] ] , { k , Length [ u ] −1}]; ListPlot [ v , PlotRange− >All ] CARTAN MAGIC FORMULA DZ LZ D L Figure 3. The matrices DZ, LZ, D, L in the case of a random com-plex with f-vector (10, 45, 120, 192, 165, 73, 15, 1). Also this was pro-duced with the code above, where Z = [X, Y ] is the commutator of two random vector fields.
7. Questions 7.1. The operators DX and LX are not symmetric in general so that complex eigen-values can appear. In that case, the solution ψ(t) = eiDXtψ(0) can grow exponen-tially. DX still often has real eigenvalues, leading to quasi-periodic solutions as the orbits eiDXtψ(0) form a subgroup of a finite dimensional torus, if the graph is finite.
Actually, if iX(k, l) ̸= 0 only for one l, then we implement a deterministic vector OLIVER KNILL field. In that case the eigenvalues of LX often non-negative integers taking values in {0, 1, . . . , dim(G) + 1}. We would like to understand the spectrum.
7.2. For D = iX + i∗ X + d + d∗we have D2 = LX + L∗ X + L. Now, if we average that over all possible vector fields using a measure which is homogeneous, we expect the LX to average out and get the wave equation governed by L. Can one make this more precise and see the wave equation ftt = −L as an average of deterministic flows ftt = −LX?
7.3. We often integer eigenvalues of LX if iX has integer values. In small dimen-sional examples, we can compute general formulas for the eigenvalues but integer eigenvalues also often appear for large random simplicial complexes. Under which conditions does LX have integer eigenvalues?
7.4. The eigenvalues of LX are most of the time real if the entries of IX are non-negative multiplies of d∗. They can become imaginary in general, if the signs are changed. Can we find conditions which assures a real spectrum?
References R. Abraham, J.E. Marsden, and T. Ratiu. Manifolds, Tensor Analysis and Applications.
Applied Mathematical Sciences, 75. Springer Verlag, New York etc., second edition, 1988.
C. Blatter. Analysis III, volume 153 of Heidelberger Taschenbuecher. Springer Verlag, 1974.
A. Chapman and M. Mesbahi. Advection on graphs. In 2011 IEE Conference on Decision and Control of European Control Conference, 2011.
H.L. Cycon, R.G.Froese, W.Kirsch, and B.Simon. Schr¨ odinger Operators—with Application to Quantum Mechanics and Global Geometry. Springer-Verlag, 1987.
´ E. Cartan. Riemannian Geometry in an Orthogonal Frame. World Scientific, 2001. Lectures delivered at the Sorbonne in 1926-1927, translated by Vladislav V. Goldberg.
C.H. Edwards. Advanced Calculus of Several Variables. Academic Press, 1973.
R. Forman. Combinatorial differential topology and geometry. New Perspectives in Geometric Combinatorics, 38, 1999.
T. Frankel. The Geometry of Physics. Cambridge University Press, second edition edition, 2004. An introduction.
O. Knill. The McKean-Singer Formula in Graph Theory.
2012.
O. Knill. An integrable evolution equation in geometry.
2013.
O. Knill. Isospectral deformations of the Dirac operator.
2013.
O. Knill. Differential equations on graphs (HCRP project with Annie Rak).
˜ knill/pde/pde.pdf, 2016.
O. Knill. The amazing world of simplicial complexes.
2018.
CARTAN MAGIC FORMULA O. Knill. Linear algebra and vector analysis.
˜ knill/teaching/math22a2018, 2018.
H.P. McKean and I.M. Singer. Curvature and the eigenvalues of the Laplacian. J. Differential Geometry, 1(1):43–69, 1967.
P. Mullen, A. McKenzie, D. Pavlov, L. Durant, Y. Tong, E. Kanso, J. E. Marsden, and M. Desbrun. Discrete Lie advection of differential forms. Found. Comput. Math., 11(2):131– 149, 2011.
P. Mullen, A. McKenzie, D. Pavlov, L. Durant, Y. Tong, E. Kanso, J. E. Marsden, and M. Desbrun. Discrete lie advection of differential forms. Journal of Foundations of Computa-tional Mathematics, 2011.
A. Rak. Advection on graphs. Senior Thesis at Harvard College, 2016.
A. Romero and F. Sergeraert. Discrete vector fields and fundamental algebraic topology. Ver-sion 6.2. University of Grenoble, 2012.
Department of Mathematics, Harvard University, Cambridge, MA, 02138 |
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2.3: Density
Last updated
: Jun 26, 2023
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2.2: Dimensional Analysis
2.4: Temperature
Page ID
: 85140
Elizabeth Gordon
Furman University
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Calculate density, mass, or volume when given 2 of these three variables.
Identify what units are required for the density equation.
Review metric conversions.
Compare densities of different chemical substances.
Classify a substance as being a heterogeneous or homogeneous mixture if solubility date is provided.
Identify where a chemical would appear in water if solubility and density data are provided.
Compare any chemical substance's density to the density of water (please memorize this value).
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Density is a physical property that is defined as a substance’s mass divided by its volume:
densityd=massvolume=mV(2.3.1)(2.3.2)
Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table 2.3.1. Memorize the density of water with its appropriate units.
Table 2.3.1: Densities of Some Common Substances
| Substance | Density (g/mL or g/cm3) |
| water | 1.0 |
| gold | 19.3 |
| mercury | 13.6 |
| air | 0.0012 |
| cork | 0.22–0.26 |
| aluminum | 2.7 |
| iron | 7.87 |
Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (Table 2.3.1), the volume units will cancel and leave you with mass units, telling you the mass of the sample:
Start with Equation 2.3.1
density=mV
and insert the relevant numbers
2.7gcm3=m7.88cm3
Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.
7.88cm3×2.7gcm3=21g of aluminum
Example 2.3.1: Mercury
What is the mass of 44.6 mL of mercury?
Solution
Use the density value for mercury from Table 2.3.1 and the definition of density (Equation 2.3.1)
density=massvolume⇒d=mV
13.6gmL=m44.6mL
Remember to cross multiply here in order to isolate variable. Then, report answer with correct units.
44.6mL×13.6gmL=607g
The mass of the mercury is 607 g.
Exercise 2.3.1
What is the mass of 25.0 cm3 of iron?
Answer
: Use the density value for iron from Table 2.3.1
density=massvolume⇒d=mV
7.87gcm3=m25.0cm3
Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.
25.0cm3×7.87gcm3=197g of iron
Another way of looking at density (some students choose to perform calculations using this method)
Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:
d=19.3g/mL=19.3gmL
Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:
d=19.3g/mL=19.3gmL
That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units but also the numbers:
19.3gmL=45gV
Cross multiplying denominators with numerators, we obtain the following algebraic equation.
19.3V=45.9mL
then you will need to isolate the variable (volume)
V=45.9mL19.3
After multiplication, the answer would be
V=2.38mL
Interactive Element
Example 2.3.2: Wine Cork
A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/mL, what is the volume of the cork? Regardless of the method that is used, you should still be able to obtain the same (and correct) answer.
Solution
To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find
0.22gmL=3.78gV
Cross multiplying denominators with numerators, we obtain the following algebraic equation.
0.22V=3.78mL
then you will need to isolate the variable (volume)
V=3.78mL0.22
so, the volume of the cork is 17.2 mL.
Exercise 2.3.2
What is the volume of 3.78 g of gold?
Answer
: Before attempting this question, be sure to obtain the density of gold in the table above. If you were to need this value on a quiz or a test, then it would be provided. Once you have this value, plug it into the density equation. Next, you will need to isolate the volume variable (basic algebra). The final answer should be 0.196 cm3.
Care must be used with density as a conversion factor. Make sure the mass units are the same or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.
Using Density in Environmental Applications
Along with solubility, density can help determine how a contaminant could affect an aquatic system. For example, imagine mercury has been spilled in Furman Lake. Looking at this element's density value and comparing it to liquid water, one could determine the location of the insoluble (you would be given solubility information) mercury layer. The more dense mercury layer would reside on the bottom of Furman Lake. If one were to take a cross-section of the lake, you could see that a heterogeneous mixture would result.
In contrast, spilling ethanol (density = 0.789g/mL) would result in the formation of a homogeneous mixture. Ethanol (grain alcohol) is soluble in water. This would make it miscible (mixable to form a solution) in water and one would not be able to denote separate layers. According to the density, an alcohol layer would remain on top, but would ultimately dissolve.
Applications
What difficulties would arise from the separation and removal of contaminants?
Hg in Furman Lake
Ethanol in Furman Lake
oil (less dense, insoluble) in Furman Lake
Watch this video and record your observations.
What component was different in the two types of beverages (mass or volume)?
How does the above-mentioned difference affect the density equation?
Which beverage is denser than water?
Need More Practice?
Turn to Section 2.E of this OER and work problems #2 and #9.
Contributors and Attributions
Elizabeth R. Gordon (Furman University)
Hayden Cox (Furman University, Class of 2018)
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Teaching fractions on a number line is tough, but the right strategies help. Explore step-by-step methods, key misconceptions, and more with Frax!
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Understanding fractions on a number line
Despite the fact that there is an increasing consensus in educational research that number lines are the most effective instructional representation for fractions, and curricula are increasingly making them an emphasis, number lines are consistently hard to teach.
There are a number of reasons for this, but they can be summed up in a simple statement: fraction number line instruction requires a high degree of just-in-time feedback to be effective, especially for struggling students. In a traditional classroom setting, it’s next to impossible to provide this level of instruction because the teacher can’t be everywhere at once.
Why number lines are a great model for teaching fractions
Fractions are often students’ first exposure to abstraction in math, and understanding is critical for future coursework, including algebra. Students who develop a strong understanding of fraction number sense—recognizing that fractions are numbers that can be expressed in different forms—will be far better prepared than those who only rely on memorized “tricks” or procedures. A key aspect of this number sense is understanding fraction magnitude (size), which allows students to reason effectively with and about fractions in various contexts.
According to research, number lines are the most effective instructional representation of fractions. But why? Teaching fractions on a number line helps students understand that fractions are numbers, too, each with a location on the number line, just like any other numbers they might encounter. However, shaded area models (like pizzas and pies) often seem more approachable for students unfamiliar with number lines because of their familiarity and visual simplicity.
In the long run, number lines will help students understand and work with fractions in far deeper contexts! For example, take fractions greater than one. If students can picture fractions as numbers on a number line, they will be much more prepared to understand fractions greater than one conceptually. With a shaded area model, it’s much harder to visualize a fraction like 7/4 if a circular area model only accounts for four slices.
“Several of my students struggled with identifying and naming fractions, but with the help of Frax, those students are now able to identify and name fractions without hesitation. The majority of my students struggled with fractions on a number line, but through constant interaction with Frax, I am seeing more and more students understand the fractions on a number line.” -Teacher
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What are effective instructional techniques for teaching with number lines?
Given the difficulties of teaching this topic in a traditional classroom, teachers often (and understandably!) feel limited and frustrated in their ability to teach fractions on a number line effectively. While there’s no magical solution to the difficulties inherent in teaching fraction number lines, a few instructional techniques can mitigate them.
When it comes to teaching fractions on a number line to students, follow these four basic but effective steps:
Start small and practice, practice, practice
Count intervals, not tick marks
Incorporate fractions greater than one into your instruction early
Create time in your lessons to provide students with immediate feedback
Number line basics #1: Start small and go slow
Number lines are difficult for students. Period. Children need a lot of time and practice successfully creating and reading fractional representations to develop reliable accuracy on various problems.
While there’s a natural temptation to see “how far you can get” with your students in a single lesson, it’s not a good idea. Number lines are a tool students use to learn several important and fairly complex concepts. Learners must be able to interpret number line representations with high fidelity and fluidity, which is only developed with repeated practice.
What’s the upshot? When first learning how to read fraction number lines, it is far better to give students multiple successes on more straightforward problems. For instance, 20-30 exercises like the below example will allow students to form increasingly solidified neural pathways around the number line representation.
On the other hand, introducing unnecessarily difficult or visually dense representations (like this example below) – before a student is confident with the number line itself – will only create “cognitive noise” and confusion. This may slow a student’s progression towards mature number line work and cost you valuable instructional time.
Bottom line: brain-based education teaches us that “practice makes permanent.” Make sure your students practice the right thing with the correct conceptualization before moving on to more complex or challenging material.
Number line basics #2: Teach students to count intervals, not tick marks
One of the main points of confusion when teaching fractions on a number line is the relationship between the denominator value and the number of tick marks. Does the following number line represent thirds or fourths?
For someone familiar with the representation, it’s clear that the number line above represents fourths – there are four intervals between 0 and 1. However, for a student who is still transitioning from the predictable world of integer number lines to the far-less-predictable world of fraction number lines, the answer is not so clear.
The tick marks are what “jump off the page” visually, so it’s completely understandable that a student would see three tick marks and think: “3… that’s thirds!”
Of course, you can teach tricks like: “add 1 to the number of tick marks between 0 and 1.” However, using this trick – as with many shortcuts in math – obscures the underlying conceptual reality that makes the number line so powerful. That is, the number line shows partitions of the space between 0 and 1 in the same way that a shaded area model shows partitions of a shape that represents 1… all without the inconvenient aspects of shaded area models (see below).
You need to train your students to look for the intervals – the actual space or distance on the number line – rather than the tick marks that divide that space.
Once students have learned to do this reliably, much of the difficulty surrounding number lines evaporates, unlocking the potential for learning in this powerful representation.
Number line basics #3: Incorporate fractions >1 as soon as possible
For many students, fractions are “easy” until they start representing quantities greater than one. At that point, metaphors like “equal sharing” become more difficult. “Share a pizza equally four ways… and then give a piece to five different people.” That's not so easy to understand...
Part of what makes the number line representation effective at building strong fraction number sense is that fractions greater than one are completely intuitive. No convoluted explanations are necessary about how you can have more pieces than a whole – fractions greater than one are just another point on the number line!
Mixing problems that require the student to work with fractions greater than one into your early instruction will help your students build surprisingly strong number sense without much fanfare. You will be amazed at the dividends this pays when you compare mixed numbers and fractions greater than one! Once students understand the number line, comparing fractions on a number line will feel natural rather than something “new” to learn.
Here’s a teaser: Which is bigger, 7/3 or 2 2/3 ? No mixed number conversion is required!
Number line basics #4: Structure your lessons to minimize the time between a student error and corrective feedback
As mentioned above, teachers tend to spend little time on number line content mainly because learning to use the number line effectively is feedback-intensive. It requires a very short gap in time between a student’s mistake and a corrective piece of feedback. Otherwise, the student may wind up inadvertently cementing erroneous ways of interpreting or working with the representation.
Any math teacher will tell you that teaching students to unlearn a given method or misconception is ten times more difficult than teaching something for the first time. We need to get number line instruction right from the beginning for our students; otherwise, a vicious cycle can occur.
The student perceives number lines as “hard” (unfamiliar and opaque to the student). Using them leads to erroneous or inefficient solutions.
The student becomes increasingly resistant to doing new work with this difficult representation.
The student’s brittle and inefficient methods are increasingly inadequate for new, more complex content.
This results in the student working harder for less success.
The perception of number lines as “difficult” is thus reinforced…and the cycle continues.
The only way to prevent this cycle from starting is never to allow it to begin in the first place.
Unfortunately, doing so requires the nearly impossible: the teacher needs to be immediately available to each student as they learn and make mistakes. When the student begins counting tick marks, the teacher must immediately redirect them to count intervals instead. When a child incorrectly counts all intervals between 0 and 2 to find the denominator value of a fraction >1, they need to be immediately redirected to count only the intervals between 0 and 1.
There are no magic solutions to this problem. However, some approaches can help.
Carefully craft student pairings when doing number line lessons.The goal is to ensure that students who are already proficient can provide feedback to students who are still building proficiency.
Unfortunately, this is often impractical or impossible. When you first introduce the topic, few students will already know enough material that they can be relied on to give accurate feedback. Students take a long time to develop their proficiency, so it will be a while before a teacher can reliably count on them to help their peers.
2. Design lessons that allow for small group instruction. Math “center time” can be helpful – have the majority of the class complete self-directed activities while you work with a small group of students, where you can immediately intervene when they make mistakes or verbalize misconceptions.
Ensuring that you have enough time dedicated to such lessons is vital. Students need a lot of practice to cement their conceptual understanding and build effective strategies for working with number lines. Only one or two of these small-group times before “moving on” will significantly increase the danger of that vicious cycle setting in for many of your students.
3. Utilize technology tools that allow the entire class to work simultaneously on number line skills. The benefit of an online resource for this topic is that it allows all students to receive just-in-time feedback as needed. If the program is well-designed, students will get the gentle scaffolding and immediate feedback they need to become proficient users of this powerful number line tool.
It’s essential that you find an appropriate, effective tool for this instructional practice. Not all online resources are created equally, and a platform that doesn’t provide appropriate feedback or adapt to a student’s growing proficiency – or recurring difficulties – will be hardly better than paper and pencil work (at best) and counterproductive (at worst).
Frax is a powerful tool that builds fraction number line skills in a very short time
ExploreLearning Frax is demonstrated to help grades 3-5 students build robust fraction skills in a remarkably small amount of time. Frax’s game-based program includes a variety of fraction number line activities to help students confidently understand fractions as numbers.
How does Frax support and enhance instruction?
Highly scaffolded instruction in each Frax mission
Offline activities that reinforce number line concepts and assess understanding
Motivating rewards to boost engagement across all ability levels
Just-in-time remediation and adaptive questioning
Progress monitoring and real-time reporting to intervene when students are stuck
Game-packed Review Room for additional practice
Classroom competitions to practice number line concepts
“Many students were engaged immediately. I love the vocabulary Frax uses. Students easily grasped the concepts without using the terms numerator and denominator. Students had a more in-depth understanding that made transitioning fractions to the placement of them on number lines easier. I also love that the written prompts can be read to them. Many of my struggling readers can still be successful by clicking the reading prompts." -Teacher
Number lines make sense with Frax
In Frax Sector 1, students develop a strong conceptual understanding of fractions as numbers (topics typically addressed in grade 3). Students then apply and deepen their understanding in Sector 2 (topics typically addressed in grade 4), where they expand their knowledge to fractions equivalency and early fractions arithmetic.
However, students in grades 3+ can use Frax at any time for intervention or extra practice. Completing Frax Sector 1 before Sector 2 leads to more efficient and beneficial fractions instruction for teachers and students in the classroom. Use Frax before your fractions unitfor the best results—you’ll accelerate instruction and build a class-wide understanding of fractions. When it comes time for your core instruction, students will already have the baseline knowledge and confidence to tackle advanced fractions work.
“Students seem to grasp the concepts and basics much faster than using manipulatives. Some students still need both the Frax program and manipulatives, but most have a more in-depth understanding with just Frax than with only manipulatives. Frax is less time consuming on a daily basis and the entire fraction unit was easier to teach than in previous years.” -Teacher
Get ready—Frax Sector 3 is coming soon!
Coming in fall 2025, NEW Frax Sector 3 will help students level up their learning with all-new adventures that extend and deepen their understanding of fractions. In Sector 3, students use the foundations built in Sectors 1 and 2 to navigate new worlds and develop a strong and durable mastery of grade 5 fraction arithmetic. No tricks or memorization required.
Request More Information
From Day 1, students work with fractions as numbers in an intuitive setting as they associate a fraction’s value with the length of a block model. With Frax, students seamlessly transition to the number line representation, with an emphasis on understanding the space between 0 and 1 and counting intervals to determine the denominator value.
Space-themed games and engaging lessons help students build confidence and competence in fraction number line representations, with adaptive logic ensuring they only see questions they are ready to tackle.
Students read points on a number line…
… and place points on number lines.
Students create number lines themselves…
…and estimate fraction and mixed number locations on unmarked number lines.
These skills are embedded throughout the Frax student experience, so even when the curriculum focuses on a different topic, students still receive the practice they need to solidify their understanding and skills. Frax unlocks your ability to teach fractions concepts in the ways you often long to but find yourself prevented from doing so because of the wide differences in student performance and the need to get all students to a baseline level of performance.
Try Frax
About the Author
Jesse Mercer, Senior Product Designer at ExploreLearning
Jesse Mercer, a Senior Product Designer at ExploreLearning, has over 20 years of experience in education and related fields. He has an MA in Math Education, is a National Board Certified teacher for secondary mathematics, and was a recipient of his district’s “Golden Apple” award for teaching. During his 13 years in the classroom, Jesse taught upper elementary and middle school mathematics courses, ranging from 5th-grade math through Algebra I. He currently works with several other designers on creating the curriculum and instructional content for Frax.
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1573 | https://knowledge-action-portal.com/en/content/package-essential-noncommunicable-pen-disease-interventions-primary-health-care-low | Package of Essential Noncommunicable (PEN) disease interventions for primary health care in low-resource settings | Knowledge Action Portal on NCDs
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Package of Essential Noncommunicable (PEN) disease interventions for primary health care in low-resource settings
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Overview
These tools of the WHO Package of Essential Noncommunicable Diseases Interventions (WHO PEN) support implementation of very cost effective interventions through an integrated approach.
Implementation of WHO PEN is a key component of the objective 4 of the Global Action Plan. These tools will enable early detection and management of cardiovascular diseases, diabetes, chronic respiratory diseases and cancer to prevent life threatening complications (e.g. heart attacks, stroke, kidney failure, amputations, blindness).
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Equitable financing of interventions in WHO PEN can be a first step for addressing prevention and control of noncommunicable diseases within the universal health coverage agenda.
The protocol for the assessment and referral of women with suspected breast cancer at primary health care for Breast and Cervical cancer is attached
AccessPEN-4 Breast & Cervix.pdf
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### World Health Organization (WHO) #### Second general meeting of the WHO Global Coordination Mechanism on the Prevention and Control of Noncommunicable Diseases: meeting report, 23-25 April 2025 Report 11 Aug 2025
### The Lancet #### Metrics for diplomats: is mortality from non-communicable diseases increasing or decreasing? Article 03 Jul 2025
### NCD Alliance #### NCD Alliance Response to the Zero Draft 2025 Political Declaration on NCDs and Mental Health Briefs & Fact Sheets 23 May 2025
### NCD Alliance #### Mobilising for action: opportunities for civil society advocacy at the HLM4 Guide 20 May 2025
### World Health Organization #### Strengthening noncommunicable disease integration in all-hazards emergency preparedness and response Briefs & Fact Sheets 12 May 2025
### World Health Organization #### Assessing national capacity for the prevention and control of noncommunicable diseases: report on 2023 global survey Report 12 May 2025
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1574 | https://fiveable.me/discrete-mathematics/unit-5/modular-arithmetic/study-guide/YOXEGeTEs2fWuTGK | printables
🧩Discrete Mathematics Unit 5 Review
5.2 Modular Arithmetic
🧩Discrete Mathematics Unit 5 Review
5.2 Modular Arithmetic
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
🧩Discrete Mathematics
Unit & Topic Study Guides
5.1 Divisibility and Prime Numbers
5.2 Modular Arithmetic
5.3 Public Key Cryptography
5.4 Cryptographic Protocols
Modular arithmetic is a powerful tool in number theory, simplifying complex calculations by working with remainders. It's the math behind clocks, calendars, and even your computer's memory.
In cryptography, modular arithmetic is the secret sauce. It enables secure communication through public key systems like RSA, where large numbers are crunched efficiently using clever modular tricks.
Modular Arithmetic Fundamentals
Understanding Modular Arithmetic and Congruence
Modular arithmetic operates on a fixed range of integers, wrapping around when reaching the modulus
Congruence relation denoted by ≡ symbol, represents numbers with the same remainder when divided by modulus
Two numbers a and b are congruent modulo m if (a−b)≡0(modm)
Clock arithmetic illustrates modular arithmetic (12-hour clock uses mod 12)
Modular addition performed by adding numbers and taking remainder when divided by modulus
Modular multiplication follows similar principle, multiply then find remainder
Properties and Applications of Modular Arithmetic
Preserves basic arithmetic properties (commutative, associative, distributive)
Useful in computer science for hashing algorithms and cryptography
Simplifies calculations with large numbers by working with smaller remainders
Modular exponentiation computed efficiently using repeated squaring method
Congruence classes group all integers with same remainder for given modulus
Modular arithmetic applies in calendar systems, digital clocks, and computer memory addressing
Multiplicative Inverses in Modular Arithmetic
Multiplicative inverse of a modulo m exists if a and m are coprime
Denoted as $a^{-1}$, satisfies equation a⋅a−1≡1(modm)
Computed using extended Euclidean algorithm
Not all numbers have multiplicative inverses in modular arithmetic
Used in cryptography for generating public and private keys
Existence of multiplicative inverse ensures unique solutions to linear congruences
Modular Exponentiation and Theorems
Efficient Modular Exponentiation Techniques
Modular exponentiation calculates $a^b \bmod m$ efficiently
Naive approach of repeated multiplication becomes impractical for large exponents
Square-and-multiply algorithm reduces number of multiplications required
Binary exponentiation method uses binary representation of exponent
Modular exponentiation crucial in RSA encryption and other cryptographic systems
Time complexity of efficient algorithms typically O(log n) where n is the exponent
Fermat's Little Theorem and Its Applications
Fermat's Little Theorem states if p is prime and a is not divisible by p, then ap−1≡1(modp)
Provides efficient method for primality testing (probabilistic)
Used in cryptographic algorithms like RSA for key generation
Generalizes to Euler's theorem for composite moduli
Helps in calculating modular inverses when modulus is prime
Foundational in understanding more advanced number theory concepts
Euler's Totient Function and Related Concepts
Euler's totient function φ(n) counts numbers coprime to n less than n
For prime p, φ(p) = p - 1
Multiplicative function, meaning φ(ab) = φ(a)φ(b) if a and b are coprime
Euler's theorem states if a and n are coprime, then aφ(n)≡1(modn)
Generalizes Fermat's Little Theorem to composite moduli
Critical in RSA cryptosystem for determining key sizes and encryption exponents
Chinese Remainder Theorem
Fundamentals of the Chinese Remainder Theorem
Chinese Remainder Theorem (CRT) solves system of simultaneous linear congruences
Applies when moduli are pairwise coprime
Provides unique solution modulo product of all moduli
Originated in ancient Chinese mathematics, rediscovered and formalized in modern times
Efficient for solving large modular equations by breaking them into smaller, manageable parts
Used in various areas including cryptography, coding theory, and computer algebra systems
Solving Simultaneous Congruences with CRT
System of congruences in form: x ≡ a₁ (mod m₁), x ≡ a₂ (mod m₂), ..., x ≡ aₖ (mod mₖ)
Calculate M = m₁ m₂ ... mₖ
For each i, compute Mᵢ = M / mᵢ
Find modular inverses yᵢ such that Mᵢyᵢ ≡ 1 (mod mᵢ)
Solution given by x ≡ (a₁M₁y₁ + a₂M₂y₂ + ... + aₖMₖyₖ) (mod M)
Verify solution by checking if it satisfies all original congruences
Applications and Extensions of Modular Equations
Modular equations arise in various mathematical and practical contexts
Linear congruences form basis for more complex modular equations
Quadratic residues and quadratic reciprocity extend modular arithmetic to square roots
Discrete logarithm problem fundamental to many cryptographic systems (ElGamal, Diffie-Hellman)
Modular equations used in error-correcting codes for digital communication
Solving systems of modular equations important in scheduling problems and computer science algorithms |
1575 | https://mitocw.ups.edu.ec/high-school/physics/exam-prep/work-energy-power/conservation-of-energy | Conservation of Energy | MIT OpenCourseWare | Free Online Course Materials
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Momentum and kinetic energy of a baseball bat; simple harmonic motion of two mass-spring systems. Solution not included.
8.01L Physics I: Classical Mechanics, Fall 2005
Dr. George Stephans
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Circular Motion and Conservation of Mechanical Energy
Motion of an object along a frictionless loop-the-loop track.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Beads Sliding Along a Ring
Forces acting on a suspended ring with sliding beads.
8.01T Physics I, Fall 2004
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Kinetic Energy and Speed
Linear speed of a streetcar on a circular track. Solution not included.
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Complete practice problem 12
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Kinetic Energy of the Earth
Conversion of the kinetic energy of the earth-sun system to potential energy. Solution not included.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Complete practice problem 19
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Ball and Spring
Work done on a ball launched vertically by a spring. Solution not included.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Complete practice problem 20
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Cart on an Inclined Track
Velocity and acceleration of a cart compressing a spring on an inclined track. Solution not included.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Complete practice problem 4
Complete practice problem 5
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Experiment 6 Pre-Lab
Collision of a cart with a fixed spring on a track.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Experiment 6 Analysis
Fitting data from Experiment 6.
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Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Block Hanging From Two Ropes
4-part problem; free-body diagram and tension while hanging; speed and tension after one rope is cut. Solutions are included after problems.
8.01 Physics I, Fall 2003
Prof. Stanley Kowalski
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Spring-Launched Ball
Finding the spring constant of a spring from the maximum height of a ball shot by the spring.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Spring and Loop
Motion of a mass propelled by a spring along a frictionless loop.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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Inclined Plane, Friction, Spring
Motion of a mass that slides down an inclined plane and compresses a spring.
8.01T Physics I, Fall 2004
Dr. Peter Dourmashkin, Prof. J. David Litster, Prof. David Pritchard, Prof. Bernd Surrow
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1576 | https://www.youtube.com/watch?v=QKuGbZph8Ok | E Z Geometric Isomers for Alkenes
Leah4sci
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Posted: 8 Oct 2018
Presents: E Z Isomers for Alkenes
Need help with Orgo? Download my free guide ’10 Secrets to Acing Organic Chemistry’ HERE:
In this video:
[0:37] When to use E/Z naming system
[2:39] Steps to ranking substituents
[7:23] Explanation of sample problem#2
[10:12] Sample problem with two unique substituents
[11:12] More Cis/Trans Practice
This video thoroughly explains E and Z geometric isomerism for alkenes. See step by step what qualifies as a E or Z or neither! Each practice problem includes how to properly name each molecule
Links & Resources Mentioned In This Video:
Cis Trans Tutorial:
Naming Alkenes:
Cahn-Ingold-Prelog Ranking in Stereochemistry Video Series:
Periodic table:
Follow along with your semester by using my Orgo Syllabus Companion:
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182 comments
Transcript:
Intro Leah here from leah4sci.com and in this video
we’re going to look at E and Z Isomers for alkenes. In the last video, we looked at cis
and trans isomers, if you haven't seen that yet, click the link below or visit leah4sci.com/cistrans And were left off with a question of what
happens when you're trying to find cis or trans for a pi bond where the substituents
are set up in a way where they could be cis to each other or they could be trans to each
other. And when that's the case you need a higher level of ranking and this is where
the E and Z configurations come in. You typically use E and Z when you have more than one carbon
coming out of the same sp2 carbon atom. For example, here we have an sp2 carbon with both
a methyl and an ethyl so we can't rank based on just one. We also use it when we have none
carbon substituents like halogens, alcohols or larger priority groups. For example if
you have to name a molecule that look like this we would have to find a way to imply
that the chlorine is on the same side as the methyl where as the methyl on the left is
opposite to methyl on the right. The first Ranking thing we have to do is rank the substituents
coming off of that carbon atom, using the Cahn - Ingold - Prelog system. This is something
I teach in detail in the stereochemistry series which you can find the linked below or visit
leah4sci.com/ranking Now we skimp on this, it's very important.
As a quick review, we are ranking based on atomic number, now you'll typically have a
periodic table on your quiz or exam but this is where those ten atoms that I have you memorized
in the beginning of orgo, this is another place where they really come in handy so if
you haven't memorize that yet, do so now. We have hydrogen, then carbon, nitrogen, oxygen
fluorine. Below that we have phosphorus, sulfur, chlorine, and then bromine and iodine. These
ten atoms are the most common that you're going to see especially in a topic like and
E and Z. As a reminder, Iodine is your highest priority, then bromine, then chlorine, sulfur,
and phosphorus, then fluorine, oxygen, nitrogen, carbon, and hydrogen is always your lowest.
If there is a deuterium thrown in, remember that it outranks hydrogen because it's just
a heavy isotope of hydrogen. If you're comfortable with ranking lets move one. Let's try the
chlorine example and if you haven't memorize chlorine example the atoms yet we'll keep it on the side for
you. Here's the method that I find to be simplest and least overwhelming, first, identify and
highlight your pi bond, make it very obvious which two carbons you're looking at and then
cover half of the molecule. This is so important because when you have a lot of substituents,
it's very easy to get overwhelmed and it's very easy to forget what you're looking at
but if you cover it by putting your hand or a piece of paper or something over half of
the molecule, you're only focusing on one sp2, one pi bond carbon at a time and keeping
it simple. Next, you want to identify the atoms that you're ranking specifically, in
this case we have a chlorine versus a carbon and that methyl group, and we see that chlorine
outranks carbon that makes chlorine priority number one and carbon priority number two.
Finally, draw an arrow perpendicular to the pi bond in the direction of your higher priority,
in this case my pi bond goes in this direction so I will draw an arrow facing up towards
the chlorine. We have one side, now we cover it so we're not distracted and do the same
thing on the other side. Here's my sp2 carbon but I only have one substituent, what am I
ranking? Don't forget that in line structure, hydrogen atoms are invisible but they're still
there. We have a hydrogen atom on this carbon of the pi bond and that means we're ultimately
comparing carbon and hydrogen. Carbon will outrank hydrogen, making the methyl group
priority number one, and the hydrogen priority number two. Since the higher priority is facing
up from the pi bond, we'll draw an arrow perpendicular to the pi bond facing priority number one
and then uncover both sides. Having the arrows allows you to avoid the groups, it allows
you to avoid being overwhelmed and all you have to do is identify, are your arrows facing
in the same direction as we have here, or do you have one arrow facing up and one down
in opposite directions. Now be careful, don't call them cis trans, this is not cis trans,
we're simply looking to see if the two higher priority groups are facing in the same or
opposite direction. Z configuration If you identify that your two arrows are facing
in the same direction, this would be the z configuration, and the way I remember that
is by thinking that the two higher priority groups are located on "ze zame zide"(the same
side) , all spelled with the Z, "ze zame zide" tells me that it's the Z configuration. If
the two high priority groups are located opposite to each other, you get the E configuration
and the way I remember it is that these two groups are located "Eeposite" (opposite) from
each other, opposite spelled with an E to remind me they are on "Eeposite" sides of
each other. Let's go back to our example, in this case, we know that the chlorine and
methyl, both higher priority are facing "ze zame zide", that makes this the Z configuration,
to name this molecule, we start with the basic naming rules where we see a total of four
carbons in the parent chain, for a first name of "but", and the pi bond sitting in the middle
of the molecule. I will number this molecule from the left so that chlorine and the pi
bond are both on number two rather than three, remember we want the total lowest set numbers
which gives me a last name of "2-ene" and then a chlorine substituent on carbon two,
for "2-chloro", and lastly, we want to include the E Z configuration. In this case, we have
Z. To put the name together, we start with the Z at the start of the name, and put it
into parenthesis followed by the substituent which is a chlorine on carbon two, we drag
the two in front of the "but", giving me "2-butene". The reason we put the Z in parenthesis is
to make sure that we don't confuse the letter Z with the number two, look at how I've drawn
them, they kind of look similar but having the Z in parenthesis reminds us "no, this
is not a designation of where something shows up, this is Z, telling me about the stereochemistry
or the geometry about that pi bond." For our final name of "(Z) - 2 - chloro - 2 - butene".
Let's go back and name our first example, Example we have a total of six bonds, giving me a
first name of hex, we have an option to number it from the right or left, if I numbered it
from the left, I hit the pi bond to three and the methyl group at four. If I number
from the right, I hit the pi bond and the methyl group at three for a total lower set
of numbers so we're gonna choose the red numbering, with a pi bond starting at carbon 3, we have
"3-ene", a methyl group at carbon 3 gives me "3-methyl", but now we have to specify
what's going on with the pi bond. We don't use cis trans because we have a methyl and
an ethyl, instead we'll use the E Z system. We'll start with the group on the left, identifying
the sp2 carbon and recognizing because the substituent is invisible, there is a hydrogen
atom. If you don't have them memorized yet, use this chart as a guide we're comparing
carbon to hydrogen, carbon outranks hydrogen, making the ethyl group priority number one,
and the hydrogen priority number two, or simply the higher priority in the up direction. Now
cover the left side and look at the sp2 carbon on the right, coming off of this carbon, we
have a methyl and an ethyl, methyl is a ch3 that's carbon, ethyl is a ch2 ch3 that's also
carbon, this is where it really helps to understand priority ranking. As a reminder we're dealing
with methyl, which is ch3 and ethyl which is ch2 ch3. We look at the very first atom
which is carbon, since it's the same, we rule it out and go to the next highest priority.
The methyl group has three hydrogens, so any one hydrogen is the next high priority. The
ethyl group has two hydrogens and one carbon but carbon is the highest priority, so now
we compare carbon to hydrogen, carbon wins, making the ethyl group priority group number
one and the methyl priority number two. We'll draw an arrow in the direction of number one,
and then uncover the molecule. Looking at the two arrows, they are away from each other
or "Eeposite" from each other, telling me that this molecule is in the E configuration.
Let's put the name together starting with E in parenthesis, followed by the substituent
"3-methyl". Pull the 3 in front of hex, giving me "3-hexene". For our final name of "(E)
- 3 methyl - 3 - hexene". Your professor may choose to leave the three where it is, in
that case the name would be "(E) - 3 methyl hex - 3 - ene", both are correct. What about
a molecule that looks like this? We have a pi bond with two unique substituents, if we
look at the sp2 carbon on the left, chlorine outranks ethyl, making the chlorine higher
priority, but on the right we don’t appear to have anything. Meaning, we have two hidden
hydrogen atoms, but once again like we saw on the last video, hydrogen is the same as
hydrogen. If you have two equal priority groups, one doesn't outrank the other and that means
there is no stereoisomerism here, because there is no way to differentiate the top or
the bottom, so this is not going to be cis trans and it's also not going to be E or Z
because of that hydrogen. The trick to remember? Terminal alkenes without anything fancy on
the end are not cis trans and not E Z. For even more examples on cis trans and E
Z, click the link below or visit my website, leah4sci.com/cistrans Are you struggling with organic chemistry?
Are you looking for resources and information to guide you through the course and help you
succeed? If so, then I have a deal for you, a free copy of my Ebook "10 Secrets to Acing
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1577 | https://www.dummies.com/article/academics-the-arts/math/basic-math/pre-algebra-practice-questions-solving-simple-algebraic-equations-249996/ | Book & Article Categories
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Solving Simple Equations in Pre-Algebra Problems
Practice questions
1. 18 – x = 12
2. 4x = 44
In the following questions, use the correct inverse operation to rewrite and solve each problem.
3. 100 – x = 58
4. 238/x = 17
In the following questions, find the value of x in each equation by guessing and checking.
5. 12x – 17 = 151
6. 19x – 8 = 600
Answers and explanations
so x = 14.
12(10) – 17 = 120 – 17 = 103
103 is less than 151, so this guess is too low. Try x = 20:
12(20) – 17 = 240 – 17 = 223
223 is greater than 151, so this guess is too high. Therefore, x is between 10 and 20. Try x = 15:
12(15) – 17 = 180 – 17 = 163
163 is a little greater than 151, so this guess is a little too high. Try x = 14:
12(14) – 17 = 168 – 17 = 151
151 is correct, so x = 14.
19(10) – 8 = 190 – 8 = 182
182 is much less than 600, so this guess is much too low. Try x = 30:
19(30) – 8 = 570 – 8 = 562
562 is still less than 600, so this guess is still too low. Try x = 35:
19(35) – 8 = 665 – 8 = 657
657 is greater than 600, so this guess is too high. Therefore, x is between 30 and 35. Try x = 32:
19(32) – 8 = 608 – 8 = 600
600 is correct, so x = 32.
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1578 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2964183/ | Nuclear Lamins - PMC
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Cold Spring Harb Perspect Biol
. 2010 Nov;2(11):a000547. doi: 10.1101/cshperspect.a000547
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Nuclear Lamins
Thomas Dechat
Thomas Dechat
1 Department of Cell and Molecular Biology, Feinberg School of Medicine, Northwestern University, Chicago, Illinois 60611
Find articles by Thomas Dechat
1, Stephen A Adam
Stephen A Adam
1 Department of Cell and Molecular Biology, Feinberg School of Medicine, Northwestern University, Chicago, Illinois 60611
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1, Pekka Taimen
Pekka Taimen
2 Department of Pathology, University of Turku, Kiinamyllynkatu 10 20520 Turku, Finland
Find articles by Pekka Taimen
2, Takeshi Shimi
Takeshi Shimi
1 Department of Cell and Molecular Biology, Feinberg School of Medicine, Northwestern University, Chicago, Illinois 60611
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1, Robert D Goldman
Robert D Goldman
1 Department of Cell and Molecular Biology, Feinberg School of Medicine, Northwestern University, Chicago, Illinois 60611
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1
Author information
Copyright and License information
1 Department of Cell and Molecular Biology, Feinberg School of Medicine, Northwestern University, Chicago, Illinois 60611
2 Department of Pathology, University of Turku, Kiinamyllynkatu 10 20520 Turku, Finland
✉
Correspondence: r-goldman@northwestern.edu
Copyright © 2010 Cold Spring Harbor Laboratory Press; all rights reserved
PMC Copyright notice
PMCID: PMC2964183 PMID: 20826548
Abstract
The nuclear lamins are type V intermediate filament proteins that are critically important for the structural properties of the nucleus. In addition, they are involved in the regulation of numerous nuclear processes, including DNA replication, transcription and chromatin organization. The developmentally regulated expression of lamins suggests that they are involved in cellular differentiation. Their assembly dynamic properties throughout the cell cycle, particularly in mitosis, are influenced by posttranslational modifications. Lamins may regulate nuclear functions by direct interactions with chromatin and determining the spatial organization of chromosomes within the nuclear space. They may also regulate chromatin functions by interacting with factors that epigenetically modify the chromatin or directly regulate replication or transcription.
Intermediate filaments form a polymeric mesh beneath the nuclear membrane that controls the shape and stability of the nucleus but also regulates chromatin positioning and gene expression.
A filamentous layer located between the inner nuclear membrane (INM) and peripheral heterochromatin was evident even in early electron microscopic studies of vertebrate cell nuclei (Fawcett 1966). This layer, later termed the nuclear lamina, is also found to be closely associated with nuclear pore complexes (NPCs) and contains three major structurally related polypeptides (Aaronson and Blobel 1975). These proteins are named nuclear lamins A, B, and C according to their molecular weights (Gerace and Blobel 1980). Further biochemical characterization and cDNA cloning of the nuclear lamins classifies them as type V intermediate filament proteins (Goldman et al. 1986; McKeon et al. 1986).
STRUCTURE AND BIOCHEMICAL PROPERTIES OF NUCLEAR LAMINS
Lamin Isoforms and Expression Patterns
Lamins are present in all metazoans examined to date ranging from hydra to human, but are not found in unicellular organisms and plants (Cohen et al. 2001; Melcer et al. 2007). Extensive characterization in several model organisms including humans, mice, frogs, fruit flies and nematodes shows that their properties are shared across species (Melcer et al. 2007; Dechat et al. 2008b). Based on their sequence homologies, expression patterns, structural features, and biochemical and dynamic properties, lamins are subdivided into A- and B-types. All metazoans express at least one B-type lamin. Typically, invertebrates have only a single lamin gene of the B-type, with some exceptions such as Drosophila, which expresses one B-type (lamin Dm0) and one A-type lamin (lamin C) encoded by two distinct genes. Most vertebrates have one A-type lamin and two B-type lamin genes except for Xenopus, which has three B-type genes (Melcer et al. 2007).
The two major isoforms of vertebrate A-type lamins, lamins A and C, are derived from a single gene (LMNA) by alternative splicing (see Fig. 1) (Lin and Worman 1993). In some vertebrates, alternative splicing can also produce two less abundant isoforms, lamins AΔ10 and C2, from LMNA (Nakajima and Abe 1995; Machiels et al. 1996). Lamins B1 and B2 are the two major B-type lamins in most vertebrates. They are encoded by the LMNB1 and LMNB2 genes, respectively (Peter et al. 1989; Vorburger et al. 1989). The latter also encodes the minor isoform lamin B3 (Furukawa and Hotta 1993). Although at least one B-type lamin is expressed in all cells throughout development, the expression of A-type lamins is developmentally regulated (Benavente et al. 1985; Schatten et al. 1985; Lehner et al. 1987). During mouse development, lamins A and C are not expressed until days 10–12 of mouse embryogenesis and then primarily in primordial muscle cells (Stewart and Burke 1987; Rober et al. 1989). Lamin A/C expression in other organs does not occur until after birth (Rober et al. 1989). Cells of hematopoietic lineage express only B-type lamins (Guilly et al. 1990; Rober et al. 1990). Similar patterns of expression of A and B-type lamins take place during the developmental progression of other vertebrates (Benavente et al. 1985; Lehner et al. 1987; Prather et al. 1989) and Drosophila (Frasch et al. 1988; Riemer et al. 1995). The regulated expression of A- and B-type lamins is also evident during differentiation of stem cells in culture. For example, undifferentiated human and mouse embryonic stem (ES) cells lack lamins A and C, but express lamins B1 and B2 (Constantinescu et al. 2006). The minor mammalian isoforms, lamins C2 and B3, are expressed exclusively in germ cells (Furukawa and Hotta 1993; Machiels et al. 1995; Nakajima and Abe 1995), whereas small amounts of lamin AΔ10 appear to be present in a variety of cell types (Machiels et al. 1995).
Figure 1.
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Structure of nuclear lamins. Schematic drawing of mature lamin A and lamin C polypeptide chains. The lamin structure consists of a short amino terminal head domain, a central α-helical rod domain (red), and the carboxy-terminal domain containing the NLS and the Ig-fold (blue; the nine β-strands of the Ig-fold motif are depicted). Modified, with permission, from (Dechat et al. 2008b).
The importance of the developmental regulation of lamin expression is evident from studies in LMNA knockout (LMNA−/−) and LMNB1 mutant (LMNB Δ/Δ) mice. In mice null for the A-type lamins, no obvious embryonic defects can be detected, but these animals show severe postnatal growth retardation and muscular dystrophy (Sullivan et al. 1999). On the other hand, mice with an insertional mutation in LMNB1 develop defects in their lungs and bones during embryogenesis and die at birth, even though they continue to express lamin B2 (Vergnes et al. 2004).
Structure and Assembly of the Nuclear Lamins
The nuclear lamins have the typical tripartite structure of intermediate filament (IF) proteins, consisting of a highly α-helical central rod domain flanked by a short globular amino-terminal “head” domain and a longer carboxy-terminal “tail” domain (Parry et al. 1986). The central rod domain is composed of four subhelical regions comprised of heptad repeats and designated as coil 1A, 1B, 2A, and 2B. These individual “coils” are separated from each other by three short linker segments, L1, L12, and L2, of which L12 is the most flexible (Parry et al. 1986). Although the head domain appears to be unstructured, the tail domain contains a highly conserved structural motif similar to a type s immunoglobulin fold (Ig-fold) (Dhe-Paganon et al. 2002; Krimm et al. 2002). A nuclear localization signal (NLS), required for transport into the nucleus, is present in all lamins between the carboxy-terminal end of the central rod domain and the Ig-fold (Loewinger and McKeon 1988) (Fig. 1).
In vitro, lamins self-assemble into higher order structures through a series of steps. The first step involves the formation of a coiled-coil dimer by the in register and in parallel association of two α-helical rod domains into a left handed superhelix (Heitlinger et al. 1991) (Fig. 2A). Next, the lamin dimers associate in a head-to-tail fashion (Heitlinger et al. 1991; Stuurman et al. 1996) and these polarized arrays interact in an antiparallel fashion to form apolar tetrameric protofilaments (Heitlinger et al. 1991) (Fig. 2B). The interaction of four lamin protofilaments leads to the formation of ∼10-nm filaments (Heitlinger et al. 1991; Stuurman et al. 1996; Ben-Harush et al. 2009) (Fig. 2C,D). Although the assembly of some invertebrate lamins appears to terminate at this stage of the assembly process, vertebrate lamins continue to assemble into well ordered paracrystalline arrays in vitro (Zackroff and Goldman 1979; Heitlinger et al. 1991; Stuurman et al. 1996; Ben-Harush et al. 2009) (Fig. 2E). The assembly of lamin proteins differs from their cytoplasmic counterparts in several ways. For example, lamin dimers associate by polar “head-to-tail” interactions to form protofilaments, which is very different from the half-staggered, antiparallel side-by-side association of cytoplasmic IF dimers, which assemble into 10 nm filaments (Strelkov et al. 2004). This difference is at least in part explained by the crystal structure of the coiled-coil dimer of the lamin A coil 2B region (Strelkov et al. 2004). Although this structure appears to be similar to the overall structure of the homologous segment of the cytoplasmic IF protein vimentin, the distribution of charged residues varies causing significant changes in the patterns of intra- and interhelical salt bridges. Additionally, vertebrate lamins have six extra heptads in the central rod domain (McKeon et al. 1986), which may also help to explain assembly differences between cytoplasmic IF and lamins. The central rod domains of some invertebrates lamins contain fewer extra heptads and these lamins stop assembling into higher order structures at the 10-nm (IF) stage of assembly, similar to cytoplasmic IF proteins (Ben-Harush et al. 2009).
Figure 2.
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Assembly of the nuclear lamins in vitro. Lamins self assemble to form dimers (A) which then join to form linear head-to-tail polymers (protofilaments) (B). Bar = 100 nm; electron micrograph of rotary shadowed chicken lamin B2. These protofilaments further assemble into “beaded” filaments or fibers (C) which in turn associate laterally into thicker fibers (D), and eventually into paracrystalline arrays (E); C,D are negatively stained electron microscope preparations. Reprinted from Stuurman et al. (1998) with permission from Elsevier.
Posttranslational Processing and Modifications of the Nuclear Lamins
Lamins A, B1, and B2 are expressed as prelamins that require extensive posttranslational modifications of their carboxy-terminal –CAAX box to become mature lamins (Rusinol and Sinensky 2006; Davies et al. 2009). Modification of the –CAAX box takes place in a highly regulated temporal sequence starting with the farnesylation of the cysteine residue (Farnsworth et al. 1990; Lutz et al. 1992) by a farnesyltransferase (Zhang and Casey 1996). This modification initiates a sequence of processing steps (Beck et al. 1990) beginning with the removal of the –AAX by a CAAX prenyl protease. Two members of this zinc metalloproteinase family have been identified in humans: Rce1 (Ras-converting enzyme 1) and Zmpste24 (Zinc metalloprotease related to the STE24 homolog in yeast), also known as FACE1 (farnesylated-proteins converting enzyme)(Boyartchuk et al. 1997; Leung et al. 2001; Corrigan et al. 2005); followed by carboxymethylation of the carboxy-terminal cysteine by isoprenylcysteine carboxyl methyltransferase (Icmt) (Winter-Vann and Casey 2005). In contrast to B-type lamins, which remain permanently farnesylated and carboxymethylated, an additional 15 amino acids are removed from the carboxyl terminus of farnesylated/carboxymethylated prelamin A by Zmpste24/FACE1 (Corrigan et al. 2005). This final processing step results in the production of mature lamin A lacking the carboxy-terminal farnesyl- and carboxymethyl-modifications (Fig. 3). This sequence of lamin processing steps is highly regulated and each step depends on the preceding modification (Kilic et al. 1997). Lamin C, which is 74 residues shorter than mature lamin A, does not possess a –CAAX box and therefore is not farnesylated or otherwise modified (see Fig. 1). The precise location of the intracellular post-translational processing sites for the lamins remains largely unknown. Although the enzymes required for processing the carboxyl terminus of the lamins are present in both the inner nuclear membrane and endoplasmic reticulum membranes, the complete processing of prelamin A can occur when the protein is confined either to the cytoplasm or to the nucleus (Barrowman et al. 2008). Even though the prelamins can be processed in the cytoplasm, it is more likely that all lamin processing occurs in the nucleus (Lutz et al. 1992), because lamins are rapidly transported into the nucleus after translation in the cytoplasm (Lehner et al. 1986).
Figure 3.
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Posttranslational processing of the carboxyl terminus of prelamins A, B1, and B2. Processing takes place in a series of steps: (1) addition of a farnesyl group to the cysteine residue of the –CAAX box of pre-lamin A, prelamin B1 and prelamin B2 by a farnesyltransferase; (2) removal of the last three residues (−AAX) by an AAX endopeptidase; (3) methylation of the terminal carboxylic acid group (−COOH) by a carboxyl methyltransferase; (4) removal of the carboxyl terminal 15 amino acids of lamin A with the farnesyl attached by the metalloprotease Zmpste24/FACE1. This last proteolysis step does not occur on B-type lamins and therefore they remain farnesylated. Modified, with permission, from Dechat et al. (2008b).
In addition to farnesylation and carboxymethylation, lamins are also posttranslationally modified by phosphorylation (Ottaviano and Gerace 1985), sumoylation (Zhang and Sarge 2008), ADP-ribosylation (Adolph 1987), and possibly by glycosylation (Ferraro et al. 1989). Phosphorylation of serine and threonine residues proximal to the NLS by protein kinase C is known to play a role in the regulated import of lamins into the nucleus (Hennekes et al. 1993; Leukel and Jost 1995). At the onset of mitosis, the phosphorylation of lamins at specific sites by cyclin-dependent kinase (Cdk) 1 and protein kinase C (PKC) is required to drive disassembly of the lamina (Gerace and Blobel 1980; Dessev and Goldman 1988; Dessev et al. 1988; Dessev et al. 1989; Heald and McKeon 1990; Peter et al. 1990; Ward and Kirschner 1990; Dessev et al. 1991; Molloy and Little 1992; Goss et al. 1994; Collas 1999). Subsequently, dephosphorylation of the mitotic sites by protein phosphatase 1a is required for lamin/lamina assembly during the telophase/early G1 transition (Thompson et al. 1997).
Lamin Structure and Dynamic Properties within the Lamina and Nucleoplasm
Very little is known about the formation, composition, and structure of lamin polymers in vivo. The structures of the lamins within the lamina that have been described to date range from a regular meshwork of ∼10–15 nm filaments observed in Xenopus oocyte germinal vesicles and in sperm pronuclei assembled in vitro (Aebi et al. 1986; Zhang and Casey 1996; Goldberg et al. 2008a) (Fig. 4A), to a more irregular filamentous meshwork seen in mammalian cells (Belmont et al. 1993; Schermelleh et al. 2008; Shimi et al. 2008) (Fig. 4B,C). These meshworks have long been assumed to be stable structures based on the resistance of the nuclear lamins within the lamina to extraction in detergent containing high salt solutions, and the propensity of purified lamins to polymerize into higher order insoluble structures in vitro at relatively low critical concentrations (Aebi et al. 1986; Lourim and Lin 1989; Glass and Gerace 1990). In further support of this assumption, measurements of lamin mobility by fluorescence recovery after photobleaching (FRAP) reveal that GFP-tagged A- and B-type lamins do not appreciably exchange subunits within the lamina for up to several hours throughout most of interphase (Broers et al. 1999; Moir et al. 2000; Dahl et al. 2006).
Figure 4.
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The nuclear lamins form a meshwork of filaments within the lamina. (A) Spread nuclear envelope from Xenopus oocytes after detergent extraction and preparation for transmission electron microscopy by freeze-drying/unidirectional metal shadowing. The micrograph shows the nuclear lamina meshwork partially studded with nuclear pore complexes. (Inset) Higher-magnification view of a particularly well-preserved area clearly shows the near-tetragonal lamina meshwork. Bars, 1 µm. Reprinted from Stuurman et al. (1998), with permission from Elsevier. (B) Structured illumination microsopy (SIM) reveals that there is an irregular meshwork of nuclear lamin B as revealed by immunofluorescence (green). This preparation is also stained with antibodies directed against nuclear pores and is stained with DAPI for DNA/chromatin. Pores (red), DAPI (blue). From Schermelleh et al. (2008). Reprinted with permission from AAAS. (C) Confocal immunofluorescence localization of lamin A/C (green) and lamin B1 (red) in HeLa cells. Lamins A/C and B1 in a single nucleus are seen in an equatorial section (left panels) and the nuclear surface (right panels). The areas indicated by white squares in the top panels are enlarged fivefold in the lower panels. These images demonstrate that lamins form mainly separate networks with some overlapping regions. Bar, 5 µm. Adapted, with permission, from Shimi et al. (2008).
Recently it has become apparent that the A- and B-type lamins form separate filamentous networks in the lamina. These structures can be identified by both high resolution light microscopy (Schermelleh et al. 2008; Shimi et al. 2008) and whole mount electron microscopy in lamin B containing Xenopus oocyte nuclei ectopically expressing A-type lamins (Goldberg et al. 2008b). Even though the A-type and B-type lamin networks appear to be mainly separate structures (Shimi et al. 2008), there is evidence that the individual networks overlap and interact to varying degrees. Evidence for interaction between these two lamin meshworks is based on studies using fluorescence resonance energy transfer (FRET) in combination with time domain fluorescence lifetime imaging and high resolution confocal immunofluorescence (Moir et al. 2000; Delbarre et al. 2006; Shimi et al. 2008).
Other evidence supporting the interactions between the A and B type lamins is derived from studies of nuclei in which either A or B type lamins are perturbed. Fibroblasts derived from LMNA−/− mouse embryos (MEFs) or a limb girdle muscular dystrophy patient with a homozygous nonsense mutation in LMNA contain abnormally shaped nuclei containing blebs or lobules (Sullivan et al. 1999; Muchir et al. 2003). A significant number of these blebbed regions lack B-type lamins and NPCs. In addition, nuclei in LMNB Δ/Δ MEFs are also highly lobulated, showing a dramatic increase in the mesh size of the lamin A/C network (Vergnes et al. 2004). HeLa cells depleted of lamin B1 by shRNA silencing frequently develop extensively enlarged lamin A/C meshworks and nuclear envelope blebs which lack lamin B2 and NPCs (Shimi et al. 2008). Many of these nuclear phenotypes are also found in cells endogenously or ectopically expressing either point mutations or truncations of lamins A/C (Ostlund et al. 2001; Vaughan et al. 2001; Vigouroux et al. 2001; Bechert et al. 2003; Favreau et al. 2003; Muchir et al. 2003; Goldman et al. 2004). It is also noteworthy that lamin A is more mobile in the nuclear lamina in nuclei with decreased amounts of lamin B1 (Tang et al. 2008).
The A- and B-type lamins have different disassembly and assembly properties during mitosis as revealed by biochemical fractionation, immunofluorescence, and live cell imaging analysis, further supporting the idea that the two types of lamins form separate networks in the lamina. As described earlier, depolymerization of the lamins is regulated by mitotic kinases and reassembly requires dephosphorylation by protein phosphatase 1a (Thompson et al. 1997; Ito et al. 2007). When the nuclear envelope is disassembled during late prophase, A-type lamins become dispersed throughout the cytoplasm in an apparently freely diffusible state whereas the majority of B-type lamins remain associated with the nuclear membranes, which appear to be mainly dispersed into the endoplasmic reticulum (Gerace and Blobel 1980; Stick et al. 1988). This association of B-type lamins with membranes during mitosis appears to be mainly attributable to their permanently farnesylated state (Rusinol and Sinensky 2006).
Further support for differences in the properties of the A- and B-type lamins is derived from studies revealing their spatial and temporal order of assembly into the nuclear lamina at anaphase/telophase and in the early stages of G1. For example, in mouse keratinocytes and hamster cells, lamin B1 accumulates along the entire periphery of the decondensing daughter chromosomes and assembles into a relatively stable polymer by mid- to late- telophase (Broers et al. 1999; Moir et al. 2000). In these cells, lamin A begins to accumulate within the nucleus mainly after the major components of the nuclear envelope including NPCs are assembled in daughter cells in late telophase. In HeLa cells, however, a small fraction of A-type lamins associates with chromosomes much earlier than in mouse cells, first assembling in the central or “core” region of chromosomes in close association with kinetochores (Dechat et al. 2004; Dechat et al. 2007; Haraguchi et al. 2008) (Fig. 5). The remainder of the A-type lamins is then transported into the nucleus after formation of an intact nuclear envelope. Later, other elements of the nuclear envelope begin to assemble in daughter cells (Dechat et al. 2004; Dechat et al. 2007; Haraguchi et al. 2008).
Figure 5.
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Association of lamin A with chromosomes during mitosis. HeLa cells expressing GFP-lamin A were followed by time-lapse microscopy from the metaphase/anaphase transition (far left panels) into early G 1 (far right panels). GFP-lamin A first associates with the core regions of chromosomes during telophase and spreads to cover the entire chromatin surface by early G1. DIC images of the same series are shown in the bottom row (Dechat et al. 2007).
Although the major fraction of the various lamin isoforms is associated with the nuclear lamina, these proteins are also present throughout the nucleoplasm, especially in interphase (Goldman et al. 1992; Lutz et al. 1992). These nucleoplasmic lamins are likely to have different functions, for example in DNA replication and transcription (Dechat et al. 2008b), and a small fraction may represent assembly intermediates that are subsequently incorporated into the nuclear lamina (Goldman et al. 1992). Interestingly, B-type lamins in the nucleoplasm appear to be relatively static, similar to those in the lamina, whereas the nucleoplasmic A-type lamins are much more dynamic. These differences in the properties of nucleoplasmic A- and B-type lamins are supported experimentally by their biochemical extractability and their dynamic properties as determined in vivo by FRAP analyses and fluorescence correlation spectroscopy (FCS) (Broers et al. 1999; Moir et al. 2000; Shimi et al. 2008). For example, FCS measurements of GFP-tagged lamins A/C and lamins B1/B2 showed that nucleoplasmic lamins A/C are highly mobile, whereas lamin B1 and lamin B2 are mainly immobile (Shimi et al. 2008). This latter observation suggests that nucleoplasmic B-type lamins are either assembled into some type of structure or are tightly associated with other unknown immobile structural components. Interestingly, decreasing the amount of nucleoplasmic lamin B1 by shRNA silencing increases the rate of mobility for a large fraction of nucleoplasmic lamin A (Shimi et al. 2008). Taken together, these FCS results suggest that, as in the case of the lamina, the A- and B-type lamins form separate yet interacting structures within the nucleoplasm. Obviously, it will be of great interest in the future to determine the specific functions of these nucleoplasmic lamins.
FUNCTIONS OF THE LAMINS
Regulation of Nuclear Shape and Mechanical Stability
It is becoming increasingly obvious that some of the functions of nuclear lamins are analogous to those of cytoskeletal intermediate filament proteins, which are known to be involved in the determination and maintenance of cell shape and mechanical properties (Goldman et al. 2008). In support of this, the nuclei of LMNA−/− MEFs display increased deformability and impaired viability under mechanical strain compared to the nuclei in control MEFs (Houben et al. 2007). Increased nuclear deformability is also observed in human ES cells lacking A-type lamins as compared to nuclei in differentiated cells expressing the A-type lamins (Pajerowski et al. 2007). In addition, cells either deficient in lamins or expressing mutant lamin proteins often contain misshapen nuclei (Ostlund et al. 2001; Vaughan et al. 2001; Vigouroux et al. 2001; Bechert et al. 2003; Favreau et al. 2003; Muchir et al. 2003; Goldman et al. 2004). Quantitative rheological measurements, particle tracking methods and differential interference contrast microscopy reveal that in vitro reconstituted lamin B1 networks are extremely porous, with elastic stiffness providing resistance to shear deformations (Panorchan et al. 2004a; Panorchan et al. 2004b). Interestingly, measurements made by atomic force microscopy show that germinal vesicles isolated from Xenopus oocytes, which normally contain only B-type lamins, display a significant increase in stiffness on the ectopic expression of lamin A (Schape et al. 2009).
Nuclear Lamins and the Regulation of Chromatin Positioning and Gene Expression
Lamins may regulate transcription by organizing chromatin into active and inactive domains. For example, electron microscopic and light microscopic observations show that there is a close association between peripherally localized heterochromatin and the nuclear lamina (Fawcett 1966; Paddy et al. 1990) (Fig. 6). This suggests that lamins may be involved in anchoring or organizing interphase chromosomes (Sullivan et al. 1999; Goldman et al. 2004; Nikolova et al. 2004; Galiova et al. 2008). Additional support for this idea comes from biochemical experiments showing that lamins A/C bind to mitotic chromosomes and to polynucleosomal particles in vitro (Burke 1990; Glass and Gerace 1990; Yuan et al. 1991). The interaction of lamins with chromatin could be mediated either by their direct binding to histones (Taniura et al. 1995; Goldberg et al. 1999; Mattout et al. 2007), or to specific DNA sequences in the matrix attachment/scaffold-associated regions (MARs/SARs) (Luderus et al. 1992; Luderus et al. 1994; Baricheva et al. 1996; Zhao et al. 1996).
Figure 6.
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The nuclear lamina is a fibrous network (fibrous lamina) located between peripheral heterochromatin and the inner nuclear membrane. A transmission electron micrograph of a thin section showing a portion of a smooth muscle nucleus in guinea pig epedidymis. Adapted from Fawcett (1966), copyright 1995, Wiley-Liss, Inc. Reprinted with permission of John Wiley & Sons, Inc. American Journal of Anatomy, Vol. 119, No. 1, pg. 140.
The Role of Lamins in Mitosis
In addition to nuclear disassembly, several lines of evidence suggest a direct role for lamins in nuclear assembly following mitosis. The microinjection of lamin antibodies into mitotic cells causes arrest in a telophase-like state with the chromosomes remaining condensed (Benavente and Krohne 1986). In addition, decreased expression of lamin Dm0 in Drosophila inhibits nuclear membrane assembly and causes an enrichment of NPCs in cytoplasmic annulate lamellae (Lenz-Bohme et al. 1997). In Caenorhabditis elegans, the down-regulation of the single lamin leads to a loss of chromosomes, defects in chromosome separation into daughter cell nuclei, and abnormal condensation of chromatin (Liu et al. 2000). In vitro nuclear assembly in frog egg extracts can be inhibited by addition of the C-terminal tail fragment of lamin B3, in part because of the Ig-fold motif in the lamin tail, which can inhibit lamin polymerization in vitro (Shumaker et al. 2005). Lastly, the expression of the mutant lamin A which causes Hutchinson-Gilford Progeria Syndrome (HGPS) leads to defects in cell division including a delay in cytokinesis, delayed completion of nuclear reassembly at the end of mitosis and an increase in mitotic cells (Cao et al. 2007; Dechat et al. 2007).
B-type lamins also play a role in the formation of the mitotic spindle. In cultured cells, a diffuse localization of B-type lamins can be observed associated with the mitotic spindle (Georgatos et al. 1997; Maison et al. 1997; Moir et al. 2000; Beaudouin et al. 2002; Tsai et al. 2006). In frog egg extracts, a mitotic spindle can be induced to form around added DNA or on beads coated with a variety of factors important for mitotic spindle formation (Tsai and Zheng 2005). In these lysates, lamin B3, the major lamin in frog eggs, is required to organize a “spindle matrix” structure. This matrix contains membrane vesicles and proteins known to regulate the assembly and dynamic properties of the microtubules comprising the mitotic spindle (Tsai et al. 2006). Depletion of lamin B3 from the egg extracts with antibodies or the addition of dominant negative fragments of lamin B3 known to disrupt lamin assembly, also disrupt the formation of the mitotic spindle. Furthermore, lamin B3 is associated with several spindle assembly factors within the spindle matrix such as NuMA and Nudel and it appears to play an important role in regulating the activity of these proteins during mitosis (Tsai et al. 2006; Ma et al. 2009).
The Role of Lamins in DNA Replication and Repair
Several lines of evidence suggest that lamins play a role in DNA replication. In cultured cells, sites of DNA replication can be visualized as discrete early or late replication foci in the nucleoplasm. These foci can be detected by incorporation of bromodeoxyuridine into sites of replication (Moir et al. 1994) or by immunolocalization of replication-associated proteins such as proliferating cell nuclear antigen (PCNA) (Shumaker et al. 2008). It has been shown that lamin B1 colocalizes with these replication foci during late S phase in mouse 3T3 cells (Moir et al. 1994) and that lamins A/C are present at sites of early replication in normal human fibroblasts (Kennedy et al. 2000). Furthermore, DNA replication in Xenopus nuclei assembled in vitro is inhibited by depletion of lamin B3 or by the addition of a dominant-negative fragment of lamin B3, which drives the disassembly of the endogenous lamin network (Lopez-Soler et al. 2001; Shumaker et al. 2005; Shumaker et al. 2008). There is evidence that the lamins play a direct role in regulating replication by their binding to DNA replication factors such as PCNA. The lamin binding site for PCNA resides within the highly conserved Ig-fold motif located in the carboxyl terminus of both the A- and B-type lamins (Shumaker et al. 2008). In addition, there is a close association between lamin B3 and PCNA at the surface of sperm head chromatin during the earliest stages of nuclear assembly in Xenopus interphase egg extracts (Shumaker et al. 2008).
Nuclear lamins have also been implicated in DNA repair, although the mechanistic basis of this remains unclear. For example, the expression of disease-causing mutant lamins impairs the formation of DNA repair foci (Liu et al. 2005; Manju et al. 2006). Furthermore, genetic instability because of defects in telomere function and DNA repair has been implicated in progeria, the premature aging syndrome (Gonzalez-Suarez et al. 2009a; Gonzalez-Suarez et al. 2009b).
The Role of Lamins in Transcription
Several lines of experimental evidence support the possibility that lamins play a role in transcription. An amino-terminally deleted dominant negative lamin A, which disassembles lamin networks, specifically inhibits RNA pol II activity in hamster cells or in nuclei isolated from Xenopus embryos (Spann et al. 2002). In addition, over-expression of lamins A/C or silencing of lamin B1 leads to a significant inhibition of pol II transcription in HeLa cells (Kumaran and Spector 2008; Shimi et al. 2008; Tang et al. 2008).
Lamins also associate with several transcription factors leading to the suggestion that they are involved in specific regulatory pathways (Heessen and Fornerod 2007; Andres and Gonzalez 2009). For example, the association of Oct-1 with lamin B1 at the nuclear envelope appears to be important for the oxidative stress response in MEFs, as lamin B1 deficiency leads to a dysregulation of Oct-1-dependent genes and to an increase in reactive oxygen species (Malhas et al. 2009). Lamins A/C also interact with the transcription factors c-Fos, MOK2, and sterol response element binding protein 1 (SREBP1) (Dreuillet et al. 2002; Lloyd et al. 2002; Ivorra et al. 2006; Dreuillet et al. 2008; Harper et al. 2009). In the case of c-Fos, its interaction with lamins A/C at the nuclear envelope appear to suppress AP-1 (activating protein 1) binding to DNA and transcriptional activity in an extracellular signal-regulated kinase (ERK) 1/2 activity-dependent fashion (Ivorra et al. 2006; Gonzalez et al. 2008). In addition to regulating transcription factor function by direct interaction, lamins are also associated with transcription factors indirectly via several lamin-binding proteins including emerin, LAP2β and pRb. Although the evidence supporting a role for the lamins in regulating transcription is convincing, it remains unclear whether this regulation involves direct or indirect binding to transcription factors.
Lamins as Regulators of the Positioning and Organization of Chromatin
The positioning of genes near or in contact with the nuclear lamina may be a mechanism for modulating gene expression. Inactive genes and gene-poor chromatin are frequently found in association with the lamina region, which suggests that the nuclear lamina may be a transcriptionally silent microdomain (Boyle et al. 2001; Cremer et al. 2001). Furthermore, lamins may play a role in establishing this microdomain by directly interacting with chromatin. The interactions of lamins with chromatin/DNA at the nuclear lamina can be identified in more detail by DamID labeling (Pickersgill et al. 2006). In this technique, an expression vector encoding a potential chromatin/DNA binding protein fused to the Escherichia coli enzyme DNA adenine methyl transferase (Dam) is expressed in cells. DNA methylated by the lamin Dam fusion protein is then analyzed by sequencing. In Drosophila Kc cells, lamin D m0 seems to mainly associate with transcriptionally inactive, mid-to-late replicating genome regions, lacking active histone marks and enriched in large intergenic regions (Pickersgill et al. 2006). In human lung fibroblasts lamin B1-associated domains (LADs) are present along chromosomes in a distinct pattern interspersed with “lamin B1 poor” regions (Guelen et al. 2008). These LADs are ≥1 Mb in size, are mostly heterochromatic. Genes present within LADs are 5–10-fold less active than genes outside LADs.
The lamina appears to have gene or chromatin-specific effects on activation or repression of transcription. Evidence for the lamina constituting a repressive environment for transcription comes from studies localizing inactive gene loci to the nuclear periphery (Kosak et al. 2002; Zink et al. 2004; Williams et al. 2006). For example, the inactive Ig-heavy and Ig-κ loci are preferentially positioned at the nuclear periphery in non-B-cell lineages, but are centrally located in B-cell nuclei where they are actively engaged in transcription (Kosak et al. 2002). In addition, transcriptionally inactive testis-specific gene clusters are frequently associated with lamin Dm0 in the lamina of Drosophila S2 cells. Upon down-regulation of lamin Dm0, these gene clusters relocate from the lamina to the nuclear interior and become transcriptionally active (Shevelyov et al. 2009). Artificial tethering of a reporter gene to the nuclear envelope using the membrane spanning domain of the inner nuclear membrane protein emerin results in the transcriptional inactivation of the gene in mouse cells (Reddy et al. 2008). However, the relocalization of chromosomes from the nuclear interior to the nuclear periphery using the inner membrane protein Lap2β inactivates some genes, but not others (Finlan et al. 2008). The idea that the nuclear lamina is not repressive to all transcription is further supported by the finding that targeting a genetic locus to the nuclear lamina by artificial tethering to lamin B1 does not interfere with its transcriptional activation (Kumaran and Spector 2008). In addition, lamin A can act as a transcriptional repressor in mammalian cells and in yeast when artificially targeted to specific promoters (Lee et al. 2009). This latter finding is intriguing because yeast do not contain lamin genes. These somewhat conflicting results demonstrate that more research is required to determine the role of lamins in transcription both in the region of the lamina and throughout the nucleoplasm.
Lamins are Involved in the Epigenetic Regulation of Chromatin
The dramatic loss of peripheral heterochromatin in the nuclei of cells expressing one of the mutant forms of lamin A, progerin, that causes the premature aging disorder HGPS emphasizes the important role of lamins in the modification and/or the organization of chromatin (Shumaker et al. 2006) (Fig. 7). A partial loss of peripheral heterochromatin is also seen in LMNA−/− MEFs (Sullivan et al. 1999; Nikolova et al. 2004). Furthermore, down-regulation of lamin B1 in HeLa cells results in lamin A/C-rich and lamin B2 deficient nuclear blebs or microdomains characterized by the absence of heterochromatin (Shimi et al. 2008). These changes in chromatin organization related to lamin expression are reflected in alterations in histone modifications including reductions in trimethylated H3K9, trimethylated H3K27 and an increase in trimethylation of H4K20 in HGPS cells (Scaffidi and Misteli 2006; Shumaker et al. 2006; Shimi et al. 2008) (Fig. 7). These studies demonstrate that the interaction of chromatin with lamins, whether direct or indirect, has strong effects on the epigenetic modification of histones, and that lamin-associated microdomains in the nucleus might regulate modification of chromatin (Pegoraro et al. 2009). One potential link between lamins and chromatin modification may involve the tumor suppressor ING1. This protein, which interacts with histone acetyltransferases and histone deacetylases, appears to be stabilized and targeted to the nucleus by its interaction with A-type lamins (Han et al. 2008). HGPS cells have reduced ING1 levels and the expression of ING1 protein lacking its lamin interacting domain causes a loss of peripheral heterochromatin in normal cells.
Figure 7.
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The association of the nuclear lamina with chromatin in normal and progeria patients' cells. (A) Thin section electron micrograph of a late passage blebbed nucleus in an HGPS patient's skin fibroblast (left and center panels) and normal human foreskin fibroblasts (right panel). A high-magnification view of the nuclear envelope in a normal human foreskin fibroblast shows a normal array of heterochromatin adjacent to the nuclear envelope, making any lamina structure difficult to detect (right panel). A higher-magnification view of a HGPS cell showing a loss of peripheral heterochromatin and a prominent electron-dense lamina region associated with the inner nuclear envelope membrane (the nucleus is to the left in center and right panels) (Goldman et al. 2004). (B) Alterations of histone methylation patterns in HGPS fibroblasts. Normal and HGPS fibroblasts from female donors were double-labeled with antibodies against lamins A/C (red) and trimethylated Lys 9 in histone H3 (H3K9me3), Lys 27 in histone H3 (H3K27me3), or Lys 20 in histone H4 (H4K20me3) (all green). Note the decrease of H3K9me3 and H3K27me3 and the increase of H4K20me3 in the lobulated HGPS nuclei compared with normal nuclei. The decrease in H3K27me3 is best observed at the inactive X chromosome, which is normally enriched in this histone modification (see arrowhead in center left panel). Bars, 10 µm. Reprinted, with permission, from Dechat et al. (2008b).
Lamins and Interphase Chromosome Organization
The organization of chromosome territories and domains is influenced by the expression of lamins. For example, in LMNB Δ/Δ MEFs and in cells derived from patients suffering from some types of laminopathies, gene-poor chromosome 18 is positioned away from its normal location at the nuclear periphery toward the nuclear interior (Malhas et al. 2007; Meaburn et al. 2007). Furthermore, in HeLa cells silenced for the expression of lamin B1, the lamin A/C rich and lamin B2-deficient nuclear envelope blebs that form are associated predominantly with gene-rich chromosome regions (Shimi et al. 2008) (Fig. 8). Together, these findings suggest that A-type lamins are preferentially associated with gene-rich chromatin regions and B-type lamins are preferentially associated with gene-poor regions of chromosomes. It must be emphasized that these studies are of a preliminary nature and that much more work is required to establish the precise relationships between the different types of lamins and gene rich/gene poor chromosome domains within the interphase nucleus.
Figure 8.
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Localization of specific gene rich chromosomal regions in nuclear blebs in LB1-silenced HeLa cells. Chromosomes are detected by fluorescence in situ hybridization. DNA is counterstained with Hoechst 33258 (blue). Chromosome 6p (or 19), and chromosome 6q (or 18) are shown in green and red, respectively. Bars, 5 µm. Adapted, with permission, from Shimi et al. (2008).
Lamins may also play a role in the localization and function of centromeres and telomeres in the nucleus. Centromeres are typically arranged near the nuclear periphery in an interphase nucleus (Solovei et al. 2004). However, the lamin A/C-rich, lamin B2-deficient nuclear envelope blebs that form in lamin B1-silenced cells are devoid of centromeres, suggesting that B-type lamins are involved in anchoring these heterochromatic structures at the nuclear lamina (Shimi et al. 2008). Telomeres are also influenced by the composition and structure of the lamina, because in LMNA−/− MEFs, the distribution, length and structure of telomeres is altered leading to increased genomic instability compared to control MEFs (Gonzalez-Suarez et al. 2009a). Human fibroblasts ectopically expressing HGPS lamin A mutant proteins show rapid telomere shortening and an accelerated replicative senescence phenotype (McClintock et al. 2006; Huang et al. 2008). Furthermore, the proliferative defects seen in normal dermal fibroblasts expressing progerin can be overcome by expression of the catalytic subunit of telomerase or the inactivation of p53 (Kudlow et al. 2008). Other support for the involvement of lamins in telomere function comes from the finding that the D4Z4 human subtelomeric repeat can localize an adjacent telomere to the nuclear periphery probably by interacting with A-type lamins and CTCF (CCCTC-binding factor) (Ottaviani et al. 2009a; Ottaviani et al. 2009b). The D4Z4 repeat acts as a CTCF and A-type lamin-dependent transcriptional insulator, suggesting that nuclear lamin architecture may organize specific regions of chromatin and influence gene expression (Ottaviani et al. 2009a).
Strong support for the role of the lamins in interphase chromosome organization comes from a study of cells obtained from patients with a form of atypical HGPS (Taimen et al. 2009b). Unlike the most common form of HGPS, which results in only defective lamin A, these patients have a mutation resulting in the expression of a point mutation, E145K, in the 1B segment of the central rod domain of both lamins A and C (see Fig. 1). The cells from these patients have a flower-shaped nucleus with centrally clustered centromeres and abnormal peripherally displaced telomeres (Fig. 9). This abnormal configuration of chromosomes requires a round of cell division and persists throughout interphase. The multilobulated nuclei are therefore unlikely to correctly establish normal interphase chromosome territory arrangements. Both the abnormal nuclear shapes and chromosome configurations in E145K cells appear to be due to the aberrant assembly of the nuclear lamina by the mutant lamin A. In vitro analysis of E145K lamin A polymerization shows that this mutant lamin forms disorganized higher-order structures, a defect not seen in the more common form of progeria resulting from a truncation near the carboxyl terminus (Taimen et al. 2009b).
Figure 9.
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Mislocalization of heterochromatin and centromeres in HGPS E145K cells. Control and E145K patient cells were stained with anti-lamin A (red), CREST antiserum (green), and Hoechst (white). Maximum projections of series of z-sections spanning the entire nucleus and side projections are shown. Centromeres are clustered in the central region of the nucleus in E145K cells, while they are either closely associated with the peripheral lamina region or elsewhere in the nucleoplasm in control cells. Inset shows an example of the close association of one centromere with the lamina in a single confocal section. In the right hand panels, side projections from merged images are shown. Centromeres are distributed throughout the nuclei in normal cells. In E145K cells centromeres are clustered in the middle of one region of the nucleus (Taimen et al. 2009a).
Lamins in Cell Proliferation and Differentiation
Deficiencies in A-type lamins or the expression of mutant A-type lamins both lead to an array of proliferative defects because of cell cycle acceleration, cell cycle arrest and/or premature senescence (Verstraeten et al. 2007; Dechat et al. 2008a). There is evidence that lamins A/C are involved in the regulation of the pRb/E2F pathway responsible for the G1/S cell cycle transition. One current model suggests that pRb in its suppressive hypophosphorylated state is in a complex with lamins A/C and LAP2α, and that the loss of lamins A/C leads to the degradation of pRb (Ozaki et al. 1994; Markiewicz et al. 2002; Johnson et al. 2004). In addition to pRb, cyclin D3, another critical regulator of G1 progression, interacts directly with lamins A/C (Mariappan et al. 2007).
The findings that lamins are involved in the regulation of cell cycle progression and that they are differentially expressed during development, suggest that they also function in the regulation of cell differentiation. In support of this, during myoblast differentiation there is a decrease in the soluble nucleoplasmic lamin A/C pool, most likely in a pRb and cyclin D3-dependent manner (Muralikrishna et al. 2001; Mariappan and Parnaik 2005). Furthermore, the down-regulation of lamins A/C or the expression of lamin A/C mutants associated with Autosomal Dominant Emery Dreifuss Muscular Dystrophy (AD-EDMD) lead to impaired myoblast differentiation possibly by disrupting the Rb-MyoD pathway (Bakay et al. 2006; Frock et al. 2006). As further evidence for the importance of the lamin A/pRb interaction, the expression of either an AD-EDMD lamin A mutant protein or the HGPS lamin A mutant protein, progerin, results in the inhibition of the phosphorylation of pRb (Markiewicz et al. 2005; Dechat et al. 2007). In the case of the AD-EDMD mutation, a further decrease in pRb levels along with the failure to hyperphosphorylate the available pRB inhibits myoblast differentiation (Favreau et al. 2004; Kandert et al. 2009). In adipocyte differentiation, the overexpression of either wild-type lamin A or a lamin A mutant associated with familial partial lipodystrophy impairs the potential of mouse fibroblasts to differentiate into mature adipocytes; and LMNA−/− mouse embryonic fibroblasts differentiate more readily into fat-containing cells compared to control cells (Boguslavsky et al. 2006). There is also evidence that lamins A/C regulate adipocyte differentiation in a complex with the inner nuclear membrane protein emerin, by influencing the nucleocytoplasmic distribution of β-catenin (Tilgner et al. 2009). In addition, lamins A/C appear to be involved in osteoblast differentiation, as silencing their expression causes impaired osteoblastogenesis and accelerated osteoclastogenesis in human bone marrow stromal cells (Akter et al. 2009; Rauner et al. 2009). Lamin A may also be involved in adult stem cell differentiation as the expression of progerin causes defects mouse stem cell populations (Espada et al. 2008) in the differentiation potential of human mesenchymal stem cells, probably by affecting the Notch-signaling pathway (Espada et al. 2008; Scaffidi and Misteli 2008).
Another aspect of the involvement of the lamins in differentiation is related to aging. In this regard, progerin can be detected in cells from healthy individuals and the protein appears to accumulate with increasing age (McClintock et al. 2006; Scaffidi and Misteli 2006; Cao et al. 2007; Rodriguez et al. 2009). Furthermore, dermal fibroblasts from older individuals (Scaffidi and Misteli 2006), and cells in aged C. elegans (Haithcock et al. 2005) display changes in nuclear shape similar to those found in premature aging.
The Lamins are Connected to the Cytoskeleton
There is evidence that the nuclear lamins are connected to the cytoskeleton through a complex of proteins called the LINC complex (linker of nucleoskeleton and cytoskeleton) (Crisp et al. 2006; Ketema et al. 2007; Stewart-Hutchinson et al. 2008; Burke and Roux 2009). In this complex the integral proteins of the inner nuclear membrane, Sun1 and Sun2, interact with the outer nuclear membrane proteins, nesprin-1, nesprin-2, and nesprin-3α,in the luminal space between the inner and outer nuclear membranes. Although the Sun proteins bind to the lamins (Crisp et al. 2006), the nesprins are associated with both the microfilament and intermediate filament cytoskeletons via direct actin binding and plectin (Tzur et al. 2006). Another lamin binding protein of the inner nuclear membrane, emerin, binds directly to nesprin isoforms, providing additional linkages between the lamins and the cytoskeleton (Wheeler et al. 2007; Zhang et al. 2007). These interactions between the nuclear lamina and the cytoskeleton may have important functional significance. For example, nuclei in fibroblasts deficient in lamin B1 rotate at a higher frequency than nuclei in control cells (Ji et al. 2007). LMNA−/− MEFs, on the other hand, are reported to show a decrease in mechanical stiffness, defects in mechanotransduction resulting in impaired strain-induced signaling, defects in cell polarization and cell migration, and a disturbed organization of microfilaments, vimentin IF networks and microtubules (Houben et al. 2007; Lee et al. 2007; Dahl et al. 2008; Houben et al. 2009).
CONCLUSIONS AND FUTURE PERSPECTIVES
Over the past decade it has become clear that the nuclear lamins are one of the key players in determining nuclear architecture and function. There is growing evidence that these Type V IF proteins provide a filamentous scaffold built from interconnecting A- and B-type lamin networks that pervade the entire nucleus. This scaffold not only determines the shape and mechanical properties of the nucleus, but also serves as a docking site for chromatin and for numerous proteins involved in chromatin organization and various nuclear functions. Besides the extensive lamin structures located within the lamina, smaller and more dynamic lamin polymers appear to be components of large protein complexes known to be involved in a wide range of nuclear housekeeping functions such as DNA replication, DNA repair and RNA PolII transcription.
We have learned much about the nuclear lamins from the analysis of the changes in nuclear form and function that take place in cells from patients suffering from the remarkably large number of diseases attributable to hundreds of mutations in the LMNA gene. However, the precise roles that normal lamins play in nuclear functions remain largely unknown. To gain more insights into their specific functions we need a better understanding of their normal structure at the highest levels of resolution. To date only a few subdomains of the lamins have been shown to be amenable to analysis by X-ray diffraction, because of the well known difficulties inherent in crystallizing IF proteins. Other challenges for future studies include determining the steps in lamin polymerization and depolymerization in vivo and determining how complex networks of the A- and B-type lamins are assembled and interact with each other both at the nuclear periphery and within the nucleoplasm throughout the cell cycle. Finally, it will be of great interest to determine the different roles of the A- and B-type lamins in determining the overall architecture of the nucleus and the mechanisms involved in their linkages to other nuclear structures, such as the nuclear membrane and pore complexes.
ACKNOWLEDGMENTS
Work in the Goldman lab on nuclear lamins is funded by the National Institute on Aging and the National Cancer Institute. PT received a fellowship from Sigrid Juselius Foundation, Orion-Farmos Research Foundation, Cancer Society of Southwestern Finland, and Finnish Cultural Foundation.
Footnotes
Editors: Tom Misteli and David L. Spector
Additional Perspectives on The Nucleus available at www.cshperspectives.org
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Abstract
STRUCTURE AND BIOCHEMICAL PROPERTIES OF NUCLEAR LAMINS
FUNCTIONS OF THE LAMINS
CONCLUSIONS AND FUTURE PERSPECTIVES
ACKNOWLEDGMENTS
Footnotes
REFERENCES
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Group 18 Elements - Characteristics of Noble Gases
Last Updated : 29 Dec, 2021
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The group's members have eight electrons in their outermost orbit (except helium which has two electrons). As a result, they have a stable configuration. Group 18 elements are gases that are chemically unreactive, meaning they do not form many compounds. Be a result, the elements are referred to as inert gases. Noble gas elements, like the other group elements, display trends in their physical and chemical properties. The noble gas family's general configuration is ns 2 np 6 (except helium which has 1s 2).
When the group's members were found and named, they were assumed to be extremely rare as well as chemically inert, and hence were dubbed the rare or inert gases. However, it is now recognised that several of these elements are relatively abundant on Earth and across the cosmos, thus the term rare is deceptive. Similarly, the term inert has the disadvantage of connoting chemical passivity, implying that compounds of Group 18 cannot be produced. The term noble has long been used in chemistry and alchemy to describe the resistance of metals such as gold and platinum to undergo a chemical reaction.
Noble (Inert) Gases or Group 18 Elements
Noble or inert gases are elements in Group 18. They are named inert because they do not participate in any chemical reaction, hence they are chemically inert.
Helium, Neon, Argon, Krypton, Xenon, and Radon are all non-metallic elements of group 18.
In the periodic table, the zero group holds an intermediate position between the strong electronegative elements of the VIIA and strong electropositive elements of the IA groups, acting as a bridge. The noble gases are found in group 18 of the periodic table, which is located on the far right of the table. The 18th group members all have 8 electrons in their outermost shell.
Trends of Noble Gases
Noble gas electronic configuration: Members of group 18 have eight valence electrons, which means they have eight electrons in their outermost orbit (except helium). As a result, they have a consistent octet configuration. Helium, on the other hand, has a duplet structure. The noble gas family's general configuration is ns 2 np 6 (except helium which has 1s 2).
Noble gas atomic radii: Members of group 18 have extremely small atomic radii. The atomic radii of noble gases grow down the group as the atomic number increases due to the addition of additional shells.
Noble gas ionisation enthalpy: Noble gases have eight valence electrons, which means they have eight electrons in their outermost orbit (except helium). As a result, they have a stable octet or duplet configuration. As a result, the elements of group 18 have extremely high ionisation enthalpies. Because of the increase in atomic size, the ionisation enthalpy of noble gases drops along with the group.
Properties of Noble Gases
Except for helium, all gases have the ns 2, np 6 arrangement. Except for helium, the differentiating electron reaches the p-subshell, and hence all of these atoms are p-block elements.
Except for helium, which has a 1s 2 structure and a totally filled 1st shell, all noble gases have an outermost shell with a complete octet.
Because of their totally filled outer shell or stable structure, these elements, also known as noble gases, have no tendency to lose or gain electrons and so do not participate in chemical processes under normal conditions.
All of these group 18 elements are gases at normal temperature and pressure conditions. Rn and Og are radioactive noble gases, and the rest are present in trace amounts in the atmosphere.
Because of their insignificant presence in the atmosphere, these gases were given the label rare gases.
Because of their insignificant presence in the atmosphere, these gases were given the label rare gases.
Because they do not participate in any chemical reactions, they have been given the name inert gases.
However, in addition to these noble gases, a number of xenon compounds and two Krypton fluorides were created.
They are all monoatomic gases.
Physical Characteristics of Noble Gases
These gases have no colour, taste, or odour.
The Van der Waals forces between these elements' particles are weak, but they become stronger as we progress down the group. This is due to an increase in the polarising capacity of the molecules.
Because of their stable nature, these are monoatomic gases in their free state.
Noble gases have low boiling and melting points. The low melting and boiling points of these gases are due to their weak Van der Waals force. These, however, rise as we progress along with the group.
At extremely low temperatures, certain elements can be condensed. The ease of liquefaction rises down the group as the atom size increases.
These are only marginally soluble in water. As we move down the group, solubility increases.
The noble gases have the biggest atomic radii in their respective eras.
The ionisation enthalpies of these noble gases are the highest in their respective periods because the electronic configuration is stable.
Chemical Characteristics of Noble Gases
This octet of electrons was assumed to be the most stable arrangement for the outermost shell of an atom in a chemical bonding theory established in 1916 by American chemist Gilbert N. Lewis and German scientist Walther Kossel. Although only noble-gas atoms had this arrangement, it was the state toward which all other element atoms tended in their chemical bonding.
Certain elements satisfied this tendency by obtaining or losing electrons outright, resulting in the formation of ions; other elements shared electrons, generating stable combinations connected together by covalent bonds. The proportions in which atoms of elements bonded to create ionic or covalent compounds (their "valences") were thus governed by the behaviour of their outermost electrons, which were dubbed "valence electrons" for this reason. This hypothesis described the chemical bonding of the reactive elements, as well as the relative inactivity of the noble gases, which came to be considered as their defining chemical property.
Sample Questions
Question 1: Group 18 Elements are not reactive? Why?
Answer:
The noble or inert gases are the group 18 elements. As the name implies, these are inert because they are chemically inert or non-reactive. Because of their totally filled outer shell, they have a stable electrical structure, which means they have no inclination to lose or gain electrons. Because these atoms have entire valence electron shells, the noble gases are exceedingly stable. These are highly unresponsive. Under typical conditions, they do not react or participate in any chemical reactions; nevertheless, there are some exceptions.
Question 2: Name the noble gases.
Answer:
Helium, argon, xenon, radon, neon, and krypton are examples of noble gases.
Question 3: How is the reactivity of noble gases under ordinary conditions?
Answer:
Under normal conditions, noble gases have no tendency to absorb or lose electrons. This is the only reason they are inert and do not participate in chemical reactions. Modern researchers have discovered that under certain conditions, noble gases can be induced to participate in a chemical reaction.
Question 4: What are noble gases why are they also called inert gases?
Answer:
Noble gases have no tendency to absorb or lose electrons under normal conditions. The only reason they are inert and do not participate in chemical reactions is because of this. Noble gases can be made to participate in a chemical process under particular conditions, according to modern researchers.
Question 5: Give reason why helium does not react with other elements.
Answer:
Helium has two electrons in its last orbital. It has a consistent electronic configuration. As a result, helium does not react with other elements.
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1580 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8807479/ | Functions and Regulation of Translation Elongation Factors - PMC
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Front Mol Biosci
. 2022 Jan 19;8:816398. doi: 10.3389/fmolb.2021.816398
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Functions and Regulation of Translation Elongation Factors
Benjin Xu
Benjin Xu
1 Department of Medical Laboratory Science, Fenyang College, Shanxi Medical University, Fenyang, China
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1,,†, Ling Liu
Ling Liu
1 Department of Medical Laboratory Science, Fenyang College, Shanxi Medical University, Fenyang, China
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1,†, Guangtao Song
Guangtao Song
2 Institute of Biophysics, Chinese Academy of Sciences, Beijing, China
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2,
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1 Department of Medical Laboratory Science, Fenyang College, Shanxi Medical University, Fenyang, China
2 Institute of Biophysics, Chinese Academy of Sciences, Beijing, China
Edited by:Taku Kaitsuka, International University of Health and Welfare (IUHW), Japan
Reviewed by:Ping Xie, Institute of Physics (CAS), China
Patrick O’Donoghue, Western University, Canada
✉
Correspondence: Benjin Xu, benjin_ibprnalab@sina.com; Guangtao Song, gsong@moon.ibp.ac.cn
This article was submitted to RNA Networks and Biology, a section of the journal Frontiers in Molecular Biosciences
†
These authors have contributed equally to this work
Received 2021 Nov 16; Accepted 2021 Dec 20; Collection date 2021.
Copyright © 2022 Xu, Liu and Song.
This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
PMC Copyright notice
PMCID: PMC8807479 PMID: 35127825
Abstract
Translation elongation is a key step of protein synthesis, during which the nascent polypeptide chain extends by one amino acid residue during one elongation cycle. More and more data revealed that the elongation is a key regulatory node for translational control in health and disease. During elongation, elongation factor Tu (EF-Tu, eEF1A in eukaryotes) is used to deliver aminoacyl-tRNA (aa-tRNA) to the A-site of the ribosome, and elongation factor G (EF-G, EF2 in eukaryotes and archaea) is used to facilitate the translocation of the tRNA 2-mRNA complex on the ribosome. Other elongation factors, such as EF-Ts/eEF1B, EF-P/eIF5A, EF4, eEF3, SelB/EFsec, TetO/Tet(M), RelA and BipA, have been found to affect the overall rate of elongation. Here, we made a systematic review on the canonical and non-canonical functions and regulation of these elongation factors. In particular, we discussed the close link between translational factors and human diseases, and clarified how post-translational modifications control the activity of translational factors in tumors.
Keywords: translation, elongation factors, regulation, expression, tumorigenesis
Introduction
With the development of structural biology, especially the rapid development of cryo-electron microscopy (cryo-EM) (Benjin and Ling, 2019), the mechanism of intracellular protein translation and its regulation have been gradually clarified. As described by the central dogma, translation is the final stage of gene expression, during which the genetic information carried by an mRNA is transformed into the amino acid sequence of a protein catalyzed by a ribosome (Voorhees and Ramakrishnan, 2013). Translation is a highly dynamic and cyclic process, which is composed of four steps: initiation, elongation, termination, and ribosome recycling (Figure 1). During translation initiation, the ribosome, with an initiator fMet-tRNA fMet in the P-site, is assembled on mRNA with the assistance of the three initiation factors (IF1-3). The ribosome complex is then ready to accept the first elongator tRNA and form the first peptide bond, which marks the beginning of the next stage, elongation. During elongation, an aa-tRNA enters the ribosome A-site with the help of EF-Tu. If proper base-pairing between the three bases of the mRNA codon and those of the aa-tRNA anticodon is established, the aa-tRNA is cognate, a peptide bond is formed with the peptide attached to the tRNA in the P-site. The peptidyl-tRNA is then moved from the A-to the P-site, and the deacylated tRNA in the P-site is moved to the E-site. The mRNA is coordinately translocated by one codon. Termination occurs when the ribosome reaches a terminator codon in an mRNA. The nascent peptide is hydrolyzed by the release factors (RF1/2) and dissociated from the ribosome. In the end, the ribosome is split into two subunits by the concerted action of EF-G and RRF, releasing the deacylated tRNA and mRNA, and preparing for a new round of translation initiation. This step is called ribosome recycling.
FIGURE 1.
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Overview of bacterial translation cycle. aa-tRNA, aminoacyl-tRNA; EF, elongation factor; IF, initiation factor; RF, release factor; RRF, ribosome recycling factor.
Translation elongation is a process of repeated decoding, peptidyl transfer and tRNA 2-mRNA translocation. It starts with the binding of the second aminoacyl-tRNA at the A-site. During elongation, an aa-tRNA is delivered to the ribosome as a ternary complex (TC) with elongation factor Tu (EF-Tu) and GTP (EF-Tu∙GTP∙aa-tRNA). Binding of cognate aa-tRNA to mRNA in the A site of the ribosome induces the crucial and generally conserved bases A1493, A1492 and G530 to flip out and interact with the minor groove of the mRNA-tRNA duplex (Schmeing and Ramakrishnan, 2009), which further induces a domain closure in the 30S subunit (Ogle et al., 2002). 30S domain closure makes the 30S shoulder towards the ternary complex (Ogle and Ramakrishnan, 2005), leading to the stimulation of the GTP hydrolysis by EF-Tu and acceleration of tRNA selection. After decoding, EF-Tu dissociates from ribosome in the form of EF-Tu∙GDP, followed by the complete accommodation of aa-tRNA into the A-site. EF-Tu∙GDP is recycled to EF-Tu∙GTP by EF-Ts, a guanosine nucleotide-exchange factor, so as to participate in multiple rounds of peptide chain elongation. The next step is peptidyl transfer and peptide bond formation, which is catalyzed by the peptidyl transfer center (PTC) of the ribosome. During this stage, the nucleophilic α-amino group of the aa-tRNA in the A-site attacks the carbonyl carbon of the peptidyl-tRNA in the P-site, yielding a pre-translocation (PRE) ribosome complex with a deacylated tRNA in the P-site and a new, one residue longer peptidyl-tRNA in the A-site (Beringer and Rodnina, 2007). The third stage is EF-G∙GTP (eEF2∙GTP in eukaryotes) catalyzed translocation. Upon the addition of EF-G into the pre-translocational ribosome (PRE) system, EF-G in complexed with GTP facilitates movements of peptidyl-tRNA on the 50S subunit, and shifts the classical pre-translocation state to the hybrid state (Holtkamp et al., 2014a; Xie, 2016). And GTP hydrolysis induces a strong conformational change of 30S subunit within the 70S ribosome, allowing the movement of tRNA 2-mRNA by one codon length inside the ribosome, shifting the tRNAs from A- and P-sites to P- and E-sites, respectively, resulting in a post-translocational ribosome (POST) (Figure 2).
FIGURE 2.
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Schematic of the bacterial elongation cycle. EF-Tu delivers aa-tRNA to the A-site of the ribosome, where the ribosome decodes for the cognate tRNA. After aa-tRNA is fully accommodated and peptide bond formation, EF-G facilitates translocation of the tRNA 2-mRNA duplex, then the next round of elongation begins.
There is a complex set of protein factors including EF-Tu/eEF1A, EF-G/eEF2, EF-P/eIF5A, and SelB/EFsec (Table 1), involved in translation elongation. Although the fundamental mechanism of the elongation cycle is very similar among three domains of life, the molecular mechanism of the elongation factors varies in different species. In the following, we will introduce in more detail the functions and regulation of the most important and well-studied canonical and non-canonical translation elongation factors that function in different stages of the elongation cycles of bacteria, archaea, and eukaryotes as well as that of organelles, including mitochondria.
TABLE1.
Translation elongation factors among bacteria, archaea, eukaryotes and mitochondria.
| Species | aa-tRNA incorporation | tRNA translocation | Other translation elongation factors |
:---: :---: |
| Bacteria | EF-Tu | EF-Ts | SelB | EF-G | EF4 | EF-P | — | Tet(O)/(M) | RelA | BipA |
| Archaea | aEF1A | — | — | aEF2 | — | aIF5A | — | — | — | — |
| Eukaryotes | eEF1A | eEF1B | EFsec | eEF2 | mtEF4 | eIF5A | eEF3 | — | — | — |
| Mitochondria | mtEF-Tu | mtEF-Ts | — | mtEF-G1 | mtEF4 | — | — | — | — | — |
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None.
Factors Involved in AA-tRNA Incorporation
EF-Tu and EF-Ts
EF-Tu is encoded by two genes tufA and tufB in E. coli (Jaskunas et al., 1975), T. thermophilus (Satoh et al., 1991) and S. typhimurium (Hughes, 1990). It is a universally conserved GTPase in all species (Caldon and March, 2003). During translation elongation, EF-Tu∙GTP transports aa-tRNA to the ribosome A-site in the form of the ternary complex (Ramakrishnan, 2002). Upon binding of the ternary complex to the ribosome, proper base pairing between the anticodon of aa-tRNA and the mRNA codon within the 30S subunit decoding region stimulates EF-Tu to hydrolyze GTP. After the hydrolysis of GTP, the conformational change following the GTP hydrolysis to GDP and a leaving phosphate group (Pi) leads to the dissociation of EF-Tu from the ribosome and accommodation of the aa-tRNA on the A-site for a peptidyl transfer (Rodnina et al., 2005). The growing peptide chain extends by one amino acid under the catalysis of the ribosome. Recycling of EF-Tu∙GDP to EF-Tu∙GTP depends on EF-Ts, another elongation factor encoded by the tsf gene (Wang et al., 1997). EF-Tu can be reversibly phosphorylated on its serine and threonine residues, and this modification has been founded in multiple organisms including E. coli (Lippmann et al., 1993), Listeria monocytogenes (Archambaud et al., 2005), Thermus thermophilus (Lippmann et al., 1993), Streptococcus pneumoniae (Sun et al., 2010), Bacillus subtilis (Lévine et al., 2006), Corynebacterium glutamicum (Bendt et al., 2003), Mycoplasma pneumoniae (Schmidl et al., 2010), Mycobacterium tuberculosis (Sajid et al., 2011), and Streptomyces collinus (Mikulík and Zhulanova, 1995). Phosphorylation inhibits the GTPase activity of EF-Tu and prevents its dissociation from the ribosome (Pereira et al., 2015). Phosphorylated EF-Tu could not bind with aa-tRNA or kirromycin (Hughes, 2013). It has been reported that phosphorylation of EF-Tu plays a vital role in bacterial dormancy, sporogenesis, virulence, and stress tolerance (Archambaud et al., 2005; Holub et al., 2007; Misra et al., 2011; Pereira et al., 2015). The exact physiological significance of EF-Tu phosphorylation is still need to be clarified while phosphorylation of EF1A in eukaryotes has been shown to be involved in maintaining a proper elongation rate under various conditions (Hughes, 2013).
eEF1A and eEF1B
The GTPase eEF1A, the homolog of EF-Tu in bacteria, is one of the most widely expressed factors in eukaryotes (Schuller and Green, 2018). Human eEF1A has 33% sequence identity with bacterial EF-Tu (Cavallius et al., 1993). Similar to EF-Tu, another guanine nucleotide-exchange factor eEF1B is required to regenerate active eEF1A∙GTP (Gromadski et al., 2007). In lower eukaryotes, eEF1B contains a guanine nucleotide exchange subunit eEF1Bα and a structural subunit eEF1Bγ, while higher eukaryotes have another guanine nucleotide exchange subunit eEF1Bδ (plants) or eIF1Bβ (mammals). The mechanism of guanine-nucleotide exchange employed by eEF1B is very different from that of EF-Ts (Rodnina and Wintermeyer, 2009). Upon binding of EF-Ts to EF-Tu, the switch I region of EF-Tu is displaced by the C-terminal helix of EF-Ts. The switch II region is moved upon binding due to pushing by subdomain N. Altogether, these changes disrupt the coordination of the Mg 2+ ion, leading to the dissociation of GDP. In contrast, eEF1Bα interacts with domains 1 and 2 of eEF1A, disrupting the binding pocket for Mg 2+ and preventing the binding of the GDP to eEF1A.
In humans, there are two eEF1A homologues, named eEF1A1 and eEF1A2 (coded by two genes: EEF1A1 and EEF1A2). The sequences of eEF1A1 and eEF1A2 have 98% similarity and 92% identity, but the expression patterns of the two proteins are different (Tomlinson et al., 2005). eEF1A1 is ubiquitously expressed, whereas the expression of eEF1A2 is switched-on in adult life in specialized tissues such as skeletal muscle, cardio-myocytes and neurons (Lee and Surh, 2009). Overexpression of eEF1A2 has been reported to be linked with a variety of tumors (Lee and Surh, 2009), and mutations in EEF1A2 are related to a new type of epilepsy syndrome and intellectual disability (Inui et al., 2016; Lam et al., 2016). In addition to its canonical functions in transporting aa-tRNA to the ribosome, eEF1A is found to be involved in cellular activities such as regulation of cytoskeleton organization (Mateyak and Kinzy, 2010), protein degradation mediated by the proteasome (Mateyak and Kinzy, 2010), viral replication and propagation (Li et al., 2009), nuclear protein export (Khacho et al., 2008), signaling transduction pathway concerning apoptosis and oncogenesis (Schulz et al., 2014; Abbas et al., 2015) (Figure 3). eEF1A1 also plays an important role in the process of heat shock stress response (Vera et al., 2014). Single-cell transcriptomic analysis revealed that the expression level of eEF1A1 in neurons was low and changed with age in glial cells (Ximerakis et al., 2019). Therefore, eEF1A may represent a potential candidate for lifespan modulation (Skariah and Todd, 2020). In addition, eEF1A, along with eEF2, has been shown to be related to neurodegenerative disorders including Alzheimer disease (AD) and Parkinson disease (PD) with an unknown mechanism. The low expression levels of these factors in the brains of AD and PD patients indicating defects in the efficiency or fidelity of translation (Li et al., 2005; Vera et al., 2014; Garcia-Esparcia et al., 2015; Beckelman et al., 2016; Skariah and Todd, 2020). eEF1A has chaperone-like activity (Lukash et al., 2004) and may also be involved in antiviral response by interaction with Sgt1, a multifunctional protein (Novosylna et al., 2015).
FIGURE 3.
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Canonical and non-canonical functions of eEF1A. In addition to its canonical functions in transporting aa-tRNA, eEF1A is also involved in cytoskeleton organization, protein degradation, viral replication and propagation, nuclear protein export, heat shock stress response, and signaling transduction pathway concerning apoptosis and oncogenesis.
Humans and yeast eEF1A are subjected to extensive methylation modifications at multiple conserved lysine residues (Hamey and Wilkins, 2018; Jakobsson et al., 2018; Robichaud et al., 2019). Methylation modification is the addition of 1∼3 (me1-me3) methyl groups to the side chains of lysine residues. Up to now, several methylation sites and corresponding methyltransferases of eEF1A have been identified in humans and yeasts. For example, human N6AMT2 (eEF1A-KMT1), METTL10 (eEF1A-KMT2), METTL21B (eEF1A-KMT3), and eEF1A-KMT4 (gene name EEF1AKMT4), which methylate eEF1A at K79 (me3), K318 (me3), K165 (me1/me2/me3), and K36 (me3), respectively (Shimazu et al., 2014; Hamey et al., 2016; Jakobsson et al., 2017; Malecki et al., 2017); In addition, two other methylation sites (N terminus and K55) have been reported in human cells, but the enzymes responsible for N-terminal trimethylation and K55 dimethylation in human eEF1A still need to be identified (Hamey et al., 2017). Saccharomyces cerevisiae Efm1, Efm4, Efm5, Efm6, and Efm7, which methylate eEF1A at K30 (me1), K316 (me2), K79 (me3), K390 (me1), and N-terminal (me3) and K3 (partially me1 and me2) respectively (Lipson et al., 2010; Dzialo et al., 2014; Jakobsson et al., 2015; Hamey et al., 2016). The abundant lysine methylations of eEF1A and the existence of multiple corresponding methyltransferases in various eukaryotes made it clear that the lysine methylation of eEF1A has important physiological significance. In METTL21B knockout (KO) human cells, the expression of proteins involved in cytoskeleton organization was downregulated, while the expression of proteins related to large ribosomal subunit biogenesis, mRNA turnover and rRNA processing were upregulated (Hamey et al., 2017). Moreover, mammalian METTL21B was found to be partially localized in the centrosome, which may reveal a non-canonical function for this protein (Malecki et al., 2017). In EEF1AKMT4 KO cells, the global translation is changed and the translation speed of codons for histidine (H), tryptophan (W), and asparagine (N) was altered compared with wild-type cells (Jakobsson et al., 2017). In EEF1AKMT1 KO cells, the expression of proteins related to tRNA aminoacylation and nuclear exosome were downregulated, while the expression of proteins related to ubiquitination regulation and small-subunit processome were upregulated (Hamey et al., 2017). Even though the physiological function of K318 methylation of eEF1A has not been clarified in mammalian, perhaps it can be speculated that K318 methylation may affect the replication of RNA virus and the migration of neural crest based on the highly conservative of this site between human and yeast (Shimazu et al., 2014). In yeast, Efm4 is involved in vesicle transport processes including secretory protein production, transfection and endocytosis (Martín-Granados et al., 2008). Efm4 is also play a vital role in Tombusvirus replication (Li et al., 2014). Similarly, the gene of EFM5 was shown to be crucial for virus replication in yeast (Kushner et al., 2003). In S. cerevisiae, EFM7 KO leads to decreased replicative lifespan (Anderson et al., 2003), which results from altered translation rate (Pan et al., 2007). Methylation of eEF1A by Efm6 occurs in its domain III, which is involved in protein translation and cytoskeleton organization (Gross and Kinzy, 2005; Liu et al., 2006). Finally, even though the five methyltransferases of eEF1A in S. cerevisiae are not absolutely necessary to its viability, the precise regulation of eEF1A function by distinct methyltransferases optimizes the cell physiology (White et al., 2019). Besides methylation, the lysine residues of eEF1A are also modified by acetylation, sumoylation and ubiquitination, as well as phosphorylation of tyrosine, threonine and serine residues (Hornbeck et al., 2012).
mtEF-Tu and mtEF-Ts
Eukaryotic cells, including those in animals and fungi, contain two translation systems, one in the cytosol and the other in the mitochondria. Mitochondria use their own translational system to synthesize proteins for respiratory chain complexes. mtEFs (mitochondrial translation elongation factors) are coded by the nuclear genome, synthesized and transported into mitochondria. These factors are more similar to their counterparts in bacteria than those from the cytoplasm of eukaryotes. Genes encoding mtEFs such as TUFM (mtEF-Tu), TFSM (mtEF-Ts), and GFM1(mtEF-G1), have been reported with mutations in cases causing down regulation of mitochondrial translation and early fatality (Hughes, 2013).
mtEF-Tu consists of 409 amino acids, and is 55∼60% identical to the homologous protein from bacteria. It was found that mtEF-Tu folded into three main domains similar to EF-Tu. One of the main differences is that the C-terminal of mtEF-Tu has an 11 amino acids extension, which may interact with aa-tRNA (Jeppesen et al., 2005). Early studies showed that, mtEF-Tu was compatible with E. coli aa-tRNAs, whereas E. coli EF-Tu was unable to catalyze polypeptide chain elongation when supplied with mitochondrial aa-tRNAs (Kumazawa et al., 1991). This is probably due to the incorrect positioning of the shorter mitochondrial aa-tRNAs on the ribosome by bacterial EF-Tu, which leads to the ineffective stimulation of the GTPase activity of EF-Tu (Christian and Spremulli, 2012). Unlike EF-Tu, the bacterial and mitochondrial EF-Ts show low (25∼30%) sequence conservation. The most widely studied form of mtEF-Ts comes from B. taurus. It consists of 283 amino acids, with a mitochondrial import signal about 55 residues reside in the N-terminal of the mature protein (Xin et al., 1995).
SelB
Selenocysteine (Sec) is a cysteine (Cys) residue analogue with a selenium-containing selenol group in place of the sulfur-containing thiol group in Cys. The selenium atom gives Sec quite different properties from Cys. Sec utilization is scattered across archaea (Mariotti et al., 2016) and bacteria (Zhang et al., 2006). In eukaryotes, selenoproteins exist in some algae and protozoa (Lobanov et al., 2009; Mariotti et al., 2015), and most metazoans (Mariotti et al., 2012). Recently, Mariotti et al. (2019) provided evidence for Sec usage in early-branching fungal phyla. In mammals, Sec exists in enzymes associated with ROS detoxification and hormone biosynthesis. It plays a vital role in many biological processes including development, reproduction, immune response, tumorigenesis, viral infections and cardiovascular diseases (Carlson et al., 2006). The terminator codon UGA behaves as the codon of Sec when the downstream of UGA possesses a selenocysteine insertion sequence (Ose et al., 2007). During the translation of selenoproteins such as glutathione peroxidase and bacterial formate dehydrogenase, SelB, a unique protein factor, is needed to deliver selenocysteinyl-tRNA Sec containing a UCA anticodon to the ribosome A-site to recognize UGA codon in proper position of an mRNA (Böck et al., 1991). Incorporation of Sec is governed by a unique mRNA hairpin that located 3′ near the Sec codon (Yoshizawa et al., 2005). This hairpin structure associates with the corresponding ternary complex that is composed of SelB, Sec-specific aa-tRNA and GTP (Yoshizawa et al., 2005). It is assumed that by this mechanism selenocysteinyl-tRNA Sec is delivered to the ribosome.
Factors Involved in Ribosome Translocation
EF-G
EF-G, which is the third most conserved trGTPase among all domains of life (Caldon and March, 2003), catalyzes the translocation of A-site peptidyl-tRNA and P-site deacylated tRNA to the P- and E-site, respectively (Ramakrishnan, 2002). Translocation of tRNA 2-mRNA during translation elongation is associated with EF-G triggered GTP hydrolysis and a series of conformational changes of the ribosomes (Savelsbergh et al., 2005). Recently, Holtkamp et al. (2014b) concluded that EF-G integrates the energy regimes of a motor protein and a GTPase, and promotes tRNA motion through the combined use of power stroke and Brownian ratchet mechanisms. After GTP hydrolysis, EF-G departs from the ribosome in the form of EF-G∙GDP. EF-G is the only canonical trGTPase that functions at two distinct phases in bacterial translation (Ero et al., 2016). In addition to facilitating translocation in elongation, it also plays an essential role in ribosome recycling, during which EF-G∙GTP works together with RRF to split the post-termination complex (PoTC) into two subunits (Song et al., 2020).
The GTPase activity of EF-G involves two hydrophobic amino acids Ile61 and Ile19 (E. coli numbers), which facilitate approaching of His92 to GTP by forming an opened hydrophobic gate. The water molecule promoted by His92 attacks the γ phosphate of GTP, resulting in an active state of the GTPase center (Yamamoto et al., 2014). EF-G is extensively modified by reversible phosphorylation. An early study reported that E. coli EF-G can be phosphorylated by a serine/threonine specific protein kinase (gp 0.7 PK) encoded by the T7 early gene 0.7 (Robertson et al., 1994). This gene was expressed early following T7 infection, leading to rapid shutdown of host RNAP at 4 min after infection, and soon thereafter, most protein synthesis began to turn off gradually. Which established favorable conditions for T7 phage growth (McAllister and Barrett, 1977). And this modification may help to increase the translation elongation rate of T7 late genes that specify T7 virion assembly and structural proteins (Robertson et al., 1994). Another in vivo and in vitro study in B. subtilis showed that EF-G can be phosphorylated on at least one threonine residue by a membrane Ser/Thr kinase PrkC and dephosphorylated by phosphatase PrpC, and the dynamic control of EF-G phosphorylation may play a regulatory role in stationary-phase B. subtilis (Gaidenko et al., 2002). Later, Shah and Dworkin (2010) proved that phosphorylation of EF-G by PrkC in B. subtilis is in response to cell wall-derived muropeptides. E. coli EF-G can also be modified at the lysine residue essential for GTP binding by pyridoxal phosphate (PLP), a selective, site-specific lysine reagent, leading to progressive loss of the EF-G activity, and destruction of its interaction with 30S subunits as well as a conformational change required for GTP hydrolysis (Giovane et al., 1982). Although these effects have been more widely studied, the physiological significance of phosphorylated EF-G still needs to be elucidated (Hughes, 2013).
eEF2
eEF2 is the eukaryotic homolog of EF-G. When cells are starved of nutrients, eEF2 is phosphorylated by the Ca 2+-activated kinase eEF2K, resulting in a lower binding affinity to the ribosome (Carlberg et al., 1990). The activity of eEF2K is regulated by nutrients through mTORC1 and AMPK (Kenney et al., 2014; Proud, 2019). During translation elongation, the active eEF2K can reduce termination read-through errors and codon-anticodon mismatches, and promote more accurate recognition of the start codon by reducing initiation at the near-AUG codons (Xie et al., 2019). Interestingly, despite the major role of eEF2’s phosphorylation in blocking whole protein translation, its phosphorylation in neurons is associated with elevated translation of Arc/Arg3.1 which plays a key role in postsynaptic endocytosis (Park et al., 2008).
Diphthamide is another conserved modification in archaeal and eukaryal eEF2, where a conserved histidine (H715 in mammals; H699 in S. cerevisiae) at the eEF2 domain IV is modified with diphthamide (Su et al., 2013; Schaffrath et al., 2014). In eukaryotes, this modification event occurs through a four-step pathway involving multiple proteins including Dph1-Dph7 (Versées, 2015). Dph1/Dph2 is a [4Fe-4S] cluster-containing heterodimeric protein complex, which is responsible for catalyzing the first step of this modification pathway. In S. cerevisiae, the highly conserved proteins Dph3 (or Kti11) and Kti13 form a heterodimer, which is involved in eEF2 modification by acting as an electron donor for the Dph1/Dph2 complex (Dong et al., 2014; Glatt et al., 2015). With the help of Dph4/5/6/7, the diphthamide group was finally added to eEF2. Moreover, Kti13 was reported to specifically binding with PIP 2 (Di Paolo and De Camilli, 2006), which might contribute to the regulation of the downstream eEF2 modification pathway (Figure 4). Lack of the diphthamide modification is fatal to mice due to severe developmental defects (Yu et al., 2014). It is worth noting that yeast lacking Dph1, an enzyme needed for diphthamide synthesis, grew normally, indicating that diphthamide likely functions in translation fidelity but not a basic mechanism of translation (Dever et al., 2018). Besides, some mammalian cells can survive in the absence of diphthamide (Stahl et al., 2015). However, related studies (Liu et al., 2012) had shown that lacking diphthamide modification results in an increase in the level of programmed -1 ribosomal frameshifting. Considering its location in eEF2, it is reasonable to speculate that diphthamide may enhance the eEF2 function by contacting with RNA in the ribosomal decoding center, thus facilitating the ribosomal translocation fidelity (Dever and Green, 2012).
FIGURE 4.
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Schematic of the diphtamide modification of eEF2. The first step in this modification pathway is catalyzed by [4Fe-4S] cluster-containing protein complex Dph1/Dph2; with the help of Dph4/5/6/7. Dph3/Kti13 heterodimer act as electron donor for Dph1/Dph2 complex. It has been reported that diphtamide group of EF2 can further be ADP-ribosylated by the bacterial diphtheria toxin, leading to a global inhibition of protein synthesis as well as an upregulated translation of mRNAs associated with oxidative stress response.
Diphthamide of eEF2 can be further ADP-ribosylated by diphtheria and cholera toxins, which catalyze the transfer of ADP-ribose from nicotinamide adenine dinucleotide (NAD+) to the diphthamide imidazole ring to yield ADP ribosyl diphthamide (Argüelles et al., 2014). ADP-ribosylation inactivates eEF2, hinders protein translation and damages cell growth (Mateyak and Kinzy, 2013). Recently, it has been reported that cells possess intrinsic abilities to modify the diphthamide group by ADP-ribosylation, and this ability will improve under certain stress conditions, causing the overall down-regulation of protein synthesis at the cost of an increased translation of IRES-containing mRNA that involved in the response of oxidative stresses (Argüelles et al., 2014). However, the molecular mechanism of how ADP-ribosylation impairs the function of eEF2 has not been fully elucidated (Mateyak and Kinzy, 2013).
mtEF-G1 and mtEF-G2
In human mitochondria, the dual function of bacterial EF-G is fulfilled by mtEF-G1 and mtEF-G2 (Tsuboi et al., 2009; Christian and Spremulli, 2012). During mitochondrial translation elongation, ribosome translocation was catalyzed by mtEF-G1. Previous studies revealed that mtEF-G1 has a strong tolerance to fusidic acid, an antibiotic that inhibits EF-G release from the ribosome without influencing on GTP hydrolysis and translocation (Gao et al., 2009; Savelsbergh et al., 2009), while the tolerance mechanism of mtEF-G1 to fusidic acid remains to be understood (Christian and Spremulli, 2012). mtEF-G1 is active not only with 55S mammalian mitochondrial ribosome but also with 70S bacterial ribosome. In contrast, E. coli EF-G can’t work with mitochondrial ribosomes (Chung and Spremulli, 1990). mtEF-G2 mediates ribosome recycling in concert with human mitochondrial RRF after termination (Achenbach and Nierhaus, 2015). However, it should be noted that overexpression of mtEF-G2 can improve the translation of respiratory chain complexes slightly in cells with mtEF-G1 mutation, indicating that mtEF-G2 perhaps plays a part in the translation elongation (Coenen et al., 2004). Unlike ribosome recycling in bacteria, mtEF-G2 catalyzed GTP hydrolysis is not essential for ribosome dissociation. Rather, it seems to be necessary for the dissociation of mtRRF and mtEF-G2 from ribosomes. Since mtEF-G2 represents a type of trGTPase participating in ribosome recycling, it has been proposed to rename this factor as mitochondrial ribosome recycling factor 2 (mtRRF2) (Tsuboi et al., 2009).
Other Elongation Factors
EF4
EF4 (LepA) was originally identified in E. coli in 1985 (March and Inouye, 1985). The high conservation of EF4 in bacteria suggests its functional importance (Margus et al., 2007). Nierhaus and co-workers reported that cell membranes behave as EF4 reservoir pool, releasing it to the cytoplasm under certain conditions such as elevated intracellular Mg 2+ concentrations or low temperature, leading to an increased rate of translation and efficient folding of newly synthesized peptides (Pech et al., 2011; Yamamoto et al., 2014). Functional studies showed that EF4 knockout affects bacterial growth under conditions of high Mg 2+ concentration (Pech et al., 2011) or low pH (Yang et al., 2014). In S. cerevisiae, lacking EF4 (Guf1) results in growth defects under conditions of starvation and low temperature, and decreased expression of cytochrome oxidase (Bauerschmitt et al., 2008). Besides, EF4 knockout E. coli cells showed a decreased translation rate and slow ribosome maturation at unfavorable conditions (Yang et al., 2014). Overexpression of EF4 in E. coli seriously affects the growth of cells (Qin et al., 2006), while EF4 knockout cells showed no obvious phenotype under culture conditions of rich LB medium (Shoji et al., 2010). Therefore, EF4 may contribute to the cell survival under adverse conditions, but the physiological role of the protein remains unclear.
A previous study proposed that EF4 acts as a ‘back-translocase’ that has a unique property of recognizing ribosomes with mistranslocated tRNAs and back-translocating them via GTP hydrolysis during elongation cycle (Qin et al., 2006). However, several subsequent studies could not confirm that EF4 did have this biochemical activity (Liu et al., 2010; Balakrishnan et al., 2014; Gibbs and Fredrick, 2018). Cooperman and coworkers have done the most detailed biochemical characterization of EF4, studying its effect on both PRE and POST state complexes (Liu et al., 2010; Liu et al., 2011). They found that the addition of EF4 to POST can promote movement of tRNA with respect to the 50S subunit but does not catalyze back-translocation. Notably, Cooperman also showed that the EF4 preferentially engages the PRE complex, and competes with EF-G rather than EF-Tu ternary complex to influence elongation. In vitro kinetic measurements showed that EF4-dependent back translocation proceeds through a four-stage kinetic route (POST→I 1→I 2→I 3→PRE), not just a reversal of translocation but exist three intermediate states, and the rate of reverse codon-anticodon movement observed in the presence of EF4 is virtually identical to that seen in its absence (Liu et al., 2010), in line with the independent work led by Fredrick and coworkers, whose ribosome profiling results revealed that EF4 contributes mainly to the initiation phase of translation in E. coli (Balakrishnan et al., 2014). This is consistent with a recent physiological study suggesting EF4 contributes to biogenesis of the 30S subunit, immature 30S particles will accumulate in cells lacking EF4 (Gibbs et al., 2017).
In summary, although the major role of EF4 is to facilitate bacteria in response to some stresses and affect protein synthesis in general, still some important issues concerning EF4 need to be resolved: What is the mechanism of EF4 release from the membrane into the cytoplasm? What is the real physiological substrate of EF4? How to reasonably explain the contradiction between the high conservation of EF4 and the obvious lack of phenotype in its deletion mutants (Figure 5)?
FIGURE 5.
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Various functions of EF4 and the issues still need to be resolved.
mtEF4
mtEF4 is the homologue of bacterial EF4 in eukaryotic mitochondria. It consists of 651 residues with a mitochondrial-targeting signal in their N termini (Gao et al., 2016). mtEF4 is located in the mitochondrial matrix, close to the inner membrane (Bauerschmitt et al., 2008). mtEF4 binds to the ribosome in a GTP-dependent manner which is similar to that of bacterial EF4. It promotes the translation of mitochondrial proteins under non optimal conditions (Bauerschmitt et al., 2008). mtEF4 also plays a quality control role in the biogenesis of mitochondrial respiratory chain complexes. Previous studies showed that mtEF4 knockout induces respiratory chain defects as well as apoptosis, whereas overexpression of the protein stimulates cancer development (Zhu et al., 2018). mtEF4 ablation in mice results in testis-specific disorder of oxidative phosphorylation, and its deletion facilitated mitochondrial protein translation in the expense of synthesis unstable proteins (Gao et al., 2016). Increased expression of mtEF4 in multiple cancers suggested that mtEF4 probably facilitates tumorigenesis through an unbalanced regulation of mitochondrial activities and cellular redox (Zhu et al., 2018). Therefore, the proper level of mtEF4 in cell is requisite for the assembly of functional respiratory chain complexes as well as mitochondrial protein synthesis.
EF-P and eIF5A
Bacterial EF-P (elongation factor P) is a homolog of eukaryotic and archaeal initiation factor 5A(e/aIF5A) (Kyrpides and Woese, 1998). It binds to the interface of the ribosome subunits and facilitates peptide bond formation through interactions with the P-site tRNA (Glick and Ganoza, 1975; Blaha et al., 2009). efp, the coding gene of EF-P, has been found throughout the bacteria (Hughes, 2013). It was reported that EF-P participates in the regulation of cell viability, growth, motility, virulence (Ude, 2013), and tolerance to multiple stresses including several classes of antibiotics, detergents, nutrient-limiting conditions and diverse growth inhibitors (Navarre et al., 2010; Zou et al., 2012). The ribosome needs the help of EF-P when successive prolines are incorporated into the nascent peptide chain (Rodnina, 2018). Without EF-P, the ribosome would stagnate at polyproline stretches, while the addition of EF-P could alleviate the translation stalling (Doerfel et al., 2013). EF-P is often subject to post-translational modification by PoxA catalyzed adding of a (R)-β-lysine to Lys34 (termed lysinylation) in bacteria like E. coli (Peil et al., 2012), S. enterica (Hersch et al., 2013) and S. typhimurium (Navarre et al., 2010; Zou et al., 2011), which could improve the catalytic activity of EF-P both in vitro and in vivo (Navarre et al., 2010; Park et al., 2012). β-lysylated EF-P also undergoes further hydroxylation, a second post-translational modification of this factor, but the function of hydroxylation does not seem to be critical for EF-P (Peil et al., 2012; Bullwinkle et al., 2013). In addition to the post-translational modifications mentioned above, other modifications of EF-P had been identified in other bacteria. For example, rhamnosylation of Arg32 in S. oneidensis, P. aeruginosa, N. gonorrhoeae, B. pertussis and N. meningitidis, which is closely related to bacterial fitness, pathogenicity and viability (Lassak et al., 2015; Rajkovic et al., 2015; Yanagisawa et al., 2016; Hummels and Kearns, 2020); 5-aminopentanolylation of Lys32 in B. subtilis, which can regulate the synthesis of diprolyl motifs of specific proteins needed for swarming motility (Rajkovic et al., 2016). There are hundreds of polyproline-containing peptides and proteins with different functions in all organisms, indicating EF-P (e/aIF5A) is necessary for the regulation of expression levels in various pathways.
eIF5A was firstly reported to promote the first peptide bond formation and was denoted as an initiation factor (Kemper et al., 1976). In archaea and eukaryotes, a conserved lysine located at the eIF5A domain I (Kim et al., 1998) is modified to hypusine post-translationally. Synthesis of hypusine is catalyzed by two consecutive enzymatic reactions involving deoxyhypusine synthase (DHPS) and deoxyhypusine hydroxylase (DOHH) (Park et al., 2010). Hydroxyl radical probing experiments revealed that the binding site of eIF5A on the ribosome is adjacent to the E-site, and the hypusine group is in proximity to the acceptor arm of the P-site tRNA (Gutierrez, 2013). An early study showed that unmodified eIF5A is unable to catalyze the formation of methionyl-puromycin (Park et al., 1991). Depletion of eIF5A leads to dysregulated translation initiation (Manjunath et al., 2019), elongation/termination (Schuller et al., 2017), and an increase in ribosomal transit times (Henderson and Hershey, 2011). In the absence of cycloheximide, inhibition of an eIF5A mutant that is sensitive to temperature leads to polysome accumulation (Saini et al., 2009), indicating eIF5A plays a vital role in the elongation phase. Moreover, if eIF5A is present, ribosome stalling will be restored when multiple prolines are to be incorporated in vitro. eIF5A depletion leads to defects in the synthesis of polyproline-containing proteins (Gutierrez, 2013), and ribosome stalling at tripeptides such as RDK, DVG, DDG, DDP, PDP and DNP (Schuller et al., 2017). Therefore, like EF-P in bacteria, eIF5A is considered to promote peptide transfer and improve the translation efficiency of poor substrates like proline (Gutierrez, 2013) (Figure 6). eIF5A was also found to play a role in the start codon selection during translation initiation. Depletion of eIF5A enhances upstream translation within 5′ UTRs across yeast and human transcriptomes (Manjunath et al., 2019).
FIGURE 6.
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Mechanism of EF-P/eIF5A in alleviating ribosome stalling at polyproline stretches. Accomodation of the third consecutive Pro-tRNA Pro into the A-site of translating ribosome leads to stalling. EF-P/eIF5A binds to a location between E- and P-sites of the stalled ribosome after E-site tRNA is dissociated, stimulating rapid proline-proline peptide bond formation. Translation can resume following dissociation of EF-P/eIF5A and binding of the next aa-tRNA.
eEF3-Fungal Specific Elongation Factor
eEF3, a third elongation factor which was reported to be important for the protein translation and viability in higher fungi including yeast and P. carinii (Triana-Alonso et al., 1995). eEF3 from S. cerevisiae is composed of 1,044 amino acids and the coding gene of this factor is YEF-3 (Qin et al., 1990). The crystal structure of S. cerevisiae eEF3 revealed that it is composed of a HEAT domain, two ATPase domains, a four-helix bundle, and a chromodomain (Andersen et al., 2006). It has a ribosome-dependent ATPase and GTPase (Dasmahapatra and Chakraburtty, 1981). eEF3 is mainly associated with cytosolic polysomes, and is needed for peptide bond formation (Kapp and Lorsch, 2004). The existence of eEF3 exclusively in fungi may be the most striking exception to the highly conserved translation elongation. It was reported that eEF2 and eEF1 from yeast can work with mammalian ribosomes to promote translation. In contrast, eEF2 and eEF1 from mammalian could function with yeast ribosomes only when eEF3 is present (Skogerson and Engelhardt, 1977), indicating eEF3 is required for yeast ribosome translation. Cryo-EM structures revealed that eEF3 has a new binding site close to the ribosome E-site, with the chromodomain stabilizing L1 stalk to facilitate tRNA release, which is consistent with the model that eEF3 may aid E-site tRNA release following translocation (Andersen et al., 2006).
The question that still needs to be answered is why do fungal ribosomes require eEF3 to promote E-site tRNA release and A-site occupancy while other ribosomes do not? Rodnina et al. (1994) reported that 80S from several higher eukaryotes can hydrolyse ATP and GTP without protein factors. This inherent ATPase activity functions as eEF3, promoting the dissociation of deacylated tRNA at the ribosome E-site (El’skaya et al., 1997), and cognate tRNA binding at the ribosome A-site seems to stimulate this activity in turn (Rodnina et al., 1994). Nevertheless, no homologues of fungal eEF3 have been found in mammalian cells. It is worth noting that a crystal structure of S. cerevisiae 80S contains an additional non-ribosomal protein, Stm1, which bound to the 40S subunit and precluded mRNA entry by placing a α-helix to the mRNA tunnel (Ben-Shem et al., 2011). In yeast, deletion of stm1 leads to increased binding affinity of eEF3 to ribosomes, whereas up-regulation of eEF3 expression in cells lacking Stm1 leads to growth defect and elevated anisomycin (a translation inhibitor) sensitivity. Moreover, a high level of Stm1 in ribosomes displays reduced eEF3 binding. Therefore, it can be concluded that Stm1 and eEF3 may jointly promote the elongation cycle (Van Dyke et al., 2009). Even though the characteristics of bacterial ribosome structures do not suggest the need for another elongation factor (Dever and Green, 2012), E. coli has an ATPase RbbA, which can bind to the ribosome and stimulates protein translation in vitro, and displays multiple biochemical properties similar to that of eEF3 (Kiel and Ganoza, 2001). More studies concerning biochemistry, structural biology and high resolution fluorescence imaging are required to fully reveal the function of eEF3 and to resolve its specific needs in fungi translation elongation (Dever and Green, 2012).
Tet(O) and Tet(M)
The tetracycline resistance proteins (Tet), which protect the bacterial ribosome from binding the antibiotic tetracycline, are another class of ribosome-associated GTPases (Margus et al., 2007). They are also called ribosomal protection proteins (RPPs). They are cytoplasmic proteins that display homology with the elongation factors EF-Tu and EF-G (Hughes, 2013). They bind to the ribosome, hydrolyze GTP and cause the release of tetracycline from the ribosome (Connell et al., 2003). RPPs are paralogs of elongation factors, and the best characterized and widely distributed RPPs are Tet(M) and Tet(O) (Hughes, 2013). Both Tet(M) and Tet(O) have ribosome-dependent GTPase activity, the hydrolysis of GTP providing the energy for the ribosomal conformational changes (Connell et al., 2003). Direct competition experiments showed that the ribosome binding site of Tet(M) is overlapping with that of EF-G (Dantley et al., 1998). To clarify the mechanism of the tetracycline resistance, Burdett purified Tet(M) protein (Burdett, 1991) and investigated its effects on several reactions that occur during protein translation (Burdett, 1996). The author found that Tet(M) could alleviate the inhibition of tetracycline on factor-dependent tRNA binding, and significantly reduce the affinity of ribosomes for tetracycline in the presence of GTP. Adding Tet(M) to the ribosome-tetracycline complex will replace the bound tetracycline. Therefore, the dissociation of tetracycline from the ribosome promoted by Tet(M) is GTP dependent. Structures of 70S∙Tet(O) complex revealed that Tet(O) really looks like EF-G and binds to the ribosome with an identical site (Spahn et al., 2001). In 2013, Li et al. identified the Tet(O) binding site on the ribosome, which involves three unique loops in Tet(O) domain IV (Li et al., 2013). The single glycine substitution of the residues in these loops leads to loss of tetracycline resistance.
RelA
In order to adapt to the changes in environment conditions, pathogens have evolved a variety of mechanisms to respond to various stresses, the most important of which is called stringent response (SR) (Winther et al., 2018). SR is a nearly universal mechanism mediated by pppGpp and ppGpp (Steinchen and Bange, 2016). Seeing that SR is related to metabolic regulation, biofilm formation, virulence gene expression (Yang et al., 2020), stress, antibiotic resistance, as well as the formation ability of bacteria retention (Hauryliuk et al., 2015), therefore, the signal pathway becomes a target for designing effective antibacterial agents (Kushwaha et al., 2019). The intracellular levels of pppGpp and ppGpp (collectively referred to as (p)ppGpp) are controlled by RelA/SpoT Homologue (RSH) proteins, which synthetize (p)ppGpp by transferring the pyrophosphate group of ATPs onto the 3′ of GDP or GTP, and degrade (p)ppGpp by removing the 3′ pyrophosphate moiety. It interacts with the ribosome to sense environmental pressure and leads to an adaptive response of pathogens. When pathogens encounter stresses like nutrition deficiency, deacylated tRNA binds to the empty ribosome A-site to form the so-called “starving” ribosome, triggering the (p)ppGpp synthetic activity of RelA (Haseltine and Block, 1973). At this time, RelA catalyzes the production of AMP and ppGpp or pppGpp from ATP and GDP or GTP, respectively (Wendrich et al., 2002), and the SR reaction is triggered.
Under stressed conditions, ppGpp, as a “global transcriptional regulator”, can up regulate the expression of many genes at the transcriptional level, and the concentration of intracellular ppGpp is largely determined by relA gene. Therefore, the function of relA gene in cell physiology is mainly reflected in controlling the concentration of intracellular ppGpp. Studies have shown that (P)ppGpp can change the physiology of bacteria from rapid growth to slow growth, so as to allow them to survive under harsh conditions (Hauryliuk et al., 2015). In E. coli, (P)ppGpp combines with RNA polymerase, up-regulating the expression of metabolic enzyme genes, especially amino acid biosynthesis genes, while reducing the transcription of tRNA, rRNA, ribosomal protein, translation factor and synthase genes (Barker et al., 2001; Lemke et al., 2011; Ross et al., 2016). Other main effects include activating the stress factor δ E and inhibiting cell wall synthesis (Costanzo et al., 2008). It was found that more than 30% of the genes in E. coli were differentially expressed by (P)ppGpp, including the up regulation of SR related genes and the down regulation of macromolecular structure related genes under isoleucine starvation conditions (Traxler et al., 2008). Several models have been proposed for the molecular mechanism of SR (Sanchez-Vazquez et al., 2019). It is reported that (P)ppGpp directly binds to RNAP with RNAP binding protein dksA, which destroys the stability of its open complex. On the other hand, (P)ppGpp indirectly regulates gene expression by δ competition (Sanchez-Vazquez et al., 2019). It is worth noting that almost all bacterial pathogens need SR, otherwise they cannot survive and infect the host during stress conditions.
BPI-Inducible Protein A
BPI-inducible protein A (BipA) is a highly conserved ribosome-dependent trGTPase that regulates multiple cellular processes including bacterial attachment and effacement, motility, virulence gene expression, avoidance of host defenses. In addition, BipA is also associated with temperature, symbiosis, antimicrobial resistance and biofilm formation (Overhage et al., 2007; deLivron and Robinson, 2008; Neidig et al., 2013). BipA is widespread in most of the bacteria and plants (Margus et al., 2007) and is similar to EF4 and EF-G except for a unique C-terminal domain (Neidig et al., 2013; Kumar et al., 2015). Despite its conservation in bacteria, BipA was regarded as an essential factor only in adverse growth conditions such as low pH and temperature, nutrient deprivation, and other stresses (Starosta et al., 2014). Mutations in BipA leads to multiple phenotypes including cold sensitivity (Pfennig and Flower, 2001), hypermotility (Farris et al., 1998), decreased capsule synthesis (Rowe et al., 2000), increased sensitivity to chloramphenicol (Duo et al., 2008), and reduced pathogenicity (Grant et al., 2003). Moreover, it also participated in regulating some mRNAs translation under stresses (Ero et al., 2016). Interestingly, deLivron reported that BipA has two different binding modes to the ribosome. It is associated with 70S in the form of GTP-bound under normal cellular conditions, whereas it interacts with 30S subunit under stress conditions (deLivron and Robinson, 2008). Therefore, they speculated that there exist two ribosome∙BipA complexes that affect the response of bacteria to environmental conditions (deLivron and Robinson, 2008). Moreover, there is a growing body of evidence that BipA also functions in the 50S ribosomal subunit biogenesis (Choudhury and Flower, 2015; Del Peso Santos et al., 2021). Choi et al. (2019) reported that BipA is vital for 50S biogenesis at a low temperature, whose expression is involved in the incorporation of L6 protein. The exogenous expression of the L20 coding gene rplT can partially repair the defects in rRNA processing and ribosomal assembly, and then restore the growth of bipA-deficient strains at low temperatures. So, the authors speculated that BipA and L20 may play a coordinating role in proper ribosomal assembly under cold-shock conditions. Another study led by Fredrick (Gibbs et al., 2020) demonstrated the flexible nature of the 50S assembly process. They also found GTP hydrolysis was crucial to the function of BipA.
Elongation Factors and Tumorigenesis
Recently, more and more elongation factors have been reported to function as oncoproteins or tumor promoters in cancer cells (Table 2). Elevated levels of protein synthesis are a critical feature of tumor cells. Alterations in protein synthesis can increase the overall translation rate and stimulate the translation of certain mRNAs to facilitate tumorigenesis, oncological progress, and survival.
TABLE 2.
Alterations of translation elongation factors in human cancer.
| Elongation factors | Observed modification | Association with cancer |
:---:
| eIF5A1 | Increased expression | eIF5A is highly overexpressed in patients with glioblastoma (Preukschas et al., 2012). eIF5A1 is a diagnostic marker of vulvar intraepithelial neoplasia (Cracchiolo et al., 2004). eIF5A1 is upregulated in colorectal adenoma (Lam et al., 2010) |
| eIF5A2 | Increased expression | eIF5A2 is amplified in ovarian carcinoma (Guan et al., 2001) and is associated with metastatic progression in colorectal cancers (Xie et al., 2008) |
| Gain of EIF5A2 results in recurrence of hepatocellular carcinoma (Chen et al., 2000) |
| Ectopic expression of EIF5A2 causes tumorigenesis in naked mice (Guan et al., 2004) |
| Overexpression of eIF5A2 results in local invasion of non-small-cell lung cancer (He et al., 2011) |
| eEF1A1 | Increased expression | eEF1A1 was upregulated in the infiltrating edge of invasive breast cancers (Zhu et al., 2003) |
| Dimethylation of eEF1A is upregulated and can serve as a diagnostic marker for poor outcomes in lung and pancreatic cancer (Liu et al., 2019) |
| eEF1A2 | Increased expression | eEF1A2 is distributed in 30% of ovarian cancers, and 26% of cancers with amplifications near the EEF1A2 |
| eEF1A2 acts as an oncoprotein which is upregulated in 67% of breast cancers (Tomlinson et al., 2005) |
| eEF2 | Increased expression | eEF2 is overexpressed in most of colorectal and gastric tumors and promotes cancer growth in vivo and in vitro (Nakamura et al., 2009) |
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eEF1A and Tumorigenesis
eEF1A is overexpressed in malignancies such as ovarian tumors. It can be inactivated by cytotoxic and anti-adipogenic ternatin and its derivatives (Carelli et al., 2015), with unclear mechanisms. Mutation in domain III of eEF1A hinders the binding of ternatin and confers resistance to its cytotoxic effects (Carelli et al., 2015). Cancers driven by the activation of PI3K-AKT axis are sensitive to the inhibitors of eEF1A (Lee and Surh, 2009). eEF1A1 was reported to be increased in the periphery of mammary cancer compared with the central region (Zhu et al., 2003), and the levels of eEF1A1 in neoplastic are relatively higher than in normal tissue (Lee and Surh, 2009). eEF1A2, an isoform of eEF1A, functions as cancer protein via a variety of mechanisms, such as facilitating cell invasion and migration by up-regulating MMP-9 and stimulating AKT in pancreatic tumors (Xu et al., 2013). It can accelerate proliferation and block apoptosis via down-regulating caspase 3 in prostate cancer tissues (Sun et al., 2014). Previous studies showed that the expression levels of eEF1A2 are elevated (Anand et al., 2002; Lee and Surh, 2009) and their genes (genomic region: 20q13) are amplified in a high proportion of solid tumors, such as ovarian (Iwabuchi et al., 1995) and breast cancers (Kallioniemi et al., 1994). For malignancy in the blood system, eEF1A2 is also up-regulated in cells of multiple myeloma (Losada et al., 2016). Therefore, eEF1A2 can be used as a diagnostic marker and target for some breast tumours (Tomlinson et al., 2005) and hematological malignancies (Shan et al., 2020).
Besides phosphorylation, other post-translational modifications on elongation factors have been found to be essential for tumorigenesis. Recently, it was reported that lysine 55 of eEF1A is dimethylated (eEF1AK55me2) by METTL13 (Methyltransferase-like 13), resulting in an increase of the inherent GTPase activity of eEF1A (Liu et al., 2019). This modification of eEF1A was utilized by RAS signal cascade to promote translational output and facilitate carcinogenesis (Figure 7). Therefore, METTL13-eEF1AK55me2 signal pathway is vital for tumors to cope with increased translational demand, and METTL13 inhibition might be an effective way to do targeted intervention for RAS driven cancers.
FIGURE 7.
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Methylation of eEF1A and the role of eEF1AK55me2 mediated translational control in tumorigenesis. Expression of METTL13 and eEF1AK55me2 are upregulated in cancer, and negatively correlate with the survival of pancreatic and lung cancer patient. Increased METTL13 and eEF1AK55me2 promotes Ras-driven tumorigenesis in vivo.
eEF2 and Tumorigenesis
eEF2 plays an essential role in many biological processes including cell cycle (Hizli et al., 2013) and genotoxic stress (Kruiswijk et al., 2012). Previous studies have shown that eEF2 is overexpressed in various tumors. Translation of eEF2 was up-regulated in most of colorectal (91.7%) and gastric (92.9%) tumors, leading to the elevation of in vivo tumorigenicity (Nakamura et al., 2009). Therefore, eEF2 is a potential target for tumor immunotherapy in multiple cancers (Oji et al., 2014).
Under conditions of energy depletion or nutrient deprivation, tumor cells redistribute energy resources through weakening overall translation, while translating specific mRNAs to cope with stresses and fight for survival. AMPK, a sensor of energy, exists in a variety of tumors. When AMPK is activated, it can stimulate eEF2K, which further inhibits eEF2 activity by phosphorylating the Thr56 of eEF2. Inhibition of eEF2 leads to slowing down of protein translation (Proud, 2015), and prevents tumor cells from growing under nutritional deficiency (Leprivier et al., 2013) (Figure 8). Therefore, eEF2K is a negative regulator of protein synthesis (Wang et al., 2017), and inhibition of eEF2K activity may have therapeutic significance in preventing the survival of tumor cells and recovering protein translation.
FIGURE 8.
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Regulation mechanism of eEF2 phosphorylation modification in cancer cells.
eEF2K is an atypical kinase, and plays an important role in the migration and survival of tumor cells. It is overexpressed in glioblastoma (Leprivier et al., 2013), medulloblastoma, mammary cancer (Liu et al., 2014) and pancreatic tumor (Ashour et al., 2014). eEF2K is activated by stresses such as nutrition deficiency, hypoxia (Moore et al., 2015), acidosis (Xie et al., 2015) and cellular energy depletion (Browne et al., 2004). Inhibiting eEF2K decreases the invasion and migration of tumor cells, while deletion of eEF2 or eEF2K overexpression promotes wound healing and invasion. These results indicated that eEF2K has a protective function in tumor cells and therefore can be used as a molecular target to prevent cancer metastasis (Xie et al., 2018). Nevertheless, a recent study showed that eEF2K can protect cells under stress conditions and make tumor cells adapt to stresses (Leprivier et al., 2013). Therefore, stimulating eEF2K-eEF2 axis to suppress tumor requires serious assessment, and more in-depth studies are required to comprehensively evaluate its effectiveness from a treatment perspective (Knight et al., 2020).
eIF5A and Tumorigenesis
eIF5A is involved in the invasive and/or metastatic processes of several types of human cancer (Mathews and Hershey, 2015). There are two subtypes of eIF5A in mammals, both of which were modified with hypusine on the same lysine (Clement et al., 2003). eIF5A1 is widely distributed in rapidly proliferating cells while eIF5A2 is expressed in a tissue-specific manner and is almost undetectable in most cases (Clement et al., 2006). It was found that both eIF5A1 and eIF5A2 are associated with several malignancies (Mathews and Hershey, 2015). Interestingly, the expression of eIF5A1 and eIF5A2, along with DOHH and DHPS, are increased in multiple tumors (Nakanishi and Cleveland, 2016). Recently, it was found that eIF5A2 promotes doxorubicin resistance of colon cancer cells by regulating EMT, suggesting that inhibition of eIF5A2 can be used as a way to reverse the drug resistance of colorectal cancer (Figure 9) (Bao et al., 2015). An early study suggested that eIF5A2 may aid tumorigenesis via promoting the translation of some mRNAs that boost DNA replication and provoke excessive proliferation of tumor cells (Hanauske-Abel et al., 1995). Nevertheless, more in-depth studies on the function and molecular mechanism of eIF5A and its hypusination modification in tumors are needed.
FIGURE 9.
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The role of eIF5A mediated translational control in tumorigenesis.
Concluding Remarks
The elongation phase of translation is an important regulatory node in health and disease. Dysregulation of this process is often related to various disorders including tumors, neurodegenerative diseases and cardiovascular diseases. The close link between translational factors and human diseases are well coincident with the concept that gene expression is accurately regulated at the translational level (Dever et al., 2018). There are multiple levels at which protein translation can be regulated to hinder disease progression. Clarifying how post-translational modifications control the activity of translational factors in tumors is helpful to reveal the regulation mechanism in tumorigenesis, and provides a rationale for the new interventional treatment (Xu and Ruggero, 2020). There are still many questions about translation elongation that need to be resolved: 1) The exact physiological function of phosphorylated EF-G and EF-Tu; 2) The molecular mechanism of how ADP-ribosylation impairs the function of eEF2; 3) The molecular mechanism of EF4 in protein translation under stress conditions; 4) The unique requirement of eEF3 in fungal and its action mechanism in translation elongation; 5) The feasibility of eEF2K as a tumor therapeutic target; 6) The molecular mechanism of eIF5A and its hypusine modification in protein translation and tumorigenesis.
Author Contributions
BX and LL: Prepared the figures and tables and wrote the original draft; GS: Revised the manuscript.
Funding
This work was supported by the Key R&D Projects of Introducing High-Level Scientific and Technological Talents in Lvliang City (2021RC-1-4), Starting Fund for Talent Introduction of Fenyang College of Shanxi Medical University (2020A01), Scientific and Technological Innovation Programs of Higher Education Institutions in Shanxi (2020L0749), National College Students’ Innovation and Entrepreneurship Training Program (202117114001), Key Projects of Innovation and Entrepreneurship Training Program for College Students in Shanxi Province (S202117114008), Project of Lvliang City Science and Technology Program (2020SHFZ29), and National Natural Science Foundation of China (31771015).
Conflict of Interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s Note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors, and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Abstract
Introduction
Factors Involved in AA-tRNA Incorporation
Factors Involved in Ribosome Translocation
Other Elongation Factors
Elongation Factors and Tumorigenesis
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1581 | https://openstax.org/books/algebra-1/pages/1-5-2-graphing-linear-functions-in-two-variables | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Algebra 1
1.5.2 Graphing Linear Functions in Two Variables
Algebra 11.5.2 Graphing Linear Functions in Two Variables
Search for key terms or text.
1.5.2 • Graphing Linear Functions in Two Variables
Activity
For questions 1 – 7, use the equation
Find the value of when .
3
Find the value of when .
4
Plot both of those points on a coordinate grid and use a ruler to draw a line.
Compare your answer:
What is the coordinate of the -intercept?
Compare your answer:
What is the coordinate of the -intercept?
Compare your answer:
What are two other names for the -intercept?
Compare your answer:
zero, solution
What is the slope of the line?
Self Check
Graph the equation . Which of the following describes the intercepts and slope of the graph?
-intercept: ; -intercept: ; Slope:
-intercept: ; -intercept: ; Slope:
-intercept: ; -intercept: ; Slope:
-intercept: ; -intercept: ; Slope:
Additional Resources
Graphing a Line and Describing Characteristics
There are many ways to graph an equation of a line. One way is to use the -intercepts and -intercepts as two points on the line.
Example
Graph using the intercepts, then identify key characteristics.
Step 1 - Find the -intercept. To find the -intercept, let .
is the -intercept The -intercept is also called a solution or zero.
Step 2 -Find the -intercept. To find the -intercept, let .
is the -intercept
Step 3 -Graph the line by connecting the -intercepts.
Step 4 - Find the slope of the line.
Try it
Graphing a Line and Describing Characteristics
For questions 1 – 2, use the equation .
Graph the equation using the intercepts.
Compare your answer:
Answer:
What is the slope of the line?
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Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis, Jay Abramson, Sharon North, Amy Baldwin, Alyssa Howell
Publisher/website: OpenStax
Book title: Algebra 1
Publication date: Jun 4, 2025
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1582 | https://www.psychiatry.org/getmedia/304e9a98-ca5c-4e04-97d7-30b4756705eb/APA-DSM5TR-Update-September-2022.pdf | DSM-5-TR™ Update Supplement to Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition, Text Revision September 2022 DSM-5-TRTM Update, September 2022, Page 2 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Copyright © 2022 American Psychiatric Association. DSM, DSM-5, and DSM-5-TR are registered trademarks of the American Psychiatric Association. Use of these terms is prohibited without permission of the American Psychiatric Association. ALL RIGHTS RESERVED. Unless authorized in writing by the APA (APA), no part of this supplement may be reproduced or used in a manner inconsistent with the APA’s copyright. This prohibition applies to unauthorized uses or reproductions in any form, including electronic applications. Correspondence regarding copyright permissions should be directed to DSM Permissions, American Psychiatric Association Publishing, 800 Maine Ave. SW, Suite 900, Washington, DC 20024-2812. DSM-5-TRTM Update, September 2022, Page 3 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. This supplement reflects updates to diagnostic criteria and related text, as well as coding updates, changes, or corrections. This supplement is intended to be used only in conjunction with DSM-5-TR, and it should not be relied upon as an independent source of information. Cautionary Statement: DSM-5 is a classification of mental disorders that was developed for use in clinical, educational, and research settings. The diagnostic categories, criteria, and textual descriptions are meant to be employed by individuals with appropriate clinical training and experience in diagnosis. It is important that DSM-5 not be applied mechanically by individuals without clinical training. The specific diagnostic criteria included in DSM-5 are meant to serve as guidelines to be informed by clinical judgment and are not meant to be used in a rigid cookbook fashion. DSM-5-TRTM Update, September 2022, Page 4 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Table of Contents I. USE OF THE MANUAL 6 I.A SUBTYPES AND SPECIFIERS (TEXT UPDATE) 6 II. SCHIZOPHRENIA SPECTRUM AND OTHER PSYCHOTIC DISORDERS 6 II.A CATATONIC DISORDER DUE TO ANOTHER MEDICAL CONDITION (CODING UPDATE TO ICD-10-CM MEDICAL CODES USED AS EXAMPLES) 6 III. BIPOLAR AND RELATED DISORDERS 6 III.A BIPOLAR I: MAJOR DEPRESSIVE EPISODE, CRITERION A9 (CRITERIA UPDATE) 6 III.B BIPOLAR II: MAJOR DEPRESSIVE EPISODE, CRITERION A9 (CRITERIA UPDATE) 6 III.C BIPOLAR SPECIFIERS, MANIC OR HYPOMANIC EPISODE, WITH MIXED FEATURES, CRITERION A6 (CRITERIA UPDATE) 6 IV. DEPRESSIVE DISORDERS 7 IV.A MAJOR DEPRESSIVE DISORDER 7 IV.A.1 MAJOR DEPRESSIVE DISORDER, CRITERION A.9 (CRITERIA UPDATE) 7 IV.A.2 MAJOR DEPRESSIVE DISORDER (TEXT UPDATE) 7 V. TRAUMA AND STRESSOR-RELATED DISORDERS 7 V.A PROLONGED GRIEF DISORDER (CODING UPDATE TO ICD-10-CM DISORDER CODE) 7 V.B OTHER SPECIFIED TRAUMA- AND STRESSOR-RELATED DISORDER (CODING UPDATE TO ICD-10-CM DISORDER CODE) 7 VI. SUBSTANCE-RELATED AND ADDICTIVE DISORDERS 8 VI.A OPIOID-INDUCED ANXIETY DISORDER (CODING UPDATE TO ICD-10-CM DISORDER CODE) 8 VII. NEUROCOGNITIVE DISORDERS 8 VII.A CHAPTER INTRODUCTION (TEXT UPDATE) 8 VII.B DELIRIUM (CODING UPDATE TO ICD-10-CM MEDICAL CODES USED AS EXAMPLES) 9 VII.C OTHER SPECIFIED DELIRIUM (CODING UPDATE TO ICD-10-CM DISORDER CODE) 9 VII.D UNSPECIFIED DELIRIUM (CODING UPDATE TO ICD-10-CM DISORDER CODE) 9 VII.E MAJOR NEUROCOGNITIVE DISORDER (CRITERIA UPDATE – SPECIFIERS, CODING UPDATE TO ICD-10-CM DISORDER CODE) 9 VII.F MILD NEUROCOGNITIVE DISORDERS (CODING UPDATE TO ICD-10-CM DISORDER CODE) 13 VII.G MAJOR OR MILD NEUROCOGNITIVE DISORDER DUE TO TRAUMATIC BRAIN INJURY (CODING UPDATE TO ICD-10-CM MEDICAL CODES USED AS EXAMPLES) 13 DSM-5-TRTM Update, September 2022, Page 5 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. VII.H MAJOR OR MILD NEUROCOGNITIVE DISORDER DUE TO UNKNOWN ETIOLOGY (NEWLY ADDED DISORDER) 14 VII.I UNSPECIFIED NEUROCOGNITIVE DISORDER (CHANGE IN THE DISORDER DEFINITION) 14 VIII. OTHER CONDITIONS THAT MAY BE A FOCUS OF CLINICAL ATTENTION 15 VIII.A CURRENT SUICIDAL BEHAVIOR, INITIAL AND SUBSEQUENT ENCOUNTERS (CODING UPDATE TO ICD-10-CM CONDITION CODE) 15 VIII.B NONADHERENCE TO MEDICAL TREATMENT (CODING UPDATE TO ICD-10-CM CONDITION CODE) 15 VIII.C IMPAIRING EMOTIONAL OUTBURSTS (NEWLY ADDED CONDITION) 15 IX. ASSESSMENT MEASURES 16 IX.A WORLD HEALTH ORGANIZATION DISABILITY ASSESSMENT SCHEDULE 2.0 (TEXT UPDATE) 16 IX.B DSM-5-TR PARENT/GUARDIAN-RATED LEVEL 1 CROSS-CUTTING SYMPTOM MEASURE—CHILD AGE 6–17 (UPDATE TO THE ASSESSMENT MEASURE) 16 IX.C DSM-5-TR SELF-RATED LEVEL 1 CROSS-CUTTING SYMPTOM MEASURE— CHILD AGE 11–17 (UPDATE TO THE ASSESSMENT MEASURE) 16 X. CONDITIONS FOR FURTHER STUDY 16 X.A DEPRESSIVE EPISODES WITH SHORT-DURATION HYPOMANIA (SECTION III): CRITERION A9 (CRITERIA UPDATE) 16 X.B SUICIDAL BEHAVIOR DISORDER (REMOVAL OF THE CONDITION) 17 XI. ALPHABETICAL LISTING OF DSM-5-TR DIAGNOSES AND ICD-10-CM CODES 17 XI.A FOOD INSECURITY (CODING UPDATE TO DSM DISORDER CODE) 17 XII. DSM-5-TR CODING UPDATE TABLE 18 XII.A LISTING OF DSM-5-TR DIAGNOSES AND NEW ICD-10-CM CODES 18 XII.B CODING CORRECTIONS (EFFECTIVE IMMEDIATELY) 27 DSM-5-TRTM Update, September 2022, Page 6 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. I. Use of the Manual I.A Subtypes and Specifiers (Text Update) Due to coding updates in the Neurocognitive Disorders chapter listed below, the coding example in the Subtypes and Specifiers section was revised as follows: Original: “(e.g., “0” in the fifth character in the F02.80 diagnostic code for major neurocognitive disorder due to Alzheimer’s disease, to indicate the absence of a behavioral disturbance versus a “1” in the fifth character of the F02.81 diagnostic code for major neurocognitive disorder due to Alzheimer’s disease to indicate the presence of a behavioral disturbance)” Updated: “(e.g., “0” in the fifth character in the F06.70 diagnostic code for mild neurocognitive disorder due to traumatic brain injury, to indicate the absence of a behavioral disturbance versus a “1” in the fifth character of the F06.71 diagnostic code for mild neurocognitive disorder due to traumatic brain injury to indicate the presence of a behavioral disturbance)” II. Schizophrenia Spectrum and Other Psychotic Disorders II.A Catatonic Disorder Due to Another Medical Condition (Coding Update to ICD-10-CM Medical Codes Used as Examples) The ICD-10-CM code for Hepatic Encephalopathy (listed under Catatonic Disorder Due to Another Medical Condition) was revised as follows: Coding note: The code for hepatic encephalopathy was revised as follows: Hepatic Encephalopathy - Original code (valid through September 30, 2022): K72.90 Hepatic Encephalopathy - Updated code (Valid on October 1, 2022): K76.82 III. Bipolar and Related Disorders III.A Bipolar I: Major Depressive Episode, Criterion A9 (Criteria Update) III.B Bipolar II: Major Depressive Episode, Criterion A9 (Criteria Update) III.C Bipolar Specifiers, Manic or hypomanic episode, with mixed features, Criterion A6 (Criteria Update) In the criteria of three disorders listed above, the term “commit suicide” was replaced with more sensitive and less stigmatizing language Original: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, or a suicide attempt or a specific plan for committing suicide.” Updated: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, a specific suicide plan, or a suicide attempt” DSM-5-TRTM Update, September 2022, Page 7 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. IV. Depressive Disorders IV.A Major Depressive Disorder IV.A.1 Major Depressive Disorder, Criterion A.9 (Criteria Update) The term “commit suicide” was replaced with more sensitive and less stigmatizing language Original: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, or a suicide attempt or a specific plan for committing suicide.” Updated: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, a specific suicide plan, or a suicide attempt” IV.A.2 Major Depressive Disorder (Text Update) Prolonged grief disorder was added as a new entry in the differential diagnosis for Major Depressive Disorder Prolonged grief disorder. Prolonged grief disorder is a persistent pervasive grief response that continues to cause clinically significant distress or impairment for more than 12 months after the death of someone close. It can be differentiated from a major depressive episode not only by the requirement for intense yearning or longing for, or preoccupation with, the deceased, but also by the requirement that the other symptoms such as emotional pain (e.g., anger, bitterness, sorrow), marked reduction in emotional experiences, feeling that life is meaningless, and difficulty reintegrating socially or feeling engaged in ongoing activities be judged to result from the significant interpersonal loss. By contrast, in a major depressive episode, there is a more generalized depressed mood that is not specifically related to the loss. It should be noted that both prolonged grief disorder and major depressive disorder should be diagnosed if criteria for both are met. V. Trauma and Stressor-Related Disorders V.A Prolonged Grief Disorder (Coding Update to ICD-10-CM Disorder Code) The ICD-10-CM code for Prolonged Grief Disorder (on DSM-5-TR Classification, the Disorder page, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: Prolonged Grief Disorder - Original code (valid through September 30, 2022): F43.8 Prolonged Grief Disorder - Updated code (Valid on October 1, 2022): F43.81 V.B Other Specified Trauma- and Stressor-Related Disorder (Coding Update to ICD-10-CM Disorder Code) The ICD-10-CM code for Other Specified Trauma- and Stressor-Related Disorder (in the DSM-5-TR Classification, the Disorder page, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: DSM-5-TRTM Update, September 2022, Page 8 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Other Specified Trauma- and Stressor-Related Disorder - Original code (valid through September 30, 2022): F43.8 Other Specified Trauma- and Stressor-Related Disorder - Updated code (Valid on October 1, 2022): F43.89 VI. Substance-Related and Addictive Disorders VI.A Opioid-Induced Anxiety Disorder (Coding Update to ICD-10-CM Disorder Code) Codes to be corrected are found in the DSM-5-TR Classification. (ICD-10-CM codes are correctly listed in all other places where this disorder appears in the manual.) . Opioid-Induced Anxiety Disorders F11.180 F11.188 With mild use disorder F11.280 F11.288 With moderate or severe use disorder F11.980 F11.988 Without use disorder VII. Neurocognitive Disorders VII.A Chapter Introduction (Text Update) The following text will be added to the chapter introduction before “Neurocognitive Domains” section: For major and mild NCDs, the diagnostic criteria for several of the etiological subtypes allow for the designation of the degree of certainty regarding the possible presence of the medical conditions, as well the strength of the causal connection between the medical condition and the NCD. For NCD due to Alzheimer’s disease, frontotemporal NCD, and NCD with Lewy bodies, establishing whether these medical conditions are present in the individual can be extremely challenging, and sometimes the etiology can only be firmly established postmortem; for these subtypes, the probable/possible designation precedes the name of the medical condition (e.g., mild NCD due to possible Alzheimer’s disease, major NCD due to probable frontotemporal degeneration). Because the diagnostic criteria for vascular NCD and NCD due to Parkinson’s disease require clear evidence of the presence of vascular disease or Parkinson’s disease, respectively, for those subtypes the uncertainty is about the causal connection between the medical condition and the NCD. For those subtypes, the designations “probably due to” and “possibly due to” apply. DSM-5-TRTM Update, September 2022, Page 9 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. VII.B Delirium (Coding Update to ICD-10-CM Medical Codes Used as Examples) The ICD-10-CM code for Hepatic Encephalopathy (listed coding note for delirium) was revised as follows: Coding note: The code for hepatic encephalopathy was revised as follows (occurs in two places): Hepatic Encephalopathy - Original code (valid through September 30, 2022): K72.90 Hepatic Encephalopathy - Updated code (Valid on October 1, 2022): K76.82 VII.C Other Specified Delirium (Coding Update to ICD-10-CM Disorder Code) The ICD-10-CM code for Other Specified Delirium (in the DSM-5-TR Classification, the Disorder page, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: Other Specified Delirium - Original code: R41.0 Other Specified Delirium - Updated code: F05 VII.D Unspecified Delirium (Coding Update to ICD-10-CM Disorder Code) The ICD-10-CM code for Unspecified Delirium (in the DSM-5-TR Classification, the Disorder page, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: Unspecified Delirium - Original code: R41.0 Unspecified Delirium - Updated code: F05 VII.E Major Neurocognitive Disorder (Criteria Update – Specifiers, Coding Update to ICD-10-CM Disorder Code) Effective October 1, 2022, the new ICD-10-CM coding scheme summarized below replaces the current coding approach for major and mild neurocognitive disorders. For major neurocognitive disorders, F01, F02, or F03 is used depending on the medical etiology, as shown in the table below. The severity specifiers mild, moderate, and severe are now coded in the 4th character (A, B, or C, respectively) as listed below in green type: DSM-5-TRTM Update, September 2022, Page 10 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Note: NCD subtypes listed in order of appearance in DSM-5-TR. Major NCD Due to Probable [Medical Etiology] Major NCD Due to Possible [Medical Etiology] Major NCD Due to Probable Vascular Disease Major NCD Due to Possible Vascular Disease Major NCD Due to [Medical Etiology] Major NCD Due to Unknown Etiology Code first the etiological medical conditiona. No medical code is neededa. No medical code is needed for vascular disease. No medical code is needed for vascular disease. Code first the etiological medical conditionb. No medical code is needed • F02.A- Mild • F02.B- Moderate • F02.C- Severe • F03.A- Mild • F03.B- Moderate • F03.C- Severe • F01.A- Mild • F01.B- Moderate • F01.C- Severe • F03.A- Mild • F03.B- Moderate • F03.C- Severe • F02.A- Mild • F02.B- Moderate • F02.C- Severe • F03.A- Mild • F03.B- Moderate • F03.C- Severe aProbable and possible medical etiologies comprise the following (with etiological codes used only for the probable diagnoses): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). Probable and possible vascular disease are listed separately in the table. bMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply, with exception of vascular disease which does not receive a medical code). Following the 4th character severity codes (A, B, C) noted above, individual 5th and 6th character codes for accompanying behavioral or psychological disturbances are then added according to the applicable specifier (e.g., major neurocognitive disorder with probable frontotemporal degeneration, severe, with agitation would be coded as F02.C11). When more than one behavioral or psychological disturbance occurs, code each of the disturbances separately as if they were separate conditions. For example, for major neurocognitive disorder with probable Alzheimer’s disease, severe, accompanied by agitation, delusions, and depression, four codes are needed: G30.9 Alzheimer’s disease; F02.C11 (major NCD with probable Alzheimer’s disease, severe, with agitation), F02.C2 (major NCD with probable Alzheimer’s disease, severe, with psychotic disturbance), and F02.C3 (major NCD with probable Alzheimer’s disease, severe, with mood symptoms). Original specifiers: Without behavioral disturbance: If the cognitive disturbance is not accompanied by any clinically significant behavioral disturbance. With behavioral disturbance (specify disturbance): If the cognitive disturbance is accompanied by a clinically significant behavioral disturbance (e.g., psychotic symptoms, mood disturbance, agitation, apathy, or other behavioral symptoms). DSM-5-TRTM Update, September 2022, Page 11 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Updated specifiers: Note: “x” in the 4th character of the codes below represents A, B, or C (for mild, moderate, or severe severity, respectively), as noted in the prior table above. .x11 With agitation: If the cognitive disturbance is accompanied by clinically significant agitation. .x4 With anxiety: If the cognitive disturbance is accompanied by clinically significant anxiety. .x3 With mood symptoms: If the cognitive disturbance is accompanied by clinically significant mood symptoms (e.g., dysphoria, irritability, euphoria). .x2 With psychotic disturbance: If the cognitive disturbance is accompanied by delusions or hallucinations. .x18 With other behavioral or psychological disturbance: If the cognitive disturbance is accompanied by other clinically significant behavioral or psychological disturbance (e.g., apathy, aggression, disinhibition, disruptive behaviors or vocalizations, sleep or appetite/eating disturbance). .x0 Without accompanying behavioral or psychological disturbance: If the cognitive disturbance is not accompanied by any clinically significant behavioral or psychological disturbance. DSM-5-TRTM Update, September 2022, Page 12 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. This table summarizes the new NCD coding approach, combining the components noted above. Note that x in the 4th character represents the severity codes A, B, or C. Finally, code any accompanying behavioral or psychological disturbance (fifth and sixth characters). Note: NCD subtypes listed in order of appearance in DSM-5-TR. Major NCD Due to Probable [Medical Etiology] Major NCD Due to Possible [Medical Etiology] Major NCD Due to Probable Vascular Disease Major NCD Due to Possible Vascular Disease Major NCD Due to [Medical Etiology] Major NCD Due to Unknown Etiology Code first the etiological medical conditiona. No additional medical code. a No additional medical code for vascular disease. No additional medical code for vascular disease. Code first the etiological medical conditionb. No additional medical code • F02.x11 With agitation • F02.x4 With anxiety • F02.x3 With mood symptoms • F02.x2 With psychotic disturbance • F02.x18 With other behavioral or psychological disturbance (e.g., apathy) • F02.x0 Without accompanying behavioral or psychological disturbance • F03.x11 With agitation • F03.x4 With anxiety • F03.x3 With mood symptoms • F03.x2 With psychotic disturbance • F03.x18 With other behavioral or psychological disturbance (e.g., apathy) • F03.x0 Without accompanying behavioral or psychological disturbance • F01.x11 With agitation • F01.x4 With anxiety • F01.x3 With mood symptoms • F01.x2 With psychotic disturbance • F01.x18 With other behavioral or psychological disturbance (e.g., apathy) • F01.x0 Without accompanying behavioral or psychological disturbance • F03.x11 With agitation • F03.x4 With anxiety • F03.x3 With mood symptoms • F03.x2 With psychotic disturbance • F03.x18 With other behavioral or psychological disturbance (e.g., apathy) • F03.x0 Without accompanying behavioral or psychological disturbance • F02.x11 With agitation • F02.x4 With anxiety • F02.x3 With mood symptoms • F02.x2 With psychotic disturbance • F02.x18 With other behavioral or psychological disturbance (e.g., apathy) • F02.x0 Without accompanying behavioral or psychological disturbance • F03.x11 With agitation • F03.x4 With anxiety • F03.x3 With mood symptoms • F03.x2 With psychotic disturbance • F03.x18 With other behavioral or psychological disturbance (e.g., apathy) • F03.x0 Without accompanying behavioral or psychological disturbance aProbable and possible medical etiologies comprise the following (with etiological codes used only for the probable diagnoses): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). Probable and possible vascular disease are listed separately in the table. bMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply, with exception of vascular disease which does not receive a medical code). Coding note: When more than one behavioral or psychological disturbance occurs, code for each of the disturbances. For example, for major neurocognitive disorder with probable Alzheimer’s disease, severe, accompanied by agitation, delusions, and depression, four codes are needed: G30.9 Alzheimer’s disease; F02.C11 (major NCD, severe, with agitation), F02.C2 (major NCD, severe, with psychotic disturbance), and F02.C3 (major NCD, severe, with mood symptoms). DSM-5-TRTM Update, September 2022, Page 13 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. VII.F Mild Neurocognitive Disorders (Coding Update to ICD-10-CM Disorder Code) In DSM-5-TR Classification, Mild Neurocognitive Disorder Coding and Recording Procedures, Mild Neurocognitive Disorder coding table, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) Coding note: Code based on medical or substance etiology. An additional code indicating the etiological medical condition must immediately precede the diagnostic code F06.7z for mild neurocognitive disorder due to a medical etiology. An additional code is not used for medical etiologies that are judged to be “possible” (i.e., mild NCD due to possible Alzheimer’s disease, due to possible frontotemporal degeneration, due to possible Lewy body disease, possibly due to vascular disease, possibly due to Parkinson’s disease). See coding table on pp. 682–683. For substance/medication-induced mild neurocognitive disorder, code based on type of substance; see “Substance/Medication-Induced Major or Mild Neurocognitive Disorder.” Note: G31.84 is used for mild neurocognitive disorder due to unknown etiology and for mild neurocognitive disorder due to a possible medical etiology (e.g., possible Alzheimer’s disease); no additional code for medical or substance etiology is used. Mild NCD Due to Probable Etiologya, Mild NCD Due to [Medical Etiology]b Mild NCD Due to Possible Etiologyc, Mild NCD Due to Unknown Etiology Code first the etiological medical condition. No additional medical code • F06.70 –Without behavioral disturbance • F06.71 –With behavioral disturbance • G31.84 aProbable medical etiologies comprise the following (with etiological codes): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), vascular disease (code first I67.9), Parkinson’s disease (code first G20). bMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply). cPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. VII.G Major or Mild Neurocognitive Disorder Due to Traumatic Brain Injury (Coding Update to ICD-10-CM Medical Codes Used as Examples) The code for Diffuse traumatic brain injury with loss of consciousness of unspecified duration, sequela (in DSM-5-TR Classification, Major Neurocognitive Disorder, Coding and Recording Procedures, Major and Mild Neurocognitive Disorders coding table, Major or Mild Neurocognitive Disorder Due to Traumatic Brain Injury, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised in DSM-5-TR as follows: Diffuse traumatic brain injury with loss of consciousness of unspecified duration, sequela – Original code (valid through September 30, 2022): S06.2X9S Diffuse traumatic brain injury with loss of consciousness of unspecified duration, sequela – Original code (Valid on October 1, 2022): S06.2XAS DSM-5-TRTM Update, September 2022, Page 14 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. VII.H Major or Mild Neurocognitive Disorder Due to Unknown Etiology (Newly Added Disorder) This new diagnosis, major or mild neurocognitive disorder due to unknown etiology, will be added after major or mild neurocognitive disorder due to multiple etiologies. Diagnostic Criteria: A. The criteria are met for major or mild neurocognitive disorder. B. There is evidence from the history, physical examination, or laboratory findings that suggest the neurocognitive disorder is the pathophysiological consequence of a presumed medical condition, a combination of medical conditions, or a combination of medical conditions and substances or medications, but there is insufficient information to establish a specific cause. C. The cognitive deficits are not better explained by another mental disorder or substance/medication-induced neurocognitive disorder and do not occur exclusively during the course of a delirium. Coding note (see coding table on pp. 682–683): For major neurocognitive disorder (NCD) due to unknown etiology: 1) code first F03 (there is no additional medical code). 2) Next, code the current severity of the cognitive disturbance (mild, moderate, severe) and 3) whether or not there is an accompanying behavioral or psychological disturbance. For example, for major NCD due to unknown etiology, moderate, with psychotic disturbance, the ICD-10-CM code is F03.B2. For major NCD with multiple clinically significant behavioral and psychological disturbances, multiple ICD-10-CM codes are needed. For example, major NCD with unknown etiology, severe, accompanied by agitation, delusions, and depression, three codes are needed: F03.C11 (with agitation); F03.C2 (with psychotic disturbance); and F03.C3 (with mood symptoms). For mild NCD due to unknown etiology, code G31.84. (Note: “With behavioral disturbance” and “Without behavioral disturbance” cannot be coded but should still be recorded.) This category is included to cover the clinical presentation of a major or mild neurocognitive disorder for which there is evidence from the history, physical examination, or laboratory findings suggestive of a medical etiology or a medical etiology in combination with use of a substance or medication, but there is insufficient information to establish a specific cause. VII.I Unspecified Neurocognitive Disorder (Change in the Disorder Definition) Unspecified neurocognitive disorder description was modified following the addition of “Major or Mild Neurocognitive Disorder Due to Unknown Etiology”: This category applies to presentations in which symptoms characteristic of a neurocognitive disorder that cause clinically significant distress or impairment in social, occupational, or other important areas of functioning predominate but do not meet the full criteria for any of the disorders in the neurocognitive disorders diagnostic class. The unspecified neurocognitive disorder category is used in situations in which the precise etiology cannot be determined with sufficient certainty to make an etiological attribution. Coding note: For unspecified major or mild neurocognitive disorder, code R41.9. (Note: Do not use additional codes for any presumed etiological medical conditions.) “With behavioral disturbance” and “without behavioral disturbance” cannot be coded but should still be recorded.) DSM-5-TRTM Update, September 2022, Page 15 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. VIII. Other Conditions That May Be a Focus of Clinical Attention VIII.A Current Suicidal Behavior, Initial and Subsequent Encounters (Coding Update to ICD-10-CM Condition Code) For T codes only, the 6th 7th character should be coded as follows: The code for Current Suicidal Behavior, Initial encounter (in DSM-5-TR Classification, Other Conditions That May Be a Focus of Clinical Attention, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: Current Suicidal Behavior, Initial encounter – As appears in DSM-5-TR: T14.91A Current Suicidal Behavior, Initial encounter – Corrected code (Valid IMMEDIATELY): T14.91XA The code for Current Suicidal Behavior, Subsequent encounter (in DSM-5-TR Classification, Other Conditions That May Be a Focus of Clinical Attention, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised as follows: Current Suicidal Behavior, Subsequent encounter – As appears in DSM-5-TR: T14.91D Current Suicidal Behavior, Subsequent encounter – Corrected code (Valid MMEDIATELY): T14.91XD VIII.B Nonadherence to Medical Treatment (Coding Update to ICD-10-CM Condition Code) The code for Nonadherence to medical treatment (in DSM-5-TR Classification, Other Conditions That May Be a Focus of Clinical Attention, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes) was revised in DSM-5-TR as follows: Nonadherence to Medical Treatment – Original code (valid through September 30, 2022): Z91.19 Nonadherence to Medical Treatment- Updated code (Valid on October 1, 2022): Z91.199 VIII.C Impairing Emotional Outbursts (Newly Added Condition) Impairing Emotional Outbursts is being added to DSM-5-TR Classification, Other Conditions That May Be a Focus of Clinical Attention, Alphabetical and Numerical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes: R45.89 Impairing Emotional Outbursts This category may be used when the focus of clinical attention is displays of anger or distress manifested verbally (e.g., verbal rages, uncontrolled crying) and/or behaviorally (e.g., physical aggression toward people, property, or self) that lead to significant functional impairment. In addition to occurring in the context of a number of different mental disorders (e.g., attention-deficit/hyperactivity disorder, autism spectrum disorder, oppositional defiant disorder, generalized anxiety disorder, posttraumatic stress disorder, mood and psychotic disorders) they can also occur independently of other conditions, as is often the case in young children. DSM-5-TRTM Update, September 2022, Page 16 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. IX. Assessment Measures IX.A World Health Organization Disability Assessment Schedule 2.0 (Text Update) A broken link in the WHODAS 2.0 Scoring Instructions provided by WHO on p. 854 and in the Online Assessment Measures was replaced. CURRENT: “WHODAS 2.0 population norms. For the population norms for IRT-based scoring of the WHODAS 2.0 and for the population distribution of IRT-based scores for WHODAS 2.0, please see www.who.int/classifications/icf/Popnorms_distrib_IRT_scores.pdf” Correction: “WHODAS 2.0 population norms. For the population norms for IRT-based scoring of the WHODAS 2.0 and for the population distribution of IRT-based scores for WHODAS 2.0, see Table 6.1 and Figure 6.1 (p. 43) in the free online PDF manual published by the World Health Organization: “Measuring Health and Disability: Manual for WHO Disability Assessment Schedule (WHODAS 2.0),” June 2012.” IX.B DSM-5-TR Parent/Guardian-Rated Level 1 Cross-Cutting Symptom Measure—Child Age 6–17 (Update to the Assessment Measure) The term “commit suicide” was replaced with more sensitive and less stigmatizing language Original: “24. In the past TWO (2) WEEKS, has he/she talked about wanting to kill himself/herself or about wanting to commit suicide?” Updated: “24. In the past TWO (2) WEEKS, has he/she talked about wanting to kill himself/herself or about wanting to end their life?” IX.C DSM-5-TR Self-Rated Level 1 Cross-Cutting Symptom Measure— Child Age 11–17 (Update to the Assessment Measure) The term “commit suicide” was replaced with more sensitive and less stigmatizing language Appears on the DSM website, Online Assessment Measures Original: “24. In the last 2 weeks, have you thought about killing yourself or committing suicide?” Updated: “24. In the last 2 weeks, have you thought about killing yourself or ending your life?” X. Conditions for Further Study X.A Depressive Episodes With Short-Duration Hypomania (Section III): Criterion A9 (Criteria Update) The term “commit suicide” was replaced with more sensitive and less stigmatizing language DSM-5-TRTM Update, September 2022, Page 17 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Original: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, or a suicide attempt or a specific plan for committing suicide.” Updated: “Recurrent thoughts of death (not just fear of dying), recurrent suicidal ideation without a specific plan, a specific suicide plan, or a suicide attempt” X.B Suicidal Behavior Disorder (Removal of the Condition) Update The DSM Steering Committee and the APA’s Assembly and Board of Trustees approved the deletion of Suicidal Behavior Disorder from Section III “Conditions for Further Study.” The decision was based on concerns that the proposed disorder did not meet the criteria for a mental disorder (required for inclusion in Section III) but constituted a behavior with diverse causes. Moreover, a history of a suicide attempt in the past two years did not necessarily indicate anything about a person’s current risk for suicide, limiting its clinical utility. Additionally, the decision was influenced by the view of clinicians working to reduce the stigma of suicidal behavior that a diagnostic label based on a single past event could lead to discrimination against a person with a history of suicidal behavior. Lastly, the retention of suicidal behavior disorder in Section III was not necessary to stimulate further research on suicidality, an area that is an intense focus of research activity. Impact Suicidal Behavior Disorder has been deleted from Section III, located in “Conditions for Further Study.” Notes ICD-10-CM codes for suicidal behavior in the chapter “Other Conditions That May Be a Focus of Clinical Attention” will NOT be affected. They will remain in DSM-5-TR and are valid ICD-10-CM codes. XI. Alphabetical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes XI.A Food insecurity (Coding Update to DSM Disorder Code) Code for correction is found in the Alphabetical Listing of DSM-5-TR Diagnoses and ICD-10-CM Codes. (ICD-10-CM codes are correctly listed in all other places in the manual.) F59.41 Z59.41 Food insecurity aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 18 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. XII. DSM-5-TR Coding update table XII.A Listing of DSM-5-TR Diagnoses and New ICD-10-CM Codes Listing of DSM-5-TR Diagnoses and New ICD-10-CM Codes Disorder ICD-10-CM Code through September 30, 2022 ICD-10-CM Code beginning October 1, 2022 Diffuse traumatic brain injury with loss of consciousness of unspecified duration, Sequela [Example under Major or Mild Neurocognitive Disorder Due to Traumatic Brain Injury] S06.2X9S S06.2XAS Hepatic encephalopathy [Examples under Catatonic Disorder Due to Another Medical Condition and Delirium] K72.90 K76.82 Impairing Emotional Outbursts [Condition newly added to DSM-5-TR] Not applicable R45.89 Other Specified Delirium R41.0 F05 Unspecified Delirium R41.0 F05 a,cMajor Neurocognitive Disorder Due to [Medical Etiology], With behavioral disturbance F02.81 See replacement codes below a,cMajor Neurocognitive Disorder Due to [Medical Etiology], Without behavioral disturbance (renamed Major neurocognitive disorder due to [Medical etiology], unspecified severity, without accompanying behavioral or psychological disturbance) F02.80 F02.80 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, With agitation Not available F02.A11 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, With anxiety Not available F02.A4 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, With mood symptoms Not available F02.A3 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, With psychotic disturbance Not available F02.A2 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, With other behavioral or psychological disturbance Not available F02.A18 aMajor Neurocognitive Disorder Due to [Medical Etiology], Mild, Without accompanying behavioral or psychological disturbance Not available F02.A0 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 19 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, With agitation Not available F02.B11 aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, With anxiety Not available F02.B4 aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, With mood symptoms Not available F02.B3 aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, With psychotic disturbance Not available F02.B2 aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, With other behavioral or psychological disturbance Not available F02.B18 aMajor Neurocognitive Disorder Due to [Medical Etiology], Moderate, Without accompanying behavioral or psychological disturbance Not available F02.B0 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, With agitation Not available F02.C11 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, With anxiety Not available F02.C4 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, With mood symptoms Not available F02.C3 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, With psychotic disturbance Not available F02.C2 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, With other behavioral or psychological disturbance Not available F02.C18 aMajor Neurocognitive Disorder Due to [Medical Etiology], Severe, Without accompanying behavioral or psychological disturbance Not available F02.C0 aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, With agitation Not available F02.811 aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, With anxiety Not available F02.84 aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, With mood symptoms Not available F02.83 aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, With psychotic disturbance Not available F02.82 aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, With other behavioral or psychological disturbance Not available F02.818 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 20 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. aMajor Neurocognitive Disorder Due to [Medical Etiology], Unspecified severity, Without accompanying behavioral or psychological disturbance F02.80 F02.80 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, With agitation Not available F03.A11 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, With anxiety Not available F03.A4 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, With mood symptoms Not available F03.A3 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, With psychotic disturbance Not available F03.A2 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, With other behavioral or psychological disturbance Not available F03.A18 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Mild, Without accompanying behavioral or psychological disturbance Not available F03.A0 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, With agitation Not available F03.B11 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, With anxiety Not available F03.B4 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, With mood symptoms Not available F03.B3 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, With psychotic disturbance Not available F03.B2 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, With other behavioral or psychological disturbance Not available F03.B18 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Moderate, Without accompanying behavioral or psychological disturbance Not available F03.B0 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, With agitation Not available F03.C11 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, With anxiety Not available F03.C4 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, With mood symptoms Not available F03.C3 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 21 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, With psychotic disturbance Not available F03.C2 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, With other behavioral or psychological disturbance Not available F03.C18 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Severe, Without accompanying behavioral or psychological disturbance Not available F03.C0 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, With agitation Not available F03.911 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, With anxiety Not available F03.94 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, With mood symptoms Not available F03.93 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, With psychotic disturbance Not available F03.92 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, With other behavioral or psychological disturbance Not available F03.918 bMajor Neurocognitive Disorder Due to Possible [Medical Etiology], Unspecified severity, Without accompanying behavioral or psychological disturbance Not available F03.90 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, With agitation Not available F02.A11 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, With anxiety Not available F02.A4 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, With mood symptoms Not available F02.A3 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, With psychotic disturbance Not available F02.A2 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, With other behavioral or psychological disturbance Not available F02.A18 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Mild, Without accompanying behavioral or psychological disturbance Not available F02.A0 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, With agitation Not available F02.B11 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 22 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, With anxiety Not available F02.B4 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, With mood symptoms Not available F02.B3 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, With psychotic disturbance Not available F02.B2 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, With other behavioral or psychological disturbance Not available F02.B18 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Moderate, Without accompanying behavioral or psychological disturbance Not available F02.B0 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, With agitation Not available F02.C11 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, With anxiety Not available F02.C4 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, With mood symptoms Not available F02.C3 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, With psychotic disturbance Not available F02.C2 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, With other behavioral or psychological disturbance Not available F02.C18 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Severe, Without accompanying behavioral or psychological disturbance Not available F02.C0 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, With agitation Not available F02.811 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, With anxiety Not available F02.84 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, With mood symptoms Not available F02.83 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, With psychotic disturbance Not available F02.82 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, With other behavioral or psychological disturbance Not available F02.818 cMajor Neurocognitive Disorder Due to Probable [Medical Etiology], Unspecified severity, Without accompanying behavioral or psychological disturbance Not available F02.80 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 23 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Major Neurocognitive Disorder Due to Vascular Disease, With behavioral disturbance F01.51 See new F01 codes below Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, With agitation Not available F01.A11 Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, With anxiety Not available F01.A4 Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, With mood symptoms Not available F01.A3 Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, With psychotic disturbance Not available F01.A2 Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, With other behavioral or psychological disturbance Not available F01.A18 Major Neurocognitive Disorder Due to Probable Vascular Disease, Mild, Without accompanying behavioral or psychological disturbance Not available F01.A0 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, With agitation Not available F01.B11 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, With anxiety Not available F01.B4 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, With mood symptoms Not available F01.B3 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, With psychotic disturbance Not available F01.B2 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, With other behavioral or psychological disturbance Not available F01.B18 Major Neurocognitive Disorder Due to Probable Vascular Disease, Moderate, Without other behavioral or psychological disturbance Not available F01.B0 Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, With agitation Not available F01.C11 Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, With anxiety Not available F01.C4 Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, With mood symptoms Not available F01.C3 Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, With psychotic disturbance Not available F01.C2 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 24 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, With other behavioral or psychological disturbance Not available F01.C18 Major Neurocognitive Disorder Due to Probable Vascular Disease, Severe, Without accompanying behavioral or psychological disturbance Not available F01.C0 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, With agitation Not available F01.511 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, With anxiety Not available F01.54 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, With mood symptoms Not available F01.53 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, With psychotic disturbance Not available F01.52 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, With other behavioral or psychological disturbance Not available F01.518 Major Neurocognitive Disorder Due to Probable Vascular Disease, Unspecified severity, Without accompanying behavioral or psychological disturbance F01.50 F01.50 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, With agitation Not available F03.A11 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, With anxiety Not available F03.A4 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, With mood symptoms Not available F03.A3 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, With psychotic disturbance Not available F03.A2 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, With other behavioral or psychological disturbance Not available F03.A18 Major Neurocognitive Disorder Due to Unknown Etiology, Mild, Without accompanying behavioral or psychological disturbance Not available F03.A0 Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, With agitation Not available F03.B11 Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, With anxiety Not available F03.B4 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 25 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, With mood symptoms Not available F03.B3 Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, With psychotic disturbance Not available F03.B2 Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, With other behavioral or psychological disturbance Not available F03.B18 Major Neurocognitive Disorder Due to Unknown Etiology, Moderate, Without accompanying behavioral or psychological disturbance Not available F03.B0 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, With agitation Not available F03.C11 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, With anxiety Not available F03.C4 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, With mood symptoms Not available F03.C3 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, With psychotic disturbance Not available F03.C2 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, With other behavioral or psychological disturbance Not available F03.C18 Major Neurocognitive Disorder Due to Unknown Etiology, Severe, Without accompanying behavioral or psychological disturbance Not available F03.C0 Major Neurocognitive Disorder Due to Unknown Etiology, Unspecified severity, With agitation Not available F03.911 Major Neurocognitive Disorder Due to Unknown Etiology, Unspecified severity, With anxiety Not available F03.94 Major Neurocognitive Disorder Due to Unknown Etiology, unspecified severity, With mood symptoms Not available F03.93 Major Neurocognitive Disorder Due to Unknown Etiology, Unspecified severity, With psychotic disturbance Not available F03.92 Major Neurocognitive Disorder Due to Unknown Etiology, Unspecified severity, With other behavioral or psychological disturbance Not available F03.918 Major Neurocognitive Disorder Due to Unknown Etiology, Unspecified severity, Without accompanying behavioral or psychological disturbance Not available F03.90 aMedical etiologies comprise the following (with etiological codes): traumatic brain injury (code first S06.2XAS), HIV infection (code first B20), prion disease (code first A81.9), Huntington’s disease (code first G10), another medical condition (code first the other medical condition), and multiple etiologies (code first all medical conditions that apply; for major neurocognitive disorder, vascular disease does not receive a medical code; if mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease). bPossible medical etiologies comprise the following (no additional medical code): Alzheimer’s disease, frontotemporal degeneration, Lewy body disease, vascular disease, Parkinson’s disease. cProbable medical etiologies comprise the following (use etiological codes for probable diagnoses only; do not use with the possible etiologies): Alzheimer’s disease (code first G30.9), frontotemporal degeneration (code first G31.09), Lewy body disease (code first G31.83), Parkinson’s disease (code first G20). If mild neurocognitive disorder due to probable vascular disease is present, code first I67.9 for vascular disease. Major neurocognitive disorder due to probable vascular disease is not included here (see separate listings); vascular disease does not receive a medical code for major neurocognitive disorder. DSM-5-TRTM Update, September 2022, Page 26 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. aMild Neurocognitive Disorder Due to [Medical Etiology], With behavioral disturbance G31.84 F06.71 aMild Neurocognitive Disorder Due to [Medical Etiology], Without behavioral disturbance G31.84 F06.70 bMild Neurocognitive Disorder Due to Possible [Medical Etiology], With behavioral disturbance G31.84 G31.84 bMild Neurocognitive Disorder Due to Possible [Medical Etiology], Without behavioral disturbance G31.84 G31.84 cMild Neurocognitive Disorder Due to Probable [Medical Etiology], With behavioral disturbance G31.84 F06.71 cMild Neurocognitive Disorder Due to Probable [Medical Etiology], Without behavioral disturbance G31.84 F06.70 Mild Neurocognitive Disorder Due to Unknown Etiology G31.84 G31.84 Nonadherence to Medical Treatment Z91.19 Z91.199 Other Specified Trauma- and Stressor-Related Disorder F43.8 F43.89 Prolonged Grief Disorder F43.8 F43.81 DSM-5-TRTM Update, September 2022, Page 27 of 27 Copyright © 2022 American Psychiatric Association. All rights reserved. XII.B Coding Corrections (Effective Immediately) Coding Corrections (Effective Immediately) Disorder Prior Code Correct Code Food insecurity F59.41 Z59.41 Opioid-Induced Anxiety Disorder, With mild use disorder F11.180 F11.188 Opioid-Induced Anxiety Disorder, With moderate or severe use disorder F11.280 F11.288 Opioid-Induced Anxiety Disorder, Without use disorder F11.980 F11.988 Current Suicidal Behavior, Initial encounter T14.91A T14.91XA Current Suicidal Behavior, Subsequent encounter T14.91D T14.91XD |
1583 | https://www.tiktok.com/@andymath.com/video/7171156770168130858?lang=en | Simplify Factorials Explained: Factorial Fun! | TikTok
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1584 | https://www.quora.com/What-is-the-dot-product-and-why-does-it-equal-to-zero-if-two-vectors-are-orthogonal-perpendicular | Something went wrong. Wait a moment and try again.
Dot Products Examples
Orthogonal Vector
Basic Linear Algebra
Vectors (mathematics)
Vector Dot Product
Perpendicular Vectors
Orthogonal Basis
Product of Vectors
5
What is the dot product and why does it equal to zero if two vectors are orthogonal (perpendicular)?
Andrew Wrigley
Studied at University of Buenos Aires (Graduated 1986)
·
Author has 3K answers and 2.2M answer views
·
6y
The scalar product of two vectors does not depend on the coordinates you use.
Therefore, if u and v are two vectors that are perpendicular to each other, you can always choose an orthogonal coordinate system such that:
u = (x,0,0)
v = (0,y,0)
Therefore, by definition:
u.v = x.0 + y.0 + 0.0 =
= 0 + 0 + 0 =
= 0
And if this quantity is 0 in one set of coordinates it is 0 in all of them.
From an intuitive perspective, the dot product sort of represents the projection of the vectors on each other. To say they are perpendicular is the same as to say they have zero length projection on each other.
Think of
The scalar product of two vectors does not depend on the coordinates you use.
Therefore, if u and v are two vectors that are perpendicular to each other, you can always choose an orthogonal coordinate system such that:
u = (x,0,0)
v = (0,y,0)
Therefore, by definition:
u.v = x.0 + y.0 + 0.0 =
= 0 + 0 + 0 =
= 0
And if this quantity is 0 in one set of coordinates it is 0 in all of them.
QED
From an intuitive perspective, the dot product sort of represents the projection of the vectors on each other. To say they are perpendicular is the same as to say they have zero length projection on each other.
Think of the length of shadows at midday on midsummer day.
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Alexander Talalai
Author has 655 answers and 423.1K answer views
·
3y
Quite often dot product of vectors X and Y in Euclidean space is denoted as (X,Y). And
(X,Y) = x1y1 + … + xnyn,
where x1, …, xn and y1, …, yn are coordinates.
The main feature if dot product - it is an invariant in respect to orthogonal linear transformations. I am not going to prove that here, too much of basic linear algebra.
You can always find a transformation that will make vector X collinear t
Quite often dot product of vectors X and Y in Euclidean space is denoted as (X,Y). And
(X,Y) = x1y1 + … + xnyn,
where x1, …, xn and y1, …, yn are coordinates.
The main feature if dot product - it is an invariant in respect to orthogonal linear transformations. I am not going to prove that here, too much of basic linear algebra.
You can always find a transformation that will make vector X collinear to the coordinate axis, i.e. vector
X = (a,0, …, 0) and (X,Y) = ay1 = product of length of vector...
Andrew Winkler
PhD in Mathematics, Courant Institute of Mathematical Sciences, NYU (Graduated 1987)
·
Author has 9.6K answers and 10.9M answer views
·
6y
Originally Answered: Why is the scalar (dot) product of perpendicular vectors equal to 0?
·
One definition of the dot product is that it is (apart from a factor of 4) the same thing as the difference between the squares of the lengths of the two diagonals of the parallelogram formed by the two vectors. When the dot product is 0, that difference is 0, which is to say the diagonals are of equal length. That makes the parallelogram a rectangle, which makes those two vectors forming two of its sides perpendicular.
The dot product is also the product of the lengths of one of the vectors, and the projection of the other vector onto it. If the product is 0, one of those vectors has length 0.
One definition of the dot product is that it is (apart from a factor of 4) the same thing as the difference between the squares of the lengths of the two diagonals of the parallelogram formed by the two vectors. When the dot product is 0, that difference is 0, which is to say the diagonals are of equal length. That makes the parallelogram a rectangle, which makes those two vectors forming two of its sides perpendicular.
The dot product is also the product of the lengths of one of the vectors, and the projection of the other vector onto it. If the product is 0, one of those vectors has length 0. A zero vector is perpendicular to everything. If the projection of one vector onto another is 0, then they are perpendicular.
Related questions
Two vectors that are orthogonal have a dot product of zero. Does this mean two parallel vectors will always have a cross product of zero?
What does it mean if the dot product of two vectors is zero?
What does the dot product of two non-orthogonal vectors equal to zero mean?
How can we use the dot product to find out if two vectors are orthogonal or perpendicular to each other?
What is the dot product of two vectors? If the dot product is zero, does this mean the vectors are orthogonal?
Assistant
Bot
·
1y
The dot product (also known as the scalar product) of two vectors is a way to multiply them that results in a scalar (a single number). For two vectors a and b in an n-dimensional space, the dot product is defined mathematically as:
a⋅b=∑ni=1aibi
where ai and bi are the components of vectors a and b.
Alternatively, the dot product can also be expressed in terms of the magnitudes (lengths) of the vectors and the cosine of the angle θ between them:
a⋅b=|a||b|cos(θ)
Why
The dot product (also known as the scalar product) of two vectors is a way to multiply them that results in a scalar (a single number). For two vectors a and b in an n-dimensional space, the dot product is defined mathematically as:
a⋅b=∑ni=1aibi
where ai and bi are the components of vectors a and b.
Alternatively, the dot product can also be expressed in terms of the magnitudes (lengths) of the vectors and the cosine of the angle θ between them:
a⋅b=|a||b|cos(θ)
Why the Dot Product Equals Zero for Orthogonal Vectors
Two vectors are considered orthogonal (or perpendicular) if the angle between them is 90∘. In this case, θ=90∘.
Using the cosine function, we have:
cos(90∘)=0
Substituting this into the dot product formula gives:
a⋅b=|a||b|cos(90∘)=|a||b|⋅0=0
Thus, the dot product of two orthogonal vectors is zero. This property is useful in various applications, including geometry, physics, and computer graphics, as it helps determine whether vectors are perpendicular to each other.
Sripad Sambrani
Knows Sanskrit
·
Author has 6.8K answers and 2.9M answer views
·
2y
Originally Answered: What is the dot product of two perpendicular vectors? Why is it equal to zero? How does it apply to physics?
·
In mutually orthogonal/perpendicular vectors, Θ=90° => cosΘ=0. Hence, the dot product is zero.
Trust this helps.
In mutually orthogonal/perpendicular vectors, Θ=90° => cosΘ=0. Hence, the dot product is zero.
Trust this helps.
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
·
Updated Sep 9
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I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend.
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Michael Quackenbush May
PhD in Plasma Physics, Princeton University (Graduated 2024)
·
Author has 68 answers and 1.1M answer views
·
6y
Originally Answered: Why is the scalar (dot) product of perpendicular vectors equal to 0?
·
What does it mean for two vectors to be perpendicular? One definition is that if we draw the two vectors and measure the angle between them with a protractor, we get a 90 degree angle.
Let’s create two general two-dimensional perpendicular vectors using this definition.
X = (rcos(t),rsin(t)), Y = (psin(t), - pcos(t))
We can vary r and t to let X be any point in the two-dimensional plane. Y will always be at 90 degrees to X for the same value of t (since sin and cos are 90 degrees out of phase with each other). Taking the dot product, we find X.Y = rpcos(t)sin(t)-rpcos(t)sin(t) = 0. This par
What does it mean for two vectors to be perpendicular? One definition is that if we draw the two vectors and measure the angle between them with a protractor, we get a 90 degree angle.
Let’s create two general two-dimensional perpendicular vectors using this definition.
X = (rcos(t),rsin(t)), Y = (psin(t), - pcos(t))
We can vary r and t to let X be any point in the two-dimensional plane. Y will always be at 90 degrees to X for the same value of t (since sin and cos are 90 degrees out of phase with each other). Taking the dot product, we find X.Y = rpcos(t)sin(t)-rpcos(t)sin(t) = 0. This parametrization can be made with vectors of arbitrary dimension, and it can be similarly shown that their dot products are zero.
Many mathematicians will be unhappy with what I’ve just said, however. This is because they would probably just start by defining the word perpendicular using, “two vectors are perpendicular when when their dot product is zero.” They would then prove that this means thy have a 90 degree angle between them. I won’t do this, but the familiar dot product law a.b= cos(theta) ||a|| ||b||, shows that this must be the case.
Related questions
Is the zero vector perpendicular (orthogonal) to every vector?
What is the relationship between perpendicular vectors and orthogonal vectors?
If two vectors are orthogonal, then what will be their dot product?
What if two vectors are orthogonal and cross product is zero?
What does it mean when two vectors are orthogonal? What is the dot product?
Roger Larson
Upvoted by
Gordon Diss
, MS Mathematics, Stanford University (1973) · Author has 5K answers and 4.7M answer views
·
1y
Originally Answered: What is the dot product of two mutually orthogonal vectors? Does it always equal zero?
·
The dot product of two mutually orthogonal vectors is 0.
The dot product is
AB = | A| |B | cos theta
where | A| is the magnitude of vector A
| B| is the magnitude of vector B
orthogonal vectors have an angle theta = 90 degrees
cosine 90 degrees = 0
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Gordon Hunter
Associate Professor of Mathematics and Computing
·
Author has 127 answers and 33.3K answer views
·
3y
One way of calculating the scalar, or “dot”, product of two vectors x and y is by multiplying the product of their magnitudes by the cosine of the angle theta between them :
x . y = |x| |y| cos(theta)
Now if the vectors are orthogonal, then the angle between them is either 90 degrees (one way) or 270 degrees (the other way), and cos(90 degrees) = cos(270 degrees) = 0.
Hence, x . y = 0, so the dot product between any two orthogonal vectors must be zero.
Although it is more difficult to visualise this in more than 3 dimensions, the same principle applies. In fact, this can be used to define orthogon
One way of calculating the scalar, or “dot”, product of two vectors x and y is by multiplying the product of their magnitudes by the cosine of the angle theta between them :
x . y = |x| |y| cos(theta)
Now if the vectors are orthogonal, then the angle between them is either 90 degrees (one way) or 270 degrees (the other way), and cos(90 degrees) = cos(270 degrees) = 0.
Hence, x . y = 0, so the dot product between any two orthogonal vectors must be zero.
Although it is more difficult to visualise this in more than 3 dimensions, the same principle applies. In fact, this can be used to define orthogonal vectors in higher dimensional spaces !
Buddha Buck
Took calculus as an undergraduate
·
Author has 5.8K answers and 16.9M answer views
·
6y
Originally Answered: Why is the scalar (dot) product of perpendicular vectors equal to 0?
·
It is convenient for it to be so, so we define the dot product (or “perpendicular”) that way.
The dot product was introduced to be able to describe the projection of one vector onto another. If you have two unit vectors, then the dot product is the magnitude of the projection of one onto the other. This works out so that for two vectors a⋅b=|a||b|cosθ where θ is the angle between them.
From this definition, it is obvious that parallel vectors (θ=0∘) have a dot product equal to the product of the magnitudes, and perpendicular vectors
It is convenient for it to be so, so we define the dot product (or “perpendicular”) that way.
The dot product was introduced to be able to describe the projection of one vector onto another. If you have two unit vectors, then the dot product is the magnitude of the projection of one onto the other. This works out so that for two vectors a⋅b=|a||b|cosθ where θ is the angle between them.
From this definition, it is obvious that parallel vectors (θ=0∘) have a dot product equal to the product of the magnitudes, and perpendicular vectors θ=±90∘) have a dot product of 0.
It’s also easy to see that for non-zero vectors a,b, you have (aa)⋅(bb)=ab(a⋅b). What isn’t as obvious is that it distributes over addition. That is, (a+b)⋅c=a⋅c+b⋅c and a⋅(b+c)=a⋅c+b⋅c. These two properties (or, four properties) together form a property called “bilinearity”. In addition, you have a⋅b=b⋅a.
But you might not have been taught that definition, but rather something like x1y1+x2y2. How does that work?
Let’s pick two basis vectors i,j such that every vector a can be expressed as a=aii+ajj.
Then you can use that bilinear property to get:
a⋅b=(aii+ajj)⋅(bii+bjj) =(aii)⋅(bii)+(aii)⋅(bjj)+(ajj)⋅(bii)+(ajj)⋅(bjj) =aibi(i⋅i)+(aibj+ajbi)(i⋅j)+ajbj(j⋅j)
If we carefully choose i,j such that i⋅i=j⋅j=1 and i⋅j=0, then that last line above collapses to a⋅b=aibi+ajbj.
Khenan Mak
Engineering mathematics
·
Author has 2.6K answers and 1.2M answer views
·
5y
There are a few different, but equivalent, explanations. A straightforward one would be that the dot (or scalar) product is the multiplication of the magnitudes of the two vectors along with the cosine of the angle between them. Since the angle between two perpendicular vectors is 90 degrees, and the cosine of 90 is zero, we get the answer that you are looking for.
Alexandru Carausu
Former Former University Associate Professor at Universitatea Tehnica "Gh. Asachi" Iasi (1966–2010)
·
Author has 3K answers and 874.7K answer views
·
2y
WRONG QUESTION !
The dot product
u • v
= 0 if :
(
i
) |
u
| =
u
= 0 , (
ii
) |
v
| =
v
= 0 , or (
iii
)
u ⊥ v
. (1)
This follows from the definition of the dot product between to free vectors :
u • v
=
u v
cos θ with θ = ∢ (
u
,
v
) , hence
u ⊥ v
<==> [ θ = π/2 ==> cos θ = 0 ] . (2)
If the two vectors are parallel / collinear, then
θ = ∢ (
u
,
v
) ∈ { 0 , π } ==> cos θ = ±1 ==> cos θ =
u • v
= ±
u v .
(3)
This relation of parallelism / collinearity between
u
,
v
excludes the trivial cases when at least one vector is = 0 <==>
u v
= 0 .
Kent McIntosh
B.S. in Civil Engineering (college major) & Applied Mathematics, California State University, Long Beach (Graduated 1996)
·
Author has 267 answers and 103K answer views
·
3y
The vector dot product between the vectors
x
and
y
is: x y cos θ, where θ is the angle between the vectors. If those vectors are orthogonal (perpendicular), then the angle θ = 90⁰, and cos 90⁰ = 0, so the dot product is zero.
Robert Toop
Author has 4.6K answers and 2.9M answer views
·
5y
Originally Answered: Why is the dot product of two perpendicular vectors zero?
·
The vector dot product is the product of the projection of one vector onto another. In other words, multiply the dimensions that are parallel. The more orthogonal the vectors, the less the dot product. The result is a signed scalar, not a vector.
Related questions
Two vectors that are orthogonal have a dot product of zero. Does this mean two parallel vectors will always have a cross product of zero?
What does it mean if the dot product of two vectors is zero?
What does the dot product of two non-orthogonal vectors equal to zero mean?
How can we use the dot product to find out if two vectors are orthogonal or perpendicular to each other?
What is the dot product of two vectors? If the dot product is zero, does this mean the vectors are orthogonal?
Is the zero vector perpendicular (orthogonal) to every vector?
What is the relationship between perpendicular vectors and orthogonal vectors?
If two vectors are orthogonal, then what will be their dot product?
What if two vectors are orthogonal and cross product is zero?
What does it mean when two vectors are orthogonal? What is the dot product?
Why is the scalar (dot) product of perpendicular vectors equal to 0?
Can two vectors be orthogonal to each other but not intersect at the zero vector?
How is the dot product of two perpendicular vectors determined?
If two vectors are orthogonal does that mean they're perpendicular?
If two vectors are equal (same), then what will be their dot product?
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1585 | https://www.youtube.com/watch?v=FsuWgpv-os8 | LINEAR COMBINATION
WOW MATH
865000 subscribers
4 likes
Description
288 views
Posted: 18 Oct 2023
1 comments
Transcript:
good evening cir Clem good evening call scholar our group was assigned to discuss practice task 4.3.8 point1 so determine if uh Vector Alpha is expressable as a linear combination of the vectors in s so V is equal to R ra to 3 space uh the vector Alpha is equal to 1 -25 and the vectors in s are 11 one so this is our a sub 1 and 1 2 3 that is our a sub 2 2 -1 1 this is our a sub 3 so in order to uh prove that this is a linear combination you Vector alet vectors contains in s so okay Vector alet scalp scal C sub one multiply a sub 1 C sub 2 a sub 2 + C sub3 a sub3 so we're going to look for the values of uh C sub 1 C sub2 and C sub3 soan varable instead C subam is X time a sub 1 so c sub2 y a sub 2 + a a sub 3 so we're going to write uh vectors into column vectors so we have so equals 1 -25 and then X is equal to 1 1 1 y equal to 1 2 3 and then Z is equal to 21 and one so next slide so to continue so X we're going to multiply X to 1 one one so that will become x x x and y multiply to 1 2 3 so that will become Y 2 y 3 y + Z multip to 21 and 1 will become 2 z z and Z so first uh we're going to write all the vectors uh that contains in our vectors system of linear equations so the first equation first R okay uh first row is x + y + 2 Z is = to 1 and then you equation number 2 is x + 2 y - Z = -2 and equation number 3 is x + 3 y + z is = 5ep is equation 2 and three so I have equation two equation two so that is x + 2 y z = to -2 equation 3 is x + 3 y + z = 5 so if we're going to add so canel so x + x that is 2x + 5 y = 3 and then equation one so equation 4 so equation 4 so next is equation one so we have x + y + 2 Z = 1 and then equation 2 multiply 2 so uh 2 X that is 2x 2 2 y that is 4 Y and then this will be 2 Z = -4 so if we're going to add x + 2x is 3x soel y + 4 Y is 5 y okay so this is 5 y = 3 okay so this is equation five so we're going to pay equation four and five to solve the Valu of X and Y okay so we have 2X + 5 y = 3 and then 3x uh + 5 y = -3 so IM multiply okay you equation 51 so1 2x + 5 y = 3 soly will become -3x so -5 y = pos3 so so will become X so cancel = 6 so divide both side by 1 so X is = to -6 so this is the value of x so to solve for y so G equation 4 we have 2X + 5 y = 3 so that is 2 -6 + 5 y = 3 so this is -12 + 5 Y is = to 3 then 5 y = to 3 okay + 12 so 5 y = 15 so y isal to 3 okay to solve for Z so SL so to solve for Z equation one so again x x isal to -6 y man is positive 3 so to solve for Z and so that is x + y + 2 Z is = to 1 so subtitute so -6 + 3 + 2 Z = 1 so this is uh -3 + 2 Z = 1 so 2 Z is = 1 + 3 soal 4 divide both side by two so Z is equal to 2 so this will be the value of so values X is = to -6 Y is = to 3 and Z is = to pos2 so to check linear combination Vector Al vectors contains in s okay so we have uh one okay2 5 is equal to so so we're going to multiply this one by X so X is -6 so multiply 1 1 1 plus okay Y is 3 1 to okay so okay so we have a 3 1 2 3 + uh 2 2 -1 1 okay so 1 -25 equals so this is6 -6 -6 + 3 6 9 + 42 positive2 so 1 -2 and 5 so this is -6 + 3 that is -3 + 4 that is uh -3 + 4 that's positive 1 okay -6 + 6 that is 0 + -2 that is -2 and then -6 + 9 that is POS 3 + 2 that is five and then they are equal so therefore we can say now uh you Alpha uh Vector Alpha is a a linear combination of the vectors in s so meaning we can express this uh Vector Alpha so all the vectors in X so the vector Alpha is a linear combination of the vectors in x |
1586 | https://helpingwithmath.com/decimals-in-a-number-line/ | Skip to content
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Home » Math Theory » Decimals » Decimals in a Number Line
Decimals in a Number Line
Introduction
We have learnt different methods to represent a number such as a number written in words, a number written in numerals etc. we have also learnt about visual representation of numbers using a number line. Another fact which we are aware of is the different types of numbers such as natural numbers, whole numbers, fractions, decimal numbers, rational numbers etc. We have leant these concepts individually such as plotting a number on a number line, or adding or subtracting decimals. Now, let us learn some concepts that combine these individual identities such plotting a decimal number on a number line. But, before that we must recall some basic definitions and concepts that are integral to this learning.
What are Decimals?
We know that a decimal indicates a part of a whole. In other words, decimals are another form of a fraction in which the whole number is separated from the fractional part using a dot.
What is a Number Line?
A number line is a straight horizontal line with numbers placed at even intervals that provides a visual representation of numbers. Primary operations such as addition, subtraction, multiplication, and division can all be performed on a number line. The numbers increase as we move towards the right side of a number line while they decrease as we move left.
Representation on a Number Line
Above is a visual representation of a standard number line. As is clearly visible, as we move from left to right, there is an increase in the value of numbers while it decreases when we move from right to left.
How are Decimal Numbers Represented on a Number Line?
Now, that we know what decimal numbers are and what is a number line, let us move to representing a decimal number on a number line. Representing decimal on a number line is defined as the plotting of decimal numbers on a number line.
An important aspect of decimals that needs to be understood for plotting them on a number line is that the decimals are place between two integers.
Let us understand this by an example.
Let us take two numbers 0 and 1 draw a number line between them. We will have,
Next, let us split the space between 0 and 1 into tenths. We will have the fractional values such as one-tenth, two-tenth, three-tenth and so on. It can be represented on the number line in the following manner.
Now, we know that
One-tenth = 0.1
Two-tenth = 0.2
Three-tenth = 0.3 and so on.
Therefore, if we want to represent the above tenth values in the form of decimals we will have,
Further, if we wish to plot a decimal number say, 0.6, we will mark it as
In the above example, we had a simple decimal number and it was quite easy to plot it. But we know according to the place value system of decimals we have decimal numbers such as one-tenth ($\frac{1}{10}$ ), one-hundredth ( $\frac{1}{100}$ ), one-thousandth ( $\frac{1}{1000}$ ) etc. so how will we plot these numbers? Let us find out.
Representing Tenths on a Number Line
We already know how to represent fractions on a number line. Let us now represent tenths of a decimal on a number line. But, let us recall what we means by tenths of a decimal?
Consider the following figure. It is divided into ten equal parts and one part is shaded. The shaded part represents one-tenth of the whole figure. It is written as $\frac{1}{10}$. $\frac{1}{10}$ is also written as 0.1 which is read as “ point one “ or “ decimal one “.
Thus the fraction $\frac{1}{10}$ is called one-tenth and is written as 0.1.
Now, let us represent tenths on a number line. We can understand this by an example.
Let us represent 0.4 on a number line. We can clearly see that there are 4 tenths in 0.4. Therefore in order to represent 0.4 on a number line we will divide the unit length between 0 and 1 into 10 equal parts and take 4 parts as shown below –
Now, we can know that 0.4 in fraction form is equal to 4/10. Hence we will mark 4/10 as 0.4 which is our desired mark on the number line.
The steps that we used above to represent a tenth on a number line can be summarised as –
We draw a number line between 0 and 1.
We then raw 10 lines dividing the total distance between 0 and 1 into 10 equal parts.
Now, one whole divided into 10 parts is equal to $\frac{1}{10}$.
$\frac{1}{10}$ in decimal form is equal to 0.1.
At each new line we are adding $\frac{1}{10}$ or 0.1.
So, between 0 and 1 we have, 0 . 1 , 0 . 2 , . 0 . 3 , 0 . 4 , 0 . 5 , 0 . 6 , 0 . 7 , 0 . 8 and 0 . 9. Similarly, between 1 and 2 we have, 1 . 1 , 1 . 2 , 1 . 3 , 1 . 4 , 1 . 5 , 1 . 6 , 1 . 7 , 1 . 8 and 1 . 9.
We can also say that the line representing $\frac{1}{2}$ or 0.5 is the half way mark between 0 and 1. Similarly, the line representing $1\frac{5}{10}$ or 1.5 is the half way mark between 1 and 2.
Ten tenths is equal to one whole.
The above process can be defined as – “To represent a tenth on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenths values. Next we plot the tenths values that lie between the whole parts.”
Representing Hundredths on a Number Line
Let us now represent hundredths of a decimal on a number line. But, let us recall what we means by hundredths of a decimal?
If an object is divided into 100 equal parts, then each part is one hundredths of the whole. This means that –
One thousandth = $\frac{1}{100}$ which in decimal form is equal to 0.01
If we take 7 parts out of 100 equal parts of an object, then 7 parts make $\frac{7}{100}$ of the whole and it is written as 0.07.
Now, let us represent hundredths on a number line. We can understand this by an example.
Let us represent the thousandth value of the number 7.45 on a number line.
To represent 7.4 on a number line, we first draw 10 lines dividing the total distance between 7 and 8 into 10 equal parts.
We can see that the arrow is four parts to the right of the whole number 7.
Similarly, to represent 7.45 on a number line, we first draw 10 lines dividing the total distance between 7.4 and 7.5 into 10 equal parts.
The steps that we used above to represent a thousandth on a number line can be summarised as –
We draw a number line between 0 and 1.
We then draw 10 lines dividing the total distance between 0 and 1 into 10 equal parts.
Now, one whole divided into 10 parts is equal to $\frac{1}{10}$.
$\frac{1}{10}$ in decimal form is equal to 0.1.
At each new line, we are adding $\frac{1}{10}$ or 0.1.
So, between 0 and 1 we have, 0 . 1 , 0 . 2 , . 0 . 3 , 0 . 4 , 0 . 5 , 0 . 6 , 0 . 7 , 0 . 8 and 0 . 9. Similarly, between 1 and 2 we have, 1 . 1 , 1 . 2 , 1 . 3 , 1 . 4 , 1 . 5 , 1 . 6 , 1 . 7 , 1 . 8 and 1 . 9.
We can also say that the line representing $\frac{1}{2}$ or 0.5 is the half way mark between 0 and 1. Similarly, the line representing $1\frac{5}{10}$ or 1.5 is the half way mark between 1 and 2.
Ten tenths is equal to one whole.
Next, we perform the same steps to draw 10 lines dividing the total distance between 0.01 and 0.02 into 10 equal parts.
In this manner, we can plot the hundredths of a decimal on a number line.
The above process can be defined as – “To represent a hundredths on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenths values. Next, we divide again the distance between two tenths to get their hundredths values.”
Representing Thousandths on a Number Line
Let us now represent thousandths of a decimal on a number line. But, let us recall what we means by thousandths of a decimal?
If an object is divided into 1000 equal parts, then each part is one thousandth of the whole. This means that –
One thousandth = $\frac{1}{1000}$ which in decimal form is equal to 0.001
If we take 7 parts out of 1000 equal parts of an object, then 7 parts make $\frac{7}{1000}$ of the whole and it is written as 0.007.
Now, let us represent thousandths on a number line. We can understand this by an example.
Let us represent the thousandth value of the number 7.456 on a number line.
To represent 7.4 on a number line, we first draw 10 lines dividing the total distance between 7 and 8 into 10 equal parts.
We can see that the arrow is four parts to the right of the whole number 7.
Similarly, to represent 7.45 on a number line, we first draw 10 lines dividing the total distance between 7.4 and 7.5 into 10 equal parts.
We can see that the arrow is five parts to the right of the decimal number 7.40.
Next, to represent 7.456 on a number line, we first draw 10 lines dividing the total distance between 7.45 and 7.46 into 10 equal parts.
We can see that the arrow is six parts to the right of the decimal number 7.45
So, in this manner, we have represented the number 7.456 on the number line.
The steps that we used above to represent a thousandth on a number line can be summarised as –
We draw a number line between 0 and 1.
We then draw 10 lines dividing the total distance between 0 and 1 into 10 equal parts.
Now, one whole divided into 10 parts is equal to $\frac{1}{10}$.
$\frac{1}{10}$ in decimal form is equal to 0.1.
At each new line, we are adding $\frac{1}{10}$ or 0.1.
So, between 0 and 1 we have, 0 . 1 , 0 . 2 , . 0 . 3 , 0 . 4 , 0 . 5 , 0 . 6 , 0 . 7 , 0 . 8 and 0 . 9. Similarly, between 1 and 2 we have, 1 . 1 , 1 . 2 , 1 . 3 , 1 . 4 , 1 . 5 , 1 . 6 , 1 . 7 , 1 . 8 and 1 . 9.
We can also say that the line representing $\frac{1}{2}$ or 0.5 is the half way mark between 0 and 1. Similarly, the line representing $1\frac{5}{10}$ or 1.5 is the half way mark between 1 and 2.
Ten tenths is equal to one whole.
Next, we perform the same steps to draw 10 lines dividing the total distance between 0.01 and 0.02 into 10 equal parts.
Again, we perform the same steps to draw 10 lines dividing the total distance between 0.001 and 0.002 into 10 equal parts.
In this manner, we can plot the thousandth of a decimal on a number line.
The above process can be defined as – “To represent a thousandth on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenths values. Next, we divide again the distance between two tenths to get their hundredths values. Finally, we divide the distance between two hundredths into 10 equal parts to get their thousandths values.”
Solved Examples
Example 1 Between what two numbers is does the decimal number 5.4 lie on the number line?
Solution We have been given the decimal number 5.4 and we need to check between which two whole numbers will it lie.
On observing the number 5.4 we can see that the number represents a tenth of a decimal as it has one digit after the decimal point.
Also, we know that 5.4 = 5 + $\frac{4}{10}$
This means that 5.4 is equal to 5 whole parts plus 4 tenths. Et us plot it on the number line. we will have,
We can clealry see that 5.4 will lie between 5 and 6. The point on the number line will be –
Hence, we can say that the number 5.4 will be between the whole numbers 5 and 6.
Example 2 Label the missing decimal numbers on the number line.
Solution We have been given four numbers marked as A , B , C and D on a number line and we need to find out which decimal numbers they represent. Let us mark them one by one.
We will start with completing the marking of the lines that have not been marked on the given number line. It can be clearly seen that there are 10 lines between two whole numbers on the number line. This means that the lines represent one tenth of the number in the decimal form. Therefore, the lines between 7 and 8 will be marked as 7 . 1 , 7 . 2 , 7 . 3 , 7 . 4 , 7 5 , 7 . 6 , 7 . 7 , 7 . 8 and d 7. 9. Similarly, We between the whole numbers 8 and 9 we have, 8 . 1 , 8 . 2 , 8 . 3 , 8 . 4 , 8 . 5 , 8 . 6 , 8 . 7 , 8 . 8 and 8 . 9. The number line so obtained will be –
Now, we shall check the position of the four points on this number line.
We can see that from the number line above, the point A lies on the decimal number 8.7. Hence A = 8 . 7
Now, let us check the position of point B.
We can see that from the number line above, point B lies on the decimal number 8.2. Hence B = 8 . 2
Now, let us check the position of point C.
We can see that from the number line above, point C lies on the decimal number 7 . 1. Hence C = 7 . 1
Now, let us check the position of point D.
We can see that from the number line above, the point D lies on the decimal number 7 . 8 . Hence D = 7 . 8
Therefore, we have,
A = 8 . 7
B = 8 . 2
C = 7 . 1
D = 7 . 8
Key Facts and Summary
A decimal indicates a part of a whole.
A number line is a straight horizontal line with numbers placed at even intervals that provides a visual representation of numbers.
Representing decimal on a number line is defined as the plotting of decimal numbers on a number line.
To represent a tenth on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenth values. Next, we plot the tenth values that lie between the whole part.
To represent a hundredth on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenth values. Next, we divide again the distance between two tenths to get their hundredths values.
To represent a thousandth on a number line, we will first have to divide the distance between two whole numbers into 10 equal parts to get their tenth values. Next, we divide again the distance between two tenths to get their hundredths values. Finally, we divide the distance between two-hundredths into 10 equal parts to get their thousandths values.
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1588 | https://www.jacr.org/article/S1546-1440(22)00651-2/fulltext | ACR Appropriateness Criteria® Acute Pyelonephritis: 2022 Update - Journal of the American College of Radiology
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Appropriate Use CriteriaVolume 19, Issue 11, SupplementS224-S239 November 2022
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ACR Appropriateness Criteria® Acute Pyelonephritis: 2022 Update
Expert Panel on Urological Imaging
Expert Panel on Urological Imaging
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∙ Andrew D.Smith, MD, PhD
Andrew D.Smith, MD, PhD
Correspondence
Corresponding author: Andrew D. Smith, MD, PhD, Attn: Dept of Radiology, University of Alabama at Birmingham, 619 19th Street South, JTN 338, Birmingham, AL 35249-6830
andrewdennissmith@uabmc.edu
Affiliations
University of Alabama at Birmingham, Birmingham, Alabama
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aandrewdennissmith@uabmc.edu ∙ Paul Nikolaidis, MD
Paul Nikolaidis, MD
Affiliations
Vice-Chair, Department of Radiology, Northwestern University Feinberg School of Medicine, Chicago, Illinois
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b ∙ … ∙ Gaurav Khatri, MD
Gaurav Khatri, MD
Affiliations
UT Southwestern Medical Center, Dallas, Texas; Chief, Division of Body MRI; Interim Chief, Division of Abdominal Imaging; Program Director, Body MRI Fellowship
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c ∙ Suzanne T.Chong, MD, MS
Suzanne T.Chong, MD, MS
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Indiana University, Indianapolis, Indiana; Committee on Emergency Radiology—General, Small, Emergency and/or Rural Practice
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d ∙ Alberto Diaz De Leon, MD
Alberto Diaz De Leon, MD
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UT Southwestern Medical Center, Dallas, Texas
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e ∙ Dhakshinamoorthy Ganeshan, MBBS
Dhakshinamoorthy Ganeshan, MBBS
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The University of Texas MD Anderson Cancer Center, Houston, Texas
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f ∙ John L.Gore, MD, MS
John L.Gore, MD, MS
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University of Washington, Seattle, Washington; American Urological Association
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g ∙ Rajan T.Gupta, MD
Rajan T.Gupta, MD
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Duke University Medical Center, Durham, North Carolina
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h ∙ Richard Kwun, MD
Richard Kwun, MD
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Swedish Medical Center, Issaquah, Washington; American College of Emergency Physicians
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i ∙ Andrej Lyshchik, MD, PhD
Andrej Lyshchik, MD, PhD
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Thomas Jefferson University Hospital, Philadelphia, Pennsylvania
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j ∙ Refky Nicola, DO, MSc
Refky Nicola, DO, MSc
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Roswell Park Cancer Institute, Jacobs School of Medicine and Biomedical Science, Buffalo, New York
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k ∙ Andrei S.Purysko, MD
Andrei S.Purysko, MD
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Cleveland Clinic, Cleveland, Ohio; ACR Learning Network, Prostate MR Image Quality Improvement Collaborative, Physician Leader
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l ∙ Stephen J.Savage, MD
Stephen J.Savage, MD
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Medical University of South Carolina, Charleston, South Carolina; American Urological Association; Professor and Vice Chairman of Urology
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m ∙ Myles T.Taffel, MD
Myles T.Taffel, MD
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New York University Langone Medical Center, New York, New York; Associate Section of Body Imaging
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n ∙ Don C.Yoo, MD
Don C.Yoo, MD
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Rhode Island Hospital/The Warren Alpert Medical School of Brown University, Providence, Rhode Island; Commission on Nuclear Medicine and Molecular Imaging
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o ∙ Erin W.Delaney, MD
Erin W.Delaney, MD
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University of Alabama at Birmingham Medical Center, Birmingham, Alabama; Primary care physician
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p ∙ Mark E.Lockhart, MD, MPH
Mark E.Lockhart, MD, MPH
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University of Alabama at Birmingham, Birmingham, Alabama; Chair UAB Department Appointments, Promotions, and Tenure Committee
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Affiliations & Notes Article Info
a University of Alabama at Birmingham, Birmingham, Alabama
b Vice-Chair, Department of Radiology, Northwestern University Feinberg School of Medicine, Chicago, Illinois
c UT Southwestern Medical Center, Dallas, Texas; Chief, Division of Body MRI; Interim Chief, Division of Abdominal Imaging; Program Director, Body MRI Fellowship
d Indiana University, Indianapolis, Indiana; Committee on Emergency Radiology—General, Small, Emergency and/or Rural Practice
e UT Southwestern Medical Center, Dallas, Texas
f The University of Texas MD Anderson Cancer Center, Houston, Texas
g University of Washington, Seattle, Washington; American Urological Association
h Duke University Medical Center, Durham, North Carolina
i Swedish Medical Center, Issaquah, Washington; American College of Emergency Physicians
j Thomas Jefferson University Hospital, Philadelphia, Pennsylvania
k Roswell Park Cancer Institute, Jacobs School of Medicine and Biomedical Science, Buffalo, New York
l Cleveland Clinic, Cleveland, Ohio; ACR Learning Network, Prostate MR Image Quality Improvement Collaborative, Physician Leader
m Medical University of South Carolina, Charleston, South Carolina; American Urological Association; Professor and Vice Chairman of Urology
n New York University Langone Medical Center, New York, New York; Associate Section of Body Imaging
o Rhode Island Hospital/The Warren Alpert Medical School of Brown University, Providence, Rhode Island; Commission on Nuclear Medicine and Molecular Imaging
p University of Alabama at Birmingham Medical Center, Birmingham, Alabama; Primary care physician
q University of Alabama at Birmingham, Birmingham, Alabama; Chair UAB Department Appointments, Promotions, and Tenure Committee
Footnotes:
The ACR seeks and encourages collaboration with other organizations on the development of the ACR Appropriateness Criteria through society representation on expert panels. Participation by representatives from collaborating societies on the expert panel does not necessarily imply individual or society endorsement of the final document.
Reprint requests to: publications@acr.org.
The authors state that they have no conflict of interest related to the material discussed in this article. Dr Yoo and Dr Chong are partners; and the others are non-partner/non-partnership track/employees.
Disclaimer: The ACR Committee on Appropriateness Criteria and its expert panels have developed criteria for determining appropriate imaging examinations for diagnosis and treatment of specified medical condition(s). These criteria are intended to guide radiologists, radiation oncologists, and referring physicians in making decisions regarding radiologic imaging and treatment. Generally, the complexity and severity of a patient’s clinical condition should dictate the selection of appropriate imaging procedures or treatments. Only those examinations generally used for evaluation of the patient’s condition are ranked. Other imaging studies necessary to evaluate other co-existent diseases or other medical consequences of this condition are not considered in this document. The availability of equipment or personnel may influence the selection of appropriate imaging procedures or treatments. Imaging techniques classified as investigational by the Food and Drug Administration have not been considered in developing these criteria; however, study of new equipment and applications should be encouraged. The ultimate decision regarding the appropriateness of any specific radiologic examination or treatment must be made by the referring physician and radiologist in light of all the circumstances presented in an individual examination.
The ACR Appropriateness Criteria documents are updated regularly. Please go to the ACR website at www.acr.org/ac to confirm that you are accessing the most current content.
DOI: 10.1016/j.jacr.2022.09.017 External LinkAlso available on ScienceDirect External Link
Copyright: © 2022 American College of Radiology
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Abstract
Acute pyelonephritis (APN) is a severe urinary tract infection (UTI) that has the potential to cause sepsis, shock, and death. In the majority of patients, uncomplicated APN is diagnosed clinically and is responsive to treatment with appropriate antibiotics. In patients who are high risk or when treatment is delayed, microabscesses may coalesce to form an acute renal abscess. High-risk patients include those with a prior history of pyelonephritis, lack of response to therapy for lower UTI or for APN, diabetes, anatomic or congenital abnormalities of the urinary system, infections by treatment-resistant organisms, nosocomial infection, urolithiasis, renal obstruction, prior renal surgery, advanced age, pregnancy, renal transplant recipients, and immunosuppressed or immunocompromised patients. Pregnant patients and patients with renal transplants on immunosuppression are at an elevated risk of severe complications. Imaging studies are often requested to aid with the diagnosis, identify precipitating factors, and differentiate lower UTI from renal parenchymal involvement, particularly in high-risk individuals. Imaging is usually not appropriate for the first-time presentation of suspected APN in an uncomplicated patient. The primary imaging modalities used in high-risk patients with suspected APN are CT, MRI, and ultrasound, although CT was usually not appropriate for initial imaging in a pregnant patient with no other complications. The ACR Appropriateness Criteria are evidence-based guidelines for specific clinical conditions that are reviewed annually by a multidisciplinary expert panel. The guideline development and revision process support the systematic analysis of the medical literature from peer-reviewed journals. Established methodology principles such as Grading of Recommendations Assessment, Development, and Evaluation or GRADE are adapted to evaluate the evidence. The RAND/UCLA Appropriateness Method User Manual provides the methodology to determine the appropriateness of imaging and treatment procedures for specific clinical scenarios. In those instances in which peer-reviewed literature is lacking or equivocal, experts may be the primary evidentiary source available to formulate a recommendation.
Key Words
Acute pyelonephritis
Appropriateness Criteria
AUC
Pyelonephritis
Urinary tract infection
ACR Appropriateness Criteria® Acute Pyelonephritis: 2022 Update. Variants 1 to 5 and Tables 1 and 2.
| Procedure | Appropriateness Category | Relative Radiation Level |
---
| US abdomen | Usually Not Appropriate | O |
| US color Doppler kidneys and bladder retroperitoneal | Usually Not Appropriate | O |
| Fluoroscopy voiding cystourethrography | Usually Not Appropriate | ☢☢ |
| Radiography abdomen and pelvis (KUB) | Usually Not Appropriate | ☢☢ |
| Fluoroscopy antegrade pyelography | Usually Not Appropriate | ☢☢☢ |
| Radiography intravenous urography | Usually Not Appropriate | ☢☢☢ |
| MRI abdomen and pelvis without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen and pelvis without IV contrast | Usually Not Appropriate | O |
| MRI abdomen without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen without IV contrast | Usually Not Appropriate | O |
| MRU without and with IV contrast | Usually Not Appropriate | O |
| MRU without IV contrast | Usually Not Appropriate | O |
| CT abdomen and pelvis with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen and pelvis without IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without IV contrast | Usually Not Appropriate | ☢☢☢ |
| DMSA renal scan | Usually Not Appropriate | ☢☢☢ |
| CT abdomen and pelvis without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CT abdomen without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CTU without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
Variant 1
Suspected acute pyelonephritis. First-time presentation. Uncomplicated patient (eg, no history of pyelonephritis, diabetes, immune compromise, history of stones or renal obstruction, prior renal surgery, advanced age, vesicoureteral reflux, lack of response to therapy, or pregnancy). Initial imaging.
Open table in a new tab
| Procedure | Appropriateness Category | Relative Radiation Level |
---
| CT abdomen and pelvis with IV contrast | Usually Appropriate | ☢☢☢ |
| US abdomen | May Be Appropriate | O |
| US color Doppler kidneys and bladder retroperitoneal | May Be Appropriate | O |
| MRI abdomen and pelvis without and with IV contrast | May Be Appropriate | O |
| MRI abdomen and pelvis without IV contrast | May Be Appropriate | O |
| CT abdomen and pelvis without IV contrast | May Be Appropriate | ☢☢☢ |
| CT abdomen with IV contrast | May Be Appropriate (Disagreement) | ☢☢☢ |
| CT abdomen and pelvis without and with IV contrast | May Be Appropriate (Disagreement) | ☢☢☢☢ |
| Fluoroscopy voiding cystourethrography | Usually Not Appropriate | ☢☢ |
| Radiography abdomen and pelvis (KUB) | Usually Not Appropriate | ☢☢ |
| Fluoroscopy antegrade pyelography | Usually Not Appropriate | ☢☢☢ |
| Radiography intravenous urography | Usually Not Appropriate | ☢☢☢ |
| MRI abdomen without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen without IV contrast | Usually Not Appropriate | O |
| MRU without and with IV contrast | Usually Not Appropriate | O |
| MRU without IV contrast | Usually Not Appropriate | O |
| CT abdomen without IV contrast | Usually Not Appropriate | ☢☢☢ |
| DMSA renal scan | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CTU without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
Variant 2
Suspected acute pyelonephritis. Complicated patient (eg, recurrent pyelonephritis, diabetes, immune compromise, advanced age, vesicoureteral reflux, or lack of response to initial therapy). Initial imaging.
Open table in a new tab
| Procedure | Appropriateness Category | Relative Radiation Level |
---
| CT abdomen and pelvis with IV contrast | Usually Appropriate | ☢☢☢ |
| CT abdomen and pelvis without and with IV contrast | Usually Appropriate | ☢☢☢☢ |
| US abdomen | May Be Appropriate | O |
| US color Doppler kidneys and bladder retroperitoneal | May Be Appropriate | O |
| MRI abdomen and pelvis without and with IV contrast | May Be Appropriate | O |
| MRI abdomen and pelvis without IV contrast | May Be Appropriate | O |
| CT abdomen and pelvis without IV contrast | May Be Appropriate | ☢☢☢ |
| CT abdomen without and with IV contrast | May Be Appropriate (Disagreement) | ☢☢☢☢ |
| Fluoroscopy voiding cystourethrography | Usually Not Appropriate | ☢☢ |
| Radiography abdomen and pelvis (KUB) | Usually Not Appropriate | ☢☢ |
| Fluoroscopy antegrade pyelography | Usually Not Appropriate | ☢☢☢ |
| Radiography intravenous urography | Usually Not Appropriate | ☢☢☢ |
| MRI abdomen without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen without IV contrast | Usually Not Appropriate | O |
| MRU without and with IV contrast | Usually Not Appropriate | O |
| MRU without IV contrast | Usually Not Appropriate | O |
| CT abdomen with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without IV contrast | Usually Not Appropriate | ☢☢☢ |
| DMSA renal scan | Usually Not Appropriate | ☢☢☢ |
| CTU without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
Variant 3
Suspected acute pyelonephritis. History of renal stones or renal obstruction. Initial imaging.
Open table in a new tab
| Procedure | Appropriateness Category | Relative Radiation Level |
---
| US abdomen | May Be Appropriate (Disagreement) | O |
| US color Doppler kidneys and bladder retroperitoneal | May Be Appropriate | O |
| MRI abdomen and pelvis without IV contrast | May Be Appropriate | O |
| MRI abdomen without IV contrast | May Be Appropriate | O |
| MRU without IV contrast | May Be Appropriate | O |
| Fluoroscopy voiding cystourethrography | Usually Not Appropriate | ☢☢ |
| Radiography abdomen and pelvis (KUB) | Usually Not Appropriate | ☢☢ |
| Fluoroscopy antegrade pyelography | Usually Not Appropriate | ☢☢☢ |
| Radiography intravenous urography | Usually Not Appropriate | ☢☢☢ |
| MRI abdomen and pelvis without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen without and with IV contrast | Usually Not Appropriate | O |
| MRU without and with IV contrast | Usually Not Appropriate | O |
| CT abdomen and pelvis with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen and pelvis without IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without IV contrast | Usually Not Appropriate | ☢☢☢ |
| DMSA renal scan | Usually Not Appropriate | ☢☢☢ |
| CT abdomen and pelvis without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CT abdomen without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CTU without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
Variant 4
Suspected acute pyelonephritis. Pregnant patient with no other complications (eg, no history of diabetes, immune compromise, history of stones or renal obstruction, prior renal surgery, vesicoureteral reflux, or lack of response to therapy). Initial imaging.
Open table in a new tab
| Procedure | Appropriateness Category | Relative Radiation Level |
---
| US duplex Doppler kidney transplant | Usually Appropriate | O |
| CT abdomen and pelvis with IV contrast | Usually Appropriate | ☢☢☢ |
| MRI abdomen and pelvis without and with IV contrast | May Be Appropriate | O |
| MRI abdomen and pelvis without IV contrast | May Be Appropriate | O |
| CT abdomen and pelvis without IV contrast | May Be Appropriate | ☢☢☢ |
| CT abdomen and pelvis without and with IV contrast | May Be Appropriate | ☢☢☢☢ |
| US abdomen | Usually Not Appropriate | O |
| US color Doppler kidneys and bladder retroperitoneal | Usually Not Appropriate | O |
| Fluoroscopy voiding cystourethrography | Usually Not Appropriate | ☢☢ |
| Radiography abdomen and pelvis (KUB) | Usually Not Appropriate | ☢☢ |
| Fluoroscopy antegrade pyelography | Usually Not Appropriate | ☢☢☢ |
| Radiography intravenous urography | Usually Not Appropriate | ☢☢☢ |
| MRI abdomen without and with IV contrast | Usually Not Appropriate | O |
| MRI abdomen without IV contrast | Usually Not Appropriate | O |
| MRI pelvis without and with IV contrast | Usually Not Appropriate | O |
| MRI pelvis without IV contrast | Usually Not Appropriate | O |
| MRU without and with IV contrast | Usually Not Appropriate | O |
| MRU without IV contrast | Usually Not Appropriate | O |
| CT abdomen with IV contrast | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without IV contrast | Usually Not Appropriate | ☢☢☢ |
| DMSA renal scan | Usually Not Appropriate | ☢☢☢ |
| CT abdomen without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
| CTU without and with IV contrast | Usually Not Appropriate | ☢☢☢☢ |
Variant 5
Suspected acute pyelonephritis. History of pelvic renal transplant with native kidneys in situ and no other complications (eg, no history of pyelonephritis, diabetes, history of stones or renal obstruction, prior renal surgery, advanced age, vesicoureteral reflux, lack of response to therapy, or pregnancy). Initial imaging.
Open table in a new tab
| Appropriateness Category Name | Appropriateness Rating | Appropriateness Category Definition |
---
| Usually Appropriate | 7, 8, or 9 | The imaging procedure or treatment is indicated in the specified clinical scenarios at a favorable risk-benefit ratio for patients. |
| May Be Appropriate | 4, 5, or 6 | The imaging procedure or treatment may be indicated in the specified clinical scenarios as an alternative to imaging procedures or treatments with a more favorable risk-benefit ratio, or the risk-benefit ratio for patients is equivocal. |
| May Be Appropriate (Disagreement) | 5 | The individual ratings are too dispersed from the panel median. The different label provides transparency regarding the panel’s recommendation. “May be appropriate” is the rating category and a rating of 5 is assigned. |
| Usually Not Appropriate | 1, 2, or 3 | The imaging procedure or treatment is unlikely to be indicated in the specified clinical scenarios, or the risk-benefit ratio for patients is likely to be unfavorable. |
Table 1
Appropriateness category names and definitions
Open table in a new tab
| RRL | Adult Effective Dose Estimate Range (mSv) | Pediatric Effective Dose Estimate Range (mSv) |
---
| O | 0 | 0 |
| ☢ | <0.1 | <0.03 |
| ☢☢ | 0.1-1 | 0.03-0.3 |
| ☢☢☢ | 1-10 | 0.3-3 |
| ☢☢☢☢ | 10-30 | 3-10 |
| ☢☢☢☢☢ | 30-100 | 10-30 |
Table 2
Relative radiation level designations
Note: Relative radiation level (RRL) assignments for some of the examinations cannot be made, because the actual patient doses in these procedures vary as a function of a number of factors (eg, region of the body exposed to ionizing radiation, the imaging guidance that is used). The RRLs for these examinations are designated as “varies.”
Open table in a new tab
Summary of Literature Review
Introduction/Background
Acute pyelonephritis (APN) is a severe urinary tract infection (UTI) that has the potential to cause sepsis, shock, and death [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
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]. The annual incidence of APN is 459,000 to 1,128,000 cases in the United States and 10.5 to 25.9 million cases globally [1
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Johnson, J.R. ∙ Russo, T.A.
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Czaja, C.A. ∙ Scholes, D. ∙ Hooton, T.M. ...
Population-based epidemiologic analysis of acute pyelonephritis
Clin Infect Dis. 2007; 45:273-280
Crossref
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]. The term “pyelonephritis” implies that there is inflammation of the renal pelvis and kidney. APN often presents with signs and symptoms of both systemic inflammation (eg, fever, chills, and fatigue) and bladder inflammation (eg, urgency, dysuria, and urinary frequency) [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
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PubMed
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]. There is a surprising lack of consensus regarding diagnostic criteria, and differentiation from infections of the lower urinary tract can be difficult [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
Scopus (101)
PubMed
Google Scholar
]. Clinical presentation of APN can range from mild flank pain with low-grade or no fever to septic shock, and up to 20% of patients lack bladder symptoms. In patients with flank pain or tenderness, without or with voiding symptoms, without or with fever, and with a urinalysis showing pyuria and/or bacteriuria, APN is an appropriate presumptive diagnosis [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
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PubMed
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]. In this setting, urine cultures yielding >10,000 colony-forming units of a uropathogen per milliliter of urine is the fundamental confirmatory diagnostic test. Positive blood cultures may assist with the diagnosis.
In young healthy women, Escherichia coli accounts for more than 90% of APN cases [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
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Stamm, W.E. ∙ Hooton, T.M. ∙ Johnson, J.R. ...
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]. However, in men, elderly women, and urologically compromised and institutionalized patients, less-virulent E.coli strains, gram-negative bacilli, gram-positive organisms, and candida are also common [1
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4.
Talan, D.A. ∙ Takhar, S.S. ∙ Krishnadasan, A. ...
Fluoroquinolone-resistant and extended-spectrum beta-lactamase-producing Escherichia coli infections in patients with pyelonephritis, United States(1)
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]. Risk factors for cystitis predispose to APN and include sexual activity, new sexual partner, spermicide exposure, personal or maternal history of UTIs, genetic predisposition, and diabetes mellitus [1
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Johnson, J.R. ∙ Russo, T.A.
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Scholes, D. ∙ Hooton, T.M. ∙ Roberts, P.L. ...
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]. Fortunately, <3% of cases of cystitis and asymptomatic bacteriuria progress to APN [1
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Ikaheimo, R. ∙ Siitonen, A. ∙ Heiskanen, T. ...
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]. Factors that disrupt urinary flow such as vesicoureteral reflux, congenital urinary tract anomalies, altered bladder function, pregnancy, renal calculi, or mechanical obstruction increase the risk of developing APN [1
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Godaly, G. ∙ Ambite, I. ∙ Svanborg, C.
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Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
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J Am Coll Radiol. 2018; 15:S232-S239
Full Text
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].
In the majority of patients, uncomplicated APN is diagnosed clinically and is responsive to treatment with appropriate antibiotics [1
1.
Johnson, J.R. ∙ Russo, T.A.
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N Engl J Med. 2018; 378:48-59
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]. In patients who are high risk or when treatment is delayed, microabscesses may coalesce to form an acute renal abscess. The renal parenchymal abscess can at times rupture into the perinephric space and lead to development of a perirenal abscess. In other cases, the infection may be confined to an obstructed collecting system causing pyonephrosis, or accumulation of purulent material in the upper urinary collecting system, that often requires decompression for treatment to be successful.
Some patients are at a high risk for developing complications from APN. High-risk patients include those with a prior history of pyelonephritis, a lack of response to therapy for lower UTI or for APN, diabetes, anatomic or congenital abnormalities of the urinary system, infections by treatment-resistant organisms, nosocomial infection, urolithiasis, renal obstruction, prior renal surgery, advanced age, and pregnancy; renal transplant recipients; and immunosuppressed or immunocompromised patients [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
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Scopus (101)
PubMed
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8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
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Velasco, M. ∙ Martinez, J.A. ∙ Moreno-Martinez, A. ...
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Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
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]. Pregnant patients and patients with renal transplants on immunosuppression are at an elevated risk of severe complications. Imaging studies are often requested to aid with the diagnosis, identify precipitating factors, and differentiate lower UTI from renal parenchymal involvement, particularly in high-risk individuals. See the ACR Appropriateness Criteria® topics on “Acute Onset Flank Pain-Suspicion of Stone Disease (Urolithiasis)” [11
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American College of Radiology
ACR Appropriateness Criteria®: acute onset flank pain—suspicion of stone disease (urolithiasis)
Available at:
Date accessed: March 31, 2022
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], “Hematuria” [12
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Wolfman, D.J. ∙ Marko, J. ∙ Nikolaidis, P. ...
ACR Appropriateness Criteria® hematuria
J Am Coll Radiol. 2020; 17:S138-S147
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], and “Recurrent Lower Urinary Tract Infections in Females” [13
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Venkatesan, A.M. ∙ Oto, A. ∙ Allen, B.C. ...
ACR Appropriateness Criteria® recurrent lower urinary tract infections in females
J Am Coll Radiol. 2020; 17:S487-S496
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] for additional information.
Special Imaging Considerations
CT urography (CTU) is an imaging study that is tailored to improve visualization of both the upper and lower urinary tracts. There is variability in the specific parameters, but it usually involves unenhanced images followed by intravenous (IV) contrast-enhanced images, including nephrographic and excretory phases acquired at least 5 minutes after contrast injection. Alternatively, a split-bolus technique uses an initial loading dose of IV contrast and then obtains a combined nephrographic-excretory phase after a second IV contrast dose; some sites include the arterial phase. CTU should use thin-slice acquisition. Reconstruction methods commonly include maximum intensity projection or 3-D volume rendering. For the purposes of this document, we make a distinction between CTU and CT abdomen and pelvis without and with IV contrast. CT abdomen and pelvis without and with IV contrast is defined as any protocol not specifically tailored for evaluation of the upper and lower urinary tracts and without both the precontrast and excretory phases.
MR urography (MRU) is also tailored to improve imaging of the urinary system. Unenhanced MRU relies upon heavily T2-weighted imaging of the intrinsic high signal intensity from urine for the evaluation of the urinary tract. IV contrast is administered to provide additional information regarding obstruction, urothelial thickening, focal lesions, and stones. A contrast-enhanced T1-weighted series should include corticomedullary, nephrographic, and excretory phase. Thin-slice acquisition and multiplanar imaging should be obtained. For the purposes of this document, we make a distinction between MRU and MRI abdomen and pelvis without and with IV contrast. MRI abdomen and pelvis without and with IV contrast is defined as any protocol not specifically tailored for evaluation of the upper and lower urinary tracts, without both the precontrast and excretory phases, and without heavily T2-weighted images of the urinary tract.
Initial Imaging Definition
Initial imaging is defined as imaging at the beginning of the care episode for the medical condition defined by the variant. More than one procedure can be considered usually appropriate in the initial imaging evaluation when:
•
There are procedures that are equivalent alternatives (ie, only one procedure will be ordered to provide the clinical information to effectively manage the patient’s care)
OR
•
There are complementary procedures (ie, more than one procedure is ordered as a set or simultaneously where each procedure provides unique clinical information to effectively manage the patient’s care).
Discussion of Procedures by Variant
Variant 1: Suspected acute pyelonephritis. First-time presentation. Uncomplicated patient (eg, no history of pyelonephritis, diabetes, immune compromise, history of stones or renal obstruction, prior renal surgery, advanced age, vesicoureteral reflux, lack of response to therapy, or pregnancy). Initial imaging
CT Abdomen and Pelvis
CT of the abdomen and pelvis is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
Scopus (101)
PubMed
Google Scholar
,10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,14
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
,15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
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]. CT imaging may be useful if symptoms persist for 72 hours [8
8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
,10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,14
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
,16
16.
Abraham, G. ∙ Reddy, Y.N. ∙ George, G.
Diagnosis of acute pyelonephritis with recent trends in management
Nephrol Dial Transplant. 2012; 27:3391-3394
Crossref
Scopus (14)
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]. Nearly 95% of patients with uncomplicated pyelonephritis become afebrile within 48 hours after appropriate antibiotic therapy, and nearly 100% become afebrile within 72 hours [8
8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
,14
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
].
CT Abdomen
CT of the abdomen is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
Scopus (101)
PubMed
Google Scholar
,10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,14
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
,15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
CTU
CTU is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
Scopus (101)
PubMed
Google Scholar
,10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,14
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
,15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
DMSA Renal Scan
Tc-99m–labeled dimercaptosuccinic acid (DMSA) renal scintigraphy is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Fluoroscopy Antegrade Pyelography
Antegrade pyelography is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Fluoroscopy Voiding Cystourethrography
Fluoroscopy voiding cystourethrography (VCUG) is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
MRI Abdomen and Pelvis
MRI of the abdomen and pelvis is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
MRI Abdomen
MRI of the abdomen is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
MRU
MRU of the abdomen and pelvis is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Radiography Abdomen and Pelvis (KUB)
Radiography of the abdomen and pelvis (KUB) is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Radiography Intravenous Urography
Intravenous urography (IVU) is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
US Abdomen
Ultrasound (US) of the abdomen is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
]. In addition, US had inferior accuracy for detection of APN compared with CT [10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,17
17.
Pierce, C. ∙ Keniston, A. ∙ Albert, R.K.
Imaging in acute pyelonephritis: utilization, findings, and effect on management
South Med J. 2019; 112:118-124
Crossref
Scopus (2)
Google Scholar
].
US Color Doppler Kidneys and Bladder Retroperitoneal
US color doppler of the kidneys, bladder, and retroperitoneum is not beneficial in the initial imaging evaluation for the first-time presentation of suspected APN in an uncomplicated patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Variant 2: Suspected acute pyelonephritis. Complicated patient (eg, recurrent pyelonephritis, diabetes, immune compromise, advanced age, vesicoureteral reflux, or lack of response to initial therapy). Initial imaging
The goal of imaging in a complicated patient with suspected APN is to identify the presence or absence of APN and identify associated complications. Patients with a history of renal calculi are separately discussed in variant 3.
CT Abdomen and Pelvis
There is widespread agreement that CT of the abdomen and pelvis with IV contrast is a useful study to diagnose APN in a complicated patient without a prior history of stone disease [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
Crossref
Scopus (101)
PubMed
Google Scholar
,8
8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
,14-21
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
Radiology. 1989; 171:703-707
Crossref
Scopus (133)
PubMed
Google Scholar
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
16.
Abraham, G. ∙ Reddy, Y.N. ∙ George, G.
Diagnosis of acute pyelonephritis with recent trends in management
Nephrol Dial Transplant. 2012; 27:3391-3394
Crossref
Scopus (14)
Google Scholar
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Pierce, C. ∙ Keniston, A. ∙ Albert, R.K.
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Contrast-enhanced CT can be used to detect signs of APN including focal or multifocal decreased parenchymal enhancement, complications of APN including a renal or perirenal abscess or emphysematous pyelonephritis, and underlying problems including hydronephrosis, obstructing stones, or congenital abnormalities [1
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CTU
There is insufficient evidence to support the use of CTU for detection of suspected APN in the complicated patient. There is variability in CTU imaging technique. The added benefit of an excretory phase from any CTU image acquisition protocol is likely negligible with respect to detection and characterization of APN in a complicated patient.
DMSA Renal Scan
According to the literature, renal scintigraphy, specifically Tc-99m DMSA scan, is not beneficial for the diagnosis of APN in adults. In contrast, renal scintigraphy is useful in the pediatric population where there is difficulty in differentiating lower UTI from APN [23
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MRU
The excretory phase of MRU does not confer additional benefit with respect to detection and characterization of APN in a complicated patient. Studies comparing MRU with DMSA renal scintigraphy for the detection of pyelonephritis and renal scaring have shown that MRU without and with IV contrast is at least equivalent or superior to DMSA for this specific purpose [33-35
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In a small retrospective study of adult patients with clinical and laboratory suspicion of APN who all underwent triphasic abdominal CT (n= 100), the accuracy of the nephrographic phase among 2 readers for diagnosis of APN was 90% to 92% and for diagnosis of urolithiasis was 96% to 99% [22
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CT Abdomen
CT of the abdomen without IV contrast has high accuracy for detection or renal stones, and CT of the abdomen with IV contrast has high accuracy for the diagnosis of APN. However, CT of the abdomen alone does not allow for a comprehensive assessment of the entire genitourinary tract [1
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Google Scholar
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Google Scholar
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Google Scholar
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Pierce, C. ∙ Keniston, A. ∙ Albert, R.K.
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Google Scholar
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Enikeev, D.V. ∙ Glybochko, P. ∙ Alyaev, Y. ...
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Crossref
PubMed
Google Scholar
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Lee, A. ∙ Kim, H.C. ∙ Hwang, S.I. ...
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PubMed
Google Scholar
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CTU
There is insufficient evidence to support the use of CTU for detection of suspected APN in the complicated patient. There is variability in the CTU imaging technique. The added benefit of an excretory phase from any CTU image acquisition protocol is likely negligible with respect to detection and characterization of APN in a complicated patient.
DMSA Renal Scan
CT may have higher accuracy than DMSA renal scintigraphy for detection of APN and certainly has higher accuracy for detection of stones, which cannot be directly visualized by renal scintigraphy [24
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Chua, M.E. ∙ Ming, J.M. ∙ Farhat, W.A.
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PubMed
Google Scholar
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Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
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].
MRI Abdomen
MRI of the abdomen has high accuracy for the diagnosis of APN in a complicated patient but does not allow for a comprehensive assessment of the entire genitourinary tract. Failure to include the pelvis could lead to a missed opportunity to detect urolithiasis in the distal ureters or urinary bladder, congenital abnormalities of the distal ureters and abnormal ureteral insertion sites, and abnormalities of the urinary bladder, among other potential sources of APN.
MRI has poor accuracy for the detection of small urothelial calculi. Another disadvantage of MRI includes its relatively reduced ability to detect gas in emphysematous pyelonephritis [31
31.
Chua, M.E. ∙ Ming, J.M. ∙ Farhat, W.A.
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MRU
The excretory phase of MRU does not confer additional benefit with respect to detection and characterization of APN in a complicated patient. Furthermore, MRU has poor accuracy for the detection of small urothelial calculi [31
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Chua, M.E. ∙ Ming, J.M. ∙ Farhat, W.A.
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Radiography Abdomen and Pelvis (KUB)
Radiography of the abdomen and pelvis (KUB) has limited utility in the imaging evaluation for suspected APN in a patient with a history of renal stones or renal obstruction [15
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Radiography Intravenous Urography
IVU is not beneficial in the imaging evaluation for suspected APN in a patient with a history of renal stones or renal obstruction [15
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Tamm, E.P. ∙ Silverman, P.M. ∙ Shuman, W.P.
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Some disadvantages of US compared with CT are a lower rate of detection of APN and renal abscess [10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
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36.
Yoo, J.M. ∙ Koh, J.S. ∙ Han, C.H. ...
Diagnosing acute pyelonephritis with CT, Tc-DMSA SPECT, and Doppler ultrasound: a comparative study
Korean J Urol. 2010; 51:260-265
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Scopus (34)
PubMed
Google Scholar
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Fontanilla, T. ∙ Minaya, J. ∙ Cortes, C. ...
Acute complicated pyelonephritis: contrast-enhanced ultrasound
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PubMed
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38.
Kim, B. ∙ Lim, H.K. ∙ Choi, M.H. ...
Detection of parenchymal abnormalities in acute pyelonephritis by pulse inversion harmonic imaging with or without microbubble ultrasonographic contrast agent: correlation with computed tomography
J Ultrasound Med. 2001; 20:5-14
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]. Of note, the sensitivity for detection of acute complicated pyelonephritis and for detection of a renal abscess is higher with contrast-enhanced US relative to unenhanced US [37
37.
Fontanilla, T. ∙ Minaya, J. ∙ Cortes, C. ...
Acute complicated pyelonephritis: contrast-enhanced ultrasound
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38.
Kim, B. ∙ Lim, H.K. ∙ Choi, M.H. ...
Detection of parenchymal abnormalities in acute pyelonephritis by pulse inversion harmonic imaging with or without microbubble ultrasonographic contrast agent: correlation with computed tomography
J Ultrasound Med. 2001; 20:5-14
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]. In one small retrospective study of adults with APN (n= 100), the accuracy of contrast-enhanced US for detection of APN approached that of contrast-enhanced CT [39
39.
Mitterberger, M. ∙ Pinggera, G.M. ∙ Colleselli, D. ...
Acute pyelonephritis: comparison of diagnosis with computed tomography and contrast-enhanced ultrasonography
BJU Int. 2008; 101:341-344
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US Color Doppler Kidneys and Bladder Retroperitoneal
US of the kidneys has nearly 100% sensitivity for detection of large stones (>5 mm) and hydronephrosis, although the accuracy for detection of small stones (<3 mm) is poor [44
44.
Tamm, E.P. ∙ Silverman, P.M. ∙ Shuman, W.P.
Evaluation of the patient with flank pain and possible ureteral calculus
Radiology. 2003; 228:319-329
Crossref
Scopus (100)
PubMed
Google Scholar
,45
45.
Moore, C.L. ∙ Scoutt, L.
Sonography first for acute flank pain?
J Ultrasound Med. 2012; 31:1703-1711
Crossref
Scopus (34)
PubMed
Google Scholar
]. Use of color Doppler has been shown to increase sensitivity for detection of APN in adults and children [40-42
40.
Kawashima, A. ∙ Sandler, C.M. ∙ Goldman, S.M.
Imaging in acute renal infection
BJU Int. 2000; 86:70-79
PubMed
Google Scholar
41.
Bykov, S. ∙ Chervinsky, L. ∙ Smolkin, V. ...
Power Doppler sonography versus Tc-99m DMSA scintigraphy for diagnosing acute pyelonephritis in children: are these two methods comparable?
Clin Nucl Med. 2003; 28:198-203
Crossref
PubMed
Google Scholar
42.
Halevy, R. ∙ Smolkin, V. ∙ Bykov, S. ...
Power Doppler ultrasonography in the diagnosis of acute childhood pyelonephritis
Pediatr Nephrol. 2004; 19:987-991
Crossref
Scopus (37)
PubMed
Google Scholar
]. Inclusion of the bladder could identify other abnormalities contributing to APN. A disadvantage of US compared with CT is a lower rate of detection of APN and renal abscess [10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,36-38
36.
Yoo, J.M. ∙ Koh, J.S. ∙ Han, C.H. ...
Diagnosing acute pyelonephritis with CT, Tc-DMSA SPECT, and Doppler ultrasound: a comparative study
Korean J Urol. 2010; 51:260-265
Crossref
Scopus (34)
PubMed
Google Scholar
37.
Fontanilla, T. ∙ Minaya, J. ∙ Cortes, C. ...
Acute complicated pyelonephritis: contrast-enhanced ultrasound
Abdom Imaging. 2012; 37:639-646
Crossref
Scopus (55)
PubMed
Google Scholar
38.
Kim, B. ∙ Lim, H.K. ∙ Choi, M.H. ...
Detection of parenchymal abnormalities in acute pyelonephritis by pulse inversion harmonic imaging with or without microbubble ultrasonographic contrast agent: correlation with computed tomography
J Ultrasound Med. 2001; 20:5-14
Crossref
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PubMed
Google Scholar
].
Variant 4: Suspected acute pyelonephritis. Pregnant patient with no other complications (eg, no history of diabetes, immune compromise, history of stones or renal obstruction, prior renal surgery, vesicoureteral reflux, or lack of response to therapy). Initial imaging
CT Abdomen and Pelvis
There is no current literature specific to the use of CT of the abdomen and pelvis in the evaluation of suspected APN in pregnant patients. CT imaging of the abdomen and pelvis is not supported as the initial imaging in pregnant patients. The main disadvantage of using CT of the abdomen and pelvis in a pregnant patient is the risk of ionizing radiation to the embryo or fetus and the mother, particularly for the pelvic portion of the examination [46
46.
American College of Radiology
ACR-SPR practice parameter for imaging pregnant or potentially pregnant adolescents and women with ionizing radiation
Date accessed: March 31, 2022
Google Scholar
].
CT Abdomen
There is no current literature to support the use of CT of the abdomen in the evaluation of suspected APN in pregnant patients. CT imaging of the abdomen does not allow for detection of pelvic abnormalities including urolithiasis in the distal ureters or urinary bladder, congenital abnormalities of the distal ureters and abnormal insertion sites, and abnormalities of the urinary bladder, among other potential sources of APN.
CTU
There is no current literature to support the use of CTU in the evaluation of suspected APN in pregnant patients. There is variability in CTU imaging technique. The added benefit of an excretory phase from any CTU image acquisition protocol is likely negligible with respect to the detection and characterization of APN in a pregnant patient. A CTU protocol that does not include unenhanced imaging and that uses a split bolus to achieve a mixed nephrographic and excretory phase could mask the presence of urolithiasis.
DMSA Renal Scan
DMSA renal scintigraphy is not beneficial in the imaging evaluation for suspected APN in a pregnant patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
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].
Fluoroscopy Antegrade Pyelography
Antegrade pyelography is not beneficial in the imaging evaluation for suspected APN in a pregnant patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Fluoroscopy Voiding Cystourethrography
VCUG is not beneficial in the imaging evaluation for suspected APN in a pregnant patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
MRI Abdomen and Pelvis
There is no current literature specific to the use of MRI of the abdomen and pelvis in the evaluation of suspected APN in pregnant patients. MRI of the abdomen and pelvis is generally safe in pregnant patients and may be useful in certain situations. MRI does do not expose the embryo, or fetus, or pregnant mother to ionizing radiation and can be useful for detecting APN, scarring, congenital anomalies of the kidneys, renal abscesses, hydronephrosis, and pyonephrosis [28-30
28.
Rathod, S.B. ∙ Kumbhar, S.S. ∙ Nanivadekar, A. ...
Role of diffusion-weighted MRI in acute pyelonephritis: a prospective study
Acta Radiol. 2015; 56:244-249
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Faletti, R. ∙ Cassinis, M.C. ∙ Fonio, P. ...
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Eur Radiol. 2013; 23:3501-3508
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De Pascale, A. ∙ Piccoli, G.B. ∙ Priola, S.M. ...
Diffusion-weighted magnetic resonance imaging: new perspectives in the diagnostic pathway of non-complicated acute pyelonephritis
Eur Radiol. 2013; 23:3077-3086
Crossref
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PubMed
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]. Although there are no known adverse effects to human fetuses and no known cases of nephrogenic systemic fibrosis linked to the use of clinical doses of gadolinium-based contrast agents (GBCAs) in pregnant patients, GBCAs should only be used if the indication is considered critical and the potential benefits justify the potential unknown risk to the fetus [47
47.
American College of Radiology
ACR Committee on Drugs and Contrast Media. Manual on contrast media
Available at:
Date accessed: March 31, 2022
Google Scholar
].
MRI of the abdomen can be useful for detecting APN, scarring, congenital anomalies of the kidneys, renal abscesses, hydronephrosis, and pyonephrosis [28-30
28.
Rathod, S.B. ∙ Kumbhar, S.S. ∙ Nanivadekar, A. ...
Role of diffusion-weighted MRI in acute pyelonephritis: a prospective study
Acta Radiol. 2015; 56:244-249
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Faletti, R. ∙ Cassinis, M.C. ∙ Fonio, P. ...
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Eur Radiol. 2013; 23:3501-3508
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PubMed
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Vivier, P.H. ∙ Sallem, A. ∙ Beurdeley, M. ...
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28.
Rathod, S.B. ∙ Kumbhar, S.S. ∙ Nanivadekar, A. ...
Role of diffusion-weighted MRI in acute pyelonephritis: a prospective study
Acta Radiol. 2015; 56:244-249
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Faletti, R. ∙ Cassinis, M.C. ∙ Fonio, P. ...
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Eur Radiol. 2013; 23:3501-3508
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De Pascale, A. ∙ Piccoli, G.B. ∙ Priola, S.M. ...
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Eur Radiol. 2013; 23:3077-3086
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]. APN, renal abscesses, and pyonephrosis have lower ADC values than normal renal cortical parenchyma [28
28.
Rathod, S.B. ∙ Kumbhar, S.S. ∙ Nanivadekar, A. ...
Role of diffusion-weighted MRI in acute pyelonephritis: a prospective study
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]. Inclusion of the pelvis could improve detection of abnormalities of the lower urinary tract.
MRI may allow for a limited evaluation of the embryo or fetus. However, traditional MRI examinations do not provide a comprehensive assessment of the urinary collecting systems [8
8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
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]. The main disadvantages of MRI are poor accuracy for the detection of small urothelial calculi and reduced accuracy for detection of emphysematous pyelonephritis, [31
31.
Chua, M.E. ∙ Ming, J.M. ∙ Farhat, W.A.
Magnetic resonance urography in the pediatric population: a clinical perspective
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32.
Leyendecker, J.R. ∙ Clingan, M.J.
Magnetic resonance urography update—are we there yet?
Semin Ultrasound CT MR. 2009; 30:246-257
Crossref
Scopus (12)
PubMed
Google Scholar
].
MRI Abdomen
There is no current literature specific to the use of MRI of the abdomen in the evaluation of suspected APN in pregnant patients. MRI of the abdomen does not allow for detection of pelvic abnormalities including urolithiasis in the distal ureters or urinary bladder, congenital abnormalities of the distal ureters and abnormal insertion sites, and abnormalities of the urinary bladder, among other potential sources of APN. MRI of the abdomen is generally safe in pregnant patients and may be useful in certain situations [28-30
28.
Rathod, S.B. ∙ Kumbhar, S.S. ∙ Nanivadekar, A. ...
Role of diffusion-weighted MRI in acute pyelonephritis: a prospective study
Acta Radiol. 2015; 56:244-249
Crossref
Scopus (44)
PubMed
Google Scholar
29.
Faletti, R. ∙ Cassinis, M.C. ∙ Fonio, P. ...
Diffusion-weighted imaging and apparent diffusion coefficient values versus contrast-enhanced MR imaging in the identification and characterisation of acute pyelonephritis
Eur Radiol. 2013; 23:3501-3508
Crossref
Scopus (53)
PubMed
Google Scholar
30.
De Pascale, A. ∙ Piccoli, G.B. ∙ Priola, S.M. ...
Diffusion-weighted magnetic resonance imaging: new perspectives in the diagnostic pathway of non-complicated acute pyelonephritis
Eur Radiol. 2013; 23:3077-3086
Crossref
Scopus (34)
PubMed
Google Scholar
]. Although there are no known adverse effects to human fetuses and no known cases of nephrogenic systemic fibrosis linked to the use of clinical doses of GBCAs in pregnant patients, GBCAs should only be used if the indication is considered critical and the potential benefits justify the potential unknown risk to the fetus [47
47.
American College of Radiology
ACR Committee on Drugs and Contrast Media. Manual on contrast media
Available at:
Date accessed: March 31, 2022
Google Scholar
].
MRI of the abdomen may allow for a limited evaluation of the embryo or fetus, although most fetuses would not be included in the field of view. The main disadvantages of MRI are poor accuracy for the detection of small urothelial calculi and reduced accuracy for detection of emphysematous pyelonephritis [31
31.
Chua, M.E. ∙ Ming, J.M. ∙ Farhat, W.A.
Magnetic resonance urography in the pediatric population: a clinical perspective
Pediatr Radiol. 2016; 46:791-795
Crossref
Scopus (11)
PubMed
Google Scholar
,32
32.
Leyendecker, J.R. ∙ Clingan, M.J.
Magnetic resonance urography update—are we there yet?
Semin Ultrasound CT MR. 2009; 30:246-257
Crossref
Scopus (12)
PubMed
Google Scholar
].
MRU
There is no current literature specific to the use of MRU without IV contrast in the evaluation of suspected APN in pregnant patients. The excretory phase of MRU does not confer additional benefit with respect to detection and characterization of APN in a pregnant patient.
Radiography Abdomen and Pelvis (KUB)
Radiography of the abdomen and pelvis is not beneficial in the imaging evaluation for suspected APN in a pregnant patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
Radiography Intravenous Urography
IVU is not beneficial in the imaging evaluation for suspected APN in a pregnant patient [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
].
US Abdomen
There is no current literature specific to the use of US of the abdomen in the evaluation of suspected APN in pregnant patients. Pregnancy increases the risk of complications from APN, although poor obstetrical outcomes are rare [48
48.
Hill, J.B. ∙ Sheffield, J.S. ∙ McIntire, D.D. ...
Acute pyelonephritis in pregnancy
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8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
]. US is often used as a screening examination in pregnancy, is sensitive and specific test for diagnosing hydronephrosis, and does not expose the patient or fetus to ionizing radiation [49
49.
McAleer, S.J. ∙ Loughlin, K.R.
Nephrolithiasis and pregnancy
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50.
Wieseler, K.M. ∙ Bhargava, P. ∙ Kanal, K.M. ...
Imaging in pregnant patients: examination appropriateness
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; discussion 30-3
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].
Physiologic hydronephrosis of pregnancy occurs in >80% of pregnant patients in the second and third trimester; therefore, hydronephrosis alone is not a reliable sign of APN in pregnant patients [51
51.
Rasmussen, P.E. ∙ Nielsen, F.R.
Hydronephrosis during pregnancy: a literature survey
Eur J Obstet Gynecol Reprod Biol. 1988; 27:249-259
Abstract
Full Text (PDF)
Scopus (141)
PubMed
Google Scholar
]. Furthermore, US has a low detection rate of APN and renal abscess [10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,36-38
36.
Yoo, J.M. ∙ Koh, J.S. ∙ Han, C.H. ...
Diagnosing acute pyelonephritis with CT, Tc-DMSA SPECT, and Doppler ultrasound: a comparative study
Korean J Urol. 2010; 51:260-265
Crossref
Scopus (34)
PubMed
Google Scholar
37.
Fontanilla, T. ∙ Minaya, J. ∙ Cortes, C. ...
Acute complicated pyelonephritis: contrast-enhanced ultrasound
Abdom Imaging. 2012; 37:639-646
Crossref
Scopus (55)
PubMed
Google Scholar
38.
Kim, B. ∙ Lim, H.K. ∙ Choi, M.H. ...
Detection of parenchymal abnormalities in acute pyelonephritis by pulse inversion harmonic imaging with or without microbubble ultrasonographic contrast agent: correlation with computed tomography
J Ultrasound Med. 2001; 20:5-14
Crossref
Scopus (60)
PubMed
Google Scholar
].
US Color Doppler Kidneys and Bladder Retroperitoneal
There is no current literature specific to the use of US color Doppler of the kidneys, bladder, and retroperitoneum in the evaluation of suspected APN in pregnant patients. US of the kidney, ureters, and bladder is safe in pregnancy, is rapid and portable, and does not require the use of contrast material [8
8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
]. US is often used as a screening examination in pregnancy, is sensitive and specific test for diagnosing hydronephrosis, and does not expose the patient or fetus to ionizing radiation [49
49.
McAleer, S.J. ∙ Loughlin, K.R.
Nephrolithiasis and pregnancy
Curr Opin Urol. 2004; 14:123-127
Crossref
Scopus (114)
PubMed
Google Scholar
,50
50.
Wieseler, K.M. ∙ Bhargava, P. ∙ Kanal, K.M. ...
Imaging in pregnant patients: examination appropriateness
Radiographics. 2010; 30:1215-1229
; discussion 30-3
Crossref
Scopus (126)
PubMed
Google Scholar
]. Use of color Doppler has been shown to increase sensitivity for detection of APN compared to US with grayscale imaging [40-42
40.
Kawashima, A. ∙ Sandler, C.M. ∙ Goldman, S.M.
Imaging in acute renal infection
BJU Int. 2000; 86:70-79
PubMed
Google Scholar
41.
Bykov, S. ∙ Chervinsky, L. ∙ Smolkin, V. ...
Power Doppler sonography versus Tc-99m DMSA scintigraphy for diagnosing acute pyelonephritis in children: are these two methods comparable?
Clin Nucl Med. 2003; 28:198-203
Crossref
PubMed
Google Scholar
42.
Halevy, R. ∙ Smolkin, V. ∙ Bykov, S. ...
Power Doppler ultrasonography in the diagnosis of acute childhood pyelonephritis
Pediatr Nephrol. 2004; 19:987-991
Crossref
Scopus (37)
PubMed
Google Scholar
].
Physiologic hydronephrosis of pregnancy occurs in >80% of pregnant patients in the second and third trimester; therefore, hydronephrosis alone is not a reliable sign of APN in pregnant patients [51
51.
Rasmussen, P.E. ∙ Nielsen, F.R.
Hydronephrosis during pregnancy: a literature survey
Eur J Obstet Gynecol Reprod Biol. 1988; 27:249-259
Abstract
Full Text (PDF)
Scopus (141)
PubMed
Google Scholar
]. Furthermore, US has a lower detection rate of APN and renal abscess than CT [10
10.
Kim, Y. ∙ Seo, M.R. ∙ Kim, S.J. ...
Usefulness of blood cultures and radiologic imaging studies in the management of patients with community-acquired acute pyelonephritis
Infect Chemother. 2017; 49:22-30
Crossref
Scopus (32)
PubMed
Google Scholar
,36-38
36.
Yoo, J.M. ∙ Koh, J.S. ∙ Han, C.H. ...
Diagnosing acute pyelonephritis with CT, Tc-DMSA SPECT, and Doppler ultrasound: a comparative study
Korean J Urol. 2010; 51:260-265
Crossref
Scopus (34)
PubMed
Google Scholar
37.
Fontanilla, T. ∙ Minaya, J. ∙ Cortes, C. ...
Acute complicated pyelonephritis: contrast-enhanced ultrasound
Abdom Imaging. 2012; 37:639-646
Crossref
Scopus (55)
PubMed
Google Scholar
38.
Kim, B. ∙ Lim, H.K. ∙ Choi, M.H. ...
Detection of parenchymal abnormalities in acute pyelonephritis by pulse inversion harmonic imaging with or without microbubble ultrasonographic contrast agent: correlation with computed tomography
J Ultrasound Med. 2001; 20:5-14
Crossref
Scopus (60)
PubMed
Google Scholar
].
Variant 5: Suspected acute pyelonephritis. History of pelvic renal transplant with native kidneys in situ and no other complications (eg, no history of pyelonephritis, diabetes, history of stones or renal obstruction, prior renal surgery, advanced age, vesicoureteral reflux, lack of response to therapy, or pregnancy). Initial imaging
Some local practice patterns do not routinely give IV contrast agents to patients with renal transplants. For this variant, we assumed there are no contraindications to IV contrast agents.
CT Abdomen and Pelvis
There is no current literature specific to the use of CT of the abdomen and pelvis in the evaluation of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. CT of the abdomen and pelvis would include imaging of the native and transplant kidneys, and CT with IV contrast is highly accurate for the diagnose APN in a complicated patient, particularly if symptoms persist or worsen after 72 hours have passed [1
1.
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ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
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PubMed
Google Scholar
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Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
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Diagnosis of acute pyelonephritis with recent trends in management
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CT Abdomen
There is no current literature to support the use of CT of the abdomen in the evaluation of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. APN of a renal allograft is more common than APN of the native kidneys, and renal transplant recipients are at high risk for complications from a variety of factors, including immunosuppression [52
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N Engl J Med. 2018; 378:48-59
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Google Scholar
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Enikeev, D.V. ∙ Glybochko, P. ∙ Alyaev, Y. ...
Imaging technologies in the diagnosis and treatment of acute pyelonephritis
Urologia. 2017; 84:179-184
Crossref
PubMed
Google Scholar
19.
Lee, A. ∙ Kim, H.C. ∙ Hwang, S.I. ...
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J Korean Med Sci. 2018; 33:e236
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Google Scholar
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Bova, J.G. ∙ Potter, J.L. ∙ Arevalos, E. ...
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J Urol. 1985; 133:375-378
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PubMed
Google Scholar
21.
Dalla-Palma, L. ∙ Pozzi-Mucelli, F. ∙ Pozzi-Mucelli, R.S.
Delayed CT findings in acute renal infection
Clin Radiol. 1995; 50:364-370
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Full Text (PDF)
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].
CTU
There is limited information on the benefit of CTU for detection of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. The added benefit of an excretory phase from any CTU image acquisition protocol is likely negligible with respect to detection and characterization of APN in this patient cohort.
DMSA Renal Scan
DMSA renal scintigraphy is not beneficial in the imaging evaluation for suspected APN in a patient with a pelvic renal transplant and native kidneys in situ [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
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Fluoroscopy Antegrade Pyelography
Antegrade pyelography is not beneficial in the imaging evaluation for suspected APN in a patient with a pelvic renal transplant and native kidneys in situ [15
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Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
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Fluoroscopy Voiding Cystourethrography
VCUG is not beneficial in the imaging evaluation for suspected APN in a patient with a pelvic renal transplant and native kidneys in situ [15
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Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
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MRI Abdomen and Pelvis
There is limited literature on the use of MRI of the abdomen and pelvis in the evaluation of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. APN is rare in native kidneys. However, APN of a renal allograft is more common, and renal transplant recipients are at high risk for complications from a variety of factors, including immunosuppression [52
52.
Fiorentino, M. ∙ Pesce, F. ∙ Schena, A. ...
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In a prospective study of renal transplant recipients with suspected APN (n= 56), contrast-enhanced MRI was positive in 66% (37/56) of patients [53
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Studies in adult patients without renal transplants have shown that DWI can be useful in the diagnosis of uncomplicated pyelonephritis [28-30
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The main disadvantages of MRI of the abdomen and pelvis are poor accuracy for the detection of small urothelial calculi and reduced accuracy for detection of emphysematous pyelonephritis, [31
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MRI Abdomen
There is limited literature on the use of MRI of the abdomen in the evaluation of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. APN is rare in native kidneys, and abdominal imaging would not include complete imaging of the pelvic transplant kidney(s), which would likely miss important pathology in the transplant kidney [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
N Engl J Med. 2018; 378:48-59
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PubMed
Google Scholar
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Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
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PubMed
Google Scholar
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14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
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Crossref
Scopus (133)
PubMed
Google Scholar
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
Radiographics. 2008; 28:255-277
quiz 327-8
Crossref
Scopus (299)
PubMed
Google Scholar
16.
Abraham, G. ∙ Reddy, Y.N. ∙ George, G.
Diagnosis of acute pyelonephritis with recent trends in management
Nephrol Dial Transplant. 2012; 27:3391-3394
Crossref
Scopus (14)
Google Scholar
17.
Pierce, C. ∙ Keniston, A. ∙ Albert, R.K.
Imaging in acute pyelonephritis: utilization, findings, and effect on management
South Med J. 2019; 112:118-124
Crossref
Scopus (2)
Google Scholar
18.
Enikeev, D.V. ∙ Glybochko, P. ∙ Alyaev, Y. ...
Imaging technologies in the diagnosis and treatment of acute pyelonephritis
Urologia. 2017; 84:179-184
Crossref
PubMed
Google Scholar
19.
Lee, A. ∙ Kim, H.C. ∙ Hwang, S.I. ...
Clinical usefulness of unenhanced computed tomography in patients with acute pyelonephritis
J Korean Med Sci. 2018; 33:e236
Crossref
Scopus (4)
Google Scholar
20.
Bova, J.G. ∙ Potter, J.L. ∙ Arevalos, E. ...
Renal and perirenal infection: the role of computerized tomography
J Urol. 1985; 133:375-378
Crossref
Scopus (28)
PubMed
Google Scholar
21.
Dalla-Palma, L. ∙ Pozzi-Mucelli, F. ∙ Pozzi-Mucelli, R.S.
Delayed CT findings in acute renal infection
Clin Radiol. 1995; 50:364-370
Abstract
Full Text (PDF)
Scopus (28)
PubMed
Google Scholar
].
MRI Pelvis
There is limited literature on the use of MRI of the pelvis in the evaluation of suspected APN in a patient with a pelvic renal transplant and native kidneys in situ. Pelvic MRI alone would not include evaluation of the native kidneys. MRI of the abdomen and pelvis would be more comprehensive for identification of the source of the APN.
MRU
There is no current literature specific to the use of MRU in the evaluation of suspected APN in patients with a pelvic renal transplant and native kidneys in situ. The excretory phase of MRU does not confer additional benefit with respect to detection and characterization of APN in a pregnant patient.
Radiography Abdomen and Pelvis (KUB)
Radiography of the abdomen and pelvis is not beneficial in the imaging evaluation for suspected APN in a patient with a pelvic renal transplant and native kidneys in situ [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
Pyelonephritis: radiologic-pathologic review
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Radiography Intravenous Urography
IVU is not beneficial in the imaging evaluation for suspected APN in a patient with a pelvic renal transplant and native kidneys in situ [15
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
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US Abdomen
APN is rare in native kidneys. Because a US of the abdomen would not include complete imaging of the pelvic transplant kidney(s), the examination is not likely to diagnose APN [1
1.
Johnson, J.R. ∙ Russo, T.A.
Acute pyelonephritis in adults
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PubMed
Google Scholar
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8.
Nikolaidis, P. ∙ Dogra, V.S. ∙ Goldfarb, S. ...
ACR Appropriateness Criteria® acute pyelonephritis
J Am Coll Radiol. 2018; 15:S232-S239
Full Text
Full Text (PDF)
Scopus (29)
PubMed
Google Scholar
,14-21
14.
Soulen, M.C. ∙ Fishman, E.K. ∙ Goldman, S.M. ...
Bacterial renal infection: role of CT
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Crossref
Scopus (133)
PubMed
Google Scholar
15.
Craig, W.D. ∙ Wagner, B.J. ∙ Travis, M.D.
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Radiographics. 2008; 28:255-277
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Crossref
Scopus (299)
PubMed
Google Scholar
16.
Abraham, G. ∙ Reddy, Y.N. ∙ George, G.
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Nephrol Dial Transplant. 2012; 27:3391-3394
Crossref
Scopus (14)
Google Scholar
17.
Pierce, C. ∙ Keniston, A. ∙ Albert, R.K.
Imaging in acute pyelonephritis: utilization, findings, and effect on management
South Med J. 2019; 112:118-124
Crossref
Scopus (2)
Google Scholar
18.
Enikeev, D.V. ∙ Glybochko, P. ∙ Alyaev, Y. ...
Imaging technologies in the diagnosis and treatment of acute pyelonephritis
Urologia. 2017; 84:179-184
Crossref
PubMed
Google Scholar
19.
Lee, A. ∙ Kim, H.C. ∙ Hwang, S.I. ...
Clinical usefulness of unenhanced computed tomography in patients with acute pyelonephritis
J Korean Med Sci. 2018; 33:e236
Crossref
Scopus (4)
Google Scholar
20.
Bova, J.G. ∙ Potter, J.L. ∙ Arevalos, E. ...
Renal and perirenal infection: the role of computerized tomography
J Urol. 1985; 133:375-378
Crossref
Scopus (28)
PubMed
Google Scholar
21.
Dalla-Palma, L. ∙ Pozzi-Mucelli, F. ∙ Pozzi-Mucelli, R.S.
Delayed CT findings in acute renal infection
Clin Radiol. 1995; 50:364-370
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Full Text (PDF)
Scopus (28)
PubMed
Google Scholar
].
US Color Doppler Kidneys and Bladder Retroperitoneal
APN is most commonly an issue with the transplant kidney. Because US Doppler of the kidneys, bladder, and retroperitoneum does not include imaging of the pelvic transplant kidney(s), the examination is not likely to diagnose APN.
US Duplex Doppler Kidney Transplant
Renal transplant US includes grayscale and color doppler imaging and is commonly used to assess transplant dysfunction. According to the American Institute of Ultrasound in Medicine, renal transplant US is the recommended imaging examination to evaluate for suspected APN of a renal transplant. Renal transplant US has high resolution and high sensitivity for detection of renal transplant hydronephrosis, calculi, vascular abnormalities, and a renal or perirenal abscess.
In a prospective study of renal transplant recipients with suspected APN (n= 56) using contrast-enhanced MRI served as the reference standard that was positive in 66% (37/56), the contrast-enhanced US was positive in 63% (35/56) of patients, with a reported sensitivity of 95%, a specificity of 100%, and a diagnostic accuracy of 96% [53
53.
Granata, A. ∙ Andrulli, S. ∙ Fiorini, F. ...
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Summary of Recommendations
•
Variant 1: Imaging is usually not appropriate for the first-time presentation of suspected APN in an uncomplicated patient.
•
Variant 2: CT abdomen and pelvis with IV contrast is usually appropriate for the initial imaging of complicated patients with suspected APN. Although the panel did not agree on recommending CT abdomen with IV contrast or CT abdomen and pelvis without and with IV contrast because there is insufficient medical literature to conclude whether these patients would benefit from the procedures, their use may be appropriate.
•
Variant 3: CT abdomen and pelvis with IV contrast or CT abdomen and pelvis without and with IV contrast is usually appropriate for the initial imaging of patients with a history of renal stones or renal obstruction with suspected APN. These procedures are equivalent alternatives (ie, only one procedure will be ordered to provide the clinical information to effectively manage the patient’s care). Although the panel did not agree on recommending CT abdomen without and with IV contrast because there is insufficient medical literature to conclude whether these patients would benefit from the procedure, its use may be appropriate.
•
Variant 4: US color Doppler kidneys and bladder retroperitoneal or MRI abdomen and pelvis without IV contrast or MRI abdomen without IV contrast or MRU without IV contrast may be appropriate for the initial imaging of pregnant patients with no other complications. These procedures are equivalent alternatives (ie, only one procedure will be ordered to provide the clinical information to effectively manage the patient’s care). Although the panel did not agree on recommending US abdomen because there is insufficient medical literature to conclude whether these patients would benefit from the procedure, its use may be appropriate.
•
Variant 5: US duplex Doppler kidney transplant or CT abdomen and pelvis with IV contrast is usually appropriate for the initial imaging of patients with a history of pelvic renal transplant with native kidneys in situ and no other complications with suspected APN. These procedures are equivalent alternatives (ie, only one procedure will be ordered to provide the clinical information to effectively manage the patient’s care).
Supporting Documents
The evidence table, literature search, and appendix for this topic are available at The appendix includes the strength of evidence assessment and the final rating round tabulations for each recommendation.
For additional information on the Appropriateness Criteria methodology and other supporting documents go to www.acr.org/ac.
Safety Considerations in Pregnant Patients
Imaging of the pregnant patient can be challenging, particularly with respect to minimizing radiation exposure and risk. For further information and guidance, see the following ACR documents:
•
ACR–SPR Practice Parameter for the Safe and Optimal Performance of Fetal Magnetic Resonance Imaging (MRI) [55
55.
American College of Radiology
ACR–SPR practice parameter for the safe and optimal performance of fetal magnetic resonance imaging (MRI)
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•
ACR-SPR Practice Parameter for Imaging Pregnant or Potentially Pregnant Adolescents and Women with Ionizing Radiation [46
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ACR-SPR practice parameter for imaging pregnant or potentially pregnant adolescents and women with ionizing radiation
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•
ACR-ACOG-AIUM-SMFM-SRU Practice Parameter for the Performance of Standard Diagnostic Obstetrical Ultrasound [56
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American College of Radiology
ACR-ACOG-AIUM-SMFM-SRU practice parameter for the performance of standard diagnostic obstetrical ultrasound
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•
ACR Manual on Contrast Media [47
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•
ACR Manual on MR Safety [57
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Relative Radiation Level Information
Potential adverse health effects associated with radiation exposure are an important factor to consider when selecting the appropriate imaging procedure. Because there is a wide range of radiation exposures associated with different diagnostic procedures, a relative radiation level (RRL) indication has been included for each imaging examination. The RRLs are based on effective dose, which is a radiation dose quantity that is used to estimate population total radiation risk associated with an imaging procedure. Patients in the pediatric age group are at inherently higher risk from exposure, because of both organ sensitivity and longer life expectancy (relevant to the long latency that appears to accompany radiation exposure). For these reasons, the RRL dose estimate ranges for pediatric examinations are lower as compared with those specified for adults (see Table 2). Additional information regarding radiation dose assessment for imaging examinations can be found in the ACR Appropriateness Criteria® Radiation Dose Assessment Introduction document [58
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1589 | https://math.stackexchange.com/questions/1599990/prove-that-that-fracx2a-fracy2b-fracz2c-geq-fracxyz2 | Stack Exchange Network
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Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$
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$\begingroup$
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.
Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you solve this using Cauchy-Schwarz?
inequality
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edited Jan 4, 2016 at 19:22
Jacob WillisJacob Willis
asked Jan 4, 2016 at 19:16
Jacob WillisJacob Willis
1,6511010 silver badges2121 bronze badges
$\endgroup$
4
1
$\begingroup$ with $a,b,c$ positive reals and $x,y,z$ real numbers $\endgroup$
Dr. Sonnhard Graubner
– Dr. Sonnhard Graubner
2016-01-04 19:19:53 +00:00
Commented Jan 4, 2016 at 19:19
2
$\begingroup$ $$(a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\ge \left(\sqrt{a}\sqrt{\frac{x^2}{a}}+ \sqrt{b}\sqrt{\frac{y^2}{b}}+\sqrt{c}\sqrt{\frac{z^2}{c}}\right)^2=(x+y+z)^2$$ $\endgroup$
user236182
– user236182
2016-01-04 19:24:01 +00:00
Commented Jan 4, 2016 at 19:24
$\begingroup$ See the proof in math.stackexchange.com/questions/1594286/… $\endgroup$
Gordon
– Gordon
2016-01-04 19:28:28 +00:00
Commented Jan 4, 2016 at 19:28
$\begingroup$ Titu's lemma (brilliant.org/wiki/titus-lemma) and the Cauchy-Schwarz inequality are essentially equivalent. It is interesting to point out that Titu's lemma for two variables (almost trivial) plus induction give a neat proof of the general Cauchy-Schwarz inequality. $\endgroup$
Jack D'Aurizio
– Jack D'Aurizio
2016-01-04 19:49:20 +00:00
Commented Jan 4, 2016 at 19:49
Add a comment |
2 Answers 2
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6
$\begingroup$
More generally, the following is called Titu's Lemma or Engel's form of Cauchy-Schwarz inequality:
For all $a_i\in\mathbb R$, $b_i\in\mathbb R^+$:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$
Proof: by Cauchy-Schwarz: $$(b_1+b_2+\cdots+b_n)\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\right)$$
$$\ge \left(\sqrt{b_1}\sqrt{\frac{a_1^2}{b_1}}+\sqrt{b_2}\sqrt{\frac{a_2^2}{b_2}}+\cdots+\sqrt{b_1}\sqrt{\frac{a_n^2}{b_n}}\right)^2$$
$$=(|a_1|+|a_2|+\cdots+|a_n|)^2\ge (a_1+a_2+\cdots+a_n)^2$$
with equality if and only if $\frac{a_1^2}{b_1^2}=\frac{a_2^2}{b_2^2}=\cdots=\frac{a_n^2}{b_n^2}$ and $(|a_1|+|a_2|+\cdots+|a_n|)^2= (a_1+a_2+\cdots+a_n)^2$, i.e. if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.
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edited Jan 4, 2016 at 19:55
answered Jan 4, 2016 at 19:36
user236182user236182
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$\endgroup$
6
$\begingroup$ Another proof is by induction in $n$. Equality is satisfied iff $\frac{a_i}{b_i}=\frac{a_j}{b_j}$ for any $i,j$ $\endgroup$
sinbadh
– sinbadh
2016-01-04 19:38:49 +00:00
Commented Jan 4, 2016 at 19:38
$\begingroup$ @sinbadh Should be $\left|\frac{a_i}{b_i}\right|=\left|\frac{a_j}{b_j}\right|$. $\endgroup$
user236182
– user236182
2016-01-04 19:45:23 +00:00
Commented Jan 4, 2016 at 19:45
$\begingroup$ That's correct. But in the last equality, it should be $(|a_1|+...+|a_n|)^2$ $\endgroup$
sinbadh
– sinbadh
2016-01-04 19:51:17 +00:00
Commented Jan 4, 2016 at 19:51
$\begingroup$ @sinbadh right, edited. $\endgroup$
user236182
– user236182
2016-01-04 19:52:51 +00:00
Commented Jan 4, 2016 at 19:52
$\begingroup$ But... if you don't want abs function, the statement $(|a_1|+\cdots+|a_n|)^2=(a_1+\cdots a_n)^2$ is equivalent to all $a_i$ have the same sign, and therefore, equality is satisfied iff $\dfrac{a_i}{b_i}=\dfrac{a_j}{b_j}$ $\endgroup$
sinbadh
– sinbadh
2016-01-04 19:55:48 +00:00
Commented Jan 4, 2016 at 19:55
| Show 1 more comment
0
$\begingroup$
HINT: the left-hand side minus the right-hand side is equal to $${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2} \geq 0$$ can you proceed? (sum of squares!)
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answered Jan 4, 2016 at 19:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
97.5k44 gold badges4242 silver badges8080 bronze badges
$\endgroup$
3
$\begingroup$ Sonnhard Grauber I thought you said this could be done by Cauchy-Schwarz. $\endgroup$
Jacob Willis
– Jacob Willis
2016-01-04 19:25:46 +00:00
Commented Jan 4, 2016 at 19:25
$\begingroup$ @JacobWillis $$(a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\ge \left(\sqrt{a}\sqrt{\frac{x^2}{a}}+ \sqrt{b}\sqrt{\frac{y^2}{b}}+\sqrt{c}\sqrt{\frac{z^2}{c}}\right)^2=(x+y+z)^2$$ $\endgroup$
user236182
– user236182
2016-01-04 19:26:08 +00:00
Commented Jan 4, 2016 at 19:26
$\begingroup$ yes you see above a proof by Cauchy Schwarz $\endgroup$
Dr. Sonnhard Graubner
– Dr. Sonnhard Graubner
2016-01-04 19:27:33 +00:00
Commented Jan 4, 2016 at 19:27
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
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8 prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$
8 Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
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Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$
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1590 | https://www.youtube.com/watch?v=-Oaa11AAMV8 | Truth-Tellers And Liars(Knights And Knaves)
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Truth-Tellers And Liars(Knights And Knaves) problems are a common appearance in math competitions. This video aims to be a comprehensive guide as to how to tackle these types of problems!
This is the 4th video of a 9 part series that explores a collection of logical potpourri!
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Intro This episode is brought to you by ThePuzzlr.com In this video, we look at problems involving Knights and Knaves! Knights and Knave problems also called true teller and liar problems are logic puzzles in which a set of statements is provided But some of the statements are true and some of the statements are false The main purpose of these problems is to determine which statements are true or false based on the information given Let us (lettuce :)) first look at a problem before we learn more There are two men. One of them is wearing a red shirt and the other is wearing a blue shirt. The two men are named Andrew and Bob, but we do not know which is Andrew and which is Bob. Identitity Crisis The guy in the blue shirt says "I am Andrew" and the guy in the red shirt says "I am Bob." (sounds pretty straightforward if you ask me) If we know that at least one of them lied, then what color shirt is Andrew wearing? It's impossible for only one of them to be lying because that would entail that they both have the same name. Which we know isn't the case. Since we know that at least one of them lied. We deduce that both must be lying. Which means we have to swap the names we were given. Bob wears a blue shirt and Andrew wears a red shirt This means that our answer is that Andrew wears a red shirt. Let's look at some general methods to tackle Knights and Knaves. 1. Using Systematic Casework We have three main ways: 1. Using systematic casework, 2. Finding shortcuts and 3. Using self referential statements Systematic casework is considering each possible "case" and deciding if the case is possible according to the given information. Let us try casework on the first problem Here are our two possible cases: Either Andrews is in the blue shirt and Bob is in the red shirt. Both of the statements would be true, which is not possible. Case 2; Bob is in the blue shirt and Andrew is in the red shirt. Both of the statements would be false, which is possible. Thus, Andrew is wearing red. Method number 2: conditioning on specific constraints There are three main techniques to consider when you're trying to find shortcuts. We can prove the correctness of our answer by contradiction... ...we can condition on specific constraints... ...or we can identify contradictory statements Technique number 1: proving the correctness of our answer by contradiction. Technique #1 A solution describing tedious casework is often neither engaging nor concise. Even worse, it is sometimes unnecessary. It is much easier to find why the incorrect answer is a contradiction. You can also prove why making your answer incorrect will be a contradiction Method number 2: conditioning on specific constraints Here is an example: After Mrs. Jacobs' five children received their test scores, she asked him what their test scores were. Her first child said that at least one of them failed algebra Her second child said that at least two of them failed algebra, her third child said at least three of them failed algebra Her fourth child said that least four of them failed algebra. If only one child is telling the truth, Can you determine how many of Mrs. Jacobs' children failed Algebra? Let us now apply technique number 2. We can now easily constrain our condition to the number of people who failed Algebra. Suppose exactly four, three, or two children fail Algebra. APPLYING TECHNIQUE #2 In each of these cases, our assumption would be incorrect. Suppose exactly one child failed Algebra. Then only her first child is telling the truth which satisfies our assumption. Suppose none of the children failed algebra. Then nobody is telling the truth which again contradicts our statement. Technique number 3: identifying contradictory statements Here is an example: Suppose there are three people: A, B, and C, each of whom make these claims Person A says B is telling the truth, Person B says A is not guilty, and person C says A is guilty. Given that only one of them is telling the truth. Can you determine who's guilty? Notice that Person B and Person C gave contradicting viewpoints So only one of them can be telling the truth and the other must be lying. Because at least one of B and C's already telling the truth a can't be telling the truth anymore So A's statement is false, which means that B is not telling the truth. Thus C is telling the truth, or in other words, A is guilty. Self-referential statements are our third tool! Sometimes the truth teller or liar problem can involve one or more self referential statements In other words, The statement makes a claim about its own truth or falseness. Here is a simple example to clarify this idea. If the statement "this statement is both true and false" is logical, is the statement true or false? As a side note, a logical statement cannot be both true and false So the statement must be false, which is consistent with the fact that it is falsely claimed to be true as well as false I hope that you learned something that will help you better tackle Knights and Knave type problems Want to test your knowledge of what you learned in this video? Here are two problems to try on your own: (this is the part where you pause the video and solve these extra problems) Thank you for watching this video! Don't forget to subscribe and hit the like button if you enjoyed or learned anything from this video Be sure to check out ThePuzzler.com for more awesome and free math content Be sure to check out our website to find out more about upcoming contests and more! |
1591 | https://acutecaretesting.org/en/articles/roc-curves-what-are-they-and-how-are-they-used | Home >
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ROC curves – what are they and how are they used?
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January 2011
ROC curves – what are they and how are they used?
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Suzanne Ekelund
Information management
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ROC curves are frequently used to show in a graphical way the connection/trade-off between clinical sensitivity and specificity for every possible cut-off for a test or a combination of tests. In addition the area under the ROC curve gives an idea about the benefit of using the test(s) in question.
ROC curves are used in clinical biochemistry to choose the most appropriate cut-off for a test. The best cut-off has the highest true positive rate together with the lowest false positive rate.
As the area under an ROC curve is a measure of the usefulness of a test in general, where a greater area means a more useful test, the areas under ROC curves are used to compare the usefulness of tests.
The term ROC stands for Receiver Operating Characteristic.
ROC curves were first employed in the study of discriminator systems for the detection of radio signals in the presence of noise in the 1940s, following the attack on Pearl Harbor.
The initial research was motivated by the desire to determine how the US RADAR "receiver operators" had missed the Japanese aircraft.
Now ROC curves are frequently used to show the connection between clinical sensitivity and specificity for every possible cut-off for a test or a combination of tests. In addition, the area under the ROC curve gives an idea about the benefit of using the test(s) in question.
HOW TO MAKE A ROC CURVE
To make an ROC curve you have to be familiar with the concepts of true positive, true negative, false positive and false negative. These concepts are used when you compare the results of a test with the clinical truth, which is established by the use of diagnostic procedures not involving the test in question.
TABLE I
: Comparing a method with the clinical truth
Before you make a table like TABLE I you have to decide your cut-off for distinguishing healthy from sick.
The cut-off determines the clinical sensitivity (fraction of true positives to all with disease) and specificity (fraction of true negatives to all without disease).
When you change the cut-off, you will get other values for true positives and negatives and false positives and negatives, but the number of all with disease is the same and so is the number of all without disease.
Thus you will get an increase in sensitivity or specificity at the expense of lowering the other parameter when you change the cut-off .
FIG. I
: Cut-off = 400 µg/L
FIG. II
: Cut-off = 500 µg/L
FIG. I and FIG. II demonstrate the trade-off between sensitivity and specificity. When 400 µg/L is chosen as the analyte concentration cut-off, the sensitivity is 100 % and the specificity is 54 %. When the cut-off is increased to 500 µg/L, the sensitivity decreases to 92 % and the specificity increases to 79 %.
An ROC curve shows the relationship between clinical sensitivity and specificity for every possible cut-off. The ROC curve is a graph with:
The x-axis showing 1 – specificity (= false positive fraction = FP/(FP+TN))
The y-axis showing sensitivity (= true positive fraction = TP/(TP+FN))
Thus every point on the ROC curve represents a chosen cut-off even though you cannot see this cut-off. What you can see is the true positive fraction and the false positive fraction that you will get when you choose this cut-off.
To make an ROC curve from your data you start by ranking all the values and linking each value to the diagnosis – sick or healthy.
TABLE II
: Ranked data with diagnosis (Yes/No)
In the example in TABLE II 159 healthy people and 81 sick people are tested. The results and the diagnosis (sick Y or N) are listed and ranked based on parameter concentration.
For each and every concentration it is calculated what the clinical sensitivity (true positive rate) and the (1 – specificity) (false positive rate) of the assay will be if a result identical to this value or above is considered positive.
TABLE III: Ranked data with calculated true positive and false positive rates for a scenario where the specific value is used as cut-off
Now the curve is constructed by plotting the data pairs for sensitivity and (1 – specificity):
FIG. III: First point on the ROC curve
FIG. IV: Second point on the ROC curve
FIG. V: Third point on the ROC curve
FIG. VI: Points #50 and #100 on the ROC curve
FIG. VII: The finalized ROC curve
AREA UNDER ROC CURVE
The area under the ROC curve (AUROC) of a test can be used as a criterion to measure the test's discriminative ability, i.e. how good is the test in a given clinical situation.
FIG. VIII: Area under ROC curve
Various computer programs can automatically calculate the area under the ROC curve. Several methods can be used. An easy way to calculate the AUROC is to use the trapezoid method. To explain it simply, the sum of all the areas between the x-axis and a line connecting two adjacent data points is calculated:
| | |
--- |
| | (Xk – Xk-1) (Yk + Yk-1)/2 |
THE PERFECT TEST
A perfect test is able to discriminate between the healthy and sick with 100 % sensitivity and 100 % specificity.
FIG. IX: No overlap between healthy and sick
It will have an ROC curve that passes through the upper left corner (~100 % sensitivity and 100 % specificity). The area under the ROC curve of the perfect test is 1.
FIG. X: ROC curve for a test with no overlap between healthy and sick
THE WORTHLESS TEST
When we have a complete overlap between the results from the healthy and the results from the sick population, we have a worthless test. A worthless test has a discriminating ability equal to flipping a coin.
FIG. XI: Complete overlap between healthy and sick
The ROC curve of the worthless test falls on the diagonal line. It includes the point with 50 % sensitivity and 50 % specificity. The area under the ROC curve of the worthless test is 0.5.
FIG. XII: ROC curve for a test with complete overlap between healthy and sick
COMPARING ROC CURVES
As mentioned above, the area under the ROC curve of a test can be used as a criterion to measure the test's discriminative ability, i.e. how good is the test in a given clinical situation. Generally, tests are categorized based on the area under the ROC curve.
The closer an ROC curve is to the upper left corner, the more efficient is the test.
In FIG. XIII test A is superior to test B because at all cut-offs the true positive rate is higher and the false positive rate is lower than for test B. The area under the curve for test A is larger than the area under the curve for test B.
FIG. XIII
: ROC curves for tests A and B
TABLE IV
: Categorization of ROC curves
As a rule of thumb the categorizations in TABLE IV can be used to describe an ROC curve.
References
CLSI/NCCLS document EP12-A2 User Protocol for Evaluation of Qualitative Test Performance; Approved Guideline 2nd Edition. Vol. 28 No. 3. 2008
References
CLSI/NCCLS document EP12-A2 User Protocol for Evaluation of Qualitative Test Performance; Approved Guideline 2nd Edition. Vol. 28 No. 3. 2008
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Suzanne Ekelund
Principal Specialist, Clinical Biochemist, MSc
Radiometer Medical ApS
Denmark
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1592 | https://programs.mcs.cmu.edu/arml/wp-content/uploads/sites/3/2023/09/algebra-09-24-17.pdf | Complex Numbers Western PA ARML Fall 2017 Page 1 Complex Numbers Andrew Kwon Problems If you are unfamiliar with any of the words or symbols below, please refer to the next page for a collection of the relevant information to help you!
1. Compute |1 + 2i|2 and (1 + 2i)2. Do the same for |2 + 3i|2, (2 + 3i)2. Do you notice anything special about the numbers you find?
2. (F06 NYCIML B19) Compute (1 −i)10.
3. (F11 NYCIML B2) Let z be a complex number that satisfies z + 6i = iz. Find z.
4. (04 AMC 12B #16) A function f is defined by f(z) = iz, where i = √−1 and z is the complex conjugate of z. How many values of z satisfy both |z| = 5 and f(z) = z?
5. (17 AMC 12A #17) There are 24 different complex numbers z such that z24 = 1. For how many of these is z6 a real number?
6. Using de Moivre’s theorem, find formulas for cos 2θ, cos 3θ in terms of cos θ.
7. (09 AMC 12A #15) For what value of n is i + 2i2 + 3i3 + · · · + nin = 48 + 49i?
8. (F09 NYCIML B23) Find the complex number c such that the equation x2+4x+6ix+c = 0 has only one solution.
9. (09 AIME I #2) There is a complex number z with imaginary part 164 and a positive integer n such that z z+n = 4i. Find n.
10. (85 AIME #3) Find c if a, b, c are positive integers which satisfy c = (a + bi)3 −107i.
(Hint: 107 is prime.) 11. (94 AIME #8) The points (0, 0), (a, 11), and (b, 37) are the vertices of an equilateral triangle. Find the value of ab. (Hint: is there a way to think of points in the plane as complex numbers?) 12. (97 AIME #14) Let v and w be distinct, randomly chosen roots of the equation z1997 − 1 = 0. Let m n be the probability that p 2 + √ 3 ≤|v + w|, where m and n are relatively prime positive integers. Find m + n.
Complex Numbers Western PA ARML Fall 2017 Page 2 Challenge Problems 1. If p is a prime number and a0, a1, . . . , ap−1 are rational numbers satisfying a0 + a1ζ + a2ζ2 + · · · + ap−1ζp−1 = 0, 2. (95 IMO) Let p > 2 be a prime number and let A = {1, 2, . . . , 2p}. Find the number of subsets of A each having p elements and whose sum is divisible by p. (Hint: Count the more general Nk, the number of subsets of A with p elements whose sum is congruent to kmod p and consider pth roots of unity.) Background A complex number is a number of the form z = a + bi, where a, b are real numbers and i is the number satisfying i2 = −1. a is referred to as the real part of z, denoted by Re z, and b is referred to as the imaginary part, denoted by Im z. Addition and multiplication of imaginary numbers works the same way as multiplication of binomials; that is, if z = a + bi, w = c + di, where a, b, c, d ∈R then z + w = (a + bi) + (c + di) = (a + c) + (b + d)i, w zw = (a + bi)(c + di) = ac + bdi2 + i(ad + bc) = (ac −bd) + i(ad + bc).
For a complex number z = a + bi, we define • the conjugate of a z, denoted z, is a −bi; • the norm or modulus of z, denoted |z|, is √ zz = √ a2 + b2.
de Moivre’s formula One useful fact is that any complex number z can be written in the form z = |z| cis θ, where 0 ≤θ < 2π, and cis θ = cos θ + i sin θ. Often, θ is referred to as the argument of z.
Then, de Moivre’s formula states that for any n ∈N, zn = |z|n cis(nθ). This provides an important connection between complex numbers and trigonometry. |
1593 | https://indico.cern.ch/event/923742/contributions/3881289/attachments/2058512/3452616/12_Activation_2020_online.pdf | Online training, June 8 -18, 2020 Induced radioactivity calculations The generation and transport of decay radiation (limited to γ, b-, b+, X-rays, and Conversion Electrons emissions for the time being) is possible during the same simulation which produces the radionuclides (one-step method). For that, a dedicated database of decay emissions is used, based mostly on information obtained from NNDC, sometimes supplemented with other data and checked for consistency.
As a consequence, results for production of residuals, their time evolution and residual doses due to their decays can be obtained in the same run, for arbitrary decay times and for a given irradiation profile.
Induced radioactivity 2 FLUKA-Implementation – Main features FLUKA-Implementation – Main features • up to 4 different decay branching for each isotope/isomer • all gamma lines down to 0.1-0.01% branching, including X-ray lines following conversion electron emissions • all beta emission spectra down to 0.1-0.01% branching: the sampling of the beta+/- spectra including screening Coulomb corrections • Auger and conversion electrons • Isomers: the present models do not distinguish among ground state and isomeric states (it would require spin/parity dependent calculations in evaporation). A rough estimate (equal sharing among states) of isomer production can be activated in the RADDECAY option. In future release branchings for isomers produced by neutrons <20 MeV will be based on JEFF • Different transport thresholds can be set for the prompt and decay radiation parts, as well as some (limited) biasing differentiation (see later) Induced radioactivity 3 Induced radioactivity 4 Input options - Overview Input card: RADDECAY requests simulation of decay of produced radioactive nuclides and allows to modify biasing and transport thresholds (defined with other cards) for the transport of decay radiation Input card: IRRPROFI definition of an irradiation profile (irradiation times and intensities) Input card: DCYTIMES definition of decay (cooling ) times measured from end of irradiation cycle (t=0) Input card: DCYSCORE associates scoring detectors (radio-nuclides, fluence, dose) with different cooling times Input card: AUXSCORE allows to associate scoring estimators with dose equivalent conversion factors or/and to filter them according to (generalized) particle identity … … 1h 8h 1d 7d -200d Induced radioactivity 5 Particle Types Name Units Description DOSE GeV/g Dose (energy deposited per unit mass) DOSE-EQ pSv Dose Equivalent (AUXSCORE) ACTIVITY Bq/cm3 Activity per unit volume ACTOMASS Bq/g Activity per unit mass SI1MEVNE cm-2 Silicon 1 MeV-neutron equivalent flux HADGT20M cm-2 Hadrons with energy > 20 MeV Induced radioactivity 6 Input option: RADDECAY[1/2] Requests the calculation of radioactive decays Decays Active radioactive decays activated for requested cooling times “activation study case”: time evolution calculated analytically for fixed (cooling) times. Daughter nuclei as well as associated radiation is considered at these (fixed) times Semi-Analogue radioactive decays activated in semi-analogue mode each radioactive nucleus is treated like all other unstable particles (random decay time, daughters and radiation), all secondary particles/nuclei carry time stamp (“age”) Patch Isom On isomer “production” activated Replicas # number of “replicas” of the decay of each individual nucleus Induced radioactivity 7 Input option: RADDECAY[2/2] Requests the calculation of radioactive decays h/µ Int .. Low-n WW switch for applying various biasing features only to prompt radiation or only to particles from radioactive decays decay cut, prompt cut 0.1 x input value is used as multiplication factors to be applied to e+/e-/gamma transport energy cutoffs (defined with EMF-CUT cards) Examples: input value for decay cut = 10 decay radiation production and transport thresholds are not modified (0.1 x 10) input value for decay cut = 200 prompt radiation threshold increased by factor of 20 (0.1 x 200) Special cases: decay cut = 99999 kill EM cascade for residual radiation prompt cut = 99999 kill EM cascade for prompt radiation Induced radioactivity 8 Input option: IRRPROFI Definition of irradiation pattern Δt # irradiation time (second) p/s # beam intensity (particles per second) Note: zero intensity is accepted and can be used, e.g., to define beam-off periods Notes: Each card has 6 inputs with 3 durations / intensities (intercalated). Several cards can be combined. Sequence order is assumed from first card (top) to last (bottom) Example (see above): 180 days 185 days 180 days 5.9 × 105 p/s 0 p/s 5.9 × 105 p/s (beam-off) Induced radioactivity 9 Input option: DCYTIMES Definition of cooling times t1 .. t6 cooling time (in seconds) after the end of the irradiation Note: Several cards can be defined.
Each cooling time is assigned an index, following the order in which it has been input. This index can be used in option DCYSCORE to assign that particular cooling time to one or more scoring detectors. A negative decay time is admitted: scoring is performed at the chosen time "during irradiation" Example: 180 days 185 days 180 days 5.9 × 105 p/s 0 p/s 5.9 × 105 p/s (beam-off) 1h 8h 1d 7d etc.
-200d Induced radioactivity 10 Input option: DCYSCORE[1/2] Association of scoring with different cooling times Cooling t # Cooling time index to be associated with the detectors Drop down list of available cooling times Det .. to Det Detector index/name of kind (SDUM/Kind) Drop down list of available detectors of kind (Kind) Step # step lengths in assigning indices Kind Type of estimator RESNUCLE, USRBIN/EVENTBIN, USRBDX, USRTRACK… Units: All quantities are expressed per unit time. For example RESNUCLE Bq USRBIN fluence rate / dose rate Induced radioactivity 11 Input option: DCYSCORE[2/2] Notes: All quantities are expressed per unit time.
For example: RESNUCLE Bq USRBIN fluence rate / dose rate In the semi-analogue decay mode, estimators can include the decay contribution (on top of the prompt one) through association by DCYSCORE with a cooling time index -1.0 Induced radioactivity 12 Input option: AUXSCORE Association of scoring with scoring with dose equivalent conversion factors Type Type of estimator to associate with drop down list of estimator types (USRBIN, USRBDX…) Part # Particle or isotope to filter for scoring Particle or particle family list. If empty then flair will prompt for Z, A, and State for filtering on specific isotopes Det .. to Det Detector range Drop down list to select detector range of type Type Step # Step in assigning indices of detector range Set Conversion factor set for dose equivalent (DOSE-EQ) scoring Drop down list of available dose conversion sets Note: This card is NOT just for activation-type scorings. It can be used for prompt radiation.
Fluence-to-dose conversion coefficients • AMB74 is the default choice for dose equivalent calculation (scoring DOSE-EQ without AUXSCORE card) • Conversion coefficients from fluence to effective dose are implemented for three different irradiation geometries: anterior-posterior rotational WORST (“Working Out Radiation Shielding Thicknesses”) is the maximum coefficient of anterior-posterior, posterior-anterior, right-lateral and left-lateral geometries. It is recommended to be used for shielding design. • Implemented for radiation weighting factors recommended by ICRP60 (e.g., SDUM=EWT74) and recommended by M. Pelliccioni (e.g., SDUM=EWTMP). • Implemented for protons, neutrons, charged pions, muons, photons, electrons (conversion coefficients for other particles are approximated by these) • Zero coefficient is applied to all heavy ions Induced radioactivity 13 Induced radioactivity 14 Fluence-to-dose conversion coefficients Examples: Ambient dose equivalent for neutrons Effective dose for WORST irradiation geometry For more information please see: Induced radioactivity 15 Input option: RESNUCLE[1/3] Scoring of residual nuclei or activity on a region basis Type Type of products to be scored 1.0 spallation products (all inelastic interactions except for low-energy neutron interactions, i.e. with multigroup treatment) 2.0 products from low-energy neutron interactions (provided the information is available) 3.0 all residual nuclei are scored (if available, see above) <= 0.0 resets the default (= 1.0) Unit Logical output unit (Default = 11.0) Max Z Maximum atomic number Z of the residual nuclei distribution Default: according to the Z of the element(s) of the material assigned to the scoring region Max M Maximum M = N - Z - NMZmin of the residual nuclei distribution (NMZmin = -5) Default: maximum value according to the A, Z of the element(s) of the material assigned to the scoring region.
Induced radioactivity 16 Input option: RESNUCLE[2/3] Scoring of residual nuclei or activity on a region basis Reg Scoring region name Default = 1.0; if set to -1.0 or @ALLREGS scoring will include all regions) Vol Volume of the region in cm3 Default = 1.0 cm3 Name Character string identifying the detector (max. 10 characters) Notes: 1.
In the case of heavy ion projectiles the default Max M, based on the region material, is not necessarily sufficient to score all the residual nuclei, which could include possible ion fragments 2.
Residual nuclei from low-energy neutron interactions are only scored if that information is available in the low-energy neutron data set (see Manual) 3.
Also protons are scored (at the end of their path) Induced radioactivity 17 Input option: RESNUCLE[3/3] Output example Isotope Yield as a function of Mass Number (nuclei / cmc / pr) A_min: 1 -A_max: 198 A: 186 1.5870372E-08 +/-9.9000000E+01 % A: 185 3.7605012E-09 +/-9.9000000E+01 % A: 184 1.4581326E-08 +/-9.9000000E+01 % A: 183 1.0712972E-08 +/-9.9000000E+01 % A: 182 7.4882118E-09 +/-9.9000000E+01 % ...
Isotope Yield as a function of Atomic Number (nuclei / cmc / pr) Z_min: 1 -Z_max: 78 Z: 74 5.2413383E-08 +/-9.9000000E+01 % Z: 42 3.0072785E-07 +/-9.9000000E+01 % Z: 41 4.7906228E-08 +/-9.9000000E+01 % Z: 40 3.7605012E-09 +/-9.9000000E+01 % Z: 38 3.7605012E-09 +/-9.9000000E+01 % ...
Residual nuclei distribution (nuclei / cmc / pr) A \ Z 68 69 70 71 72 73 74 75 76 77 78 186 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 1.59E-08 0.00E+00 0.00E+00 0.00E+00 0.00E+00 +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % 185 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 3.76E-09 0.00E+00 0.00E+00 0.00E+00 0.00E+00 +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % 184 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 1.46E-08 0.00E+00 0.00E+00 0.00E+00 0.00E+00 +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % 183 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 1.07E-08 0.00E+00 0.00E+00 0.00E+00 0.00E+00 +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % ...
Induced radioactivity 18 Input option: PHYSICS Please activate the following cards if scoring of residual nuclei is of interest: Use of PEANUT model at all energies (now part of default settings) Evaporation of heavy fragments Activation of coalescence treatment Induced radioactivity 19 Input option: PHYSICS The evaporation of heavy fragments produces deuterons, which need to be transported!
Please activate the RQMD and DPMJET packages.
All ions (including deuterons) are treated with RQMD (>150 MeV/n) and DPMJET (> 5 GeV/n); All ions (excluding deuterons) < 150 MeV/n are treated with BME; Deuterons < 150 MeV/n need to be treated as individual nucleons with IONSPLIT option of the PHYSICS card.
Activation of ion splitting of (only!) deuterons into p+n, from 5 to 150 MeV/n.
Induced radioactivity 20 Input option: BEAM / ISOTOPE Simulation of a radioactive source Example: Radioactive source of 60Co (two main γ-emissions: 1332.5 keV and 1173.2 keV) cylindrical shape, 2cm diameter, 2mm height along z, centre of base of cylinder at origin Note: Do not forget switching on radioactive decays with the RADDECAY card!
Induced radioactivity 21 Geometry modifications Remove concrete shield for transport of radioactive decay radiation (e.g., for calculation of residual dose rate from target only) Remove target for transport of radioactive decay radiation (e.g., for calculation of residual dose rate from concrete shield only) Note: The material for the second step (transport of radioactive decay radiation) can only be set to VACUUM or BLCKHOLE. The run stops if other materials are selected.
Induced radioactivity 22 Geometry modifications target shielding total Induced radioactivity 23 Summary of main input cards RADDECAY requests simulation of decay of produced radioactive nuclides and allows to modify biasing and transport thresholds (defined with other cards) for the transport of decay radiation IRRPROFI definition of an irradiation profile (irradiation times and intensities) DCYTIMES definition of decay (cooling) times DCYSCORE associates scoring detectors (radio-nuclides, fluence, dose equivalent) with different cooling times AUXSCORE allows to associate scoring estimators with dose equivalent conversion factors or/and to filter them according to (generalized) particle identity PHYSICS switch to activate the evaporation of heavy fragments (up to A=24) and the simulation of coalescence Benchmarks Induced radioactivity 24 Induced radioactivity 25 Benchmark experiment Irradiation of samples of different materials to the stray radiation field created by the interaction of a 120 GeV positively charged hadron beam in a copper target 25 Cu target 120GeV pos. hadrons Reference: M. Brugger, S. Roesler, et al., Nuclear Instruments and Methods A 562 (2006) 814-818 Induced radioactivity 26 Benchmark experiment - Instrumentation Low-background coaxial High Precision Germanium detector (Canberra) • use of two different detectors (90 cm3 sensitive volume, 60% and 40% relative efficiency) Genie-2000 (Ver. 2.0/2.1) spectroscopy software by Canberra and PROcount-2000 counting procedure software • include a set of advanced spectrum analysis algorithms, e.g., nuclide identification, interference correction, weighted mean activity, background subtraction and efficiency correction • comprise well-developed methods for peak identification using standard or user-generated nuclide libraries. HERE: use of user-generated nuclide libraries, based on nuclides expected from the simulation and material composition Efficiency calibration with LABSOCS • allows the creation of a corrected efficiency calibration by modelling the sample taking into account self-absorption inside the sample and the correct detector geometry Portable spectrometer Microspec NaI detector, cylindrical shape, 5 x 5 cm folds spectrum with detector response (“calibrated” with 22Na source) physical centre of detector determined with additional measurements with known sources (60Co, 137Cs, 22Na) to be 2.4 cm Induced radioactivity 27 27 0.8 < R < 1.2 0.8 < R ± Error < 1.2 R = Ratio FLUKA/Exp R + Error < 0.8 or R – Error > 1.2 Exp/MDA < 1 Reference: M. Brugger, S. Roesler et al., Nuclear Instruments and Methods A 562 (2006) 814-818 Induced radioactivity 28 Benchmark experiment - Results Dose rate as function of cooling time for different distances between sample and detector Reference: M. Brugger, S. Roesler et al., Radiat. Prot. Dosim. 116 (2005) 12-15 Induced radioactivity 29 Benchmark experiment - Results Dose rate as function of cooling time for different distances between sample and detector Reference: M. Brugger, S. Roesler et al., Radiat. Prot. Dosim. 116 (2005) 12-15 Induced radioactivity 30 Benchmark experiment - Results Dose rate as function of cooling time for different distances between sample and detector Reference: M. Brugger, S. Roesler et al., Radiat. Prot. Dosim. 116 (2005) 12-15 Induced radioactivity 31 Benchmark experiment - Results 6th FLUKA Course - CERN 2008 31 tcool < 2 hours : beta emitter (11C, t1/2 = 20.38min) 2 hours < tcool < 1 day : gamma emitter (24Na, t1/2 = 14.96hrs) |
1594 | https://proofwiki.org/wiki/Order_of_5_in_Units_of_Ring_of_Integers_Modulo_2%5En | Order of 5 in Units of Ring of Integers Modulo 2^n - ProofWiki
Order of 5 in Units of Ring of Integers Modulo 2^n
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Contents
1 Theorem
2 Proof 1
2.1 Induction Hypothesis
2.2 Basis for the Induction
2.3 Induction Step
3 Proof 2
3.1 Lemma
3.2 Proof of lemma
4 Generalization
5 Sources
Theorem
Let n∈Z≥0 n∈Z≥0 be a positive integer and n≥2 n≥2.
Let (Z/2 n Z,+,×)(Z/2 n Z,+,×) be the ring of integers modulo 2 n 2 n.
Let U=((Z/2 n Z)×,×)U=((Z/2 n Z)×,×) denote the group of units of (Z/2 n Z,+,×)(Z/2 n Z,+,×).
The order of 5 5 in U U is 2 n−2 2 n−2.
Proof 1
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Proof by induction:
Induction Hypothesis
For all k≥3 k≥3, we can write
(1):(1):5 2 k−2 5 2 k−2==1+a k⋅2 k 1+a k⋅2 k
where a k a k is odd.
Basis for the Induction
For n=3 n=3, it means 5 2=1+3⋅8 5 2=1+3⋅8
Induction Step
This is our induction step:
Suppose for some k≥3 k≥3, (1) is satisfied.
Consider n=k+1 n=k+1.
5 2 k−1 5 2 k−1==(5 2 k−2)2(5 2 k−2)2
==(1+a k⋅2 k)2(1+a k⋅2 k)2
==1+a k 2 k+1+a 2 k 2 2 k 1+a k 2 k+1+a k 2 2 2 k
==1+a k(1+a k 2 k−1)odd 2 k+1 1+a k(1+a k 2 k−1)⏟odd 2 k+1
This proves the inductive step.
Hence the result by induction.
□◻
Now we prove the order of 5 5 in U U is 2 n−2 2 n−2.
By (1)
5 2 n−2≡1(mod 2 n)5 2 n−2≡1(mod 2 n)
so the order of 5 5 in U U divides 2 n−2 2 n−2.
Aiming for a contradiction, suppose it is less than 2 n−2 2 n−2, then it divides 2 n−3 2 n−3, so:
5 2 n−3=1+m⋅2 n 5 2 n−3=1+m⋅2 n
for some integer m m. Then:
5 2 n−2=(1+m⋅2 n)2=1+(2 m+m 2 2 n)2 n 5 2 n−2=(1+m⋅2 n)2=1+(2 m+m 2 2 n)2 n
where the coefficient 2 m+m 2 2 n 2 m+m 2 2 n is even, contradicting lemma.
■◼
Proof 2
Lemma
For all k≥3 k≥3,
5 2 k−3≡1+2 k−1(mod 2 k)5 2 k−3≡1+2 k−1(mod 2 k)
Proof of lemma
Write ν 2(n)ν 2(n) for the highest exponent of a power of 2 2 dividing n n.
For all i≥2 i≥2,
i≤2 i−1 i≤2 i−1
so
ν 2(i)≤i−1 ν 2(i)≤i−1
so
ν 2(i)ν 2(i)≤≤i−1 i−1
(1):(1):≤≤2 i−3 2 i−3 By i≥2 i≥2
From definition of binomial coefficient:
(2 k−3 i)=2 k−3(2 k−3−1 i−1)i(2 k−3 i)=2 k−3(2 k−3−1 i−1)i
we have
ν 2((2 k−3 i))ν 2((2 k−3 i))==ν 2(2 k−3(2 k−3−1 i−1))−ν 2(i)ν 2(2 k−3(2 k−3−1 i−1))−ν 2(i)
≥≥k−3−ν 2(i)k−3−ν 2(i)
(2):(2):≥≥k−2 i k−2 i By (1)
so
5 2 k−3 5 2 k−3==(2 2+1)2 k−3(2 2+1)2 k−3
==1+2 k−1+∑i≥2(2 k−3 i)2 2 i 1+2 k−1+∑i≥2(2 k−3 i)2 2 iBinomial Theorem for Integral Index
≡≡1+2 k−1(mod 2 k)1+2 k−1(mod 2 k)By (2)
□◻
By the lemma,
5 2 n−3≢1(mod 2 n)5 2 n−3≢1(mod 2 n)
so the order of 5 5 in U U does not divide 2 n−3 2 n−3.
Squaring both sides of the lemma:
5 2 n−2≡(1+2 n−1)2≡1(mod 2 n)5 2 n−2≡(1+2 n−1)2≡1(mod 2 n)
so the order of 5 5 in U U divides 2 n−2 2 n−2.
Hence the order of 5 5 in U U is 2 n−2 2 n−2.
■◼
Generalization
This page or section has statements made on it that ought to be extracted and proved in a Theorem page.
In particular: Put this in its own page
You can help P r∞f W i k i P r∞f W i k i by creating any appropriate Theorem pages that may be needed.
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Let U U be the same as above.
Let u u be any odd number of the form 8 k+3 8 k+3 or 8 k+5 8 k+5.
The order of u u in U U is 2 n−2 2 n−2.
Sources
1801:Carl Friedrich Gauss: Disquisitiones Arithmeticae, arts. 90 90
Bernard ( Prove that ord 2 k 5=2 k−2 ord 2 k 5=2 k−2 where k k is any integer ≥3≥3, URL (version: 2016-10-05):
Retrieved from "
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Thermal Shock Resistance and why Borosilicate Glass Won’t Crack
August 14, 2025
Image Source: Pixabay
Borosilicate glass resists cracking because it has low thermal expansion and a special composition. You see borosilicate in items like pyrex, which stay strong when exposed to high heat. When you put regular glass through rapid temperature change, it often cracks. This happens due to thermal shock. Borosilicate contains boron trioxide, making its coefficient of thermal expansion about one-third of soda-lime glass. That means borosilicate glass handles sudden shifts much better. You can trust borosilicate for safe use in high heat and environments with rapid temperature change.
Borosilicate glass offers Thermal Shock Resistance.
Soda-lime glass, with higher expansion, breaks more easily under thermal shock.
Key Takeaways
Borosilicate glass resists cracking because it expands very little when heated or cooled quickly.
Its special mix, including boron oxide, makes it strong against sudden temperature changes and chemicals.
You can safely use borosilicate glass in labs, kitchens, and industries where heat changes fast.
Regular glass breaks easily from quick heat changes, but borosilicate glass stays durable and reliable.
To keep borosilicate glass safe, avoid sudden temperature jumps and handle it gently to prevent cracks.
Thermal Shock Resistance
What Is Thermal Shock?
You might wonder why some glass shatters when you pour hot water into it or move it from a freezer to an oven. This happens because of thermal shock. Thermal shock occurs when a material experiences a quick change in temperature. When you expose glass to sudden temperature changes, the inside and outside of the glass expand or contract at different rates. This difference creates stress inside the glass.
Borosilicate glass stands out because it has high thermal shock resistance. You can use borosilicate in situations where rapid temperature changes are common. For example, you might use it in a laboratory or a kitchen. The unique composition of borosilicate, which includes boron trioxide, gives it low thermal expansion. This means the glass does not expand or contract much when heated or cooled. As a result, borosilicate glass can handle sudden temperature changes without cracking.
Tip: If you need glassware for boiling, baking, or scientific experiments, choose borosilicate glass. Its thermal resistance keeps it safe and reliable.
Why Glass Cracks
Regular glass, like soda-lime glass, does not have the same thermal shock resistance as borosilicate. When you expose standard glass to rapid temperature changes, it often cracks. This happens because the glass cannot handle the stress caused by uneven expansion and contraction. If you heat or cool regular glass quickly, the temperature difference between the surface and the inner layers becomes too great.
Standard glass usually cracks when the temperature changes by about 60°F (30°C) or more. In most cases, glass can only tolerate temperature shifts up to around 45°C (81°F) before it starts to break. When you go beyond these limits, stress fractures form, and the glass can shatter.
Borosilicate glass resists these problems. Its high thermal shock resistance comes from its special composition. Borosilicate contains more boron and less soda and lime than regular glass. This gives it greater thermal resistance and makes it much less likely to crack under stress. You can rely on borosilicate for tasks that involve sudden temperature changes, such as moving glassware from a hot oven to a cold countertop.
Key reasons why borosilicate glass does not crack easily:
Low thermal expansion reduces internal stress.
High thermal resistance allows it to survive rapid temperature changes.
Borosilicate structure stays stable even in extreme conditions.
If you want glass that can handle high thermal shock resistance, borosilicate is the best choice. You will find it in scientific labs, kitchens, and industrial settings where safety and durability matter most.
Borosilicate Glass Science
Composition and Boron Oxide
You might wonder what makes borosilicate glass so different from regular glass. The answer lies in its unique composition. Borosilicate glass contains about 12–13% boron oxide (B2O3) and over 80% silicon dioxide (SiO2). This special mix gives borosilicate its exceptional thermal resistance. The boron oxide forms strong bonds with silicon, creating a network that holds up under extreme heat. You can see this in the way borosilicate glass stays stable even when you expose it to sudden temperature changes.
The presence of boron oxide does more than just lower the melting point. It also connects the borate and silicate parts of the glass, making the structure more stable. This stability helps borosilicate glass resist both thermal shock and chemical attack. When you use borosilicate in the lab or kitchen, you get glass that stands up to both heat and harsh chemicals.
Note: The typical boron oxide content in borosilicate glass is critical for its low expansion rate and high thermal shock resistance.
Low Thermal Expansion
Borosilicate glass stands out because of its low thermal expansion. This means it does not change size much when heated or cooled. The thermal expansion and contraction rate for borosilicate glass is much lower than that of regular soda-lime glass. You can see this in the numbers. The linear thermal expansion coefficient for clear and amber borosilicate glass is about 5.4 to 5.7 ×10⁻⁶ per degree Celsius. Type 1 borosilicate glass can go as low as 3.3 ×10⁻⁶ per degree Celsius. These values are the lowest among commercial glasses.
Here is a quick comparison:
| Borosilicate Glass Type | Linear Thermal Expansion Coefficient (×10⁻⁶ /°C) |
--- |
| Amber Borosilicate | 5.4 to 5.7 |
| Clear Borosilicate | 5.4 to 5.7 |
| Type 1 Borosilicate | 3.3 to 5.1 |
A low expansion rate means borosilicate glass can handle extreme heat without cracking. When you heat or cool it quickly, the glass does not expand or contract much. This property keeps internal stress low and prevents shattering. You can trust borosilicate glass for tasks that involve sudden temperature changes, like moving a beaker from a freezer to a hot plate.
Key benefits of low thermal expansion in borosilicate glass:
Reduces internal stress during heating and cooling
Prevents cracking from thermal shock
Maintains shape and strength under extreme heat
Stability in Sudden Temperature Changes
You need glass that stays strong when you expose it to sudden temperature changes. Borosilicate glass delivers this stability. The low thermal expansion and contraction rate means the glass does not build up much stress, even when you move it from one extreme temperature to another. This is why you see borosilicate glass in laboratories, kitchens, and industrial settings.
Borosilicate glass can withstand temperature changes of around 170°C without fracturing.
Regular soda-lime glass can only handle about 40°C before it cracks.
The secret to this stability lies in the boron trioxide content. Boron trioxide lowers the thermal expansion, so the glass expands and contracts less. This keeps the structure intact, even during rapid heating or cooling. The manufacturing process also helps. Annealing and tempering remove flaws and add strength, making borosilicate glass even more reliable.
Tip: If you want glassware that will not crack when you pour boiling water or move it from oven to countertop, choose borosilicate glass. Its exceptional thermal resistance and low expansion rate make it the safest choice.
Borosilicate glass gives you peace of mind. You can use it for high-temperature experiments, baking, or any task that involves sudden temperature changes. The glass will keep its shape and strength, even under extreme heat.
Comparing Glass Types
Image Source: pexels
Borosilicate vs. Soda-Lime
When you start comparing borosilicate glass and soda-lime glass, you notice big differences in how each type handles heat and chemicals. Borosilicate glass contains boron oxide, which gives it much better thermal shock resistance and thermal resistance. Soda-lime glass, found in most windows and bottles, has a higher thermal expansion rate. This means soda-lime glass cracks more easily when you expose it to sudden temperature changes.
Here is a table that highlights the key differences:
| Aspect | Borosilicate Glass | Soda-Lime Glass |
---
| Key Composition | Contains at least 5% boric oxide | Contains about 70% silica, soda, and lime |
| Thermal Expansion | Lower; more resistant to thermal shock | Higher; less resistant to thermal shock |
| Durability & Strength | Stronger, more durable under extreme conditions | Softer, less scratch-resistant |
| Common Uses | Laboratory glassware, cookware, high-performance | Windows, bottles, tableware, everyday use |
| Cost & Manufacturing | More expensive, harder to produce | Cheaper, easier to produce |
You see borosilicate glass in labs and kitchens because it survives rapid temperature changes. Soda-lime glass works well for everyday items but fails in high-heat or chemical environments.
Chemical Durability
You need glass that stands up to harsh chemicals and heat. Borosilicate glass offers superior chemical resistance and chemical durability. It resists acids, alkalis, and aggressive chemicals much better than soda-lime glass. Borosilicate forms a protective layer that slows down degradation. In fact, borosilicate glass degrades ten times slower than soda-lime glass in acidic or alkaline environments at high temperatures.
Soda-lime glass contains more alkali, so it reacts with chemicals and loses clarity over time. You will notice etching and cloudiness if you use soda-lime glass with strong chemicals. Borosilicate glass keeps its strength and clarity, even after repeated exposure to tough conditions.
Tip: Choose borosilicate glass for chemical processing, laboratory work, or any job that demands high chemical durability and thermal resistance.
Other Specialty Glasses
You may wonder about other types of glass. Some specialty glasses, like aluminosilicate and fused quartz, offer even higher thermal resistance and chemical durability than borosilicate glass. Aluminosilicate glass resists heat and chemicals but costs more. Fused quartz handles extreme temperatures and chemicals, making it perfect for scientific uses.
Borosilicate glass remains the best choice for most people because it balances cost, thermal shock resistance, and chemical durability. You get reliable performance in labs, kitchens, and industrial settings. When you need glass that lasts, borosilicate stands out.
Borosilicate glass: Best for thermal shock resistance, thermal resistance, and chemical durability.
Soda-lime glass: Good for everyday use, but not for high-heat or chemical environments.
Specialty glasses: Used for extreme conditions, but often more expensive.
Comparing borosilicate glass and soda lime glass helps you see why borosilicate is the top choice for demanding tasks. You get longer lifespan, better safety, and less maintenance.
Real-World Uses
Lab and Scientific Equipment
You see borosilicate glass everywhere in laboratories. Scientists trust borosilicate for beakers, flasks, test tubes, and pipettes. The low thermal expansion and durable and heat resistant nature of borosilicate glass make it perfect for experiments that involve rapid heating and cooling. You can heat a borosilicate flask over a Bunsen burner, then cool it quickly without worrying about cracks. Pyrex is a common brand in labs because it uses borosilicate glass. You also find borosilicate in precision instruments like sight glasses and red line tubes. These tools let you monitor chemical reactions and fluid levels safely. Borosilicate glass resists chemical corrosion, so you can use strong acids and bases without damaging your equipment.
Tip: Always check for ISO 3585 or “borosilicate glass 3.3” labels when buying lab glassware. These standards guarantee safety and quality.
Cookware and Oven Safety
You use borosilicate glass in cookware every day. Baking dishes, measuring cups, and mixing bowls made from borosilicate glass handle high oven temperatures and sudden changes. Pyrex cookware made with borosilicate glass lets you move dishes from the oven to the countertop without shattering. Borosilicate glass in cookware tolerates temperature differences up to 330°F before fracturing. Soda-lime glass cookware only tolerates about 100°F, so it shatters more often. Consumer Reports found that borosilicate glass cookware rarely fails in home ovens, even at temperatures as high as 500°F. You get peace of mind knowing your cookware will not explode or break easily.
Key safety standards for borosilicate glass cookware:
ISO 3585 covers chemical composition, thermal resistance, and durability.
FDA approval ensures chemical stability and non-toxicity for food use.
Products labeled “borosilicate glass 3.3” meet these standards.
You should choose borosilicate glass for baking, roasting, and serving. Pyrex dishes made from borosilicate glass last longer and keep you safe in the kitchen.
Industrial Applications
You find borosilicate glass in many industrial settings. Its thermal shock resistance makes it ideal for sight glasses, level gauge glasses, and high-pressure tubes. You see borosilicate glass in boiler glass tubes and heavy wall sight glasses, which endure high mechanical stress and temperature changes. These applications of borosilicate glass protect workers and equipment in factories, chemical plants, and power stations.
| Aspect | Details |
--- |
| Market Size 2023 | USD 2.75 billion |
| Market Size 2033 | USD 4.6 billion |
| CAGR (2025-2033) | 5.3% |
| Key Industrial Applications | Pharmaceuticals, laboratory equipment, electronics, construction, chemicals |
| Regional Highlight | Asia-Pacific largest market driven by pharmaceutical, electronics, and laboratory sectors |
Borosilicate glass supports industries that need reliable, heat resistant materials. You see it in pharmaceutical packaging, electronics, and chemical processing. The market for borosilicate glass keeps growing because companies need safe, durable solutions.
Preventing Exploding Pyrex
Safe Usage Tips
You want to avoid accidents in your kitchen, especially when using glass cookware. Preventing exploding pyrex starts with understanding why glass breaks. Most incidents happen because of thermal shock. When you move borosilicate glass or pyrex from the freezer straight into a hot oven, or pour boiling water into a cold dish, you risk shattering. Microcracks from rough handling or dropping glassware also make it more likely to break under heat.
Follow these safety tips for oven use and general care:
Always preheat your oven before placing borosilicate glass or pyrex inside.
Never put frozen or cold glassware directly into a hot oven.
Let hot borosilicate glass cool before rinsing or adding cold liquids.
Inspect your cookware for chips or cracks. Replace damaged pieces.
Avoid using borosilicate glass on stovetops or under broilers.
Store glassware carefully to prevent accidental bumps.
Clean with mild detergents and avoid harsh chemicals.
Handle glassware gently, supporting it with both hands.
Hundreds of millions of glass dishes are used safely each year, but following proper handling and maintenance reduces your risk even more.
Identifying Borosilicate Glass
You can spot borosilicate glass by looking for certain features. Authentic borosilicate often has a slight blue-green tint at the edges and feels lighter than soda-lime glass. When you tap it, you hear a clear, ringing sound. The surface feels smooth and sometimes slippery. Manufacturers may etch logos or codes on the bottom, and packaging should say “borosilicate.” For cookware, look for a flat, thick bottom and even heat distribution.
| Feature | Borosilicate Glass | Soda-Lime Glass |
---
| Tint | Blue-green at edges | Clear or greenish |
| Weight | Lighter | Heavier |
| Sound | Clear “ping” | Dull “thud” |
| Markings | Etched or stamped | Often none |
Always check labels and manufacturer information to confirm you have borosilicate, especially if you want the best thermal shock resistance.
Myths and Misconceptions
Many people believe all pyrex is made from borosilicate glass, but that is not true. In the United States, most modern pyrex uses soda-lime glass, which does not resist thermal shock as well. Only borosilicate pyrex, often found in Europe, handles rapid temperature changes safely. Another myth says all pyrex is oven safe. In reality, only bakeware labeled for oven use is safe at high temperatures. Storage containers can shatter if heated.
Some think borosilicate glass is unbreakable. While it is much stronger than soda-lime glass, it can still break if you ignore proper handling and maintenance. Placing hot pyrex on a cold surface or adding cold liquids can cause thermal shock and breakage. Always follow manufacturer guidelines and treat all glassware with care.
Remember: Borosilicate glass is durable, but not indestructible. Use proper handling and maintenance to keep your cookware safe and long-lasting.
You can trust borosilicate glass for both safety and performance. Scientific studies show that borosilicate glass resists thermal shock and chemical attack better than soda lime glass. The table below highlights key differences:
| Property | Borosilicate Glass | Soda Lime Glass | Quartz Glass |
--- --- |
| Thermal Shock Resistance | Excellent | Poor | Excellent |
| Melting Temperature (°C) | 1,400 | 1,000 | 1,700 |
| Chemical Resistance | High | Moderate | Very High |
| Cost | Moderate | Low | High |
| Common Uses | Labware, packaging | Food, windows | Optics, semiconductors |
You gain peace of mind when you use borosilicate glass in your oven, lab, or kitchen. The benefits of using borosilicate glass include durability, safety, and reliability. Choose borosilicate glass for your daily and professional needs.
FAQ
What makes borosilicate glass different from regular glass?
You find borosilicate glass contains boron trioxide. This ingredient lowers its thermal expansion. Regular glass, like soda-lime, does not have this. Borosilicate glass resists cracking when you heat or cool it quickly.
Can borosilicate glass go from freezer to oven safely?
You can move borosilicate glass between extreme temperatures more safely than regular glass. Always let it adjust for a few minutes. Avoid placing hot glass on cold surfaces or adding cold liquids to hot glass.
How do you know if your glassware is borosilicate?
Look for labels like “borosilicate” or “3.3” on the bottom.
You may notice a blue-green tint at the edges.
Packaging or product descriptions often mention borosilicate.
Is borosilicate glass safe for food and drinks?
Yes, you can use borosilicate glass for food and drinks. It does not react with acids or bases. The FDA approves it for food contact. You get a safe, non-toxic material for your kitchen.
Why does some Pyrex glass break in the oven?
Some Pyrex glass uses soda-lime instead of borosilicate. Soda-lime glass cannot handle rapid temperature changes. Always check the label before using Pyrex in the oven.
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Table of Contents
Key Takeaways
Thermal Shock Resistance
What Is Thermal Shock?
Why Glass Cracks
Borosilicate Glass Science
Composition and Boron Oxide
Low Thermal Expansion
Stability in Sudden Temperature Changes
Comparing Glass Types
Borosilicate vs. Soda-Lime
Chemical Durability
Other Specialty Glasses
Real-World Uses
Lab and Scientific Equipment
Cookware and Oven Safety
Industrial Applications
Preventing Exploding Pyrex
Safe Usage Tips
Identifying Borosilicate Glass
Myths and Misconceptions
FAQ
What makes borosilicate glass different from regular glass?
Can borosilicate glass go from freezer to oven safely?
How do you know if your glassware is borosilicate?
Is borosilicate glass safe for food and drinks?
Why does some Pyrex glass break in the oven?
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1596 | https://www.greenemath.com/College_Algebra/30/Simplifying-Radical-ExpressionsPracticeTest.html | GreeneMath.com Master Math & Ace Your Exam!
Simplifying Radical Expressions Practice Test
About Simplifying Radical Expressions:
We simplify radicals using the product/quotient rule for radicals. This rule allows us to break the radicand up and pull out rational numbers. For example, we can break up the square root of 20 into: the square root of 5 times the square root of 4. The square root of 4 represents a rational number 2. We report our simplified radical as 2 times the square root of 5.
Test Objectives
Demonstrate the ability to use the product rule for radicals
Demonstrate the ability to use the quotient rule for radicals
Demonstrate the ability to simplify a radical
Simplifying Radical Expressions Practice Test:
1:
Instructions: Simplify each.
Assume all variables are positive real numbers.
$$a)\hspace{.2em}\sqrt{96x}$$
$$b)\hspace{.2em}\sqrt{180x^3}$$
Watch the Step by Step Video Solution | View the Written Solution
2:
Instructions: Simplify each.
Assume all variables are positive real numbers.
$$a)\hspace{.2em}\sqrt{512x^2}$$
$$b)\hspace{.2em}\sqrt{80x^7}$$
Watch the Step by Step Video Solution | View the Written Solution
3:
Instructions: Simplify each.
Assume all variables are positive real numbers.
$$a)\hspace{.2em}4\sqrt{320x^8y^3}$$
$$b)\hspace{.2em}{-}6\sqrt{64xy^5}$$
Watch the Step by Step Video Solution | View the Written Solution
4:
Instructions: Simplify each.
Assume all variables are positive real numbers.
$$a)\hspace{.2em}{-}2\sqrt{375x^2y^4z^8}$$
$$b)\hspace{.2em}{-}5\sqrt{72x^4yz^2}$$
Watch the Step by Step Video Solution | View the Written Solution
5:
Instructions: Simplify each.
Assume all variables are positive real numbers.
$$a)\hspace{.2em}\sqrt{\sqrt{2x}}$$
$$b)\hspace{.2em}\sqrt{\sqrt{x + 1}}$$
Watch the Step by Step Video Solution | View the Written Solution
Written Solutions:
1:
Solutions:
$$a)\hspace{.2em}4 \sqrt{6x}$$
$$b)\hspace{.2em}6x \sqrt{5x}$$
Watch the Step by Step Video Solution
2:
Solutions:
$$a)\hspace{.2em}2 \sqrt{4x^2}$$
$$b)\hspace{.2em}2x \sqrt{5x^3}$$
Watch the Step by Step Video Solution
3:
Solutions:
$$a)\hspace{.2em}16x^2y\sqrt{5x^2}$$
$$b)\hspace{.2em}{-}24y\sqrt{xy^2}$$
Watch the Step by Step Video Solution
4:
Solutions:
$$a)\hspace{.2em}{-}10yz^2\sqrt{3x^2yz^2}$$
$$b)\hspace{.2em}{-}30x^2z\sqrt{2y}$$
Watch the Step by Step Video Solution
5:
Solutions:
$$a)\hspace{.2em}\sqrt{2x}$$
$$b)\hspace{.2em}\sqrt{x + 1}$$
Watch the Step by Step Video Solution |
1597 | https://staff.buffalostate.edu/nazareay/che112/ex7.htm | Determination of Common Anions
| | | | | | | | |
--- --- --- --- |
| Introduction. To establish the presence of a compound in an unknown mixture, you need to run a reaction with clear visual effect. The best kind of effect is the change of color of the solution or formation of the colored precipitate. The formation of colorless (white) precipitate is a little more difficult to detect. The formation of gas (some bulbs inside the solution) can be a possible option as a visual effect. In this laboratory you wll learn how to make simple but reliable tests on several anions which are among the most common in our environment. Qualitative tests on anions: Cl, Br, I, SO42-, PO43-, CO32-, NO3-. Preliminary notes: no only one of Cl, Br, I is to be expected in an unknown. Otherwise the problem is too complex for CHE112. Preliminary tests. 1. To the 5-10 drops of the unknown, add some HNO3 + AgNO3. Formation of the precipitate indicates the presence of at least one of halogenide anions (Cl, Br, I). Ag+ + Cl- ® AgCl or AgNO3 + NaCl ® AgCl + NaNO3 Ag+ + Br- ® AgBr or AgNO3 + KBr ® AgBr + KNO3 Ag+ + I- ® AgI or AgNO3 + NaI ® AgI + NaNO3 AgCl is white, AgBr is slightly yellowish, and AgI is light yellow: | | | | --- | Cl | Br | I | 2. To the 5-10 drops of the unknown, add some BaCl2. The formation of the white precipitate indicates the presence of at least one of SO42-, PO43-, CO32-. Ba2+ + SO42- ® BaSO4 Ba2+ + PO43- ® Ba3(PO4)2 Ba2+ + CO32- ® BaCO3 Determination of individual ions: Br and I: To the 5-10 drops of the unknown, add some aqueous Cl2 , some organic solvent (toluene, chloroform, etc) and shake. Br2 give yellow to orange color while I2 is purple. An excess of aqueous Cl2 will cause I2 oxidation to colorless iodate ion thus masking iodine. 2 I- + Cl2 ® I2 + 2 Cl- 2 Br- + Cl2 ® Br2 + 2 Cl- | | | --- | | I2 | Br2 | Cl In this case (CHE112) Cl- is detected by the preliminary test 1 only. Ag+ + Cl- ® AgCl CO32- . To the 5-10 drops of the unknown, add some acid (HCl, for example). Formation of the gaseous CO2 is a good indication of carbonate. CO32- + 2 H+ ® CO2 ( ) + H2O SO42- . To the 5-10 drops of the unknown, add 1-2 mL of HCl solution and some BaCl2. Formation of the white precipitate indicates the presence of SO42-. Ba2+ + SO42- ® BaSO4 Phosphate and carbonate do not react in the presence of HCl. PO43-: To the 3-5drops of the unknown, add some nitric acid and ammonium molibdate solution. Warm the solution gently. Formation of the yellow precipitate indicates the presence of PO43-. PO43-+12 MoO42- + 24H+ + 3NH4+ ® (NH4)3[PMo12O40] + 12H2O | | | (NH4)3[PMo12O40] | NO3-: To a one(!) drop of unknown, slowly add some diphenylamine solution in concentrated sulfuric acid (caution!). A deep blue color at the spot of contact of these two solutions indicates nitrate. | | | .. | OH- in solution can be indicated simply by pH sensitive indicator paper strip. | |
1598 | https://www.savemyexams.com/dp/chemistry/ib/23/sl/revision-notes/models-of-the-particulate-nature-of-matter/counting-particles-by-mass-the-mole/avogadros-law/ | Avogadro's Law - IB Chemistry Revision Notes
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Revision Notes
Models of the Particulate Nature of Matter
5 Topics · 20 Revision Notes
1.
Introduction to the Particulate Nature of Matter
1. ##### Chemical Elements, Compounds & Mixtures
2. ##### Separating Mixtures
3. ##### Changes of State
4. ##### Average Kinetic Energy
2.
The Nuclear Atom
1. ##### Nuclear Model of the Atom
2. ##### Subatomic Particles
3. ##### Isotopes
3.
Electronic Configurations
1. ##### The Electromagnetic Spectrum
2. ##### Emission Spectra
3. ##### Energy Levels, Sublevels & Orbitals
4. ##### Writing Electron Configurations
4.
Counting Particles by Mass: The Mole
1. ##### The Mole Unit
2. ##### Molar Mass
3. ##### Empirical Formula
4. ##### Molar Concentration
5. ##### Avogadro's Law
5.
Ideal Gases
1. ##### Ideal Gases
2. ##### Molar Gas Volume
3. ##### The Ideal Gas Equation
4. ##### Real Gases
Models of Bonding & Structure
4 Topics · 21 Revision Notes
1.
The Ionic Model
1. ##### Forming Ions
2. ##### Binary Ionic Compounds
3. ##### Ionic Lattices
2.
The Covalent Model
1. ##### Covalent Bonds
2. ##### Lewis Formulas
3. ##### Multiple Bonds
4. ##### Coordinate Bonds
5. ##### Shapes of Molecules
6. ##### Bond Polarity
7. ##### Molecular Polarity
8. ##### Giant Covalent Structures
9. ##### Intermolecular Forces
10. ##### Physical Properties of Covalent Substances
11. ##### Chromatography
3.
The Metallic Model
1. ##### Properties of Metals & Their Uses
2. ##### s & p Block Elements
4.
From Models to Materials
1. ##### Bonding Models
2. ##### Bonding & Properties
3. ##### Properties of Alloys
4. ##### Polymers
5. ##### Addition Polymers
Classification of Matter
2 Topics · 12 Revision Notes
1.
The Periodic Table: Classification of Elements
1. ##### The Periodic Table
2. ##### Electron Configurations & the Periodic Table
3. ##### Periodic Trends
4. ##### Group 1 Metals with Water
5. ##### Group 17 Elements with Halide Ions
6. ##### Metallic & Non-Metallic Oxides
7. ##### Oxidation States
2.
Functional Groups: Classification of Organic Compounds
1. ##### Representing Formulas of Organic Compounds
2. ##### Functional Groups
3. ##### Homologous Series
4. ##### IUPAC Nomenclature
5. ##### Structural Isomers
What Drives Chemical Reactions?
3 Topics · 14 Revision Notes
1.
Measuring Enthalpy Change
1. ##### Difference Between Heat & Temperature
2. ##### Exothermic & Endothermic Reactions
3. ##### Energy Profiles
4. ##### Standard Enthalpy Change
5. ##### Calorimetry Experiments
2.
Energy Cycles in Reactions
1. ##### Bond Enthalpy Calculations
2. ##### Hess's Law
3. ##### Hess's Law Calculations
3.
Energy from Fuels
1. ##### Combustion Reactions
2. ##### Incomplete Combustion
3. ##### The Amount of Carbon Dioxide Produced When Fuels Burn
4. ##### Carbon Dioxide Levels & the Greenhouse Effect
5. ##### Biofuels
6. ##### Fuel Cells
How Much, How Fast & How Far?
3 Topics · 18 Revision Notes
1.
How Much? The Amount of Chemical Change
1. ##### Balancing Equations
2. ##### Reacting Mass Calculations
3. ##### Avogadro's Law & Molar Volume of Gas
4. ##### Concentration Calculations
5. ##### Limiting & Excess Reactants
6. ##### Percentage Yield Calculations
7. ##### Atom Economy
2.
How Fast? The Rate of Chemical Change
1. ##### Rate of Reaction
2. ##### Measuring Rates of Reaction
3. ##### Collision Theory
4. ##### Factors Affecting Rates of Reaction
5. ##### Activation Energy
6. ##### Energy Profiles With & Without Catalysts
7. ##### Maxwell-Boltzmann Distribution Curves
3.
How Far? The Extent of Chemical Change
1. ##### The Characteristics of Dynamic Equilibrium
2. ##### The Equilibrium Law
3. ##### The Equilibrium Constant
4. ##### Le Chatelier's Principle
What are the Mechanisms of Chemical Change?
4 Topics · 24 Revision Notes
1.
Proton Transfer Reactions
1. ##### Brønsted–Lowry Acids & Bases
2. ##### Conjugate Acids & Bases
3. ##### Amphiprotic Species
4. ##### The pH Scale
5. ##### The Ion Product of Water
6. ##### Strong & Weak Acids
7. ##### Neutralisation Reactions
8. ##### pH Curves
2.
Electron Transfer Reactions
1. ##### Oxidation & Reduction
2. ##### Half Equations
3. ##### Relative Ease of Oxidation & Reduction
4. ##### Acids with Reactive Metals
5. ##### Primary Cells
6. ##### Secondary Cells
7. ##### Electrolytic Cells
8. ##### Oxidation of Alcohols
9. ##### Reduction of Carboxylic Acids, Aldehydes & Ketones
10. ##### Reduction of Unsaturated Compounds
3.
Electron Sharing Reactions
1. ##### Radicals
2. ##### Homolytic Fission
3. ##### Halogenation of Alkanes
4.
Electron Pair Sharing Reactions
1. ##### Nucleophilic Substitution
2. ##### Heterolytic Fission
3. ##### Electrophilic Addition Reactions
Tools
3 Topics · 9 Revision Notes
1.
Tool 1: Experimental Techniques
1. ##### Safety, Ethical & Environmental Issues in Chemistry
2. ##### Measuring Variables in Chemistry
3. ##### Applying Techniques in Chemistry
2.
Tool 2: Technology
1. ##### Applying Technology to Collect Data in Chemistry
2. ##### Applying Technology to Process Data in Chemistry
3.
Tool 3: Mathematics
1. ##### Applying General Mathematics in Chemistry
2. ##### Using Units, Symbols & Numerical Values in Chemistry
3. ##### Processing Uncertainties in Chemistry
4. ##### Graphing in Chemistry
Inquiry Process
1 Topic · 2 Revision Notes
1.
Inquiry 3: Concluding and Evaluating in Chemistry
1. ##### Concluding in Chemistry
2. ##### Evaluating in Chemistry
IBChemistryDPSLRevision NotesModels of the Particulate Nature of Matter Counting Particles by Mass: The Mole Avogadro's Law
Avogadro's Law(DP IB Chemistry):Revision Note
Download PDF
Written by:Alexandra Brennan
Reviewed by:Philippa Platt
Updated on 14 May 2025
Exam board:
DP
DP Chemistry: SLDP Chemistry: HL
Avogadro's Law
Volumes of gases
In 1811, Amedeo Avogadro proposed that equal volumes of gases (under the same conditions of temperature and pressure) contain the same number of molecules
This is known as Avogadro’s Law, and it allows us to determine mole ratios from the volumes of reacting gases
At standard temperature and pressure (STP):
One mole of any gas occupies 22.7 dm3
Units are written as dm3mol-1
STP conditions are defined as:
0 °C (273 K)
100 kPa pressure
Stoichiometric relationships
The volume ratio of gaseous reactants and products in a balanced chemical equation is the same as the mole ratio
You can use this ratio to calculate unknown gas volumes in a reaction at STP
Example: Combustion of propane
In the reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (l)
If 50 cm³ of propane is burned:
Volume of O2needed = 5 × 50 = 250 cm3
Volume of CO2formed = 3 × 50 = 150 cm3
The mole ratio from the equation (1:5:3) is directly applied to gas volumes
Examiner Tips and Tricks
These gas volume relationships are only valid if all gases are measured under the same conditions of temperature and pressure
If gases are not in the same ratio as the balanced equation, use limiting reactant principles to determine the actual amount of product formed
Worked Example
What is the total volume of gases remaining when 70 cm 3 of ammonia is combusted completely with 50 cm 3 of oxygen according to the equation shown?
4NH3(g) + 5O2(g) → 4NO (g) + 6H2O (l)
Answer:
Step 1
From the equation deduce the molar ratio of the gases (water is in the liquid state so is not included
NH 3 :O 2 :NO or 4:5:4
Step 2
Oxygen will run out first (the limiting reactant) and so 50 cm 3 of O 2 requires 4/5 x 50 cm 3 of NH 3 to react= 40 cm 3
Step 3
Using Avogadro's Law, 40 cm 3 of NO will be produced
Step 4
There will be of 70-40 = 30 cm 3 of NH 3 left over
Total remaining = 40 + 30 = 70 cm3of gases
Examiner Tips and Tricks
Since gas volumes work in the same way as moles, we can use the 'lowest is limiting' technique in limiting reactant problems involving gas volumes. This can be handy if you are unable to spot which gas reactant is going to run out first. Divide the volumes of the gases by the coefficients and whichever gives the lowest number is the limiting reactant
E.g. in the previous problem we can see that
For NH 3 gives 17.5
For O 2 gives 10, so oxygen is limiting
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Author
Reviewer
Author:Alexandra Brennan
Expertise:Chemistry Content Creator / Senior Marketing Executive
Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.
Reviewer:Philippa Platt
Expertise:Chemistry Content Creator
Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener
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Elops
Elops machnata
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Elops machnata Tenpounder, Australian Giant Herring, Banana Fish, Giant Herring, Lady Fish, Pincushion Fish
Elops machnata is commonly referred to as Tenpounder, Australian Giant Herring, Banana Fish, Giant Herring, Lady Fish, Pincushion Fish. Difficulty in the aquarium: suitable for large display tanks (public aquarium or zoo) only. Toxicity: Toxic hazard unknown.
Bernard Dupont, Frankreich
Foto: Sea World Aquarium, Durban, KwaZulu-Natal, Süd-Afrika
Courtesy of the author Bernard Dupont, Frankreich Copyright Bernard Dupont. Please visit www.flickr.com for more information.
Uploaded by AndiV.
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lexID:15083 AphiaID:212252Scientific:Elops machnata German:Indischer Frauenfisch English:Tenpounder, Australian Giant Herring, Banana Fish, Giant Herring, Lady Fish, Pincushion Fish Category:LadyfishesFamily tree:Animalia (Kingdom) > Chordata (Phylum) > Actinopteri (Class) > Elopiformes (Order) > Elopidae (Family) >Elops (Genus) > machnata (Species)Initial determination:(Forsskål, ), 1775Occurrence:Djibouti, Hong Kong, Suez-Kanal, Andaman and Nicobar Islands, Arabian Sea, Australia, Bali, China, East Africa, Eastern Indian Ocean, Egypt, Gulf of Oman / Oman, Hawaii, India, Indian Ocean, Indonesia, Japan, Kenya, Lesser Sunda Islands, Madagascar, Malaysia, Mascarene Islands, Mauritius, Mozambique, New Caledonia, Northern Territory (Australia), Oceanodromous, Pakistan, Papua New Guinea, Philippines, Red Sea, Réunion, Socotra Arch, Somalia, South-Africa, Sri Lanka, Taiwan, Tansania, the Seychelles, Timor, Timor Sea, West Indies, Western Australia, Yemen, ZanzibarSea depth:0 - 50 Meter Habitats:Brackish water, Estuaries (river mouths), Seawater, Sea water Size:up to 46.46" (118 cm) Weight:11.8 kg Temperature:69.26 °F - 80.42 °F (20.7°C - 26.9°C) Food:Carnivore, Crabs, Crustaceans, Edible crab, Fish (little fishes), Predatory, Schrimps, Sepia, Zoobenthos Difficulty:suitable for large display tanks (public aquarium or zoo) only Offspring:Not available as offspring Toxicity:Toxic hazard unknown CITES:Not evaluated Red List:Least concern (LC) Related species at
Catalog of Life:
Elops affinis Pazifischer Marienfisch, Machete, Zehnpfünder
Elops hawaiensis Hawaii-Frauenfisch
Elops lacerta
Elops machnata Indischer Frauenfisch
Elops saurus Frauenfisch
Elops senegalensis
Elops smithi
Author:Meerwasser-Lexikon Team Publisher:Meerwasser-Lexikon.de Created:2022-08-08 11:12:05 Last edit:2025-02-25 21:30:30
Info
The Indian Ladyfish is a bony fish from the genus Elopida , which currently contains only 7 species.
Elops machnata is a very slender and fast predator and also looks like an elongated giant herring with a bluish-gray back, silvery sides with a yellow tint, and pale yellow fins.
This species lives pelagically in shallow coastal waters and frequently enters estuaries and lagoons.
Juveniles use warm, turbid estuaries as nursery grounds.
Voracious predators, they feed on fish, squid and crustaceans.
The specific name is derived from the Arabic name "machnat", the common name for this species at the type locality of Jeddah (English, Jeddah), Saudi Arabia.
In Lake St. Lucia, South Africa, Elops machnata feeds primarily on the orangemouth anchovy (Thryssa vitrirostris), the herring Gilchristella aestuarius, and to a lesser extent the half-billed pike Hyporhamphus knysnaensis (Whitfield, 1977).
Synonyms:
Argentina machnata Forsskål, 1775
Elops capensis Smith, 1838-47
Elops indicus Swainson, 1839
Elops purpurascens Richardson, 1846
External links
Annotated checklist of the fish species (Pisces) of La Réunion, including a Red List of threatened and declining species (de). Abgerufen am 08.08.2022.
Annotated checklist of the fishes of Madagascar, southwestern Indian Ocean, with 158 new records (en). Abgerufen am 08.08.2022.
Authorship, availability and validity of fish names described by Peter (Pehr) Simon Forsskål and Johann Christian Fabricius in the ‘Descriptiones animalium’ by CarSten niebuhr in 1775 (Pisces) (en). Abgerufen am 09.08.2022.
CAAB - Codes for Australian Aquatic Biota (en). Abgerufen am 08.08.2022.
Coastal and estuarine resources of Bangladesh: management and conservation issues (en). Abgerufen am 08.08.2022.
FAO CATALOGUE OF CULTIVATED AQUATIC ORGANISMS (en). Abgerufen am 08.08.2022.
FAQ (en). Abgerufen am 08.08.2022.
Field identification guide to the living marine resources of Myanmar (en). Abgerufen am 08.08.2022.
Fishes of Australia (en). Abgerufen am 08.08.2022.
Fishes of the Kosi system (en). Abgerufen am 08.08.2022.
Global conservation status and research needs for tarpons (Megalopidae), ladyfishes (Elopidae) and bonefishes (Albulidae) (en). Abgerufen am 08.08.2022.
IUCN Red List of Threatened Species (multi). Abgerufen am 08.08.2022.
NATURAL DISASTER AND ECOSYSTEM CHANGE: CASE STUDY OF A LAKE (en). Abgerufen am 08.08.2022.
Proceedings of the 4th Technical Seminar on Marine Fishery Resources Survey in the South China Sea, Area IV: Vietnamese Waters, 18-20 September 2000 (en). Abgerufen am 08.08.2022.
RECORDS OF THE ZOOLOGICAL SURVEY OF INDIA (en). Abgerufen am 08.08.2022.
Seasonal Abundance Distribution and Catch Per Unit Effort of Fishes in the Swartkops Estuary (en). Abgerufen am 08.08.2022.
Temperature and salinity as primary determinants influencing the biogeography of fishes in South African estuaries (en). Abgerufen am 08.08.2022.
The influence of turbidity on juvenile marine fishes in estuaries. part 1. field studies at Lake St. Lucia on the southeastern coast of Africa (en). Abgerufen am 08.08.2022.
Turbidity-induced changes in feeding strategies of fish in estuaries (en). Abgerufen am 08.08.2022.
World Register of Marine Species (WoRMS) (en). Abgerufen am 08.08.2022.
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