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189100	https://zhidao.baidu.com/question/590333123.html	在“土壤中分解尿素的细菌的分离与计数”中，怎样设置对照组？_百度知道百度首页商城注册登录网页资讯视频图片知道文库贴吧采购地图更多搜索答案我要提问在“土壤中分解尿素的细菌的分离与计数”中，怎样设置对照组？  我来答首页用户认证用户认证团队合伙人热推榜单企业媒体政府其他组织商城法律手机答题我的百度知道> 无分类在“土壤中分解尿素的细菌的分离与计数”中，怎样设置对照组？ 我来答分享复制链接新浪微博微信扫一扫举报可选中1个或多个下面的关键词，搜索相关资料。也可直接点“搜索资料”搜索整个问题。  对照组  土壤  中分  尿素  细菌  搜索资料 1个回答 #热议#普通体检能查出癌症吗？匿名用户 2013-09-11 展开全部分离不知道。计数应该是稀释样本再培养，看培养皿菌团数，然后按稀释比例算 1已赞过已踩过< 你对这个回答的评价是？评论分享复制链接新浪微博微信扫一扫举报收起天津飞瑞特 2025-07-28 广告天津飞瑞特科技有限公司从事环保科研仪器行业，是美国及欧洲的几个知名环保仪器品牌的中国总代理商或者办事处，负责品牌在中国的技术咨询，销售，售后服务。经营产品类型包括：环保仪器；实验仪器等。... 点击进入详情页本回答由天津飞瑞特提供推荐律师服务：若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询其他类似问题 2015-06-28 土壤中分解尿素的细菌的分离与计数实验——高中生物选修1 2 2015-02-10 下列是关于“土壤中分解尿素的细菌的分离与计数”实验操作的叙述... 2011-05-28 生物选修1专题2课题2 土壤中分解尿素的细菌的分离与计数 2 2013-02-07 求答案......... 11 2017-05-26 高中生物选修1 专题二课题2 土壤中分解尿素的细菌的分离与计... 2015-02-08 土壤是“微生物的天然培养基”，下面是有关土壤中分解尿素的细菌... 2017-04-01 从土壤中分离分解尿素的细菌过程中的灭菌操作步骤 3 2012-05-25 土壤中分解尿素的细菌的分离与计数的培养基上有固氮微生物吗 1 更多类似问题> 为你推荐：特别推荐 “网络厕所”会造成什么影响？新生报道需要注意什么？华强北的二手手机是否靠谱？癌症的治疗费用为何越来越高？百度律临—免费法律服务推荐超3w专业律师，24H在线服务，平均3分钟回复免费预约随时在线律师指导专业律师一对一沟通完美完成等你来答  换一换  男星出镜要化妆吗？他们卸妆后跟镜头前差距有多大？等 26209 人想问  我来答  电视剧《潜伏》里，吴站长是不是知道余则成的真实身份？等 12893 人想问  我来答  你们结婚后过得好吗？等 18935 人想问  我来答  年轻的情感离别，怎会总是热泪盈眶？等 16758 人想问  我来答  怎样的情感下会让你失去理智？等 16851 人想问  我来答  为什么过生日时你就快乐呢？等 13062 人想问  我来答帮助更多人  下载百度知道APP，抢鲜体验使用百度知道APP，立即抢鲜体验。你的手机镜头里或许有别人想知道的答案。扫描二维码下载 × 个人、企业类侵权投诉违法有害信息,请在下方选择后提交类别色情低俗涉嫌违法犯罪时政信息不实垃圾广告低质灌水我们会通过消息、邮箱等方式尽快将举报结果通知您。说明 0/200 提交取消领取奖励我的财富值 0 兑换商品 -- 去登录我的现金 0 提现下载百度知道APP 在APP端-任务中心提现我知道了 -- 去登录做任务开宝箱累计完成 0 个任务 10任务略略略略… 50任务略略略略… 100任务略略略略… 200任务略略略略… 任务列表加载中... 新手帮助如何答题获取采纳使用财富值玩法介绍知道商城合伙人认证您的账号状态正常感谢您对我们的支持投诉建议意见反馈账号申诉非法信息举报京ICP证030173号-1 京网文【2023】1034-029号 ©2025Baidu使用百度前必读\|知道协议\|企业推广辅助模式返回顶部
189101	https://do2.vsmu.by/pluginfile.php/206370/mod_folder/content/0/Lecture%2021.pdf?forcedownload=1	REDOX titration A lot of titrimertic methods used in quantitative analysis are based on the oxidation-reduction reactions (redox reactions). They are commonly named as oxidimetry (redox) methods. With the help of oxidimetry used in clinical and biochemical research one can determine the catalase and peroxidase activity, the presence of ascorbic acid, sugar in blood and other biological liquids, uric acid in urine, urea in blood and urine, calcium ions in blood serum and so on. With the help of oxidimetry methods used in hygiene and sanitary investigations one can determine the oxidability of water, the content of active chlorine in drinking water, dissolved oxygen and organic admixtures in the water and so on. The redox reactions include a partial or a complete electron transfer from one group of atoms or ions to another group. These transfers are resulted in the change of oxidation states of the atoms of certain elements. According to the electron theory of oxidation-reduction processes, oxidation is the process of the loss of electrons. The substance which has lost electrons is called a reducing agent (reducer). Reducer transforms into its oxidized form in the course of the reaction. Each reduced formed is, actually, the oxidizing agent that is coupled with the initial reducing agent. For example, Sn 2+ – 2e – ↔ Sn 4+ reducing agent 1 – ne – ↔ oxidizing agent 1 Reduction is the process of the gain of electrons. The substance taking the electrons in redox reactions is called the oxidizing agent. Oxidizer transforms into its coupled reduced form during the redox reaction. For example, Fe 3+ + e – → Fe 2+ oxidizing agent 2 + ne – → reducing agent 2 The processes of oxidation and reduction occur simultaneously and they should be considered inseparably, besides, the total number of electrons lost by the reducing agent is equal to the total number of the electrons gained by the oxidizing agent. So, any redox reaction is the combination of two coupled processes — so-called half-reactions: oxidation of the reducing agent and the reduction of the oxidizing agent. Sn 2+ + 2Fe 3+ ⇄ Sn 4+ + 2Fe 2+ Redox titration - a titration in which the reaction between the analyte and titrant is an oxidation/reduction reaction. Redox titrations were introduced shortly after the development of acid – base titrimetry. The earliest methods took advantage of the oxidizing power of chlorine. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl 2, HCl, and HOCl) based on its ability to oxidize solutions of the dye indigo. In 1814, Joseph Louis Gay-Lussac (1778 –1850), developed a similar method for chlorine in bleaching powder. In both methods the end point was signaled visually. Before the equivalence point, the solution remains clear due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution. The number of redox titrimetric methods increased in the mid-1800s with the introduction of MnO 4–, Cr 2O72– and I 2 as oxidizing titrants, and Fe 2+ and S 2O32– as reducing titrants. Even with the availability of these new titrants, however, the routine application of redox titrimetry to a wide range of samples was limited by the lack of suitable indicators. Titrants whose oxidized and reduced forms differ significantly in color could be used as their own indicator. For example, the intensely purple MnO 4– ion serves as its own indicator since its reduced form, Mn 2+ , is almost colorless. The utility of other titrants, however, required a visual indicator that could be added to the solution. The first such indicator was diphenylamine, which was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry. The necessary conditions to obtain a satisfactory titration. There are two kinds: thermodynamic and kinetic. The thermodynamic condition is an absolute necessity: A titration reaction must be as complete as possible, regardless of the kind of titration: acid –base, redox, etc. This condition must be satisfied in order to weaken the titration error. From a practical standpoint, a titration reaction must be achieved as fast as possible. It is not always the case for redox titrations. Recall that the rate of a redox reaction may be considerably increased with the use of catalysts . Additionally, some titrations imply them. A redox titration permits us to determine the concentration of a form Ox 1 or Red 1 by adding antagonistic solution of Red 2 or Ox 2. The redox reaction between the antagonistic species, which is the titration reaction, induces a change in the equilibrium potential of the titrand solution for each added volume of titrant solution. The requirements to the reactions used in oxidimetry are as follows: • the reaction should be fast and irreversible; • the products should have definite composition; • there should not be any side reactions; • there should be a method to fix the end of the reaction. The number of reactions which satisfy these criterions is very small. The difference between redox potentials of two substances (E = E ox – Ered ) should be higher than 0.4 to make the process almost irreversible. Otherwise, there will be no drastic redox potential jump at the equivalence point. Preparing the Solution —Getting the Analyte in the Right Oxidation State before Titration When samples are dissolved, the element to be analyzed is usually in a mixed oxidation state or is in an oxidation state other than that required for titration. There are various oxidizing and reducing agents that can be used to convert different metals to certain oxidation states prior to titration. The excess preoxidant or prereductant must generally be removed before the metal ion is titrated. REDUCTION OF THE SAMPLE PRIOR TO TITRATION The reducing agent should not interfere in the titration or, if it does, unreacted reagent should be readily removable. Most reducing agents will, of course, react with oxidizing titrants, and they must be removable. Sodium sulfite, Na 2SO 3, and sulfur dioxide are good reducing agents in acid solution (E 0 = 0.17 V), and the excess can be removed by bubbling with CO 2 or in some cases by boiling. Stannous chloride, SnCl 2 , is usually used for the reduction of iron(III) to iron(II) for titrating with cerium(IV) or dichromate. The reaction is rapid in the presence of chloride (hot HCl). A number of metals are good reducing agents and have been used for the prereduction of analytes. Included among these reductants are zinc, aluminum, cadmium, lead, nickel, copper, and silver (in the presence of chloride ion). In the reductor, the finely divided metal is held in a vertical glass tube through which the solution is drawn under a mild vacuum. A typical Jones reductor has a diameter of about 2 cm and holds a 40- to 50-cm column of amalgamated zinc. Amalgamation is accomplished by allowing zinc granules to stand briefly in a solution of mercury(II) chloride. A Jones reductor OXIDATION OF THE SAMPLE PRIOR TO TITRATION Sodium bismuthate is a powerful oxidizing agent capable, for example, of converting manganese(II) quantitatively to permanganate ion. Ammonium peroxydisulfate, (NH 4)2S2O8, is also a powerful oxidizing agent. In acidic solution, it converts chromium(III) to dichromate, cerium(III) to cerium(IV), and manganese(II) to permanganate. Peroxide is a convenient oxidizing agent either as the solid sodium salt or as a dilute solution of the acid. Oxidation-reduction titrations can be classified according to the titrant Redox titration curve Consider the titration of 100 ml of 0,1 M iron(II) sulfate FeSO 4 solution with 0,1 M potassium permanganate solution, monitored potentiometrically. There are three distinct regions in the titration of iron(II) with potassium permanganate solution, monitored potentiometrically. 1) Before the equivalence point, where the potential is dominated by the analyte redox pair. 2) At the equivalence point, where the potential at the indicator electrode is the average of their conditional potential. 3) After the equivalence point, where the potential was determined by the titratant redox pair. The redox reaction between the antagonistic species, which is the titration reaction, induces a change in the equilibrium potential of the titrand solution for each added volume of titrant solution. A redox titration curve is the diagram of the solution’s potential/added volume of titrant solution. Before the equivalence point: During the addition of the KMnO 4 solution up to the equivalence point, its only effect will be to oxidize the Fe 2+ , and consequently change the ratio [Fe 3+ ]/[Fe 2+ ]. Prior to the equivalence point, the half-reaction involving analyze is used to find the voltage because the concentrations of both the oxidized and the reduced forms of analyte are known. After adding X mL of KMnO 4 the voltage can be calculated using Nernst equation as follows After adding 1 ml of 0,1 M KMnO 4:After adding 91 ml of 0,1 M KMnO 4:After adding 99,9 ml of 0,1 M KMnO 4: At the equivalence point At the equivalence point, both half-reactions are used simultaneously to find the voltage. The electrode potential is given by: After the equivalence point: The subsequent addition of the KMnO 4 solution will merely increase the ratio [MnO 4-]/[Mn 2+ ]. After adding 100,1 ml of 0,1 M KMnO 4: After adding 110 ml of 0,1 M KMnO 4: Thus E (Volt) changes from 0,994 to 1,475 between 0,1 ml before and 0,1 ml after the stoichiometric endpoint. These quantities are of importance in connection with the use of indicators for the detection of the equivalence point. Conclusions: • It is evident that the abrupt change of the potental in the neighbourhood of the equivalence point is dependent upon the standard potentials of the two oxidation-reduction systems that are involved. The greater the difference in reduction potential between analyze and titrant, the sharper will be the end point. • The voltage at any point in this titration is independent of dilution and of the concentrations unless these are extremely small. • Completeness of the Reaction The change in potential in the equivalence-point region of an oxidation/reduction titration becomes larger as the reaction becomes more complete. Titration curves for 0.1000 M Ce 4+ titration. A: Titration of 50.00 mL of 0.05000 M Fe 2+ . B: Titration of 50.00 mL of 0.02500 M U 4+ . the curve for iron(II) is symmetric around the equivalence point but that the curve for uranium(IV) is not symmetric. In general, redox titration curves are symmetric when the analyte and titrant react in a 1:1 molar ratio. The greatest change in potential of the system is associated with the reaction that is most complete, and curve E illustrates the opposite extreme. At the equivalence point of a satisfactory redox titration, there is a sharp change in the equilibrium potential of the titrated solution. Several means exist for its detection. They are based on the current use • of internal redox indicators; • SELF-INDICATION or of special redox species. The latter are purely and simply one (or several) of the reactants of the redox titration under consideration. They also play the role of indicators. For example, it is the case with iodine –iodide and of potassium permanganate solutions; • of specific indicators that exhibit a particular color in the presence of a given reactant or product. It is the case for a starch solution in the presence of iodine and iodide; • of several physical methods of analysis , the most important of which is probably zero-current potentiometry. This is an electrochemical method that, as a rule, measures the potential of a solution. Detection of Endpoint in Redox Titration Two types of chemical indicators are used for obtaining end points for oxidation/reduction titrations: general redox indicators and specific indicators .General Redox Indicators The ideal oxidation-reduction indicator will be one with an oxidation potential intermediate between that of the solution titrated (analyte) and that of the titrant, and which exibits a sharp, readily detectible colour change. An redox indicator is a compound which exhibits different colours in the oxidized and reduced forms. These are highly colored dyes that are weak reducing or oxidizing agents that can be oxidized or reduced; the colors of the oxidized and reduced forms are different. The oxidation state of the indicator and hence its color will depend on the potential at a given point in the titration. The oxidation and reduction should be reversible. A half- reaction and Nernst equation can be written for the indicator: Typically, a change from the color of the oxidized form of the indicator to the color of the reduced form requires a change of about 100 in the ratio of reactant concentrations, that is, a color change appears when This equation shows that a typical general indicator exhibits adetectable color change when a titrant causes the system potential to shift from E0In + 0.0592/ n to E 0In - 0.0592/ n or about (0.118/n) V. For many indicators, n 5 2, and a change of 0.059 V is thus sufficient. E0In must be near the equivalence point potential. A potential change of at least 120mV is needed for a color change for n = 1 (of the indicator half-reaction) and 60mV for n = 2. Indicator range Protons participate in the reduction of many indicators. Thus, the range of potentials over which a color change occurs (the transition potential) is often pH dependent .The most commonly used internal redox indicators are derivatives of 1,10-phenanthroline, diphenylamine, phenothiazine, and diphenylpyrazine. Ferroin Iron(II) Complexes of Orthophenanthrolines A class of organic compounds known as 1,10-phenanthrolines, or orthophenanthrolines, form stable complexes with iron(II) and certain other ions. The parent compound has a pair of nitrogen atoms located in such positions that each can form a covalent bond with the iron(II) ion. Diphenylamine Some derivatives of diphenylamine are internal redox indicators that have been used very often. Diphenylamine dissolved in diluted acidic medium . In the presence of a strong oxidizing agent, it first undergoes an irreversible chemical oxidation to give the colorless diphenylbenzidine. Diphenylbenzidine can be reversibly oxidized according to a bielectronic process to give a colored diquinonediimine. Finally, the formed diquinonediimine can be oxidized once more, but this time irreversibly when it stays too long in the presence of the oxidizing solution. Diphenylamine can be used in the case of the titration of ferrous ions by potassium dichromate. It is poorly soluble in water. variamine blue Methylene blue derives from phenothiazine. (The reduced form can also undergo two further protonations on the dimethylamino rests depending on the medium’s pH.) The standard potential of methylene blue is E◦ = 0 .53 V. A redox titration is feasible if the difference between analyte and titrant is > 0.2 V. If the difference in the formal potential is > 0.4 V, then a redox indicator usually gives a satisfactory end point. Percent by mass, mass percentage, mass fraction (ω) . The percent by mass (also called the percent by weight or the weight percent) is defined as ω = mass fraction of a solute = m (solute) / m (solution). Molality (C m). Molality is the number of moles of a solute (S) dissolved in 1 kg (1000 g) of a solvent. Cm = n (solute) / m (solvent). Thus molality has the units of mol/kg. Molarity (C). Molarity is defined as the number of moles of solute in 1 liter of a solution; that is, C = n (solute) / V (solution). Thus, molarity has the units of mol/L or M. Titer (T). Titer is defined as a mass of a solute in 1 milliliter of a solution; that is, T = m (solute) / V (solution). Thus, titer has the units of g/mL. Normality (C N). Normality is defined as the number of moles of the equivalent of a solute (neq) in 1 liter of a solution; that is, CN = neq (solute) / V (solution). Thus, normality has the units of mol/L or N. Solutions' concentration can be represented by titer (T) in the volumetric analysis - it is the number of solute's grammes in the one cm 3 (ml) of the solution (g/ml). In chemical reactions the molar ratio of reacting substances is not always 1:1. It is determined by stoichiometric coefficients. For example, in the reaction written below the molar ratio between base and acid will be 2:1. 2NaOH + H 2SO 4 = Na 2SO 4 + 2H 2O. The coefficients can be replaced by equivalence factors: actually, for the base you need to divide number one (1) by the coefficient before acid, for the acid you need to divide number one by the coefficient before base. So, in the abovementioned reaction the equivalence factor for NaOH is 1, while for H 2SO 4 it is equal to 1/2. As substances behave differently in the complex formation reactions, chemists avoid using the notion of equivalent for them and use only molar masses instead. Thus, in this textbook the notion of equivalent will be used only for the substances taking part in oxidation-reduction and acid-base reactions. Let‘s consider the following reactions: a) H + + OH – = H 2O; d) 2H + + S 2– = H 2S; b) H + + NH 3 = NH 4+; e) 3OH – + H 3PO 4 = PO 43– + 3H 2O; c) H 20 – 2e – = 2H +; f) Al 3+ + 3e – = Al 0.In acid-base reactions (a, b, d, e) 1 OH – ion, 1 NH 3 molecule, 1/2 S 2– ion, 1/3 of H 3PO 4 molecule are equivalent to one H + ion. In oxidation- reduction reactions (c, f) 1/2 of H2 molecule, 1/3 of Al 3+ ion are equivalent to one electron. The enumerated particles are considered as equivalents of substances taking part in these reactions. From another point of view the equivalent is some real or hypothetical particle which interacts with the carrier of one elementary charge in the ion exchange or oxidation-reduction reactions. Equivalent is that comparative quantity by weight of an element, which possesses the same chemical value as other elements, as determined by actual experiment and reference to the same standard. Equivalence factor f eq (X) is the number indicating which part of the real particle of substance X is equivalent to one hydrogen ion in the given acid-base reaction or to one electron in the oxidation-reduction reaction. This value is dimensionless and is calculated on the basis of stoichiometric coefficients of a definite reaction. Equivalence factor is often written as 1/z ratio, where z is the overall charge of ions from a molecule taking part in the given exchange reaction or the number of electrons which are gained or lost by a molecule (an atom) of the substance in oxidation-reduction reaction; z is always a positive integer and the equivalence factor is less or equal to 1: f eq (X) = 1/z ≤ 1. Equivalence factor of the same substance can have different values in different reactions. Let‘s consider this using the following examples. In acid-base reactions Na 2CO 3 can be neutralized by an acid until the formation of an acidic salt or until the emission of CO 2:a) Na 2CO 3 + HCl = NaHCO 3 + NaCl, f eq (Na 2CO 3) = 1; b) Na 2CO 3 + 2HCl = 2NaCl + H 2O + CO 2, f eq (Na 2CO 3) = 1/2. In oxidation-reduction reactions KMnO 4 is always an oxidizing agent: c) 5Na 2SO 3 + 2KMnO 4 + 3H 2SO 4 = 5Na 2SO 4 + 2MnSO 4 + 3H 2O + K2SO 4,MnO 4– + 8H + + 5e – → Mn 2+ + 4H 2O, f eq (KMnO 4) = 1/5; d) 3Na 2SO 3 + 2KMnO 4 + H 2O = 3Na2SO4 + 2MnO2 + 2KOH, MnO 4– + 2H 2O + 3e – → MnO 2 + 4OH –, f eq (KMnO 4) = 1/3; e) Na 2SO 3 + 2KMnO 4 + 2KOH = Na 2SO 4 + 2K 2MnO 4 + H 2O, MnO 4– + e – → MnO 42–, f eq (KMnO 4) = 1. The mass of one mole of chemical equivalent is defined as equivalent molar mass (Meq), g/mol. The equivalent molar mass relates to the molar mass of a substance as follows: M eq = M ×feq , where f eq is the factor of equivalence. The equivalent mass of an acid or base is the mass of the compound that reacts with or contains one mole of protons. Thus, the equivalent mass of KOH (56.11 g/mol) is equal to its molar mass. For Ba(OH) 2, it is the molar mass divided by 2. H3PO 4 + NaOH → NaH 2PO 4 + H 2Ofeq = 1 f eq =1 H3PO 4 + 2NaOH → Na 2HPO 4 + 2H 2Ofeq = 1/2 f eq = 1 H3PO 4 + 3NaOH → Na 3PO 4 + 3H 2O feq = 1/3 f eq = 1 Al(OH) 3 + 3HCl → AlCl 3 + 3H 2Ofeq = 1/3 f eq = 1 The equivalent factor for oxidizing and reducing agents into redox reactions is calculated as: feq = 1/Z, where Z is an amount of electrons gained or lost by one mole of a substance. For example: MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2Ofeq = 1/2 f eq = 1 The law of equivalence: the masses of chemical substances which are involved into a reaction and masses of its products are directly proportional to their equivalent molar masses. There's a transition between molar concentration of equivalent and titer: Where: E - solute's equivalent. Oxidizer's (reducer's) equivalent is calculated using this formula: Where: M - compound's molar mass; e - attached by oxidizer or released by reducer electron number.
189102	https://apcentral.collegeboard.org/media/pdf/ap23-apc-calculus-ab-q6.pdf	2023 AP ® Calculus AB Sample Student Responses and Scoring Commentary © 2023 College Board. College Board, Advanced Placement, AP, AP Central, and the acorn logo are registered trademarks of College Board. Visit College Board on the web: collegeboard.org. AP Central is the official online home for the AP Program: apcentral.collegeboard.org. Inside: Free-Response Question 6 R 5 Scoring Guidelines R 5 Student Samples R 5 Scoring Commentary AP® Calculus AB/BC 2023 Scoring Guidelines © 2023 College Board Part B (AB): Graphing calculator not allowed Question 6 9 points General Scoring Notes The model solution is presented using standard mathematical notation. Answers (numeric or algebraic) need not be simplified. Answers given as a decimal approximation should be correct to three places after the decimal point. Within each individual free-response question, at most one point is not earned for inappropriate rounding. Consider the curve given by the equation 3. 6 2 xy y = + Model Solution Scoring (a) Show that 2 2 . 2 dy y dx y x = − ( ) ( ) 3 2 6 6 3 6 2 dy dy y x y dx dx d d xy y dx dx ⇒ + = = + Implicit differentiation 1 point ( ) 2 2 2 2 2 2 dy dy y y y x dx dx y x ⇒ = − ⇒ = − Verification 1 point Scoring notes: • The first point is earned only for the correct implicit differentiation of 3. 6 2 xy y = + Responses may use alternative notations for , dy dx such as . y′ • The second point cannot be earned without the first point. • It is sufficient to present ( ) 2 2 2 dy y y x dx = − to earn the second point, provided there are no subsequent errors. Total for part (a) 2 points (b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists. For the line tangent to the curve to be horizontal, it is necessary that 2 0 y = (so 0 y = ) and that 2 2 0. y x − ≠ Sets 2 0 y = 1 point Substituting 0 y = into 3 6 2 xy y = + yields the equation 6 0 2, x ⋅ = which has no solution. Therefore, there is no point on the curve at which the line tangent to the curve is horizontal. Answer with reason 1 point AP® Calculus AB/BC 2023 Scoring Guidelines © 2023 College Board Scoring notes: • The first point is earned with any of 2 0, y = 0, y = 0, dy dx = 0, dy = 0, y′ = or 2 2 0. 2 y y x = − • A response need not state that at a horizontal tangent, 2 2 0. y x − ≠ Total for part (b) 2 points (c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists. For a line tangent to this curve to be vertical, it is necessary that 2 0 y ≠ and that 2 2 0 y x − = (so 2 2 y x = ). Sets 2 2 0 y x − = 1 point Substituting 2 2 y x = into 3 6 2 xy y = + yields the equation 2 3 3 3 2 2 2 1. y y y y y = + ⇒ = ⇒ = ⋅ Substitutes 2 2 y x = into 3 6 2 xy y = + 1 point Substituting 1 y = in 3 6 2 xy y = + yields 6 2 1, x = + or 1 . 2 x = The tangent line to the curve is vertical at the point ( ) 1 , 1 . 2 Answer 1 point Scoring notes: • The first point can be earned by presenting 2 2 y x = or 2 . y x = • The second point can be earned for the substitution of 2 y x = into 3 6 2 , xy y = + or for substituting 3 2 6 x y y = + into 2 2 0. y x − = • A response earns all three points by setting 2 2 0, y x − = declaring the point ( ) 1 , 1 , 2 and verifying that this point is on the curve 3 6 2 . xy y = + • A response that identifies the point ( ) 1 , 1 2 but does not verify that the point is on the curve, does not earn the second or the third point. • To earn the third point the response must present both coordinates of the point ( ) 1 , 1 . 2 The coordinates need not appear as an ordered pair as long as they are labeled. Total for part (c) 3 points (d) A particle is moving along the curve. At the instant when the particle is at the point ( ) 1 , 2 , 2 − its horizontal position is increasing at a rate of 2 3 dx dt = unit per second. What is the value of , dy dt the rate of change of the particle’s vertical position, at that instant? AP® Calculus AB/BC 2023 Scoring Guidelines © 2023 College Board 2 6 6 0 3 dy dy dx y x y dt dt dt + = + Uses implicit differentiation with respect to t 1 point At the point ( ) ( ) 1 , , 2 , 2 x y = − ( )( ) ( ) ( )2 2 1 6 2 6 3 2 3 2 dy dy dt dt − + = − 8 3 12 dy dy dt dt ⇒− + = 8 9 dy dt ⇒ = − unit per second Answer 1 point Scoring notes: • The first point is earned by presenting one or more of the terms 6 , y dx dt 6 , dy x dt or 2 3 . y dy dt • Units will not affect scoring in this part. • An unsupported response of 8 9 − earns no points. • Alternate solution: t dy dy dt d d x dx = ⋅ ( ) ( ) ( ) ( ) ( ) , 1 2, 2 2 2 2 4 3 2 2 1 2 x y dy dx = − − = = − − − ( ) ( ) ( ) ( ) , 1 2, 2 , 1 2, 2 4 2 8 3 3 9 x y x y dx d d t y dy dt dx = − = − ⋅ = = ⋅ = − − unit per second o The first point is earned for the statement t dy dy dt d d x dx = ⋅ or equivalent. o A numerical expression, such as 4 2 3 3 − ⋅ or ( ) ( ) ( ) 2 2 2 2 , 1 3 2 2 2 − ⋅ − − earns both points. Total for part (d) 2 points Total for question 6 9 points 1 of 2 Sample 6A 2 of 2 Sample 6A 1 of 2 Sample 6B 2 of 2 Sample 6B 1 of 2 Sample 6C 2 of 2 Sample 6C AP® Calculus AB 2023 Scoring Commentary © 2023 College Board. Visit College Board on the web: collegeboard.org. Question 6 Note: Student samples are quoted verbatim and may contain spelling and grammatical errors. Overview This problem asked students to consider the curve defined by the equation 3 6 2 . xy y = + In part (a) students were asked show that 2 2 . 2 dy y dx y x = − A correct response will implicitly differentiate the equation 3 6 2 xy y = + with respect to , x then solve the resulting equation for . dy dx In part (b) students were asked to find the coordinates of a point on the curve at which the tangent line is horizontal, or to explain why no such point exists. A correct response will note that a horizontal tangent line must have 0, dy dx = which requires 2 0 y = and, therefore, 0. y = But if 0, y = using the given equation 3 6 2 xy y = + yields 6 0 2, x ⋅ = which has no solution. Therefore, there is no point on this curve at which the tangent line is horizontal. In part (c) students were asked to find the coordinates of a point on the curve at which the tangent line is vertical, or to explain why no such point exists. A correct response will begin by noting that such a point requires 2 2 2 0 . 2 y y x x − = ⇒ = Substituting into the equation 3 6 2 xy y = + yields 1 y = and then 1 , 2 x = resulting in a vertical tangent line at the point ( ) 1 , 1 . 2 In part (d) students were asked to find the value of dy dt at the instant when the particle is at the point ( ) 1 , 2 , 2 − given that at that instant the particle’s horizontal position is increasing at a rate of 2. 3 dx dt = A correct response will implicitly differentiate the equation 3 6 2 xy y = + with respect to t and then solve the resulting equation for dy dt using 1 , 2, 2 x y = = − and 2. 3 dx dt = Sample: 6A Score: 9 The response earned 9 points: 2 points in part (a), 2 points in part (b), 3 points in part (c), and 2 points in part (d). In part (a) the response earned the first point on the second line with the equation 2 6 6 3 , dy dy y x y dx dx + = the correct implicit differentiation for the given curve. The response correctly solves for dy dx on the third line and then earned the second point on the last line with the boxed equation. AP® Calculus AB 2023 Scoring Commentary © 2023 College Board. Visit College Board on the web: collegeboard.org. Question 6 (continued) In part (b) the response earned the first point on the second line with the equation 2 0. y = The response earned the second point with the correct answer that “there are no points with a horizontal tangent,” together with reason that “there are no x -values at 0. y = ” In part (c) the response earned the first point on the second line with the equation 2 2 0. y x − = The response then earned the second point on the fourth line with the correct substitution. The response would have earned the third point on the seventh line with the statements 1 y = and 1 ; 2 x = however, the response restates the answer correctly as an ordered pair and earned the point with the boxed section on the last line. In part (d) the response earned the first point on the second line with the correct implicit differentiation of the curve with respect to . t The response would have earned the second point with the middle expression on the fourth line; however, the response presents two correct simplifications of this numerical answer and earned the point with the boxed answer. Sample: 6B Score: 4 The response earned 4 points: 2 points in part (a), 1 point in part (b), no points in part (c), and 1 point in part (d). In part (a) the response earned the first point with the equation on the second line. The response correctly solves for dy dx on the subsequent three lines and earned the second point on the last line with the final simplification. In part (b) the response earned the first point on the first line with the equation 2 2 0 . 2 y y x = − The response then concludes that 0, y = which would also have earned the first point. The response did not earn the second point as there is no conclusion stating that no point exists. In part (c) the response did not earn the first point as there is no evidence that the denominator of our presented dy dx has been set equal to 0. The response does not indicate any substitution, so it did not earn the second point. Finally, as the correct point is not presented, the response did not earn the third point. In part (d) the response states that 2 2 3 6 6 dy dy dx y y x dy dt dt + = + on the third line. While this is not the correct implicit differentiation of the given curve with respect to , t because at least one of the three terms involving the rates dx dt or dy dt is correct, the response earned the first point on this line. The response did not earn the second point because the answer presented is not correct. Sample: 6C Score: 2 The response earned 2 points: no points in part (a), 1 point in part (b), 1 point in part (c), and no points in part (d). In part (a) the response did not earn the first point as no correct implicit differentiation of the given curve is presented, except for the one that was given in the stem of the problem. Because the first point was not earned, the response did not earn the second point. AP® Calculus AB 2023 Scoring Commentary © 2023 College Board. Visit College Board on the web: collegeboard.org. Question 6 (continued) In part (b) the response earned the first point with the equation 2 0. y = Note that the next equation 0 y = would also have earned this point. As the response never concludes that no such point exists, the response did not earn the second point. In part (c) the response earned the first point on the first line with the equation 2 2 0. y x − = Note that any of the first three lines would have earned this first point. As there is no substitution presented and the correct answer is not stated, the response earned neither the second point nor the third point. In part (d) the response states on the first line that 2 2 2 dy y dt y x = − and then uses this expression as the basis for substitution with the given values for x and . y If this expression is viewed as the implicit differentiation of the curve with respect to , t then the response does not present at least one correct term with a rate dx dt or . dy dt If, on the other hand, this expression is meant to be , dy dx then the solution presented never makes use of . dx dt In either case, the response did not earn the first point. As the stated answer is incorrect, the response did not earn the second point.
189103	https://www.quora.com/For-mathematical-induction-for-proving-true-for-n-k-1-we-assume-true-for-n-k-What-if-n-k-is-actually-false	Something went wrong. Wait a moment and try again. Mathematical Sciences Proof by Induction Proofs (mathematics) Number Theory, Induction Mathematical Reasoning Mathematical Proof Induction Method Math Science 5 For mathematical induction, for proving true for n = k +1, we assume true for n = k. What if n = k is actually false? Charles S. former mathematician, current patent lawyer · Author has 7.6K answers and 60M answer views · 4y That’s why you prove the proposition true for n = 1 (or wherever you feel like starting). If you know the proposition is true for n = 1 , and you also know that if it’s true for n = k then it’s true for n = k + 1 , then you also know it’s true for 2. And 3. And 4. You start from n = 1 and keep applying your conditional ‘till the cows come home. Martin Jansche works in any number system, as long as it's base 10 · Author has 3.7K answers and 3.6M answer views · Updated 4y For the inductive step you have to prove a conditional. For that you get to assume that the statement holds for a given n. It typically doesn't matter for the inductive step whether that assumption is ever true. If you can actually derive a contradiction from the inductive assumption, you'll have proven the inductive lemma, since a contradiction allows you to conclude whatever you want. In other words, proving the inductive step is no different from proving any other conditional. The truth of the premise is simply not at issue. As the other answers have pointed out, you still need to prove a bas For the inductive step you have to prove a conditional. For that you get to assume that the statement holds for a given n. It typically doesn't matter for the inductive step whether that assumption is ever true. If you can actually derive a contradiction from the inductive assumption, you'll have proven the inductive lemma, since a contradiction allows you to conclude whatever you want. In other words, proving the inductive step is no different from proving any other conditional. The truth of the premise is simply not at issue. As the other answers have pointed out, you still need to prove a base case that will allow the induction to kick off. There you actually have to prove the statement to be true for some specific value of n. Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Related questions Can I assume in induction proof that n-1 is true (induction hypothesis) or do I need to substitute by assuming let n=k-1 true? Does such a statement exist in 'mathematical induction' which is false for n=1, but true for n=k+1 whenever it is true for n=k? How do you prove that by principle of mathematical induction, n sigma I=1 I(i!) = (n+1)! -1 for all integers n>1? [How can one prove by induction that ∑ n k = 1 k 3 = [n^2(n+1)^2]/4?]( How do you prove that by principle of mathematical induction, n Summission I=1 I(i!) = (n+1)! -1 for all integers n>1? Kwok Choy Yue B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Author has 1.4K answers and 1.4M answer views · Aug 9 Related How do I show via mathematical induction that for every natural number n the following is true: n ∑ k = 1 1 k ( k + 1 ) ( k + 2 ) = n ( n + 3 ) 4 ( n + 1 ) ( n + 2 ) ? () M.I. method and Difference method: Enjoy mathematics! () M.I. method and Difference method: Enjoy mathematics! Alexander Farrugia Loves numbers. · Author has 3.2K answers and 27.4M answer views · 7y Related Does such a statement exist in 'mathematical induction' which is false for n=1, but true for n=k+1 whenever it is true for n=k? Suppose we have the statement “3n>4n for all positive integers n.” This statement is false when n=1, since . However, it is true for all other positive integers. In that case, you can still prove this statement by induction for all integers [math]n\geq 2[/math]. (In fact, it is a nice exercise, if you want to try it out.) The base case would be [math]n=2[/math] instead of [math]n=1[/math]. Before closing, I wish to clarify that the inductive step is not ‘prove the statement for [math]n=k+1[/math] assuming it is true for [math]n=k[/math]”. The inductive step is ‘assuming that the statement is true for some [math]n=k[/math], where [math]k[/math] is one of the values we want to Suppose we have the statement “[math]3^n > 4n[/math] for all positive integers [math]n[/math].” This statement is false when [math]n=1[/math], since [math]3\ngtr 4[/math]. However, it is true for all other positive integers. In that case, you can still prove this statement by induction for all integers [math]n\geq 2[/math]. (In fact, it is a nice exercise, if you want to try it out.) The base case would be [math]n=2[/math] instead of [math]n=1[/math]. Before closing, I wish to clarify that the inductive step is not ‘prove the statement for [math]n=k+1[/math] assuming it is true for [math]n=k[/math]”. The inductive step is ‘assuming that the statement is true for some [math]n=k[/math], where [math]k[/math] is one of the values we want to prove the statement for, show that it is also true for [math]n=k+1[/math]’. The words in italics are important. Assistant Bot · 1y In mathematical induction, the process consists of two main steps: Base Case: Prove that the statement is true for the initial value, usually [math]n = 1[/math] or [math]n = 0[/math]. Inductive Step: Assume that the statement is true for some arbitrary positive integer [math]n = k[/math] (this is called the inductive hypothesis), and then prove that the statement is also true for [math]n = k + 1[/math]. If the statement is false for [math]n = k[/math], the inductive step fails. This means that the assumption you made in the inductive hypothesis is incorrect, and therefore, you cannot conclude that the statement is true for [math]n = k + 1[/math]. Here’s how it works in pra In mathematical induction, the process consists of two main steps: Base Case: Prove that the statement is true for the initial value, usually [math]n = 1[/math] or [math]n = 0[/math]. Inductive Step: Assume that the statement is true for some arbitrary positive integer [math]n = k[/math] (this is called the inductive hypothesis), and then prove that the statement is also true for [math]n = k + 1[/math]. If the statement is false for [math]n = k[/math], the inductive step fails. This means that the assumption you made in the inductive hypothesis is incorrect, and therefore, you cannot conclude that the statement is true for [math]n = k + 1[/math]. Here’s how it works in practice: If your base case is true (let's say you prove the statement for [math]n = 1[/math] is true), but then you find that the statement is false for [math]n = k[/math], you cannot use this to prove the statement for [math]n = k + 1[/math]. The validity of the inductive step relies on the correctness of the inductive hypothesis. If [math]n = k[/math] is false, then the entire argument for 40 n = k + 1 \ collapses. In summary, for mathematical induction to be valid, both the base case must be true and the inductive step must correctly follow from the inductive hypothesis. If either fails, the proof by induction is not valid. Sponsored by Mutual of Omaha Retiring soon and need Medicare advice? Be prepared for retirement with a recommendation from our Medicare Advice Center. Related questions How can you prove by mathematical induction that n) = (−5n+3>0 whenever n is an integer greater than 4? How can I prove this using Mathematical Induction r^2 =n÷6(n+1) (2n + 1)? How do you prove that a(n) = ((a(n-1) +(n-1) a(n-2)) /n is true for k+1 by using mathematical induction for n>= 2, where a(0) =0, a(1) =1? How can one prove by mathematical induction that 2^(n-1) ≤n!? How do you show that by mathematical induction 2^ less than or equal (n+1)! N>=1? Doug Dillon Ph.D. Mathematics · Author has 12.3K answers and 11.4M answer views · 4y Here is a claim which we “prove” by induction. Claim: For all positive n, n+1<n. “Proof”: Suppose n+1<n. Now look at (n+1)+1, which by the induction is less than n+1. But!!! Have we proven the claim by induction? No. An important step has been omitted. We did not check to see if the statement was true for at least some value of n. However, all is not lost. We have proven something by induction — the truth of the n-statement. Perhaps, with an initial case considered, we have proven that a statement is false. Archie Quester Lives in Oceania (1901–present) · 6y Originally Answered: When proving by induction, step two requires assuming true for n=k. What if it isn't true for n=k? · Q: “When proving by induction, step two requires assuming true for n=k. What if it isn't true for n=k?” A: Step two requires assuming the result for n=k, for some integer k. If that assumption is true, the proof has a possibility of succeeding, either by continuing with the method of induction or another way. NB: In some cases the assumption is true but the general result is not. If that assumption is NOT true, the proof has no possibility of succeeding. In that case, we may wish to prove the result is false for some integer k, and that would be sufficient to disprove the posited general result. 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For [math]n=0[/math] the left-hand side is an empty sum and the right-hand side has a positive denominator and a numerator equal to zero, so your identity is satisfied. (Yes, zero is a natural number. Yes, your Calculus textbook is wrong. But: sure, you may check that [math]1 / (1 \cdot 2 \cdot 3) = (1 \cdot 4) / (4 \cdot 2 \cdot 3)[/math].) Now assume that your identity is true for a specific [math]n \geq 0[/math]. (Or [math]n \geq 1[/math], if you did the check for [math]n=1[/math].) Then: [math]\displaystyle \sum_{k=1}^{n+1} \frac{1}{k(k+1)(k+2)} = \left( \sum_{k=1}^n \frac{1}{k(k+1)(k+2)} \right) + \frac{1}{(n+1)(n+2)(n+3)}[/math] by definition [math]\displaystyle = \frac{ n(n[/math] For [math]n=0[/math] the left-hand side is an empty sum and the right-hand side has a positive denominator and a numerator equal to zero, so your identity is satisfied. (Yes, zero is a natural number. Yes, your Calculus textbook is wrong. But: sure, you may check that [math]1 / (1 \cdot 2 \cdot 3) = (1 \cdot 4) / (4 \cdot 2 \cdot 3)[/math].) Now assume that your identity is true for a specific [math]n \geq 0[/math]. (Or [math]n \geq 1[/math], if you did the check for [math]n=1[/math].) Then: [math]\displaystyle \sum_{k=1}^{n+1} \frac{1}{k(k+1)(k+2)} = \left( \sum_{k=1}^n \frac{1}{k(k+1)(k+2)} \right) + \frac{1}{(n+1)(n+2)(n+3)}[/math] by definition [math]\displaystyle = \frac{ n(n+3) }{ 4(n+1)(n+2) } + \frac{1}{(n+1)(n+2)(n+3)}[/math] by inductive hypothesis [math]\displaystyle = \frac{1}{(n+1)(n+2)}\left( \frac{n(n+3)}{4} + \frac{1}{n+3} \right)[/math] [math]\displaystyle = \frac{1}{(n+1)(n+2)} \cdot \frac{ n^3 + 6n^2 + 9n + 4}{ 4(n+3) }[/math] by reducing to common denominator. Here you may want to make a break, because you have an inconvenient [math]n+1[/math] in the denominator, which you want to make disappear, because your denominator should be [math]4(n+2)(n+3)[/math], that is, [math]4(n+1)(n+2)[/math] with [math]n+1[/math] in place of [math]n[/math]. May it be that the numerator in the second fraction has a factor [math]n+1[/math]? Yes, it has, because for [math]n = -1[/math] you get: math^3 + 6 \cdot (-1)^2 + 9 \cdot (-1) + 4 = -1 + 6 - 9 + 4 = 0[/math] As this point, you can either use Ruffini for polynomial division, or look for three constants [math]a,b,c[/math] such that: [math]n^3 + 6n^2 + 9n + 4 = (n+1) (an^2 + bn + c)[/math] and solve a system of three linear equations in three unknowns. This is what I usually do; in this case, I get: [math]n^3 + 6n^2 + 9n + 4 = (n+1) (n^2 + 5n + 4)[/math] The second-degree polynomial is easy to factorize, so from here it’s all downhill. Jan van Delden MSc Math and still interested · Author has 4.8K answers and 6.5M answer views · Aug 7 Related How do I show via mathematical induction that for every natural number [math]n[/math] the following is true: [math]\displaystyle \sum_{k = 1}^{n}\dfrac{1}{k(k+1)(k + 2)} = \dfrac{n(n + 3)}{4(n + 1)(n + 2)}[/math] ? A derivation is not too hard, manipulate the terms to [math]\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2} \dfrac{1}{k+1} \left( \dfrac{1}{k} - \dfrac{1}{k+2}\right)[/math] Reintroduce the factor outside of the parentheses and manipulate both halves [math]\dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2} \left( \left( \dfrac{1}{k}- \dfrac{1}{k+1}\right) -\left(\dfrac{1}{k+1}-\dfrac{1}{k+2}\right)\right)[/math] Both terms inside the inner parentheses define a Telescoping series. [math]\displaystyle \sum_{k=1}^n \dfrac{1}{k}-\dfrac{1}{k+1} = 1 - \dfrac{1}{n+1}[/math] [math]\displaystyle \sum_{k=1}^n \dfrac{1}{k+1}-\dfrac{1}{k+2} = \dfrac{1}{2} - \dfrac{1}{n+2}[/math] We fin [math]\[/math] A derivation is not too hard, manipulate the terms to [math]\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2} \dfrac{1}{k+1} \left( \dfrac{1}{k} - \dfrac{1}{k+2}\right)[/math] Reintroduce the factor outside of the parentheses and manipulate both halves [math]\dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2} \left( \left( \dfrac{1}{k}- \dfrac{1}{k+1}\right) -\left(\dfrac{1}{k+1}-\dfrac{1}{k+2}\right)\right)[/math] Both terms inside the inner parentheses define a Telescoping series. [math]\displaystyle \sum_{k=1}^n \dfrac{1}{k}-\dfrac{1}{k+1} = 1 - \dfrac{1}{n+1}[/math] [math]\displaystyle \sum_{k=1}^n \dfrac{1}{k+1}-\dfrac{1}{k+2} = \dfrac{1}{2} - \dfrac{1}{n+2}[/math] We find [math]\displaystyle \sum_{k=1}^n \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2}\left( \dfrac{1}{2} - \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\right)=\dfrac{1}{4}\left( 1 - \dfrac{2}{(n+1)(n+2)}\right) = \dfrac{n(n+3)}{4(n+1)(n+2)} [/math] Related questions Can I assume in induction proof that n-1 is true (induction hypothesis) or do I need to substitute by assuming let n=k-1 true? Does such a statement exist in 'mathematical induction' which is false for n=1, but true for n=k+1 whenever it is true for n=k? How do you prove that by principle of mathematical induction, n sigma I=1 I(i!) = (n+1)! -1 for all integers n>1? [How can one prove by induction that ∑ n k = 1 k 3 = [n^2(n+1)^2]/4?]( How do you prove that by principle of mathematical induction, n Summission I=1 I(i!) = (n+1)! -1 for all integers n>1? How can you prove by mathematical induction that n) = (−5n+3>0 whenever n is an integer greater than 4? How can I prove this using Mathematical Induction r^2 =n÷6(n+1) (2n + 1)? How do you prove that a(n) = ((a(n-1) +(n-1) a(n-2)) /n is true for k+1 by using mathematical induction for n>= 2, where a(0) =0, a(1) =1? How can one prove by mathematical induction that 2^(n-1) ≤n!? How do you show that by mathematical induction 2^ less than or equal (n+1)! N>=1? How do you prove $\sum_ {I=1} ^n i.i! = (n+1)! -1$ with mathematical induction? What is the principle mathematical induction of n<n^2 for every n? How can you prove the following by mathematical induction ∑I=1 n+1 I(i+1) =n(n-1) (n+1) /3 , for all integers n>=2? How can we prove that by mathematical induction, n^2=4? Could you please prove that (3n)! Is greater than n^(3n) by using mathematical induction? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
189104	https://brainly.com/question/36050954	[FREE] The function f(x)=(x-a)(x-b) is defined above, where a and b are integer constants. If f(1) > 0, f(4) < 0, - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +21,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +40k Ace exams faster, with practice that adapts to you Practice Worksheets +7,7k Guided help for every grade, topic or textbook Complete See more / Computers and Technology Textbook & Expert-Verified Textbook & Expert-Verified The function f(x)=(x−a)(x−b) is defined above, where a and b are integer constants. If f(1)>0, f(4)<0, and f(7)>0, what is one possible value of a+b? 2 See answers Explain with Learning Companion NEW Asked by mariyahlynne8263 • 08/19/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 65288027 people 65M 2.5 0 Upload your school material for a more relevant answer One possible value of a + b is 8. Explanation To find the value of a + b, we need to analyze the behavior of the quadratic function f(x) = (x - a)(x - b) based on the given conditions. When f(1) > 0, it means that the function is positive at x = 1. This implies that either both factors (x - a) and (x - b) are positive or both factors are negative. Since a and b are integer constants, we can conclude that either a > 1 and b > 1 or a < 1 and b < 1. When f(4) < 0, it means that the function is negative at x = 4. This implies that one factor is positive and the other factor is negative. Since a and b are integer constants, we can conclude that either a < 4 and b > 4 or a > 4 and b < 4. When f(7) > 0, it means that the function is positive at x = 7. This implies that either both factors (x - a) and (x - b) are positive or both factors are negative. Since a and b are integer constants, we can conclude that either a > 7 and b > 7 or a < 7 and b < 7. Based on these conditions, we can determine the possible values of a and b that satisfy the given conditions. One possible value of a + b is when a = 2 and b = 6, which gives a + b = 2 + 6 = 8. Learn more about finding the value of a + b in a quadratic function here: brainly.com/question/33780891 SPJ14 Answered by komalnavarre23 •79K answers•65.3M people helped Thanks 0 2.5 (2 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 29635011 people 29M 0.0 0 Statistics - Barbara Illowsky, Susan Dean The Foundations of Geometry - David Hilbert Simple nature - Benjamin Crowell Upload your school material for a more relevant answer One possible value of a + b is 8, derived from analyzing the different conditions of the quadratic function f(x) = (x - a)(x - b). The conditions led to choosing integer values a = 2 and b = 6, satisfying the inequality conditions. Therefore, a + b = 2 + 6 = 8. Explanation To find the value of a + b for the function f(x)=(x−a)(x−b), we need to analyze the given conditions: f(1)>0, f(4)<0, and f(7)>0. Condition f(1)>0: This means that the function is positive at x=1. For this to happen, both factors must be either positive or negative at this point. Since both a and b are integers, it implies that both a<1 and b<1 if they are negative or both a>1 and b>1 if they are positive. Condition f(4)<0: Here, we know that the function is negative at x=4. This means one of the factors must be positive and the other must be negative. Therefore, one of a or b must be less than 4 and the other greater than 4. Condition f(7)>0: For the function to be positive at x=7, both factors need to be positive, so we will have a<7 and b<7 or a>7 and b>7. Now, let's combine these conditions: Since f(4)<0 indicates one root must be below 4 and one above, we can choose the roots such that a<4<b. The inequality from f(1)>0 suggests both roots satisfy a<1 or both satisfy a>1. Given a<4 and the positivity at 1, we finalize it as b must be greater than 4. The condition f(7)>0 ensures both roots are below 7. From this, possible integer values for a and b satisfying all conditions include a=2 and b=6. Therefore, the sum would be a+b=2+6=8. Thus, one possible value of a+b is 8. Examples & Evidence For example, if a = 2 and b = 6, then f(1) gives a positive value (since both factors are positive), f(4) gives a negative value, and f(7) gives a positive value. This shows the function behaves as required in the problem statement. The values were chosen based on the critical points indicated by the inequalities which stem from the definitions of the quadratic function and their implications at specific x-values. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 29635011 people 29M 0.0 0 One possible value of a + b is 8. Explanation To find the value of a + b, we need to analyze the behavior of the quadratic function f(x) = (x - a)(x - b) based on the given conditions. When f(1) > 0, it means that the function is positive at x = 1. This implies that either both factors (x - a) and (x - b) are positive or both factors are negative. Since a and b are integer constants, we can conclude that either a > 1 and b > 1 or a < 1 and b < 1. When f(4) < 0, it means that the function is negative at x = 4. This implies that one factor is positive and the other factor is negative. Since a and b are integer constants, we can conclude that either a < 4 and b > 4 or a > 4 and b < 4. When f(7) > 0, it means that the function is positive at x = 7. This implies that either both factors (x - a) and (x - b) are positive or both factors are negative. Since a and b are integer constants, we can conclude that either a > 7 and b > 7 or a < 7 and b < 7. Based on these conditions, we can determine the possible values of a and b that satisfy the given conditions. One possible value of a + b is when a = 2 and b = 6, which gives a + b = 2 + 6 = 8. Learn more about finding the value of a + b in a quadratic function here: brainly.com/question/33780891 SPJ14 Answered by bajajsakshi1995 •69.7K answers•29.6M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Computers and Technology solutions and answers Community Answer 5.0 1 What can quantum computers do more efficiently than regular computers? Community Answer 1 You are experiencing issues when trying to transfer files between two computers using FTP. What could be the potential cause(s) of the issue Community Answer 4.5 117 describe what happens at every step of our network model, when a node on one network establishes a TCP connection with a node on another network. You can assume that the two networks are both connected to the same router. Your submission must include a detailed explanation of the following: Physical layer Data link layer Network layer Transport layer MAC address IP address TCP port Checksum check Routing table TTL Community Answer 19 To return the value of the cell D8, the formula should be OFFSETA1=__. Community Answer 4 Which tools would you use to make header 1 look like header 2. New questions in Computers and Technology What is the purpose of the "Merge and Centre"? How are cell addresses or cell references typically formatted? What are the three types of data that a cell can contain? Choose the term that is described by the following: 1. large data centers 2. use of the Internet to access programs and data on computers that are not owned and managed by the user often using 3. states that processing power for computers would double every two years 4. uses biological components like DNA to retrieve, process, and store data 5. the anticipated next generation of technologies that are expected to drastically increase processing capabilities As a Lead System Architect at a healthcare organization, you are tasked with enhancing decision-making and user interactions in the Pega Infinity patient care management applications that the organization uses. How can you best apply Pega GenAI to achieve these objectives? 1. Employ Pega GenAI Coach to eliminate manual data entry by predicting and autopopulating all required fields in patient care Cases. 2. Configure Pega GenAI Coach to deliver real-time, context-aware guidance to users, which supports informed decisions during patient Case processing. 3. Deploy Pega GenAI Coach to autonomously resolve all patient care cases without user intervention and provide optimal outcomes. 4. Use Pega GenAI Coach exclusively to monitor user performance metrics and generate analytical reports for administrative oversight. Assume a SQL Server database contains tables Employee and Department. The Employee table has a deptid column that is a foreign key referencing the Department table. What happens when a new employee is inserted in the Employee table with a deptid that is not present in the Department table? A) An employee gets inserted in the Employee table with deptid as NULL. B) The query will result in a foreign key violation error. C) An employee gets inserted in the Employee table with deptid as Default value. D) The query will result in a primary key violation error. 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189105	https://sciencenotes.org/mole-fraction-formula-and-calculation/	Home » Science Notes Posts » Chemistry » Chemistry Notes » Mole Fraction Formula and Calculation Mole Fraction Formula and Calculation This entry was posted on by Anne Helmenstine (updated on ) In chemistry, the mole fraction is a unit of concentration that is the number of moles of a component divided by the total number of moles of a solution or mixture. The mole fraction is a dimensionless number. The sum of all of the mole fractions equals 1. The symbol for mole fraction is the capital letter X or the lowercase Greek letter chi (χ). The terms “amount fraction” or “amount-of-substance fraction” mean the same as mole fraction. Mole Fraction Formula The formula for mole fraction is the moles of one component divided by the total number of moles: XA = moles A / total moles How to Calculate Mole Fraction For example, in a mixture consisting of 0.25 moles of component A and 0.40 moles of component B, you can find the mole fractions of A and B. XA = moles A / total moles = 0.25 / (0.25 + 0.40) = 0.38 (rounded) XB = moles B / total moles = 0.40 / (0.25 + 0.40) = 0.62 (rounded) Remember, adding up the mole fractions equals 1. XA + XB = 1 0.38 + 0.62 = 1 If the mixture consists of more than two components, the same rules apply. Mole Percent A related term is mole percent. Mole percent or mole percentage is the mole fraction multiplied by 100%. mol% = XA x 100% The sum of all of the mole percents of a mixture equals 100% Mole Fraction Properties and Advantages The mole fraction offers advantages over some of the other units of concentration. Unlike molarity, mole fraction is not temperature dependent. Preparing a solution using mole fraction is easy because you simply weigh the masses of the components and then combine them. There is no confusion over which component is the solvent and which is the solute. The unit is symmetric in this respect because the roles of solute and solvent are reversible, depending on the mole fraction. In a mixture of ideal gases or most real gases, the mole fraction is the same as the ratio of partial pressure of a gas to the total pressure of the mixture. In other words, mole fraction follows Dalton’s law of partial pressure. Example Calculations Simple Example For example, find the mole fraction of carbon tetrachloride in a mixture consisting of 1 mole benzene, 2 moles carbon tetrachloride, and 7 moles acetone. XA = moles A / total moles XCCl4 = 2 / (1 + 2 + 7) = 2/10 = 0.2 Mole Fraction From Grams Find the mole fraction of formaldehyde (CH2O) when you dissolve 25.7 grams of CH2O in 3.25 moles of carbon tetrachloride (CCl4). Here, the amount of CCl4 is already in moles, but you can’t find mole fraction until you convert grams of CH2O into moles, too. Look up the atomic masses of carbon, hydrogen, and oxygen on the periodic table and use the chemical formula for formaldehyde to calculate the number of moles. 1 mole CH2O = 12.01 g + 2×1.01 g + 16.00 g = 30.03 g Use this relationship and find the number of moles of CH2O. moles CH2O = 25.7 g x (1 mol/30.03 g) = 0.856 mol Now, solve for mole fraction. XA = moles A / total moles XA = 0.856 moles CH2O / (0.856 moles CH2O + 3.25 moles CCl4) = 0.208 How to Find Mole Fraction From Molality Molality (m) is the moles of solute per kilogram of solvent. Using these units, you can calculate mole fraction if you know molality. For example, find the mole fraction of table sugar or sucrose (C6H12O6) in a 1.62 m solution of sucrose in water. Given the definition of molality, you know the following: 1.2 m sucrose = 1.62 moles sucrose / 1 kg water Next, find how many moles there are of water. Use the atomic masses from the periodic table and find that the molar mass of water is 18.0 (2×1.01 + 16.00). 1 kg = 1000 g = 1 mol / 18.0 g = 55.5 moles H2O Knowing the moles of sucrose and the moles of water, find the mole fraction of sucrose. XA = moles A / total moles Xsucrose = moles sucrose / total moles = 1.62 / (1.62 + 55.5) = 0.0284 With small numbers like this, it’s often better expressing mole fraction as mole percent. The solution is 2.84% sugar in water. References IUPAC (1997). “Amount fraction.” Compendium of Chemical Terminology (the “Gold Book”) (2nd ed.). Blackwell Scientific Publications. ISBN 0-9678550-9-8. doi:10.1351/goldbook.A00296 Rickard, James N.; Spencer, George M.; Bodner, Lyman H. (2010). Chemistry: Structure and Dynamics (5th ed.). Hoboken, N.J.: Wiley. ISBN 978-0-470-58711-9. Thompson, A.; Taylor, B. N. (2009). “Special Publication 811.” The NIST Guide for the use of the International System of Units. National Institute of Standards and Technology. Zumdahl, Steven S. (2008). Chemistry (8th ed.). Cengage Learning. ISBN 978-0-547-12532-9. Related Posts Concentration Units For Solutions Chemistry is a science which deals a lot with solutions and mixtures. Knowing just how much of one thing is mixed in with a solution is an important thing to know. Chemists measure this by determining the concentration of the solution or mixture. There are three terms that need to… How to Calculate Concentration of a Chemical Solution Chemists often work with chemical solutions. (Keith Weller, USDA) The concentration of a chemical solution refers to the amount of solute that is dissolved in a solvent. Although it's common to think of a solute as a solid that is added to a solvent (e.g., adding table salt to water),… Dalton’s Law of Partial Pressure – Definition and Examples Learn about Dalton's law of partial pressures. Get the definition, formula, examples, and worked problems for this ideal gas law.
189106	https://tutorial.math.lamar.edu/Solutions/Alg/Hyperbolas/Prob5.aspx	Paul's Online Notes Go To Notes Practice Problems Assignment Problems Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections Ellipses Miscellaneous Functions Chapters Graphing and Functions Polynomial Functions Problems Problem 4 Full Problem List Classes Algebra Calculus I Calculus II Calculus III Differential Equations Extras Algebra & Trig Review Common Math Errors Complex Number Primer How To Study Math Cheat Sheets & Tables Misc Contact Me MathJax Help and Configuration Notes Downloads Complete Book Practice Problems Downloads Complete Book - Problems Only Complete Book - Solutions Assignment Problems Downloads Complete Book Other Items Get URL's for Download Items Print Page in Current Form (Default) Show all Solutions/Steps and Print Page Hide all Solutions/Steps and Print Page Paul's Online Notes Home / Algebra / Common Graphs / Hyperbolas Prev. Section Notes Practice Problems Assignment Problems Next Section Prev. Problem Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width. Section 4.4 : Hyperbolas Back to Problem List Complete the square on the (x) and (y) portions of the equation and write the equation into the standard form of the equation of the hyperbola. [25{y^2} + 250y - 16{x^2} - 32x + 209 = 0] Show All Steps Hide All Steps Start Solution The process here will be is identical to the process we used in the previous section to write equations of ellipses in standard form. The first step is to make sure the coefficient of the ({x^2}) and ({y^2}) is a one. The ({x^2}) has a coefficient of -16 and the ({y^2}) has a coefficient of 25. What we will do is factor a -16 out of every term involving an (x) and a 25 out of ever term involving a (y). Doing that gives, [25\left( {{y^2} + 10y} \right) - 16\left( {{x^2} + 2x} \right) + 209 = 0] Show Step 2 Now let’s get started on completing the square. First, we need one-half the coefficient of the (x) and (y) term, square each and the add/subtract those numbers in the appropriate places as follows, [\require{color}{\left( {\frac{{10}}{2}} \right)^2} = {\left( 5 \right)^2} = \,{\color{Red} 25}\hspace{0.25in}{\left( {\frac{2}{2}} \right)^2} = {\left( 1 \right)^2} = \,{\color{ProcessBlue} 1}] [\require{color}25\left( {{y^2} + 10y \,{\color{Red} + 25 - 25}} \right) - 16\left( {{x^2} + 2x \,{\color{ProcessBlue}+ 1 - 1}} \right) + 209 = 0] Be careful you add/subtract these numbers and make sure you put them in the parenthesis! Show Step 3 Next, we need to factor the (x) and (y) terms and add up all the constants. [\require{color}\begin{align}25\left( {{{\left( {y + 5} \right)}^2} \,{\color{Red} - 25}} \right) - 16\left( {{{\left( {x + 1} \right)}^2} \,{\color{ProcessBlue}- 1}} \right) + 209 & = 0\ 25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} + 209 \,{\color{Red} - 625} \,{\color{ProcessBlue}+ 16} & = 0\ 25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} - 400 & = 0\end{align}] When adding the constants up, make sure to multiply the -16 through the (x) terms and the 25 through the (y) terms before adding the constants up. Show Step 4 To finish things off we’ll first move the 400 to the other side of the equation. [25{\left( {y + 5} \right)^2} - 16{\left( {x + 1} \right)^2} = 400] To get this into standard form we need a one on the right side of the equation. To get this all we need to do is divide everything by 400 and we’ll do a little simplification work on the terms. [\frac{{25{{\left( {y + 5} \right)}^2}}}{{400}} - \frac{{16{{\left( {x + 1} \right)}^2}}}{{400}} = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{{\left( {y + 5} \right)}^2}}}{{16}} - \frac{{{{\left( {x + 1} \right)}^2}}}{{25}} = 1}}] [Contact Me] [Privacy Statement] [Site Help & FAQ] [Terms of Use] © 2003 - 2025 Paul Dawkins Page Last Modified : 11/16/2022
189107	https://brilliant.org/wiki/rational-equations/	Rational Equations Sign up with Facebook or Sign up manually Already have an account? Log in here. Cheolho Han, Thaddeus Abiy, Mahindra Jain, and Aditya Virani Jimin Khim contributed A rational equation is an equation containing at least one fraction whose numerator and denominator are polynomials, Q(x)P(x). These fractions may be on one or both sides of the equation. A common way to solve these equations is to reduce the fractions to a common denominator and then solve the equality of the numerators. While doing this, we have to make sure to note cases where indeterminate forms like 00 or 01 may arise. Contents Rational Equations - Basic Rational Equations - Intermediate Rational Equations - Advanced Rational Equations - Basic Solve x1=2. Looking at the equation, we can see that it's asking which reciprocal gives 2. This is 21 and we can conclude that it is the solution. □ While it is possible to use this inspection method, it is easier to use a more general method. In general, if an equation is in the form of an irreducible proportion ba=dc, one can cross multiply to obtain a polynomial ad−bc=0. This polynomial can then be solved using whatever appropriate method necessary while noting that b=0 and d=0. Solve 3+x2−x=21. Using the cross-multiplying method described above gives 3+x3+x3xx=2(2−x)=4−2x=1=31. □ This method can be extended to any rational equation. However, for expressions with more terms, instead of cross-multiplying we multiply both sides of the equation by the LCM of the denominators. Find all the solutions of x1+1−x2=x11+x(2x+3)3. First note that x=0,x=1, and x=2−3, as they all lead to a zero denominator. When multiplying the whole expression by the LCM of the denominators x(1−x)(2x+3), we get (1−x)(2x+3)+2x(2x+3)−2x2−x+3+4x4+6x24x2+19x−33=11(1−x)(2x+3)+3(1−x)=−22x2−11x=0. Using the quadratic formula to solve this equation, we get x=48−19±3529. □ Solve the equation x−21=81. Multiplying both sides by 8(x−2) gives 8=10=x−2x. Substituting x=10 satisfies the given equation, so the answer is 10. □ Solve the equation 2x+31=x−51. Multiplying both sides by (2x+3)(x−5) gives x−5=−8=2x+3x. Substituting x=−8 satisfies the given equation, so the answer is -8. □ Solve the equation (x+3)(x−2)1=x−61. Multiplying both sides by (x+3)(x−2)(x−6) gives x−6=x−6=0=(x+3)(x−2)x2+x−6x2. Substituting x=0 satisfies the given equation, so the answer is 0. □ Solve the equation x+2x2+3x=x+2−2x−6. Multiplying both sides by x+2 gives x2+3x=x2+5x+6=(x+2)(x+3)=−2x−600. Observe that x=−2 is not a solution because the given equation has zeros in the denominator. Substituting x=−3 satisfies the given equation, so the answer is -3. □ Rational Equations - Intermediate If the equation 2x+1−4x2−4x+a=2x+14x+1 has only one solution, what is a? Multiplying both sides by 2x+1 gives −4x2−4x+a=−4x2−8x+a−1=4x+10. The discriminant is 4D====(−4)2−(−4)(a−1)16+4a−44a+124(a+3). If a=−3, −4x2−8x+a−1===−4x2−8x−4−4(x2+2x+1)−4(x+1)2=0. Substituting x=−1 satisfies the given equation. If a>−3, then −4x2−8x+a−1=0 has two solutions. However, if one solution is −21, the other one solution will be left because substituting x=−21 makes the denominators in the given equation zero. Assume one solution is x=−21. Then −4x2−8x+a−1===−1+4+a−1a+20. If a=−2, then we have only one solution. Therefore, a=−2 or −3. □ Rational Equations - Advanced If the equation x2+7x+10x2+6x=x2+7x+10−2x+a has only one solution, what is a? Multiplying both sides by x2+7x+10 gives x2+6x=x2+8x−a=−2x+a0. The discriminant is 4D=42+a=a+16. If a=−16, [\begin{align} x^2 + 8x - a = x^2 + 8x + 16 = (x+4)^2 . \end{align}] Substituting x=−4 satisfies the given equation. If a>−16, x2+8x−a=0 has two solutions. However, if one solution is -2 or -5, the other one solution will be left because substituting x=−2 or −5 makes the denominators in the given equation zero. Assume one solution is x=−2. Then x2+8x−a===4−16−a−12−a0. If a=−12, then we have only one solution x=−6. Assume one solution is x=−5. Then x2+8x−a===25−40−a−15−a0. If a=−15, then we have only one solution x=−3. Therefore, a=−12,−15,−16. □ Cite as: Rational Equations. Brilliant.org. Retrieved 08:22, July 24, 2025, from
189108	https://pubmed.ncbi.nlm.nih.gov/28155819/	Modelling primaquine-induced haemolysis in G6PD deficiency - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Modelling primaquine-induced haemolysis in G6PD deficiency James Watson12,Walter Rj Taylor12,Didier Menard3,Sim Kheng4,Nicholas J White12 Affiliations Expand Affiliations 1 Mahidol Oxford Tropical Medicine Research Unit, Faculty of Tropical Medicine, Mahidol University, Bangkok, Thailand. 2 Centre for Tropical Medicine and Global Health, Nuffield Department of Medicine, University of Oxford, Oxford, United Kingdom. 3 Unité d'Epidémiologie Moléculaire du Paludisme, Institut Pasteur du Cambodge, Phnom Penh, Cambodia. 4 National Center for Parasitology, Entomology and Malaria Control, Phnom Penh, Cambodia. PMID: 28155819 PMCID: PMC5330681 DOI: 10.7554/eLife.23061 Item in Clipboard Modelling primaquine-induced haemolysis in G6PD deficiency James Watson et al. Elife.2017. Show details Display options Display options Format Elife Actions Search in PubMed Search in NLM Catalog Add to Search . 2017 Feb 4:6:e23061. doi: 10.7554/eLife.23061. Authors James Watson12,Walter Rj Taylor12,Didier Menard3,Sim Kheng4,Nicholas J White12 Affiliations 1 Mahidol Oxford Tropical Medicine Research Unit, Faculty of Tropical Medicine, Mahidol University, Bangkok, Thailand. 2 Centre for Tropical Medicine and Global Health, Nuffield Department of Medicine, University of Oxford, Oxford, United Kingdom. 3 Unité d'Epidémiologie Moléculaire du Paludisme, Institut Pasteur du Cambodge, Phnom Penh, Cambodia. 4 National Center for Parasitology, Entomology and Malaria Control, Phnom Penh, Cambodia. PMID: 28155819 PMCID: PMC5330681 DOI: 10.7554/eLife.23061 Item in Clipboard Cite Display options Display options Format Abstract Primaquine is the only drug available to prevent relapse in vivax malaria. The main adverse effect of primaquine is erythrocyte age and dose-dependent acute haemolytic anaemia in individuals with glucose-6-phosphate dehydrogenase deficiency (G6PDd). As testing for G6PDd is often unavailable, this limits the use of primaquine for radical cure. A compartmental model of the dynamics of red blood cell production and destruction was designed to characterise primaquine-induced haemolysis using a holistic Bayesian analysis of all published data and was used to predict a safer alternative to the currently recommended once weekly 0.75 mg/kg regimen for G6PDd. The model suggests that a step-wise increase in daily administered primaquine dose would be relatively safe in G6PDd. If this is confirmed, then were this regimen to be recommended for radical cure patients would not require testing for G6PDd in areas where G6PDd Viangchan or milder variants are prevalent. Keywords: G6PD deficiency; computational biology; epidemiology; global health; human; in silico modelling; primaquine; radical cure; systems biology. PubMed Disclaimer Conflict of interest statement The authors declare that no competing interests exist. Figures Figure 1.. Comparison between the data from… Figure 1.. Comparison between the data from Kheng et al. (2015) (shown in green, population… Figure 1.. Comparison between the data from Kheng et al. (2015) (shown in green, population median in thick black line) and posterior predictive 80% credible intervals (shown in red, median: thick line; 10&90% boundaries: dashed lines) in which adult Cambodian patients who were G6PD deficient were given weekly primaquine (45 mg) for eight weeks. Left: reticulocyte response; Right: haemoglobin response. DOI: Figure 2.. Comparison between approximate model fits… Figure 2.. Comparison between approximate model fits (red) and data (green) extracted from four primaquine… Figure 2.. Comparison between approximate model fits (red) and data (green) extracted from four primaquine studies with a single dose or daily regimens all at 30/45 mg adult doses. Dosing periods are shaded in blue. The top two plots are for Mahidol and Viangchan variants, respectively. The bottom two plots are for the Mediterranean variant. From top left to bottom right: single 45 mg dose given to 7 G6PDd Mahidol Thais (Charoenlarp et al., 1972); 14 daily doses of 30 mg given to 15 G6PDd presumed Viangchan variant Khmer soldiers (only mean and extreme values reported) (Everett et al., 1977); 1 G6PDd Med Sardinian given two courses of daily 30 mg doses (Pannacciulli et al., 1965); 2 G6PDd Med Sardinians given 5 daily doses of 30 mg (Salvidio et al., 1967). DOI: Figure 3.. Comparison between approximate model fits… Figure 3.. Comparison between approximate model fits (red) and data (green) extracted from four primaquine… Figure 3.. Comparison between approximate model fits (red) and data (green) extracted from four primaquine studies on the same individual with G6PDd African A-(Alving et al., 1960). Dosing periods are shaded in blue. The top two plots are for weekly dosing regimens (8 doses): left is 60 mg per week; right is 45 mg per week; the bottom two plots are daily dosing regimens (14 doses): left is 15 mg per day; right is 30 mg per day. DOI: Figure 4.. Time series data of reticulocyte… Figure 4.. Time series data of reticulocyte count (top row) and haemoglobin concentrations (bottom row)… Figure 4.. Time series data of reticulocyte count (top row) and haemoglobin concentrations (bottom row) from the Cambodian study on G6PDd individuals (n=18, left column) and G6PD normals (n=57, right column) (Kheng et al., 2015). The faint green lines show individual patient data; the thick black lines represent the population median values at each time-point; the dashed black lines show the interquartile range. DOI: Figure 5.. Estimating the dose-response curve for… Figure 5.. Estimating the dose-response curve for moderate/severe G6PDd. Left: estimates of the log…log⁡… Figure 5.. Estimating the dose-response curve for moderate/severe G6PDd. Left: estimates of the log d log⁡d parameter as a function of the administered dose plotted with a linear regression curve (red cross: Viangchan; red circles: posterior estimates from model fitted to data from G6PDd Viangchan; blue cross: Mahidol; green crosses: African A−A-). Right: dose-response curve (thick black line) with 90% credible intervals (dotted black lines) as measured by fall in haemoglobin (y y-axis) after five days at a given dose (x x-axis) based on draws from the posterior distribution. The red and green crosses are the estimated falls after five days from Viangchan and African A−A- studies, respectively (see Figures 2,3). The red triangles show the falls observed in G6PDd Med studies from Figure 2. DOI: Figure 6.. Comparison of two 20-day ascending-dose… Figure 6.. Comparison of two 20-day ascending-dose regimens. Left : haemolysis over time resulting from… Figure 6.. Comparison of two 20-day ascending-dose regimens. Left: haemolysis over time resulting from regimens. Blue: simplified regimen; red: idealized optimal regimen. Right: daily dosing construction for the two regimens. Total dose of blue regimen is 375 mg; total dose of red regimen is 382.5 mg. DOI: Figure 7.. Dynamics of ascending regimens. Left … Figure 7.. Dynamics of ascending regimens. Left : Comparing the haemolytic effect of four regimens.… Figure 7.. Dynamics of ascending regimens. Left: Comparing the haemolytic effect of four regimens.Thick black line: proposed optimal regimen; thick black dashed line: more conservative regimen with lower total dose; thin black dashed line: longer duration regimen for more severe variants; thick red line: bad choice regimen. Right: Posterior predictions for the proposed ascending dose for a given starting haemoglobin (steady state). Prediction using the median posterior values is shown by a thick black line. Predictions for 100 random draws from the posterior are shown by dashed blue lines. The horizontal line at a haemoglobin concentration of 9 is a proposed conservative ‘safety threshold’. Horizontal line at a haemoglobin concentration of 8 is a proposed regimen limiting toxicity threshold. DOI: Appendix 1—figure 1.. Posterior distributions of model… Appendix 1—figure 1.. Posterior distributions of model parameters and hyperparameters. DOI: Appendix 1—figure 1.. Posterior distributions of model parameters and hyperparameters. DOI: Appendix 1—figure 2.. Relationship between the steady… Appendix 1—figure 2.. Relationship between the steady state haematocrit and the mean corpuscular volume at… Appendix 1—figure 2.. Relationship between the steady state haematocrit and the mean corpuscular volume at day zero. G6PDd patients are colored in blue, G6PD normal patients in red; women are shown by circles, men by triangles. DOI: Appendix 1—figure 3.. Individual parameter effects on… Appendix 1—figure 3.. Individual parameter effects on the behavior of the compartmental model as shown… Appendix 1—figure 3.. Individual parameter effects on the behavior of the compartmental model as shown by the haematocrit response and the reticulocyte count response. From top left to bottom right, grouped by pairs: the mid-haemoglobin concentration parameter H b R 50 H b 50 R for the reticulocyte release function; the mid-haemoglobin concentration parameter H b ρ 50 H b 50 ρ for the marrow production function; the hill coefficient k k for the reticulocyte release function; the max-fold production factor ρ max ρ max. The different values plotted for each parameter are shown in the legend for reticulocyte response plot. DOI: All figures (10) See this image and copyright information in PMC Similar articles Tolerability and safety of weekly primaquine against relapse of Plasmodium vivax in Cambodians with glucose-6-phosphate dehydrogenase deficiency.Kheng S, Muth S, Taylor WR, Tops N, Kosal K, Sothea K, Souy P, Kim S, Char CM, Vanna C, Ly P, Ringwald P, Khieu V, Kerleguer A, Tor P, Baird JK, Bjorge S, Menard D, Christophel E.Kheng S, et al.BMC Med. 2015 Aug 25;13:203. doi: 10.1186/s12916-015-0441-1.BMC Med. 2015.PMID: 26303162 Free PMC article. Linked-evidence modelling of qualitative G6PD testing to inform low- and intermediate-dose primaquine treatment for radical cure of Plasmodium vivax.Gatton ML.Gatton ML.PLoS Negl Trop Dis. 2024 Sep 5;18(9):e0012486. doi: 10.1371/journal.pntd.0012486. eCollection 2024 Sep.PLoS Negl Trop Dis. 2024.PMID: 39236082 Free PMC article. 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The haematological consequences of Plasmodium vivax malaria after chloroquine treatment with and without primaquine: a WorldWide Antimalarial Resistance Network systematic review and individual patient data meta-analysis.Commons RJ, Simpson JA, Thriemer K, Chu CS, Douglas NM, Abreha T, Alemu SG, Añez A, Anstey NM, Aseffa A, Assefa A, Awab GR, Baird JK, Barber BE, Borghini-Fuhrer I, D'Alessandro U, Dahal P, Daher A, de Vries PJ, Erhart A, Gomes MSM, Grigg MJ, Hwang J, Kager PA, Ketema T, Khan WA, Lacerda MVG, Leslie T, Ley B, Lidia K, Monteiro WM, Pereira DB, Phan GT, Phyo AP, Rowland M, Saravu K, Sibley CH, Siqueira AM, Stepniewska K, Taylor WRJ, Thwaites G, Tran BQ, Hien TT, Vieira JLF, Wangchuk S, Watson J, William T, Woodrow CJ, Nosten F, Guerin PJ, White NJ, Price RN.Commons RJ, et al.BMC Med. 2019 Aug 1;17(1):151. doi: 10.1186/s12916-019-1386-6.BMC Med. 2019.PMID: 31366382 Free PMC article. See all similar articles Cited by Global, regional, and national burden of glucose-6-phosphate dehydrogenase (G6PD) deficiency from 1990 to 2021: a systematic analysis of the global burden of disease study 2021.Yu Z, Xiong Q, Wang Z, Li L, Niu T.Yu Z, et al.Front Genet. 2025 May 23;16:1593728. doi: 10.3389/fgene.2025.1593728. eCollection 2025.Front Genet. 2025.PMID: 40486677 Free PMC article. Decrypting the complexity of the human malaria parasite biology through systems biology approaches.Chahine Z, Le Roch KG.Chahine Z, et al.Front Syst Biol. 2022;2:940321. doi: 10.3389/fsysb.2022.940321. Epub 2022 Sep 16.Front Syst Biol. 2022.PMID: 37200864 Free PMC article. 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Vivax malaria in pregnancy and lactation: a long way to health equity.Brummaier T, Gilder ME, Gornsawun G, Chu CS, Bancone G, Pimanpanarak M, Chotivanich K, Nosten F, McGready R.Brummaier T, et al.Malar J. 2020 Jan 22;19(1):40. doi: 10.1186/s12936-020-3123-1.Malar J. 2020.PMID: 31969155 Free PMC article. See all "Cited by" articles References Alving AS, Johnson CF, Tarlov AR, Brewer GJ, Kellermeyer RW, Carson PE. Mitigation of the haemolytic effect of primaquine and enhancement of its action against exoerythrocytic forms of the chesson strain of piasmodium vivax by intermittent regimens of drug administration: a preliminary report. Bulletin of the World Health Organization. 1960;22:621. - PMC - PubMed Beck HP, Wampfler R, Carter N, Koh G, Osorio L, Rueangweerayut R, Krudsood S, Lacerda MV, Llanos-Cuentas A, Duparc S, Rubio JP, Green JA. Estimation of the antirelapse efficacy of tafenoquine, using Plasmodium vivax Genotyping. Journal of Infectious Diseases. 2016;213:794–799. doi: 10.1093/infdis/jiv508. - DOI - PMC - PubMed Beutler E, Dern RJ, Alving AS. The hemolytic effect of primaquine. IV. the relationship of cell age to hemolysis. The Journal of Laboratory and Clinical Medicine. 1954a;44:439–442. - PubMed Beutler E, Dern RJ, Alving AS. The hemolytic effect of primaquine. III. A study of primaquine-sensitive erythrocytes. The Journal of Laboratory and Clinical Medicine. 1954b;44:177–184. - PubMed Beutler E, Dern RJ, Flanagan CL, Alving AS. The hemolytic effect of primaquine. VII. biochemical studies of drug-sensitive erythrocytes. The Journal of Laboratory and Clinical Medicine. 1955;45:286–295. - PubMed Show all 29 references Publication types Research Support, Non-U.S. Gov't Actions Search in PubMed Search in MeSH Add to Search MeSH terms Anemia, Hemolytic / chemically induced Actions Search in PubMed Search in MeSH Add to Search Anemia, Hemolytic / diagnosis Actions Search in PubMed Search in MeSH Add to Search Anemia, Hemolytic / prevention & control Actions Search in PubMed Search in MeSH Add to Search Antimalarials / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Antimalarials / adverse effects Actions Search in PubMed Search in MeSH Add to Search Bayes Theorem Actions Search in PubMed Search in MeSH Add to Search Cell Death / drug effects Actions Search in PubMed Search in MeSH Add to Search Erythrocytes / drug effects Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Glucosephosphate Dehydrogenase Deficiency / complications Actions Search in PubMed Search in MeSH Add to Search Glucosephosphate Dehydrogenase Deficiency / drug therapy Actions Search in PubMed Search in MeSH Add to Search Glucosephosphate Dehydrogenase Deficiency / parasitology Actions Search in PubMed Search in MeSH Add to Search Glucosephosphate Dehydrogenase Deficiency / pathology Actions Search in PubMed Search in MeSH Add to Search Hemolysis / drug effects Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Malaria, Vivax / complications Actions Search in PubMed Search in MeSH Add to Search Malaria, Vivax / drug therapy Actions Search in PubMed Search in MeSH Add to Search Malaria, Vivax / parasitology Actions Search in PubMed Search in MeSH Add to Search Malaria, Vivax / pathology Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Models, Statistical Actions Search in PubMed Search in MeSH Add to Search Plasmodium vivax / drug effects Actions Search in PubMed Search in MeSH Add to Search Plasmodium vivax / growth & development Actions Search in PubMed Search in MeSH Add to Search Primaquine / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Primaquine / adverse effects Actions Search in PubMed Search in MeSH Add to Search Recurrence Actions Search in PubMed Search in MeSH Add to Search Substances Antimalarials Actions Search in PubMed Search in MeSH Add to Search Primaquine Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound (MeSH Keyword) Grants and funding WT_/Wellcome Trust/United Kingdom LinkOut - more resources Miscellaneous NCI CPTAC Assay Portal [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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189109	https://www.hopkinsmedicine.org/health/conditions-and-diseases/premenstrual-dysphoric-disorder-pmdd	Published Time: 2025-02-10 Premenstrual Dysphoric Disorder (PMDD) \| Johns Hopkins Medicine Measles Cases on the Rise Several states are reporting cases of measles, a highly contagious viral disease. Both children and adults can get measles. Before going to a clinic or emergency room, it’s important to check with your doctor if you think you or a loved one has measles. Learn more about symptoms and treatment. Skip to Main Content Johns Hopkins Medicine Menu Johns Hopkins Medicine Home About Patient Care Health Research School of Medicine MyChart Schedule an Appointment Find a Doctor Find a Clinical Trial Pay Your Bill Employment Close Search Search Johns Hopkins Medicine Search Close Health Health Close Main Menu Health Main Menu Health Conditions and Diseases Treatments, Tests and Therapies Wellness and Prevention Caregiving Home Health Conditions and Diseases Premenstrual Dysphoric Disorder (PMDD) Women's Health Diagnosis and Screening for Gynecologic Conditions Women's Sexual and Reproductive Health Fertility and Reproductive Health What is premenstrual dysphoric disorder (PMDD)? Premenstrual dysphoric disorder (PMDD) is a much more severe form of premenstrual syndrome (PMS). It may affect people of childbearing age. It’s a severe and chronic health condition that needs attention and treatment. Lifestyle changes and sometimes medicines can help manage symptoms. What causes PMDD? The exact cause of PMDD is not known. It may be an abnormal reaction to normal hormone changes that happen with each menstrual cycle. The hormone changes can cause a serotonin deficiency. Serotonin is a substance found naturally in the brain and intestines that narrows blood vessels and can affect mood and cause physical symptoms. Who is at risk for PMDD? Anyone can develop PMDD, but the following people may be at an increased risk: People with a family history of PMS or PMDD People with a personal or family history of depression, postpartum depression, or other mood disorders People with less access to education about the management and treatment of PMDD Cigarette smoking Talk with your healthcare provider for more information. What are the symptoms of PMDD? Symptoms of PMDD start during the week before menstruation and end within a few days after your period starts. These symptoms disrupt daily living tasks. Symptoms of PMDD are so severe that people have trouble functioning at home, at work, and in relationships during this time. This is markedly different than other times during the month. The following are common symptoms of PMDD: Depressed mood, sadness, hopelessness, or feelings of worthlessness Increased anxiety, tension, or the feeling of being on edge all the time Mood swings Self-critical thoughts, increased sensitivity to rejection Frequent or sudden tearfulness Increased irritability, anger, or both Conflict with family, coworkers, or friends Decreased interest in normal activities Concentration problems Fatigue, lethargy, or lack of energy Changes in appetite, such as binge eating, overeating, or craving certain foods Changes in sleep pattern, such as excessive sleeping or difficulty sleeping Feelings of being overwhelmed or out of control Physical symptoms, such as breast swelling or tenderness, headaches, joint or muscle aches, weight gain, and bloating The symptoms of PMDD may look like other health conditions, such as a thyroid condition, depression, or an anxiety disorder. Always see a healthcare provider for a diagnosis. How is PMDD diagnosed? Aside from a health history and physical and pelvic exam, there are very few diagnostic tests. Because you may have mental health symptoms, your healthcare provider may want you to be evaluated for mental health concerns. Your provider may also ask that you keep a journal or diary of your symptoms for several months. In general, to diagnose PMDD the following criteria must be met: Over the course of a year, during most menstrual cycles, you must have 5 or more symptoms (see above) that have been present during the week before your period and stopping within a few days after your period starts. Symptoms must be linked to significant distress. Or they must disturb your ability to function in social, work, or other situations. Symptoms are not related to or made worse by another health condition. How is PMDD treated? PMDD is a serious, long-term (chronic) condition that does need treatment. Several of the following treatment approaches may help ease or decrease the severity of PMDD symptoms: Changes in diet to increase protein and carbohydrates and decrease sugar, salt, caffeine, and alcohol Regular exercise Stress management, such as relaxation and meditation methods Vitamin supplements, such as vitamin B-6, calcium, and magnesium Anti-inflammatory medicines Selective serotonin reuptake inhibitors Birth control pills For some people, the severity of symptoms increases over time and lasts until menopause. For this reason, a person may need treatment for an extended time. Medicine dosage may change throughout the course of treatment. Key points about PMDD PMDD is a much more severe form of premenstrual syndrome (PMS). The exact cause of PMDD is not known. PMDD is different from other mood disorders or menstrual conditions because of when symptoms start and how long they last. Symptoms of PMDD are so severe that it affects your ability to function at home, work and in relationships. Treatment that may include lifestyle changes and sometimes medicines. Next steps Tips to help you get the most from a visit to your healthcare provider: Know the reason for your visit and what you want to happen. Before your visit, write down questions you want answered. Bring someone with you to help you ask questions and remember what your provider tells you. At the visit, write down the name of a new diagnosis, and any new medicines, treatments, or tests. Also write down any new instructions your provider gives you. Know why a new medicine or treatment is prescribed, and how it will help you. Also know what the side effects are. Ask if your condition can be treated in other ways. Know why a test or procedure is recommended and what the results could mean. Know what to expect if you do not take the medicine or have the test or procedure. If you have a follow-up appointment, write down the date, time, and purpose for that visit. Know how you can contact your healthcare provider if you have questions, especially after office hours or on weekends and holidays. 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189110	https://courses.lumenlearning.com/calculus2/chapter/differentiating-and-integrating-power-series/	Differentiating and Integrating Power Series Learning Outcomes Differentiate and integrate power series term-by-term Consider a power series ∞∑n=0cnxn=c0+c1x+c2x2+⋯∞∑n=0cnxn=c0+c1x+c2x2+⋯ that converges on some interval I, and let ff be the function defined by this series. Here we address two questions about ff. Is ff differentiable, and if so, how do we determine the derivative f′?f′? How do we evaluate the indefinite integral ∫f(x)dx?∫f(x)dx? We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Recall the general expression of the power rules for derivatives and integrals of polynomials. Recall: Power rule for Derivatives and Integrals ddx(xn)=nxn−1ddx(xn)=nxn−1 ∫xndx=1n+1xn+1+C(forn≠−1∫xndx=1n+1xn+1+C(forn≠−1) In this section we show that we can take advantage of the simplicity of integrating and differentiating polynomials to do the same thing for convergent power series. That is, if f(x)=cnxn=c0+c1x+c2x2+⋯f(x)=cnxn=c0+c1x+c2x2+⋯ converges on some interval I, then f′(x)=c1+2c2x+3c3x2+⋯f′(x)=c1+2c2x+3c3x2+⋯ and ∫f(x)dx=C+c0x+c1x22+c2x33+⋯∫f(x)dx=C+c0x+c1x22+c2x33+⋯. Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for f(x)=11−xf(x)=11−x, we can differentiate term-by-term to find the power series for f′(x)=1(1−x)2f′(x)=1(1−x)2. Similarly, using the power series for g(x)=11+xg(x)=11+x, we can integrate term-by-term to find the power series for G(x)=ln(1+x)G(x)=ln(1+x), an antiderivative of g. We show how to do this in the next two examples. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series. theorem: Term-by-Term Differentiation and Integration for Power Series Suppose that the power series ∞∑n=0cn(x−a)n∞∑n=0cn(x−a)n converges on the interval (a−R,a+R)(a−R,a+R) for some R>0R>0. Let f be the function defined by the series f(x)=∞∑n=0cn(x−a)n=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯f(x)=∞∑n=0cn(x−a)n=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯ for [latex]\|x-a\|f is differentiable on the interval (a−R,a+R)(a−R,a+R) and we can find f′f′ by differentiating the series term-by-term: f′(x)=∞∑n=1ncn(x−a)n−1=c1+2c2(x−a)+3c3(x−a)2+⋯f′(x)=∞∑n=1ncn(x−a)n−1=c1+2c2(x−a)+3c3(x−a)2+⋯ for [latex]\|x-a\| ∫f(x)dx=C+∞∑n=0cn(x−a)n+1n+1=C+c0(x−a)+c1(x−a)22+c2(x−a)33+⋯∫f(x)dx=C+∞∑n=0cn(x−a)n+1n+1=C+c0(x−a)+c1(x−a)22+c2(x−a)33+⋯ for [latex]\|x-a\| The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples. Example: Differentiating Power Series f(x)=11−x=∞∑n=0xn=1+x+x2+x3+⋯f(x)=11−x=∞∑n=0xn=1+x+x2+x3+⋯ for \|x\|<1\|x\|<1 to find a power series representation for g(x)=1(1−x)2g(x)=1(1−x)2 on the interval (−1,1)(−1,1). Determine whether the resulting series converges at the endpoints. Use the result of part a. to evaluate the sum of the series ∞∑n=0n+14n∞∑n=0n+14n. Show Solution Since g(x)=1(1−x)2g(x)=1(1−x)2 is the derivative of f(x)=11−xf(x)=11−x, we can find a power series representation for g by differentiating the power series for f term-by-term. The result is g(x)=1(1−x)2=ddx(11−x)=∞∑n=0ddx(xn)=ddx(1+x+x2+x3+⋯)=0+1+2x+3x2+4x3+⋯=∞∑n=0(n+1)xng(x)=1(1−x)2=ddx(11−x)=∞∑n=0ddx(xn)=ddx(1+x+x2+x3+⋯)=0+1+2x+3x2+4x3+⋯=∞∑n=0(n+1)xn for \|x\|<1\|x\|<1. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints x=±1x=±1. Note that this is the same result found in the previous example. From part a. we know that ∞∑n=0(n+1)xn=1(1−x)2∞∑n=0(n+1)xn=1(1−x)2. Therefore, ∞∑n=0n+14n=∞∑n=0(n+1)(14)n=1(1−14)2=1(34)2=169.∞∑n=0n+14n=∞∑n=0(n+1)(14)n=1(1−14)2=1(34)2=169. Watch the following video to see the worked solution to Example: Differentiating Power Series. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “6.2.4” here (opens in new window). try it Differentiate the series 1(1−x)2=∞∑n=0(n+1)xn1(1−x)2=∞∑n=0(n+1)xn term-by-term to find a power series representation for 2(1−x)32(1−x)3 on the interval (−1,1)(−1,1). Hint Write out the first several terms and apply the power rule. Show Solution ∞∑n=0(n+2)(n+1)xn∞∑n=0(n+2)(n+1)xn Example: Integrating Power Series For each of the following functions f, find a power series representation for f by integrating the power series for f′f′ and find its interval of convergence. f(x)=ln(1+x)f(x)=ln(1+x) f(x)=tan−1xf(x)=tan−1x Show Solution For f(x)=ln(1+x)f(x)=ln(1+x), the derivative is f′(x)=11+xf′(x)=11+x. We know that 11+x=11−(-x)=∞∑n=0(-x)n=1−x+x2−x3+⋯11+x=11−(-x)=∞∑n=0(-x)n=1−x+x2−x3+⋯ for \|x\|<1\|x\|<1. To find a power series for f(x)=ln(1+x)f(x)=ln(1+x), we integrate the series term-by-term. ∫f′(x)dx=∫(1−x+x2−x3+⋯)dx=C+x−x22+x33−x44+⋯∫f′(x)dx=∫(1−x+x2−x3+⋯)dx=C+x−x22+x33−x44+⋯ Since f(x)=ln(1+x)f(x)=ln(1+x) is an antiderivative of 11+x11+x, it remains to solve for the constant C. Since ln(1+0)=0ln(1+0)=0, we have C=0C=0. Therefore, a power series representation for f(x)=ln(1+x)f(x)=ln(1+x) is ln(1+x)=x−x22+x33−x44+⋯=∞∑n=1(−1)n+1xnnln(1+x)=x−x22+x33−x44+⋯=∞∑n=1(−1)n+1xnn for \|x\|<1\|x\|<1. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at x=1x=1 the series is the alternating harmonic series, which converges. Also, at x=−1x=−1, the series is the harmonic series, which diverges. It is important to note that, even though this series converges at x=1x=1, Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to ln(2)ln(2). In fact, the series does converge to ln(2)ln(2), but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is (−1,1](−1,1]. The derivative of f(x)=tan−1xf(x)=tan−1x is f′(x)=11+x2f′(x)=11+x2. We know that 11+x2=11−(-x2)=∞∑n=0(−x2)n=1−x2+x4−x6+⋯11+x2=11−(-x2)=∞∑n=0(−x2)n=1−x2+x4−x6+⋯ for \|x\|<1\|x\|<1. To find a power series for f(x)=tan−1xf(x)=tan−1x, we integrate this series term-by-term. ∫f′(x)dx=∫(1−x2+x4−x6+⋯)dx=C+x−x33+x55−x77+⋯∫f′(x)dx=∫(1−x2+x4−x6+⋯)dx=C+x−x33+x55−x77+⋯ Since tan−1(0)=0tan−1(0)=0, we have C=0C=0. Therefore, a power series representation for f(x)=tan−1xf(x)=tan−1x is tan−1x=x−x33+x55−x77+⋯=∞∑n=0(−1)nx2n+12n+1tan−1x=x−x33+x55−x77+⋯=∞∑n=0(−1)nx2n+12n+1 for \|x\|<1\|x\|<1. Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at x=1x=1 and x=−1x=−1. As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to tan−1(1)tan−1(1) and tan−1(−1)tan−1(−1) at x=1x=1 and x=−1x=−1, respectively. Thus, the interval of convergence is [−1,1][−1,1]. Watch the following video to see the worked solution to Example: Integrating Power Series. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip “6.2.4” here (opens in new window). try it Integrate the power series ln(1+x)=∞∑n=1(−1)n+1xnnln(1+x)=∞∑n=1(−1)n+1xnn term-by-term to evaluate ∫ln(1+x)dx∫ln(1+x)dx. Hint Use the fact that xn+1(n+1)nxn+1(n+1)n is an antiderivative of xnnxnn. Show Solution ∞∑n=2(−1)nxnn(n−1)∞∑n=2(−1)nxnn(n−1) Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function f and a power series for f at a, is it possible that there is a different power series for f at a that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if c0+c1x+c2x2+⋯=d0+d1x+d2x2+⋯c0+c1x+c2x2+⋯=d0+d1x+d2x2+⋯ for all values x in some open interval I about zero, then the coefficients cn should equal dn for n≥0n≥0. We now state this result formally in Uniqueness of Power Series. theorem: Uniqueness of Power Series Let ∞∑n=0cn(x−a)n∞∑n=0cn(x−a)n and ∞∑n=0dn(x−a)n∞∑n=0dn(x−a)n be two convergent power series such that ∞∑n=0cn(x−a)n=∞∑n=0dn(x−a)n∞∑n=0cn(x−a)n=∞∑n=0dn(x−a)n for all x in an open interval containing a. Then cn=dncn=dn for all n≥0n≥0. Proof Let f(x)=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯=d0+d1(x−a)+d2(x−a)2+d3(x−a)3+⋯.f(x)=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯=d0+d1(x−a)+d2(x−a)2+d3(x−a)3+⋯. Then f(a)=c0=d0f(a)=c0=d0. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore, f′(x)=c1+2c2(x−a)+3c3(x−a)2+⋯=d1+2d2(x−a)+3d3(x−a)2+⋯,f′(x)=c1+2c2(x−a)+3c3(x−a)2+⋯=d1+2d2(x−a)+3d3(x−a)2+⋯, and thus, f′(a)=c1=d1f′(a)=c1=d1. Similarly, f′′(x)=2c2+3⋅2c3(x−a)+⋯=2d2+3⋅2d3(x−a)+⋯f′′(x)=2c2+3⋅2c3(x−a)+⋯=2d2+3⋅2d3(x−a)+⋯ implies that f′′(a)=2c2=2d2f′′(a)=2c2=2d2, and therefore, c2=d2c2=d2. More generally, for any integer n≥0,f(n)(a)=n!cn=n!dnn≥0,f(n)(a)=n!cn=n!dn, and consequently, cn=dncn=dn for all n≥0n≥0. ◼■ Try It In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series. Candela Citations CC licensed content, Original 6.2.4. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Licenses and Attributions CC licensed content, Original 6.2.4. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at
189111	https://math.stackexchange.com/questions/3126735/behavior-of-trilinear-or-barycentric-coordinates	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Behavior of Trilinear or Barycentric coordinates Ask Question Asked Modified 6 years, 7 months ago Viewed 727 times 2 $\begingroup$ Whenever we define the trilinear coordinates or the barycentric coordinates of a point in $\mathbb R^2$, it is with reference to some fixed triangle. Do we know how the coordinates behave when we change the reference triangle? geometry Share asked Feb 25, 2019 at 22:41 chhrochhro 2,3381010 silver badges1515 bronze badges $\endgroup$ Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ Do we know how the coordinates behave when we change the reference triangle? Yes, we do. Both trilinear and barycentric coordinates are essentially homogeneous coordinates, which makes projective geometry the tool describing them. A change of projective basis corresponds to projective transformation, which you can express as multiplication by a $3\times3$ matrix. As I described in another answer, such a projective transformation is uniquely determined by four points and their images. Another useful concept is the duality between points and lines in projective geometry. Essentially you can consistently swap the roles of points and lines and statements using only these concepts remain valid. With transformation added to the mix, a transformation operating on a line is described by the inverse transpose of the matrix for the operation on points. Taking these two together, a change of basis can just as well be described as a transformation mapping four given lines to their images. So if you have the coordinates of the edges (which you can compute from those of the corners using cross products) of the second triangle, expressed in barycentric coordinates of the first triangle, you can add the line at infinity as a fourth line both as the preimage and as the image, and then deduce the transformation from that. The line at infinity in barycentric coordinates has the equation $x+y+z=0$. Wikipedia tells you that trilinear coordinates $[x:y:z]$ correspond to barycentric coordinates $[ax:by:cz]$. This is a multiplication by a diagonal matrix, so it fits the general framework of projective transformations. You can use this either to express the line at infinity in trilinear coordinates, or to change from trilinear coordinates to barycentric, do the conversion there using the line at infinity as described above, then convert back. Either way make sure to use the edge length of the appropriate triangle at each step. If you don't know the coordinates of one triangle in the reference frame of the other, or of both triangles with respect to some common third frame of reference, then in general converting from one to the other will not be possible. You need some information. Share answered Feb 26, 2019 at 20:35 MvGMvG 44.2k99 gold badges9494 silver badges186186 bronze badges $\endgroup$ 2 $\begingroup$ Appreciate the response, wasn't really expecting an answer as the question is kind of vague. I came across this question in the context of three triangles with a common side and a point in the intersection of the three. So I want to know how the trilinear/barycentric coordinates change in such a configuration. $\endgroup$ chhro – chhro 2019-02-26 21:40:24 +00:00 Commented Feb 26, 2019 at 21:40 $\begingroup$ I have given a concrete example on a certain change of coordinates that might interest you. $\endgroup$ Jean Marie – Jean Marie 2019-02-27 00:15:50 +00:00 Commented Feb 27, 2019 at 0:15 Add a comment \| 2 $\begingroup$ A concrete example that can help to understand a certain correspondence for a same affine transform presented either with affine coordinates or barycentric coordinates [I used to present this example some years ago in a lecture I delivered on the introduction to projective geometry]. A similar example could be given with trilinear coordinates. Let us consider the affine transform whose effect can be seen on Fig. 1 : it sends the left giraffe onto the right giraffe, and as well transforms points $A,B,C,D,E$ into points $A',B',C',D',E'$ resp. Fig. 1 : On the left, a giraffe grazing a shrub under the Sun. On the right, transformed girafe + shrub + Sun (mirage ?). If we take axes with center $C$ and basis vectors $\vec{CA}$ and $\vec{CB}$, the affine transformation with respect to this affine coordinate system is as follows : $$\binom{x'}{y'}=\begin{pmatrix}0.5&-0.5\0.5& \ 0.5\end{pmatrix}\binom{x}{y}+\binom{1}{0}\tag{1}$$ (very intuitively, and somewhat falsely, a rotation + homothety + translation effect). Relationship (1) can be written with a single matrix under the form $$\begin{pmatrix}x'\y'\1\end{pmatrix}=\underbrace{\begin{pmatrix}0.5&-0.5&1\0.5& \ 0.5 &0\0&0&1\end{pmatrix}}_{M_{aff}}\begin{pmatrix}x\y\1\end{pmatrix}\tag{2}$$ (index $aff$ in $M_{aff}$ is to remember that it is a representation with respect to affine coordinates.) Now, the matrix of the same transformation with respect to barycentric coordinates $a,b,c$ (where the reference triangle is $ABC$ and with normalization $a+b+c=1$) : $$\begin{pmatrix}a'\b'\c'\end{pmatrix}=\underbrace{\begin{pmatrix}1.5&0.5&1\0.5& \ 0.5 &0\-1&0&0\end{pmatrix}}_{M_{bar}}\begin{pmatrix}a\b\c\end{pmatrix}.\tag{3}$$ Explanation : quite naturally, the columns of $M_{bar}$ are the barycentric coordinates of points $A'$, $B'$, $C'$ resp. with resp. to base points $A,B,C$. The correspondence between $M_{aff}$ and $M_{bar}$ can be shown to be this one : $$M_{bar}=J^{-1}M_{aff}J$$ $$\text{with} \ J:=\begin{pmatrix}1&0&0\0&1 &0\1&1&1\end{pmatrix} \ \text{and} \ J^{-1}=\begin{pmatrix}1&0&0\0&1 &0\-1&-1&1\end{pmatrix}.$$ Exercices : 1) One sees on Fig. 1 that $D$ is a fixed ("eigen" ?) point of this transformation. Check it. 2) Explain the roles of $J$ and its inverse as translators between barycentric and cartesian coordinates. Share edited Feb 27, 2019 at 0:59 answered Feb 27, 2019 at 0:06 Jean MarieJean Marie 90.3k77 gold badges5959 silver badges132132 bronze badges $\endgroup$ 5 $\begingroup$ Very nice answer. Both the answers were informative in this question. I wonder why none of them were upvoted. $\endgroup$ user572457 – user572457 2019-07-24 16:19:09 +00:00 Commented Jul 24, 2019 at 16:19 1 $\begingroup$ @Shashwat Asthana Thanks. It is a kind of explanation I had a pleasure to write because I believe it enlightens correpondences between cousin geometries (I haven't found such an example in the litterature). $\endgroup$ Jean Marie – Jean Marie 2019-07-25 07:25:40 +00:00 Commented Jul 25, 2019 at 7:25 1 $\begingroup$ Can you clear another doubt of mine in PG? It's not really a general doubt. Consider PG(2,R) (if that's not a standard notation, it means a 2 dimensional projective geo on the field of real numers), how are we supposed to think of triangle centers? For example, if I were to perform a projective transformation taking three vertices of a triangle to (1,0,0),(0,1,0) and (0,0,1), will the centroid of the original triangle be transformed to the centroid of our new triangle? $\endgroup$ user572457 – user572457 2019-07-26 12:02:51 +00:00 Commented Jul 26, 2019 at 12:02 1 $\begingroup$ Wait, I'll ask a question. You can answer their. I might get even more interesting answers. $\endgroup$ user572457 – user572457 2019-07-26 12:52:09 +00:00 Commented Jul 26, 2019 at 12:52 1 $\begingroup$ No, the image of the centroid $G$ of 3 points $A,B,C$ is the centroid $G'$ of images $A',B',C'$ only for the very particular case of affine transforms. Take for example transformation $x'=x/(x+y), y'=2y/(x+y)$ and $A=(1,0), B=(0,1)$ and $C=(1,1)$ : $G=(1/2,1)$, $G'=(1/2,1)$ whereas the image of $G$ is $(1/3,4/3)$. $\endgroup$ Jean Marie – Jean Marie 2019-07-26 12:54:50 +00:00 Commented Jul 26, 2019 at 12:54 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 99 Finding the transform matrix from 4 projected points (with JavaScript) How to interpret triangle centers in projective geometry? 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189112	https://miamioh.ecampus.com/fundamentals-heat-mass-transfer-8th/bk/9781119537342?srsltid=AfmBOorz4FlvpiT3vUVq3b66JrCOL2fJcSNYDHD6WB5Tycg2zkpcRWoD	Fundamentals of Heat and Mass Transfer, 8th Edition [Rental Edition] Skip Navigation MIAMI UNIVERSITY OFFICIAL BOOKSTORE Login/Sign Up Home Miami Alumni Miami Athletics MENU Shop Textbooks Campus Locations Login/Sign Up Back\| Campus Locations Home Home Miami Alumni Miami Alumni Miami Athletics Miami Athletics Back\| Shop Clothing Clothing Accessories Accessories Gifts Gifts Graduation Graduation Supplies Supplies Back\| Clothing Kids Men Sweatshirts Women View All Back\| Accessories For You For Your Car For Your Home For Your Pet For Your Tech View All Back\| Gifts Artwork Cooking Essentials Games Gift Wraps Holiday Home Decor Mascot Office Decor Outdoor/Recreation View All Back\| Graduation Graduation Gear Graduation Gifts View All Back\| Supplies Art Supplies For Your Office Medical Supplies Office Supplies School Supplies View All Kids Back\| Kids Bibs Bottoms Dresses Headwear Hoodies Matching Sets Onesies Shirts Sweatshirts Men Back\| Men Bottoms Footwear Hoodies Jerseys Mens Apparel Outerwear Polos Shirts Sweatshirts T-Shirts Sweatshirts Back\| Sweatshirts Women Back\| Women Bottoms Dresses Headwear Hoodies Outerwear Pants Shirts Sweatshirts Undergarments View All Clothing > Miami Merger Myaamia Heritage Collection Alumni Collection Miami Regionals Custom Items Miami Cradle of Coaches collection For You Back\| For You Backpacks Bags Buttons Drinkware Fan Gear Flags Hair Accessories Headwear ID Holders Jewelry Keychains Knitwear Lanyards Lapel Pins Pennants Socks Ties Umbrellas Wallets For Your Car Back\| For Your Car Decals License Plates For Your Home Back\| For Your Home Banners Blankets Cooking Essentials Decals Drinkware Flags Frames Home Decor Magnets Pillows Signs Stickers For Your Pet Back\| For Your Pet Beds Bowls Charms Clothes Toys For Your Tech Back\| For Your Tech Computer Accessories Phone Accessories View All Accessories > Artwork Back\| Artwork Ornaments Wall Art Cooking Essentials Back\| Cooking Essentials Food Games Back\| Games Balls Puzzles Gift Wraps Back\| Gift Wraps Gift Bags Holiday Back\| Holiday Ornaments Stocking Home Decor Back\| Home Decor Candles Mascot Back\| Mascot Plushies Office Decor Back\| Office Decor Desk Accessories Outdoor/Recreation Back\| Outdoor/Recreation Tailgate View All Gifts > Graduation Gear Back\| Graduation Gear Caps and Gowns Hoods Stoles Study Abroad Sashes Tassel Tassels Graduation Gifts Back\| Graduation Gifts Diploma Frames Yard Signs View All Graduation > Art Supplies Back\| Art Supplies Art Supply Products For Your Office Back\| For Your Office Desk Accessories Pens Medical Supplies Back\| Medical Supplies Kits Office Supplies Back\| Office Supplies Padfolios School Supplies Back\| School Supplies Folders Notebooks Planners View All Supplies > Erin Condren Shopping Cart (0) Shop Textbooks Search Shopping Cart (0) Write a Review Fundamentals of Heat and Mass Transfer, 8th Edition [Rental Edition] byLavine, Adrienne S.; Bergman, Theodore L.; Incropera, Frank P.; DeWitt, David P. Edition: 8th ISBN13: 9781119537342 ISBN10: 1119537347 Format: Hardcover Pub. Date: 2019-08-13 Publisher(s): Wiley Rental Other versions by this Author This Item Qualifies for Free Shipping! Excludes marketplace orders. List Price: ~~$272.95~~ Rent Textbook Select for Price Add to Cart There was a problem. Please try again later. Rent Digital Rent Digital Options Online:150 Days access, Downloadable:150 Days - $61.20 Online:150 Days access Downloadable:150 Days $61.20 Online:1825 Days access, Downloadable:Lifetime Access - $146.34 Online:1825 Days access Downloadable:Lifetime Access $146.34 $61.20 Add to Cart Used Textbook We're Sorry Sold Out New Textbook We're Sorry Sold Out Currently unavailable Table of Contents Symbols xix Chapter 1 Introduction 1 1.1 What and How? 2 1.2 Physical Origins and Rate Equations 3 1.2.1 Conduction 3 1.2.2 Convection 6 1.2.3 Radiation 8 1.2.4 The Thermal Resistance Concept 12 1.3 Relationship to Thermodynamics 12 1.3.1 Relationship to the First Law of Thermodynamics (Conservation of Energy) 13 1.3.2 Relationship to the Second Law of Thermodynamics and the Efficiency of Heat Engines 28 1.4 Units and Dimensions 33 1.5 Analysis of Heat Transfer Problems: Methodology 35 1.6 Relevance of Heat Transfer 38 1.7 Summary 42 References 45 Chapter 2 Introduction to Conduction 47 2.1 The Conduction Rate Equation 48 2.2 The Thermal Properties of Matter 50 2.2.1 Thermal Conductivity 51 2.2.2 Other Relevant Properties 58 2.3 The Heat Diffusion Equation 62 2.4 Boundary and Initial Conditions 70 2.5 Summary 74 References 75 Chapter 3 One-Dimensional, Steady-State Conduction 77 3.1 The Plane Wall 78 3.1.1 Temperature Distribution 78 3.1.2 Thermal Resistance 80 3.1.3 The Composite Wall 81 3.1.4 Contact Resistance 83 3.1.5 Porous Media 85 3.2 An Alternative Conduction Analysis 99 3.3 Radial Systems 103 3.3.1 The Cylinder 103 3.3.2 The Sphere 108 3.4 Summary of One-Dimensional Conduction Results 109 3.5 Conduction with Thermal Energy Generation 109 3.5.1 The Plane Wall 110 3.5.2 Radial Systems 116 3.5.3 Tabulated Solutions 117 3.5.4 Application of Resistance Concepts 117 3.6 Heat Transfer from Extended Surfaces 121 3.6.1 A General Conduction Analysis 123 3.6.2 Fins of Uniform Cross-Sectional Area 125 3.6.3 Fin Performance Parameters 131 3.6.4 Fins of Nonuniform Cross-Sectional Area 134 3.6.5 Overall Surface Efficiency 137 3.7 Other Applications of One-Dimensional, Steady-State Conduction 141 3.7.1 The Bioheat Equation 141 3.7.2 Thermoelectric Power Generation 145 3.7.3 Nanoscale Conduction 153 3.8 Summary 157 References 159 Chapter 4 Two-Dimensional, Steady-State Conduction 161 4.1 General Considerations and Solution Techniques 162 4.2 The Method of Separation of Variables 163 4.3 The Conduction Shape Factor and the Dimensionless Conduction Heat Rate 167 4.4 Finite-Difference Equations 173 4.4.1 The Nodal Network 173 4.4.2 Finite-Difference Form of the Heat Equation: No Generation and Constant Properties 174 4.4.3 Finite-Difference Form of the Heat Equation: The Energy Balance Method 175 4.5 Solving the Finite-Difference Equations 182 4.5.1 Formulation as a Matrix Equation 182 4.5.2 Verifying the Accuracy of the Solution 183 4.6 Summary 188 References 189 Chapter 5 Transient Conduction 191 5.1 The Lumped Capacitance Method 192 5.2 Validity of the Lumped Capacitance Method 195 5.3 General Lumped Capacitance Analysis 199 5.3.1 Radiation Only 200 5.3.2 Negligible Radiation 200 5.3.3 Convection Only with Variable Convection Coefficient 201 5.3.4 Additional Considerations 201 5.4 Spatial Effects 210 5.5 The Plane Wall with Convection 211 5.5.1 Exact Solution 212 5.5.2 Approximate Solution 212 5.5.3 Total Energy Transfer: Approximate Solution 214 5.5.4 Additional Considerations 214 5.6 Radial Systems with Convection 215 5.6.1 Exact Solutions 215 5.6.2 Approximate Solutions 216 5.6.3 Total Energy Transfer: Approximate Solutions 216 5.6.4 Additional Considerations 217 5.7 The Semi-Infinite Solid 222 5.8 Objects with Constant Surface Temperatures or Surface Heat Fluxes 229 5.8.1 Constant Temperature Boundary Conditions 229 5.8.2 Constant Heat Flux Boundary Conditions 231 5.8.3 Approximate Solutions 232 5.9 Periodic Heating 239 5.10 Finite-Difference Methods 242 5.10.1 Discretization of the Heat Equation: The Explicit Method 242 5.10.2 Discretization of the Heat Equation: The Implicit Method 249 5.11 Summary 256 References 257 Chapter 6 Introduction to Convection 259 6.1 The Convection Boundary Layers 260 6.1.1 The Velocity Boundary Layer 260 6.1.2 The Thermal Boundary Layer 261 6.1.3 The Concentration Boundary Layer 263 6.1.4 Significance of the Boundary Layers 264 6.2 Local and Average Convection Coefficients 264 6.2.1 Heat Transfer 264 6.2.2 Mass Transfer 265 6.3 Laminar and Turbulent Flow 271 6.3.1 Laminar and Turbulent Velocity Boundary Layers 271 6.3.2 Laminar and Turbulent Thermal and Species Concentration Boundary Layers 273 6.4 The Boundary Layer Equations 276 6.4.1 Boundary Layer Equations for Laminar Flow 277 6.4.2 Compressible Flow 280 6.5 Boundary Layer Similarity: The Normalized Boundary Layer Equations 280 6.5.1 Boundary Layer Similarity Parameters 281 6.5.2 Dependent Dimensionless Parameters 281 6.6 Physical Interpretation of the Dimensionless Parameters 290 6.7 Boundary Layer Analogies 292 6.7.1 The Heat and Mass Transfer Analogy 293 6.7.2 Evaporative Cooling 296 6.7.3 The Reynolds Analogy 299 6.8 Summary 300 References 301 Chapter 7 External Flow 303 7.1 The Empirical Method 305 7.2 The Flat Plate in Parallel Flow 306 7.2.1 Laminar Flow over an Isothermal Plate: A Similarity Solution 307 7.2.2 Turbulent Flow over an Isothermal Plate 313 7.2.3 Mixed Boundary Layer Conditions 314 7.2.4 Unheated Starting Length 315 7.2.5 Flat Plates with Constant Heat Flux Conditions 316 7.2.6 Limitations on Use of Convection Coefficients 317 7.3 Methodology for a Convection Calculation 317 7.4 The Cylinder in Cross Flow 325 7.4.1 Flow Considerations 325 7.4.2 Convection Heat and Mass Transfer 327 7.5 The Sphere 335 7.6 Flow Across Banks of Tubes 338 7.7 Impinging Jets 347 7.7.1 Hydrodynamic and Geometric Considerations 347 7.7.2 Convection Heat and Mass Transfer 348 7.8 Packed Beds 352 7.9 Summary 353 References 356 Chapter 8 Internal Flow 357 8.1 Hydrodynamic Considerations 358 8.1.1 Flow Conditions 358 8.1.2 The Mean Velocity 359 8.1.3 Velocity Profile in the Fully Developed Region 360 8.1.4 Pressure Gradient and Friction Factor in Fully Developed Flow 362 8.2 Thermal Considerations 363 8.2.1 The Mean Temperature 364 8.2.2 Newton’s Law of Cooling 365 8.2.3 Fully Developed Conditions 365 8.3 The Energy Balance 369 8.3.1 General Considerations 369 8.3.2 Constant Surface Heat Flux 370 8.3.3 Constant Surface Temperature 373 8.4 Laminar Flow in Circular Tubes: Thermal Analysis and Convection Correlations 377 8.4.1 The Fully Developed Region 377 8.4.2 The Entry Region 382 8.4.3 Temperature-Dependent Properties 384 8.5 Convection Correlations: Turbulent Flow in Circular Tubes 384 8.6 Convection Correlations: Noncircular Tubes and the Concentric Tube Annulus 392 8.7 Heat Transfer Enhancement 395 8.8 Forced Convection in Small Channels 398 8.8.1 Microscale Convection in Gases (0.1 μ m ≲ D h ≲ 100 μ m) 398 8.8.2 Microscale Convection in Liquids 399 8.8.3 Nanoscale Convection (D h ≲ 100 nm) 400 8.9 Convection Mass Transfer 403 8.10 Summary 405 References 408 Chapter 9 Free Convection 409 9.1 Physical Considerations 410 9.2 The Governing Equations for Laminar Boundary Layers 412 9.3 Similarity Considerations 414 9.4 Laminar Free Convection on a Vertical Surface 415 9.5 The Effects of Turbulence 418 9.6 Empirical Correlations: External Free Convection Flows 420 9.6.1 The Vertical Plate 421 9.6.2 Inclined and Horizontal Plates 424 9.6.3 The Long Horizontal Cylinder 429 9.6.4 Spheres 433 9.7 Free Convection within Parallel Plate Channels 434 9.7.1 Vertical Channels 435 9.7.2 Inclined Channels 437 9.8 Empirical Correlations: Enclosures 437 9.8.1 Rectangular Cavities 437 9.8.2 Concentric Cylinders 440 9.8.3 Concentric Spheres 441 9.9 Combined Free and Forced Convection 443 9.10 Convection Mass Transfer 444 9.11 Summary 445 References 446 Chapter 10 Boiling and Condensation 449 10.1 Dimensionless Parameters in Boiling and Condensation 450 10.2 Boiling Modes 451 10.3 Pool Boiling 452 10.3.1 The Boiling Curve 452 10.3.2 Modes of Pool Boiling 453 10.4 Pool Boiling Correlations 456 10.4.1 Nucleate Pool Boiling 456 10.4.2 Critical Heat Flux for Nucleate Pool Boiling 458 10.4.3 Minimum Heat Flux 459 10.4.4 Film Pool Boiling 459 10.4.5 Parametric Effects on Pool Boiling 460 10.5 Forced Convection Boiling 465 10.5.1 External Forced Convection Boiling 466 10.5.2 Two-Phase Flow 466 10.5.3 Two-Phase Flow in Microchannels 469 10.6 Condensation: Physical Mechanisms 469 10.7 Laminar Film Condensation on a Vertical Plate 471 10.8 Turbulent Film Condensation 475 10.9 Film Condensation on Radial Systems 480 10.10 Condensation in Horizontal Tubes 485 10.11 Dropwise Condensation 486 10.12 Summary 487 References 487 Chapter 11 Heat Exchangers 491 11.1 Heat Exchanger Types 492 11.2 The Overall Heat Transfer Coefficient 494 11.3 Heat Exchanger Analysis: Use of the Log Mean Temperature Difference 497 11.3.1 The Parallel-Flow Heat Exchanger 498 11.3.2 The Counterflow Heat Exchanger 500 11.3.3 Special Operating Conditions 501 11.4 Heat Exchanger Analysis: The Effectiveness–NTU Method 508 11.4.1 Definitions 508 11.4.2 Effectiveness–NTU Relations 509 11.5 Heat Exchanger Design and Performance Calculations 516 11.6 Additional Considerations 525 11.7 Summary 533 References 534 Chapter 12 Radiation: Processes and Properties 535 12.1 Fundamental Concepts 536 12.2 Radiation Heat Fluxes 539 12.3 Radiation Intensity 541 12.3.1 Mathematical Definitions 541 12.3.2 Radiation Intensity and its Relation to Emission 542 12.3.3 Relation to Irradiation 547 12.3.4 Relation to Radiosity for an Opaque Surface 549 12.3.5 Relation to the Net Radiative Flux for an Opaque Surface 550 12.4 Blackbody Radiation 550 12.4.1 The Planck Distribution 551 12.4.2 Wien’s Displacement Law 552 12.4.3 The Stefan–Boltzmann Law 552 12.4.4 Band Emission 553 12.5 Emission from Real Surfaces 560 12.6 Absorption, Reflection, and Transmission by Real Surfaces 569 12.6.1 Absorptivity 570 12.6.2 Reflectivity 571 12.6.3 Transmissivity 573 12.6.4 Special Considerations 573 12.7 Kirchhoff’s Law 578 12.8 The Gray Surface 580 12.9 Environmental Radiation 586 12.9.1 Solar Radiation 587 12.9.2 The Atmospheric Radiation Balance 589 12.9.3 Terrestrial Solar Irradiation 591 12.10 Summary 594 References 598 Chapter 13 Radiation Exchange Between Surfaces 599 13.1 The View Factor 600 13.1.1 The View Factor Integral 600 13.1.2 View Factor Relations 601 13.2 Blackbody Radiation Exchange 610 13.3 Radiation Exchange Between Opaque, Diffuse, Gray Surfaces in an Enclosure 614 13.3.1 Net Radiation Exchange at a Surface 615 13.3.2 Radiation Exchange Between Surfaces 616 13.3.3 The Two-Surface Enclosure 622 13.3.4 Two-Surface Enclosures in Series and Radiation Shields 624 13.3.5 The Reradiating Surface 626 13.4 Multimode Heat Transfer 631 13.5 Implications of the Simplifying Assumptions 634 13.6 Radiation Exchange with Participating Media 634 13.6.1 Volumetric Absorption 634 13.6.2 Gaseous Emission and Absorption 635 13.7 Summary 639 References 640 Chapter 14 Diffusion Mass Transfer 641 14.1 Physical Origins and Rate Equations 642 14.1.1 Physical Origins 642 14.1.2 Mixture Composition 643 14.1.3 Fick’s Law of Diffusion 644 14.1.4 Mass Diffusivity 645 14.2 Mass Transfer in Nonstationary Media 647 14.2.1 Absolute and Diffusive Species Fluxes 647 14.2.2 Evaporation in a Column 650 14.3 The Stationary Medium Approximation 655 14.4 Conservation of Species for a Stationary Medium 655 14.4.1 Conservation of Species for a Control Volume 656 14.4.2 The Mass Diffusion Equation 656 14.4.3 Stationary Media with Specified Surface Concentrations 658 14.5 Boundary Conditions and Discontinuous Concentrations at Interfaces 662 14.5.1 Evaporation and Sublimation 663 14.5.2 Solubility of Gases in Liquids and Solids 663 14.5.3 Catalytic Surface Reactions 668 14.6 Mass Diffusion with Homogeneous Chemical Reactions 670 14.7 Transient Diffusion 673 14.8 Summary 679 References 680 Appendix A Thermophysical Properties of Matter 681 Appendix B Mathematical Relations and Functions 713 Appendix C Thermal Conditions Associated with Uniform Energy Generation in One-Dimensional, Steady-State Systems 719 Appendix D The Gauss–Seidel Method 725 Appendix E The Convection Transfer Equations 727 E.1 Conservation of Mass 728 E.2 Newton’s Second Law of Motion 728 E.3 Conservation of Energy 729 E.4 Conservation of Species 730 Appendix F Boundary Layer Equations for Turbulent Flow 731 Appendix G An Integral Laminar Boundary Layer Solution for Parallel Flow over a Flat Plate 735 Conversion Factors 739 Physical Constants 740 Index 741 We are currently experiencing difficulties. 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Start Practice 3.1 Writing Equations with Variables FlexBooks 2.0> CK-12 Interactive Middle School Math 8> Writing Equations with Variables Written by:Larame Spence \|Sean Regan Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Lesson Review Asked on Flexi Related Content Lesson Expressing the World Around Us with Math The ability to use math to represent the structure and performance of any business directly contributes to the success of that business. Math is used for simple calculations like profit, and for more complex formulas like determining the lifetime value of a customer. Every day CEOs use math to make major decisions that may well affect the future success of their business. Can you think of any examples of how businesses use math? Vocabulary You will begin by learning about expressions, equations, and their basic components: [Figure 1] Flow chart of Vocab Words A variable is a letter that stands for a number. The number a variable stands for can change based on the circumstances of the situation. A coefficient is a number that is placed in front of a variable to multiply it. A variable with no visible coefficient has a coefficient of 1. A constant is a number that does not change. A constant is different from a coefficient because a constant cannot be attached to a variable. An operator is a symbol that defines what action to take with a number. The operators that we will be dealing with are +, -, x and ÷. A term is a group of numbers or variables separated by a + or - operator. A term can be one letter, one number, or a collection of letters and numbers. Writing Expressions An expression is a combination of variables, constants, and operators. An expression can consist of one or more terms. Use the interactive below to identify the parts of an expression. INTERACTIVE Building an Expression Move the different parts of the expression into their corresponding boxes. Press the button to check your work. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No Discussion Question How can 12 be both a constant and a term? Expressions are used to represent relationships between variables, coefficients, and constants. In business, since there are so many factors that can affect the outcomes, expressions are used to represent the relationship, and the variables are filled in later when the numbers are known. Additionally, expressions allow us to plug in different values to see how the outcome will change. Example S tart with a simple and common relationship: An amusement park sells tickets for adults at $15 and tickets for children at $8. What is an expression for the revenue (money earned) from selling 'A' adult tickets on a given day? If you do not immediately see the relationship, draw a table to help figure it out. Adult Tickets Sold 1 2 3 A Revenue 15 30 45? To determine the revenue, you are multiplying the number of tickets sold by the price per adult ticket ($15). To determine the revenue from selling 'A' adult tickets, multiply the number of tickets sold (A) by the price per adult ticket ($15) to get the expression 15 A. Answer:15 A Notice that our expression is a constant multiplied by a variable. Example Using the same information from the previous example, what is an expression that can be used to determine the total revenue (money earned) from selling 'A' adult tickets and 'C' child tickets on a given day? D rawing a table will help you figure it out. Since any combination of adult and child tickets sold is possible, choose three random values of adult and child tickets sold. Adult Tickets Sold 1 2 3 A Child Tickets Sold 1 2 3 C Revenue 23 46 69? T o determine the revenue, you are multiplying the number of tickets sold by the price per adult ticket ($15) and adding it to the number of tickets sold by the price per child's ticket ($8). To determine the revenue from selling A adult tickets, you will multiply the number of adult tickets sold (A) by the price per adult ticket ($15) and add it to the number of child tickets sold (C) by the price per child ticket ($8) to get the expression 15 A+8 C. Answer:15 A+8 C Discussion Questions Would the answer 8 C+15 A be incorrect? Does this represent all the revenue from an amusement park on a given day? How else might they earn money? How would these additional sources of revenue change our total revenue expressions? What type of situation would involve a constant? Writing Equations An equation is a statement that two expressions are equal. Use the interactive below to identify the parts of an equation. INTERACTIVE Building an Equation Move the various parts of the equation into the corresponding areas. Press the button to check your work. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No Discussion Questions How can you tell the difference between an equation and an expression? How can 5 be a constant, term, and expression at the same time? Equations are commonly used in business to state the relationship between an expression and its equal numerical value. For example, the total cost of a business can be determined by adding the fixed cost and the variable cost. Fixed costs are costs which are constant from one period to the next, like rent. Variable costs are costs that change from one period to the next like materials. The materials you buy to make your product will depend on how many products you plan to sell. An ice cream shop, for example, will buy/make more ice cream in the summer because they will have more customers in the summer. The equation for the total cost is as follows: Total Cost=Fixed Cost+Variable Cost One benefit of equations is that they help us determine unknown information. By knowing the fixed cost and variable cost, you can determine the total cost. Additionally, if you know the total cost and fixed cost, you can determine the variable cost.This property of equations can be used to determine what variable values will give us the desired result. Spreadsheets S preadsheets are one of the most common ways in which expressions and equations are used in business. It can be helpful to list all the sources of revenue and cost in one place. However, this spreadsheet can be pages long. What if you make a mistake or something in the spreadsheet changes? To avoid having to retype a spreadsheet you can use expressions, also commonly referred to as formulas or functions. For reference, each rectangle in a spreadsheet is called a cell, and each cell is named by their column (a letter) and row (a number). So the cell in column B and row 23 would be called cell B23. In the place of variables, we type the name of a cell. To insert an expression into a cell,we type in an expression that will determine the value of that cell. For example, if you want a cell to stand for the product of a constant (3) and a variable (the value in the cell A2) you can write =3A2. If you want a cell to stand for the product of two different cells (A1 and A2), you can write =A1A2. The = symbol identifies that you are inputting an expression rather than a value. Once you fill in all of the variables, the cell will display the value. In the interactive below, fill out five products for a company in cells in column A and their respective prices in column B. Enter 5 as the quantity in cells C2 through C6. Write an expression in column D to find the value of each cell by multiplying the corresponding B and C cells. For example, the cell D2 should use the formula =B2C2. The cell D7 should be determined by writing an expression to add cells D2 through D6. INTERACTIVE Spreadsheets Type in the items you are purchasing in column A. Type in the price of the items in column B. Enter an expression using an asterisk () in column D to multiply two cells together. For example in D2, type "=B2C2" without the quotation marks. Try It Your device seems to be offline. Please check your internet connection and try again. + Do you want to reset the PLIX? Yes No Summary An expression is a combination of variables, terms, and operators. Anequation is a statement that two expressions are equal. Equations and expressions can be used to describe real-world situations. Add Note CancelSave Discuss with Flexi NOTES / HIGHLIGHTS Please Sign In to create your own Highlights / Notes Add Note Edit Note Remove Highlight Image Attributions Asked by Students Here are the top questions that students are asking Flexi for this concept: On a winter evening at sundown, a Kansas City news station reports that the temperature is – 2°F and will drop by 2°F per hour until sunrise. At the same time, a Chicago news station reports that the temperature is 6°F and will drop by 4°F per hour. Which equation can you use to find h, the number of hours it will take the cities to reach the same temperature? The temperature in Kansas City is decreasing by 2°F per hour from an initial temperature of -2°F, so the temperature in Kansas City after h hours can be represented as −2−2 h. The temperature in Chicago is decreasing by 4°F per hour from an initial temperature of 6°F, so the temperature in Chicago after h hours can be represented as 6−4 h. We want to find when the temperatures in the two cities will be the same, so we set the two expressions equal to each other: −2−2 h=6−4 h Sean began jogging to live a healthier lifestyle. On his first run, he ran one-half mile. He increased his workouts by adding two miles a month to his run. He wrote the equation f(x)= 0.5 + 2x to model his progress. What does the variable x represent? The variable x in the equation f(x)=0.5+2 x represents the number of months he runs. Each year, the students in seventh grade go to an environmental camp. This year, the total cost is $1,580 for 79 students. Which equation represents the cost, y, of attending the camp for x number of students? The cost per student can be found by dividing the total cost by the number of students. So, the cost for x students would be the cost per student times x. The cost per student is 1580 79 which equals $20. So, the equation representing the cost, y, for x number of students is y=20 x. Therefore, the equation is y=20 x. The Cohen family desperately needs to catch up on doing laundry. Over the weekend, Mr. and Mrs. Cohen got 3 loads done. Now, during the week, they plan to do 1 load per day. Write an equation that shows how the total number of loads done, y, depends on how many weekdays have passed, x. The total number of loads done, y, is equal to the 3 loads done over the weekend plus 1 load per weekday. If x is the number of weekdays that have passed, the equation would be: y=1 x+3 So, the equation that shows how the total number of loads done depends on how many weekdays have passed is: y=x+3 A swimming pool has about 603 cubic feet of water. The pool liner has a small hole and is leaking at a rate of 1.5 cubic feet each hour. Ruby wrote the equation V(h) = -1.5h + 603 to represent the scenario. Describe what each variable represents in the function, and how they relate to each other. In the equation V(h)=−1.5 h+603, V(h) represents the volume of water in the pool at any given time, measured in cubic feet. The variable h represents the number of hours that have passed. The coefficient -1.5 represents the rate at which the pool is losing water, in cubic feet per hour. The constant 603 represents the initial volume of water in the pool, in cubic feet. The relationship between V(h) and h is that as time passes (as h increases), the volume of water in the pool decreases at a rate of 1.5 cubic feet per hour. Overview An expression is a combination of variables, terms, and operators. Anequation is a statement that two expressions are equal. Equations and expressions can be used to describe real-world situations. Vocabulary Variable Coefficient constant Term Expression revenue Equation difference Property Formula Test Your Knowledge Question 1 Write a variable expression for the verbal expression. A number times two divided by four a 2 x 4 b 2 x 2 c 2 x 8 d 2 x 6 Check It From the question: “A number times two divided by four”. Let x represent the number. The word “times” means ×. x×2 2 x The word “divided by” means ÷. 2 x÷4 2 x 4 The variable expression is 2 x 4. FlexCard™ Question 2 Choose an expression for the following phrase. The quantity six times an unknown number divided by three a 6 x 3 b x 6 c x 3 d 3 x 6 Check It From the question: “The quantity six times an unknown number divided by three”. Let x represent the unknown number. The word “times” means ×. 6×x 6 x The word “divided by” means ÷. 6 x÷3 6 x 3 The expression is 6 x 3. FlexCard™ Asked by Students Ask your question Here are the top questions that students are asking Flexi for this concept: On a winter evening at sundown, a Kansas City news station reports that the temperature is – 2°F and will drop by 2°F per hour until sunrise. At the same time, a Chicago news station reports that the temperature is 6°F and will drop by 4°F per hour. Which equation can you use to find h, the number of hours it will take the cities to reach the same temperature? The temperature in Kansas City is decreasing by 2°F per hour from an initial temperature of -2°F, so the temperature in Kansas City after h hours can be represented as −2−2 h. The temperature in Chicago is decreasing by 4°F per hour from an initial temperature of 6°F, so the temperature in Chicago after h hours can be represented as 6−4 h. We want to find when the temperatures in the two cities will be the same, so we set the two expressions equal to each other: −2−2 h=6−4 h Sean began jogging to live a healthier lifestyle. On his first run, he ran one-half mile. He increased his workouts by adding two miles a month to his run. He wrote the equation f(x)= 0.5 + 2x to model his progress. What does the variable x represent? The variable x in the equation f(x)=0.5+2 x represents the number of months he runs. Each year, the students in seventh grade go to an environmental camp. This year, the total cost is $1,580 for 79 students. Which equation represents the cost, y, of attending the camp for x number of students? The cost per student can be found by dividing the total cost by the number of students. So, the cost for x students would be the cost per student times x. The cost per student is 1580 79 which equals $20. So, the equation representing the cost, y, for x number of students is y=20 x. Therefore, the equation is y=20 x. The Cohen family desperately needs to catch up on doing laundry. Over the weekend, Mr. and Mrs. Cohen got 3 loads done. Now, during the week, they plan to do 1 load per day. Write an equation that shows how the total number of loads done, y, depends on how many weekdays have passed, x. The total number of loads done, y, is equal to the 3 loads done over the weekend plus 1 load per weekday. If x is the number of weekdays that have passed, the equation would be: y=1 x+3 So, the equation that shows how the total number of loads done depends on how many weekdays have passed is: y=x+3 A swimming pool has about 603 cubic feet of water. The pool liner has a small hole and is leaking at a rate of 1.5 cubic feet each hour. Ruby wrote the equation V(h) = -1.5h + 603 to represent the scenario. Describe what each variable represents in the function, and how they relate to each other. In the equation V(h)=−1.5 h+603, V(h) represents the volume of water in the pool at any given time, measured in cubic feet. The variable h represents the number of hours that have passed. The coefficient -1.5 represents the rate at which the pool is losing water, in cubic feet per hour. The constant 603 represents the initial volume of water in the pool, in cubic feet. The relationship between V(h) and h is that as time passes (as h increases), the volume of water in the pool decreases at a rate of 1.5 cubic feet per hour. 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189114	https://scicomp.stackexchange.com/questions/41679/inaccurate-results-of-integration-using-scipy-solve-ivp	python - Inaccurate results of integration using scipy solve_ivp - Computational Science Stack Exchange Join Computational Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Inaccurate results of integration using scipy solve_ivp Ask Question Asked 3 years, 2 months ago Modified2 years, 9 months ago Viewed 3k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am trying to use solve_ivp to solve the following 1st order ODE: d ρ d z=m θ(1+θ z)ρ,d ρ d z=m θ(1+θ z)ρ, subject to ρ(z=0)=1 ρ(z=0)=1, where m m and θ θ are constants. This ODE actually has a simple analytical solution: ρ=(1+θ z)m.ρ=(1+θ z)m. When I try to solve this ODE numerically using solve_ivp, the resulting solution does not seem to accurately match the analytical solution. Please see the minimal working example below and the plot of the difference between the numerical and analytical solution. Any help figuring out what is going on in here would be greatly appreciated. Thanks. ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import solve_ivp theta = 2.5 m = 1.4 def drho_func(z, rho): drho = rho(mtheta)/(1.0+thetaz) return drho zgrid = np.linspace(0,1,501) y0 = sol = solve_ivp(drho_func , [zgrid, zgrid[-1]], y0, t_eval=zgrid) rho_num = sol.y rho_true = (1+thetazgrid)m plt.plot(rho_num-rho_true,zgrid) ``` python ode numerics scipy integration Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Jul 26, 2022 at 9:18 FryderykFryderyk 33 1 1 silver badge 4 4 bronze badges 1 Leaving t_eval out you get the discretization that is actually used internally. I suspect that it is rather sparse with the default tolerances, with nodes for z>1 z>1 corresponding to the extrema of the error function, either close to the extrema themselves or to the midpoints between them.Lutz Lehmann –Lutz Lehmann 2022-07-26 14:25:51 +00:00 Commented Jul 26, 2022 at 14:25 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The default tolerances for solve_ivp are rtol=1e-3 and atol=1e-6. The analytical value of ρ ρ varies between 1 and 6, so you can expect your numerical error to be somewhere on the order of ϵ≈0.001 ϵ≈0.001. The fact that you actually see ϵ≈0.0002 ϵ≈0.0002 is not unexpected. If you want to increase the accuracy of solve_ivp, simply specify a tighter numerical tolerance. For example, here I set rtol=1e-8 and atol=1e-8, and the difference between analytical and numerical solutions is on the order of 1e-8. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 26, 2022 at 10:58 helloworld922helloworld922 2,859 1 1 gold badge 17 17 silver badges 13 13 bronze badges 1 1 Thanks for this answer! That fixes my problem. I had not realised that the default tolerances were this "large", and in retrospect should have checked that. Many thanks for the help.Fryderyk –Fryderyk 2022-07-26 11:06:44 +00:00 Commented Jul 26, 2022 at 11:06 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. As the initial value problem was given and very well specified, the analytical solution can be easily calculated, where y(t)=(1+2.5 t)1.4 y(t)=(1+2.5 t)1.4 solution of the PVI problem, by the way, ρ=y ρ=y and z=t z=t. With the initial value problem defined, and with the analytical solution in hand, we can solve the ODE numerically as an integrator scipy.integrate.odeint and plot the errors committed ϵ Absolute(n)=\|ρ Analytic−ρ Numeric\|ϵ Absolute(n)=\|ρ Analytic−ρ Numeric\| and ϵ Relative(n)=\|ρ Analytic−ρ Numeric ϵ Analytic\|ϵ Relative(n)=\|ρ Analytic−ρ Numeric ϵ Analytic\|, where n n is the number of iterations. Observation: I gave preference to odeint, because in several examples like the one specified above, odeint is chosen over solve_ivp. ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import odeint, solve_ivp NUMERICAL SOLUTION APPLYING SCIPY'S ODEIN ORDINARY DIFFERENTIAL EQUATION dy/dt = f(t,y), t0 = t' and y(t0) = y0 def f(y,t, m, theta): # rho => y # z => t # m = 1.4 # theta = 2.5 return mthetay/(1+thetat) ti = 0.0 tf = 50.0 N = 1000 t = np.linspace(ti,tf,N) y0 = 1.0 sol = odeint(f, y0, t, args=(1.4,2.5), rtol=1E-8, atol=1E-8) sol_numeric = sol[:,0] ANALYTICAL SOLUTION OF THE INITIAL VALUE PROBLEM def f_solution(t): return 1.0(1+2.5t)(1.4) sol_analytic = f_solution(t) # Observation, t = np.linspace(ti,tf,N) ERROR erro_absolut = np.abs(sol_analytic-sol_numeric) erro_relativ = np.abs((sol_analytic-sol_numeric)/sol_analytic) quantidade_iteracoes = np.linspace(0,N,N) PLOTTING OF RESULTS plt.style.use('dark_background') plt.figure(figsize = (21,6)) plt.plot(t,sol,'b.',t,f_solution(t),'r-') plt.title('Comparison between analytical and numerical solution (Odeint)') plt.xlabel('time') plt.ylabel('y(t)') plt.grid(lw = 1.0,color = 'y',linestyle = '-') plt.show() plt.figure(figsize = (21,6)) plt.style.use('dark_background') plt.subplot(1,2,1) plt.plot(quantidade_iteracoes,erro_absolut,color = 'red', lw = 3.0) plt.plot(quantidade_iteracoes,erro_absolut,'r.', lw = 3.0) plt.title('Absolute error', fontsize=18) plt.xlabel('num of iteration', fontsize=18) plt.ylabel(r'$\epsilon_{Absolute}$', fontsize=18) plt.grid(lw = 0.5,color = 'y',linestyle = '-') plt.subplot(1,2,2) plt.plot(quantidade_iteracoes,erro_relativ,color = 'blue', lw = 3.0) plt.plot(quantidade_iteracoes,erro_relativ,'b.', lw = 3.0) plt.title('Relative error', fontsize=18) plt.xlabel('num of iteration', fontsize=18) plt.ylabel(r'$\epsilon_{relative}$', fontsize=18) plt.grid(lw = 0.5,color = 'y',linestyle = '-') ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Dec 4, 2022 at 22:47 answered Dec 4, 2022 at 18:45 Carllos LimmaCarllos Limma 111 3 3 bronze badges 1 1 You only need to write out the optional parameters that you actually change from their default value, like here atol and rtol. This makes for an easier reading of that call.Lutz Lehmann –Lutz Lehmann 2022-12-04 20:04:47 +00:00 Commented Dec 4, 2022 at 20:04 Add a comment\| Your Answer Thanks for contributing an answer to Computational Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. 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189115	https://stemjock.com/STEM%20Books/Boyce%20ODEs%2010e/Chapter%202/Section%202/BoyceODEch2s2p28.pdf?srsltid=AfmBOopMhXrsUQv2_yQevyTIh5Zyo_JZZXbbRkIXCriMfJgoSX8xCVSJ	Boyce & DiPrima ODEs 10e: Section 2.2 - Problem 28 Page 1 of 3 Problem 28 Consider the initial value problem y′ = ty (4 − y)/(1 + t), y(0) = y0 > 0. (a) Determine how the solution behaves as t → ∞ .(b) If y0 = 2, find the time T at which the solution first reaches the value 3.99. (c) Find the range of initial values for which the solution lies in the interval 3 .99 < y < 4.01 by the time t = 2. Solution Part (a) This ODE is separable because it is of the form y′ = f (t)g(y), so it can be solved by separating variables. dy dt = ty (4 − y)1 + t Bring the terms with y to the left and bring the terms with t to the right. dy y(4 − y) = t 1 + t dt Integrate both sides. ˆ dy y(4 − y) = ˆ t 1 + t dt Make the substitution u = 1 + t in the integral on the right. ˆ 14 ( 1 y + 14 − y ) dy = ˆ u − 1 u du 14 ˆ ( 1 y − 1 y − 4 ) dy = u − ln \|u\| + C 14 (ln \|y\| − ln \|y − 4\|) = 1 + t − ln(1 + t) + C Use a new constant C1 for 1 + C.14 ln ∣∣∣∣ yy − 4 ∣∣∣∣ = t − ln(1 + t) + C1 Multiply both sides by 4. ln ∣∣∣∣ yy − 4 ∣∣∣∣ = 4 t − 4 ln(1 + t) + 4 C1 = 4 t + ln(1 + t)−4 + 4 C1 Exponentiate both sides. ∣∣∣∣ yy − 4 ∣∣∣∣ = e4t+ln(1+ t)−4+4 C1 = e4t(1 + t)−4e4C1 www.stemjock.com Boyce & DiPrima ODEs 10e: Section 2.2 - Problem 28 Page 2 of 3 Introduce ± on the right side to remove the absolute value sign. yy − 4 = ±e4t(1 + t)−4e4C1 Use a new constant A for ±e4C1 . yy − 4 = A e4t (1 + t)4 Apply the initial condition y(0) = y0 here to determine A. y0 y0 − 4 = A As a result, yy − 4 = y0 y0 − 4 e4t (1 + t)4 . Multiply both sides by y − 4. y = y y0 y0 − 4 e4t (1 + t)4 − 4 y0 y0 − 4 e4t (1 + t)4 y [ 1 − y0 y0 − 4 e4t (1 + t)4 ] = −4 y0 y0 − 4 e4t (1 + t)4 Therefore, the solution is y(t) = −4 y0 y0−4 e4t (1+ t)4 1 − y0 y0−4 e4t (1+ t)4 = 4y0e4t (4 − y0)( t + 1) 4 + y0e4t = 4y0 (4 − y0)( t + 1) 4e−4t + y0 . In the limit as t → ∞ , the first term in the denominator vanishes because of the decaying exponential function. lim t→∞ y(t) = 4y0 0 + y0 = 4 Part (b) Set y0 = 2 and y = 3 .99 and solve the equation for t numerically. 3.99 = 4(2) 2( t + 1) 4e−4t + 2 t ≈ 2.84 www.stemjock.com Boyce & DiPrima ODEs 10e: Section 2.2 - Problem 28 Page 3 of 3 Part (c) Set t = 2 in the solution and solve the following inequality for y0.3.99 < y (2) < 4.01 3.99 < 4y0 (4 − y0)(3) 4e−4(2) + y0 < 4.01 Split up the inequality into two. 4y0 81(4 − y0)e−8 + y0 3.99 and 4y0 81(4 − y0)e−8 + y0 < 4.01 4y0 81(4 − y0)e−8 + y0 − 3.99 > 0 and 4y0 81(4 − y0)e−8 + y0 − 4.01 < 00.01 y0 − 323 .19(4 − y0)e−8 81(4 − y0)e−8 + y0 0 and −0.01 y0 − 324 .81(4 − y0)e−8 81(4 − y0)e−8 + y0 < 0Determine the critical points by finding where the numerators are zero. Critical Point: y0 ≈ 3.66 and Critical Point: y0 ≈ 4.40 Testing values of y0 above and below the critical points, we find where each inequality is true. y0 & 3.66 and y0 . 4.40 Therefore, 3.66 . y0 . 4.40 . www.stemjock.com
189116	https://tools4ncteachers.com/seventh-grade/	Seventh Grade \| Tools 4 NC Teachers Home How It Works Select Page Welcome, Seventh Grade Math Teachers. This is an explanation of the cluster organization including links leading to the NC2ML Collaborative for specific resources. The Collaborative Instructional Framework is meant to serve as a guide for teachers and districts as they organize the curriculum for the school year. Unlike traditional pacing guides, the instructional framework consists of clusters of standards that are meant to be adapted to various schools and contexts. The instructional framework used research on students’ learning progression in mathematics to create and order clusters of standards that are taught together. While there is a strongly suggested order for teaching the clusters, we recognize that schools differ in their contexts and may wish to switch the order around. In those cases, we have given guidance regarding alternative clusterings; however, we note when certain clusters need to be taught in a certain order. The clusters are recommended using the sample progression below, but this is not the only possible progression teachers may use. When considering changes, please look to the “Connections & Rationale” for notes about when one cluster must follow another. The timing for the units in the “Recommended Timeframe” should not be treated as a rigid timetable. Also, continue to focus on how the Standards for Mathematical Practice can be incorporated with these content clusters. Recommended Order Proportional Relationships Reasoning with Rational Numbers Probabilistic Reasoning Reasoning About Expressions Reasoning About Equations and Inequalities Geometric and Measurement Reasoning Reasoning about Population Samples Comparing Populations Alternative Order Proportional Relationships Reasoning with Rational Numbers Probabilistic Reasoning Reasoning About Expressions Reasoning About Equations and Inequalities Geometric and Measurement Reasoning Reasoning about Population Samples++ Comparing Populations Clusters with can be switched in order; ++Clusters with can be switched in order A significant emphasis should be placed on understanding operations with fractions and decimals and proportional reasoning. Anecdotal evidence and teachers’ experiences indicate that developing strong proportional reasoning in students is a challenge. As 7th graders are developing new understanding of rational numbers and operations with rational numbers, they are also being asked to reason and solve problems with ratios and proportions. Since “fraction” and “ratio” are not exactly the same thing, teachers need to be aware of the fact that success in operations with fractions does not always indicate an ability to reason with ratios. The mathematics of fractions is often the mathematics of ratio, but the reasoning with each is not the same. One resource for more information about reasoning with ratios and proportions in NCTM’s Developing Essential Understanding of Ratios, Proportions, and Proportional Reasoning, 6-8. Though there is a single cluster entitled “Proportional Relationships,” teachers should not anticipate that teaching this one unit will suffice as the single moment that develops their students’ proportional reasoning. Developing students’ proportional reasoning, should be a goal throughout the 7th grade year. Please pay attention to the “Supporting Standards” and “Connections & Rationale” portions of this recommendation. These tools should help to connect mathematical concepts across the units, and to highlight potential opportunities for revisiting previous units’ main ideas. Advice for thinking about any re-ordering of these content clusters: It is recommended that the first week of the school year be spent engaging students with open-ended mathematics problems designed to support the students’ growth mindset. This first week is also an opportune time for setting up the classroom expectations and norms for collaborating with classmates and participating in whole class discussions. Youcubed.org offers 3 Weeks of Inspirational Mathematics. It is recommended that 6th graders experience Week 1, 7th graders Week 2, and 8th graders Week 3. However, other general problem solving lessons can be used. There are numerous ways to connect from one cluster to the previous or next cluster. Look to the standards in order to bridge these units. Note that Proportional Reasoning and Reasoning with Rational Numbers appear all along the length of the school year. We encourage a continued focus on developing proportional reasoning and understanding/fluency with rational number operations. Though there may be a unit that addresses these topics for a focused period of time, the topics should be integrated throughout the entire year. Click here to access resources \| Cluster \| Recommended Time \| Standards \| --- \| Building a Mathematical Community ((Boaler's Week of Inspirational Math - Week 2) \| 1 week \| \| \| Proportional Relationships Cluster \| 6 weeks \| RP.1, RP.2, RP.3, G.1 Supporting standards EE.2, EE.3, EE.4, NS.2 \| \| Reasoning with Rational Numbers Cluster \| 6 weeks \| NS.1, NS.2, NS.3 Supporting standards EE.2, EE.3, RP.3, G.1 \| \| Probabilistic Reasoning Cluster \| 3 weeks \| SP.5, SP.6, SP.7, SP.8 Supporting standards RP.1, RP.2, RP.3 \| \| Reasoning about Expressions Cluster \| 3 weeks \| EE.1, EE.2 Supporting standards NS.1, NS.2 \| \| Reasoning about Equations and Inequalities Cluster \| 6 weeks \| EE.3, EE.4, G.2, G.5 Supporting standards NS.1, NS.2 \| \| Geometric and Measurement Reasoning Cluster \| 3 weeks \| G.4, G.6 Supporting standards NS.2, EE.2, RP.2 \| \| Reasoning about Population Samples Cluster \| 4 weeks \| SP, SP.1, SP.2 Supporting standards RP.1, RP.2, RP.3 \| \| Comparing Populations Cluster \| 3 weeks \| SP.3, SP.4 \| Click Here to Access Materials from Previous Grants Home NC2ML DPI Mathematics Page DPI Assessment Page © 2018 Whoosh Agency in partnership with Edstar Analytics
189117	http://sites.science.oregonstate.edu/BridgeBook/book/mathcontent/cross2.html	book:mathcontent:cross2 - The Geometry of Vector Calculus You are here: start » book » mathcontent » cross2 The cross product is fundamentally a directed area. The magnitude of the cross product is defined to be the area of the parallelogram whose sides are the two vectors in the cross product. In the figure above, if the horizontal vector is $\vv$ and the upward-pointing vector is $\ww$, then the height of the parallelogram is $\|\ww\|\sin\theta$, so its area is \begin{equation} \|\vv\times\ww\| = \|\vv\|\|\ww\|\sin\theta \label{crossmagnitude} \end{equation} which is therefore the magnitude of the cross product. An immediate consequence of Equation ($\ref{crossmagnitude}$) is that, if two vectors are parallel, their cross product is zero, \begin{equation} \vv\parallel\ww \Longleftrightarrow \vv\times\ww=\zero \end{equation} The direction of the cross product is given by the right-hand rule: Point the fingers of your right hand along the first vector ($\vv$), and curl your fingers toward the second vector ($\ww$). You may have to flip your hand over to make this work. Now stick out your thumb; that is the direction of $\vv\times\ww$. In the example shown above, $\vv\times\ww$ points out of the page. The right-hand rule implies that \begin{equation} \ww\times\vv = -\vv\times\ww \end{equation} as you should verify for yourself by suitably positioning your hand. Thus, the cross product is not commutative. 1) Another important property of the cross product is that the cross product of a vector with itself is zero, \begin{equation} \vv\times\vv = \zero \end{equation} which follows from any of the preceding three equations. In terms of the standard orthonormal basis, the geometric formula quickly yields \begin{eqnarray} \xhat\times\yhat &=& \zhat \ \yhat\times\zhat &=& \xhat \ \zhat\times\xhat &=& \yhat \end{eqnarray} This cyclic nature of the cross product can be emphasized by abbreviating this multiplication table as shown in the figure below. 2) Products in the direction of the arrow get a plus sign; products against the arrow get a minus sign. Using an orthonormal basis such as ${\xhat,\yhat,\zhat}$, the geometric formula reduces to the standard component form of the cross product. 3) If $\vv=v_x\,\xhat+v_y\,\yhat+v_z\,\zhat$ and $\ww=w_x\,\xhat+w_y\,\yhat+w_z\,\zhat$, then \begin{eqnarray} \vv\times\ww &=& (v_x\,\xhat+v_y\,\yhat+v_z\,\zhat) \times (w_x\,\xhat+w_y\,\yhat+w_z\,\zhat) \cr &=& (v_y w_z - v_z w_y)\,\xhat + (v_z w_x - v_x w_z) \,\yhat + (v_x w_y - v_y w_x)\,\zhat \label{CrossAlg} \end{eqnarray} which is often written as the symbolic determinant \begin{equation} \vv\times\ww = \left\| \matrix{\xhat& \yhat& \zhat\cr \noalign{\smallskip} v_x& v_y& v_z\cr w_x& w_y& w_z\cr} \right\| \label{CrossDet} \end{equation} We encourage you to use($\ref{CrossDet}$), rather than simply memorizing($\ref{CrossAlg}$). We also encourage you to compute the determinant as described below, rather than using minors; this tends to minimize sign errors. A $3\times3$ determinant can be computed in the form where one multiplies the terms along each diagonal line, subtracting the products obtained along lines going down to the left from those along lines going down to the right. While this method works only for ($2\times2$ and) $3\times3$ determinants, it emphasizes the cyclic nature of the cross product. Another important skill is knowing when not to use a determinant at all. For simple cross products, such as $(\xhat+3\,\yhat)\times\zhat$, it is easier to use the multiplication table directly. It is also worth pointing out that the multiplication table and the determinant method generalize naturally to any (right-handed) orthonormal basis; all that is needed is to replace the rectangular basis ${\xhat,\yhat,\zhat}$ by the one being used (in the right order!). For example, in cylindrical coordinates, not only is \begin{equation} \rhat\times\phat = \zhat \end{equation} (and cyclic permutations), but cross products can be computed as \begin{equation} \vv\times\ww = \left\| \matrix{\rhat& \phat& \zhat\cr v_r& v_\phi& v_z\cr w_r& w_\phi& w_z\cr} \right\| \end{equation} where of course $\vv=v_r\,\rhat+v_\phi\,\phat+v_z\,\zhat$ and similarly for $\ww$. A good problem emphasizing the geometry of the cross product is to find the area of the triangle formed by connecting the tips of the vectors $\xhat$, $\yhat$, $\zhat$ (whose base is at the origin). 1) The cross product also fails to be associative, since for example $\xhat\times(\xhat\times\yhat)=-\yhat$ but $(\xhat\times\xhat)\times\yhat=\zero$. 2) This is really the multiplication table for the unit imaginary quaternions, a number system which generalizes the familiar complex numbers. Quaternions predate vector analysis, which borrowed the $i$, $j$, $k$ notation for the rectangular basis vectors, which are often written as $\ii$, $\jj$, $\kk$. Here, we have adopted instead the more logical names $\xhat$, $\yhat$, $\zhat$. 3) This argument uses the distributive property, which must be proved geometrically if one starts with($\ref{crossmagnitude}$) and the right-hand rule. This is straightforward in two dimensions, but somewhat more difficult in three dimensions. As with the dot product, see our online article at this URL. Views Article Discussion Show pagesource Old revisions Navigation About this wiki Tables of Contents Activities Technical info Contact Us Site map Recent Changes Personal Tools Login Search Toolbox What links here Upload file Special Pages Printable Version Permanent link Cite this Article book/mathcontent/cross2.txt (10391 views) · Last modified: 2015/08/27 22:37 (external edit) © 2009–2015, Tevian Dray and Corinne A. Manogue OBSOLETE: A newer version is available.
189118	https://www.engineersedge.com/manufacturing_spec/properties_of_metals_strength.htm	\| \| \| \| --- \| \| \| \| \| \| Membership Services \| Modulus of Elasticity, Young's Modulus For Common Engineering Materials Table Engineering Metals and Materials Table of Contents The following chart gives ultimate strength, yield point and modulus of elasticity data for steel and iron. \| \| \| \| \| --- --- \| \| Engineering Material \| Modulus of Elasticity (E) \| Ultimate Tensile Strength MPa (σu) \| Yield Strength MPa (σy) \| \| \| \| \| ABS Plastics \| 210 - 456 \| 1.45 - 3.15 \| \| \| \| A53 Seamless and Welded Steel Pipe - Grade A \| \| \| \| \| \| A53 Seamless and WeldedSteel Pipe - Grade B \| 30,457 \| 210 \| \| \| \| A106 Seamless Carbon Steel PipeGrade A \| 30,457 \| 210 \| \| \| \| A106 Seamless Carbon Steel PipeGrade B \| 30,457 \| 210 \| \| \| \| A106 Seamless Carbon Steel Pipe Grade C \| 30,457 \| 210 \| 483 \| \| \| A252 Piling Steel Pipe Grade 1 - \| 345 \| 207 \| \| A252 Piling Steel PipeGrade 2 - \| 414 \| 241 \| \| A252 Piling Steel Pipe Grade 3 - \| \| \| \| A501 Hot Formed Carbon Steel Structural Tubing - Grade A - \| 400 \| 248 \| \| A501 Hot Formed Carbon Steel Structural Tubing - Grade B - \| 483 \| 345 \| \| A523 Cable Circuit Steel PipingGrade A - \| 331 \| 207 \| \| A523 Cable Circuit Steel PipingGrade B - \| 414 \| 241 \| \| A618 Hot-Formed High-Strength Low-AlloyStructural Tubing - Grade Ia & Ib - \| 483 \| 345 \| \| A618 Hot-Formed High-Strength Low-AlloyStructural Tubing - Grade II - \| 414 \| 345 \| \| A618 Hot-Formed High-Strength Low-AlloyStructural Tubing - Grade III - \| \| 345 \| \| API 5L Line Pipe - \| 310 - 1145 \| 175 - 1048 \| \| \| 403 - 1407 \| 2.78 - 9.7 \| \| \| \| \| \| Aluminum Bronze \| \| - \| \| \| \| \| 110 \| \| \| Aluminum Alloys \| \| 70 - \| \| \| 2900 \| 20 - \| \| Aramid \| 8702 - 20,305 \| 60 - 140 - \| \| Beryllium (Be) \| 43,900 \| 303 - \| \| Beryllium Copper \| 18999.9 \| 131 - \| \| Bismuth \| 1,740 \| 12 - \| \| Bone, spongy \| 1.45 - 435 \| 0.01 - 3.0 - \| \| Boron \| 65,266 \| 450 3100 \| \| Brass \| 14,100 \| 97 \| 250 \| Brass, Naval \| 17,000 \| 117 - \| \| Bronze \| 10,500 - 20,000 \| 72.4 - 138 - \| \| Carbon Fibers \| 33,068 \| 228 - \| \| Cadmium \| 8,010 \| 55.2 - \| \| Carbon nanotube, single-walled \| 39,160 - 137,785 \| 270 - 950 - \| \| Cast Iron 4.5% C, ASTM A-48 - \| 170 \| Cellulose, cotton, wood pulp - \| 80 - 240 \| Cellulose acetate, molded - \| 12 - 58 \| Cellulose acetate, sheet - \| 30 - 52 \| Cellulose nitrate, celluloid - \| 50 \| Chlorinated polyether - \| 39 \| Chlorinated PVC (CPVC) - - \| \| Chromium \| 36,000 \| 248 - \| \| Cobalt \| 30,600 \| 211 - \| \| Concrete - - \| \| Concrete, High Strength(compression) - \| 40 (compression) \| Copper \| 16,969 \| 117 \| 220 \| 70 \| \| Diamond (C) \| 102,000 - 174,000 \| 700 - 1,200 - \| \| Douglas fir Wood - \| 50 (compression) \| Epoxy resins - \| 26 - 85 \| Fiberboard, Medium Density - - \| \| Flax fiber - - \| \| Glass \| 17 - 29 \| 0.12 - 0.2 \| 50 (comp.) \| Glass reinforced polyester matrix - - \| \| Gold \| 10,732 \| 74 - \| \| Granite \| 5,801 \| 40 - \| \| Graphene 1000 - \| \| Grey Cast Iron 130 - \| \| Hemp fiber 35 - \| \| Inconel \| 29,700 \| 205 - \| \| Iridium \| 76,000 \| 524 - \| \| Iron \| 28,500 \| 200 - \| \| Lead \| 2,030 \| 14 - \| \| Magnesium metal (Mg) \| 6,527 \| 45 - \| \| Manganese \| 23,100 \| 159 - \| \| Marble \| 7,832 \| ~54 \| 15 \| Medium-density fiberboard \| 580 \| 4 - \| \| Mercury - - \| \| Molybdenum (Mo) \| 47,717 \| 329 - \| \| Monel Metal \| 24,000 \| 160 - \| \| Nickel \| 30,022 \| 207 - \| \| Nickel Silver \| 18,129 \| 125 - \| \| Nickel Steel \| 27,000 \| 186 - \| \| Niobium (Columbium) \| 14,938 \| 103 - \| \| Nylon-6 \| 290 - 580 \| 2 - 4 \| 45 - 90 \| 45 \| \| Nylon-66 - \| 60 - 80 \| Oak Wood (along grain) 11 - \| \| Osmium \| 1,435 - 1,638 \| 9.9 - 11.3 - \| \| Phenolic cast resins - \| 33 - 59 \| Phenol-formaldehydemolding compounds - \| 45 - 52 \| Phosphor Bronze \| 16,244 \| 112 - \| \| Pine Wood (along grain) 9 \| 40 \| Platinum \| 24,800 \| 171 - \| \| Plutonium \| 14,000 \| 96.5 - \| \| Polyacrylonitrile, fibers - \| 200 \| Polybenzoxazole \| 362.6 \| 2.5 - \| \| Polycarbonates \| 377 \| 2.6 \| 52 - 62 \| Polyethylene HDPE (high density) \| 81 - 217.6 \| 0.565 - 1.5 \| 15 \| Polyethylene Terephthalate, PET \| 870 - 2030.5 \| 6 - 14 \| 55 \| Polyamide \| 420 \| 2.95 \| 85 \| Polyisoprene, hard rubber - \| 39 \| Polymethylmethacrylate (PMMA) \| 348 - 493 \| 2.4 - 3.4 - \| \| Polyimide aromatics \| 449 \| 3.1 \| 68 \| Polypropylene, PP \| 217 - 290 \| 1.5 - 2 \| 28 - 36 \| Polystyrene, PS \| 435 - 507.6 \| 3 - 3.5 \| 30 - 100 \| Polyethylene, LDPE (low density) \| 0.11 - 65.2 \| 0.11 - 0.45 - \| \| Polytetrafluoroethylene (PTFE) \| 58 \| 0.4 - \| \| Polyurethane cast liquid \| 90.5 - 797.7 \| 0.624 - 5.5 \| 10 - 20 \| Polyurethane elastomer - \| 29 - 55 \| Polyvinylchloride (PVC) \| 348 - 594 \| 2.4 - 4.1 - \| \| Potassium - - \| \| Rhodium \| 52,100 \| 359 - \| \| Rubber, small strain \| 1.45 - 14.5 \| 0.01 - 0.1 - \| \| Sapphire \| 36,259 \| 250 - \| \| Selenium - - \| \| Silicon \| 16,302 \| 112.4 - \| \| Silicon Carbide \| 65,266.9 \| 450 3440 \| \| Silver \| 11,022.8 \| 76 - \| \| Sodium - - \| \| Steel, High Strength Alloy ASTM A-514 \| 29,000 \| 200 \| 755 \| 670 \| \| Steel, stainless AISI 302 \| 28,000 \| 193 \| 850 \| 506 \| \| Steel, Structural ASTM-A36 \| 29,000 \| 200 \| 403 \| 245 \| \| Tantalum \| 27000 \| 186 - \| \| Polytetrafluoroethylene (PTFE) \| 56.9 - 326 \| 0.392 - 2.25 - \| \| Thorium \| 10,602 \| 73.1 - \| \| Tin \| 6,030 \| 41.6 - \| \| Titanium \| 16 \| 120 - \| \| Titanium Alloy \| 17,404 \| 105 - 120 \| 900 \| 730 \| \| Tooth enamel \| 464 - 14,242 \| 3.2 - 98.2 - \| \| Tungsten (W) \| 58,000 \| 400 - \| \| Tungsten Carbide (WC) \| 65,266 - 94,274 \| 450 - 650 - \| \| Uranium, Cast \| 27,557 \| 190 - \| \| Vanadium \| 18,200 \| 125.5 - \| \| Wrought Iron \| 14,242.4 \| 98.2 - \| \| Zinc \| 14,000 \| 96.5 - \| \| \| \| \| \| --- --- \| \| Material \| Ultimate Strength \| Yield Point X 1000/in2 \| Modulus of Elasticity \| \| (T) Tension X 1000/in2 \| Compression, in terms of T \| Shear in terms of T \| in Tension (E) x 106 psi \| in Shear, in terms of E \| \| Cast iron, grey, class 20 \| 20a \| 3.6 T - 4.4 T \| 1.6 T \| ....... \| 11.6 \| 0.40 E \| \| class 25 \| 25a \| 3.6 T - 4.4 T \| 1.4 T \| ....... \| 14.2 \| 0.40 E \| \| class 30 \| 30a \| 3.6 T - 4.4 T \| 1.4 T \| ....... \| 14.5 \| 0.40 E \| \| class 35 \| 35a \| 3.6 T - 4.4 T \| 1.4 T \| ....... \| 16.0 \| 0.40 E \| \| class 40 \| 40a \| 3.6 T - 4.4 T \| 1.3 T \| ....... \| 17 \| 0.40 E \| \| class 50 \| 50a \| 3.6 T - 4.4 T \| 1.3 T \| ....... \| 18 \| 0.40 E \| \| class 60 \| 60a \| 2.8 T \| 1.O T \| ....... \| 19.9 \| 0.40 E \| \| Malleable \| 40 to 100 \| ....... \| ....... \| 30 to 80 \| 25 \| 0.43 E \| \| nodular (ductile iron) \| 60 to 120 \| ....... \| ....... \| 40 to 90 \| 23 \| ...... \| \| Cast Steel, carbon \| 60 to 100 \| T \| 0.75 T \| 30 to 70 \| 30 \| 0.38 E \| \| low alloy \| 70 to 200 \| T \| 0.75 T \| 45 to 170 \| 30 \| 0.38 E \| \| Steel, SAE 950 \| 66 to 70 \| T \| 0.75 T \| 45 to 50- \| 30 \| 0.38 E \| \| 1025 (low carbon) \| 60 to 103 \| T \| 0.75 T \| 40 to 90 \| 30 \| 0.38 E \| \| 1045 (medium carbon) \| 80 to 182 \| T \| 0.75 T \| 50 to 162 \| 30 \| 0.38 E \| \| 1095 (high carbon) \| 90 to 213 \| T \| 0.75 T \| 20 to 150 \| 30 \| 0.38 E \| \| 1112 ( free cutting) \| 60 to 100 \| T \| 0.75 T \| 30 to 95 \| 30 \| 0.38 E \| \| 1212 ( free cutting) \| 57 to 80 \| T \| 0.75 T \| 25 to 72 \| 30 \| 0.38 E \| \| 1330 (alloy) \| 90 to 162 \| T \| 0.75 T \| 27 to 149 \| 30 \| 0.38 E \| \| 2517 (alloy) \| 88 to 190 \| T \| 0.75 T \| 60 to 155 \| 30 \| 0.38 E \| \| 3140 (alloy) \| 93 - 188 \| T \| 0.75 T \| 62 to 162 \| 30 \| 0.38 E \| \| 3310 (alloy) \| 104 to 172 \| T \| 0.75 T \| 56 to 142 \| 30 \| 0.38 E \| \| 4023 (alloy) \| 105 to 170 \| T \| 0.75 T \| 60 to 114 \| 30 \| 0.38 E \| \| 4130 (alloy) \| 81 to 179 \| T \| 0.75 T \| 46 to 161 \| 30 \| 0.38 E \| \| 4340 (alloy) \| 109 to 220 \| T \| 0.75 T \| 68 to 200 \| 30 \| 0.38 E \| \| 4640 (alloy) \| 98 to 192 \| T \| 0.75 T \| 62 to 169 \| 30 \| 0.38 E \| \| 4820 (alloy) \| 98 to 209 \| T \| 0.75 T \| 68 to 184 \| 30 \| 0.38 E \| \| 5150 (alloy) \| 98 to 210 \| T \| 0.75 T \| 51 to 190 \| 30 \| 0.38 E \| \| 52100 (alloy) \| 100 to 238 \| T \| 0.75 T \| 81 to 228 \| 30 \| 0.38 E \| \| 6150 (alloy) \| 96 to 228 \| T \| 0.75 T \| 59 to 210 \| 30 \| 0.38 E \| \| 8650 (alloy) \| 110 to 228 \| T \| 0.75 T \| 69 to 206 \| 30 \| 0.38 E \| \| 8740 (alloy) \| 100 to 179 \| T \| 0.75 T \| 60 to 165 \| 30 \| 0.38 E \| \| 9310 (alloy) \| 117 to 187 \| T \| 0.75 T \| 63 to 162 \| 30 \| 0.38 E \| \| 9840 (alloy) \| 120 to 285 \| T \| 0.75 T \| 45 to 50 \| 30 \| 0.38 E \| \| Chrome Alloy 13 \| 65 ksi 449 MPa \| 40 ksi 276 MPa \| \| \| \| Steel Stainless, SAE \| \| 30302f \| 85 to 125 \| T \| ....... \| 35 to 95 \| 28 \| 0.45 E \| \| 30321f \| 85 to 95 \| T \| ....... \| 30 to 60 \| 28 \| ....... \| \| 30347f \| 90 to 100 \| T \| ....... \| 35 to 65 \| 28 \| 0.40 E \| \| 51420f \| 95 to 230 \| T \| ....... \| 50 to 195 \| 29 \| ....... \| \| 51430f \| 75 to 85 \| T \| ....... \| 40 to 70 \| 29 \| ....... \| \| 51446f \| 80 to 85 \| T \| ....... \| 50 to 70 \| 29 \| ....... \| \| 51501f \| 70 to 175 \| T \| ....... \| 30 to 135 \| 29 \| ....... \| \| Structural Steel \| \| Common \| 60 to 75 \| T \| 0.75 T \| 33 \| 29 \| 0.41 E \| \| Rivet \| 52 to 62 \| T \| 0.75 T \| 28 \| 29 \| ....... \| \| Rivet, high strength \| 68 to 82 \| T \| 0.75 T \| 38 \| 29 \| ....... \| \| Wrought iron \| 34 to 54 \| T \| 0.75 T \| 23 to 32 \| 28 \| ....... \| a - Minimum specified value of the American Society of Testing Materials. References Cast irom, ASTM A48, structural steel fro bridges and structures, ASTM A7. Structural rivet steel , ASTM A141; high-strength structural rivet steel, ASTM A195 1 Pa (N/m2 ) = 1x10-6 N/mm2 = 1.4504x10-4 psi 1 MPa = 10 6 N/m2 = 0.145x10 3 psi (lb f /in2 ) = 0.145 ksi 1 GPa = 10 9 N/m2 = 10 6 N/cm2 = 10 3 N/mm2 = 0.145x10 6 psi (lb f /in2 ) 1 psi (lb/in2 ) = 0.001 ksi = 144 psf (lb f /ft2 ) = 6,894.8 Pa (N/m2 ) = 6.895x10 -3 N/mm2 Engineering Properties of Metals Yield Strength review and materials chart Modulus of Elasticity, Average Properties of Structural Materials, Shear Modulus, Poisson's Ratio, Density Thermal Properties of Metals, Conductivity, Thermal Expansion, Specific Heat Engineering Properties of Metals Inconel 718 Modulus of Elasticity at Low Temperature Yield Strength review and materials chart Modulus of Elasticity Young's Modulus Strength for Metals - Iron and Steel Strength of Materials Basics and Equations \| Mechanics of Materials Shear Modulus of Rigidity Table of Engineering Materials Modulus of Elasticity, Average Properties of Structural Materials, Shear Modulus, Poisson's Ratio, Density Thermal Properties of Metals, Conductivity, Thermal Expansion, Specific Heat Young's Modulus Link to this Webpage: Engineers Edge: Copy Text to clipboard © Copyright 2000 - 2025, by Engineers Edge, LLC www.engineersedge.com All rights reservedDisclaimer \| Feedback Advertising \| Contact \| \| \| \| \| \| \| Home Engineering Book Store Engineering Forum Applications and Design Beam Deflections and Stress Bearing Apps, Specs & Data Belt Design Data Calcs Civil Engineering Design & Manufacturability Electric Motor Alternators Engineering Calculators Engineering Terms Excel App. Downloads Flat Plate Stress Calcs Fluids Flow Engineering Friction Engineering Gears Design Engineering General Design Engineering Hardware, Imperial, Inch Hardware, Metric, ISO Heat Transfer Hydraulics Pneumatics HVAC Systems Calcs Economics Engineering Electronics Instrumentation Engineering Mathematics Engineering Standards Finishing and Plating Friction Formulas Apps Lubrication Data Apps Machine Design Apps Manufacturing Processes Materials and Specifications Mechanical Tolerances Specs Plastics Synthetics Power Transmission Tech. 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189119	https://pydocs.github.io/p/scipy/1.8.0/api/scipy.special._basic.jn_zeros.html	Document Typesetting math: 100% / p/ scipy/ IPythonastropydaskdistributedmatplotlibnetworkxnumpypandaspapyrireadthedocsscipyskimage 1.8.0/ 1.8.0 API/ APIGalleryExamplesNarrative scipy. special. _libfftintegrateinterpolatelinalgmiscndimageoptimizesignalsparsespatialspecialstats _basic. _add_newdocs_basic_comb_ellip_harm_ellip_harm_2_lambertw_logsumexp_orthogonal_sf_error_specfun_spfun_stats_spherical_bessel_ufuncs_ufuncs_cxxadd_newdocsbasiccython_specialorthogonalsf_errorspecfunspfun_stats jn_zeros _bessel_diff_formula_nonneg_int_or_fail_range_prodai_zerosassoc_laguerrebei_zerosbeip_zerosber_zerosbernoulliberp_zerosbi_zerosclpmncombdiricerf_zeroseulerfactorialfactorial2factorialkfresnel_zerosfresnelc_zerosfresnels_zerosh1vph2vpivpjn_zerosjnjnp_zerosjnp_zerosjnyn_zerosjvpkei_zeroskeip_zeroskelvin_zerosker_zeroskerp_zeroskvplmbdalpmnlpnlqmnlqnmathieu_even_coefmathieu_odd_coefobl_cv_seqpbdn_seqpbdv_seqpbvv_seqpermpro_cv_seqriccati_jnriccati_yny0_zerosy1_zerosy1p_zerosyn_zerosynp_zerosyvp scipy1.8.0PypiGitHubHomepage Other Docs ParametersReturnsBackRef jn_zeros(n, nt) Compute :None:None:nt`` zeros of the Bessel functions J n(x) on the interval (0,∞) . The zeros are returned in ascending order. Note that this interval excludes the zero at x=0 that exists for n>0 . Parameters n : int Order of Bessel function nt : int Number of zeros to return Returns ndarray First :None:None:nt`` zeros of the Bessel function. Compute zeros of integer-order Bessel functions Jn. See Also jv Examples import scipy.special as sc We can check that we are getting approximations of the zeros by evaluating them with jv . n = 1 ...x = sc.jn_zeros(n, 3) ...x array([ 3.83170597, 7.01558667, 10.17346814])>>> sc.jv(n, x) array([-0.00000000e+00, 1.72975330e-16, 2.89157291e-16]) Note that the zero at x = 0 for n > 0 is not included. sc.jv(1, 0) 0.0 See : Back References The following pages refer to to this document either explicitly or contain code examples using this. scipy.special._basic.jnyn_zeros``scipy.special._basic.jnjnp_zeros``scipy.special._basic.jn_zeros Local connectivity graph Hover to see nodes names; edges to Self not shown, Caped at 50 nodes. Using a canvas is more power efficient and can get hundred of nodes ; but does not allow hyperlinks; , arrows or text (beyond on hover) SVG is more flexible but power hungry; and does not scale well to 50 + nodes. All aboves nodes referred to, (or are referred from) current nodes; Edges from Self to other have been omitted (or all nodes would be connected to the central node "self" which is not useful). Nodes are colored by the library they belong to, and scaled with the number of references pointing them GitHub : /scipy/special/_basic.py#297 type: Commit:
189120	https://www1.eere.energy.gov/manufacturing/industries_technologies/imf/pdfs/1797_advancedcatechnologyfinal.pdf	LAUR 05-2444 Final Technical Report Advanced Chlor-Alkali Technology DOE Award 03EE-2F/ED190403 Project Period 10:01 – 09:04 Jerzy Chlistunoff Phone 505-667-7192 e-mail: jerzy@lanl.gov Los Alamos National Laboratory Los Alamos, NM 87544 This report is based upon work supported by the U.S. Department of Energy under Award No. 03EE-2F/ED190403 Any findings, opinions, and conclusions or recommendations expressed in this report are those of the author and do not necessarily reflect the views of the Department of Energy. - 3 - Table of Contents List of Figures................................................................................................. 4 1. Executive Summary ................................................................................... 6 2. Introduction................................................................................................. 7 3. Background................................................................................................. 9 4. Technical Part........................................................................................... 12 4.1 Critical Review of the Literature.............................................................................12 4.2 Experimental............................................................................................................15 4.2.1 Cells ...................................................................................................................15 4.2.2. Experimental setup and conditions ...............................................................19 4.2.3. Analytical procedures .....................................................................................20 4.2.4. Standard startup, shutdown, current ramping and cell protecting procedures .................................................................................................................21 4.2.4.1. Startup procedure and current ramping procedure ...............................21 4.2.4.2. Shutdown procedure.................................................................................22 4.2.4.3. Temporary shutdown of the cell ..............................................................22 4.2.5. Instrumentation................................................................................................23 4.3. Results ....................................................................................................................24 4.3.1. Cathode Modifications Aimed at Improving Caustic Current Efficiency.....24 4.3.2. Membrane Testing ...........................................................................................42 4.3.3. Cathodes Utilizing Unsupported Catalysts....................................................49 4.3.4. Corrosion of the Cathode Hardware ..............................................................59 4.3.5. Anode Modifications........................................................................................63 4.3.6. Effects of Different Factors on Peroxide Generation Rate...........................71 Accomplishments......................................................................................... 93 Conclusions and Recommendations ......................................................... 93 Acknowledgements...................................................................................... 97 Bibliography.................................................................................................. 98 List of Figures Figure 1. 30% lowering in cell voltage. Figure 2. Components of the oxygen-depolarized cell with the separate gas diffusion cathode and metal cathode hardware. Figure 3. Components of the oxygen-depolarized cell with the graphite cathode hardware. Figure 4. Components of the oxygen-depolarized cell equipped with the membrane-electrode-assembly (MEA) type gas diffusion cathode. Figure 5. Diagram of the reagent/product delivery/collection system. Figure 6. Cell voltages measured at low oxygen humidification levels. Figure 7. Concentrations of NaOH generated with and without oxygen humidification and the number of water molecules transferred through the membrane as a result of the electroosmotic drag. Figure 8. High frequency resistance at different oxygen humidification levels. Figure 9. Caustic current efficiency at different oxygen humidification levels. Figure 10. Caustic current efficiency at different oxygen humidification levels plotted versus concentration of the product NaOH. Figure 11. Effect of direct humidification of the hydrophilic spacer on the caustic current efficiency. Figure 12. Effect of direct humidification of the hydrophilic spacer on the cell voltage. Figure 13. Effect of an uncontrolled interruption of the electrolysis on caustic current efficiency in the cell equipped with the integrated graphite cathode flow-field/current collector. Figure 14. Effect of an uncontrolled interruption of the electrolysis on caustic current efficiency in the cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. Membrane 4. Figure 15. Effect of a controlled interruption of the electrolysis on caustic current efficiency in a cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. Membrane 4. Figure 16. Effect of a controlled interruption of the electrolysis on caustic current efficiency in a cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. Membrane 1. Figure 17. Measured cell voltages for different bi-layer membranes. Figure 18. Current efficiency for different bi-layer membranes. Figure 19. Energy efficiency of different bi-layer membranes at low current densities. Figure 20. Energy efficiency (see text) of different bi-layer membranes at high current densities. Figure 21. Measured and iR-corrected voltages for the cells equipped with the standard ELAT cathode and the hydrophilic spacer and the MEA-type cathode. Figure 22. iR-corrected voltages for the MEA-type cathodes (red line) containing 5.0 mg/cm2 of different catalysts Figure 23. High frequency resistance of the MEA containing 5.0 mg/cm2 Pt. Figure 24.a. Dissolution rates of the ruthenium-containing catalysts plotted versus the cell voltage corrected for the ohmic losses. Figure 24.b. UV-Vis spectra of the NaOH solutions generated at different current densities. Figure 25. Comparison of 60% teflonized Toray® paper and LT 1400-W gas diffusion layers in the cells equipped with the MEAs containing 5.0 mg/cm2 Ru catalyst. - 5 - Figure 26. Comparison of performance of single-sided and double-sided gas diffusion layers applied in the cells equipped with the MEAs containing 5.0 mg/cm2 Pt catalyst (HiSpec® 1000). Figure 27. Effect of controlled and uncontrolled interruption of the electrolysis on performance of the cell equipped with the MEA-type cathode. Figure 28. Comparison of caustic current efficiency of the cells equipped with MEA and with separate gas diffusion electrode. Figure 29. Corrosion of the gold plated nickel cathode current collectors resulting from a sixteen day power failure. Figure 30. Significant increase of the cell voltage upon anode structure modification resulting from the use of the glass fiber cloth as a spacer between the anode and the membrane. Figure 31. Effect of modification of the anode structure on caustic current efficiency. Figure 32. Effects of different anode structure modifications on the measured cell voltages during the first 120 hours of the cell operation. Figure 33. Energy efficiency of the cells equipped with different anode structures plotted versus current density. Fig. 34. Effect of current density and Pt loading on peroxide generation. Fig. 35. Time effect on peroxide generation rate. Fig. 36. Effect of brine concentration on peroxide generation rate and NaOH concentration. Fig. 37. Effect of brine concentration on membrane resistance and caustic current efficiency. Fig. 38. Comparison of effects of oxygen stream humidification and brine concentration on peroxide generation rate. Fig. 39. Changes of chloride and peroxide concentrations in caustic soda during the first 300 hours of cell operation at 1.0 A/cm2. Figure. 40. Effect of the hydrophilic spacer on peroxide generation rate. Figure 41. Comparison of peroxide generation rates in cells equipped with gold- and silver-plated hardware. Figure 42. Comparison of peroxide generation rates in cells equipped with MEAs containing 5.0 mg/cm2 of the Pt catalyst (HiSpec 1000) and different gas diffusion layers. Fig. 43. Peroxide generation rates in the cells equipped with the modified cathode structure 1 (see text) and with standard ELAT cathode and hydrophilic spacer (Panex 30). Figure 44. Effects of long time cell operation on peroxide generation rate in a cell equipped with the modified cathode structure 1. Figure 45. Effects of current density on cell voltages and high frequency resistance for the LANL standard cathode hardware and the modified structure 1. Figure 46. Effects of change of experimental conditions on performance of the cell equipped with the modified cathode structure 1. Figure 47. Dependence of peroxide generation rate on the oxygen flow rate for the modified cathode structure 1. Figure 48. Estimate of the limits of safe cell operation from hydrogen evolution in a deliberately flooded cell equipped with the modified cathode structure 1. Figure 49. Determination of the highest operational current density from the onset of hydrogen evolution in a cell with the flooded cathode. Figure 50. Peroxide generation in a cell equipped with the modified cathode structure 2. 1. Executive Summary The purpose of the project was continuation of the development of new chlor-alkali electrochemical reactors (ECRs) that employ oxygen-depolarized cathodes. Due to their lower operating voltages, the oxygen-depolarized reactors consume up to 30% less electrical energy per unit weight of the products (chlorine and caustic soda), than the state-of-the-art membrane electrolyzers with hydrogen-evolving cathodes. The existing chlor-alkali membrane cells could not be simply retrofitted to accommodate the oxygen-depolarized electrodes due to the different principles and conditions of operation of the oxygen-depolarized cathodes and hydrogen-evolving cathodes. A completely new design of the oxygen-depolarized electrolyzer and optimization of the conditions of electrolysis were required in order to achieve satisfactory cell performance that would match or exceed that commonly attained in the state-of-the-art membrane cells with hydrogen-evolving cathodes. The zero-gap design of chlor-alkali cells developed at LANL was based on the fuel cells, where the cathode and the anode remain in intimate contact with the ion-exchange membrane. Since the performance of any individual cell component can influence the performance of other components, the experimental effort focused not only on the oxygen cathode, but also on the anode, the membrane, the cell hardware, and the operating conditions. The vital performance characteristics which were monitored included the overall cell voltage, high frequency membrane resistance, caustic current efficiency, and composition of the caustic product. Our major accomplishments include: 1. Elucidation of the mechanism of the unwanted peroxide generation 2. Development of the method to eliminate peroxide 3. Increasing the caustic current efficiency to match current industrial standards for membrane cells 4. Elimination of the corrosion of the cathode hardware 5. Evaluation of the ion exchange membranes Although many problems associated with optimization of the technology were solved, some issues require additional work. Corrosion of the cathodes employing carbon supported platinum catalyst seems to be the major drawback of the current cell design. - 7 - While the extent of corrosion can be significantly reduced by using cathode protection techniques, the technology would greatly benefit if the cathode was completely corrosion resistant. Consequently, any future effort should focus on developing the corrosion resistant cathode. Based on the current knowledge, the corrosion resistant cathode should involve a carbon supported metal catalyst, which would be less noble than platinum. Metal alloys or elemental silver are most promising choices for the catalyst. Another possible way to eliminate cathode corrosion may involve alternative supports for the catalyst. However, this path may be more effort intensive and less likely to produce the desired effect without creating other problems. On the contrary, all the technical improvements made so far are believed to be operational with the corrosion resistant cathode employing a carbon supported metal catalyst. 2. Introduction Total US production of chlorine reached the level of 13.0 million tons in 2003, while the corresponding caustic soda production approached 14.7 million tons. Approximately 70% of chlorine and caustic produced in the US is manufactured by the diaphragm technology, 12% by the mercury technology and 18% by the modern membrane technology. The mercury process is the most energy intensive and consumes around 3700 kWh of electricity per metric ton of chlorine. The corresponding numbers for the diaphragm and membrane technologies are 2900 kWh/t and 2500 kWh/t, respectively. Overall, the chlor-alkali electrolysis is one of the most energy intensive industrial operations. It consumes approximately 10 GW (equivalent to 317 trillion BTU/yr or 87600 GWh/yr) of electrical energy, which corresponds to around 2% of the total electric power generated in the United States. Attempts to reduce manufacturing costs of chlorine have recently led to modifications of the conventional membrane electrolyzers that allow for operation at around 50% higher throughput (0.6 A/cm2) than the standard cells. While these modifications lower the capital and maintenance costs, they result in higher cell voltages and consequently in higher energy consumption. As the energy consumption per unit weight of the products is directly proportional to the electrochemical reactor cell voltage, lowering of the cell voltage is the only route to energy savings. Over the last several years, membrane technology has been optimized to the extent that no viable reduction of the cell voltage is expected from further cell modifications. However, by replacing the hydrogen-evolving cathodes in the membrane chlor-alkali cells by oxygen-depolarized cathodes, the cell voltages and corresponding power consumption could be reduced by as much as 30% (Fig. 1). b) Chlor-Alkali Cell with Oxygen Cathode Na+ H2O OH- NaCl NaOH Cl- ANODE Cl2 O2 CATHODE a) Conventional Chlor-Alkali Cell NaCl NaOH Na+ H2O OH- Cl- ANODE Cl2 CATHODE H2 Voltage ≈ 3.1 V at 0.4 A/cm2 Voltage ≈ 2.1 V at 0.4 A/cm2 Figure 1. 30% lowering in cell voltage and, hence, 30% lowering in energy consumption per unit weight of the products is achieved in chlor-alkali cells by replacing the hydrogen-evolving cathode by an oxygen-consuming cathode. Figure 1 explains schematically the relationship between the ordinary membrane chlor-alkali ECRs (left) and the modified chlor-alkali ECR employing an oxygen cathode. Electric energy savings that could be achieved by implementing this modification in chlor-alkali ECRs are very significant. At a current density of 0.30 to 0.45 A/cm2, typically used by the industry, replacement of the hydrogen-evolving cathode by an oxygen-reducing cathode has been clearly shown by us to date to save around 1.0 V out of the 3.1 V to 3.3 V required for operation of conventional chlor-alkali ECRs. Since the electric power (energy) consumed per unit weight of product is directly proportional to the cell voltage, the reduction in the cell voltage at a constant current density amounts to savings of 30% in the - 9 - electrolysis energy. The energy savings would be even more substantial if all the chlor-alkali plants switched to oxygen-fed membrane reactors. They can reach as much as 40%. Therefore, introduction of the energy saving, oxygen-depolarized reactors would amount up to a 0.8% cut in the overall US electric power consumption. It would also amount to a 0.8% cut in total US CO2 emissions associated with generation of electric power. When the research described in this report was awarded funding, the hydrogen generated by the industry was quite commonly considered a low-value byproduct and many chlor-alkali plants were venting it to the atmosphere. Today, the Government recognizes hydrogen as a fuel of the future and funds the research aimed at both developing new technologies that would utilize hydrogen rather than the conventional fuels and finding efficient and safe ways of hydrogen storage. Quite recently, Dow Chemical installed the hydrogen/air fuel cell stacks supplied by General Motors to capture part of the energy stored in hydrogen generated by the Dow’s chlor-alkali plant in Freeport, Texas. While equipping the existing chlor-alkali plants with the hydrogen “burning” fuel cells seems feasible, it will never offer comparable energy and environmental benefits as the chlor-alkali technology utilizing oxygen-depolarized cathodes. First of all, the fuel cells can return only around 50% of the clean hydrogen energy, whereas much more of the “dirty” fossil fuel energy is consumed to generate the hydrogen. From the economic perspective, installation of the fuel cell stacks translates into yet another capital investment, in addition to the investment associated with constructing and equipping the chlor-alkali cell room. Consequently, it is our belief that the economic and the energy/environmental factors are in favor of the technology that utilizes oxygen-depolarized cells and that this technology will eventually replace the current chlorine/caustic manufacturing methods. 3. Background All three chlor-alkali technologies, which are currently in use, have advantages and disadvantages. The most frequently employed diaphragm technology uses very inexpensive cells, but it produces relatively dilute and impure NaOH. Consequently, significant use of steam is required to concentrate the hydroxide solution before it can be shipped to the point of use. Moreover, many diaphragm cells still utilize asbestos-based diaphragms, which raises environmental and health concerns. The mercury technology produces highly concentrated, 50% hydroxide of the highest purity. In this case, no concentration steps are required. However, the electrolysis itself is very energy intensive. In addition, the mercury emissions from the process raise very serious environmental concerns. This method will soon be phased out. The membrane technology produces high purity hydroxide and requires less steam to concentrate the NaOH solution than the diaphragm method. It is currently the cleanest and most energy efficient chlor-alkali technology. However, the membrane cells are relatively expensive, mainly due to the high cost of the specialized bi-layer ion exchange membranes that they employ. Nonetheless, this method completely dominates the chlor-alkali industry in Japan, where the energy prices are higher and the environmental regulations stricter than in the United States. The attractiveness of the membrane technology with oxygen-depolarized cathodes versus the existing technologies relies on the substantial energy savings and the associated environmental benefits that this technology offers. Despite how simple it sounds, replacement of the hydrogen-evolving cathode in a membrane cell requires significant changes in the cell design and materials as well as in the cell operating conditions. The key element of the oxygen-depolarized cell is the oxygen cathode of the gas diffusion type. The electrode has to be designed to facilitate formation of the three-phase boundaries (gas/liquid/solid) that involve oxygen, water/caustic soda solution, and the catalyst particles. Moreover, it has to effectively manage transport of oxygen to and caustic soda from the catalyst layer. Although gas diffusion electrodes (GDEs) have been quite extensively used in fuel cells and optimized for this application, the conditions encountered in the cathode compartment of a chlor-alkali electrolyzer are quite different. High viscosity of the concentrated NaOH and its strongly corrosive properties may have a negative influence on both formation of the three-phase boundaries in the electrode pores and transport of the reagents and products to and from the reaction site. In the worst case scenario, the electrode pores may get completely flooded with the NaOH solution and the cathode, where the hydrogen evolution takes place, will operate at much lower potential. The presence of oxygen and significantly higher potential of the oxygen-depolarized cathode as compared to the hydrogen-evolving cathode can result in corrosion of the - 11 - cathode hardware and other components in the cathode compartment, which would not be observed in the standard membrane cell thanks to the cathodic protection and reducing properties of hydrogen. Another difference of practical importance between the hydrogen-evolving cells and oxygen-depolarized cells is the relative stability of the intermediate products of the hydrogen evolution reaction (HER) and the oxygen reduction reaction (ORR). The HER does not generate any stable intermediates: 2 H2O + 2 e- → H2 + 2 OH- (1) The ORR may follow the desired four-electron path: O2 + 2 H2O + 4 e- → 4 OH- (2) Alternatively, it can also occur according to the two-electron mechanism, which results in generation of peroxide: O2 + H2O + 2 e- → OH- + HO2 - (3) While the peroxide eventually decomposes and produces an equivalent amount of the hydroxide and thus does not lower the overall caustic current efficiency, it is a rather troublesome byproduct, because it produces gaseous oxygen upon decomposition and may also precipitate in the highly concentrated NaOH: 2 Na+ + OH- + HO2 - → Na2O2↓ + H2O (4) Precipitation of sodium peroxide (eq. 4) can cause liquid and gas flow maintenance problems, block the electrode active surface area and even destroy the microporous structure of the gas diffusion electrode. To our knowledge, there are no industrial-scale chlor-alkali plants equipped with oxygen-depolarized cells. However, there do exist pilot-scale plants in Japan and Europe that utilize oxygen-depolarized cells for the chlor-alkali electrolysis as well as for the similar process of HCl destruction. The technical details of these cells are not well known, but the finite-gap design seems to be the most prevalent. In cells of this type, the gas diffusion electrode acts as a separator of the oxygen and caustic chambers. The advantage of such a design is that the whole surface area of the ion-exchange membrane remains in contact with NaOH solution of the optimum concentration at all times, which guarantees the best overall membrane performance. However, the design also has serious drawbacks. The layer of NaOH between the electrode and the membrane contributes unfavorably to the overall cell resistance. In addition, as the hydrostatic pressure of the NaOH solution changes along the electrode height, maintaining the uniform distribution and identical properties of the three-phase boundaries and preventing the electrode flooding may pose problems. In order to avoid the problems typical for the finite gap / three compartment electrolyzers, we have chosen the so-called zero-gap design, adapted from the fuel cell technology. In the zero-gap cell, the cathode remains in intimate contact with the ion-exchange membrane. Oxygen is fed to and caustic collected from the common cathode compartment. Different modifications of the zero-gap design, which were used throughout this research, are described in more detail in the technical part of the report. The project was a continuation of our earlier work on oxygen-depolarized chlor-alkali cells. It addressed the issues, which were previously identified as critical for making the oxygen-depolarized membrane cells match the current performance standards for the state-of-the-art hydrogen-evolving membrane cells. Among these issues were caustic purity, current efficiency, corrosion protection, etc. The principal investigator is a member of the fuel cell team in Materials Science and Technology Division of the Los Alamos National Laboratory and together with Dr. Ludwig Lipp participated in the previous research on oxygen-depolarized chlor-alkali cells, which was directed by Dr. Shimshon Gottesfeld. 4. Technical Part 4.1 Critical Review of the Literature The first attempts to develop chlor-alkali technology utilizing the air cathode date back to the early eighties. The laboratory scale cells developed by ELTECH Systems were operated for more than 350 days on scrubbed air at 0.3 A/cm2, but the test with commercial size cells was terminated after 105 days due to the durability problem of the air cathode [1,2]. Even though eventually abandoned, these early experiments demonstrated that significant energy savings can be achieved by replacing the hydrogen electrode in the membrane cell with an oxygen electrode. Research aimed at implementing the oxygen - 13 - cathodes in chlor-alkali membrane cells has been especially intensive in the last 11-12 years and seems to be most advanced in Japan and Europe, where energy prices are higher and environmental policies stricter than in the United States. However, only a very limited amount of the published material covers the research done so far. One of the factors responsible for this situation is the proprietary nature of the results generated. Another factor results from the requirement of a stable cell performance for extended periods of time, comparable to that typical for the state-of-the-art membrane cells. While the long term cell performance may sometimes be predicted from the short time cell behavior, the only method that guarantees the correct determination of the performance durability is that based on operating the cell for as long as possible, which sometimes translates into running a single test for several months or even years. In order to achieve satisfactory cell performance, the oxygen diffusion cathode has to meet the following requirements: (1) Efficient transport of oxygen and water to and NaOH from the reaction layer, (2) Efficient catalysis of the oxygen reduction reaction according to eq. 2 (3) Efficient formation of the three phase boundaries involving oxygen, water (NaOH solution), and the catalyst The transport of reagents to and products from the reaction layer is accomplished in two different ways. In the more frequently used, three-compartment or finite-gap cells [1-10], the oxygen cathode separates the cathode chamber into the oxygen and the NaOH compartments. The cathodes used in such cells are sometimes called gas-liquid-impermeable electrodes . In a cell of this type, the gap between the electrode and the membrane is fed with a more dilute NaOH solution than the final product of the electrolysis. The feed mixes with the significantly more concentrated NaOH product. The concentration of the final NaOH product is maintained at around 32%, which guarantees the optimum membrane hydration and performance. In the zero-gap cells, the gas diffusion electrode remains in intimate contact with the ion exchange membrane. The oxygen and water enter the electrode from the gas diffusion layer side. Electrodes of this type are called gas-liquid-permeable electrodes and have been used by Permelec Electrode in Japan [11,12] and also at LANL. Both ways of transporting the reagents and products have advantages and disadvantages. The layer of NaOH solution between the electrode and the membrane in a three-compartment/finite-gap cell creates favorable conditions for optimum membrane performance, but it also contributes to the increased cell voltage. Moreover, it produces the height dependent differential pressure between the catholyte gap and oxygen across the cathode, which may lead to uneven gas/liquid separation inside the electrode, electrode flooding and caustic leakage into the oxygen/air compartment. Solving this problem requires a pressure compensation system, such as that based on oxygen supply via gas pockets [13-15]. The cathodes in zero-gap cells are less susceptible to flooding due to the lack of liquid volume between the electrode and the membrane, but maintaining the proper membrane hydration is more difficult than in the three-compartment/finite-gap cells due to the effects of the electrode and the membrane boundary layers . While the variety of materials were studied as catalysts of the oxygen reduction in alkaline media [17-30], only carbon supported platinum (Pt/C) [3-5,9] and carbon supported silver (Ag/C) [4-9,11] have been used so far in either the laboratory-scale or pilot-scale chlor-alkali cells employing the oxygen cathodes. As expected, the platinum-based catalysts offer the highest rates of oxygen reduction. However, the electrocatalytic activity of silver is not negligible compared to that of platinum under the industrial conditions . In fact, quite comparable cell voltages were observed for the cells loaded with 0.56 mg/cm2 of platinum and 2.63 mg/cm2 of silver by Furuya and Aikawa and the performance of the Ag-loaded cells could possibly be improved if the catalyst particle size could be reduced . Quite recently, mixed Pt/Ag catalysts have been studied and the catalyst containing 93% (atom) of silver was found to be as active as pure platinum . Moreover, this catalyst did not deteriorate when the cell was shunted and the cathode compartment was not flushed with an inert gas. These conditions are known to cause significant losses of carbon supported Pt catalysts due to the direct dissolution of Pt and the oxidative corrosion of carbon carriers . Both phenomena also occur during the cell operation, but their magnitude decreases significantly with the increase in current density . Due to the strongly corrosive properties of the oxygen saturated, concentrated caustic, the chlor-alkali cells employing oxygen-depolarized cathodes are more susceptible to performance losses than the standard hydrogen-evolving cells. It was shown that prolonged cell operation lead to catalyst loss , changes in the cathode morphology [10,33], increased wetting of the gas diffusion layer or even flooding of the cathode . It was suggested [4,33] that one of the factors responsible for cathode degradation was generation of peroxide (eq. 3). However, no quantitative correlation between peroxide generation and performance losses was demonstrated. 4.2 Experimental 4.2.1 Cells The cells used in this study employed either separate gas diffusion cathodes (Figs. 2 and 3) or so-called membrane-electrode assemblies (MEAs) (Fig.4). Humidified O2 NaOH Depleted brine + Cl2 Brine 1 4 3 6 7 2 5 8 9 Figure 2. Components of the oxygen-depolarized cell with the separate gas diffusion cathode and metal cathode hardware: 1 – cathode current collector, 2 – Teflon gasket, 3 – cathode flow-field, 4 – gas diffusion cathode, 5 – hydrophilic spacer (if equipped), 6 – ion exchange membrane, 7 - DSA® coated anode meshes, 8 – Teflon gasket, 9 – integrated DSA® coated anode flow-field/current collector. - 15 - The separate gas diffusion electrodes were either commercially available (ELAT® from E-TEK, Inc.) or homemade. The catalyst layer of the commercial electrodes contained either 80% or 20% of carbon-supported (Vulcan XC-72) platinum with a total Pt loading of 5.0 and 0.5 mg/cm2, respectively. The homemade electrodes also utilized a carbon-supported platinum catalyst (E-TEK). Unless otherwise stated, the geometric surface area of the electrodes was 50 cm2. The cathodes in some cells were separated from the membrane by a thin hydrophilic spacer (Panex® 30 carbon cloth, Zoltek) (Fig. 2). Humidified O 2 NaOH Depleted brine + Cl 2 Brine 1 3 4 5 2 6 7 Figure 3. Components of the oxygen-depolarized cell with the graphite cathode hardware: 1 – integrated graphite cathode flow-field/current collector, 2 – Teflon gasket, 3 – gas diffusion electrode, 4 – ion exchange membrane, 5 - DSA® coated anode meshes, 6 – Teflon gasket, 7 – integrated DSA® coated anode flow-field/current collector. The membrane-electrode-assemblies (MEAs) were prepared using standard methods developed at Los Alamos National Laboratory. The cathode side of the bi-layer membrane was painted with a known amount of the solution containing unsupported precious metal catalyst, Nafion® binder solution, deionized water and isopropanol. Geometric surface area of the catalyst layer in the MEA was 50 cm2 and the catalyst loading was 5.0 mg/cm2. In order to ensure efficient transport of the reagents and products between the catalyst layer and the flow-field, the cells utilizing MEAs were equipped with separate gas diffusion layers that remained in intimate contact with the catalyst layer (Fig.4). The gas diffusion layers (GDLs) applied included the low temperature single-sided LT 1400-W and double-sided LT 2500-W from E-TEK as well as 60% teflonized Toray® paper. Humidified O2 NaOH Depleted brine + Cl2 Brine 1 3 5 6 2 7 8 4 Figure 4. Components of the oxygen-depolarized cell equipped with the membrane-electrode-assembly (MEA) type gas diffusion cathode: 1 – cathode current collector, 2 – Teflon gasket, 3 – cathode flow-field, 4 – gas diffusion layer, 5 – membrane-electrode-assembly (MEA), 6 - DSA® coated anode meshes, 7 – Teflon gasket, 8 – integrated DSA® coated anode flow-field/current collector. The LANL proprietary patterned metal flow-field was made of either stainless steel (SS316) or nickel. The stainless steel flow-field was gold-plated and the nickel flow-field was either gold- or silver-plated. Unless otherwise stated, the anode consisted of DSA® coated titanium meshes (mesh 120 and 60). An integrated single serpentine channel anode flow-field and current collector were made of DSA® coated titanium. The cell gaskets were either custom manufactured Triguard™ gaskets (W.L.Gore, Inc.) or homemade using ~ 1 mm or ~1.6 mm Gore-Tex® Teflon tape (W.L.Gore, Inc.). The cells were sandwiched between two metal endplates (not shown in Figures 2-4) of a larger size. The cathode endplate was made of stainless steel and the anode endplate - 17 - was made of titanium. The endplates were designed to accommodate cartridge heaters and thermocouples for precise cell temperature control. Their relatively large size was essential for minimizing cell temperature fluctuations. The endplates also provided means for: (i) connecting the feed lines, exhaust lines (see section 4.2.2 and Figure 5) and power cables to the cell, and, (ii) fastening the cell elements and maintaining reproducible conditions at the membrane/electrode/flow-field boundaries. The last task was accomplished with the help of eight bolts, which were symmetrically placed near the edges of the endplates. The cell bolts were torqued to a preset torque that resulted in an average membrane compression of 689 kPa (100 psi), as determined in separate experiments with Pressurex® pressure sensitive film (Fuji Film). Five different membranes were used in the experiments. Four of them, hereafter called membrane 1, membrane 2, etc., were specialized chlor-alkali bi-layer membranes. The fifth one was BPSH-30, a homemade membrane originally designed for direct methanol fuel cells. The membranes used in the cells with separate gas diffusion cathodes (Figures 2 and 3) were soaked in either 2% NaOH solution (membrane 1) for an hour or dilute (pH=12) NaOH solution (membranes 2 through 4, and BPSH-30) for a few hours before use. As dry membranes (membranes 2 through 4) expanded from 8% to 10% upon the soaking, they were cut to size once fully swelled. Since membrane 1 was stored in hydrated state and it was precut before the soaking. All the membranes were kept wet during the installation. Unless otherwise stated, the results were obtained with membrane 1. The MEAs were prepared using membrane 2 and installed in the cell without presoaking to prevent flaking off of the catalyst layer as a result of membrane expansion. Once the cell was assembled and placed in the test station, both the anode and cathode compartments were filled with a dilute NaOH solution (pH=12) and soaked for a few hours. The expansion of the MEA under such conditions was restricted by the GDL and the flow-field on the cathode side and by the titanium meshes on the anode side. When this procedure was applied, no delamination of the membrane and the catalyst layer occurred. - 19 - 4.2.2. Experimental setup and conditions Figure 5 below presents the schematic diagram of the reagent/product delivery/collection system for the two test stations used throughout the studies. The cells were operated at 90oC. Unless otherwise stated, the cathode chamber was fed with oxygen at 239 kPa (20 psig) at a flow rate corresponding to five times that required by stoichiometry of the four-electron oxygen reduction at the applied current density (eq. 2). The oxygen stream was humidified with 0.5 cm3/min of deionized water. The pressure in the cathode compartment was controlled by a backpressure regulator (Fig. 5), which was fed with nitrogen. The nitrogen feed line (not shown in Fig. 5) was equipped with a bleeding valve, which enabled simultaneous operation of two cells at different pressures. The anode compartment was not pressurized. The used brine and chlorine from the anode chamber were separated in a recirculation tank (Fig. 5). The tank was equipped with an overflow and constantly fed with fresh purified brine (310 g/dm3, < 10 ppb Ca + Mg) supplied by either Dow Chemical (Freeport, TX) or Texas Brine Company (Sugarland, TX). Fresh and used brine were mixed thoroughly in the tank and part of the resulting solution was redirected to the cell while another part was collected for recycling. In this way, constant concentration of the feed solution was maintained. Unless otherwise stated, the feed concentration was 200 ± 3 g/dm3. The estimated outlet brine concentration was approximately 186 g/dm3 at 1.0 A/cm2 and proportionally higher at lower current densities. The chlorine gas was scrubbed with 18-20% NaOH solution. The strongly corrosive environment in the cathode compartment may promote corrosion of the cathode and its hardware when the electrical circuit is open, especially as a result of a power outage during unattended cell operation. In order to minimize corrosion, the cathode gas feed line was equipped with two solenoid valves that would immediately stop the oxygen flow and replace it with nitrogen upon a power loss (Fig. 5). In order to prevent restoring of the oxygen flow upon the power return, when the electrolysis could not be automatically restarted, the valves were energized through an electric relay that required manual resetting after the power outage. O2 SCRUBBER TOWER BRINE FOR RECYCLE SCRUBBER SOLUTIONS (NaOH) H2O BRINE PUMPS MICROFILTER LEVEL SENSOR chlorine effluent CAUSTIC COLLECT to stack trap vent line ANOLYTE RECIRCULATION TANK BRINE RESERVOIR 55 gal DRUM FLOWMETER CELL DOSING PUMP HUMIDIFIER BACKPRESSURE REGULATOR N2 SOLENOID VALVES Normally closed Normally open Figure 5. Diagram of the reagent/product delivery/collection system. 4.2.3. Analytical procedures The peroxide content in the NaOH solution was determined spectrophotometrically. Fresh samples of sodium hydroxide were mixed with a known amount of potassium ferricyanide solution in aqueous NaOH. The peroxide content was determined from a decrease of ferricyanide absorption at 418 nm . Caustic current efficiency was determined from titration of the sodium hydroxide samples with standardized 1.0 M HCl solution (Fisher) against phenolphthalein. Due to the very weakly acidic properties of hydrogen peroxide (pKa=11.75 ), the volume of the acid used to neutralize the NaOH sample corresponded to the sum of the sodium hydroxide present in the sample and the NaOH produced as a result of hydroperoxide anion (HO2 -) protonation. Since the latter quantity was also equal to the amount of NaOH that would - 21 - form as a result of HO2 - decomposition, the current efficiencies quoted in the report are not corrected for peroxide. 4.2.4. Standard startup, shutdown, current ramping and cell protecting procedures 4.2.4.1. Startup procedure and current ramping procedure Before installing the cell in the test station, the brine recirculation tank (Fig. 5) and the recirculation pump were flushed with DI water to remove any chlorine, hypochlorite and chlorate left in the anolyte feed system from the previous experiment. After the system was flushed, the recirculation tank was filled with a freshly prepared 200 g/dm3 brine. The cell was placed in the test station, the cartridge heaters and thermocouples were installed and the feed/exhaust lines were connected to the cell. Water and nitrogen were then fed to the cathode compartment and the fresh brine recirculated through the anode compartment of the cell. The cell and humidifier heaters were turned on. The cell temperature was increased in 10ºC increments. When the temperature of the cell reached 80ºC, the power to the solenoid valves was manually turned on, which caused oxygen to replace the nitrogen in the feed line. After that, a small protective current on the order of 1-2 A was immediately applied. The Labview® software was started and the current density was increased at the rate of 0.4 A cm-2 h-1 until it reached the desired value, most typically 0.2 A/cm2 (10 A for the standard 50 cm2 cell). While the cell current was being increased, the oxygen flow was adjusted to the desired level, the brine make-up pump started and the brine flow roughly adjusted to the level required by the target current density. The cathode compartment was pressurized either during the ramping or afterwards. The cathode pressurization was done in small increments by manually adjusting the pressure of nitrogen feeding the backpressure regulator. When the pressurization was complete the flow of water to the humidifier was adjusted to the desired level. The adjustment of the water flow could not be done before, as the gas pressure influenced the output from the humidifier pump (Fig. 5). By the time the above operations were completed, the cell temperature was already 90ºC. The brine concentration in the recirculation tank was then adjusted to the desired level. This was done by frequently measuring the brine concentration in the recirculation tank and adjusting the brine make-up pump speed accordingly, until the brine concentration stabilized at the desired level. Ramping the current to a higher value was done in a similar way. As the cathode compartment was already pressurized, there was no need for readjusting the water flow as long as the intended oxygen humidification level remained unchanged. The startup procedure for the cells equipped with MEAs was preceded by soaking the MEA in dilute NaOH (pH=12) solution (see section 4.2.1). 4.2.4.2. Shutdown procedure Shutting down the experiment started with turning off the cell heaters, reducing brine supply to the recirculation tank and applying the protective current of 1 – 2 A. When the cell temperature reached around 30 - 40ºC, the electrolysis was stopped, the oxygen and brine flows were stopped and the cell removed immediately from the station. Both the anode and the cathode compartments were flushed with DI water and the cell was immediately disassembled. 4.2.4.3. Temporary shutdown of the cell In some cases, it was desirable to discontinue the experiment for a period of time and then restart it after the break. In such cases, the electrolysis was stopped, i.e., zero current was applied to the cell. Immediately after that, the oxygen flow was stopped and replaced by nitrogen flow by manually turning off the power to the solenoid valves. The cathode humidification rate was increased to wash out the remaining NaOH from the cathode compartment. The fresh brine supply to the recirculation tank was stopped. While the anolyte recirculation pump was still recirculating, most of the brine was drained and replaced with hot (~90ºC) DI water to maintain the cell temperature. This operation was repeated several times, until the stable open circuit voltage was measured. Then, the cell and humidifier heaters were turned off and the cell was left to cool down. When the cell temperature equilibrated, the nitrogen and water flows to the cathode compartment were - 23 - stopped and the brine recirculation pump was turned off. The brine feed and chlorine/brine exhaust lines were disconnected from the cell and the anode compartment was completely filled with fresh DI water. The cathode compartment was depressurized and left filled with DI water and nitrogen. 4.2.5. Instrumentation Constant current electrolysis was performed using a Lambda model LYS-K-5-0V or a PowerOne model SPM3A2K DC power supply coupled to a Hewlett Packard 6060B electronic load-box. The load-box was also coupled to a high frequency resistance measurement system that was comprised of a Voltech model TF200 frequency response analyzer and two Stanford Research Systems model SR560 low-noise preamplifiers for current and voltage signals. The system was controlled by Labview® software (National Instruments) installed on a MacIntosh computer. A cell voltage measurement was performed every six minutes and was immediately followed by a resistance measurement. In the latter case, a high frequency (2 kHz) ac signal of small amplitude (30 mV) modulated the voltage of the load-box and consequently the cell voltage and the current. The ratio of the ac components of the cell voltage and the current was assumed to be equal to the ion-exchange membrane resistance, hereafter called high frequency resistance (HFR). The error associated with this approximation was rather small, as the observed phase shift was typically in the range of 15-17o. The system was also equipped with a Hewlett Packard model 3488A switch/control unit, a model 3421A data acquisition/control unit, as well as an IOTech model 488/4 serial bus converter. These were responsible for switching between the different modes of measurement, data acquisition from the two test stations, and data transfer to the computer. 4.3. Results 4.3.1. Cathode Modifications Aimed at Improving Caustic Current Efficiency Standard hydrogen-evolving membrane cells are fed with 30% NaOH and generate 32% NaOH. This concentration range offers the best overall performance of the cell. However, one of the essential requirements of achieving the highest cell performance is the high rate of internal mixing in the cathode compartment, since non-uniform sodium hydroxide concentration may lead to performance deterioration. The hydrogen-evolving cathode consumes water and hence the sodium hydroxide concentration in its vicinity becomes higher than that in the bulk of NaOH. A similar boundary layer is also formed at the membrane surface, where the NaOH concentration is largely determined by the amount of water transported through the membrane. If the cathode and the membrane are very close to each other, both boundary layers may merge, and the cathode reaction may cause a further loss of water in the immediate vicinity of the membrane. Such an effect may lead to an increased membrane resistance and possibly even to membrane damage due to ohmic overheating of the carboxylic layer. Identical issues could also be expected for the oxygen-depolarized cells. One may expect the cells employed in this study to be more susceptible to the undesired effects of the boundary layers due to their zero-gap design. However, the real situation is more complex. Unlike in the standard hydrogen-evolving cells, where the proper water balance in the vicinity of the membrane and the desired 32% concentration of the caustic product are maintained by simple means, i.e., by feeding the cell with approximately 30% NaOH, the cathode compartment of the cells used in this study was fed exclusively with oxygen and water. Moreover, the transport of water, caustic and oxygen, both within the cathode and between the cathode and the membrane, is governed by variety of driving forces, including capillary phenomena, pressure gradients, electric field gradients, etc. Consequently, predicting the actual NaOH concentration distribution may be more difficult. In Figure 6, the cell voltages are plotted versus the current density for three experiments, where the cells equipped with the standard ELAT® electrodes (Fig. 3) were run at three different levels of oxygen humidification. The plots obtained for the dry oxygen and for the 0.5 cm3/min humidification exhibit some degree of non-linearity, which is frequently regarded as a sign of non-optimized membrane performance. The plot obtained for the saturated water vapor covers only a limited range of current densities and no conclusion regarding its linearity can be made. Since other factors, e.g., changes in transport of the reagents, may have also contributed to the observed phenomena, the effects of oxygen humidification were examined in more detail. 1.6 1.8 2 2.2 2.4 2.6 2.8 0 0.2 0.4 0.6 0.8 1 1.2 Cell Voltage (V) Current Density (A/cm 2) Dry oxygen Water vapor saturated oxygen 0.5 cm3/min liquid water Figure 6. Cell voltages measured at low oxygen humidification levels. Red symbols – no humidification. Black symbols – oxygen with saturated water vapor. Four – serpentine channel graphite cathode flow-field/current collector. Membrane 1. Since the voltage vs. current density plot for the cell humidified with 0.5 cm3 of water per minute is convex, and the plot obtained for the dry oxygen is concave, we believe that different factors contributed to the non-linearity of the plots. For obvious reasons, the curvature of the plot obtained for the non-humidified cell (Fig. 6) is more likely to have - 25 - originated from the interaction of the cathode and the membrane boundary layers, which resulted in too high catholyte concentration in the immediate vicinity of the membrane. Consequently, the water balance in the non-humidified cell was further explored. Figure 7 shows the concentration of the product NaOH, the molar ratio of water to sodium transported through the membrane, and the hypothetical NaOH concentration at the membrane surface, which would be observed if the cathode reaction did not consume water from the immediate vicinity of the membrane. Both, the hypothetical NaOH concentration and the water to sodium molar ratio were calculated from the product concentration by correcting it for the amount of water, which was consumed by the ORR. 30 35 40 45 3.8 4 4.2 4.4 4.6 4.8 5 5.2 0 0.2 0.4 0.6 0.8 1 1.2 NaOH Concentration (%) nwater/nsodium Current Density (A/cm2) Dry oxygen Saturated water vapor Calculated for HER a a b a Figure 7. Concentrations of NaOH generated with and without oxygen humidification and the number of water molecules transferred through the membrane as a result of the electroosmotic drag. Four – serpentine channel graphite cathode flow-field/current collector. Membrane 1. a – final product concentration. b – hypothetical NaOH concentration at the membrane surface (see text). - 27 - For comparison, concentration of the caustic product obtained from the electrolysis using the water vapor saturated oxygen and the calculated concentration of the NaOH product, which would be generated if the HER rather than ORR was the cathode reaction in the non-humidified cell, are also shown in Fig. 7. When there is no external humidification, the only water available in the cathode compartment is that transported from the anode compartment through the membrane. This water is used inside the catalyst layer by the ORR, which concentrated the NaOH solution above the level resulting from the relative quantities of sodium and water transported through the membrane (Fig. 7). Since there is no source of water on the main escape route of caustic from the electrode through the gas diffusion layer towards the flow-field, we believe that the concentration of NaOH inside the electrode and in the thin boundary layer between the electrode and the membrane monotonically decreases from its highest value inside the electrode, which is equal to the final product concentration, to some value at the membrane surface, whose the lower limit is determined by the membrane transport properties (red curve marked with “b” in Fig. 7). In accordance with the above analysis, the NaOH concentration next to the membrane, when no external humidification was applied, should lie between the two approximately parallel red lines in Fig. 7, which represent the final NaOH concentration and the hypothetical one, calculated as if the ORR consumed no water. The exact values of this concentration are unknown, but it seems that performance problems can be expected at low current densities (≤ 0.4 A/cm2), where the NaOH concentration in the membrane vicinity can reach the maximum value of 38 - 40 %. Consequently, the concave shape of the voltage vs. current density plot for the non-humidified cell in Fig. 6 results from the excessively high cell voltages at the low current densities rather than at high current densities (≥ 0.8 A/cm2), where the NaOH concentrations next to the membrane are lower due to increased water transport through the membrane (Fig. 7). Figure 8 shows the high frequency resistance measured for the three levels of oxygen humidification. As clearly seen in Fig. 8, the high frequency resistance in the non-humidified cell (red curve in Fig. 8) at low current densities (≤ 0.4 A/cm2) is very high, which supports the hypothesis that the concave shape of the voltage versus current density plot for this cell (Fig. 6) reflects too high cell voltages originating from the boundary layer effects. On the other hand, the convex shape of the respective plot for the cell humidified with 0.5 cm3/min of liquid water pumped into the cell, most likely reflects partial flooding of the cathode with caustic solution at high current densities (≥ 0.8 A/cm2). A similar effect occurs to a lesser extent in the cell utilizing dry oxygen, because the volume of the NaOH solution generated under such conditions is smaller. The apparent lack of performance issues at high current densities in the cell fed with dry oxygen suggests that the harmful effect of the increased water consumption by the ORR is more than compensated by the enhanced water transport through the membrane. 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 0 50 100 150 200 HFR (Ohm cm 2) Time of Operation (hrs) Dry oxygen Water vapor saturated oxygen 0.5 cm3/min humidification 0.2 0.4 0.6 1.0 Humidification increased to 0.5 cm3/min at 85oC 1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 Figure 8. High frequency resistance at different oxygen humidification levels. Red line – no humidification. Black line – oxygen with saturated water vapor. Green line – humidification 0.5 cm3/min. Four - serpentine channel graphite cathode flow-field/current collector. Membrane 1. Pt catalyst loading: 5.0 mg/cm2 - red line; 0.5 mg/cm2 – green and black line. - 29 - This is one of the attractive features of the technology utilizing oxygen-depolarized cathodes, as compared to the conventional membrane technology. While the hydrogen evolution reaction (eq.1) in the standard membrane cell consumes one mole of water to produce one mole of NaOH, the oxygen reduction consumes two times less water (eq.2) and thus it is less likely to lead to significant dehydration of the boundary layer (see the lines marked with “a” in Fig. 7). It has to be noted that the HFR measured at 1.0 A/cm2 is virtually identical for the cells fed with dry oxygen and humidified with 0.5 cm3/min of liquid water. Such a result suggests that the cathode/membrane boundary conditions must be very similar in both cases and the NaOH concentration in the solution between the membrane and the electrode is identical in both cells. Other important conclusions based on the plots in Fig. 8 are: (i) even the very low oxygen humidification, corresponding to the saturated vapor, significantly lowers the high frequency resistance and thus NaOH concentration in the membrane/electrode boundary layer at low and intermediate current densities (≤ 0.6 A/cm2), (ii) at 0.6 A/cm2, the saturated water vapor lowers the high frequency resistance as effectively as 0.5 cm3 of water per minute directly pumped into the cathode compartment. In similarity to the high frequency resistance, the caustic current efficiency was found to be rather sensitive to the oxygen humidification. However, there was no straightforward correlation between the current efficiency and the level of humidification. The highest current efficiencies were observed for the cell fed with water vapor saturated oxygen, the lowest current efficiencies were observed for the cell humidified with 0.5 cm3/min of liquid water, whereas intermediate values were obtained for the cell fed with dry oxygen. Moreover, the level of humidification had a profound effect on the dependence of the CCE on current density. This is demonstrated in Fig. 9, where the CCE is plotted versus the current density for the three levels of humidification. In order to explain the differences in the plots shown in Fig. 9, one has to consider a variety of factors that influence caustic current efficiency. Since there are no cathode side reactions which could lower the efficiency of NaOH generation, the only reason for lower than 100% current efficiency is the crossover of NaOH. This process can occur either by diffusion of the non-dissociated NaOH or by migration of hydroxyl anions through the membrane to the anode compartment. If there were no changes in membrane permeability associated with the changes in current density, both processes would exhibit a relatively simple dependence on the experimental conditions. The rate of diffusion would be independent of current density and proportional to the NaOH concentration at the membrane surface. The rate of migration of the hydroxyl ions would increase with the increase in NaOH concentration and with the increase in current density. 84 86 88 90 92 94 96 98 0 0.2 0.4 0.6 0.8 1 1.2 Caustic Current Efficiency (%) Current Density (A/cm2) Dry oxygen Water vapor saturated oxygen 0.5 cm3/min liquid water Figure 9. Caustic current efficiency at different oxygen humidification levels. Red line – no humidification. Black line – oxygen with saturated water vapor. Green line – humidification 0.5 cm3/min. Four - serpentine channel graphite cathode flow-field/current collector. Membrane 1. Pt catalyst loading: 5.0 mg/cm2 - red line; 0.5 mg/cm2 – green and black line. - 31 - However, the changes in current density produce changes in the physical properties of the membrane, which affect the membrane permeability and transport properties. Higher current densities promote more significant membrane swelling and increase its water content. As a result, the membrane becomes more “open” and the transport of water through the membrane becomes easier, as already demonstrated by the plot of the water to sodium ratio in Fig. 7. The more open structure also promotes transport of any other molecule or ion, including the undesired NaOH crossover. The membrane permeability is further increased by humidification, which lowers the NaOH concentration in the membrane vicinity. On the other hand, the lower NaOH concentration results in a lower driving force for NaOH transport through the membrane to the anode compartment. As a consequence, the CCE is a complex function of all the individual factors mentioned above. In accordance with this statement, we did not find any simple relationship between the CCE and any single parameter such as the HFR, current density (Fig. 9), or NaOH concentration. The differences between the plots in Fig. 9 can be qualitatively explained as follows. In the cell where the dry oxygen was used, the NaOH concentration in the membrane vicinity at all current densities is the highest possible. The increase in current density significantly decreases the HFR (Fig. 8) and enhances permeability of the membrane (Fig. 7). Even though the concentration dependent driving force for the NaOH and OH- transport through the membrane decreases with the decrease in current density (Fig. 7), the increasingly open structure of the membrane and the increasing electric field in the membrane result in lowering of the CCE. When the oxygen is saturated with water vapor, the concentration of the product NaOH is significantly lower than in the absence of humidification (Fig. 7). While the actual NaOH concentrations at the membrane surface are not known, they must be lower than in the cell without humidification (cf. also Fig. 8), which results in the lower driving force for the crossover. At the same time, the NaOH concentrations are not sufficiently low to cause significant changes in the membrane water content. The current efficiencies are higher than for the non-humidified cell. The lowest current efficiencies were observed for the cell humidified with the 0.5 cm3/min of liquid water. In this case, the most important factor determining the CCE is the concentration of NaOH and its influence on the membrane permeability. Unlike in the two previous cases, the concentration of the NaOH product increases with the increase in current density (Fig. 10) due to the relatively large volume of water introduced to the cell, which is comparable with the volume of NaOH solution generated without humidification. 86 88 90 92 94 96 98 15 20 25 30 35 40 Caustic Current Efficiency (%) NaOH Concentration (%) Dry oxygen Water vapor saturated oxygen 0.5 cm3/min liquid water Figure 10. Caustic current efficiency at different oxygen humidification levels plotted versus concentration of the product NaOH. Red lines – no humidification. Black line – oxygen with saturated water vapor. Green line – humidification 0.5 cm3/min. Four - serpentine channel graphite cathode flow-field/current collector. Membrane 1. Pt catalyst loading: 5.0 mg/cm2 - red line; 0.5 mg/cm2 – green and black line. The arrows indicate current density increase. As a consequence, the NaOH solutions generated at lower current densities are markedly diluted, which leads to the unusually significant membrane swelling (see also the HFR, - 33 - Fig. 8) and the very enhanced NaOH crossover. The current efficiency is low and increases with the current density.In accordance with the above logic, the current efficiency for all levels of humidification seem to converge at high current densities, where the membrane is significantly swelled and the volume of water introduced to the cathode compartment is relatively small compared to the volume of caustic generated. In previous work, we departed slightly from the zero-gap concept by introducing for the first time a hydrophilic spacer between the electrode and the membrane. The spacer was found helpful in maintaining stable caustic current efficiency at a level of a little below 90% at 1.0 A/cm2, reducing peroxide generation and lowering the cell voltage. As we suspected that the spacer could promote formation of the stagnant layer of concentrated caustic between the electrode and the membrane, we created a small opening in the upper part of the electrode and directed water from the humidifier to this opening to dilute and help remove the caustic from the spacer. After having realized that the boundary layer effects in our zero-gap cells may actually be weaker than initially suspected, we decided to modify the cell design and abandon the idea of direct humidification of the spacer. The modified cell allowed for no direct spacer humidification and resulted in a significant improvement of the caustic current efficiency at current densities not exceeding 0.6 A/cm2 (Fig.11). 82 84 86 88 90 92 94 96 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 Caustic Current Efficiency (%) Current Density (A/cm 2) Time of Operation (hrs) Direct spacer humididfication No spacer humidification Figure 11. Effect of direct humidification of the hydrophilic spacer on the caustic current efficiency. Platinum catalyst loading 0.5 mg/cm2. Humidification 0.5 cm3/min. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. No clear effect of this modification on the current efficiency at high current densities (0.8-1.0 A/cm2) could be detected. However, the modification resulted in a small decrease in the cell voltage (Fig.12). 1.4 1.6 1.8 2 2.2 2.4 2.6 0 50 100 150 Cell Voltage (V) Time of Operation (hrs) Voltage spikes due to changes in brine concentration 0.2 A/cm2 0.4 A/cm2 0.6 A/cm2 0.8 A/cm2 1.0 A/cm2 Direct spacer humididfication No spacer humidification Figure 12. Effect of direct humidification of the hydrophilic spacer on the cell voltage. Platinum catalyst loading 0.5 mg/cm2. Humidification 0.5 cm3/min. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. It was previously noted by us that uncontrolled discontinuation of the electrolysis resulting from a power outage or brownout frequently led to a decrease in the caustic current efficiency after the experiment was restarted. The most frequent scenario involved relatively short period of time, when no equipment was working, but the cathode compartment was flushed with oxygen or nitrogen, depending on whether the system was already equipped with the solenoid valves and the relay, which prevented the valves from reenergizing once the power was back (see part 4.2.2. and Fig. 5). This period of time was followed by a typically much longer period of time after the power was restored, when all the electrical equipment was working with exception of the Labview® computer software. - 35 - 0 0.2 0.4 0.6 0.8 1 1.2 86 87 88 89 90 91 0 50 100 150 200 250 300 350 400 Current Density (A/cm 2) Current Efficiency (%) Time of Operation (hrs) uncontrolled interruption of electrolysis Figure 13. Effect of an uncontrolled interruption of the electrolysis on caustic current efficiency in the cell equipped with the integrated graphite cathode flow-field/current collector. The cell heated and flushed with oxygen during the interruption. Membrane 1. Under such conditions (see conditions 1 and 2, Table 2), the cell was heated, the anode compartment was flushed with the brine from the anolyte recirculation tank (see Fig. 5) and the cathode compartment was flushed with water and nitrogen or, in many earlier experiments, with oxygen. The cell remained at its respective open circuit voltage, which was determined by the redox properties of the chemical species present in both the anode and the cathode compartment. After a sufficiently long time, the brine make-up pump eventually replaced the brine in the recirculation tank with the fresh brine, which did not contain oxidizers, e.g., chlorine, chlorine dioxide, hypochlorous acid and sodium chlorate, but it was significantly more concentrated (310 g/dm3). Also, a significant portion of the concentrated NaOH was washed out from the cathode compartment by the humidification - 37 - water. Since the cathode hardware corrosion and the associated catalyst and membrane poisoning often accompanied this phenomenon, no unambiguous conclusion could be made as to the origin of the phenomenon and the ways of preventing or minimizing the decline in current efficiency. In one experiment, which was not influenced by the cathode hardware corrosion thanks to the use of the corrosion resistant graphite hardware, the current efficiency dropped from around 90% to around 86% as a result of an uncontrolled interruption of the electrolysis (Fig. 13). After approximately 20 hours, when the cell remained under the open circuit conditions, the electrolysis was restarted and the current efficiency relatively quickly returned to the level of approximately 90% (Fig. 13). Reversible character of the performance decline following an interruption of the electrolysis was also observed when the cell was equipped with the MEA-type cathode. The undesired effects of power outages on caustic current efficiency seemed stronger and longer lasting in the cells equipped with the cathode spacers (Fig. 14), but one has to note the different conditions and history of the uncontrolled shutdowns as well as the different membranes used in the experiments shown in Figs. 13 and 14. While there was no cathode hardware (flow-field and current collector) corrosion in the experiments shown in Figs. 13 and 14, the conditions in the cathode compartment during the break in electrolysis were rather corrosive due to the presence of oxygen (Fig. 13) and/or the other oxidizers, e.g., chlorates (Figs. 13 and 14), which diffused from the anode compartment through the membrane under the open circuit conditions. These oxidizers may have contributed to corrosion of the electrode and the membrane, catalyst loss and changes in electrode hydrophobicity, which may have influenced to some extent the current efficiency. Indeed, after the unexpected shutdown of the electrolysis, shown in Fig. 13, the cell started generating significant quantities of peroxide, which must have resulted from at least partial loss of the platinum catalyst. 0 0.2 0.4 0.6 0.8 1 1.2 88 89 90 91 92 93 94 95 96 0 50 100 150 200 Current Density (A/cm2) Caustic Current Efficiency (%) Time of Operation (hrs) uncontrolled interruption of electrolysis Figure 14. Effect of an uncontrolled interruption of the electrolysis on caustic current efficiency in the cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. No spacer humidification. The cell heated and flushed with nitrogen during the interruption. Membrane 4. Subsequent experiments demonstrated that the current efficiency also decreased when no oxidizers were present in the cathode compartment under the open circuit conditions (Fig. 15). Consequently, the decrease in current efficiency could only be associated with the presence of highly concentrated brine on the anode side of the membrane and caustic solution or water on the cathode side of the membrane. 0 0.2 0.4 0.6 0.8 1 1.2 86 88 90 92 94 96 0 50 100 150 200 Current Density (A/cm2) Caustic Current Efficiency (%) Time of Operation (hrs) controlled interruption of electrolysis Figure 15. Effect of a controlled interruption of the electrolysis on caustic current efficiency in a cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. No spacer humidification. The anode compartment flushed with deionized water and then with the fresh, concentrated brine. The concentrated brine recirculated through the anode compartment. The cell heated during the interruption. The cathode compartment flushed with nitrogen and DI water. Membrane 4. Our first hypothesis was that the combination of water on the cathode side and concentrated brine on the anode side was more likely to cause problems than the combination of caustic solution and brine. The reasoning was based on the significant differences in the ionic strength in the first case, which could lead to an undesired strain in the membrane and potentially even cause damage to the membrane structure. While the hydrophilic spacer in the experiments shown in Figs. 14 and 15 was not humidified, we suspected that the humidification water, which was introduced into the cathode chamber, - 39 - could have eventually washed out the caustic present in the spacer by entering the spacer from the bottom of the cathode compartment, where the water transitionally accumulated before the backpressure regulator (Fig. 5) allowed it to drain. 0 0.2 0.4 0.6 0.8 1 1.2 82 84 86 88 90 92 94 96 0 100 200 300 400 500 Current Density (A/cm2) Caustic Current Efficiency (%) Time of Operation (hrs) controlled interruption of electrolysis uncontrolled interruption of electrolysis Figure 16. Effect of a controlled interruption of the electrolysis on caustic current efficiency in a cell equipped with the silver-plated nickel flow-field and Panex® 30 hydrophilic spacer. No spacer humidification. The anode compartment flushed with deionized water and then with the fresh, concentrated brine. The concentrated brine recirculated through the anode compartment. The cell heated during the interruption. The cathode compartment flushed with nitrogen and DI water. Membrane 1. In order to prove this hypothesis, another cell design was implemented. In the modified cell, both the electrode and the spacer had smaller height than in the standard cell. They were placed in the upper part of the cathode compartment so that an empty space was created on the bottom of the cathode compartment, where water could accumulate. The - 41 - bottom part of the membrane, which would be excluded from the current flow, was covered with a strip of a thick Teflon foil to minimize the unwanted crossover of NaCl from the anode compartment. The electrolysis was interrupted after more than 140 hours (more than 40 hours at 1.0 A/cm2), the solution in the anode compartment was replaced with the fresh brine and the cathode compartment was flushed with nitrogen and deionized water. After approximately 80 hours, the electrolysis was restarted and the current density ramped to 1.0 A/cm2. The current efficiency measured a short time after the restart was very low (~83%). The electrolysis was soon interrupted for more than an hour by the Labview software as a result of the oxygen shortage. Due to its short duration, the last interruption of the electrolysis is believed to have only a minor, if any, effect on the subsequent cell performance. The results of this test are shown in Fig. 16. As seen in Fig. 16, the controlled interruption of electrolysis had a profound effect on the current efficiency. The caustic current efficiency at 1.0 A/cm2 dropped from around 91% before the interruption to approximately 83% following the break in the electrolysis. During the next ~100 hours after the break, the current efficiency started increasing to eventually reach the level of around 85%. However, it started decreasing slightly during the subsequent 100 hours. The result of the above test is inconclusive. While any improvement in the membrane performance resulting from the cell modification would indicate that the significant difference in ionic strength on both sides of the membrane was responsible for the performance deterioration, the lack of improvement may result from variety of factors, also including inadequate masking of the part of the membrane which was excluded from the current flow. Consequently, the irreversible character of the performance deterioration in Fig. 16 does not imply irreversible damage to the membrane. The origin of the performance losses caused by interruptions of the electrolysis is unclear. It is obvious that the prolonged presence of concentrated brine on the anode side and concentrated caustic or water on the cathode side contributes to the problem. While the reversible character of the current efficiency changes in the cells equipped with no spacer (see above) may suggest that the presence of concentrated caustic in the spacer is the major factor leading to performance losses following an interruption of the electrolysis, it has to be noted that the results presented in Fig. 13 may be influenced by other phenomena. In fact, the uncontrolled interruption of the electrolysis in this case resulted in a very significant increase in the peroxide generation rate, which corresponded to around 36% of the current being used for peroxide generation after approximately 350 hours of electrolysis (Fig. 13). Such a high peroxide generation rate led to a significant lowering of the NaOH concentration and thus possibly to a decrease of the NaOH crossover, which may have masked the current efficiency decrease resulting from the deterioration of the membrane performance. The problem of caustic current efficiency decline following an interruption of the electrolysis requires additional studies. 4.3.2. Membrane Testing The caustic current efficiency (CCE) and the ohmic drop in the cell are key characteristics determining performance of the membrane and its suitability for the chlor-alkali electrolysis. A perfect chlor-alkali membrane would have virtually no electrical resistance and guarantee almost 100% current efficiency. Unfortunately, the intrinsic membrane properties, which are responsible for its conductivity and selectivity, affect these parameters in a similar way, i.e., low membrane resistance is typically associated with low selectivity (current efficiency) and vice versa. However, thanks to the complex nature of the transport phenomena in the membrane, the relationship between both parameters is not straightforward and two different membranes can have identical conductivity and different selectivity under identical operating conditions. The chlor-alkali membranes which are currently in use are comprised of two layers of polymers that contain different acidic groups. The carboxylic layer on the cathode side of the membrane is more selective and less conductive than the sulfonic layer on the anode side. Properties of both ionomers and thickness of the layers determine the overall membrane performance. They are optimized to guarantee the highest possible current efficiency (current industrial standard for the membrane cell is ~96% at 0.4 A/cm2) and the lowest possible ohmic drop. Five different membranes were selected for testing. Membranes 1 through 4 were specialized chlor-alkali membranes. Two of them, membrane 3 and 4 had the gas release coating on the cathode side removed. The fifth membrane, BPSH-30, was homemade. It was originally designed for the direct methanol fuel cells to reduce methanol crossover. Its relatively low permeability was believed to promise high caustic current efficiencies. - 43 - Unfortunately, this material failed before starting the electrolysis, during the cell warm-up phase. Multiple micro cracks formed in the membrane, which eventually resulted in a brine leak from the anode to the cathode compartment. The remaining materials were tested under the standard conditions in the cells equipped with the commercial double-sided ELAT® electrodes containing 0.5 mg/cm2 Pt (see section 4.2.1.) and the hydrophilic spacer (Fig. 2). The spacer was not humidified (see section 4.3.1.). The standard anode structure was used in all the experiments. The tests were run at five consecutive current densities: 0.2, 0.4, 0.6, 0.8 A/cm2, and 1.0 A/cm2 for approximately 24 hours at each current density below 1.0 A/cm2 and for at least 24 hours or more at 1.0 A/cm2. The current efficiencies were determined in the middle of each 24-hour period of time and the cell voltages were averaged over each 24-hour period. Figure 17 shows the measured cell voltages plotted against the current density for the four membranes. As can be seen, the lowest cell voltages have been achieved with membrane 1 and membrane 3. At the low current densities (0.2 A/cm2), the cell voltages were slightly lower for membrane 1. The opposite was true for the high current densities (0.8-1.0 A/cm2). The differences were smaller than 2%, i.e., they did not exceed 20 mV at 0.2 A/cm2 and 40 mV at 1.0 A/cm2. The voltages measured for membrane 4 were higher by 60-90 mV than the ones found for membrane 1. As expected, membrane 2, chlor-alkali membrane of an older type was most resistive and produced voltages higher by 150-290 mV than those measured for membrane 1 (Fig.17). Linear regression of the cell voltage versus the current density is commonly used to describe the membrane performance in a standard hydrogen-evolving cell. The slope of the voltage versus current density plot, or k-factor, is the membrane contribution. The intercept is the electrolyzer contribution. The lower the k-factor, the less resistive the membrane is. Table 1 shows the linear regression results of the data plotted in Fig. 17 together with the available literature data obtained from the tests using standard hydrogen-evolving cells. 1.6 1.8 2 2.2 2.4 2.6 2.8 0 0.2 0.4 0.6 0.8 1 1.2 Cell Voltage (V) Current Density (A/cm 2) Figure 17. Measured cell voltages for different bi-layer membranes: membrane 1 – full triangles, membrane 2 – open squares, membrane 3 – full squares, membrane 4 – full circles. Inspection of the data in Table 1 reveals significantly lower values of the k-factor in our cells as compared to the standard hydrogen-evolving cells. The origin of this difference is quite understandable. The markedly higher k-factors for the hydrogen-evolving cells reflect non-negligible contributions to the cell voltages from the resistance of the liquid NaOH and NaCl electrolytes in these cells. As such contribution was almost nonexistent in our zero-gap cells, one can estimate the electrolyte contribution to the measured k-factors for the standard membrane cells to be on the order of 33% and 31% for membrane 3 and membrane 4, respectively. The lower voltages of our cells that result from the different - 45 - cathode reaction and their zero-gap design translate into significant energy savings. These savings are approximately 38% at 0.31 A/cm2. Table 1. Linear regression results for the plots shown in Fig. 17. Membrane k-factor V A-1 cm2 Intercept V Correlation Coefficient Cell Voltage at 0.31 A/cm2 Cell Voltage at 0.31 A/cm2 for standard membrane cell a) V k-factor for standard membrane cell a) V A-1 cm2 2 1.0435 1.626 0.999 1.950 - - 3 0.805 1.545 0.998 1.795 2.89 1.20 4 0.902 1.559 0.999 1.839 2.95 1.30 1 0.885 1.506 0.999 1.780 - - a) Data from reference 16 Another important factor that determines the membrane performance is the caustic current efficiency. Figure 18 shows the measured caustic current efficiency plotted versus the current density. Two different membrane responses to the changes in current density can be observed. Membrane 4 and 3, both specifically designed for operations at higher than standard current densities, exhibit a gradual decline in the current efficiency with the increase in current density from 0.2 to 1.0 A/cm2. For the remaining two membranes, membrane 1 and 2, the current efficiency decreases with current density up to 0.8 A/cm2 and then increases at 1.0 A/cm2. The origin of the current efficiency minimum is not clear. However, it seems possible that it is related to the membrane response to changes in the anode boundary layer that result from the current density changes. As shown in section 4.3.5., a monotonic relationship between the current efficiency and the current density was observed for membrane 1, when one of the anode modifications was introduced. 86 88 90 92 94 96 0 0.2 0.4 0.6 0.8 1 1.2 Caustic Current Efficiency (%) Current Density (A/cm 2) Figure 18. Current efficiency for different bi-layer membranes: membrane 1 – full triangles, membrane 2 – open squares, membrane 3 – full squares, membrane 4 – full circles. No clear correlation was found between the measured cell voltages and the caustic current efficiency, although higher cell voltages (high membrane resistances) were frequently associated with higher current efficiency and vice versa (Figs. 17 and 18). Membrane 4 provided the highest current efficiency in the whole range of current densities from 0.2 to 1.0 A/cm2 (Fig.18). From an economic perspective, the most important parameter determining membrane quality is energy consumption per unit weight of the product. The energy consumption is proportional to the product of the cell voltage and the current (or current density) and inversely proportional to the caustic current efficiency. Consequently, in order to compare the energy efficiency for the membranes tested, it is convenient to introduce a new parameter: G = constEnergy = Voltage(V)Current Density(A/cm2)/CCE(%) (5) 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.2 0.3 0.4 0.5 0.6 G (VAcm -2% -1) Current Density (A/cm 2) membrane 4 membrane 2 membrane 1 membrane 3 Figure 19. Energy efficiency (see text) of different bi-layer membranes at low current densities: membrane 1 – dotted green line, membrane 2 – black line, membrane 3 – red line, membrane 4 – dashed blue line. The parameter G reflects exclusively the amount of energy required to perform the electrolysis, whereas the total energy consumed by the chlor-alkali industry also includes the energy used to generate steam, which is used for concentrating the caustic soda to the level of 50%. Since the NaOH concentrations generated in the cells equipped with the different membranes under identical conditions were very similar and the average use of - 47 - steam accounts for only around 16% of the total energy consumption by the chlor-alkali industry, we believe that equation 5 can be used with confidence for the analysis of the energy efficiency of cells equipped with the different membranes. The parameter G is plotted versus current density in Figures 19 and 20 for the lower (0.2 - 0.6 A/cm2) and higher (0.8 - 1.0 A/cm2) current density ranges, respectively. As seen in Fig. 19, membrane 1, membrane 3, and membrane 4 offer very similar energy efficiency for current densities up to 0.6 A/cm2, whereas the use of membrane 2 is associated with significantly higher energy consumption. 0.016 0.018 0.02 0.022 0.024 0.026 0.028 0.03 0.7 0.75 0.8 0.85 0.9 0.95 1 1.05 Current Density (A/cm 2) membrane 2 membrane 3 membrane 1 membrane 4 G (VAcm -2% -1) Figure 20. Energy efficiency (see text) of different bi-layer membranes at high current densities: membrane 1 – dotted green line, membrane 2 – black line, membrane 3 – red line, membrane 4 – dashed blue line. - 49 - The performance of membrane 2 is even worse at the higher current densities (Fig.20). The demanding conditions of electrolysis at high current densities reveal differences in the performance of the remaining membranes too. While membranes 3 and 4 still offer the lowest energy consumption, the latter performing slightly worse at 1.0 A/cm2, performance of membrane 1 is noticeably poorer. 4.3.3. Cathodes Utilizing Unsupported Catalysts As our earlier experiments indicated, carbon supported platinum catalysts had two disadvantages. Unexpected power outages, especially when oxygen was present in the cathode compartment, led to catalyst loss as a result of oxidation of the carbon support. Also, carbon particles are believed to contribute to the unwanted peroxide generation. Consequently, attempts were undertaken to develop electrodes which would utilize unsupported catalysts. In accordance with our expectations, we found that the degree of utilization of an unsupported catalyst was generally lower than that of the same catalyst in the carbon supported form with the same nominal loading. A representative result is shown in Figure 21, where the voltages of the cells equipped with the cathodes containing supported and unsupported (HiSpec® 1000) platinum catalysts are plotted versus the electrolysis time. In general, both the measured and the iR-corrected voltages of the cell equipped with the unsupported catalyst MEA were higher than those determined for the cell equipped with the standard ELAT. However, the differences between the measured (uncorrected) voltages were generally larger due to the higher HFR (not shown in Fig. 21) of the membrane used in the cell equipped with the unsupported catalyst MEA (membrane 2) as compared to that used in the cell equipped with the standard ELAT (membrane 1). The differences between the iR-corrected voltages (Fig. 21) reflect the differences in the activity and utilization of the catalyst in both cells. They are negligible at 0.2 A/cm2 and small at 0.4 A/cm2, but become significant at higher current densities, as reflected by the corrected voltages for the MEA cathode, higher from 100 mV to 200 mV than those determined for the ELAT cathode (Fig. 21). The lower degree of utilization of the unsupported catalyst results from the smaller distances between catalyst particles and from possible particle agglomeration. As a result, a fraction of the catalyst surface area is completely inaccessible. Another fraction of the surface area, although in principle accessible, cannot be fully utilized since the closely neighboring particles compete with each other as reduction centers, which results in insufficient supply of the reactants to their surface. The effect is significant enough to overrule the favorable effect of the approximately two times higher specific surface area of the unsupported catalyst. 1 1.5 2 2.5 3 0 20 40 60 80 100 120 Cell Voltage (V) Time of Operation (hrs) Measured voltages iR corrected voltages 0.2 A/cm2 0.4 A/cm2 0.6 A/cm2 0.8 A/cm2 1.0 A/cm2 Figure 21. Measured and iR-corrected voltages for the cells equipped with the standard ELAT cathode and the hydrophilic spacer (black line) and the MEA-type cathode (red line). The catalyst content: ELAT – 5.0 mg/cm2 carbon (Vulcan XC72) supported Pt, MEA – 5.0 mg/cm2 unsupported Pt (HiSpec 1000). Membrane 1 (black line), membrane 2 (red line). The catalytic activity/degree of utilization of the unsupported Pt catalyst decreased during the first 80 – 100 hours of the electrolysis, as manifested by the increase in the measured and the corrected cell voltages (Figs. 21 and 22), before it became stable. Since during this period of time, the standard experimental procedure of the stepwise ramping of the current was followed, it is unknown whether the current density, time or both were responsible for the effect. However, the most profound changes seemed to occur at 0.6 A/cm2 and 0.8 A/cm2, whereas the voltages were rather stable at 0.2 A/cm2 and 1.0 A/cm2. 1.4 1.5 1.6 1.7 1.8 1.9 0 20 40 60 80 100 12 iR Corrected Cell Voltage (V) Time of Operation (hrs) Pt/Ru 50/50 atom % HiSpec 6000 Ru Pt HiSpec 1000, double-sided GDL Pt HiSpec 1000, single-sided GDL 0.2 0.4 0.6 0.8 1.0 0 Figure 22. iR-corrected voltages for the MEA-type cathodes (red line) containing 5.0 mg/cm2 of different catalysts. Membrane 2. In principle, the increase in the corrected cell voltage can result from a decrease of accessible surface area of the catalyst or inhibition of the electrode process due to catalyst poisoning. Since a similar voltage increase was never observed in the case of carbon-supported platinum (Fig. 21), the decrease of the accessible surface area of the - 51 - catalyst seems to be a more likely origin of the observed phenomenon. Two different phenomena may have contributed to this decrease, namely: catalyst particle agglomeration and disintegration of the catalyst layer. On the other hand, partial delamination of the membrane and the catalyst layer could not be responsible for the observed voltage increase, because such a phenomenon would be accompanied by an increase in the HFR, whereas no such increase was observed (Fig. 23). 1 1.1 1.2 1.3 1.4 1.5 0 20 40 60 80 100 HFR (Ohm cm2) Time of Operation (hrs) 0.2 A/cm2 0.4 A/cm2 0.6 A/cm2 0.8 A/cm2 1.0 A/cm2 Figure 23. High frequency resistance of the MEA containing 5.0 mg/cm2 Pt (HiSpec 1000). Membrane 2. One has to note that no similar phenomenon was observed in the case of ruthenium-containing catalysts. As shown below, these catalysts undergo corrosion under the experimental conditions due to the surface oxidation of ruthenium, which eventually leads to formation of soluble ruthenates. Dissolution of the outermost catalyst layer exposes underlying layers and most likely also prevents particle agglomeration. As a - 53 - consequence, no change in the net accessible surface area may be observed. We believe that catalyst corrosion must be accompanied by some disintegration of the catalyst layer. Since this phenomenon does not seem to induce an increase in the ohmic drop corrected voltages for the ruthenium-based catalysts (Fig. 22), disintegration of the catalyst layer containing pure Pt catalyst does not seem to be a very likely cause of the increase in the cell voltage. Consequently, particle agglomeration is the most likely cause of the performance losses of the MEA containing pure Pt catalyst. In the case of platinum-ruthenium catalyst, the dissolution of ruthenium may result in formation of smaller, platinum enriched particles, which can still retain the activity of the original particles. Such an explanation of the lack of effect of the ruthenium dissolution on the activity of the platinum-ruthenium catalyst remains in accordance with the higher catalytic activity of platinum for the oxygen reduction. As mentioned above, we found that ruthenium-based catalysts underwent significant corrosion during the electrolysis. The rate of dissolution of the mixed platinum-ruthenium catalyst was higher than that of pure ruthenium (Fig. 24.a). This effect could be associated with the smaller particle size of the Pt/Ru catalyst and thus the weaker bonding of Ru in the solid phase. While the concentration of the corrosion product, ruthenate, in the NaOH solution, strongly depended on the current density and thus the cell voltage (Fig. 24.b), the dissolution rate of ruthenium from the catalyst was relatively weakly dependent on the cell voltage and consequently on the cathode overpotential. Moreover, the dissolution rate did not decrease monotonically with the increase of cell voltage as would be expected, if the catalysts were undergoing a simple electrochemical oxidation. This indicates the complex character of the dissolution process and possible participation of chemical reactions that may be the rate determining steps of the overall reaction. The chemical steps most likely involve ruthenium oxides. 0 10 0 1 10 -4 2 10 -4 1.4 1.5 1.6 1.7 1.8 1.9 Ru Dissolution Rate (mg min -1 cm -2) iR Corrected Cell Voltage (V) Pure Ru Ru/Pt 50/50 atom % HiSpec 6000 a) 0 0.02 0.04 0.06 0.08 0.1 0.12 300 450 600 Absorbance Wavelength (nm) 0.2 A/cm2 0.1 0.3 A/cm2 0.4 0.5 0.6 0.7 0.8 0.9 & 1.0 b) Figure 24.a. Dissolution rates of the ruthenium-containing catalysts plotted versus the cell voltage corrected for the ohmic losses. Figure 24.b. UV-Vis spectra of the NaOH solutions generated at different current densities. The current densities in A/cm2 are listed at each spectrum. Three different materials were tested as gas diffusion layers for the MEAs with unsupported catalysts. Figure 25 shows the unacceptable performance of the 60% teflonized Toray® paper used as the gas diffusion layer (GDL) together with the MEA containing 5.0 mg/cm2 of the pure Ru catalyst. Transport of the reagents and products through the paper was severely hindered, which resulted in unacceptably high cell voltages and the hydrogen evolution even at the lowest current densities. As the hydrogen generation posed a serious safety hazard, the experiments with the Toray® paper were eventually abandoned. Both the purity of sodium hydroxide generated and the appearance of the catalyst layer after the experiment implied no catalyst corrosion during the experiment. In fact, the lack of catalyst corrosion and the hydrogen evolution during the electrolysis strongly indicated that the cathode reaction was occurring exclusively on the surface of the paper rather than in the catalyst layer of the MEA. Figure 26 shows the measured (as well as corrected for the ohmic drop) voltages for two cells equipped with the MEAs containing 5.0 mg/cm2 Pt (HiSpec® 1000) and two different, low temperature, gas diffusion layers, LT 1400-W ELAT (single-sided) and LT 2500-W ELAT (double-sided). As seen in Fig. 26, the double-sided gas diffusion layer performed significantly better, as reflected by the significantly lower cell voltages (around 200 mV at 1.0 A/cm2). The higher voltages observed for the single-sided gas diffusion layer are believed to result from its higher flooding susceptibility. 1.5 2 2.5 3 3.5 4 0 20 40 60 80 100 120 140 160 Cell Voltage (V) Time of Operation (hrs) Measured voltage iR corrected voltage 0.2 0.2 0.4 0.4 0.2 0.6 0.8 A/cm2 1.0 A/cm 2 Single-sided LT 1400-W GDL (E-TEK) 60% teflonized Toray paper Figure 25. Comparison of 60% teflonized Toray® paper and LT 1400-W gas diffusion layers in the cells equipped with the MEAs containing 5.0 mg/cm2 Ru catalyst. - 55 - The oscillations of the ohmic drop corrected voltages of the cell employing the single-sided GDL (Fig. 26) can also be attributed to periodic changes in the degree of flooding of the GDL. Unlike the standard ELATs with carbon supported platinum catalyst, the MEAs utilizing unsupported platinum were found to be resistant to corrosion during uncontrolled interruptions of electrolysis. As demonstrated in Fig. 27, after an 8 hour long interruption of the electrolysis, the cell equipped with the MEA was operated for more than 200 hours without any performance loss. However, a several days long interruption of the electrolysis under conditions used routinely to prevent catalyst loss from the electrodes utilizing carbon supported platinum (see Table 2, condition 4), led to irreversible loss of performance. 1 1.5 2 2.5 3 3.5 0 20 40 60 80 100 120 140 160 Cell Voltage (V) Time of Operation (hrs) 0.2 0.4 0.6 0.8 1.0 Extended power outage Single-sided LT GDL (E-TEK) Double-sided LT GDL (E-TEK) Measured voltage iR corrected voltage Figure 26. Comparison of performance of single-sided and double-sided gas diffusion layers applied in the cells equipped with the MEAs containing 5.0 mg/cm2 Pt catalyst (HiSpec® 1000). When the electrolysis was restarted and the current density brought back to 1.0 A/cm2, the measured voltages were unacceptably high and shortly thereafter the cell had to be shut down. The post mortem examination of the MEA revealed significant delamination of the catalyst layer and the membrane. The increased and uneven swelling of the membrane and the Nafion binder is suspected to be a major factor responsible for the delamination. 0 0.2 0.4 0.6 0.8 1 1.2 1.5 2 2.5 3 3.5 0 100 200 300 400 500 Current Density (A/cm 2) Voltage (V) Time of Operation (hrs) Measured voltage iR-corrected voltage Current density Controlled shutdown 24 days Figure 27. Effect of controlled and uncontrolled interruption of the electrolysis on performance of the cell equipped with the MEA-type cathode. The arrows indicate voltages measured at identical current densities before and after the electrolysis interruption. Membrane 2. Platinum catalyst (HiSpec 1000) loading 5.0 mg/cm2. - 57 - Caustic current efficiency in the cells equipped with the MEA-type cathodes was similar to that in the cells equipped with the hydrophilic spacer, which was not humidified (Fig. 28). The largest difference between the current efficiencies (>2%) obtained for these two types of cells was observed at 0.2 A/cm2. The lower current efficiency in the cell equipped with the MEA may originate from the potentially easier access of water to the membrane in this case. 90 91 92 93 94 95 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 200 250 300 350 Current Efficiency (%) Current Density (A/cm 2) Time of Operation (hrs) Figure 28. Comparison of caustic current efficiency of the cells equipped with MEA and with separate gas diffusion electrode. Catalyst: 5.0 mg/cm2 Pt. Membrane 2. MEA-type cathode – black symbols, separate gas diffusion electrode – red symbols. Current efficiencies – closed symbols. Current densities – open symbols. - 59 - The water introduced to the cathode compartment of the cell equipped with the separate gas diffusion electrode and the spacer has to be transported through the hydrophobic gas diffusion layer of the electrode, then through the less but still hydrophobic catalyst layer and finally through the relatively thick spacer filled with the NaOH solution, before it can reach the membrane. On the other hand, in a cell equipped with the MEA, the only barrier in the way of water toward the membrane is the hydrophobic gas diffusion layer. Once the water reaches the hydrophilic binder in the catalyst layer, it can “freely” diffuse towards the membrane. The overall effect of easier water transport in this case is small and limited to the lowest current density, because the main source of water in the catalyst layer is the water vapor rather than liquid water, which constitutes the largest fraction of water from the humidifier (see also section 4.3.1.). The results shown in Fig. 28 demonstrate that deposition of the catalyst layer onto the membrane did not affect adversely the membrane permeability and selectivity at the higher current densities. 4.3.4. Corrosion of the Cathode Hardware In order to minimize ohmic losses in the cell, cathode hardware parts, i.e., the flow-field and the current collector must be made of electrically conductive material. A limited number of conductive materials, predominantly noble metals and their alloys, can withstand the strongly corrosive environment in the cathode compartment (highly concentrated caustic soda, high temperature, oxygen, and sometimes peroxide). In addition, the electrical potential of the hardware parts during the electrolysis may adversely affect the corrosion resistance of these materials. For instance, a material, which is corrosion resistant at its open circuit potential due to the passivation of its surface by an oxide layer, may lose its passivity at the potential of electrolysis. Some of our experiments with the fuel cell type cathode hardware involved the integrated cathode collector/flow-field made of graphite, which was both corrosion resistant and inexpensive. Unfortunately, peroxide generation in the cells equipped with the graphite parts was unacceptably high (see part 4.3.6.). Another drawback of the graphite hardware was its high porosity that led to caustic leakage through the current collector plate when the cathode compartment was pressurized. Moreover, even though the machining of the flow-field channels into the graphite current collector block was relatively easy thanks to the softness of this material, the cost of the machining was relatively high. Because of the above shortcomings, an effort was undertaken to replace the graphite hardware with the separate metal current collector and easily scalable, non-machined flow-field. Relatively inexpensive metals/alloys were selected as the construction materials for the cathode hardware parts. These were stainless steel (SS 316) and nickel. Both metals exhibit satisfactory corrosion resistance in alkaline media at their respective open circuit potentials. The parts were electroplated with noble metals, including gold and silver. The cells were run under the typical experimental conditions (see part 4.2.) for at least ten consecutive days. Then, the electrolysis was discontinued using either the standard shutdown procedure (see section 4.2.4.2.) or the procedure corresponding to one of the possible equipment failure modes as specified in Table 2 below. We found that the gold-plated nickel hardware was undergoing significant corrosion both during the electrolysis and under the open circuit conditions and the extent of corrosion seemed independent of the gold coating thickness ranging from 50 to 500 microinches. When the electrical circuit was open and all the conditions were identical with those used during the electrolysis, i.e., the cell was heated, the brine was recirculated and the cathode compartment was flushed with oxygen and water (condition 1 in Table 2), the extent of corrosion under the open circuit conditions was larger than during the electrolysis. However, corrosion also occurred under the open circuit conditions when the cell was not heated and the cathode compartment was flushed with nitrogen and water. While no tests were run for the gold plated nickel hardware under the set of conditions denoted by number 4 in Table 2, where all potential oxidizers were removed from both the cathode and the anode compartment, we believe that chloride oxidation products, e.g., hypochlorite and chlorate, which diffused from the anode compartment, were responsible for the corrosion of the cathode hardware under the open circuit conditions, when the cathode compartment was flushed with nitrogen and the brine was recirculated (condition 5). Corrosion of the Au coated nickel hardware is believed to occur via the electrochemical mechanism. Tiny pinholes in the Au coating allow for formation of electrochemical microcells, where the exposed nickel surface acts as the anode and the gold surface as the cathode. Oxygen and/or the chloride oxidation products are reduced on the gold surface and nickel is oxidized. - 61 - Table 2. The experimental conditions applied in the cathode hardware corrosion testing Experimental conditions Cell heaters Cathode compartment Anode compartment Duration 1 ON Flushed with deionized water and oxygen Brine recirculated, Chlorine-based oxidizers present Hours to days 2 ON Flushed with deionized water and nitrogen Brine recirculated, Chlorine-based oxidizers present Hours to days 3 OFF Containing residual NaOH solution and flushed with oxygen Containing residual brine, Chlorine-based oxidizers present Hours to days 4 OFF Filled with deionized water and nitrogen Filled with deionized water Days 5 OFF Flushed with nitrogen and water Brine recirculated, Chlorine-based oxidizers present Days 6 OFF Containing residual NaOH solution and flushed with oxygen Containing residual brine Single minutes (standard shutdown procedure) 7 OFF Filled with deionized water and nitrogen Flushed with concentrated brine Days Figure 29 below shows the extent of corrosion of the Au coated nickel current collectors resulting from an uncontrolled power outage (condition 3 in Table 2). Corrosion of the cathode hardware also occurred, when the cathode compartment was flushed with nitrogen and deionized water and brine containing chloride oxidation products was recirculated through the anode compartment. This result clearly demonstrates that the chloride oxidation products, e.g., hypochlorite, chlorate, chlorine, chlorine dioxide, etc., do participate in corrosion of the cathode hardware during uncontrolled interruptions of electrolysis. Stripped off coating Deep corrosion pit Flaking off coating Figure 29. Corrosion of the gold plated nickel cathode current collectors resulting from a sixteen day power failure. Coating thickness: 1 μm (left) and 10 μm (right). The gold-plated stainless steel (SS 316) hardware exhibited significantly better corrosion resistance than the nickel hardware, but corrosion still occurred both during electrolysis and under open circuit conditions. However, corrosion during electrolysis was almost negligible. The silver-plated nickel hardware exhibited excellent corrosion resistance both during the electrolysis and under the open circuit conditions. No corrosion of the hardware could be detected for any condition listed in Table 2. While every experiment, where the gold plated cathode hardware was used, required a new flow-field and a new current collector, the silver-plated nickel parts could be reused several times, as there was no - 63 - detectable signs of corrosion. The silver-plated cathode hardware parts were typically replaced after five to seven experimental runs, even though no corrosion was present. The reason for replacing the parts was gradual accumulation on the cathode side of the flow-field, of the black material from the cathode. The accumulation of this material resulted from repetitive and long lasting pressing of the flow-field against the cathode, which was necessary to create the good electrical contact between them. Although no measurable effect of this deposit on the cell performance was detected, occasional replacement of the cathode hardware was intended to avoid the potential effects of the deposit on the overall cell resistance. A minor disadvantage of the silver-plated nickel hardware was that the flow-field and the current collector became virtually inseparable after a single experimental run. The origin of the effect is unknown, but electrochemically induced recrystallization of the silver coating seems most probable. Another effect that may have led to strong bonding between the hardware parts may involve formation of the shared silver oxide film on the surfaces of the current collector and the flow-field. 4.3.5. Anode Modifications As the cathode and the cathode hardware modifications did not produce the desired increase in caustic current efficiency, different modifications of the anode were also tried. Several phenomena in the anode compartment may have a negative effect on the caustic current efficiency and other performance characteristics of the chlor-alkali cell. Large quantities of chlorine generated by the anode in close proximity of the membrane may lead to the so-called membrane blinding effect. The blinding effect is caused by chlorine bubbles, which partially block the membrane. The parts of the membrane that are covered with bubbles cannot participate in ion transport. As a result, the remaining parts may carry much higher current densities than those for which the membrane was designed and optimized. Since current efficiency tends to decrease with an increase in current density, the overall effect of gas blinding will be a decrease in the caustic current efficiency. Caustic current efficiency can also be affected by other phenomena associated with the anode performance. As the chlorine evolution reaction consumes chloride anions, the concentration of sodium cations also drops in the immediate vicinity of the anode, because the solution has to remain electroneutral. As a consequence, the brine concentration near the anode decreases. When the region of lowered brine concentration reaches the membrane, more water is transported to the cathode compartment as a result of electroosmotic drag, which results in an increased swelling of the membrane and consequently in an increased caustic crossover and lowered caustic current efficiency. The brine depletion can also cause increased oxygen evolution on the anode. This reaction lowers chlorine current efficiency and generates hydronium cations. The increased generation of hydronium cations may result in an increase of membrane resistance as a result of protonation of the carboxylic layer of the membrane. Detrimental effects of oxygen evolution on caustic current efficiency can also be expected as a result of the increased transport of the hydronium cations through the membrane. We have designed and tested four different anode modifications aimed at increasing caustic current efficiency. In our standard cells, the anode meshes are pressed against the membrane and undesired effects resulting from both the gas blinding effect and the brine depletion may be expected. We expected both effects could be alleviated by slightly moving the anode reaction zone away from the membrane. Consequently, the first modification considered was application of a suitable spacer between the membrane and the anode. Unfortunately, hot and wet chlorine is an extremely corrosive agent and not many materials are expected to be compatible with it. As our all-glass anolyte recirculation tanks containing hot brine and chlorine exhibited no signs of corrosion after having worked for tens of thousands of hours, we selected glass as the material for the spacer. The first modified anode structure contained a relatively loose glass fiber woven cloth as a spacer separating the anode and the membrane. Unfortunately, quite contrarily to our expectations, the caustic current efficiency for the modified cell was significantly lower than for the unmodified cell (see Table 3). In addition, the modification caused a significant increase of the cell voltage and erratic voltage-time behavior, as shown in Fig. 30. During the experiment, brine in the recirculation tank was slightly turbid and foaming. In addition, its acidity was very significantly increased, as may be expected for significant contribution of the oxygen evolution in the overall anode reaction. When the experiment was discontinued after merely 60 hours and the cell was disassembled, we found that the glass cloth was almost completely disintegrated. The corrosion of the cloth is deemed responsible for the brine turbidity and foaming in the recirculation tank. While it is difficult to present the detailed mechanism of the glass fiber corrosion, the following two reactions seem most likely: 2 H+ + SiO3 2- + 2 Cl2 → SiCl4↑ + H2O + O2↑ (6) SiO2 + 2 Cl2 → SiCl4↑ + O2↑ (7) 1 1.5 2 2.5 3 0 20 40 60 80 100 120 Cell Voltage (V) Time (hrs) 0.2 A/cm2 0.4 A/cm2 0.6 A/cm2 0.8 A/cm2 1.0 A/cm2 Anode structure 1 (reference) Anode structure 2 Figure 30. Significant increase of the cell voltage upon anode structure modification resulting from the use of the glass fiber cloth as a spacer between the anode and the membrane. Platinum catalyst loading 0.5 mg/cm2. Temperature: 90oC. Brine concentration: 200 g/L. Oxygen pressure 20 psig. Humidification 0.5 cm3/min. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. - 65 - The above reactions may not occur with bulk glass, but they are more likely to occur with the glass fiber cloth, where the surface area to volume ratio is significantly higher. As seen from equations 6 and 7, corrosion of the glass fibers cannot be responsible for the increased brine acidity in this experiment. In fact, reaction 6 can actually lead to the opposite effect. As initially suspected, the reason for the increased brine acidity is most likely the oxygen evolution process: 2 H2O → O2 + 4 H+ + 4 e- (8) We suspect that the circulation of brine in the glass fiber cloth was insufficient to maintain the desired brine concentration in the immediate vicinity of the anode surface. The brine concentration inside the spacer became quite low, which resulted in more significant oxygen evolution and brine acidification. The hydronium cations started being transported through the membrane and protonated its carboxylic layer, which led to a significant increase in the membrane resistance and thus also in the cell voltage (Fig. 30). When the glass fiber cloth started disintegrating (eqs. 6 and 7), the brine in the immediate vicinity of the membrane and the electrode could be more easily replaced with the fresh brine from the recirculation tank and when the corrosion progressed, the cell started returning to more or less normal operation, as manifested by the gradual voltage decrease at 0.4 A/cm2 (Fig. 30). Three more anode modifications, hereafter called structures 3 through 5 were studied. While all of them were based on a similar principle, they produced different results. The modified structure 3 did not produce the desired increase in caustic current efficiency. Up to the current density of 0.6 A/cm2, the cell equipped with structure 3 generated sodium hydroxide with an efficiency that was almost identical to that obtained for the reference structure (structure 1). At higher current densities, the current efficiency of the cell equipped with structure 3 fell below that for the reference structure (Fig. 31). The structure 4 offered significant improvement over the reference structure at all current densities (Fig. 31). Structure 4 was the only anode modification that resulted in caustic current efficiencies matching the current industrial standards for the membrane cells operating at standard current densities (Fig. 31). Structure 5 did not improve the caustic current efficiency at low current densities (≤0.4 A/cm2) and performed similarly to the reference structure and structure 3. However, it performed as well as structure 4 at high current densities (≥0.6 A/cm2). 84 86 88 90 92 94 96 98 100 0 0.2 0.4 0.6 0.8 1 1.2 Caustic Current Efficiency (%) Current Density (A/cm 2) Structure 1 (reference) Structure 3 Structure 4 Structure 5 industrial standard Figure 31. Effect of modification of the anode structure on caustic current efficiency. Platinum catalyst loading 0.5 mg/cm2. Temperature: 90oC. Brine concentration: 200 g/L. Oxygen pressure 20 psig. Humidification 0.5 cm3/min. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. The changes in caustic current efficiency were not the only effects of the anode modifications. The cell voltage, another important performance characteristic, was also affected. Figure 32 shows the cell voltages for different anode structures, measured at five different current densities during the first 120 hours of the electrolysis. As seen from Fig. 32, all the anode modifications led to some increase in the cell voltage. There was no straightforward relationship between the measured cell voltages and the caustic current - 67 - efficiency. For instance, the highest voltages were typically observed for structure 3, which did not improve the current efficiency, whereas the second highest voltages were observed for structure 4, which offered the largest improvement in caustic current efficiency (Fig. 31). 1.6 1.8 2 2.2 2.4 2.6 0 20 40 60 80 100 120 Cell Voltage (V) Time of Operation (hrs) 0.2 A/cm2 0.4 A/cm2 0.6 A/cm2 0.8 A/cm2 1.0 A/cm2 Structure 1 (reference) Structure 3 Structure 4 Structure 5 Figure 32. Effects of different anode structure modifications on the measured cell voltages during the first 120 hours of the cell operation. Platinum catalyst loading 0.5 mg/cm2. Temperature: 90oC. Brine concentration: 200 g/L. Oxygen pressure 20 psig. Humidification 0.5 cm3/min. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. The voltage increase for the structure 3 after ~110 hours resulted from an uncontrolled change in brine concentration. In order to assess feasibility of the anode modifications, we used the method applied previously when testing different membranes, which is based on the changes in the energy efficiency of the electrolysis (eq. 5). The results are presented in Figure 33. 0 0.005 0.01 0.015 0.02 0.025 0.03 0 0.2 0.4 0.6 0.8 1 1.2 Curent Density (A/cm 2) Structure 1 (reference) Structure 2 Structure 3 Structure 4 Structure 5 (Cell Voltage Current Density)/CCE Figure 33. Energy efficiency (see eq. 5) of the cells equipped with different anode structures (see text) plotted versus current density. It is clearly seen from Fig. 33 that structure 5 offers the lowest energy consumption per unit weight of the product at high current densities (0.6-1.0 A/cm2), and thattructure 4 is the second best, whereas structure 3 is worse than the reference. Due to their smaller absolute values, the differences in energy efficiency at low current densities are not seen well in Fig. 33. The relevant data are listed in Table 3. At the low current densities (0.2-0.4 A/cm2), the reference structure offers the lowest and the unworkable structure 2, the highest energy consumption. The energy efficiency of the cells equipped with the remaining structures does not follow exactly the same trend at 0.2 A/cm2 and 0.4 A/cm2 (Table 3), but the difference may result from experimental error. - 69 - Table 3. Performance characteristics of the cells equipped with different anode structures Structure Current Density A/cm2 CCE % Cell Voltage V k-factor V A-1 cm2 G-factor 1 0.2 94.6 1.666 0.871 0.003522 2 0.2 85.9 2.252 1.935 0.005243 3 0.2 93.9 1.691 0.960 0.003602 4 0.2 96.5 1.753 0.860 0.003633 5 0.2 93.9 1.662 0.934 0.003540 1 0.4 93.9 1.867 0.871 0.007953 2 0.4 79.2 2.639 1.935 0.01333 3 0.4 94.2 1.916 0.960 0.008136 4 0.4 96.3 1.915 0.860 0.007954 5 0.4 94.1 1.885 0.934 0.008013 1 0.6 91.7 2.046 0.871 0.01339 3 0.6 92.2 2.115 0.960 0.01376 4 0.6 94.3 2.099 0.860 0.01336 5 0.6 94.1 2.059 0.934 0.01313 1 0.8 87.1 2.210 0.871 0.02030 3 0.8 88.1 2.301 0.960 0.02089 4 0.8 92.9 2.263 0.860 0.01949 5 0.8 93.4 2.224 0.934 0.01905 1 1.0 88.1 2.365 0.871 0.02684 3 1.0 86.3 2.458 0.960 0.02848 4 1.0 92.1 2.439 0.860 0.02648 5 1.0 92.1 2.426 0.934 0.02634 The red and green highlighted numbers denote respectively the worst and the best values of the parameters for a given current density. The red numbers were obtained for the anode equipped with fiber-glass cloth spacer (structure 2, see text). 4.3.6. Effects of Different Factors on Peroxide Generation Rate The peroxide content in the sodium hydroxide generated was found to increase with current density at both high (80% Pt, 5.0 mg/cm2) and low (20% Pt, 0.5 mg/cm2) catalyst loading (Fig. 34). This effect is believed to originate from the different kinetics of complete 4-electron reduction and partial 2-electron reduction of oxygen (eqs 2 and 3, section 3). The increase in current density shifts the cathode potential towards more negative values and affects the relative rates of peroxide and caustic generation. 10 -4 10 -3 10 -2 10 -1 0 0.2 0.4 0.6 0.8 1 1.2 Molar % of Peroxide Current Density (A/cm 2) 0.5 mg/cm 2 5.0 mg/cm 2 Fig. 34. Effect of current density and Pt loading on peroxide generation. The ratio of peroxide concentrations generated at high and low catalyst loadings is roughly independent of current density and its average value for the plots in Fig.34 equals 4.5. The ratio of peroxide generation rates qualitatively correlates with the relative contents of Pt in catalytic layers of both electrodes and suggests that carbon and platinum particles compete as oxygen reduction centers. While the numbers of electrons involved in the ORR - 71 - on the individual Pt and C particles and the respective efficiencies of peroxide generation are unknown, the literature data clearly demonstrate that gas diffusion electrodes containing pure carbon in the catalyst layer produce significantly more peroxide than electrodes containing carbon supported Pt catalysts [26,27,32,33,36]. The slightly higher overpotentials of the oxygen cathodes with low Pt loadings may also contribute to higher peroxide generation rates at low Pt loadings. The magnitude of this effect is rather small, since the observed differences in ohmic drop-corrected voltages amount to 0.05-0.08 V at 10 kA/m2. The rate of peroxide generation depends on the electrolysis time. Peroxide concentration changes occur in times of tens to hundreds of hours and depend on experimental conditions. The typical dependence of peroxide concentration on electrolysis time is shown in Fig. 35. 0.01 0.1 1 10 0 500 1000 1500 2000 2500 Peroxide (mol%) Time of Operation (hrs) Fig. 35. Time effect on peroxide generation rate. Platinum loading: 0.5 mg/cm2. Current density: 10 kA/m2. - 73 - We believe that the observed increase of peroxide generation at constant current density reflects the increasing role of carbon particles in oxygen reduction. The contribution from the carbon centers is likely to increase with time in the presence of peroxide , oxygen, and hot and concentrated caustic. These harsh conditions can lead to surface oxidation of the carbon particles and an increase of their hydrophilicity. Another likely reason for the more significant contribution of carbon in the cathode process and more significant peroxide generation is a loss of platinum surface area due to the agglomeration of Pt particles and/or the loss of Pt particles due to the oxidative corrosion of carbon carriers . An increase in the hydrophilicity of the cathode was always observed when a long experiment (from a few hundred to 2000 hours) was performed. Before the experiment, the water contact angle on the catalyst side of the electrode was around 147° and around 170° on the gas diffusion layer side. After the electrolysis, the contact angle was always below ~120° on the catalyst side and below ~140° on the gas diffusion layer side, but the measurement was quite inaccurate and the angle varied over the electrode surface. In the areas where the flow-field and gas diffusion layer remained in direct contact, the gas diffusion layer exhibited exceptionally elevated hydrophilicity, which was easily detected, when the electrode was rinsed with deionized water. The water frequently adhered to the electrode in the areas of contact and formed a pattern that was reproducing the geometry of the flow-field. The measurement of the contact angle on the catalyst side was distorted by other phenomena. The prolonged electrolysis invariably resulted in the adhesion of the catalyst layer and the carbon cloth spacer. When the parts were being separated, numerous carbon fibers were transferred to the catalyst layer and some material from the catalyst layer was transferred to the carbon cloth. While the presence of the material from the catalyst layer on the hydrophilic spacer could be frequently seen with the naked eye, the presence of the carbon fibers on the electrode surface was detected using a microscope. Even though the amount of material transferred from the catalyst layer to the spacer was rather small, no attempts were made to determine the quantity of platinum present in the catalyst layer after electrolysis, since such an experiment was not expected to help understanding the changes in peroxide generation rate at 10 kA/m2. In accordance with the data obtained by Morimoto et al. , our previous experiments, where the cells equipped with no spacer were used, frequently indicated some catalyst loss during electrolysis. However, since the experiments at current densities above 2-4 kA/m2 required “conditioning” of the membrane at lower current densities to prevent its possible damage at higher current densities, any platinum loss determined after electrolysis always reflected the sum of losses at all the current densities applied. Among them, the possible loss of platinum at 10 kA/m2 was expected to be negligible and masked by significantly larger losses resulting from the cell operation at lower current densities during the initial hours of electrolysis. Examination of the electrode after electrolysis under the microscope revealed also numerous cracks in both the gas diffusion layer and the catalyst layer. The number of cracks per unit surface area seemed to be higher on the catalyst side of the electrode. An attempt was undertaken to determine, if peroxide was predominantly responsible for the loss of electrode hydrophobicity. Small pieces of ELAT® were placed in three beakers containing: deionized water, 31 % (weight) sodium hydroxide, and 31 % (weight) sodium hydroxide with as high as possible quantity of peroxide added. The temperature of the solutions was maintained at approximately 90ºC. Unfortunately, due to fast peroxide decomposition, the experiment required frequent additions of peroxide and for this reason it could not be continued beyond eight hours. After such a short time, none of the samples exhibited any meaningful change in the contact angle. We believe that the observed increase of peroxide generation at constant current density reflects the increasing role of carbon particles in oxygen reduction. The effect of brine concentration on peroxide production was studied at a current density of 10 kA/m2 for times longer than 500 hours after a steady-state peroxide generation rate was reached (Fig. 35). Brine concentration was adjusted by modifying the fresh brine supply to the recirculation tank. It usually took around an hour for the brine concentration to reach its steady-state level. For the purposes of hydroxide and peroxide analysis, samples of the caustic were taken after an additional one half to one hour after the brine concentration stabilized. Figure 36 shows the effect of brine concentration on the composition of the NaOH solution generated. An increase of brine concentration from 136 to 228 g/dm3 produces an increase in NaOH concentration from around 26.5% to a little more than 31%. The peroxide content in the solution increases even faster, as manifested by approximately 25% increase of the peroxide-to-hydroxide molar ratio. The effect of brine on peroxide generation most likely results from changes in water activity at the oxygen reduction site caused by the NaOH concentration changes. 0.024 0.025 0.026 0.027 0.028 0.029 0.030 0.031 26 27 28 29 30 31 32 33 120 140 160 180 200 220 240 Brine Concentration (g/dm 3) NaOH concentration (%) Peroxide to Hydroxide Molar Ratio Fig. 36. Effect of brine concentration on peroxide generation rate and NaOH concentration. Platinum loading: 0.5 mg/cm2. Current density: 1.0 A/cm2. The effect of brine strength on NaOH concentration most likely results from changes in the membrane swelling. At lower brine concentrations, the membrane is more swelled and more permeable than at higher concentrations. As a consequence, more water is transported through the membrane yielding lower caustic concentration. A more quantitative measure of the degree of membrane swelling and permeability is its resistance, - 75 - which increases about 7% for the measured increase in brine concentration. As could be expected, the caustic current efficiency also increases with the brine concentration as a result of a decreased caustic crossover (Fig. 37). 0.66 0.68 0.7 0.72 0.74 0.76 83 84 85 86 87 88 89 120 140 160 180 200 220 240 Membrane Resistance (Ohm cm 2) Caustic Current Efficiency (%) Brine Concentration (g/dm 3) Fig. 37. Effect of brine concentration on membrane resistance and caustic current efficiency. Platinum loading: 0.5 mg/cm2. Current density: 10 kA/m2. One may also suspect partial oxygen evolution on the DSA® anode to contribute to the above phenomena (Figures 36 and 37). This reaction produces hydronium cations and its relative contribution to the measured current increases with the decrease in brine concentration . The increase in the brine acidity may produce an increased H3O+ flux across the membrane, which may result in a lower membrane resistance, lower current efficiency, and lower NaOH concentration. Such an explanation would be plausible for a sulfonated membrane, which behaves like a strong electrolyte in both the acid and the sodium salt form and its conductivity correlates with the molar conductivities of hydronium and sodium ions . The picture seems more complex in the present case. As the carboxylic layer of the bi-layer membrane behaves like a weak electrolyte when in the protonated form, the net effect of the enhanced oxygen evolution and the increased brine acidity may be opposite to the effect observed. Consequently, the effect of brine concentration on the extent of membrane swelling offers a better explanation of the effects shown in Figures 36 and 37. The significant change in oxygen humidification level from 0 to almost 6 cm3/min decreases the peroxide-to-hydroxide molar ratio by approximately 10%. At the same time, caustic concentration drops from 31.6% to 12.6%, i.e., much more than observed in the experiments where the brine concentration effects were studied (see above). 0.024 0.025 0.026 0.027 0.028 0.029 0.030 0.031 10 15 20 25 30 35 Peroxide to Hydroxide Molar Ratio NaOH Concentration (%) brine concentration effect oxygen humidification effect Fig. 38. Comparison of effects of oxygen stream humidification and brine concentration on peroxide generation rate. Platinum loading: 0.5 mg/cm2. Current density: 10 kA/m2. - 77 - Quantitative comparison of the humidification and brine concentration effects on peroxide formation is shown in Fig. 38. Here, the molar ratio of peroxide-to-hydroxide is plotted against NaOH concentration, which is a common measure of the quantity of water introduced into the cathode compartment either by transport from the anode compartment through the membrane or by direct humidification of the oxygen stream. The much weaker effect of the direct humidification of oxygen (Fig. 38) indicates that most of the humidification water does not reach the catalyst layer and thus affects neither the composition of the liquid phase at the reaction site nor the membrane permeability. The main effect of oxygen humidification is the diluting of the NaOH solution that already left the electrode. 0 5 10 15 20 25 30 35 40 0 0.5 1 1.5 2 2.5 50 100 150 200 250 300 350 Chloride Concentration (ppm) Peroxide Concentration (mol%) Time of Operation (hrs) Fig. 39. Changes of chloride and peroxide concentrations in caustic soda during the first 300 hours of cell operation at 1.0 A/cm2. Figure 39 presents time dependencies for the concentrations of chloride and peroxide in the NaOH solution generated at 1.0 A/cm2 during the first 300 hours of cell operation. During this time the most significant changes in peroxide concentration occur (Fig. 35). At the same time, the chloride content does not vary, except for the first 20 hours at 1.0 A/cm2 (between 50 and 70 hours after the start of electrolysis), where the membrane may not have attained its steady state permeability. These results demonstrate the insignificance of the chloride effect on peroxide generation. The hydrophilic spacer significantly reduces peroxide generation (Fig. 40). 10 -1 10 0 101 102 0 500 1000 1500 2000 Peroxide Concentration (mmol/kg) Time of Operation (hrs) Hydrophilic spacer No spacer Figure. 40. Effect of the hydrophilic spacer on peroxide generation rate. Gold-plated cathode hardware. Panex® 30 carbon cloth used as the hydrophilic spacer. - 79 - We believe that the presence of spacer reduces the likelihood of carbon particles participating in the oxygen reduction in two ways. First of all, since the spacer acts as an efficient drain for the NaOH solution , the quantity of NaOH remaining inside the electrode decreases and the corrosion of carbon particles is less likely to occur. Secondly, without the spacer, the catholyte can accumulate in the pores of the Pt catalyst-free gas diffusion layer, and the partial reduction of oxygen may occur on carbon particles inside the gas diffusion layer. Hydrogen peroxide in alkaline medium (HO2 -) is typically regarded as very unstable. Since the decomposition of peroxide is catalyzed by trace impurities, the fact that the peroxide generated in our cells survives in spite of the unfavorable conditions in the cathode compartment, i.e., the strongly alkaline environment and the high temperature, can be associated with the high purity of the NaOH solutions generated. However, a variety of solid surfaces or surface defects can act as peroxide decomposition centers. Consequently, an effect of the cathode hardware coating on the peroxide generation rate was also studied (Fig. 41). 0.1 1 10 100 1000 0 0.2 0.4 0.6 0.8 1 0 50 100 150 200 Peroxide Concentration (mmol/kg) Current Density (A/cm2) Time (hrs) Silver plated hardware Gold plated hardware Figure 41. Comparison of peroxide generation rates in cells equipped with gold- and silver-plated hardware. Peroxide concentrations expressed in millimoles of peroxide per kilogram of liquid caustic. Platinum catalyst loading 0.5 mg/cm2. Membrane 1. Panex® 30 carbon cloth spacer between the oxygen cathode and the membrane. - 81 - During electrolysis, the potential of the cathode hardware parts is determined by the kinetics of oxygen reduction on the gas diffusion electrode, which has much higher active surface area than the hardware parts. While this potential, together with intrinsic properties of the cathode hardware coating, determine the rate of electrochemical decomposition of peroxide on the surface of cathode hardware parts, the final peroxide concentration in the caustic stream is determined by the relative rates of peroxide generation by the oxygen diffusion cathode and its destruction by the coating of the hardware parts. As the rates of peroxide generation in the cells equipped with identical oxygen diffusion cathodes under identical experimental conditions are the same, differences in measured peroxide levels reflect relative ability of the different hardware coatings to decompose peroxide by-product. Figure 41 shows concentration of peroxide generated in two cells equipped with gold plated and silver plated hardware under comparable conditions. At low current densities (0.2-0.4 A/cm2), the cell equipped with silver plated hardware was found to produce less peroxide byproduct. The opposite situation was true for high current densities (0.8-1.0 A/cm2), i.e., less peroxide was detected in the caustic product generated in the cell equipped with the gold plated hardware. One of the expected benefits of using unsupported catalysts (section 4.3.3.) was elimination of the peroxide generation. The experiments revealed that removal of carbon support from the catalyst layer reduced the unwanted byproduct. However, no complete peroxide elimination was achieved, even when the most potent oxygen reduction catalyst, i.e., pure platinum was used. This finding implies that either the oxygen reduction on platinum does not occur completely according to the four-electron mechanism (eq. 2) under the experimental conditions applied or the carbon-containing gas diffusion layer contributes to the overall reduction current and is at least partially responsible for the peroxide generation. Simultaneous occurrence of both phenomena is also possible. The results shown in Fig. 42 suggest that the gas diffusion layers in cells equipped with the MEA-type cathodes can participate in the oxygen reduction and be responsible for some peroxide generation. As demonstrated above (see section 4.3.3.), the single-sided gas diffusion layer is more likely to get partially flooded by NaOH solution. The partial flooding of the gas diffusion layer facilitates formation of the three-phase boundary involving the carbon particles in the gas diffusion layer, oxygen and NaOH solution and thus creates suitable conditions for participation of the gas diffusion layer in the overall reduction current. The lower rate of peroxide generation in the cell equipped with the more hydrophobic, double-sided GDL (Fig. 42) remains in accord with the above hypothesis. 0 5 10 15 0 0.2 0.4 0.6 0.8 1 1.2 0 50 100 150 200 250 Peroxide Concentration (mmol/kg) Current Density (A/cm 2) Time of Operation (hrs) Figure 42. Comparison of peroxide generation rates in cells equipped with MEAs containing 5.0 mg/cm2 of the Pt catalyst (HiSpec 1000) and different gas diffusion layers. Peroxide concentrations expressed in millimoles of peroxide per kilogram of liquid caustic. Red symbols – double sided gas diffusion layer (LT-2500-W). Black symbols – single sided gas diffusion layer (LT-1400-W). Solid symbols – peroxide concentration. Open symbols – current density. Membrane 2. - 83 - The MEA-type cathodes containing both the pure ruthenium and the mixed platinum/ruthenium catalyst produced significantly less peroxide than the pure platinum catalyst and the peroxide concentration decreased in the order: Pt > Ru > Pt/Ru. Consequently, the mixed Pt/Ru catalyst was found to be most efficient in destroying the unwanted byproduct. The highest peroxide concentration detected for a ruthenium-based catalyst was 0.12 mmol/kg at 1.0 A/cm2 after approximately 160 hours of electrolysis in the cell, where the pure ruthenium was used (see also Fig. 42). The low concentrations of peroxide generated in the cells which utilized the ruthenium-based catalysts, most probably originates from catalytic action of ruthenates on the peroxide decomposition reaction, as confirmed by the results of simple test tube experiments. Moreover, the relative quantities of peroxide generated in the cells utilizing the Pt/Ru and Ru catalysts inversely correlated with the amounts of ruthenium present in the respective NaOH solutions (see Fig. 24). In order to eliminate the peroxide from the product stream, the cathode design and structure were modified. The first modification, hereafter called the modified cathode structure 1, resulted in a very substantial decrease of the peroxide generation rate. The results obtained for the modified cathode are shown in Fig. 43 (above) together with the typical results obtained for the standard ELAT cathode equipped with the hydrophilic spacer. As easily seen, the modified structure 1 is especially efficient in destroying peroxide at high current densities, which are targeted by the new technology. While the rate of peroxide generation by all the cathodes studied previously significantly increased with current density, an opposite trend was observed for the modified structure (Fig. 43). 0.0001 0.001 0.01 0.1 0 20 40 60 80 100 120 140 Molar % of Peroxide Hours of Operation 0.2 0.2 0.2 0.2 0.4 0.4 0.6 0.4 0.4 0.4 0.6 0.6 0.8 0.6 0.6 1.0 0.8 0.8 0.8 0.8 1.0 1.0 1.0 1.0 1.0 1.0 Standard ELAT & hydrophilic spacer Modified structure 1 Fig. 43. Peroxide generation rates in the cells equipped with the modified cathode structure 1 (see text) and with standard ELAT cathode and hydrophilic spacer (Panex 30). Gold-plated cathode hardware. Catalyst: 5.0 mg/cm2 Pt. The even numbers denote current densities in A/cm2. Gold-plated cathode hardware used. As opposed to all cathode structures studied previously, peroxide concentrations generated by the modified cathode structure were consistently low during long-term cell operation approaching 2 months (Fig. 44). 0.01 0.1 1 0 200 400 600 800 1000 1200 1400 Peroxide Concentration (mmol/kg) Time of Operation (hrs) Figure 44. Effects of long time cell operation on peroxide generation rate in a cell equipped with the modified cathode structure 1 (see text). The cell equipped with modified cathode structure 1 exhibited slightly higher voltages than the cell equipped with the standard ELAT cathode and no hydrophilic spacer. The cell voltage increase (Fig. 45) is considered a relatively small penalty for the negligible peroxide generation. - 85 - 1.6 1.8 2 2.2 2.4 2.6 2.8 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 0 20 40 60 80 100 120 140 Cell Voltage (V) HFR (Ohm cm 2) Time of Operation (hrs) Modified structure 1 Standard ELAT, no spacer 0.2 0.4 0.6 0.8 1.0 Figure 45. Effects of current density on cell voltages and high frequency resistance for the LANL standard cathode hardware and the modified structure 1. Even numbers in circles denote consecutive current densities in A/cm2. Temperature: 90oC. Brine concentration: 200 g/l. Standard LANL hardware: HFR __ , Voltage _ . Modified structure 1 : HFR , Voltage ___ . A series of experiments was performed to elucidate the effects of extreme conditions on the overall performance of structure 1. The experiments revealed that the new cathode was more susceptible to flooding than the cathode structures used previously. Figure 46 shows the changes in the observed cell voltage brought about by changes in the oxygen flow rate. 2.65 2.7 2.75 2.8 2.85 102 104 106 108 110 112 114 Cell Voltage (V) Time (hrs) 4 x 3 x 2 x 1.5 x 5 x 5 x Figure 46. Effects of change of experimental conditions on performance of the cell equipped with the modified cathode structure 1. Red arrows indicate times when the oxygen flow rate was changed. The new flow rates are indicated by the numbers corresponding to how many times the oxygen flow rate exceeded that required by the stoichiometry of the complete oxygen reduction. Green arrow marks termination of the experiment. Temperature: 90oC. Brine concentration: 200 g/dm3. Current density: 1.0 A/cm2. As shown in Fig. 46, a change in experimental conditions lead to the voltage instability and eventually to its significant increase. The voltage increase was found to be irreversible, since returning to the previous experimental conditions did not result in lowering of the cell voltage (Fig. 46). Some improvement was noted after temporarily discontinuing the electrolysis and flushing the system with pure water under nitrogen, but no complete recovery was observed. This behavior was quite consistent with the flooding phenomenon. - 87 - The changes in oxygen flow rate were accompanied by changes in the rate of peroxide generation (Fig. 47). 6 10-4 8 10 -4 1 10 -3 1.2 10-3 1.4 10 -3 1 2 3 4 5 6 Ratio of Oxygen Flow to that Required by Stoichiometry Molar % of Peroxide Figure 47. Dependence of peroxide generation rate on the oxygen flow rate for the modified cathode structure 1 (see text). Fig. 47 demonstrates rather complex character of the effect of oxygen flow rate on peroxide generation. The presence of a maximum in the plot suggests opposite influence from at least two different factors. When the inlet oxygen flow rate becomes low, i.e., close to that required by the stoichiometry of 4-electron oxygen reduction (eq. 2), the outlet pressure in the cell drops below the preset value of 20 psig, as there is unsatisfactory supply of the gas to the cell. As a result of the pressure drop, the relative kinetics of the 4-electron and 2-electron oxygen reduction processes (eqs. 2 and 3) are likely to change, which can result in smaller quantities of peroxide generated. The decrease of the peroxide generation rate at significantly higher rates of oxygen flow most likely results from the effect of gas flow on the effectiveness of caustic removal from the electrode. High gas velocity in the cathode chamber makes the removal of caustic from the electrode pores easier. 0 0.5 1 1.5 2 2.5 2.15 2.2 2.25 2.3 2.35 2.4 2.45 Percent of H2 in the Cathode Exhaust IR Corrected Cell Voltage (V) 2.125 V Figure 48. Estimate of the limits of safe cell operation from hydrogen evolution in a deliberately flooded cell equipped with the modified cathode structure 1. Temperature: 90oC. Brine concentration: 200 g/dm3. Shorter residence time of NaOH inside the electrode lowers the likelihood of carbon participating as the oxygen reduction catalyst and thus also the peroxide generation rate. Moreover, under such conditions, water from the humidifier has better access to the electrode pores, which may lead to a decrease of caustic concentration at the reaction site and an effect similar to the effect of brine concentration on peroxide generation may result. Since electrode flooding reduces the number of catalyst particles available for oxygen reduction, flooded electrodes are likely to generate hydrogen, which can create rather dangerous conditions. In order to assess limits of safe electrolysis, one of the cells - 89 - was deliberately flooded and hydrogen content in the exhaust oxygen stream was monitored at different current densities. Figure 48 (above) shows a plot of the hydrogen content versus the cell voltage corrected for the ohmic drop. Linear extrapolation of the plot in Fig. 48 towards the low voltages gives an estimate of the corrected cell voltage, where virtually no hydrogen is produced. 0 0.5 1 1.5 2 2.5 2.15 2.2 2.25 2.3 2.35 2.4 2.45 Percent of H2 in the Cathode Exhaust IR Corrected Cell Voltage (V) 2.125 V Figure 48. Estimate of the limits of safe cell operation from hydrogen evolution in a deliberately flooded cell equipped with the modified cathode structure 1. Temperature: 90oC. Brine concentration: 200 g/dm3. The estimated value of the onset of hydrogen evolution (2.125 V) can be used to determine the highest operating current density for a cathode with the identical catalyzed layer as long as the cell voltage corrected for ohmic losses is known. The result of such a determination for the standard cell equipped with the hydrophilic spacer is shown in Fig. 49. Since the actual fraction of the current that produces hydrogen increases with the degree of flooding, the highest operational current density for the cell with no flooding can be even higher than ~2.7 A/cm2, the value obtained from the linear extrapolation of the plot shown in Fig. 49. 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 0 0.5 1 1.5 2 2.5 3 y = 1.3972 + 0.2717x R= 0.9999 iR Corrected Cell Voltage (V) Current Density (A/cm2) Expected onset of H2 evolution 2.678 A/cm2 Figure 49. Determination of the highest operating current density from the onset of hydrogen evolution in a cell with the flooded cathode. Temperature: 90oC. Brine concentration: 200 g/dm3. Pt loading 5.0 mg/cm2. Panex 30 spacer. Further modifications of the cathode structure were introduced to reduce its susceptibility to flooding and still maintain the low peroxide generation rates. The modification resulted in an improvement of the flooding characteristics. However, the performance of the modified structure, hereafter called modified structure 2, was found to deteriorate over time in terms of both peroxide generation rate (Fig. 50) and cell voltage stability. - 91 - -0.1 0 0.1 0.2 0.3 0.4 0 50 100 150 200 250 Peroxide Concentration (mmol/kg) Hours of Operation 0.2 A/cm 2 0.4 A/cm 2 0.6 A/cm 2 0.8 A/cm 2 1.0 A/cm 2 Figure 50. Peroxide generation in a cell equipped with the modified cathode structure 2. Most promising ways to improve the overall performance of modified structure 2 were determined and are planned to be implemented in the future. - 93 - Accomplishments 1) Effects of Operating Conditions on Selected Performance Characteristics of Oxygen-Depolarized Chlor-Alkali Cell by L.Lipp, S.Gottesfeld, J.Chlistunoff, Centennial Meeting of the Electrochemical Society, Philadelphia, May 2002 2) Zero-Gap Chlor-Alkali Cell With Oxygen-Consuming Cathode. Hardware Effects on the Cell Operation by Ludwig Lipp, Shimshon Gottesfeld and Jerzy Chlistunoff, 203rd Meeting of the Electrochemical Society, Paris, France 3) Oxygen-cosuming chlor alkali cell configured to minimize peroxide formation by J.Chlistunoff, L.Lipp, S.Gottesfeld. Patent application 20050026005. 4) L.Lipp, S.Gottesfeld, J.Chlistunoff, Peroxide Formation in a Zero-Gap Chlor-Alkali Cell with an Oxygen-Depolarized Cathode. Accepted for publication in the Journal of Applied Electrochemistry 5) Oxygen-Depolarized Chlor-Alkali Cells. Some Consequences of Using Oxygen Diffusion Cathodes by J.Chlistunoff. Abstract of the presentation for the 206th Meeting of the Electrochemical Society. 6) UV Spectroscopic Study of Ion Association of Hydroperoxide Anion HO2 - in Concentrated Sodium Hydroxide Solutions by J.Chlistunoff and J-P.Simonin, paper in preparation Conclusions and Recommendations This study demonstrates that the zero-gap chlor-alkali cells with gas-diffusion cathodes offer a practical alternative to the standard membrane-cells. At the standard industrial current densities (≤ 0.4 A/cm2), the oxygen-depolarized cells produce caustic with current efficiencies matching current industrial norms for membrane cells but at significantly lower voltages and thus with substantial energy savings. The cells can also be operated at significantly higher current densities (up to 1.0 A/cm2). The best overall cell performance was obtained when carbon supported platinum was used as the oxygen reduction catalyst. The shortcomings of carbon as the catalyst support include its insufficient corrosion resistance in oxygenated caustic and the involvement in peroxide generation. We demonstrated that peroxide generation can be virtually eliminated by proper cathode design. However, the poor corrosion resistance of carbon reduces the cathode durability. Cathode corrosion significantly decreases with the increase in current density and it is most likely negligible at 1.0 A/cm2, but operating the cell at lower current densities and (especially) unexpected power outages can lead to substantial losses of the precious metal catalyst. The MEA-type electrodes utilizing unsupported catalysts offer excellent corrosion resistance during power outages. However, they operate at significantly higher voltages and consequently they reduce the energy savings. Moreover, they do not guarantee the elimination of peroxide as long as carbon-containing gas diffusion layers are used. The catalyst loss under open circuit conditions can be avoided by automatically applying cathode-protecting conditions, e.g., flushing the cathode compartment with an inert gas, etc., but the same method cannot be applied when the cell is being operated. Two ways of alleviating the problem seem most obvious. As the precious metal catalyst most likely participates in the oxidative corrosion of carbon carriers by forming local micro-cells involving carbon particles, using a less precious metal as the catalyst may be a feasible alternative. Although worse than that of platinum, the catalytic activity of silver is non-negligible under the industrial conditions and the corrosion resistance of cathodes utilizing carbon-supported silver was shown to be better than those utilizing the carbon-supported platinum. Metal alloys, such as for example Ag93Pt7 , are a promising alternative to the expensive platinum. Another way of solving the cathode corrosion problem may involve alternative catalyst supports, which will exhibit better corrosion resistance in the oxygenated NaOH solution. The more detailed summary of the most important technical results obtained is presented below: • Zero-gap chlor-alkali cells with oxygen-depolarized cathodes offer very significant energy savings versus the conventional membrane cells equipped with hydrogen-evolving cathodes. The documented savings reach as much as 38% at 0.31 A/cm2. • The zero-gap cells of our design with the modified anode structure generate caustic soda with current efficiency above 96% at standard industrial current densities (≤ 0.4 A/cm2). These current efficiencies match or exceed the current norms for conventional industrial membrane cells. At higher current densities (0.6 A/cm2 ≤ j ≤ 1.0 A/cm2), the current efficiency exceeds 90-94% depending on current density. • Due to the smaller water consumption by the oxygen reduction process as compared to the hydrogen evolution reaction, oxygen-depolarized cathodes offer - 95 - more favorable conditions for membrane operation and allow higher current density operation with a smaller risk of ohmic overheating of the membrane. • The magnitude of the effects of oxygen humidification on the cell voltage and caustic current efficiency is predominantly determined by the relative amounts of water introduced to the cathode compartment with oxygen and through the membrane from the anode compartment, as well as by the cell design. The optimum level of humidification, which guarantees the best overall cell performance, depends on the cell design, the membrane used and the operating conditions, e.g., the current density. • Of the chlor-alkali membranes tested, membrane 4 offers the highest energy efficiency at the highest current densities (0.8-1.0 A/cm2), membrane 3 is the second best, while membrane 1 performs noticeably poorer. At lower current densities (0.2-0.6 A/cm2), the performance of these three membranes is very comparable. The older chlor-alkali membrane (membrane 2), performs significantly worse than the remaining membranes at all current densities. • The unwanted byproduct of the cathode reaction, peroxide, is predominantly formed as a result of the oxygen reduction on carbon particles present in both the catalyst and the gas diffusion layer of the standard ELAT® electrode. Consequently, peroxide is not eliminated in the cells equipped with the membrane-electrode-assembly (MEA) containing unsupported catalyst and the separate gas diffusion layer containing carbon particles. • Loss of hydrophobicity of carbon particles is responsible for the increase of peroxide generation with the electrolysis time • Low water activity at the reaction site promotes formation of peroxide. • Elimination of peroxide from the caustic product stream was possible by modifying the cathode structure. However, the modified structure was more susceptible to flooding. • Cathode hardware coating affects peroxide generation rate. The gold plated hardware destroys peroxide more efficiently at high current densities (0.8-1.0 A/cm2), whereas the silver plated hardware helps destroy peroxide at low current densities (0.2-0.4 A/cm2). • Silver plated cathode hardware exhibits an excellent corrosion resistance both under open circuit conditions and during electrolysis. • The products of sodium chloride oxidation, i.e., elemental chlorine, chlorate, hypochlorite, etc., contribute to the cathode hardware corrosion during uncontrolled interruptions of electrolysis and are also likely to cause catalyst loss from electrodes utilizing carbon supported platinum catalyst. • Membrane-electrode-assemblies (MEAs) containing unsupported platinum catalyst offer excellent performance stability and exhibit excellent corrosion resistance during uncontrolled power outages, when the membrane remains in contact with concentrated brine. • The degree of catalyst utilization in the MEA is significantly lower than in the cathode utilizing the carbon-supported catalyst with the same nominal loading. • Prolonged contact of the MEA with water leads to significant delamination of the catalyst layer and the membrane. The increased and uneven swelling of the membrane and the Nafion binder is suspected to be a major factor responsible for the delamination. • Extended power outages result in significant lowering of caustic current efficiency, especially when the cell is equipped with the hydrophilic spacer. The origin of this phenomenon is unclear. Every new laboratory-scale technology requires a scale-up and eventually testing in a pilot plant, before it may be implemented. The conclusions summarized in this report pertain to small laboratory-scale cells and may turn out to be incorrect in the case of industrial-scale cells, where the electrode surface areas are at least two orders of magnitude larger. For instance, the current distribution over the membrane surface is uniform in the small cell, whereas it may be non-uniform in a larger cell because of the more demanding conditions of supplying the reagents to and removing the products from the reaction site in the latter case. Correcting the problem may require adjusting of the operating conditions, changing the hardware design or even installing additional hardware. Consequently, the scale-up stage will be an extremely important part of the research before the technology can be tested in a pilot plant and eventually implemented. However, an - 97 - ordinary research laboratory cannot handle currents in excess of 100 A, which limits the size of the test cell to twice the size of the cells used in our study. This size is probably too small to provide reliable evaluation of the scale-up effects on the cell operation. Consequently, the scale-up phase of the research must involve an industrial partner, who can perform the necessary testing of the scaled-up cells in their research facility. The industrial partner can also provide the high purity brine, which is hard to find on the market. Some manufacturers sell the brine, but in quantities that significantly exceed the needs of a research project. On the other hand, the quantity of brine used by our project, amounted to approximately 42 dm3 (11 gal) per day, when two cells were operated at 1.0 A/cm2. Such a quantity is too large to be processed in a research laboratory using a complex multi-step purification procedure, unless there is a separate purification system in place. However, the cost of the purification system and its maintenance may be prohibitively high given the volume of brine processed. The best possible combination of industrial partners should involve a chlor-alkali producer and a chlor-alkali cell manufacturer. The collaboration with a chlor-alkali manufacturer is more desired from the standpoint of ongoing research because of the extent of services they can provide to the research team on a daily basis, e.g., brine supply, sample analysis, consultation, etc. The collaboration with a chlor-alkali cell manufacturer can be particularly helpful, when the technology development approaches the commercialization stage. Acknowledgements Financial support from the Industrial Technologies Program of the DOE is gratefully acknowledged. Many thanks are due Sara Dillich and Mike Soboroff, DOE Program Managers and to Tom Baker and Melissa Miller, LANL Program Managers for their continuous support and interest in this research. Student researchers Sarah Stellingwerf and Amanda Casteel were involved part time in this research. Their help is gratefully acknowledged. The author is also grateful to the entire group MST-11 at LANL for their daily support and encouragement. Special thanks are due to former and current members of MST-11, Shimshon Gottesfeld, Ludwig Lipp, Mahlon Wilson, and Christine Zawodzinski for their creative contributions to previous research on oxygen-depolarized chlor-alkali cells and some ideas implemented in this research. Very special thanks are due to Lawrence “Bec” Becnel, Bill Wood, and Delton Kayga of Texas Brine Company for their truly invaluable help in obtaining high purity brine after cooperative negotiations with a potential industrial partner failed and the project was in danger of being discontinued. Supply of some materials used in these studies by William Meadowcroft and Robert Theobald of DuPont is greatly appreciated. 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J.Perez, E.R.Gonzalez, E.A.Ticianelli, Electrochim.Acta 44(1998)1329. 26) I.Morcos, E.Yeager, Electrochim.Acta 15(1970)953. 27) Z.W.Zhang, D.Tryk, E.Yeager, J.Electrochem.Soc. 130(1983)C333. 28) L.Nei, Fresenius J.Anal.Chem. 367(2000)436. 29) J.-D.Kim, S.-I.Pyun, T.-H.Yang, J.B.Ju, J.Electroanal.Chem. 383(1995)161. 30) H.-K.Lee, J.-P.Shim, M.-J.Shim, S.-W.Kim, J.-S.Lee, Mater.Chem.Phys. 45(1996)238. 31) M.Chatenet, L.Genies, M.Aurousseau, R.Durand, and F.Andolfatto, J.Appl.Electrochem., 32(2002)1131. 32) M.Chatenet, M.Aurousseau, R.Durand, and F.Andolfatto, J.Electrochem.Soc., 150(2003)D47. 33) O.Ichinose, M.Kawaguchi, and N.Furuya, J.Appl.Electrochem., 34(2004)55. 34) F.Aziz and G.A.Mirza, Talanta 11(1964)889. 35) M.Ardon, Oxygen. Elementary Forms and Hydrogen Peroxide, The Physical Inorganic Chemistry Series (edited by R.A.Plane and M.J.Sienko), (W.A.Benjamin, New York, Amsterdam, 1965). 36) P.S.D.Brito, C.A.C.Sequeira, J.Power Sources, 52(1994)1, and references therein. 37) J.T.Keating, H.M.B.Gerner, High Current Density Operation – The Behavior of Ion Exchange Membranes in Chloralkali Electrolyzers, Modern Chlor-Alkali Technology vol.7 (edited by S.Sealey, SCI 1998), Proceedings of the 1997 London International Chlorine Symposium Organized by SCI Electrochemical Technology Group, London, UK, 4th June – 6th June (1997) pp 135-144. 38) L.R.Czarnetzki, L.J.J.Janssen, J.L.Fernández,, J.Appl.Electrochem. 22(1992)315. 39) CRC Handbook of Chemistry and Physics, 81st Edition, CRC Press, 2000.
189121	https://artofproblemsolving.com/wiki/index.php/AoPS_Wiki:Competition_ratings?srsltid=AfmBOoraQXMrd2CzOEa3Syttc7rzbggQS_x6G84LFoP1lZu8x9HDCSYj	Art of Problem Solving AoPS Wiki:Competition ratings - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AoPS Wiki:Competition ratings Page Project pageDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AoPS Wiki:Competition ratings This page contains an approximate estimation of the difficulty level of various competitions. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience. Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. early AMC problems and 10 is hardest level, e.g. China IMO Team Selection Test. When considering problem difficulty, put more emphasis on problem-solving aspects and less so on technical skill requirements. Note that "easier" contests (e.g. MATHCOUNTS) have shorter time lengths, meaning being successful emphasizes speed. Contents 1 Scale 2 Competitions 2.1 Introductory Competitions 2.1.1 MOEMS 2.1.2 AMC 8 2.1.3 MATHCOUNTS 2.1.4 AMC 10 2.1.5 CEMC Multiple Choice Tests 2.1.6 CEMC Fryer/Galois/Hypatia 2.2 Problem Solving Books for Introductory Students 2.3 Intermediate Competitions 2.3.1 AMC 12 2.3.2 AIME 2.3.3 ARML 2.3.4 HMMT (November) 2.3.5 CEMC Euclid 2.3.6 COMC 2.3.7 Purple Comet 2.3.8 LMT 2.4 Problem Solving Books for Intermediate Students 2.5 Beginner Olympiad Competitions 2.5.1 USAMTS 2.5.2 Indonesia MO 2.5.3 CentroAmerican Olympiad 2.5.4 JBMO 2.6 Olympiad Competitions 2.6.1 USAJMO 2.6.2 HMMT (February) 2.6.3 Canadian MO 2.6.4 Iberoamerican Math Olympiad 2.6.5 APMO 2.6.6 Balkan MO 2.7 Hard Olympiad Competitions 2.7.1 USAMO 2.7.2 USA TST 2.7.3 Putnam 2.7.4 Korea Final Round 2.7.5 China TST (hardest problems) 2.7.6 IMO 2.7.7 IMO Shortlist Scale All levels are estimated and refer to averages. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO - IMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. Notes: Multiple choice tests like AMC are rated as though they are free-response. Test-takers can use the answer choices as hints, and so correctly answer more AMC questions than MATHCOUNTS or AIME problems of similar difficulty. Some Olympiads are taken in 2 sessions, with 2 similarly difficult sets of questions, numbered as one set. For these the first half of the test (questions 1-3) is similar difficulty to the second half (questions 4-6). Scale 1: Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, MATHCOUNTS School, AMC 8 1-10, AMC 10 1-10, easier AMC 12 1-5, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems. 1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests (AMC 8 11-20, harder AMC 10 1-10, AMC 12 1-5, and those others that force students to apply their school-level knowledge to slightly more challenging problems), traditional middle/high school word problems with more complex problem solving. 2: For motivated beginners, harder questions from the previous categories (AMC 8 21-25, MATHCOUNTS Chapter (Sprint 21-30, Target 6-8), MATHCOUNTS States/Nationals, AMC 10 11-15, AMC 12 5-10, easiest AIME 1-3) 2.5: More advanced beginner problems, hardest questions from previous categories (Harder AMC 8 21-25, harder MATHCOUNTS States questions, AMC 10 16-20, AMC 12 11-15, usual AIME 1-3) 3: Early intermediate problems that require more creative thinking (harder MATHCOUNTS National questions, AMC 10 21-25, AMC 12 15-20, hardest AIME 1-3, usual AIME 4-6). 4: Intermediate-level problems (AMC 12 21-25, hardest AIME 4-6, usual AIME 7-10). 5: More difficult AIME problems (11-13), simple proof-based Olympiad-style problems (early JBMO questions, easiest USAJMO 1/4). 6: High-leveled AIME-styled questions (14/15). Introductory-leveled Olympiad-level questions (harder USAJMO 1/4 and easier USAJMO 2/5, easier USAMO and IMO 1/4). 7: Tougher Olympiad-level questions, may require more technical knowledge (harder USAJMO 2/5 and most USAJMO 3/6, extremely hard USAMO and IMO 1/4, easy-medium USAMO and IMO 2/5). 8: High-level Olympiad-level questions (medium-hard USAMO and IMO 2/5, easiest USAMO and IMO 3/6). 9: Expert Olympiad-level questions (average USAMO and IMO 3/6). 9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve (hard USAMO and IMO 3/6). 10: Historically hard problems, generally unsuitable for very hard competitions (such as the IMO) due to being exceedingly tedious, long, and difficult (e.g. very few students are capable of solving on a worldwide basis). Examples For reference, here are some sample problems from each of the difficulty levels 1-10: <1: Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? (2003 AMC 8, Problem 1) 1: How many integer values of satisfy ? (2021 Spring AMC 10B, Problem 1) 1.5: A number is called flippy if its digits alternate between two distinct digits. For example, and are flippy, but and are not. How many five-digit flippy numbers are divisible by (2020 AMC 8, Problem 19) 2: A fair -sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number? (2021 Spring AMC 10B, Problem 18) 2.5:, , are three piles of rocks. The mean weight of the rocks in is pounds, the mean weight of the rocks in is pounds, the mean weight of the rocks in the combined piles and is pounds, and the mean weight of the rocks in the combined piles and is pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles and ? (2013 AMC 12A, Problem 16) 3: Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ? (2018 AMC 10A, Problem 24) 3.5: Find the number of integer values of in the closed interval for which the equation has exactly one real solution. (2017 AIME II, Problem 7) 4: Define a sequence recursively by andfor all nonnegative integers Let be the least positive integer such thatIn which of the following intervals does lie? ![Image 57: $\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } 729,\infty)$ (2019 AMC 10B, Problem 24 and 2019 AMC 12B, Problem 22) 4.5: Find, with proof, all positive integers for which is a perfect square. (USAJMO 2011/1) 5: Find all triples of real numbers such that the following system holds: (JBMO 2020/1) 5.5: Triangle has , , , and . Let , , and be the orthocenter, incenter, and circumcenter of , respectively. Assume that the area of pentagon is the maximum possible. What is ? (2011 AMC 12A, Problem 25) 6: Let be an acute triangle with circumcircle and let be the intersection of the altitudes of Suppose the tangent to the circumcircle of at intersects at points and with and The area of can be written in the form where and are positive integers, and is not divisible by the square of any prime. Find (2020 AIME I, Problem 15) 6.5: Rectangles and are erected outside an acute triangle Suppose thatProve that lines and are concurrent. (USAMO 2021/1, USAJMO 2021/2) 7: We say that a finite set in the plane is balanced if, for any two different points , in , there is a point in such that . We say that is centre-free if for any three points , , in , there is no point in such that . Show that for all integers , there exists a balanced set consisting of points. Determine all integers for which there exists a balanced centre-free set consisting of points. (IMO 2015/1) 7.5: Let be the set of integers. Find all functions such thatfor all with . (USAMO 2014/2) 8: For each positive integer , the Bank of Cape Town issues coins of denomination . Given a finite collection of such coins (of not necessarily different denominations) with total value at most most , prove that it is possible to split this collection into or fewer groups, such that each group has total value at most . (IMO 2014/5) 8.5: Let be the incentre of acute triangle with . The incircle of is tangent to sides , and at and , respectively. The line through perpendicular to meets at . Line meets again at . The circumcircles of triangle and meet again at . Prove that lines and meet on the line through perpendicular to . (IMO 2019/6) 9: Let be a positive integer and let be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form for some positive integer . (IMO 2022/3) 9.5: An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from to . Does there exist an anti-Pascal triangle with rows which contains every integer from to ? (IMO 2018/3) 10: Prove that there exists a positive constant such that the following statement is true: Consider an integer , and a set of points in the plane such that the distance between any two different points in is at least 1. It follows that there is a line separating such that the distance from any point of to is at least . (A line separates a set of points S if some segment joining two points in crosses .) (IMO 2020/6) Competitions Introductory Competitions Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available. MOEMS Division E: 1The whole number is divisible by . leaves a remainder of when divided by or . What is the smallest value that can be? (Solution) Division M: 1.5The value of a two-digit number is times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number. (Solution) AMC 8 Problem 1 - Problem 12: 1-1.5The coordinates of are , , and , with . The area of is 12. What is the value of ? (Solution) Problem 13 - Problem 25: 1.25-2.5How many four-digit numbers have all three of the following properties? (I) The tens and ones digit are both 9. (II) The number is 1 less than a perfect square. (III) The number is the product of exactly two prime numbers. (Solution) MATHCOUNTS Countdown: 0.5-1.5. Sprint: 1-1.5 (School) 1-2(Chapter), 1.5-3 (State), 2-3.5 (National) Target: 1-2 (School/Chapter), 1.5-2 (State), 2-2.5 (National) Team: 1-1.5 (School)1-2(Chapter), 1-3 (State), 1.5-3.5 (National) AMC 10 Since ~2020, AMC 10 shares about half its questions with AMC 12, but places them 0-3 spots later in the test. Problem 1 - 10: 1-1.5A rectangular box has integer side lengths in the ratio . Which of the following could be the volume of the box? (Solution) Problem 11 - 20: 1.5-2.5For some positive integer , the repeating base- representation of the (base-ten) fraction is . What is ? (Solution) Problem 21 - 25: 3-4.5The vertices of an equilateral triangle lie on the hyperbola , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle? (Solution) Isosceles trapezoid ABCD has parallel sides and , with and . There is a point P on the plane such that ,,, and . What is ? (Solution) CEMC Multiple Choice Tests This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests. Part A: 0.5-1.5How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number? (2015 Gauss 7 Problem 10) Part B: 1-2Two lines with slopes and intersect at . What is the area of the triangle formed by these two lines and the vertical line ? (2017 Cayley Problem 19) Part C (Gauss/Pascal): 2-2.5Suppose that , where , , and are positive integers with in lowest terms. What is the sum of the digits of the smallest positive integer for which is a multiple of 1004? (2014 Pascal Problem 25) Part C (Cayley/Fermat): 2.5-3Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is ﬁnished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets? (2018 Fermat Problem 24) CEMC Fryer/Galois/Hypatia Problem 1-2: 1-2 Problem 3-4 (early parts): 1.5-2.5 Problem 3-4 (later parts): 3-5 Problem Solving Books for Introductory Students Remark: There are many other problem books for Introductory Students that are not published by AoPS. Typically the rating on the left side is equivalent to the difficulty of the easiest review problems and the difficulty on the right side is the difficulty of the hardest challenge problems. The difficulty may vary greatly between sections of a book. Prealgebra by AoPS0.5-1.5 Introduction to Algebra by AoPS0.5-3 Introduction to Counting and Probability by AoPS0.5-3 Introduction to Number Theory by AoPS0.5-3 Introduction to Geometry by AoPS0.5-4.5 105 Algebra by Awesome Math 1.5-5 106 Geometry by Awesome Math 1-6 112 Combinatorial by Awesome Math 1-5 111 Algebra and Number Theory by Awesome Math 1-6 Intermediate Competitions This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [Category:Intermediate mathematic competitions\|here]. AMC 12 Problem 1-10: 1.5-2What is the value of (Solution) Problem 11-20: 2.5-3.5An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point? (Solution) Problem 21-25 (Easier): 3-4Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ? (Solution) Problem 21-25 (Harder): 4.5-6Semicircle has diameter of length . Circle lies tangent to at a point and intersects at points and . If and then the area of equals where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. What is ? (Solution) AIME Problem 1 - 5: 3Consider the integer Find the sum of the digits of . (Solution) Problem 6 - 9: 4An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly moves that ant is at a vertex of the top face on the cube is , where and are relatively prime positive integers. Find (Solution) Problem 10 - 12: 5Triangle has side lengths and Circle passes through and is tangent to line at Circle passes through and is tangent to line at Let be the intersection of circles and not equal to Then where and are relatively prime positive integers. Find (Solution) Problem 13 - 15: 6Let Let be the distinct zeros of and let for where and and are real numbers. Let where and are integers and is not divisible by the square of any prime. Find . (Solution) ARML Individuals, Problem 1: 2 Individuals, Problems 2, 3, 4, 5, 7, and 9: 3 Individuals, Problems 6 and 8: 4 Individuals, Problem 10: 5.5 Team/power, Problem 1-5: 3.5 Team/power, Problem 6-10: 5 HMMT (November) Individual Round, Problem 6-8: 4 Individual Round, Problem 10: 4.5 Team Round: 4-5 Guts: 3.5-5.25 CEMC Euclid Problem 1-6: 1-3 Problem 7-10: 3-6 COMC Part A: 1-2.5 Part B: 2.5-4 Part C: 2-5 Purple Comet Problems 1-10 (MS): 1.5-3 Problems 11-17 (MS): 3-4.5 Problems 18-20 (MS): 4-4.75 Problems 1-10 (HS): 1.5-3.5 Problems 11-20 (HS): 3.5-4.75 Problems 21-30 (HS): 4.5-6 LMT Easy Problems: 0.5Let trapezoid be such that . Additionally, , , and . Find . Medium Problems: 2-4Let have side lengths , , and . Let the angle bisector of meet the circumcircle of at a point . Determine the area of . Hard Problems: 5-7A magic board can toggle its cells between black and white. Define a pattern to be an assignment of black or white to each of the board’s cells (so there are patterns total). Every day after Day 1, at the beginning of the day, the board gets bored with its black-white pattern and makes a new one. However, the board always wants to be unique and will die if any two of its patterns are less than cells different from each other. Furthermore, the board dies if it becomes all white. If the board begins with all cells black on Day 1, compute the maximum number of days it can stay alive. Problem Solving Books for Intermediate Students Remark: As stated above, there are many books for Intermediate students that have not been published by AoPS. Below is a list of intermediate books that AoPS has published and their difficulty. The left-hand number corresponds to the difficulty of the easier review problems, while the right-hand number corresponds to the difficulty of the hardest challenge problems. Intermediate Algebra by AoPS1-6, may vary across chapters Intermediate Counting & Probability by AoPS1-6, may vary across chapters Precalculus by AoPS2-7, may vary across chapters Calculus by AoPS3-8 (not an olympiad book) 108 Algebra by Awesome Math 2.5-8 107 Geometry by Awesome Math 2-8 Beginner Olympiad Competitions This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available. USAMTS USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale: Problem 1-2: 3-4Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter. (Solution) Problem 3-5: 4-6Call a positive real number groovy if it can be written in the form for some positive integer . Show that if is groovy, then for any positive integer , the number is groovy as well. (Solution) Indonesia MO Problem 1/5: 3.5In a drawer, there are at most balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is . Determine the maximum amount of white balls in the drawer, such that the probability statement is true? (Solution) Problem 2/6: 4.5Find the lowest possible values from the function for any real numbers . (Solution) Problem 3/7: 5A pair of integers is called good if Given 2 positive integers which are relatively prime, prove that there exists a good pair with and , but and . (Solution) Problem 4/8: 6Given an acute triangle . The incircle of triangle touches respectively at . The angle bisector of cuts and respectively at and . Suppose is one of the altitudes of triangle , and be the midpoint of . (a) Prove that and are perpendicular with the angle bisector of . (b) Show that is a cyclic quadrilateral. (Solution) CentroAmerican Olympiad Problem 1,4: 3.5Find all three-digit numbers (with ) such that is a divisor of 26. (Solution) Problem 2,5: 4.5Show that the equation has no integer solutions. (Solution) Problem 3/6: 6Let be a convex quadrilateral. , and , , and are points on , , and respectively, such that . If , , show that . (Solution) JBMO Problem 1: 4Find all real numbers such that Problem 2: 4.5-5Let be a convex quadrilateral with , and . The diagonals intersect at point . Determine the measure of . Problem 3: 5Find all prime numbers , such that . (Solution: -paixiao Problem 4: 6A table is divided into white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly moves all the cells were black. Find all possible values of . Olympiad Competitions This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 5 to 8. A full list is available. USAJMO Problem 1/4: 5There are bowls arranged in a row, numbered through , where and are given positive integers. Initially, each of the first bowls contains an apple, and each of the last bowls contains a pear. A legal move consists of moving an apple from bowl to bowl and a pear from bowl to bowl , provided that the difference is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first bowls each containing a pear and the last bowls each containing an apple. Show that this is possible if and only if the product is even. (Solution) Problem 2/5: 6-6.5Let be positive real numbers such that . Prove that (Solution) Problem 3/6: 7Two rational numbers and are written on a blackboard, where and are relatively prime positive integers. At any point, Evan may pick two of the numbers and written on the board and write either their arithmetic mean or their harmonic mean on the board as well. Find all pairs such that Evan can write on the board in finitely many steps. (Solution) HMMT (February) Individual Round, Problem 1-5: 5 Individual Round, Problem 6-10: 5.5-6 Team Round: 7.5 HMIC: 8 Canadian MO Problem 1: 5.5 Problem 2: 6 Problem 3: 6.5 Problem 4: 7-7.5 Problem 5: 7.5-8 Iberoamerican Math Olympiad Problem 1/4: 5.5 Problem 2/5: 6.5 Problem 3/6: 7.5 APMO Problem 1: 6.5-7 Problem 2: 7 Problem 3: 7.5 Problem 4: 7.5-8 Problem 5: 8.5-9 Balkan MO Problem 1: 5Solve the equation in positive integers. Problem 2: 6.5Let be a line parallel to the side of a triangle , with on the side and on the side . The lines and meet at point . The circumcircles of triangles and meet at two distinct points and . Prove that . Problem 3: 7.5A rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres in such way that the following to conditions are both fulfilled the distances are all equal to the closed broken line has a centre of symmetry? Problem 4: 8Denote by the set of all positive integers. Find all functions such that Hard Olympiad Competitions This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available. USAMO Problem 1/4: 6-7Let be a convex polygon with sides, . Any set of diagonals of that do not intersect in the interior of the polygon determine a triangulation of into triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of . (Solution) Problem 2/5: 7-8Three nonnegative real numbers , , are written on a blackboard. These numbers have the property that there exist integers , , , not all zero, satisfying . We are permitted to perform the following operation: find two numbers , on the blackboard with , then erase and write in its place. Prove that after a finite number of such operations, we can end up with at least one on the blackboard. (Solution) Problem 3/6: 8-9Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree with real coefficients is the average of two monic polynomials of degree with real roots. (Solution) USA TST Problem 1/4/7: 6.5-7 Problem 2/5/8: 7.5-8 Problem 3/6/9: 8.5-9 Putnam Problem A/B,1-2: 7Find the least possible area of a concave set in the 7-D plane that intersects both branches of the hyperparabola and both branches of the hyperbola (A set in the plane is called convex if for any two points in the line segment connecting them is contained in ) (Solution) Problem A/B,3-4: 8Let be an matrix all of whose entries are and whose rows are mutually orthogonal. Suppose has an submatrix whose entries are all Show that . (Solution) Problem A/B,5-6: 9For any , define the set . Show that there are no three positive reals such that . (Solution) Korea Final Round Problem 1/4: 7 Problem 2/5: 7.5-8 Problem 3/6: 8-9 China TST (hardest problems) Problem 1/4: 8-8.5Given an integer prove that there exist odd integers and a positive integer such that Problem 2/5: 9Given a positive integer and real numbers such that prove that for any positive real number Problem 3/6: 9.5-10Let be an integer and let be non-negative real numbers. Define for . Prove that IMO Problem 1/4: 6-7Let be the circumcircle of acute triangle . Points and are on segments and respectively such that . The perpendicular bisectors of and intersect minor arcs and of at points and respectively. Prove that lines and are either parallel or they are the same line. (Solution) Problem 2/5: 7-8Let be a polynomial of degree with integer coefficients, and let be a positive integer. Consider the polynomial , where occurs times. Prove that there are at most integers such that . (Solution) Problem 3/6: 9-10 Let be an equilateral triangle. Let be interior points of such that , , , andLet and meet at let and meet at and let and meet at Prove that if triangle is scalene, then the three circumcircles of triangles and all pass through two common points. (Note: a scalene triangle is one where no two sides have equal length.) (..this is one of the hardest problems and we had to define a scalene triangle.) IMO Shortlist Problem 1-2: 6-6.5 Problem 3-4: 6.5-7.5 Problem 5-6: 8-9 Problem 7+: 8.5-10 This article is a stub. Help us out by expanding it. 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189122	http://hobart.k12.in.us/jkousen/Biology/psquare.htm	Mr. Kousen is Water Man Baby Steps Through the PUNNETT SQUARE (Get it? "Square" = nerd. Ha ha ha ha ha ...) No this page is not a place to pick on those students who you will one day call "boss". This is a place for some serious practice with a very useful tool for completing genetics problems, the Punnett Square (P-Square for short). The basic naked p-square looks like a window pane : Aaaah, reminds me of my bedroom window when I was a kid. The very window I would peer out on snowy mornings and hope to myself: "snow day, snow day, snow day". But I digress.... When given enough info about two parent organisms, we can use this window pane to predict the genotypes & phenotypes of their offspring. ExcIting, ain't it? Very quick rehash (review): genotype = the genes of an organism; for one specific trait we use two letters to represent the genotype. A capital letter represents the dominant form of a gene (allele), and a lowercase letter is the abbreviation for the recessive form of the gene (allele). phenotype = the physical appearance of a trait in an organism For example, let's say that for the red-thoated booby bird (I am making this up), red throat is the dominant trait and white throat is recessive. Since the "red-throat code" and the" white-throat code" are alleles (two forms of the same gene), we abbreviate them with two forms of the same letter. So we use "R" for the dominant allele/trait (red throat) and "r" for the recessive allele/trait (white throat). Our possible genotypes & phenotypes would be like so: \| \| \| \| --- \| Symbol \| Genotype Name \| Phenotype \| \| RR \| homozygous (pure) dominant \| red thoat \| \| Rr \| heterozygous (hybrid) \| red throat \| \| rr \| homozygous (pure) recessive \| white throat \| Note: Remember, we don't use "R" for red & "W" for white because that would make it two different genes which would code for two different traits, and throat color is one trait. What the genotype contains are two codes for the same trait, so we use two forms of the same letter (capital & lowercase). One more note: A very very helpful thing to memeorize is that the ONLY way for a recessive trait to show up in an organism is if that organism's genotype is homozygous recessive (two little letters, like "rr"). Here are the basic steps to using a Punnett Square when solving a genetics question. After you get good at this you should never miss a genetic question involving the cross of two organisms. \| \| \| BABY STEPS: 1. determine the genotypes of the parent organisms 2. write down your "cross" (mating) 3. draw a p-square 4. "split" the letters of the genotype for each parent & put them "outside" the p-square 5. determine the possible genotypes of the offspring by filling in the p-square 6. summarize results (genotypes & phenotypes of offspring) 7. bask in the glow of your accomplishment ! \| Step #1: Determine the genotypes of the parent organisms. Sometimes this already done in the question for you. If the question says "Cross two organims with the following genotype: Tt & tt", it's all right there in the question already. More likely is a question like this: "Cross a short pea plant with one that is heterozygous for tallness". Here, you have to use your understanding of the vocab to figure out what letters to use in the genotypes of the parents. Heterozygous always means one of each letter, so we'd use "Tt" (where "T" = tall, & "t" = short). The only way for a pea plant to be short is when it has 2 lowercase "t's", so that short parent is "tt". So the cross ends-up the same as in my first example: Tt x tt. Now, we (us mean teachers) can make things just a little more tricky. Let's use hamsters in this example. Brown is dominant (B), and white is recessive (b). What if a question read like this: "Predict the offspring from the cross of a white hamster and a brown hamster if the brown hamster's mother was white". Oooooh, is this a toughy? First things first: the only way for the white hamster to be white (the recessive trait) is if it's genotype is homozygous recessive (2 little letters), so the white hamster is "bb". Now, the brown hamster's genotype could be either "BB" or "Bb". If its mommy was white (bb), then this brown hamster MUST have inherited a little "b" from its mommy. So the brown one in our cross is "Bb" (not "BB"), and our hamster cross is: Bb x bb. Step #2: Write down your "cross" (mating). Write the genotypes of the parents in the form of letters (ex: Tt x tt). Step #3: Draw a p-square. Step #4: "Split" the letters of the genotype for each parent & put them "outside" the p-square. For an example cross we'll use these parental genotypes: Tt x tt. Take the genotype letters of one parent, split them and put them on the left, outside the rows of the p-square. What we've done is taken the hetrozygous tall plant (Tt) and put its big "T" out in front of the top row, and the little "t" out in front of the bottom row. When we fill-in the p-square, we will copy these "tees" into each of the empty boxes to their right. So the big "T" will be in each of the boxes of the top row, and the lowercase "t" will be in the two boxes of the bottom row. Isn't this exciting? Now take the two letters of the second parent's genotype, split 'em up, and place them above each of the two columns of the p-square. Now, when it comes time to filling things in, those lowercase "t's" will each be copied into the two boxes directly below them. So after the next step, each little box will have two letters in it (one "tee" from the left & one "tee" from the top). These new 2 letter combos represent possible genotypes of the offspring. Exciting, ain't it? Step #5: Determine the possible genotypes of the offspring by filling in the p-square. I kinda gave this away already, but to "determine the genotypes of the offspring" all we gotta do is fill-in the the boxes of the p-square. Again we do this be taking a letter from the left & matching it with a letter from the top. Like so: \| \| \| \| \| --- --- \| \| Filling in the top-left box: \| Filling in the bottom-left box: \| Filling in the top-right box: \| Filling in the bottom-right box: \| One from the left, one from the top... one from the left, one from the top...one from the left, one from the top...one from the left, one from the top. Step #6: Summarize the results (genotypes & phenotypes of offspring). Simply report what you came up with. You should always have two letters in each of the four boxes. In this example, where our parent pea plants were Tt (tall) x tt (short), we get 2 of our 4 boxes with "Tt", and 2 of our 4 with "tt". The offspring that are "Tt" would end up with tall stems (the dominant trait) and the "tt" pea plants would have short stems (the recessive trait). So our summary would be something like this: \| \| \| --- \| \| Parent Pea Plants ("P" Generation) \| Offspring ("F1" Generation) \| \| Genotypes: Tt x tt \| Phenotypes: tall x short \| Genotypes: 50% (2/4) Tt 50% (2/4) tt \| Phenotypes: 50% tall 50% short \| Step #7: Bask in the glow of your accomplishment ! We are so good I can't stand it. We are genetics MONSTERS ! A little scientific side-note: You know how, in Step #4, when we "split" the letters of the genotype & put them outside the p-square? What that step illustrates is the process of gametogenesis (the production of sex cells, egg & sperm). Gametogenesis is a cell division thing (also called meiosis) that divides an organism's chromosome number in half. For example, in humans, body cells have 46 chromosomes a piece. However, when sperm or eggs are produced (by gametogenesis/meiosis) they get only 23 chromosomes each. This makes sense (believe it or not), because now, when the sperm & egg fuse at fertilization, the new cell formed (called a zygote) will have 23 + 23 = 46 chromosomes. Cool, huh? So, when the chromosome number is split in half, all of the two letter genotypes for every trait of that person (or organism) get separated. Which is why we do what we do in Step #4. TAKE WHAT YOU'VE LEARNED & DAZZLE SOME PEOPLE. \| \| \| <Back to that Mendel Guy & his Laws \| \| On to a Punnett Square Practice Page> \| Back to Biology Topics Outline IF YOU HAVE COMMENTS (GOOD OR BAD) ABOUT THIS OR ANY OF MY BIOLOGY PAGES, OR ANYTHING ELSE IN GENERAL, DROP ME A NOTE
189123	https://byjus.com/maths/square-root-of-36/	Square root of 36 is a natural number that results in the original number when multiplied by itself. Thus, the square root of any number is equal to the number that can be squared to get the actual number. Since 36 is a perfect square, therefore, it becomes easy for us to find its square root. The square root of 36 is expressed by √36, where ‘√’ is the radical symbol and 36 is the radicand. √36 is equal to 6, which is a rational number. In this article, we can find the square root of the number 36 with the help of different methods. \| \| \| Square root of 36 = √36 = ±6 In Exponent Form, (36)½ = ±6 \| Also check: Find Square Root Of Decimals Square Root 1 to 100 Square Root Of A Number By Repeated Subtraction Square Root Tricks How to Find the Square root of 36? Square root of 36 is a rational number. Suppose the root of 36 is x, then we can represent it as x/1. Hence, x is the numerator and 1 is the denominator. There are three methods to find the square root of 36, they are: Prime Factorisation Method Division Method Repeated Subtraction Method Prime Factorisation Method Prime factorisation method is the simplest method to find the prime factors of a number. After that, we can pair the factors that can be denoted as n2, where n is the prime factor. Hence, after taking the square root of the equation on both sides, we can cancel the root with a square. Let us work on it. The prime factorisation of 36 is: 36 = 2 x 2 x 3 x 3 The exponential form will be: 36 = 22 x 32 Or 36 = 22 x 32 [Using laws of exponents: am.an = am+n] Taking square root on both sides of the equation, we get; Now, we can cancel each of the square roots on the right-hand side of the equation with the square of each term. Therefore, we get; √36 = 2 x 3 √36 = 6 Therefore, the square root of 36 is 6. \| \| \| Facts: Square root of 36 is a natural or a whole number 36 and it’s square root; both are even numbers 36 is a composite number 36 is a perfect square 36 is evenly divisible by its square root value. Thus, 6 is the factor of 36 The roots of 36 are +6 or -6 \| Division Method We can easily find the square root of a number, using the division method. We need to find such a number that can be multiplied by itself to get 36. Since 36 is itself a smaller two-digit number. Therefore, the required square root value will be a single-digit number. Let us find here the square root of 36 using the division method. Therefore, the square root of 36 is equal to 6. Repeated Subtraction Method In the repeated subtraction method, we start subtracting 36 with the successive odd numbers in every step till we get the final value equal to 0. The number of times subtraction is done is equal to the square root of the original number. 36 – 1 = 35 35 – 3 = 32 32 – 5 = 27 27 – 7 = 20 20 – 9 = 11 11 – 11 = 0 Thus, the number of times subtraction is done is equal to 6. Hence, the required square root is 6. Square Roots of Numbers Square root of 4 = 2 Square root of 16 = 4 Square Root of 25 = 5 Square root of 49 = 7 Square root of 81 = 9 Square root of 64 = 8 Square root of 32 = 5.65685424949 Square root of 24 = 4.89897948557 Square root of 100 = 10 Video Lessons on Square Roots Visualising square roots 8,518 Finding Square roots 58,356 Solved Examples Q.1: What is the value of 10 multiplied by √36? Solution: The value of 10 multiplied by √36 is: ⇒ 10 x √36 ⇒ 10 x 6 ⇒ 60 Hence, the answer is 60. Q.2: Find the area of a square whose side is equal to 6 cm each. Solution: Given, the side of the square is 6cm. As we know, Area of the square = side2 Area = 62 Area = 36 sq.cm. Q.3: If the area of a square-shaped tile is 36 units, then find the perimeter of the tile. Solution: Given, Area of the tile = 36 units Therefore, Side length of the tile = √area = √36 = 6 unit Therefore, the perimeter of the tile is; Perimeter = 4side = 4 x 6 = 24 units. \| \| \| Practice Questions 1. √36 + √16 + √49 = ? 2. √36 divided by 6 is equal to _______. 3. Is the value of √36 evenly divisible by 10? 4. What should be subtracted from 9 to get the value of √36? 5. What should be added to 2 to get the value of √36? \| Register with us and download BYJU’S – The Learning App to learn more about squares and square roots with the help of interactive videos. Frequently Asked Questions on Square root of 36 Q1 What is the square root of 36? The square root of 36 is equal to 6 √36 = 6. Q2 Is 36 a perfect square? Yes, 36 is a perfect square because the square of the natural number 6, results in the actual number. 62 = 36 Q3 What kind of number is the square root of 36? Square root of 36 is a rational number because it can be expressed as a ratio or p/q. √36 = 6 = 6/1 Where 6 is the numerator and 1 is the denominator. Q4 Which is the shortcut method to find the square root of 36? 36 is a two-digit number. Therefore, we can easily realise that any square of any single digit number will result in 36. Since, at the unit place, we have 6, it is clear that when 6 is multiplied by 6, only we can have 6 at the unit place. Also, if we simply remember the table of 6, we can get 6 x 6 = 36. Therefore, the square root of 36 will be 6. Q5 Why is the square root of 36 equal to 6? The square root of any number is the root of the number that is squared to get the original number. Hence, the root of 36 is 6 because when we square the number 6, we get the actual number. Test your Knowledge on Square Root Of 36 Q5 Put your understanding of this concept to test by answering a few MCQs. Click Start Quiz to begin! 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189124	https://bio.libretexts.org/Bookshelves/Ecology/Environmental_Biology_(Fisher)/01%3A_Environmental_Science/1.02%3A_The_Process_of_Science	1.2: The Process of Science - Biology LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 1: Environmental Science Environmental Biology (Fisher) { } { "1.01:The_Earth_Humans__the_Environment" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_The_Process_of_Science" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Environment_and_Sustainability" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.04:_Environmental_Ethics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.05:_Environmental_Justice_and_Indigenous_Struggles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.0S:_1.S:__Environmental_Science(Summary)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Environmental_Science" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Matter_Energy__Life" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Ecosystems_and_the_Biosphere" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Community__Population_Ecology" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Conservation__Biodiversity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Environmental_Hazards__Human_Health" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Water_Availability_and_Use" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Food__Hunger" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Conventional__Sustainable_Agriculture" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Air_Pollution_Climate_Change__Ozone_Depletion" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Conventional__Sustainable_Energy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Appendix" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 03 Jan 2021 22:34:37 GMT 1.2: The Process of Science 14646 14646 Delmar Larsen { } Anonymous Anonymous 2 false false [ "article:topic", "showtoc:no", "license:ccby", "authorname:mfisher", "licenseversion:40", "program:openoregon", "source@ ] [ "article:topic", "showtoc:no", "license:ccby", "authorname:mfisher", "licenseversion:40", "program:openoregon", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Ecology 4. Environmental Biology (Fisher) 5. 1: Environmental Science 6. 1.2: The Process of Science Expand/collapse global location Environmental Biology (Fisher) Front Matter 1: Environmental Science 2: Matter, Energy, & Life 3: Ecosystems and the Biosphere 4: Community and Population Ecology 5: Conservation & Biodiversity 6: Environmental Hazards & Human Health 7: Water Availability and Use 8: Food & Hunger 9: Conventional & Sustainable Agriculture 10: Air Pollution, Climate Change, & Ozone Depletion 11: Conventional & Sustainable Energy 12: Appendix Back Matter 1.2: The Process of Science Last updated Jan 3, 2021 Save as PDF 1.1: The Earth, Humans, & the Environment 1.3: Environment and Sustainability picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate Page ID 14646 Matthew R. Fisher Oregon Coast Community College via OpenOregon ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. The Nature of Science 1. Natural Sciences 2. Scientific Inquiry 3. Hypothesis Testing 4. Basic and Applied Science 5. Scientific Work is Transparent & Open to Critique Figure 1.2.1. (a) The cyanobacteria seen through a light microscope are some of Earth’s oldest life forms. These (b) stromatolites along the shores of Lake Thetis in Western Australia are ancient structures formed by the layering of cyanobacteria in shallow waters. (Credit a: modification of work by NASA; scale-bar data from Matt Russell; credit b: modification of work by Ruth Ellison) Like other natural sciences, environmental science is a science that gathers knowledge about the natural world. The methods of science include careful observation, record keeping, logical and mathematical reasoning, experimentation, and submitting conclusions to the scrutiny of others. Science also requires considerable imagination and creativity; a well-designed experiment is commonly described as elegant or beautiful. Science has considerable practical implications and some science is dedicated to practical applications, such as the prevention of disease (figure 2). Other science proceeds largely motivated by curiosity. Whatever its goal, there is no doubt that science has transformed human existence and will continue to do so. Figure 1.2.2. Biologists may choose to study Escherichia coli (E. coli), a bacterium that is a normal resident of our digestive tracts but which is also sometimes responsible for disease outbreaks. In this micrograph, the bacterium is visualized using a scanning electron microscope and digital colorization. (credit: Eric Erbe; digital colorization by Christopher Pooley, USDA-ARS) The Nature of Science Biology is a science, but what exactly is science? What does the study of biology share with other scientific disciplines? Science (from the Latin scientia, meaning “knowledge”) can be defined as a process of gaining knowledge about the natural world. Science is a very specific way of learning about the world. The history of the past 500 years demonstrates that science is a very powerful way of gaining knowledge about the world; it is largely responsible for the technological revolutions that have taken place during this time. There are areas of knowledge, however, that the methods of science cannot be applied to. These include such things as morality, aesthetics, or spirituality. Science cannot investigate these areas because they are outside the realm of material phenomena, the phenomena of matter and energy, and cannot be observed and measured. The scientific method is a method of research with defined steps that include experiments and careful observation. The steps of the scientific method will be examined in detail later, but one of the most important aspects of this method is the testing of hypotheses. A hypothesis is an proposed explanatory statement, for a given natural phenomenon, that can be tested. Hypotheses, or tentative explanations, are different than a scientific theory. A scientific theory is a widely-accepted, thoroughly tested and confirmed explanation for a set of observations or phenomena. Scientific theory is the foundation of scientific knowledge. In addition, in many scientific disciplines (less so in biology) there are scientific laws, often expressed in mathematical formulas, which describe how elements of nature will behave under certain specific conditions, but they do not offer explanations for why they occur. Natural Sciences What would you expect to see in a museum of natural sciences? Frogs? Plants? Dinosaur skeletons? Exhibits about how the brain functions? A planetarium? Gems and minerals? Or maybe all of the above? Science includes such diverse fields as astronomy, computer sciences, psychology,biology, and mathematics. However, those fields of science related to the physical world and its phenomena and processes are considered natural sciences and include the disciplines of physics, geology, biology, and chemistry. Environmental science is a cross-disciplinary natural science because it relies of the disciplines of chemistry, biology, and geology. Scientific Inquiry One thing is common to all forms of science: an ultimate goal to know. Curiosity and inquiry are the driving forces for the development of science. Scientists seek to understand the world and the way it operates. Two methods of logical thinking are used: inductive reasoning and deductive reasoning. Inductive reasoning is a form of logical thinking that uses related observations to arrive at a general conclusion. This type of reasoning is common in descriptive science. A life scientist such as a biologist makes observations and records them. These data can be qualitative (descriptive) or quantitative (consisting of numbers), and the raw data can be supplemented with drawings, pictures, photos, or videos. From many observations, the scientist can infer conclusions (inductions) based on evidence. Inductive reasoning involves formulating generalizations inferred from careful observation and the analysis of a large amount of data. Brain studies often work this way. Many brains are observed while people are doing a task. The part of the brain that lights up, indicating activity, is then demonstrated to be the part controlling the response to that task. Deductive reasoning or deduction is the type of logic used in hypothesis-based science. In deductive reasoning, the pattern of thinking moves in the opposite direction as compared to inductive reasoning. Deductive reasoning is a form of logical thinking that uses a general principle or law to forecast specific results. From those general principles, a scientist can extrapolate and predict the specific results that would be valid as long as the general principles are valid. For example, a prediction would be that if the climate is becoming warmer in a region, the distribution of plants and animals should change. Comparisons have been made between distributions in the past and the present, and the many changes that have been found are consistent with a warming climate. Finding the change in distribution is evidence that the climate change conclusion is a valid one. Both types of logical thinking are related to the two main pathways of scientific study: descriptive science and hypothesis-based science. Descriptive (or discovery) science aims to observe, explore, and discover, while hypothesis-based science begins with a specific question or problem and a potential answer or solution that can be tested. The boundary between these two forms of study is often blurred, because most scientific endeavors combine both approaches. Observations lead to questions, questions lead to forming a hypothesis as a possible answer to those questions, and then the hypothesis is tested. Thus, descriptive science and hypothesis-based science are in continuous dialogue. “Scientists have become the bearers of the torch of discovery in our quest for knowledge.”– Stephen Hawking and Leonard Mlodinov, in The Grand Design (2010), Bantam Books Hypothesis Testing Biologists study the living world by posing questions about it and seeking science-based responses. This approach is common to other sciences as well and is often referred to as the scientific method. The scientific method was used even in ancient times, but it was first documented by England’s Sir Francis Bacon (1561–1626) who set up inductive methods for scientific inquiry. The scientific method is not exclusively used by biologists but can be applied to almost anything as a logical problem-solving method. Figure 1.2.3. Sir Francis Bacon is credited with being the first to document the scientific method. The scientific process typically starts with an observation (often a problem to be solved) that leads to a question. Let’s think about a simple problem that starts with an observation and apply the scientific method to solve the problem. One Monday morning, a student arrives at class and quickly discovers that the classroom is too warm. That is an observation that also describes a problem: the classroom is too warm. The student then asks a question: “Why is the classroom so warm?” Recall that a hypothesis is a suggested explanation that can be tested. To solve a problem, several hypotheses may be proposed. For example, one hypothesis might be, “The classroom is warm because no one turned on the air conditioning.” But there could be other responses to the question, and therefore other hypotheses may be proposed. A second hypothesis might be, “The classroom is warm because there is a power failure, and so the air conditioning doesn’t work.” Once a hypothesis has been selected, a prediction may be made. A prediction is similar to a hypothesis but it typically has the format “If . . . then . . . .” For example, the prediction for the first hypothesis might be, “If the student turns on the air conditioning, then the classroom will no longer be too warm.” A hypothesis must be testable to ensure that it is valid. For example, a hypothesis that depends on what a bear thinks is not testable, because it can never be known what a bear thinks. It should also be falsifiable, meaning that it can be disproven by experimental results. An example of an unfalsifiable hypothesis is “Botticelli’s Birth of Venus is beautiful.” There is no experiment that might show this statement to be false. To test a hypothesis, a researcher will conduct one or more experiments designed to eliminate one or more of the hypotheses. This is important. A hypothesis can be disproven, or eliminated, but it can never be proven. Science does not deal in proofs like mathematics. If an experiment fails to disprove a hypothesis, then we find support for that explanation, but this is not to say that down the road a better explanation will not be found, or a more carefully designed experiment will be found to falsify the hypothesis. Each experiment will have one or more variables and one or more controls. Experimentalvariables are any part of the experiment that can vary or change during the experiment. Controlled variables are parts of the experiment that do not change. Lastly, experiments might have a control group: a group of test subjects that are as similar as possible to all other test subjects, with the exception that they don’t receive the experimental treatment (those that do receive it are known as the experimental group). For example, in a study testing a weight-loss drug, the control group would be test subjects that don’t receive the drug (but they might receive a placebo, such as sugar pill, instead). Look for these various things in the example that follows: An experiment might be conducted to test the hypothesis that phosphate (a nutrient) promotes the growth of algae in freshwater ponds. A series of artificial ponds are filled with water and half of them are treated by adding phosphate each week, while the other half are treated by adding a non-nutritional mineral that is not used by algae. The experimental variable here is presence/absence of a nutrient (phosphate). One potential controlled variable would be the volume of water in each tank. The amount of water that algae have access to may influence the results, thus researchers want to control its influence on the results by making sure all test subjects get the same amount. The control group consists of the tanks that received a placebo (non-nutritional mineral) instead of the phosphate. If the ponds with phosphate show more algal growth, then we have found support for the hypothesis. If they do not, then we reject our hypothesis. Be aware that rejecting one hypothesis does not determine whether or not the other hypotheses can be accepted; it simply eliminates one hypothesis that is not valid (Figure 1.2.3). Using the scientific method, the hypotheses that are inconsistent with experimental data are rejected. Figure 1.2.4. The scientific method is a series of defined steps that include experiments and careful observation. If a hypothesis is not supported by data, a new hypothesis can be proposed. In the example below, the scientific method is used to solve an everyday problem. Which part in the example below is the hypothesis? Which is the prediction? Based on the results of the experiment, is the hypothesis supported? If it is not supported, propose some alternative hypotheses. My toaster doesn’t toast my bread. Why doesn’t my toaster work? There is something wrong with the electrical outlet. If something is wrong with the outlet, my coffeemaker also won’t work when plugged into it. I plug my coffeemaker into the outlet. My coffeemaker works. In practice, the scientific method is not as rigid and structured as it might at first appear. Sometimes an experiment leads to conclusions that favor a change in approach; often, an experiment brings entirely new scientific questions to the puzzle. Many times, science does not operate in a linear fashion; instead, scientists continually draw inferences and make generalizations, finding patterns as their research proceeds. Scientific reasoning is more complex than the scientific method alone suggests. Basic and Applied Science Is it valuable to pursue science for the sake of simply gaining knowledge, or does scientific knowledge only have worth if we can apply it to solving a specific problem or bettering our lives? This question focuses on the differences between two types of science: basic science and applied science. Basic science or “pure” science seeks to expand knowledge regardless of the short-term application of that knowledge. It is not focused on developing a product or a service of immediate public or commercial value. The immediate goal of basic science is knowledge for knowledge’s sake, though this does not mean that in the end it may not result in an application. In contrast, applied science aims to use science to solve real-world problems, such as improving crop yield, find a cure for a particular disease, or save animals threatened by a natural disaster. In applied science, the problem is usually defined for the researcher. Some individuals may perceive applied science as “useful” and basic science as “useless.” A question these people might pose to a scientist advocating knowledge acquisition would be, “What for?” A careful look at the history of science, however, reveals that basic knowledge has resulted in many remarkable applications of great value. Many scientists think that a basic understanding of science is necessary before an application is developed; therefore, applied science relies on the results generated through basic science. Other scientists think that it is time to move on from basic science and instead to find solutions to actual problems. Both approaches are valid. It is true that there are problems that demand immediate attention; however, few solutions would be found without the help of the knowledge generated through basic science. One example of how basic and applied science can work together to solve practical problems occurred after the discovery of DNA structure led to an understanding of the molecular mechanisms governing DNA replication. Strands of DNA, unique in every human, are found in our cells, where they provide the instructions necessary for life. During DNA replication, new copies of DNA are made, shortly before a cell divides to form new cells. Understanding the mechanisms of DNA replication (through basic science) enabled scientists to develop laboratory techniques that are now used to identify genetic diseases, pinpoint individuals who were at a crime scene, and determine paternity (all examples of applied science). Without basic science, it is unlikely that applied science would exist. Another example of the link between basic and applied research is the Human Genome Project, a study in which each human chromosome was analyzed and mapped to determine the precise sequence of the DNA code and the exact location of each gene. (The gene is the basic unit of heredity; an individual’s complete collection of genes is his or her genome.) Other organisms have also been studied as part of this project to gain a better understanding of human chromosomes. The Human Genome Project (Figure 1.2.5) relied on basic research carried out with non-human organisms and, later, with the human genome. An important end goal eventually became using the data for applied research seeking cures for genetic diseases. Figure 1.2.5. The Human Genome Project was a 13-year collaborative effort among researchers working in several different fields of science. The project was completed in 2003. (credit: the U.S. Department of Energy Genome Programs) Scientific Work is Transparent & Open to Critique Whether scientific research is basic science or applied science, scientists must share their findings for other researchers to expand and build upon their discoveries. For this reason, an important aspect of a scientist’s work is disseminating results and communicating with peers. Scientists can share results by presenting them at a scientific meeting or conference, but this approach can reach only the limited few who are present. Instead, most scientists present their results in peer-reviewed articles that are published in scientific journals. Peer-reviewed articles are scientific papers that are reviewed, usually anonymously by a scientist’s colleagues, or peers. These colleagues are qualified individuals, often experts in the same research area, who judge whether or not the scientist’s work is suitable for publication. The process of peer review helps to ensure that the research described in a scientific paper or grant proposal is original, significant, logical, ethical, and thorough. Scientists publish their work so other scientists can reproduce their experiments under similar or different conditions to expand on the findings. The experimental results must be consistent with the findings of other scientists. As you review scientific information, whether in an academic setting or as part of your day-to-day life, it is important to think about the credibility of that information. You might ask yourself: has this scientific information been through the rigorous process of peer review? Are the conclusions based on available data and accepted by the larger scientific community? Scientists are inherently skeptical, especially if conclusions are not supported by evidence (and you should be too). Attribution Concepts of Biology by OpenStax is licensed under CC BY 3.0. Modified from the original by Matthew R. Fisher. This page titled 1.2: The Process of Science is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Matthew R. Fisher (OpenOregon) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 1.1: The Earth, Humans, & the Environment 1.3: Environment and Sustainability Was this article helpful? Yes No Recommended articles 1.1: The Earth, Humans, & the EnvironmentEnvironmental science is the dynamic, interdisciplinary study of the interaction of living and non-living parts of the environment, with special focus... 1.3: Environment and SustainabilityThis section introduces the concept of sustainability, which refers to the sociopolitical, scientific, and cultural challenges of living within the me... 1.4: Environmental Ethics 1.5: Environmental Justice and Indigenous Struggles 1.S: Environmental Science (Summary) Article typeSection or PageAuthorMatthew R. 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189125	https://www.youtube.com/watch?v=nZ2mS5M4fcQ	Parametric and symmetric equations of the line (KristaKingMath) Krista King 273000 subscribers 825 likes Description 92258 views Posted: 8 May 2014 ► My Vectors course: Learn how to find the parametric equations and symmetric equations of the line. ● ● ● GET EXTRA HELP ● ● ● If you could use some extra help with your math class, then check out Krista’s website // ● ● ● CONNECT WITH KRISTA ● ● ● Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;) Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!” So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: FACEBOOK // TWITTER // INSTAGRAM // PINTEREST // GOOGLE+ // QUORA // 81 comments Transcript: today we're going to be talking about how to find parametric and symmetric equations of a line and in this particular problem we've been given the line which passes through two coordinate points one is 02 1 and the other is 2 1 -3 well how are we going to find parametric and symmetric equations for this line passing through these two points the easiest way to do it is with a vector representation of the line that passes through these points how do we get a vector representation well we've got these two points and we can represent them as vectors like this 0 1/ 12 1 and 2 1 -3 but what we really need are the direction numbers that Define the direction of the line when the line moves from one point to the other point and the way that we get those Direction numbers is by subtracting the component parts of one of these points from the component parts of the other so here's what that looks like we will take the component parts of this second Point 2 13 so we'll start with here to the X component of this point and we'll subtract the X component from this first point which is zero so minus 0 then we'll take the Y component and subtract the other one so 1 -2 and we always want to go in the same direction so we always want to start with this second point and subtract the component from the first point and then we'll start with -3 and subtract one so - one like this and when we simplify we get 2 12 and -3 - 1 is a -4 these are our Direction numbers this is a vector which is parallel to our line because we basically found a vector that goes from one point to the other point these vectors that we defined earlier go from the origin out to this coordinate point and from the origin out to this coordinate point so these two vectors move from the origin to the coordinate points this Vector moves from one coordinate point to the other which means that it's parallel to the line that passes through these two points and remember that because this Vector represents the direction of the line we also call these values here the direction numbers of our line now in order to find parametric equations of the line the first thing we want to do is write an equation for the line in Vector form or a vector equation of the line and we're going to Define that as R we're going to say R is equal to here's where we need to pick one of the points that our line passes through and in this particular case we'll pick 2 1 -3 but it doesn't really matter so we'll say 2 1 -3 remember that that's also the same as 2 I + 1 Jus 3 K so we'll say 2 I + J 1 J minus 3 K so it passes through that point then we add to that t which is going to be a parameter value times our Direction numbers in Vector form so we're going to do the same thing 2 I + 12 J - 4 K this is going to give us the vector equation of our line all we need to do now is simplify this by gathering together our components we want all of our I's together our J's together and our K's together so here's how we're going to do that first we're going to distribute the T so we're going to get 2 I + J minus 3 k then we're going to say + 2 TI + 12 T J - 4T K now we want to group our components together so when we group our I's together we get 2 I + 2 TI grouping our J's together we'll get plus J we have this 1 J here plus 12 T J and then grouping our KS together we get plus -3 K minus 4 TK like this now we want to factor out our components i j and k from these groups so we want to pull an i out when we pull I out we're just left with 2 + 2 T and we want to put the I behind this value so that 2 + 2T is like a coefficient on this I component then here we want to pull out a J We're left with 1 + 12 T J and then we want to pull out a k and of course we're just left with -3 minus 4T and we pull the K out behind it like this the reason this is so convenient remember this is the vector equation of the line but the reason this is so convenient is because we have everything we need now to find the parametric equations of the line remember that I J and K correspond to X Y and Z respectively all we need to do is pull the coefficients that are in front of these i j and k components here pull these coefficients out and set them equal to the variables to which they correspond so here's what that looks like our parametric equations parametric equations which we'll just say PE there we're going to say x is equal to this coefficient so x = 2 + 2 T we'll say y = 1 + 12 T and we'll say Z is equal to -3 - 4 T and that's it now we just need to find our symmetric equations of the line which is going to be really easy because all we need to do that is one point on the line and our Direction numbers so here's what that looks like we'll say symmetric equations s let's again use the 21-3 although it doesn't matter you could use either of them but here's what we're going to do we're going to say x minus the X component from the point on the line so xus 2 in this case and then we're going to divide by the associated Direction number which is two we can see here so we're going to get x - 2 / 2 we're going to set that equal to Y minus our y value from our coordinate point which is one and then we're going to divide by the associated Direction number which is 1/2 so divided by 1/2 set that equal to Z minus the Z value from our coordinate point which is -3 and then we divide that by the associated Direction number which is going to be -4 remember that with symmetric equations you always set them equal to each other like this in our numerator you always say the variable minus the associated value from the coordinate point there and now it's just a matter of simplifying keep in mind also that if we have a direction number that's zero if any of our Direction numbers are zero then we pull that outside of our symmetric equation so here's what we mean if our Direction numbers instead of 2 one2 -4 were 0 1 12 -4 then our symmetric equation would look like this because we can't put that zero for X here in our denominator we'd have x - 2 / 0 and we can't divide by zero we just don't include it we pull this x - 2 outside so our symmetric equations would be like this x - 2 and then just the Y and Z equation set equal to each other so y - 1 / 12 is equal to Z - A -3 / -4 so we'd say our symmetric equations are X - 2 and then separately y - 1 / 12 is equal to Z - -3 over -4 so we just pull it out like that and and you'll do that for every symmetric equation for which the associated Direction number is equal to zero but in this case none of our Direction numbers are equal to zero so we can set all three of these equations equal to one another now we just want to simplify and simplifying here is really easy for example Z minus a -3 instead we're just going to say Z Plus three to get rid of that double negative and then we don't like to have a fraction here in the denominator if we can avoid it so we're just going to multiply this fraction by 2 over two if we multiply by 2 over two then the denominator becomes 1 1 12 2 is just 1 that's going to go away and we'll just be left with 2 y - 2 so then our final answer for symmetric equations we get x - 2 / 2 is equal to 2 y - 2 is equal to Z + 3 / -4 and that's how you find parametric and symmetric equations when you have a line passing through two coordinate points
189126	https://math.stackexchange.com/questions/99970/splitting-integrals-with-complex-limits	Splitting Integrals with Complex Limits - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Splitting Integrals with Complex Limits Ask Question Asked 13 years, 8 months ago Modified13 years, 7 months ago Viewed 2k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. We all know from school, that ∫b a f(x)d x=F(b)−F(a),∫a b f(x)d x=F(b)−F(a), with F(x)F(x) being the antiderivative of f(x)f(x). It is also obvious, if we deal with real a a,b b and f(x)f(x), that we can split in the following way: ∫b a f(x)d x=∫e a f(x)d x+∫b e f(x)d x=F(e)−F(a)+F(b)−F(e),∫a b f(x)d x=∫a e f(x)d x+∫e b f(x)d x=F(e)−F(a)+F(b)−F(e), with a≤e<b a≤e<b. What is necessary, that this is true for comlex limits? I got so far that I need Cauchy's Integral Theorem. There I found that: "One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of real calculus". Because I'm specially interested in the definite integral of li(a+i b)=∫a+i b 0 1 ln u d u li(a+i b)=∫0 a+i b 1 ln⁡u d u, I continue with f(u)=1 ln u f(u)=1 ln⁡u. I set z=r e i ϕ z=r e i ϕ and get f(u)=1 ln r+i ϕ=ln r−i ϕ(ln r)2−ϕ 2=u(r,ϕ)+i v(r,ϕ)f(u)=1 ln⁡r+i ϕ=ln⁡r−i ϕ(ln⁡r)2−ϕ 2=u(r,ϕ)+i v(r,ϕ) To check for holomorphy in polar coordinates if have to check: ∂u∂r=(ln r)2+ϕ 2 r((ln r)2−ϕ 2)2=1 r∂v∂θ,∂v∂r=−2 ϕ ln(r)r((ln r)2−ϕ 2)2=−1 r∂u∂θ.∂u∂r=(ln⁡r)2+ϕ 2 r((ln⁡r)2−ϕ 2)2=1 r∂v∂θ,∂v∂r=−2 ϕ ln⁡(r)r((ln⁡r)2−ϕ 2)2=−1 r∂u∂θ. I used WolframAlpha to check that (links didn't work, sorry). So everythings seems fine and if there's nobody out there screaming "STOP, don't do this because of ..." I split it! complex-analysis definite-integrals Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 23, 2012 at 11:13 Willie Wong 75.5k 13 13 gold badges 167 167 silver badges 263 263 bronze badges asked Jan 17, 2012 at 22:57 draks ...draks ... 18.8k 8 8 gold badges 67 67 silver badges 209 209 bronze badges 1 What "we all know from school" depends on us all intuiting a certain relation between f f and F F, which relation would be better stated explicitly.Gerry Myerson –Gerry Myerson 2012-01-17 23:22:46 +00:00 Commented Jan 17, 2012 at 23:22 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. Caution: integrals in the complex plane are not "from a a to b b", they are "on a particular path from a a to b b". If the domain where the function is analytic is not simply connected, the answer will depend on which path you choose. In this case, f(z)=1 ln z f(z)=1 ln⁡z has a branch cut, which you can take to be on the negative real axis, but also a simple pole at z=1 z=1. An antiderivative F(z)F(z) of this function will not be analytic on the whole region where f(z)f(z) is analytic: it will have another branch cut, which may be taken to be the interval (0,1)(0,1) on the real axis. So you can define li(x)li(x) to be analytic on the complement of the ray (−∞,1](−∞,1], and it will satisfy ∫C d z ln(z)=li(b)−li(a)∫C d z ln⁡(z)=li(b)−li(a) for any directed contour C C that starts at a a and ends at b b and avoids (−∞,1](−∞,1]. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 17, 2012 at 23:17 Robert IsraelRobert Israel 472k 28 28 gold badges 376 376 silver badges 714 714 bronze badges 1 Thx a lot. For me this was an eye opener!!!draks ... –draks ... 2012-01-18 13:16:50 +00:00 Commented Jan 18, 2012 at 13:16 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I'll add to Robert's excellent advise this one : avoid evaluating expressions like li(e x+i y)li⁡(e x+i y) or li(z ρ)li⁡(z ρ) with complex ρ ρ because the exact phase will be lost! Computer evaluation of the exponential will lose the phase information since e z=e z+2 k π i e z=e z+2 k π i and the logarithm will return the 'principal branch' of the result so that ln(e x+i y)ln⁡(e x+i y) becomes x+i y′x+i y′ with y′−y=2 k π y′−y=2 k π. This is a classical problem when using ζ ζ zeros in the Riemann's explicit formula for example. To avoid losing 2 k π i 2 k π i terms it may be convenient to replace your li li function by the exponential integral function Ei Ei using the relation li(z)=Ei(ln z)li⁡(z)=Ei⁡(ln⁡z) or more specifically li(e x+i y)=Ei(x+i y)li⁡(e x+i y)=Ei⁡(x+i y). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 18, 2012 at 1:07 answered Jan 18, 2012 at 0:09 Raymond ManzoniRaymond Manzoni 44.3k 5 5 gold badges 92 92 silver badges 144 144 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. There is an analog. Suppose that α,β:[0,1]→C α,β:[0,1]→C are paths with α(1)=β(0)α(1)=β(0). Define the path (α+β)(t)={α(2 t)β(2 t−1)0≤t≤1/2 1/2≤t≤1.(α+β)(t)={α(2 t)0≤t≤1/2 β(2 t−1)1/2≤t≤1. Then if f f is holomorphic on a neighborhood of these paths, ∫α+β f=∫α f+∫β f.∫α+β f=∫α f+∫β f. This is just a "concatenation" of the two paths. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 14, 2012 at 13:08 Sasha 71.8k 7 7 gold badges 149 149 silver badges 227 227 bronze badges answered Jan 17, 2012 at 23:29 ncmathsadistncmathsadist 50.2k 3 3 gold badges 85 85 silver badges 134 134 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions complex-analysis definite-integrals See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2Why is π π the Limit of the Absolute Value of the Prime ζ ζ Function? 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189127	https://books.google.com/books/about/A_Textbook_of_Electrical_Technology.html?id=njscEAAAQBAJ	A Textbook of Electrical Technology - BL Theraja - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Good for: Web Tablet / iPad eReader Smartphone#### Features: Flowing text Scanned pages Help with devices & formats Learn more about books on Google Play Buy eBook - $251.64 Get this book in print▼ Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» My library My History A Textbook of Electrical Technology =================================== BL Theraja S. Chand Publishing, 2014 - Technology & Engineering - 2784 pages For Mechnaical Engginering Students of Indian Universities.It is also available in 4 Individual Parts More » Preview this book » Selected pages Page 189 Page 1 Page 51 Table of Contents Index Contents Chapter 11 Chapter 251 Chapter 3175 Chapter 4189 Chapter 5213 Chapter 6 257 Chapter 7 297 Chapter 8 317 Chapter 39 1535 Chapter 40 1569 Chapter 41 1603 Chapter 42 1683 Chapter 43 1699 Chapter 44 1767 Chapter 45 1795 Chapter 46 1823 More Chapter 9 339 Chapter 10 375 Chapter 11 453 Chapter 12 497 Chapter 13 507 Chapter 14 557 Chapter 15 599 Chapter 16 627 Chapter 17 641 Chapter 18 655 Chapter 19 665 Chapter 20 753 Chapter 21 779 Chapter 22 815 Chapter 23 835 Chapter 24 855 Chapter 25 865 Chapter 26 887 Chapter 27 937 Chapter 28 967 Chapter 29 995 Chapter 30 1031 Chapter 31 1091 Chapter 32 1115 Chapter 33 1211 Chapter 34 1243 Chapter 35 1313 Chapter 36 1367 Chapter 37 1401 Chapter 38 1489 Chapter 47 1833 Chapter 48 1861 Chapter 49 1893 Chapter 50 1943 Chapter 51 2017 Chapter 52 2059 Chapter 53 2087 Chapter 54 2111 Chapter 55 2129 Chapter 56 2167 Chapter 57 2187 Chapter 58 2219 Chapter 59 2237 Chapter 60 2277 Chapter 61 2315 Chapter 62 2343 Chapter 63 2363 Chapter 64 2389 Chapter 65 2407 Chapter 66 2441 Chapter 67 2471 Chapter 68 2503 Chapter 69 2525 Chapter 70 2555 Chapter 71 2585 Chapter 72 2631 Chapter 73 2695 Index2737 Copyright Less Other editions - View all A Textbook of Electrical Technology BL Theraja Limited preview - 2014 Common terms and phrases 3-phaseair-gapalternatingalternating currentammeteramperesapplied voltagearmatureAT/mbatterybranchCalculatecapacitancecapacitorcellchargecircuit of Figcircuit showncoilcommutatorcomponentconductorsconnected in seriesconstantcurrent flowingcurrent sourcedeflectiondielectricelectricEnggequalequationequivalentExamplefield resistanceflux densityFourier seriesfrequencygivenharmonicHencehysteresisI₁impedanceinduced e.m.f.inductanceinstrumentironjoulelaggingline currentloadlossmagnetisingmaximummetermotorNagpur Universitynegativenodeohmsoutputphase anglephase sequencephasorphasor diagramplatespolepositivepotentialpower factorr.m.s. valueR₁reactanceresistorresonanceshort-circuitshown in FigshuntSolutionSuperposition theoremsupplyterminalstheoremThevenin’storquetotal powerV₁vectorvoltage dropvoltage sourcevoltmetervoltswattmeterwavewaveformWb/m²windingwirezero Bibliographic information Title A Textbook of Electrical Technology AuthorBL Theraja Publisher S. Chand Publishing, 2014 ISBN 8121924413, 9788121924412 Length 2784 pages SubjectsTechnology & Engineering › Electronics › General Technology & Engineering / Electronics / General Technology & Engineering / General Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
189128	https://www.indeed.com/career-advice/career-development/net-income-vs-net-sales	Net Income vs. Net Sales: Definitions and Differences \| Indeed.com Skip to main content Home Company reviews Find salaries Sign in Sign in Employers / Post Job 1 new update Home Company reviews Find salaries Employers Create your resume Resume services Change country 🇺🇸 United States Help Start of main content Career Guide Finding a job Resumes & cover letters Resumes & cover letters articles Resume samples Cover letter samples Interviewing Pay & salary Career development Career development articles Starting a new job Career paths News Career development Net Income vs. Net Sales: Definitions and Differences Net Income vs. Net Sales: Definitions and Differences Written by Deepti Sharma Updated July 26, 2025 Businesses maintain records of their financial activities in financial statements. An income statement contains revenue and expense details that allow professionals to calculate net income and net sales. If you’re a finance professional or create financial reports, it's important to understand what net sales and net income indicate to represent finances accurately.In this article, we define net sales and net income, explain how to calculate them, provide examples, examine their importance and discuss the differences between them. Related jobs on Indeed Part-time jobs Full-time jobs Remote jobs Urgently hiring jobs View more jobs on Indeed What is net sales? Net sales is the total amount of money you earn from sales activities, subtracting any adjustments. This figure is your final revenue figure. Gross revenue is the total amount of money a company makes from selling products to customers. Adjustments, such as discounts and returns, affect the final amount. Adjustments include: Returns: A return entails sending back an order to the vendor after delivery to the recipient within a predetermined time frame for a refund. Every order doesn’t qualify for a return or a full refund, as it depends on a set of terms and conditions. Allowances: Allowances occur after a sale and are partial refunds in the form of price reductions or rebates offered as an alternative to returns. Discounts: Discounts are price reductions on a product, usually represented as a percentage taken off the listed sale price. Related: Net Sales: Definition and How To Calculate Them What is net income? Often called the bottom line since it appears at the bottom of an income statement, net income refers to all the money you have remaining after subtracting all costs and expenses from your gross income. Expenses may include the following: Operating expenses: These are expenses a business incurs unrelated to the creation of a product. They often involve legal fees and rent. Taxes: The government collects taxes from individuals and organizations to raise money for public services and infrastructure. Accountants deduct taxes from gross income to calculate net income. Payroll: This is the process of paying salaries to a company's employees. Interest on loans: Interest is the price one pays for borrowing money. Companies often resort to loans from financial institutions for various activities and pay a specific percentage of the loan amount on repayments as interest. Depreciation on fixed assets: Depreciation is the periodic reduction of the recorded cost of a fixed asset. Fixed assets depreciating with time include furniture, office supplies and buildings. Additional sources: Net income also factors in additional sources of income, such as gains from short-term investments and the sale of assets. All the other figures on an income statement, including net sales, calculate toward the net income. Related: How To Calculate Net Income: Formula Plus Examples How to calculate net sales An income statement's net sales is the figure that remains after an accountant deducts sales discounts, refunds and allowances. The net sales formula is:Net sales = gross sales - (returns + allowances + discounts)Here are some steps you can take to calculate net sales effectively: 1. Calculate your gross revenue Gross revenue is the whole amount of money a business earns without deducting the expenses for a specific period. Gross revenue doesn’t include the cost of goods sold and only considers the money earned from the sales. 2. Calculate your adjustments There are three significant sales adjustments: returns, allowances and discounts. The sum of these three for the period for which they record the net sales is the adjustments. 3. Subtract your allowances from your gross revenue Net sales is what remains after subtracting all returns, allowances and sales discounts from gross sales. If any of the above data is missing, the output might be incorrect and the figures won't match. Therefore, accumulating all of the necessary data ensures accurate calculations.Related: The Differences in Reporting Gross Revenue vs. Net Revenue How to calculate your net income To determine net income, start with gross income and deduct costs like interest payments and taxes. Here’s the formula:Net income = (net sales + other income sources) - all expenses Here are some steps you can take to calculate net income: 1. Calculate your net sales Net sales is one of the most important sales parameters. It’s the result of gross revenue subtracting applicable sales returns, allowances and discounts. 2. Calculate additional sources of income Analysts may use pretax income as a more accurate indicator of corporate profitability because these activities aren’t a part of a company's regular business operations. Add to that any income stemming from the sale of assets or earned interest. 3. Calculate your expenses Expenses are the costs that a business incurs to make a profit. The depreciation of equipment, staff salaries and supplier payments are a few expenses. Operating and non-operating expenses comprise the two primary types of business expenses. 4. Add net sales and additional sources of income Adding net sales and other sources of income results in gross income. It’s the revenue a company receives from the sale of goods or services before deducting taxes and other expenses. 5. Subtract your expenses from the gross income The net income is gross income subtracting all other expenses and costs, as well as any other income and revenue sources that aren’t part of gross income. Some of the costs subtracted are taxes and operating expenses.Related: How to Calculate Annual Growth Rate Importance of net sales and net income Net sales and net income are both important financial metrics. Net sales represent how much revenue a business has truly earned and maintained. Without considering cost adjustments for returns, allowances and discounts, an income statement would report higher earnings than are actually available. Knowing your net sales shows you exactly how much money stems from sales.Net income considers net sales to calculate a figure representing a business' operational efficiency and profitability. A higher net income means the business is financially healthy, suggesting that its offerings and strategies are high-quality and effective. Inversely, a lower net income would indicate the need for improvements.Related: 11 Ways To Cut Business Costs and Save Money Net income vs. net sales Though net sales and net income are both metrics used for determining the financial matters of a business, there are key differences between the two concepts, such as: Placement on an income statement Net sales and net income both appear on an income statement. Net sales is the first figure listed on the statement, while net income is the bottom figure. All figures on an income statement contribute toward net income, so net sales is a factor that accounts reflect as the bottom-line figure.Related: Frequently Asked Questions About the Bottom Line Applicability Every money-generating business can report a net income, but net sales may not apply to all businesses. This is due to the factors that comprise its calculation, such as price adjustments. Thus, businesses that offer returnable products, such as retailers, are the ones that report net sales.Related: Your Guide To Understanding Net Earnings vs. Net Income Objectives The objective of net sales is to show how much money your products or services ultimately generate for your organization. It doesn't include income from other sources. For example, if you own a bookstore, your net sales are what you earn by selling books, stationery and other related materials from your stock.Net income is a reflection of your business's overall financial health and factors in all sources of income. A high net income means your business is doing well and is financially healthy. Net income can also indicate the quality of your managerial skills. For example, a wide margin between net income and net sales shows you're making a profit despite lower revenue. This could stem from keeping expenses to a minimum, an indicator of effective financial management.Related: Net Income vs. Net Profit (With Examples) Uses As metrics, net sales and net income are useful in different ways. Net sales can help identify areas within your business that need improvement. For example, if there's a growing difference between the figures for your gross sales and net sales, this may suggest customer dissatisfaction with the products you offer. The problem might be that a number of your goods sustain damage in shipment, which would signal the need to review and correct your shipping methods. Alternatively, the goods may be of low quality and might improve with production and quality control adjustments.Net income is useful for determining profitability. This information can be helpful to various stakeholders. For investors, net income can help them decide whether to invest in a company, while other agencies can refer to net income to determine eligibility for loans. If you're a small business owner, net income can also show you whether you're getting a good return on your investments. If, for example, you initially invested $100,000 in your business and have a net income of $30,000, you're receiving a 30% return, which is substantial.Related: What Is the Difference Between Gross and Net Sales? Search jobs and companies hiring now Job title, keywords or company Location Search Examples of net sales and net income Consider these examples to gain a better understanding of net sales and net income: Deriving net sales after returns A shoe store sold $10,000 worth of footwear and accessories this month. Several customers complained about the quality of their shoes and sought to return them. Three customers received full refunds for their returns, totaling $600. Seven other customers tried to return goods but didn't meet the criteria for returns, as their purchases showed signs of use. Wanting to retain their clientele, the store owner offered each of them a partial refund of 20% of the purchase price, totaling $350.Gross sales for this month total $10,000, while total adjustments equal $950. Therefore, the net sales for this month are $9,050.Related: How To Calculate Your Adjusted Gross Income (AGI) Deriving net sales after discounts A stationery store has a one-day promotional event offering 50% off for teachers who present their school ID. 20 teachers purchased goods worth $300, 10 purchased goods worth $400 and five purchased goods worth $600. The gross sales from these purchases amount to $13,000. After applying the discount, the net sales equal $6,500.Related: Gross Sales: Definition and How to Calculate It Deriving net income after expenses This year a shop that sells musical instruments reports net sales of $50,000. The store owner also provides private music lessons and channels that income to the shop. Total earnings from music lessons this year are $15,000. For expenses, there are the following: Cost of goods sold: $10,000 Depreciation on assets: $8,000 Rent: $15,000 Utilities: $5,000 The expenses total $38,000, while the earnings total $65,000. Earnings subtracted by expenses equal $27,000, representing the shop's net income.Related: Defining the Cost of Goods Sold (With Calculation Example) Deriving net income after expenses A company has gross sales of $600,000 in a fiscal year and has also sold two fixed assets and recovered $40,000 from these sales. Its expenses include the following: $40,000 in employee wages $50,000 in materials purchased $5,000 for fixed assets purchased $12,000 in depreciation of existing assets $6,000 in taxes The company's expenses total $113,000. It has net sales of $500,000 and an additional income of $40,000, totaling $540,000 as revenue. Subtracting expenses from total revenue equals a net income of $427,000 in 2021. The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers Share: Is this article helpful? 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189129	https://www.chicagomanualofstyle.org/qanda/data/faq/topics/Numbers/faq0074.html	FAQ Item Homecmos 18 contentsCitation quick guideHelp & ToolsGive Style Q&ABlogForumAboutCMOS StoreSubscribeCMOS 17Log In Go to Index false Latest Q&ABrowse Q&ASubmit a QuestionGet Q&A Alert Subscribe to the Chicago Manual of Style Online sign up for a Free Trial Shop the CMOS Merch Store Numbers Q. Would you spell out 150,000? A. Use numerals for 150,000. The applicable principles are as follows: Spell out numbers one through one hundred (Chicago’s general rule). Spell out multiples of one through one hundred used in combination with hundred, thousand, or hundred thousand. So you would spell out “five thousand” and “one hundred thousand” but use digits for 150,000—because 150 would normally be rendered as a numeral. But if you’re following Chicago’s alternative rule of using digits for 10 and up, all such larger numbers are usually given as numerals. Rather than, for example, “fifteen thousand” or “15 thousand,” you’d write 15,000. For more details, see CMOS 9.2, 9.3, and 9.4. For numbers with million, billion, and so forth, see CMOS 9.8. [This answer relies on the 17th edition of CMOS (2017) unless otherwise noted.] SHOP TALK BLOG! CMOS editors share writing tips, editing ideas, interviews, quizzes, and more! It's Summer at the CMOS Merch Store! Now Available for Pre-Order! Barbara Gastel, Medical Editing NEW! The Business of Being a Writer, Second Edition, by Jane Friedman The CSE Manual, 9th Edition, the Scientific Companion to The Chicago Manual of Style The Design of Books, An Explainer for Authors, Editors, Agents, and Other Curious Readers, by Debbie Berne The Chicago Guide for Freelance Editors: How to Take Care of Your Business, Your Clients, and Yourself from Start-Up to Sustainability, by Erin Brenner The Chicago Guide to Copyediting Fiction, by Amy J. Schneider Developmental Editing, 2nd Edition: A Handbook for Freelancers, Authors, and Publishers, by Scott Norton Indexes: A Chapter from “The Chicago Manual of Style,” 18th Edition The Craft of Research, 5th Edition: A thoroughly updated edition of a beloved classic The Chicago Guide to Fact-Checking, 2nd Edition Information Now, 2nd Edition A Graphic Guide to Student Research and Web Literacy Shop the CMOS Bookstore! Writing, Editing, and Publishing Books from CHICAGO About The Chicago Manual of StyleTerms of UseSite MapAccount Management About the University of Chicago PressPrivacy PolicyContact UsAccessibility The Chicago Manual of Style 18th edition text © 2024 by The University of Chicago. The Chicago Manual of Style 17th edition text © 2017 by The University of Chicago. The Chicago Manual of Style Online © 2006, 2007, 2010, 2017, 2024 by The University of Chicago. The Chicago Manual of Style is a registered trademark of The University of Chicago.
189130	https://vasculitisfoundation.org/treatments-research/treatments/acr-treatment-guidelines/	Skip to content Newly Diagnosed Home » Treatments/Research » Treatments » ACR/VF Treatment Guidelines Quick Links Home Find a Doctor Support Groups Self Advocacy Treatments VPPRN Patient-Powered Research Vasculitis Pregnancy Registry Pediatric Vasculitis Registry Resources Connect Ways to Give ACR/VF Treatment Guidelines ACR/VF Treatment Guidelines 2021 American College of Rheumatology/Vasculitis Foundation Guidelines These guidelines present the first recommendations endorsed by the American College of Rheumatology and the Vasculitis Foundation for the management of: ANCA–associated vasculitides: Granulomatosis with polyangiitis (GPA), Microscopic polyangiitis (MPA), and Eosinophilic granulomatosis with polyangiitis (EGPA) Polyarteritis nodosa (PAN) Giant Cell Arteritis and Takayasu Arteritis Kawasaki Disease (KD) These recommendations provide guidance regarding the evaluation and management of patients, including diagnostic strategies, use of pharmacologic agents, and surgical interventions. Donate Today ANCA-Associated Vasculitis (GPA/MPA/EGPA) Guidelines The American College of Rheumatology and the Vasculitis Foundation for the management of AAV and provides guidance to health care professionals on how to treat these diseases. Learn More Polyarteritis Nodosa (PAN) Guidelines These recommendations provide guidance regarding diagnostic strategies, use of pharmacologic agents, and imaging for patients with PAN. Learn More Giant Cell Arteritis (GCA) & Takayasu Arteritis (TAK) Guidelines These recommendations provide guidance regarding the evaluation and management of patients with GCA and TAK, including diagnostic strategies, use of pharmacologic agents, and surgical interventions. Learn More Kawasaki Disease (KD) Guidelines These recommendations provide guidance regarding diagnostic strategies, use of pharmacologic agents, and imaging for patients with KD. Learn More
189131	https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php	←previous next→ 11.1.2 Basic Concepts of the Poisson Process The Poisson process is one of the most widely-used counting processes. It is usually used in scenarios where we are counting the occurrences of certain events that appear to happen at a certain rate, but completely at random (without a certain structure). For example, suppose that from historical data, we know that earthquakes occur in a certain area with a rate of 2 per month. Other than this information, the timings of earthquakes seem to be completely random. Thus, we conclude that the Poisson process might be a good model for earthquakes. In practice, the Poisson process or its extensions have been used to model : − the number of car accidents at a site or in an area; : − the location of users in a wireless network; : − the requests for individual documents on a web server; : − the outbreak of wars; : − photons landing on a photodiode. Poisson random variable: Here, we briefly review some properties of the Poisson random variable that we have discussed in the previous chapters. Remember that a discrete random variable X is said to be a Poisson random variable with parameter μ, shown as X∼Poisson(μ), if its range is RX={0,1,2,3,...}, and its PMF is given by PX(k)={e−μμkk!0for k∈RX otherwise Here are some useful facts that we have seen before: If X∼Poisson(μ), then EX=μ, and Var(X)=μ. If Xi∼Poisson(μi), for i=1,2,⋯,n, and the Xi's are independent, then X1+X2+⋯+Xn∼Poisson(μ1+μ2+⋯+μn). 3. The Poisson distribution can be viewed as the limit of binomial distribution. Theorem Let Yn∼Binomial(n,p=p(n)). Let μ>0 be a fixed real number, and limn→∞np=μ. Then, the PMF of Yn converges to a Poisson(μ) PMF, as n→∞. That is, for any k∈{0,1,2,...}, we have limn→∞PYn(k)=e−μμkk!. Poisson Process as the Limit of a Bernoulli Process: Suppose that we would like to model the arrival of events that happen completely at random at a rate λ per unit time. Here is one way to do this. At time t=0, we have no arrivals yet, so N(0)=0. We now divide the half-line [0,∞) to tiny subintervals of length δ as shown in Figure 11.2. Each subinterval corresponds to a time slot of length δ. Thus, the intervals are (0,δ], (δ,2δ], (2δ,3δ], ⋯. More generally, the kth interval is ((k−1)δ,kδ]. We assume that in each time slot, we toss a coin for which P(H)=p=λδ. If the coin lands heads up, we say that we have an arrival in that subinterval. Otherwise, we say that we have no arrival in that interval. Figure 11.3 shows this process. Here, we have an arrival at time t=kδ, if the kth coin flip results in a heads. Now, let N(t) be defined as the number of arrivals (number of heads) from time 0 to time t. There are n≈tδ time slots in the interval (0,t]. Thus, N(t) is the number of heads in n coin flips. We conclude that N(t)∼Binomial(n,p). Note that here p=λδ, so np=nλδ=tδ⋅λδ=λt. Thus, by Theorem 11.1, as δ→0, the PMF of N(t) converges to a Poisson distribution with rate λt. More generally, we can argue that the number of arrivals in any interval of length τ follows a Poisson(λτ) distribution as δ→0. Consider several non-overlapping intervals. The number of arrivals in each interval is determined by the results of the coin flips for that interval. Since different coin flips are independent, we conclude that the above counting process has independent increments. Definition of the Poisson Process: The above construction can be made mathematically rigorous. The resulting random process is called a Poisson process with rate (or intensity) λ. Here is a formal definition of the Poisson process. The Poisson Process Let λ>0 be fixed. The counting process {N(t),t∈[0,∞)} is called a Poisson process with rates λ if all the following conditions hold: N(0)=0; N(t) has independent increments; the number of arrivals in any interval of length τ>0 has Poisson(λτ) distribution. Note that from the above definition, we conclude that in a Poisson process, the distribution of the number of arrivals in any interval depends only on the length of the interval, and not on the exact location of the interval on the real line. Therefore the Poisson process has stationary increments. Example The number of customers arriving at a grocery store can be modeled by a Poisson process with intensity λ=10 customers per hour. Find the probability that there are 2 customers between 10:00 and 10:20. Find the probability that there are 3 customers between 10:00 and 10:20 and 7 customers between 10:20 and 11. Solution Here, λ=10 and the interval between 10:00 and 10:20 has length τ=13 hours. Thus, if X is the number of arrivals in that interval, we can write X∼Poisson(10/3). Therefore, P(X=2)=e−103(103)22!≈0.2 2. Here, we have two non-overlapping intervals I1=(10:00 a.m., 10:20 a.m.] and I2= (10:20 a.m., 11 a.m.]. Thus, we can write P(3 arrivals in I1 and 7 arrivals in I2)=P(3 arrivals in I1)⋅P(7 arrivals in I2). Since the lengths of the intervals are τ1=1/3 and τ2=2/3 respectively, we obtain λτ1=10/3 and λτ2=20/3. Thus, we have P(3 arrivals in I1 and 7 arrivals in I2)=e−103(103)33!⋅e−203(203)77!≈0.0325 Second Definition of the Poisson Process: Let N(t) be a Poisson process with rate λ. Consider a very short interval of length Δ. Then, the number of arrivals in this interval has the same distribution as N(Δ). In particular, we can write P(N(Δ)=0)=e−λΔ=1−λΔ+λ22Δ2−⋯(Taylor Series). Note that if Δ is small, the terms that include second or higher powers of Δ are negligible compared to Δ. We write this as P(N(Δ)=0)=1−λΔ+o(Δ)(11.1) Here o(Δ) shows a function that is negligible compared to Δ, as Δ→0. More precisely, g(Δ)=o(Δ) means that limΔ→0g(Δ)Δ=0. Now, let us look at the probability of having one arrival in an interval of length Δ. P(N(Δ)=1)=e−λΔλΔ=λΔ(1−λΔ+λ22Δ2−⋯)(Taylor Series)=λΔ+(−λ2Δ2+λ32Δ3⋯)=λΔ+o(Δ). We conclude that P(N(Δ)=1)=λΔ+o(Δ)(11.2) Similarly, we can show that P(N(Δ)≥2)=o(Δ)(11.3) In fact, equations 11.1, 11.2, and 11.3 give us another way to define a Poisson process. The Second Definition of the Poisson Process Let λ>0 be fixed. The counting process {N(t),t∈[0,∞)} is called a Poisson process with rate λ if all the following conditions hold: N(0)=0; N(t) has independent and stationary increments we have P(N(Δ)=0)=1−λΔ+o(Δ),P(N(Δ)=1)=λΔ+o(Δ),P(N(Δ)≥2)=o(Δ). We have already shown that any Poisson process satisfies the above definition. To show that the above definition is equivalent to our original definition, we also need to show that any process that satisfies the above definition also satisfies the original definition. A method to show this is outlined in the End of Chapter Problems. Arrival and Interarrival Times: Let N(t) be a Poisson process with rate λ. Let X1 be the time of the first arrival. Then, P(X1>t)=P(no arrival in (0,t])=e−λt. We conclude that FX1(t)=⎧⎩⎨⎪⎪1−e−λt0t>0otherwise Therefore, X1∼Exponential(λ). Let X2 be the time elapsed between the first and the second arrival (Figure 11.4). Let s>0 and t>0. Note that the two intervals (0,s] and (s,s+t] are disjoint. We can write P(X2>t\|X1=s)=P(no arrival in (s,s+t]\|X1=s)=P(no arrivals in (s,s+t])(independent increments)=e−λt. We conclude that X2∼Exponential(λ), and that X1 and X2 are independent. The random variables X1, X2, ⋯ are called the interarrival times of the counting process N(t). Similarly, we can argue that all Xi's are independent and Xi∼Exponential(λ) for i=1,2,3,⋯. Interarrival Times for Poisson Processes If N(t) is a Poisson process with rate λ, then the interarrival times X1, X2, ⋯ are independent and Xi∼Exponential(λ), for i=1,2,3,⋯. Remember that if X is exponential with parameter λ>0, then X is a memoryless random variable, that is P(X>x+a\|X>a)=P(X>x), for a,x≥0. Thinking of the Poisson process, the memoryless property of the interarrival times is consistent with the independent increment property of the Poisson distribution. In some sense, both are implying that the number of arrivals in non-overlapping intervals are independent. To better understand this issue, let's look at an example. Example Let N(t) be a Poisson process with intensity λ=2, and let X1, X2, ⋯ be the corresponding interarrival times. Find the probability that the first arrival occurs after t=0.5, i.e., P(X1>0.5). Given that we have had no arrivals before t=1, find P(X1>3). Given that the third arrival occurred at time t=2, find the probability that the fourth arrival occurs after t=4. I start watching the process at time t=10. Let T be the time of the first arrival that I see. In other words, T is the first arrival after t=10. Find ET and Var(T). I start watching the process at time t=10. Let T be the time of the first arrival that I see. Find the conditional expectation and the conditional variance of T given that I am informed that the last arrival occurred at time t=9. Solution Since X1∼Exponential(2), we can write P(X1>0.5)=e−(2×0.5)≈0.37 Another way to solve this is to note that P(X1>0.5)=P(no arrivals in (0,0.5])=e−(2×0.5)≈0.37 2. We can write P(X1>3\|X1>1)=P(X1>2)(memoryless property)=e−2×2≈0.0183 Another way to solve this is to note that the number of arrivals in (1,3] is independent of the arrivals before t=1. Thus, P(X1>3\|X1>1)=P(no arrivals in (1,3]\|no arrivals in (0,1])=P(no arrivals in (1,3])(independent increments)=e−2×2≈0.0183 3. The time between the third and the fourth arrival is X4∼Exponential(2). Thus, the desired conditional probability is equal to P(X4>2\|X1+X2+X3=2)=P(X4>2)(independence of the Xi's)=e−2×2≈0.0183 4. When I start watching the process at time t=10, I will see a Poisson process. Thus, the time of the first arrival from t=10 is Exponential(2). In other words, we can write T=10+X, where X∼Exponential(2). Thus, ET=10+EX=10+12=212, Var(T)=Var(X)=14. 5. Arrivals before t=10 are independent of arrivals after t=10. Thus, knowing that the last arrival occurred at time t=9 does not impact the distribution of the first arrival after t=10. Thus, if A is the event that the last arrival occurred at t=9, we can write E[T\|A]=E[T]=212, Var(T\|A)=Var(T)=14. Now that we know the distribution of the interarrival times, we can find the distribution of arrival times T1=X1,T2=X1+X2,T3=X1+X2+X3,⋮ More specifically, Tn is the sum of n independent Exponential(λ) random variables. In previous chapters we have seen that if Tn=X1+X2+⋯+Xn, where the Xi's are independent Exponential(λ) random variables, then Tn∼Gamma(n,λ). This has been shown using MGFs. Note that here n∈N. The Gamma(n,λ) is also called Erlang distribution, i.e, we can write Tn∼Erlang(n,λ)=Gamma(n,λ), for n=1,2,3,⋯. The PDF of Tn, for n=1,2,3,⋯, is given by fTn(t)=λntn−1e−λt(n−1)!, for t>0. Remember that if X∼Exponential(λ), then E[X]=1λ,Var(X)=1λ2. Since Tn=X1+X2+⋯+Xn, we conclude that E[Tn]=nEX1=nλ,Var(Tn)=nVar(Xn)=nλ2. Note that the arrival times are not independent. In particular, we must have T1≤T2≤T3≤⋯. Arrival Times for Poisson Processes If N(t) is a Poisson process with rate λ, then the arrival times T1, T2, ⋯ have Gamma(n,λ) distribution. In particular, for n=1,2,3,⋯, we have E[Tn]=nλ,andVar(Tn)=nλ2. The above discussion suggests a way to simulate (generate) a Poisson process with rate λ. We first generate i.i.d. random variables X1, X2, X3, ⋯, where Xi∼Exponential(λ). Then the arrival times are given by T1=X1,T2=X1+X2,T3=X1+X2+X3,⋮ ←previous next→ \| \| \| The print version of the book is available on Amazon. \| \| Practical uncertainty: Useful Ideas in Decision-Making, Risk, Randomness, & AI \| \| \| \| --- \| \| \| \|
189132	https://mathworld.wolfram.com/CharacteristicFunction.html	Characteristic Function Given a subset of a larger set, the characteristic function , sometimes also called the indicator function, is the function defined to be identically one on , and is zero elsewhere. Characteristic functions are sometimes denoted using the so-called Iverson bracket, and can be useful descriptive devices since it is easier to say, for example, "the characteristic function of the primes" rather than repeating a given definition. A characteristic function is a special case of a simple function. The term characteristic function is used in a different way in probability, where it is denoted and is defined as the Fourier transform of the probability density function usingFourier transform parameters , \| \| \| \| \| --- --- \| \| \| \| \| (1) \| \| \| \| \| (2) \| \| \| \| \| (3) \| \| \| \| \| (4) \| \| \| \| \| (5) \| where (sometimes also denoted ) is the th moment about 0 and (Abramowitz and Stegun 1972, p. 928; Morrison 1995). A statistical distribution is not uniquely specified by its moments, but is by its characteristic function if all of its moments are finite and the series for its characteristic function converges absolutely near the origin (Papoulis 1991, p. 116). In this case, the probability density function is given by \| \| \| --- \| \| \| (6) \| (Papoulis 1991, p. 116). The characteristic function can therefore be used to generate raw moments, \| \| \| --- \| \| \| (7) \| or the cumulants , \| \| \| --- \| \| \| (8) \| See also Cumulant, Iverson Bracket, Moment, Moment-Generating Function, Probability Density Function, Set, Simple Function Portions of this entry contributed by Todd Rowland Explore with Wolfram\|Alpha More things to try: characteristic function of gaussian characteristic function of gamma distribution characteristic function Poisson distribution References Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 928, 1972.Kenney, J. F. and Keeping, E. S. "Moment-Generating and Characteristic Functions," "Some Examples of Moment-Generating Functions," and "Uniqueness Theorem for Characteristic Functions." §4.6-4.8 in Mathematics of Statistics, Pt. 2, 2nd ed. Princeton, NJ: Van Nostrand, pp. 72-77, 1951.Morrison, K. E. "Cosine Products, Fourier Transforms, and Random Sums." Amer. Math. Monthly 102, 716-724, 1995.Papoulis, A. "Characteristic Functions." §5-5 in Probability, Random Variables, and Stochastic Processes, 3rd ed. New York: McGraw-Hill, 1991. Referenced on Wolfram\|Alpha Characteristic Function Cite this as: Rowland, Todd and Weisstein, Eric W. "Characteristic Function." From MathWorld--A Wolfram Resource. Subject classifications Find out if you already have access to Wolfram tech through your organization
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189134	https://www.youtube.com/watch?v=xkFsBKf_Mbg	Work Expressions for POLYTROPIC Processes in 10 Minutes! Less Boring Lectures 101000 subscribers 639 likes Description 36703 views Posted: 6 Jun 2022 Polytropic Processes Isobaric, Isothermal, Isochoric, Adiabatic Ideal Gas Equation Work Equations for Polytropic Process 0:00 Work Equations So Far 0:28 What Has Been Covered 1:13 Polytropic Process Definition 1:30 Polytropic Relation Derivation 3:49 Implication for Two States 3:58 Isobaric, Adiabatic, and Isochoric 4:58 PV Diagrams for Iso-Processes 5:43 Work Equations Derivation 7:03 Isothermal Case 7:39 Example for Isothermal Case 8:20 Solution Example 1: Example 2: Example 3: coming August 2022 Previous Lecture: 12. Specific Heat and Incompressible Substances: Next Lecture: 14. Mass and Volumetric Flow Rate: Other Thermodynamics Lectures: Thermodynamics Intro and Units Review: Zeroth Law of Thermodynamics: Closed vs. Open Systems & Intensive vs. Extensive Properties: First Law and Energy Types: Efficiency in Thermodynamic Systems: Five Regions of T-v Diagrams: Interpolating and Quality Derivation: Enthalpy and Internal Energy: Compressed Liquid Properties: Ideal Gas and Compressibility Factor: Specific Heats: Specific Heat Relationships of Ideal Gases and Incompressible Substances: Polytropic Process Work: Mass and Volumetric Flow Rate: Flow Work and Energy Conservation in Open Systems: Pipe Flow, Nozzles, Diffusers, and Throttling Devices: Turbines, Compressors, and Pumps: Mixing Chambers and Heat Exchangers: Transient Systems: Second Law Intro and Power Cycles: Refrigerators and Heat Pumps: Carnot Cycles and Reversible Processes: Entropy as a Thermodynamic Property: Reference Entropy and Specific Heats: Isentropic Efficiency: Brayton Cycle: Intercooling, Reheating, and Regenerators: Otto Cycle: Standard Diesel Cycle: Rankine Cycle: . . . . Other Engineering Courses (Playlists): Statics: Mechanics of Materials: Machine Design: SolidWorks (CAD): Fluid Mechanics: 24 comments Transcript: Work Equations So Far Up to this point, we only know how to calculate the work of a thermodynamics process for two scenarios: the first one, if we know or calculate the internal energy and heat transfer from our conservation of energy equation (and KE and PE terms if any), and solve for W, and the second, W = PΔV, in the cases where the pressure in our process is constant. In this lecture, we’ll learn how to calculate work in variable pressure systems. What Has Been Covered In the Enthalpy and Internal Energy 12-minute lecture, link below, we concluded that for constant pressure processes, since the change in internal energy is equal to the heat transfer minus the work done (when neglecting other types of energy), the work is equal to the integral between states of P dV. And we went over how, for these constant pressure processes, that integral results in just P (V2-V1) or Pm(v2-v1), and how we can even skip this, if we’re calculating the heat transfer, and find the total heat transfer with the enthalpy values, H2-H1 (or m(h2-h1)). But this is true for constant pressure or non-constant pressure processes. Polytropic Process Definition When pressure varies during a process, we call it a polytropic process, and the expression that allows us to solve for unknown properties of an ideal gas that undergoes a polytropic process is PV to the n equal to a constant. The n variable is known as the polytropic index. Polytropic Relation Derivation Let’s see where this equation comes from, and we’ll use several of the statements we made in the previous lecture, so make sure to watch that one before watching this: We know that for infinitesimal changes, our energy conservation equation, or 1st law of thermodynamics, results in dQ = dU + dW. The heat can be written as mc∙dT, with c being the specific heat at constant pressure, cp, or constant volume, cv (depending on the process, and therefore indetermined for now). And dU is equal to mcv∙dT from the definition of what cv is. The infinitesimal work is just PdV. We can simplify this equation as m(c-cv)dT= PdV. For an ideal gas, PV = mRT. The differential form of this equation would be PdV + VdP = mRdT. We can solve for dT and substitute it in the conservation of energy equation. Since the mass term cancels, we can now multiply by R on both sides, then distribute the parentheses, and since in the last lecture we proved that cp = cv + R, we can substitute these terms by a cp. We can now divide by (c-cv) to have a coefficient next to PdV, and also divide all terms by PV. Since the coefficient of c’s is constant, we can integrate (with those terms outside of the integral), to get the parenthesis times natural log of V plus natural log of P equal to a constant. If you remember basic exponentiation and log rules, you might remember the log-of-a-power rule where y ln(x) is the same as ln (x^y). We can therefore write the first term as ln(V^parenthesis). And you might also recall the product rule, where ln(x) + ln(y) = ln(xy). This means that the left hand side is ln of PV^parenthesis, and therefore this is equal to a constant. The parenthesis is what we call the constant n, and therefore, for a polytropic process of an ideal gas, PV^n is constant. Implication for Two States This also means that P1V1^n is equal to P2V2^n, equal to P3V3^n, etc. We’ll take a look at other cases in just a second, but what happens if we try to apply Isobaric, Adiabatic, and Isochoric this equation to an isobaric process? An isobaric process, or in other words, a process at constant pressure means that our general specific heat c, is cp. This would mean that n = 0. What about an adiabatic process, meaning that no heat comes in or out. In this case, since dQ is zero, and temperature changes can still occur, we would say that c is equal to zero. Substituting this zero, we see that n becomes cp over cv, or k, the specific heat ratio that we established in the previous lecture. For adiabatic processes this equation becomes PV^k = constant. For an isochoric process, constant volume, we would have the general c be cv, the specific heat at constant volume. In this case, the denominator would approach zero, and therefore n would approach infinity. PV Diagrams for Iso-Processes This is not very useful in terms of solving for values, but it does allow us to plot these processes in a PV diagram, where P is on the y axis, V on the x axis, and for n = 0 (isobaric process), the slope is zero, n = infinity is an infinite slope (or a vertical line), and an adiabatic process has a logarithmic slope equal to k. The first two make sense: n=0 is isobaric, meaning that the pressure never changes, and n=infinity is isochoric, meaning that the volume is always the same. We’ll look at an isothermal process in just a second. Just remember that depending on the path from state 1 to state 2, the shape of this curve can be described by different values of n. Work Equations Derivation The reason we use this relationship is because we need to know P in terms of V, so that we can integrate Pdv, and calculate work. Substituting P in terms of V, in the integral, looks like this, or with the V in the numerator and the constant outside of the integral, we get V to the minus n. After integrating this expression, we find an equation for W in terms of the constant. And since the constant is P1V1^n or P2V2^n, we can distribute it in our expression and substitute it for P2V2 in the first term, and P1V1 in the second term, and use the product exponentiation rule to find the expression for work of a polytropic process in terms of P and V only . This expression is only true for n different than one; and not because we would run into some mathematical issues dividing by zero, but because if it had been one from the beginning, the integral would have been PV ln of V2 over V1. This is the expression for work of a polytropic process when n is equal to 1. And notice that P and V can be that for any state, P1V1, P2V2, P3V3… since the constant is the same for any state. Isothermal Case What does it mean that n is equal to 1. Well, if P1V1 is equal to P2V2, from our ideal gas equation this is only true if the right hand side of the P times V equation is constant. The right hand side is Rbar T1 for the first one, and Rbar T2 for the second one. If these two are equal, it means that T1 and T2 are the same. Therefore, the big conclusion here is that when isotropic index, n, is equal to one, the process is an isothermal process, or a process where temperature is kept constant. Example for Isothermal Case Let’s make use of these expressions with a simple example, and if you want to see more complex examples, make sure you check the example videos for this topic in the description below. Two kg of air at 130 kPa and 22°C is contained within a piston cylinder. It is compressed to a final pressure of 550 kPa. During compression, heat is transferred from the air and the temperature remains constant. Determine the work input required. The molecular mass of air is 28.97 kg/kmol. As always, pause here and try to solve this problem before watching the solution up next. Solution To solve for the work, we know we have the two options we derived today. One for when the polytropic index is not one, and one for when the index is in fact one. And since we’re being told that the temperature remains constant, meaning an isothermal process, we know n = 1. We have P1, P2, and T1, and again, we’re trying to find the work, which from these equations we see that we first need V1 and V2. To find V1 first, we can multiply the specific volume v1 times the given mass, and to find v1, we use the ideal gas equation. Remembering that R is R bar over M, with M for air, and that the temperature has to be input in Kelvin, not Celsius, we substitute the pressure to find the specific volume of state 1 in m3/kg. And like I said, since we were given the mass, we find V1. Now, since we do know that n = 1, then P1V1 = P2V2, and V2 is equal to P1V1/P2. With this, we can find the volume at state 2. And finally, we use the expression for work that we derived today. And it doesn’t matter if we use P1V1, or P2V2 here; Since n=1 they are both the same. This negative value we get for the work is consistent with what we know about heat and work. Since it’s a compression process, we need to put work into the system to compress it. And our convention was that work “in” must be negative. If you want to check out other examples on this topic where we use the other expression we derived here today, or the other lectures of the thermodynamics course and lectures of other engineering courses, make sure to check out the links in the description of this video. Thanks for watching!
189135	https://teachingintheheartofflorida.com/2022/09/how-to-teach-division-with-equal-groups.html	How to Teach Division With Equal Groups - Teaching in the Heart of Florida Skip to consent choices Privacy preferences We use cookies and similar technologies on our website and process your personal data (e.g. IP address), for example, to personalize content and ads, to integrate media from third-party providers or to analyze traffic on our website. Data processing may also happen as a result of cookies being set. We share this data with third parties that we name in the privacy settings. The data processing may take place with your consent or on the basis of a legitimate interest, which you can object to in the privacy settings. You have the right not to consent and to change or revoke your consent at a later time. For more information on the use of your data, please visit our privacy policy. You are under 16 years old? Then you cannot consent to optional services. Ask your parents or legal guardians to agree to these services with you. Continue without consentSet privacy settings individuallyAccept all Privacy policy • GDPR Cookie Consent with Real Cookie Banner Top Teaching in the Heart of Florida Home Blog Shop Freebies About Contact How to Teach Division With Equal Groups Differentiating instruction is a key part of teaching and is especially important when teaching division with equal groups. Division can be tricky for students to master, but with the right approach, you can help them understand and learn this important skill. In this post, we’ll discuss ways to differentiate division instruction so all students can succeed. Keep reading to learn more! Make sure you grab the FREE resource at the end of this post! The Basics of Division: Equal Groups Dividing into equal groups is a core division strategy that students should learn. When dividing using equal groups, the division problem is represented by dividing objects into equal parts. For example, if there are ten students in a classroom and twenty apples, each student would get two apples. In other words, when the division problem is represented by dividing into equal groups, the answer will always be the number of groups. Differentiating when teaching students about equal groups can be as easy as giving students different numbers of objects to divide. Hands-On with Manipulatives Differentiating division instruction can also be as easy as allowing students to use math manipulatives whenever needed. Division is about equal groups, so manipulatives like pattern blocks or even Legos can help illustrate this concept. For students struggling with division, offering the chance to use manipulatives can make a big difference. With manipulatives, they can physically see and touch the division process, which can help them to better understand what division is and how it works. Allowing students to use manipulatives is a simple way to differentiate division instruction, but it can be incredibly effective. Using Arrays to Make Equal Groups If you’re teaching division, there’s a good chance you’ll be using arrays at some point. An array is a way of visually representing division, and it can be helpful for students to see division in this way. An array is a division problem that has been turned into a picture. An array is a math model that can be helpful for students who are having trouble understanding division or for those who are struggling to see division as anything other than finding the answer to a problem. Arrays can also be helpful for differentiation. Using arrays, you can create division problems tailored to your students’ needs. For example, you can create arrays with larger numbers of objects for your more advanced students or with smaller numbers of objects for your struggling students. No matter your students’ needs, arrays can be a helpful tool for division. Assessing Student Mastery with Equal Groups Teachers can assess student mastery with division and equal groups in a few different ways: Give students exit tickets with division problems to solve and have them explain their thinking. Have students work in pairs or small groups to solve division problems and then discuss their solutions with the class Have students create math models or drawings to solve division problems. By using a variety of assessment methods, teachers can get a clear sense of how well their students understand division and equal groups. Using Math Groups to Differentiate for Learners Teachers know that our students don’t learn at the same pace. That’s where math small groups come in – they help ensure that students are given a chance to work on division concepts that they are struggling with. As you move through your division instruction, these small groups can provide extra support and practice for the students who need it. At the same time, the rest of the class continues learning new skills and strategies like division error analysis. Two things always come in handy when working with small groups—manipulatives and these Division Worksheets. The first thing is the most obvious: manipulative such as blocks or counting aids. Some students need this concrete way of understanding how to divide. Plus, they keep them engaged while they are learning! When students are ready to apply their division learning on paper, these Division Worksheets have a variety of ways for them to practice any skills-level gaps. Plus, it is easy for you to assist students and then assess them afterward. Grab these FREE Division Mats for your Small Groups! The most valuable resource that all teachers have is each other. Without collaboration our growth is limited to our own perspectives. 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189136	https://brainly.com/question/52830502	[FREE] Use your graphing calculator to solve the equation graphically for all real solutions. x^3 - 5x^2 + 2x + - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +34k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +22,2k Ace exams faster, with practice that adapts to you Practice Worksheets +8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Use your graphing calculator to solve the equation graphically for all real solutions. x 3−5 x 2+2 x+12=0 Solutions: x=□ Make sure your answers are accurate to at least two decimal places. 1 See answer Explain with Learning Companion NEW Asked by misswright918 • 10/04/2024 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1183155682 people 1183M 0.0 0 Upload your school material for a more relevant answer To solve the equation x 3−5 x 2+2 x+12=0 graphically, you can use a graphing calculator. Here’s how you can do it step-by-step: Enter the Equation: Turn on your graphing calculator and go to the function mode, usually labeled as "Y=". Enter the equation y=x 3−5 x 2+2 x+12 into one of the function slots. Graph the Equation: After entering the equation, press the graph button, commonly labeled as "GRAPH", to display the graph of the equation on the screen. Find the X-Intercepts: The solutions to the equation x 3−5 x 2+2 x+12=0 are the x-values where the graph crosses the x-axis. These are also known as the roots or zeros of the function. Zoom and Adjust: If necessary, use the zoom feature to adjust the viewing window so that the x-axis intercepts are clearly visible. You can use "Zoom In" or "Zoom Out" to help see where the graph crosses the x-axis better. Calculate the Solutions: Use the calculator’s feature, often called “Calc” or something similar, and select “zero” or “root” to find the precise x-values where the graph intersects the x-axis. Follow the prompts to move the cursor near each intercept and confirm to find exact x-values. Record the Real Solutions: The graph should show the x-intercepts. Based on the calculations, these solutions are approximately: x=3.0 x=−1.2 x=3.2 Therefore, the real solutions to the equation x 3−5 x 2+2 x+12=0 are x=3.0,−1.2,3.2. These are the points where the graph of the function crosses the x-axis. Answered by GinnyAnswer •8M answers•1.2B people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) 0.0 0 Celestial Mechanics - Jeremy Tatum Chemical Process Dynamics and Controls - Peter J. Woolf Statistics - Barbara Illowsky, Susan Dean Upload your school material for a more relevant answer To solve the equation x 3−5 x 2+2 x+12=0 graphically using a graphing calculator, input the equation, graph it, and find the x-intercepts. The real solutions are approximately x=3.00,−1.20,−2.80. Use the calculator's features to pinpoint these roots accurately. Explanation To solve the equation x 3−5 x 2+2 x+12=0 graphically using a graphing calculator, follow these steps: Enter the Equation: Turn on your graphing calculator and access the function mode, usually labeled 'Y='. Input the equation as y=x 3−5 x 2+2 x+12. Graph the Equation: After entering the equation, press the graph button to display the function's graph on the screen. Find the X-Intercepts: The solutions to the equation are the x-values where the graph crosses the x-axis (where y=0). Zoom and Adjust: If needed, use the zoom feature to adjust the viewing window until the x-intercepts are clear and visible. Calculate the Solutions: Use the calculator’s calculation feature (often called 'Calc'), and select 'zero' or 'root' to find the x-values. Move the cursor near each intercept and confirm to find the precise values. Record the Real Solutions: The graph should indicate the x-intercepts. For the equation x 3−5 x 2+2 x+12=0, the approximate solutions are: x≈3.0 x≈−1.2 x≈−2.8 Therefore, the real solutions to the equation are approximately x=3.00,−1.20,−2.80. Examples & Evidence An example of using this method includes entering other polynomial equations into the graphing calculator and applying the same process to find their roots, which helps develop problem-solving skills. For instance, if you have a different cubic equation, simply follow the same graphing steps. Graphing calculators reliably show polynomial equations, and the technique of finding x-intercepts for roots is standard in mathematics. Using graphical methods is widely accepted in solving polynomial equations. Thanks 0 0.0 (0 votes) Advertisement misswright918 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 5.0 Use your graphing calculator to solve the equation graphically for all real solutions x^3-4x+2x+4=0 Solutions: x = Make sure your answers are accurate to at least two decimals Community Answer Use your graphing calculator to solve the equation graphically for all real solutions: x^3 - 6x^2 + 2x + 24 = 0 Solutions: x = Community Answer Use your graphing calculator to solve the equation graphically for all real solutions 23 6.x2 + 4x + 16 = 0 Solutions: 2 = Make sure your answers are accurate to at least two decimals Community Answer Use your graphing calculator to solve the equation graphically for all real solutions x3-6x2+2x+24=0 Solutions: x = Make sure your answers are accurate to at least two decimals? Community Answer Use your graphing calculator to solve the equation graphically for all real solutions 3 _ 52 +2+15 = 0 Solutions: a = Make sure your answers are accurate to at least two decimals Video Message instructor Question Help: Calculator Submit Question Community Answer Use your graphing calculator to solve the equation graphically for all real solutions. x³-7x² +9x+5=0 Solutions: x= Community Answer You can retry this question below Use your graphing calculator to solve the equation graphically for all real solutions 2³ +0.82²- 21.35 - 21.15 = 0 Solutions = -4.03290694 X Make sure your answers are accurate to at least two decimals Question Help: Message instructor Post to forum Submit Question Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD New questions in Mathematics Evaluate the integral ∫1−3 c o s x s i n xd x Find the distance between the pair of points (2 2,5) and ( −2,3 5 ). Which of the following is an odd function? A. f(x)=3 x 2+x B. f(x)=4 x 3+7 C. f(x)=5 x 2+9 D. f(x)=6 x 3+2 x Find f+g,f−g,f g and g f. Determine the domain for each function. f(x)=5 x−6,g(x)=x+4 Find the distance between the pair of points (1.5, 8.2) and (-0.5, 6.2). 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189137	https://www.vaia.com/en-us/textbooks/math/geometry-1-edition/chapter-4/problem-38-prove-that-a-triangle-can-have-at-most-one-obtuse/	Problem 38 Prove that a triangle can have, ... [FREE SOLUTION] \| Vaia Find study content Learning Materials Discover learning materials by subject, university or textbook. 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LoginSign up Log outGo to App Go to App Learning Materials Explanations Explanations Anthropology Archaeology Architecture Art and Design Bengali Biology Business Studies Greek Chemistry Chinese Combined Science Computer Science Economics Engineering English English Literature Environmental Science French Geography German History Hospitality and Tourism Human Geography Italian Japanese Law Macroeconomics Marketing Math Media Studies Medicine Microeconomics Music Nursing Nutrition and Food Science Physics Polish Politics Psychology Religious Studies Sociology Spanish Sports Sciences Translation Textbooks Textbooks Biology Business Studies Chemistry Combined Science Computer Science Economics English Environmental Science Geography History Math Physics Psychology Sociology Features Flashcards Vaia AI Notes Study Plans Study Sets Discover Find a job Find a degree Student Deals Magazine Mobile App All Textbooks Math Geometry Chapter 4 Problem 38 Geometry StemEZ Math 1 Edition Chapter 1 13 Chapter 2 2 Chapter 3 1 Chapter 4 13 Chapter 5 5 Chapter 6 11 Chapter 7 8 Chapter 8 11 Chapter 9 2 Chapter 10 19 Chapter 11 10 Chapter 12 31 Chapter 13 16 Chapter 14 9 Chapter 15 8 Chapter 16 7 Chapter 17 12 Chapter 18 8 Chapter 19 11 Chapter 20 13 Chapter 21 18 Chapter 22 12 Chapter 23 29 Chapter 24 12 Chapter 25 3 Chapter 26 4 Chapter 27 15 Chapter 28 12 Chapter 29 22 Chapter 30 10 Chapter 31 2 Chapter 32 3 Chapter 33 3 Chapter 34 6 Chapter 35 12 Chapter 36 11 Chapter 37 20 Chapter 38 5 Chapter 39 10 Chapter 40 10 Chapter 41 15 Chapter 42 26 Chapter 43 18 Chapter 44 42 Chapter 48 19 Chapter 49 17 Chapter 50 14 Chapter 51 12 Chapter 52 9 Short Answer Step-by-step Solution Key Concepts Chapter 4: Problem 38 URL copied to clipboard! Now share some education! Prove that a triangle can have, at most, one obtuse angle. Short Answer Expert verified To prove that a triangle can have, at most, one obtuse angle, we assume that a triangle has two obtuse angles, say angle A and angle B. Since both angle A and angle B are obtuse, their sum will be greater than 180 degrees. Adding angle C to the sum of angle A and angle B, we have: A+B+C>180°+C. However, we know that the sum of the angles in a triangle is equal to 180 degrees, which means A+B+C=180°. But from our calculation, we concluded that A+B+C>180°+C, which is a contradiction. Therefore, our assumption that a triangle can have two obtuse angles is false, and a triangle can have, at most, one obtuse angle. Step by step solution 01 Sum of angles in a triangle Recall that the sum of the interior angles of a triangle is always equal to 180 degrees. This will be an important fact to help us prove that a triangle cannot have more than one obtuse angle. 02 Assume a triangle has two obtuse angles We will use proof by contradiction. Let's assume that a triangle has two obtuse angles, say angle A and angle B. 03 Add the two obtuse angles Now, let's add the two obtuse angles together. An obtuse angle is an angle greater than 90 degrees. Since we have assumed that both angle A and angle B are obtuse, their sum will be greater than 180 degrees. Mathematically, this can be represented as: A+B>180° 04 Add the third angle Now, let's consider the third angle of the triangle, angle C. It must also be greater than 0 degrees, as it is an interior angle of a triangle. Adding angle C to the sum of angle A and angle B, we have: A+B+C>180°+C 05 Sum of angles in triangle contradiction However, we know that the sum of the angles in a triangle is equal to 180 degrees. This means that: A+B+C=180° But from step 4, we concluded that A+B+C>180°+C, which is a contradiction. Therefore, our assumption that a triangle can have two obtuse angles is false. 06 Conclusion After proving that a triangle cannot have two obtuse angles, it directly follows that a triangle can have, at most, one obtuse angle. Unlock Step-by-Step Solutions & Ace Your Exams! Full Textbook Solutions Get detailed explanations and key concepts Unlimited Al creation Al flashcards, explanations, exams and more... Ads-free access To over 500 millions flashcards Money-back guarantee We refund you if you fail your exam. Start your free trial Over 30 million students worldwide already upgrade their learning with Vaia! Key Concepts These are the key concepts you need to understand to accurately answer the question. Sum of Interior Angles Understanding the sum of interior angles is fundamental when studying triangles in geometry. Every triangle has three angles, and if you add these angles together, the total will always be 180 degrees. This rule is not just a random fact; it is a cornerstone of triangular geometry and is true no matter what type of triangle you're dealing with—whether it's scalene, isosceles, or equilateral. The angle sum property is essential for solving many problems involving triangles, including finding missing angles and proving potential relationships between angles. One common exercise is to demonstrate that a triangle cannot have more than one obtuse angle, which relies heavily on this angle sum property. This rule provides structure to our understanding of triangles and limits what is geometrically possible within their boundaries. Proof by Contradiction Proof by contradiction is a powerful mathematical tool used to establish the truth of a statement by showing that assuming the opposite leads to an inconsistency or a paradox. This approach is also known in mathematics as a reductio ad absurdum proof. To apply this method, we first assume the negation of what we want to prove. If this assumption leads to an impossible situation or contradicts a well-established fact, we can conclude that the original statement must be true. In the context of geometry, this technique helps us show that certain geometric conditions are not feasible, such as the impossibility of a triangle having more than one obtuse angle, which we'll explore in relation to other key concepts here. Obtuse Angle Definition An obtuse angle is one of the several types of angles found in geometry, characterized by its measure being greater than 90 degrees but less than 180 degrees. Here's a quick overview: An obtuse angle creates what is known as an obtuse-angled triangle when it's one of the three interior angles of a triangle. Since the total sum of angles in any triangle is 180 degrees, having an obtuse angle significantly restricts the range of possible measures for the other two angles. The concept of an obtuse angle is essential not just in triangle geometry but also in understanding more complex shapes and the behavior of angles in various contexts. Interior Angles of a Triangle Diving deeper into the properties of a triangle, the interior angles are those found between the sides of a triangle. Here are two essential points to remember: The interior angles of a triangle always add up to 180 degrees, as established earlier. Each of these angles can be acute (less than 90 degrees), right (exactly 90 degrees), or obtuse (greater than 90 degrees), but not more than one angle can be obtuse as that would violate the angle sum property of triangles. When examining or proving properties related to triangles, this understanding is critical in determining the shape and nature of the triangle in question. Geometry Contradiction A geometry contradiction occurs when an assumption leads to a conclusion that defies established geometric principles or previous findings. This often signals an error in reasoning or in the initial assumption. Contradictions play a crucial role in proof by contradiction; they allow us to invalidate a false supposition and thereby confirm the truth of the proposition in question. In our specific case of a triangle with more than one obtuse angle, the contradiction arises when this assumption results in a total angle sum that exceeds 180 degrees—something that is impossible for a triangle. Recognizing and understanding contradictions helps build solid logical foundations in geometry and is crucial for mastering geometric proofs. Chapters Chapter 1 13 Chapter 2 2 Chapter 3 1 Chapter 4 13 Chapter 5 5 Chapter 6 11 Chapter 7 8 Chapter 8 11 Chapter 9 2 Chapter 10 19 Chapter 11 10 Chapter 12 31 Chapter 13 16 Chapter 14 9 Chapter 15 8 Chapter 16 7 Chapter 17 12 Chapter 18 8 Chapter 19 11 Chapter 20 13 Chapter 21 18 Chapter 22 12 Chapter 23 29 Chapter 24 12 Chapter 25 3 Chapter 26 4 Chapter 27 15 Chapter 28 12 Chapter 29 22 Chapter 30 10 Chapter 31 2 Chapter 32 3 Chapter 33 3 Chapter 34 6 Chapter 35 12 Chapter 36 11 Chapter 37 20 Chapter 38 5 Chapter 39 10 Chapter 40 10 Chapter 41 15 Chapter 42 26 Chapter 43 18 Chapter 44 42 Chapter 48 19 Chapter 49 17 Chapter 50 14 Chapter 51 12 Chapter 52 9 One App. One Place for Learning. All the tools & learning materials you need for study success - in one app. Get started for free Most popular questions from this chapter Prove that the base angles of an isosceles right triangle have measure 45∘.Prove that if a triangle has no 2 angles congruent, then it is scalene.The measure of the vertex angle of an Isosceles triangle exceeds the measure of each base angle by 30∘. Find the value of each angle of the triangle.Show that the angle bisectors of a triangle are concurrent at a point equidistant from the sides of the triangle.Prove that an equilateral triangle has three equal angle. See all solutions Recommended explanations on Math Textbooks Discrete Mathematics Read Explanation Theoretical and Mathematical Physics Read Explanation Pure Maths Read Explanation Calculus Read Explanation Applied Mathematics Read Explanation Decision Maths Read Explanation View all explanations What do you think about this solution? We value your feedback to improve our textbook solutions. 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189138	https://www.youtube.com/watch?v=Vt0yAVd4Uj4	When Sudoku Segments Sum To 10... Cracking The Cryptic 665000 subscribers 1169 likes Description 31650 views Posted: 14 Oct 2024 TODAY'S PUZZLE After the trauma of the Young Sheldon finale, Simon pulls himself together enough to attempt Black_Doom's outstanding puzzle, Tenfold. This sits at a 99% approval rating (and average difficulty) and it features some absolutely lovely logic. Play the puzzle at the link below: Rules: Normal sudoku rules apply. Draw a one-cell-wide loop of orthogonally connected cells. The loop can't touch itself, not even diagonally. Additionally, the loop consists of non-overlapping segments, each of which sums to 10. Digits may repeat on the loop and even within an individual segment, if allowed by the other rules. The digit in a caged cell shows how many (of the up to 8) surrounding cells are on the loop. The caged cells cannot be on the loop. The number in the top-left corner of a caged cell shows the sum of digits in the loop cells around it. (Tonight's music intro is Homeward Bound by Simon & Garfunkel.) NEW SUDOKU HUNT COMPETITION RELEASED 1 OCTOBER October's sudoku hunt competition launched yesterday and many of you have already told us you're loving it! An approachable, Halloween-themed, Japanese Sum-fest from The Paint By Numbers Institute called Caroline's Halloween Adventure. Check it out for as little as $2/month here: NEW STREAM COMING UP TOMORROW We solved more puzzles from our 600k Subscriber Pack on Tuesday. If you missed it, check out the link below: STRATFORD LITERARY FESTIVAL: LIVE CRACKING THE CRYPTIC If you're interested in watching a live episode of Cracking The Cryptic and meeting us in real life then there is a link to the live show we're doing on 27 October at 10am at the Stratford Literary Festival below. We aren't sure how many tickets are available so please do book early if you are keen to go. ▶ SUDOKU PAD - Use Our Software For Your Puzzles ◀ You can input classic sudoku puzzles into our software and help support Sven, the programmer responsible for the wonderful user interface we all use to play these puzzles everyday. The app also comes with 12 handmade puzzles from us: iOS: Steam: Android: ALSO on Amazon: Search for “SudokuPad” ▶ Contents Of This Video ◀ 0:00 Intro music 0:25 Young Sheldon 1:05 Puzzle intro 1:51 Simon's T-Shirt 2:36 Tametsi stream tomorrow 3:06 The Pinball Hall Of Fame 4:28 Work colleagues 5:20 Fog-Of-War yesterday 5:48 Rules 9:08 Start of Solve: Let's Get Cracking ▶ Contact Us ◀ Twitter: @Cracking The Cryptic email: crackingthecryptic@gmail.com Our PO Box address: Simon Anthony & Mark Goodliffe Box 102 56 Gloucester Road London SW7 4UB (Please note to use our real names rather than 'Cracking The Cryptic'.) ▶ CTC FAN DISCORD SERVER◀ ▶ OUR BACK CATALOGUE – ALL CATEGORISED WITH LINKS!◀ ▶ NEW CRACKING THE CRYPTIC MERCHANDISE◀ ▶SEND US PUZZLES TO SOLVE/CONTACT US◀ crackingthecryptic@gmail.com ▶FOLLOW US◀ Twitter: @crypticcracking Instagram (for how to solve daily clues from The Times): We also post the Wordle In A Minute videos on TikTok. sudoku #puzzle #maths 117 comments Transcript: Intro music [Music] I'm sitting in a station got a ticket for my destination on a of one night stands my suitcase and guitar every is hello and welcome to Monday's edition Young Sheldon of cracking the cryptic um where I V I've V just recovered to do this video after the trauma the emotional trauma of last night when it was the premiere on British TV I think I think it's been out in America for a while but it was the premiere of the last two episodes ever of young Sheldon of which I'm a big fan uh fell in love with the Big Bang Theory a few years ago um but I wasn't expecting a sitcom to have such an emotional effect on me no spoilers I won't I won't reveal anything about what happens in those last two episodes but Goodness Me Goodness Me it left me in pieces anyway anyway I'm I'm sort of recovered now and I'm going to be trying to solve a puzzle for you this is called tenfold by black doom and I don't think Puzzle intro I've ever done a black Doom puzzle on the channel I think Mark has I think Mark's done three or four black Doom puzzles but this puzzle has been recommended many times because it has a 99% approval rating on logic Masters Germany um only three stars out of five for difficulty and apparently it's rather wonderful the idea is we're going to draw a loop in the grid and then the loop we draw we should be capable of dividing it into segments Each of which sums to 10 I think that's basically the idea um so that is what I'm going to try and do in a couple of moments time um but before I do that let me tell you a couple of things um oh the first thing's a little bit embarrassing I I mean I have to tell you Simon's T-Shirt it um Mark has been busy of you asked believe it or not for a t-shirt um saying that you're one of my favorite people everybody who watches the channel is one of my favorite people occasionally I've been known to share a secret on that basis um but there is now a t-shirt available um and um yeah it's available in different colors I'll put a link under the video and just in case anyone is interested um I know some of you have asked for it so I I feel I feel I feel um strange about it being there but it is it is there if anyone wants to see it um then tosy we're going to be having a Tametsi stream tomorrow go at a game tomorrow night on stream uh 10 p.m. UK time we'd love to have your company um and let me show you a picture this game looks really cool um it's sort of a hexels variant I think but with different sizes of shapes um and um yeah apparently it's really good so um yeah we'd love to see you I'll try and remember to put a link on the screen to that um I've got no birthdays to do today which is very peculiar but I've The Pinball Hall Of Fame got a couple of shoutouts I'll start off by with a thank you to Jim and Veronica um Veronica wrote to me and I I said happy birthday to Jim uh on in Friday's video uh and I mentioned that they're from Las Vegas I mentioned one of the my great Joys whenever I go to Vegas is this pinball h of Fame Museum and literally you go into this great big room and it is just covered everywhere with pinbal machines like really early pimble machines and the latest pimble machines and everything in between so um I have grown up loving pinball since I was a small child um I I do actually own a pinball machine Star Trek the Next Generation Um the greatest pinball machine of all time I will broke it no argument on that I mean there's The Adams Family is good I I I will you know I will accept that that is good but the Star Trek the Next Generation is the best pinball machine of all time um and but you can play all of these wonderful games there and uh look they they sent me a picture they were going past it on their way home which is um it must be great gosh if I if I live there i' I probably go there every day anyway thank you Jim thank you Veronica and then um not a birth but a shout out um a shout out to Work colleagues Arie from your work colleague Kelsey um and I think the two of them were discussing a love of sodoku and found that they both watched cracking the cryptic and indeed AR's wife is is um one of our patrons as well so many thanks to all of you for watching and um uh Kelsey mentioned Ari that uh she's very grateful for your support at work so lovely to hear that and um Yeah a different sort of announcement sounds like sounds like an announcement or sounds like there could be some cake involved just to celebrate the fact that people have found cracking the cryptic independently and it then touches on their real lives anyway that's all the news I think I think I think that's all the news let's let's have a go at solving a puzzle oh no there was one other thing I was meant to Fog-Of-War yesterday mention which is yesterday's puzzle is proving very popular uh it's a sort of fog of War uh by ket's dad restricted access is its name and lots of you have been trying that and having a lot of fun with it I I loved solving it um last night so do have a go at that as you can see there is some fog in the grid which is a very popular Innovation or mechanic in the world of soku now but anyway let's have a look at tenfold by black Doom these are the rules of today's Rules puzzle what we've got is normal sidoku rules apply so we've got to put the digits 1 to n once each in every row every column and every 3x3 box then it says 10 Loop draw a one cell wide Loop of orthogonally connected cells what does that mean well a loop uh that's orthogonally connected would not look like this because orthogonally connected means shares an edge and you can see that these two cells don't share an edge we can make the loop orthogonally connected by adding those squares in and creating a fisl ring but we were but uh if this does turn out to be the loop we're looking for that is slightly slightly fortunate but anyway so we've got to make a loop like that uh the loop can't touch itself not e oh not even diagonally so that means you can't or The Loop that we draw can't look something like that because although that Loop is an orthogonally connected Loop you can see these two squares do touch one another diagonally and that's too naughty for words so we'd have to shift it round to be something more like that um yeah and then the loop consists of one or more contiguous groups of cells Each of which sums to 10 in fact let's go back so it ought to if this was our Loop oh it says these groups of cells cannot overlap well that's the same thing and digits oh digits May repeat within sums so let's try and make it so that if we let's say that we divide the loop up like I don't know this for example so the idea would be that each stretch of red each Red Line contains uh digits that sum to 10 and this it is possible for those two squares to be the same number and that's fine providing normal sidoku rules are obeyed so this is a possible Loop and then there's one more rule which says sweeper uh digits in a caged cell so there are a few of those sprinkled around um show how many the surrounding cells up to eight are on the loop so what you do so in this case this would be a two because in those eight cells two of the cells are on the loop the cage cells cannot be on the loop so that's going to be important and the number in the top left corner of a cage shows the sum of the digits in the loop cells around it okay so this is this is lovely so if if this was a two then these two cell oh well then then I've broken the puzzle because these two squares are meant to sum to 10 and therefore that would have to be a zero according to my red linage so that that defin this definitely isn't the correct Loop um but that's how it would work so let's lose a different one if this this would be saying those three oh in fact this isn't meant to be on the loop but let's say it could be these three cells would be summing to 17 is how it works but this what I've managed to draw here is absolutely definitely wrong so let's get rid of it all of it um but do have a go the way to Start of Solve: Let's Get Cracking play is to click the link under the video as usual but now I get to play Let's get cracking now the first thing we oh it doesn't work I was going to try and doubleclick the cages because I wanted to indicate somehow that these are all off the loop so off the loop we'll have light green shall we um so okay so there's some things we can do if this cell was on the loop that it wouldn't be a loop anymore it would be a cold attack it would go in there and it would stay there so that can't be on the loop yes and the same thing okay the same thing is going to be true down here uh so this can't be on the loop or The Loop would be cold disect and this requirement that the loop can't touch itself um which I'm sure I red I'm just looking at yeah it can't touch itself not even diagonally so what that means is if the loop ever does hit the edge of the grid it has to do what I'll call a wide turn it's got a wide turning Circle because if it does hit the edge say this was the loop and it went and it turned could it turn again could it sort of do a very quick turn no because if it did do that you can see that the loop the loop is touching itself here and that's not it's a one cell wide loop it's not allowed to to do that so if it does come into the edge and turn it must sort of do a wide turn it's going to be something more like the these patterns that I'm drawing um and what that means is the loop can't come down here look because it would have to do a tight turn and it's not it's not allowed to [Music] um now I get a given digit that's very peculiar so maybe that's affecting this is it um this yes okay that there's a maximum of three cells that could be on the loop that are connected to this cell but it can't be a three because of sidoku so this digit is a one or a two now can it be a one is that actually maybe it can be a one uh if it was a one we would be saying one exactly one of these three cells is on the loop so it couldn't be this one because that would be a turning and there would be two cells on the loop and it couldn't be this one because that would be a turning so it would have to be that one and the loop would have to do that so it didn't take a second cell around the five clue and that would be a five and that would presumably then contribute towards the 14 Clues total okay what about the opposite then if this is a two so isn't this always on the loop that might be wrong but I that feels right because if this is a one this is the only cell around these three cells that's on the loop if it's a two we can't construct the two by making it those two squares with that one not being on the loop because these would both be C dect and that's naughty we can't do that so this is always on the loop and the reason that might be important is that tells us that the loop is going through this little you know these are sort of Siler and cibus aren't they and the loop is going through the through the middle [Music] um it's not actually the 10 or even the 21 that's that's okay the 21 looks that looks forced actually right sorry I should have started here okay there are a maximum of four cells around that 21 so this could be a four but it can't be a three because of sodoku and it can't be a two because then say it was only those two cells around the 21 added up to 21 you can't make two different sedu numbers add up to 21 the maximum would be a 9 and an8 which would sum to 17 so what that means is this is a four and we can write four in and remember the loop the Lop a loop by its nature is continuous so we can never stop the loop the loop can never be cldis Act and the loop can never touch itself oh whoa whoa whoa whoa whoa in fact the loop can never touch itself not even diagonally so all of those squares have to be green don't they because the loop couldn't even do that that would be naughty so we've got to be very look we've got loads of loop all of a sudden well yes that's got to go there that's doing something with this um but let me let me just let me just think about this for a second or two um this is four so these four digits sum to 21 that that's that that's actually weird that is weird I tell you why that feels weird to me is because we've got this requirement that somehow or other we're going to break this Loop into pieces Each of which sums to 10 now H I think that means that the the pieces well I think that means that what we can't do is that or that in in generating our pieces no I realize this is that's that's a slightly strange way of putting it so so we're going to divide the loop into segments now let's imagine that one segment so this this sort of divided one segment from another so there's some segment coming down here and there's a segment starting here can that be possible and I think the answer must be no because a 10 sum by its nature cannot be less than two cells long because soku numbers they the biggest one is nine so even if we put a nine here this would have to extend one cell to create a 10 sum wouldn't it so around this 21 clue if if we did have a if we did try and start the 21 from here we'd get a 10 sum which must be at least two so we could stop there we'd get another 10 sum and we could stop there but we can't create the extra one that we need we can't you cannot get to 21 in segments of 10 10 1 around this 21 clue there's no way to do that so that means that I think that means that's a 10 sum so because because having having their therefore deduced that no sum begins either in this square or this square the 10 sum that this Square belongs to must extend into box two and the 10 sum that this cell belongs to must extend on here we don't know how long these are yet but now we have a different question which is whoo how on Earth do we get to 21 at all then and the only way we're going to be able to get to 21 is if we Max out the value of those two squares isn't it I.E they are a 10 sum yeah we can't we can't extend this in here because then we're not going to have any you know the the maximum size of those two squares would then be nine the maximum size of those two squares would then be nine and n and n is not 21 fact you couldn't even do that anyway so you must have a 10 sum here which means okay that means those two squares summ to 11 by mathematics because 21 - 10 is 11 so these are not 47 well okay so now I actually do want to think so this is this is surprised me because um earlier on I wrote this or I put this digit in and I noticed this was this seemed to be defining a 10 sum and now it's not well it is but it's it's not so let imagine that the loop just continued to hear for a second and say this 10 sum was applying to those three squares they cannot be the 10 sum we're looking for can they because although they sum to 10 we know that you know we can't chop this up into a segment like that because that's going to break the the 21 clue so that is wrong even though there's going to be some digit that do some oh I see so maybe we could have no that doesn't work oh so okay sorry what I was thinking then is could I do something like this so that this wouldn't be a contiguous stretch of digits that would add to 10 I could put a digit here then these might only up to 9 or eight or something but that doesn't work either does it that doesn't work for a very pretty reason so in fact I yeah I think these all have to be green this is a very clever puzzle actually it's it's far deeper than you think um BEC and the reason I think these have to be green is let's say that let's say one of these or more were not Green Let's let's say this one was gray now in box two I have definitely got four different digits at least that sum to 10 well there is only one way of achieving that in sidoku if they're different digits and they would be they're all in box two they'd have to be 1 2 3 and four that is the only way of doing it so you certainly couldn't have any more cells that were gray because then we' could be getting into you know we'd have to add five and six to those 1 2 3 and four and we certainly couldn't make those out up to 10 but you could do it with just four digits except that I have to write in here the number of digits that are gray in that surrounding area and that would be four so I can't put 1 2 3 and four into here because this cannot then be the four that we're looking for so I think what that means is that those are all green and this is a three and that's very odd indeed because now I've got to be careful here because I don't know how long this segment is but I think those three digits do add up to 10 and they don't include four and they don't include three so I think they have to be 1 27 let me just think about that so 712 7 would be the biggest digit you can put in three cells that add up to 10 10 if you put 8 in 8 + 1 + 2 is 11 so it's too big if you put seven in it has to be 712 if you put six in it has to be 613 if you put five in it's either 523 or 514 and that doesn't work if you put four in you've got to repeat a digit like 442 and that's just silly so so it's 127 but what is this doing well ah ah this is very good I really like this okay there's many things that this is doing but I think the main thing it's doing is what we're saying is this is a segment this is a two cell segment remember this could have been longer well it's not longer now because imagine it was a three cell segment well what would this digit be and by my mathematics it would be the same as that digit because if if what we're saying is that these three digits add up to 10 and those three digits add up to 10 the implication there seems to be that those two have to be the same number because if this was a one and a two that would be a seven and that would be a seven for example it it just doesn't work and obviously if we extend if we extend this further and say oh well it's a four cell segment that sums to 10 well one of these digit I mean these he's already add up to 10 I'd have to write zero in there that won't work so I think I think what we're looking for is this and then well then these won't add up to 10 because we know all three of them add up to 10 so so something's going on oh hang on oh well that's that's very interesting as well so the loop goes here but this can't be a three cell segment for the same reason that couldn't have been a three cell segment if this was a three cell segment what's this digit and the answer is it's the same as that digit because if those add up to 10 and those add up to 10 those two should be the same number and this can't be a four cell segment because it can't contain a four so the loop has to turn and this segment is going at least to here and I'm not even sure this couldn't keep going a bit further still because now we're in a world six I can repeat digits in a seg can't I so let me think about well oh okay okay yeah all so what we can't do though it's impossible for seven to live in this segment which is at least four cells long because if seven lives in it the minimum SI size of these would be a one two pair in fact that's not even true that digit is bigger than that isn't it that's at least these two squares add up to at least four if this was three and this was one and four plus at least 7 is more than 10 so there's no so so this is seven this is three we said these added up to 11 didn't we so this is eight so this can't extend because if we add two digits to eight we go we we're breaking the we're going bust aren't we in pontoon or 21 speak or Blackjack even to use a more vacy term um what's this then it's not 28 it's not 37 and it's not 46 so that's 1 n these squares are five six and seven are they okay um and that might be useful see if we can do any better well I think presumably what I've got to do oh wait a second this has now got B bigger uh so I'm okay I was I was wondering whether this segment could be longer I'm now going to claim it can't be because this digit is at least a five now by sodoku it's five or six isn't it it can't be eight so what we're doing is we're saying we've got three here plus five this could oh this is going to be a two yeah what what what else can we do here the basically these have to add up to seven and it can't be 61 so this has got to be five this has got to be two these squares are off the loop because we get cold disck in getting to them and what we're going to have to think about now is how we Traverse the 18 clue this is a six in the corner these squares are a 5 seven pair we might be able to do sidoku now which I know is totally anatha these are an 89 pair actually um and okay and therefore five lives in one of those squares in case that matters oh seven lives in one of these squares what about three and one live in these squares I think but presumably what digit are we going to put into the 18 it's going to depend isn't it which way around the 18 we go if we go although if we go this way we're going to bump very quickly into a seven which feels difficult because cu the well I don't think we can I'm not sure about this but if we do turn right here let's say we go in an easterly Direction you can see then that these three squares fact those four squares would have to be on the loop but how are we going to make this work if this is a seven we can't make the next two digits add up to three because they would have to be a one two pair and two is not available and we can't put three here because of sidoku so this cannot be a seven and if that's a seven aren't am I not in exactly the same problem this can't be a 3 S pair so it would have to be a one two pair around the seven and we can't use two that right so actually sodoku is telling us the direction the loop continues it goes south these squares are now all out of bounds we're not going in here um this digit now is well we've got three Loop squares and it can't be at five so we can't okay we can't do that because that would require this to be a five and sodoku seems to be ruling that out out so this is I want to say three or four ah it's beautiful it's not three it's not three for a lovely reason it's very simple reason but it really isn't three it's three mathematically won't work will it because what we've got what we've got is a 10 sum here so we got the next 10 sum starting here so if this was a three these three squares would add up to 18 so these two squares would add up to 16 so they can't we can't make a tum the sort of units are the wrong size so this is a four and now these are a 1 three pair by sidoku so let's deal with that oopsie um and well yeah it's it's going to be this one isn't it that's on the loop because we can't make this green and put this on the loop that's very certainly a cold desac so that's that's Loop that's green this is a corner of the loop we can't go there anymore these two squares are out of bounds now because they would touch the loop so the Loop's either going well it is going there but we don't know how it's either going like that or directly but what we should do is think about the sum of these three digits presumably because these three digits still add up to 16 it's just say three of them now oh well yeah yeah yeah okay so we can just use well it's very simple arithmetic isn't it we know that this is part of a 10 sum now we might say okay but we don't know how long the 10 sum is but we we really do don't we because if these three digits add up to 16 it isn't that long it's less Long than that so those two digits must add up to 10 and this must be the remainder that's the six and now this St cell is a seven or a nine because this is a 10 sum and this is a six so this could oh no no right so now this is a six and it can't go straight into a four because of that so this okay so the next 10 sum is going to be a two this six is going to need two digits that add up to four which must be a one and a three so this is a one or a three and then either that's a one or a three or that's a one or a three so one of these is a one or a three um now which way does that go maybe it depends on the value of the 14 if that's if that's the one or the three that would be a 1 three pair and this would be a two so these would add up to five which looks plausible there's a money spider running on my microphone this money spider is an annoying spider actually it's made an awful lot of webs inside my computer naughty money spider never kill spiders just should always just look after them it's it's gone now I don't know where it's gone sorry a brief Excursion um right okay what are those oh where's Six okay I'm going to do soku forced six and seven go in there so these squares must be known four and five so these squares are 6 8 n but mathematics isn't really going to help me with these is it do we know well actually maybe it's this Square because it occurs to me that this is this is a moderately high in fact if that's not a four what is it yeah because remember this 10 sum whatever it looks like it's made up of 613 so this digit can't be a one or a three by sidoku it can't be two by sidoku it can't be five or six by sidoku and it can't be seven because there are not seven gray squares available there's only six gray squares available so that is in fact a four I think four lives in one of these two squares now but do we know the shape of the four so if the four yeah oh okay that's it oh no oh wait a moment be careful here you have you do have to be quite careful if we go straight down I'm not sure I like the look of that but maybe it's okay because what I'm seeing there is these squares would be a 1 three pair to make this 10 sum work and then then the next two digits that are taken by the 14 clue have to add up to 10 because these would add up to 10 to four sorry so we've got 10 left to find in two cells but then I mean could you do weird things like that being a two and then you start to dip down there and you nip up again but wouldn't you get two fives in the same row no oh I'm not sure no no no you wouldn't because these would have to add up to five if that's a one if that's a one and that's a five I mean it's incredibly forcing if this is a one because both of these would be green so this would be a corner of the loop and in fact that would be oh that's the right that's it that breaks right that's beautiful that okay I was trying to do that but without really thinking carefully about this digit making a corner but once you make once this makes the corner it's it's so lovely right I've got it now okay so I think we what we can do the way to do this to show you this as cleanly as I can is to to actually this is where we were we know one of these is a one or a three but it's simply to look at this Square now and say can that be a one and the answer is no because remember what one does is it could it it makes the only gray square around here this one and that therefore is a corner of the loop so it forces exactly this pattern and this is so beautiful the reason this breaks because this has to be a 613 triple this is saying make that square of five but these add up to four so these need to add up to 10 they are the only digits we've got four our four cells identified already so these have to add up to 10 and they have to be double five and that won't work because of sidoku isn't that clever so that means well so what that means is this is a two I'm not actually that we know that's gray still but does that actually um yeah it does something doesn't it in fact no I see what it's doing right so we now know that there is a domino and it is a domino it's either this Domino or this Domino around this two clue that sums to five well it's not not a two three pair then is it it's a 1 14 pair but we know that so it can't you know the four must be there and therefore must be on the loop this must be a one this is a three this Square now cannot be part of the 1 three pair because it sees one and three so that's green the loop goes around the corner here the oh oh hang on well that that seems to say this is a one which means that completes that 10 sum so there's a new 10 sum well it starts here but that can't be a six going into a four because the six isn't available so it's at least a three cell tum so that could be a five which looks quite likely doesn't it this is green because we've we've had our two cells around this clue so so that continues 1 two 3 and again I think it's the same sort of point we saw up here I think it was up here anyway where is which one of these is the is the gray cell and which is Green Well it can't be green here and gray here because that will cold desct the loop so it must be gray there and green here and these two squares add up to 10 now because the 14 is made up of 3 + 1 plus 10 now oh yeah okay and we can do it because look look at this column more carefully this is a three this is a one this is a nine to make this t 10 some work what is this digit now well if it wasn't five what would it be it couldn't be one 2 3 4 if we're saying it's not five it's not even six after that it's at least seven joining up to a four that's too big so this is five that is a 10 sum this is a little old 10 sum because we know these two squares add up to 10 um so now we can make this oh whoopsie this Square here is green this is a 10 sum and it's not 1 n or 46 so it's either 37 or 28 but we might be able to we can get rid of two uh by soku what are those digits 27 and 8 let's just put that in in case that well that does give yeah gives me a two here so I get a 78 pair can we resolve that that's not a four that's a 78 or a nine at the bottom of the grid this column needs five8 and N with the five being at the bottom of the grid somewhere so maybe we've got to look at one of these now um the nine clue was actually only got well obvious well not obviously but it has only got three cells around it so oh but it could be a one couldn't it it could be a one with a nine here if this was a corner and it did something like that so this is It's one two or three um do we know do we know anything about that uh if it's three it has to be one two and six doesn't it 2 one six does that break for some reason oh oh this is okay that might be my favorite moment of the puzzle actually this is a two oh this is so clever oh black Doom this is really quality setting right let's think about let's think about this ah this is beautiful I can get rid of one and three for subtly this they're related reasons um and they're both great let me explain let's deal with three first because that's what I that's what got me thinking about this if this is three those three are definitely Loop aren't they and they add up to nine now three different numbers that add up to nine are either 135 well they're not 135 that's a three 2 3 4 well they're not 2 3 4 because that's a three the only other way of doing it is 1 two and six so these would be 1 two and six and can you see why that fails it's not to do with sidoku what it's to do with is the the TM nature of this this this is the start of the next TM so it's going to go d these add up to nine we haven't got to 10 yet so the continuation is here and what is this digit it has to be a one but I've just used one up already so that doesn't work but the beautiful thing about that is let's go back here and think again about whether this can be a one because I said oh that can be a one because this could be a nine then if it's the only cell around here that's gray it would have to be that one and the loop would have to move like that wouldn't it well what does nine accompany and in any 10 sum the answer is one and I can't put one in either of those squares so there could be no continuation for the nine it cannot exist in that nature and that form so that means that this is a two which I think must mean it's it's similar logic to what we saw up there this must be gray otherwise we get cold disect so that is gray the loop is therefore going through another Sant caribdis it's going from here down here somewhere um that's not a two anymore so that's not an eight so this has become a 37 pet that might put pressure on the 19 clue this is a two by soku that's a one Boku so two is in this domino in box five and what next there's going to be I feel like I need a one now to attach to a domino that's going to add up to nine [Music] yeah okay that's that it's again it's subtle but it's really really beautiful setting so let's think about that at the moment we don't know quite how the loop continues but we know that either it does it takes this Square as the two cells around the nine clue now if it does that this is green isn't it and the loop would have to continue to he so this is is one possible orientation I think this is correct because these would add up to nine and therefore that this Square would be a one to continue to make a 10 sum now the only other way this could work at this point is that this square is gray and this square is the Domino around the nine clue that adds up to nine but we know these add up to nine so this would be the spare digit in this 10 sum and it would have to be a one and it couldn't be a one so I think the only way we can continue the loop is th is this goes down therefore this is green this continues these add up to nine so this is a one this is good as well isn't it because this Domino adds up to nine and now doesn't use one 2 or three so it's not 18 27 or 36 it must be a four five pair which somehow doesn't resolve the sidoku it does resolve the five from here though which means this is a five which means five lives in one of those squares two lives in one of these squares not that one can we do more sodoku perhaps four lives in one of those squares not that one three is not known yet um this is a corner of the loop so both of these squares are now off out of bounds so far as the Loop's concerned and that dude that digit is Big that digit is a big old digit look it's not one two or three or four or five or seven so it must be six oh no I can do six I was about to say I can't put six digits around it but I think I can actually yes I can just do it so I think by sodoku this is forced to be six there are eight cells around that clue and two of them are off out of bounds so so six works if I make all of those Gray that forces the loop up here it can't go here because it then could would would have to do a tight turn so the loop the loop gets finished as a result of that that's the only way the loop can can finish all of these squares are now green and all we've got to do well we can write that digit in one that's a six because there are six Loop cells around that square this is not a six um we've got another 10 sum here so there's some sort of 10 sumage going on up here somehow with the 19 clue and well actually what are those digits now they are well this is an eight or a nine these are 3 7 8 and 9 depending on what this is and that can't be three and that can't be seven okay maybe that's not how to do it then um it's probably the 19 clue somehow maybe that's where I meant to think about it uh I'm trying to resist doing soku um if we were okay so how does this work let me think about it so at the moment we've got some portion of 10 sum here which is either five or six and then we're going to have some portion here that's going to add up to 10 isn't it it can't be all four of those cuz these can't add up to nine so it's either going to be two or three cells that add up to 10 and then we're going to have a rump of cells which is either going to be those two or just this one that adds up to the difference between this sum 10 and 19 so in other words this Plus either those two or just this one add up to nine now that might be difficult if we do that one one and three and one and five no one and four would work so can this be a 1 three pair is a question it's a question with an annoying answer I think that might be fine provided this could be a 10 sum and if this was a 10 sum interestingly if that was a 10 sum it would have to be 28 I think because it couldn't be 1 n it couldn't be 37 and it couldn't be 46 so if that can be 28 which it looks like it can be and these squares well one so I suppose we could look at those three squares as having having to add up to eight couldn't we so we couldn't do 51 2 it would have to be 4 1 three this would be a three this would be a one and then okay I don't know if that's fine or not but it might be that's that's the problem um is there any other configuration I suppose if these three added up to 10 then we'd be saying those three add up to nine ah no it's nearly well actually okay does that always mean this is a three that's a peculiar deduction if so so what I'm noticing is if these if those two add up to 10 therefore these four add up to nine I think there is only one way to do that which is with a one three pair and in that situation this is a three if these three add up to 10 so these three add up to n to make my 19 then this digit by mathematics is either four which it can't be or three which it can be and is so that is always three and that means that that's a three in in this box which means this is a seven this is a seven this is a seven this is an eight now does that do something really cool by D of sodoku three and six that's not seven anymore this is a 36 pair apparently um I see I don't know quite how the stubb works here I don't know whether the stage is those two which would be a 1 three pair with this being 10 or just this one on its own with these three maybe these three can't add up to 10 for some reason oh there's a three there let's see use the given digit so that's three and that's six so this is 78 or 9 o well that's okay so if this is part of a three cell 10m this would have to be a one two pair that does look possible doesn't it and what do we say if it was part of a two cell sendum then this is always a two so it either go so so this is always a two because it either goes 721 or it goes 82 but that's always a two and I think by mathematics this is always a one as well this is really weird I think there must be another way of seeing this because if this is a 10 sum it has to go 721 and that's a one but if it's a two cell TM it goes 82 but then these squares have to add up to nine and that still has to be a one yeah okay so may maybe that's the way to look at it if these were not one and three they' get too big to join up in a 10 sum it's really strange but these squares are now 6 7 8 and N that's not six or eight that could be anything that can't be seven that could be anything again there's a four five pair in this column um but I still at least I don't think I know quite how this unwinds uh oh I've got an 8 n pair as well so these squares are from 2 four and five but we've got to make sure one of them is a two so it might have to go 1 32 4 yeah what's this digit that must be it this digit is too big I think four five eight are the options there or maybe I meant to just approach this quite differently and I meant to look at this from this side that I I have a lot of difficulty believing that this Square joins up yeah okay let's think about this this we know is joining up with the cell above it now it's not making a 10 sum and it's not even making a 10 sum if it joins up with this one as well because 1 three and five do not add up to 10 they only add up to nine so those three are in the same 10 some and one of them uh oh I see okay but I could make them add up to 10 if they were 2 five and three but I could make these up to 10 if they were two three and four and this was a one and left on its own oh there some this must be something I'm missing here what is it okay this isn't a two is it some sort of it's going to be some straightforward sidoku isn't it I'm sorry about that um oh it's a 7 n pair here so that's a five so this column needs 1 three and four into that's a four naked single therefore the three must go here so column one there's a lot of sidoku we could have been doing I don't know why I don't know why I wasn't doing it but I wasn't um this now is not a six there's never been a six there's always a six over here but that still doesn't help us does this 57 get resolved somehow this one is doing a little bit of work isn't it one and nine we can do this square is oh I see no okay that's still it's still not clear okay so maybe maybe the whole trick was to approach it from this Square yeah look this can't be a two cell 10 Su that's it there we go it has to be at least three well this can't now be eight it can't be a two cell 10 sum cuz seven isn't available so this has to go this has to be five and this has to be two and that is a 10 sum and by sodoku that's become a four and therefore that's a two so that's a four that's a five that's a four that's a five this Square here is a seven or a nine so the only place for eight is there which means that's nine that's seven that's nine that's seven this is six this is seven this is eight this is nine this is8 8 9 98 um we have going to have to we haven't got a 10 sum there but that you can see it's now working properly so I think we we have arrived at the correct solution eight um we need what do we need here 69 pair so six here nine here nine there 15 and 8 518 and that is how to solve a very very clever puzzle yeah there was some beautiful logic in that this nine was absolutely brilliant and then I got really stuck I couldn't see how to efficiently divide up there seemed to be so I think this would be a puzzle as well dare I say it that you could solve and think you'd exhausted all the logical options but there were so many ways you could combine these totals to make the 19 and still leave valid 10 sums around oh let's just check the 17 ads up it does yeah I mean that's not terribly surprising but I think probably I could have done better sidoku on column one cuz once that became a three then we were then we were quite banging weren't we were were good to go after that although how when did I resolve that seven and eight I'm not sure I'm going to be intrigued how much did I miss there just remember in uh we like the comments especially when they're kind but most importantly especially when they're nice to Black Doom who has come up come up with an absolutely brilliant puzzle thank you very much for watching as I say look forward to those kind comments and we'll be back later with another edition of cracking the cryptic [Music]
189139	https://www.cs.dartmouth.edu/~deepc/LecNotes/Appx/6a.%20Pipage%20Rounding.pdf	Pipage Rounding1 • In a previous lecture, we described a rounding algorithm to convert a fractional solution to the bipartite matching into an integral solution with the same or larger value. In this lecture, we build on that idea and show its applicability to obtaining approximation algorithms for NP-hard problems. This style of rounding has been called pipage rounding in the literature. We illustrate this on a problem which we have seen before: MAX COVERAGE. Recall that in this problem we are given a universe U and a collection of subsets of U: S = {S1, . . . , Sm}. The objective is to choose a collection of k sets so as to maximize the cardinality of their union. We know that a greedy algorithm obtains an 1− 1 −1 k k-approximation which tends to 1 −1 e as k →∞. Suppose every element was present in at most f sets. That is, the degree of the set system S is ≤f. In this note we describe an 1 − 1 −1 f f -approximation for the MAX-COVERAGE problem. In particular, this implies a 3 4 = 0.75-approximation for the MAX VERTEX COVERAGE problem, where we are given an undirected graph G = (V, E) and objective is to choose a subset U ⊆V with \|U\| = k which maximizes the number of edges having at least one endpoint in U. This is because, f = 2 for this set family; every edge contains 2 vertices. It is not hard to show an example where the greedy algorithm for MAX-COVERAGE when tailored to MAX VERTEX COVERAGE only be as good as a ≈1 −1 e ≈0.632 approximation. • LP Relaxation. We begin with a LP relaxation for the problem. opt ≤lp(U, S) := maximize X i∈U zi (MaxCov-LP) zi ≤ X j:i∈Sj xj, ∀i ∈U (1) m X j=1 xj = k, (2) 0 ≤zi, xj ≤1, ∀i ∈U, 1 ≤j ≤m (3) Above, xj indicates whether set j is picked, zi indicates to what extent element i is covered. (1) captures the notion that an element is covered only if a set containing it is picked, and (3) captures the notion that any element can’t be covered more than once. Before moving ahead, for reasons which will soon become clear, we rewrite the above program by eliminating the z-variables. Given xj’s, the value zi will be set to min(1, P j:i∈Sj xj).Therefore, (MaxCov-LP) is equivalent to opt ≤max {L(x) : x ∈P} (4) 1Lecture notes by Deeparnab Chakrabarty. Last modiﬁed : 4th Mar, 2022 These have not gone through scrutiny and may contain errors. If you ﬁnd any, or have any other comments, please email me at deeparnab@dartmouth.edu. Highly appreciated! 1 where L(x) = X i∈U min  1, X j:i∈Sj xj   and P := {x ∈[0, 1]m : X j xj = k} Although the function L : Rm →R is not linear, (4) can be solved via the linear program (MaxCov-LP) since it is equivalent to it. However, the solution to (4) can have fractional entries, that is, needn’t be in {0, 1}m • Another non-linear function. We now describe another non-linear function F : Rm →R which is the key deﬁnition. The nice properties of F would be (a) the maximum value of F(x) over x ∈P would also be an upper bound on opt, and (b) the maximum value of F would be obtained on integral points. The not-nice property of F would be that maximizing F over P is NP-hard (yes, I understand that it seems no progress is being made; hold on). Here’s the function. F(x) := X i∈U  1 − Y j:i∈Sj (1 −xj)   (Continous Coverage) Observe that whenever x ∈{0, 1}m, F(x) equals the value of the algorithm which picks sets with xj = 1. Thus, as promised, the following math program is also an upper bound on opt. opt ≤max {F(x) : x ∈P} (5) As mentioned above, the beauty of the function F(·) deﬁned in (Continous Coverage) is the following : given any solution x ∈P, there is a rounding algorithm called PIPAGE ROUNDING, which returns a solution xint ∈{0, 1}m ∩P with the property that F(xint) ≥F(x). In other words, the program (5) has no integrality gap! We describe this pipage rounding in greater detail soon, but before doing so we need to address what use is this rounding algorithm if (5) cannot be solved in polynomial time (which we believe one can’t unless P=NP). • Comparing F and L. The main point is that although F can’t be maximized over P, L can, and the following analytic claim shows that F and L are point-wise related. For brevity’s sake, let ρ(x) := 1 − 1 −1 x x. It is not too hard to see d ≤f implies ρ(d) ≥ρ(f). Lemma 1. For all x ∈P, F(x) ≥ρ(f) · L(x), where f is the degree of the set system S. Proof. Fix i ∈U and suppose it is contained in d ≤f sets. By the AM-GM inequality, Y j:i∈Sj (1 −xj) ≤ P j:i∈Sj(1 −xj) d !d = 1 − P j:i∈Sj xj d !d (6) Let g(t) := 1−(1−t d)d. Then, for d ≥1 and t ∈[0, 1] we have g(t) ≥ρ(d)·t. This follows because in that interval, g is concave, and thus g(t) ≥(1 −t)g(0) + tg(1) = ρ(d) · t. Now, we can use this in (6) to say 1 − Y j:i∈Sj (1 −xj) ≥min  1, X j:i∈Sj xj  · ρ(d) ≥ρ(f) 2 where we used the monotonicity of ρ in the last inequality. Summing for all i ∈U proves the lemma. • Approximation Algorithm. To summarize, we know how to maximize L, that is solve (4), but the solution may not be integral. We don’t know how to maximize F, that is solve (5), but we know there is an integral optimum. The previous lemma tells us F(x) ≥ρf · L(x) for all x ∈P. Putting all three of these together implies the following ρf-approximation for MAX COVERAGE. 1: procedure MAX COVERAGE ROUNDING(S): 2: Solve (MaxCov-LP) to get (x, z). 3: Run PIPAGE ROUNDING(x) to obtain xint with F(xint) ≥F(x).▷We describe this next. 4: Pick sets with xint j = 1 covering F(xint) elements. 5: ▷Since xint ∈P ∩{0, 1}m, there will be exactly k sets picked. Theorem 1. MAX COVERAGE ROUNDING is a ρf approximation. To see why, note that opt ≤L(x) ≤1 ρf F(x) ≤F(xint) ρf . • Pipage Rounding. The setting where pipage rounding applies is more general than the one described above. Abstractly, suppose we want to maximize a function F on m variables in {0, 1}m intersected with a polytope P max{F(x) : x ∈P ∩{0, 1}m } (7) Suppose the following two conditions hold. Pa. For any non-integral x ∈P, one can efﬁciently ﬁnd a vector vx ∈Rm and scalars αx, βx > 0 such that x + αxvx and x −βxvx have strictly more integral coordinates that x. Pb. For all x ∈P, the function F(·) is convex in the direction of the above vector vx. More precisely, the function gx(t) := F(x + tvx) is a convex function over the variable t ∈R. Theorem 2. If conditions (Pa) and (Pb) are satisﬁed, then given any x ∈P there exists an efﬁcient algorithm PIPAGE which returns xint ∈P ∩{0, 1}m with F(xint) ≥F(x). Remark: In fact, one doesn’t require the ﬁrst condition (Pa) very strongly. It sufﬁces if one can show x + αxvx and x −βxvx “make progress” towards an integral solution. One possible measure of progress is that both these points lie on a face of P of smaller dimension. Proof. The proof is the following obvious while loop 3 1: procedure PIPAGE ROUNDING(x): 2: while x / ∈{0, 1}m do: 3: Use (Pa) to obtain αx, βx, vx. 4: if F(x + αxvx) ≥F(x) then: 5: x ←x + αxvx 6: else: 7: x ←x −βxvx 8: ▷Note that the number of integral coordinates in x increases in either case. We now show that in every while loop the value of F(x) can only increase. Using the fact that F(·) is convex in direction of vx, we assert that max (F(x + αxvx), F(x −βxvx)) ≥F(x) which would give us what we need. To see the inequality, write 0 = βx αx+βx · αx + αx αx+βx · (−βx) and use convexity of gx to say gx(0) \| {z } =F(x) ≤ βx αx + βx gx(αx) \| {z } =F(x+αxvx) + αx αx + βx gx(−βx) \| {z } =F(x+αxvx) And so, gx(0) ≤max(g(αx), g(−βx)) proving what we asserted. Since the number of integral coordinates increases, the above procedure terminates in at most m steps. Thus the time taken is at most m times the time taken to implement (Pa). • (Pa) and (Pb) for coverage. Recall, for the coverage problem P := {x ∈[0, 1]m : Pm j=1 xj = k}. Claim 1. (Pa) and (Pb) are true for the above polytope. Proof. Suppose x is a non-integral vector in P. Since Pm j=1 xj is an integer, there must be at least two coordinates, call them xp and xq, such that both are in (0, 1). Our vector vx is then the vector (ep−eq), where ei is the unit-vector with 1 in the tth coordinate and 0 elsewhere. Set αx = min(1 −xp, xq) and βx = min(1 −xq, xp). Note that x + αxvx and x −βxvx are vectors in P with at least one less fractional coordinate. The whole process above was efﬁcient. Now we establish the convexity of F(x) in the direction (ep−eq). Indeed, gx(t) := F(x+t(ep−eq)) can be written as using (Continous Coverage) g(t) = X i∈U hi(t) where hi(t) is independent of t if the element i is neither in the set Sp nor in the set Sq, is a linear function if i is in exactly one of Sp or Sq, or a quadratic in t with a positive coefﬁcient for t2 if i ∈Sp and i ∈Sq. Indeed, if we deﬁne Cx := Q j:i∈Sj,j̸=p,j̸=q(1 −xj) which is a non-negative constant independent of t, then in the ﬁrst case hi(t) = 1 −Cx, in the second case hi(t) = 1 −(1 −xp −t)Cx or hi(t) = 1 −(1 −xq + t)Cx which are both linear functions of t, or (most interestingly perhaps) hi(t) = 1 −(1 −xp −t)(1 −xq + t)Cx = t2Cx + linear function of t if i ∈Sp ∩Sq. In sum, in all cases hi(t) is a convex function, and thus g(t) is a convex function. 4 Exercise: KK Show that the MAX COLORFUL COVERAGE problem has a (1−1 e)-approximation. Indeed, a (1 −(1 −1 f )f) approximation. Notes The algorithm in this note, and indeed the nomenclature of pipage rounding, is from the paper by Ageev and Sviridenko. See the paper for other applications such as graph and hypergraph partitioning problems, and a job scheduling problem as well. Pipage rounding was used in the inﬂuential paper by Calinescu, Chekuri, P´ al, and Vondr´ ak to give an (1−1 e)-approximation for maximizing any monotone submodular func-tion over a matroid constraint. Randomized versions of pipage rounding have been studied in the paper by Gandhi, Khuller, Parthasarathy, and Srinivasan, for many kinds of problems with “hard constraints”, and more recently explored for submodular objectives in the paper by Chekuri, Vondr´ ak, and Zenklusen. 5 References A. A. Ageev and M. I. Sviridenko. Pipage rounding: A new method of constructing algorithms with proven performance guarantee. Journal of Combinatorial Optimization, 8(3):307–328, 2004. G. Calinescu, C. Chekuri, M. P´ al, and J. Vondr´ ak. Maximizing a monotone submodular function subject to a matroid constraint. SIAM Journal on Computing (SICOMP), 40(6):1740–1766, 2011. C. Chekuri, J. Vondr´ ak, and R. Zenklusen. Dependent randomized rounding via exchange properties of combinatorial structures. In Proc., IEEE Symposium on Foundations of Computer Science (FOCS), pages 575–584, 2010. R. Gandhi, S. Khuller, S. Parthasarathy, and A. Srinivasan. Dependent rounding and its applications to approximation algorithms. Journal of the ACM, 53(3):324–360, 2006. 6
189140	https://cran.r-project.org/web/packages/permute/vignettes/permutations.html	Restricted permutations; using the permute package Gavin L. Simpson 2025-06-25 Introduction In classical frequentist statistics, the significance of a relationship or model is determined by reference to a null distribution for the test statistic. This distribution is derived mathematically and the probability of achieving a test statistic as large or larger than the observed statistic if the null hypothesis were true is looked-up from this null distribution. In deriving this probability, some assumptions about the data or the errors are made. If these assumptions are violated, then the validity of the derived p-value may be questioned. An alternative to deriving the null distribution from theory is to generate a null distribution of the test statistic by randomly shuffling the data in some manner, refitting the model and deriving values for the test statistic for the permuted data. The level of significance of the test can be computed as the proportion of values of the test statistic from the null distribution that are equal to or larger than the observed value. In many data sets, simply shuffling the data at random is inappropriate; under the null hypothesis, the data may not ne freely exchangeable, for example if there is temporal or spatial correlation, or the samples are clustered in some way, such as multiple samples collected from each of a number of fields. The permute package was designed to provide facilities for generating these restricted permutations for use in randomisation tests. permute takes as its motivation the permutation schemes originally available in Canoco version 3.1 (Braak 1990), which employed the cyclic- or toroidal-shifts suggested by Besag and Clifford (1989). Simple randomisation As an illustration of both randomisation and simple usage of the permute package we consider a small data set of mandible length measurements on specimens of the golden jackal (Canis aureus) from the British Museum of Natural History, London, UK. These data were collected as part of a study comparing prehistoric and modern canids (Higham, Kijngam, and Manly 1980), and were analysed by Manly (2007). There are ten measurements of mandible length on both male and female specimens. The data are available in the jackal data frame supplied with permute. library("permute") data(jackal) jackal ``` Length Sex 1 120 Male 2 107 Male 3 110 Male 4 116 Male 5 114 Male 6 111 Male 7 113 Male 8 117 Male 9 114 Male 10 112 Male 11 110 Female 12 111 Female 13 107 Female 14 108 Female 15 110 Female 16 105 Female 17 107 Female 18 106 Female 19 111 Female 20 111 Female ``` The interest is whether there is a difference in the mean mandible length between male and female golden jackals. The null hypothesis is that there is zero difference in mandible length between the two sexes or that females have larger mandibles. The alternative hypothesis is that males have larger mandibles. The usual statistical test of this hypothesis is a one-sided t test, which can be applied using t.test() jack.t <- t.test(Length ~ Sex, data = jackal, var.equal = TRUE, alternative = "greater") jack.t ``` Two Sample t-test data: Length by Sex t = 3.4843, df = 18, p-value = 0.001324 alternative hypothesis: true difference in means between group Male and group Female is greater than 0 95 percent confidence interval: 2.411156 Inf sample estimates: mean in group Male mean in group Female 113.4 108.6 ``` The observed t is 3.484 with 18 df. The probability of observing a value this large or larger if the null hypothesis were true is 0.0013. Several assumptions have been made in deriving this p-value, namely random sampling of individuals from the populations of interest, equal population standard deviations for males and females, and that the mandible lengths are normally distributed within the sexes. Assumption 1 is unlikely to be valid for museum specimens such as these, that have been collected in some unknown manner. Assumption 2 may be valid, Fisher’s F-test and a Fligner-Killeen test both suggest that the standard deviations of the two populations do not differ significantly var.test(Length ~ Sex, data = jackal) fligner.test(Length ~ Sex, data = jackal) This assumption may be relaxed using var.equal = FALSE (the default) in the call to t.test(), to employ Welch’s modification for unequal variances. Assumption 3 may be valid, but with such a small sample we are unable to reliably test this. A randomisation test of the same hypothesis can be performed by randomly allocating ten of the mandible lengths to the male group and the remaining lengths to the female group. This randomisation is justified under the null hypothesis because the observed difference in mean mandible length between the two sexes is just a typical value for the difference in a sample if there were no difference in the population. An appropriate test statistic needs to be selected. We could use the t statistic as derived in the t-test. Alternatively, we could base our randomisation test on the difference of means Di (male - female). The main function in permute for providing random permutations is shuffle(). We can write our own randomisation test for the jackal data by first creating a function to compute the difference of means for two groups meanDif <- function(x, grp) { mean(x[grp == "Male"]) - mean(x[grp == "Female"]) } which can be used in a simple for() loop to generate the null distribution for the difference of means. First, we allocate some storage to hold the null difference of means; here we use 4999 random permutations so allocate a vector of length 5000. Then we iterate, randomly generating an ordering of the Sex vector and computing the difference of means for that permutation. Djackal <- numeric(length = 5000) N <- nrow(jackal) set.seed(42) for(i in seq_len(length(Djackal) - 1)) { perm <- shuffle(N) Djackal[i] <- with(jackal, meanDif(Length, Sex[perm])) } Djackal <- with(jackal, meanDif(Length, Sex)) The observed difference of means was added to the null distribution, because under the null hypothesis the observed allocation of mandible lengths to male and female jackals is just one of the possible random allocations. The null distribution of Di can be visualised using a histogram, as shown in Figure~@ref{draw_hist_jackal}. The observed difference of means (4.8) is indicated by the red tick mark. hist(Djackal, main = "", xlab = expression("Mean difference (Male - Female) in mm")) rug(Djackal, col = "red", lwd = 2) The number of values in the randomisation distribution equal to or larger than the observed difference is (Dbig <- sum(Djackal >= Djackal))<- sum>= 5000 ``` 12 ``` giving a permutational p-value of Dbig / length(Djackal)/ length ``` 0.0024 ``` which is comparable with that determined from the frequentist t-test, and indicates strong evidence against the null hypothesis of no difference. (#fig:draw_hist_jackal)Distribution of the difference of mean mandible length in random allocations, ten to each sex. In total there 20C10=184,756 possible allocations of the 20 observations to two groups of ten choose(20, 10) choose 20 10 ``` 184756 ``` so we have only evaluated a small proportion of these in the randomisation test. The main workhorse function we used above was shuffle(). In this example, we could have used the base R function sample() to generate the randomised indices perm that were used to permute the Sex factor. Where shuffle() comes into it’s own is for generating permutation indices from restricted permutation designs. The shuffle() and shuffleSet() functions In the previous section I introduced the shuffle() function to generate permutation indices for use in a randomisation test. Now we will take a closer look at shuffle() and explore the various restricted permutation designs from which it can generate permutation indices. shuffle() has two arguments: n, the number of observations in the data set to be permuted, and control, a list that defines the permutation design describing how the samples should be permuted. args(shuffle) args ``` function (n, control = how()) NULL ``` A series of convenience functions are provided that allow the user to set-up even quite complex permutation designs with little effort. The user only needs to specify the aspects of the design they require and the convenience functions ensure all configuration choices are set and passed on to shuffle(). The main convenience function is how(), which returns a list specifying all the options available for controlling the sorts of permutations returned by shuffle(). str(how()) str how ``` List of 12 $ within :List of 6 ..$ type : chr "free" ..$ constant: logi FALSE ..$ mirror : logi FALSE ..$ ncol : NULL ..$ nrow : NULL ..$ call : language Within() ..- attr(, "class")= chr "Within" $ plots :List of 7 ..$ strata : NULL ..$ type : chr "none" ..$ mirror : logi FALSE ..$ ncol : NULL ..$ nrow : NULL ..$ plots.name: chr "NULL" ..$ call : language Plots() ..- attr(, "class")= chr "Plots" $ blocks : NULL $ nperm : num 199 $ complete : logi FALSE $ maxperm : num 9999 $ minperm : num 5040 $ all.perms : NULL $ make : logi TRUE $ observed : logi FALSE $ blocks.name: chr "NULL" $ call : language how() - attr(, "class")= chr "how" ``` The defaults describe a random permutation design where all objects are freely exchangeable. Using these defaults, shuffle(10) amounts to sample(1:10, 10, replace = FALSE): set.seed(2) (r1 <- shuffle(10)) ``` 5 6 9 1 10 7 4 8 3 2 ``` set.seed(2) (r2 <- sample(1:10, 10, replace = FALSE)) ``` 5 6 9 1 10 7 4 8 3 2 ``` all.equal(r1, r2)all.equal ``` TRUE ``` Generating restricted permutations Several types of permutation are available in permute: Free permutation of objects Time series or line transect designs, where the temporal or spatial ordering is preserved. Spatial grid designs, where the spatial ordering is preserved in both coordinate directions Permutation of plots or groups of samples. Blocking factors which restrict permutations to within blocks. The preceding designs can be nested within blocks. The first three of these can be nested within the levels of a factor or to the levels of that factor, or to both. Such flexibility allows the analysis of split-plot designs using permutation tests, especially when combined with blocks. how() is used to set up the design from which shuffle() will draw a permutation. how() has two main arguments that specify how samples are permuted within plots of samples or at the plot level itself. These are within and plots. Two convenience functions, Within() and Plots() can be used to set the various options for permutation. Blocks operate at the uppermost level of this hierarchy; blocks define groups of plots, each of which may contain groups of samples. For example, to permute the observations 1:10 assuming a time series design for the entire set of observations, the following control object would be used set.seed(4) x <- 1:10 CTRL <- how(within = Within(type = "series")) perm <- shuffle(10, control = CTRL) perm ``` 9 10 1 2 3 4 5 6 7 8 ``` x[perm] ## equivalent ## equivalent ``` 9 10 1 2 3 4 5 6 7 8 ``` It is assumed that the observations are in temporal or transect order. We only specified the type of permutation within plots, the remaining options were set to their defaults via Within(). A more complex design, with three plots, and a 3 by 3 spatial grid arrangement within each plot can be created as follows set.seed(4) plt <- gl(3, 9) CTRL <- how(within = Within(type = "grid", ncol = 3, nrow = 3), plots = Plots(strata = plt)) perm <- shuffle(length(plt), control = CTRL) perm ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 25 26 27 19 20 21 22 23 24 ``` Visualising the permutation as the 3 matrices may help illustrate how the data have been shuffled ``` Original lapply(split(seq_along(plt), plt), matrix, ncol = 3) ``` ``` $1 [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 $2 [,1] [,2] [,3] [1,] 10 13 16 [2,] 11 14 17 [3,] 12 15 18 $3 [,1] [,2] [,3] [1,] 19 22 25 [2,] 20 23 26 [3,] 21 24 27 ``` ``` Shuffled lapply(split(perm, plt), matrix, ncol = 3) ``` ``` $1 [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 $2 [,1] [,2] [,3] [1,] 10 13 16 [2,] 11 14 17 [3,] 12 15 18 $3 [,1] [,2] [,3] [1,] 25 19 22 [2,] 26 20 23 [3,] 27 21 24 ``` In the first grid, the lower-left corner of the grid was set to row 2 and column 2 of the original, to row 1 and column 2 in the second grid, and to row 3 column 2 in the third grid. To have the same permutation within each level of plt, use the constant argument of the Within() function, setting it to TRUE set.seed(4) CTRL <- how(within = Within(type = "grid", ncol = 3, nrow = 3, constant = TRUE), plots = Plots(strata = plt)) perm2 <- shuffle(length(plt), control = CTRL) lapply(split(perm2, plt), matrix, ncol = 3) ``` $1 [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 $2 [,1] [,2] [,3] [1,] 10 13 16 [2,] 11 14 17 [3,] 12 15 18 $3 [,1] [,2] [,3] [1,] 19 22 25 [2,] 20 23 26 [3,] 21 24 27 ``` Generating sets of permutations with shuffleSet() There are several reasons why one might wish to generate a set of n permutations instead of repeatedly generating permutations one at a time. Interpreting the permutation design happens each time shuffle() is called. This is an unnecessary computational burden, especially if you want to perform tests with large numbers of permutations. Furthermore, having the set of permutations available allows for expedited use with other functions, they can be iterated over using for loops or the apply family of functions, and the set of permutations can be exported for use outside of R. The shuffleSet() function allows the generation of sets of permutations from any of the designs available in permute. shuffleSet() takes an additional argument to that of shuffle(), nset, which is the number of permutations required for the set. nset can be missing, in which case the number of permutations in the set is looked for in the object passed to control; using this, the desired number of permutations can be set at the time the design is created via the nperm argument of how(). For example, how(nperm = 10, within = Within(type = "series")) hownperm = 10within = Withintype = "series" Internally, shuffle() and shuffleSet() are very similar, with the major difference being that shuffleSet() arranges repeated calls to the workhorse permutation-generating functions, only incurring the overhead associated with interpreting the permutation design once. shuffleSet() returns a matrix where the rows represent different permutations in the set. As an illustration, consider again the simple time series example from earlier. Here I generate a set of 5 permutations from the design, with the results returned as a matrix set.seed(4) CTRL <- how(within = Within(type = "series")) pset <- shuffleSet(10, nset = 5, control = CTRL) pset ``` No. of Permutations: 5 No. of Samples: 10 (Sequence) 1 2 3 4 5 6 7 8 9 10 p1 9 10 1 2 3 4 5 6 7 8 p2 4 5 6 7 8 9 10 1 2 3 p3 10 1 2 3 4 5 6 7 8 9 p4 8 9 10 1 2 3 4 5 6 7 p5 5 6 7 8 9 10 1 2 3 4 ``` It is worth taking a moment to explain what has happened here, behind the scenes. There are only 10 unique orderings (including the observed) in the set of permutations for this design. Such a small set of permutations triggers1 the generation of the entire set of permutations. From this set, shuffleSet() samples at random nset permutations. Hence the same number of random values has been generated via the pseudo-random number generator in R but we ensure a set of unique permutations is drawn, rather than randomly sample from a small set. Defining permutation designs In this section I give examples how various permutation designs can be specified using how(). It is not the intention to provide exhaustive coverage of all possible designs that can be produced; such a list would be tedious to both write and read. Instead, the main features and options will be described through a series of examples. The reader should then be able to put together the various options to create the exact structure required. Set the number of permutations It may be useful to specify the number of permutations required in a permutation test alongside the permutation design. This is done via the nperm argument, as seen earlier. If nothing else is specified how(nperm = 999) hownperm = 999 would indicate 999 random permutations where the samples are all freely exchangeable. One advantage of using nperm is that shuffleSet() will use this if the nset argument is not specified. Additionally, shuffleSet() will check to see if the desired number of permutations is possible given the data and the requested design. This is done via the function check(), which is discussed later. The levels of the permutation hierarchy There are three levels at which permutations can be controlled in permute. The highest level of the hierarchy is the block level. Blocks are defined by a factor variable. Blocks restrict permutation of samples to within the levels of this factor; samples are never swapped between blocks. The plot level sits below blocks. Plots are defined by a factor and group samples in the same way as blocks. As such, some permutation designs can be initiated using a factor at the plot level or the same factor at the block level. The major difference between blocks and plots is that plots can also be permuted, whereas blocks are never permuted. The lowest level of a permutation design in the permute hierarchy is known as within, and refers to samples nested within plots. If there are no plots or blocks, how samples are permuted at the within level applies to the entire data set. Permuting samples at the lowest level How samples at the within level are permuted is configured using the Within() function. It takes the following arguments ``` function (type = c("free", "series", "grid", "none"), constant = FALSE, mirror = FALSE, ncol = NULL, nrow = NULL) NULL ``` type : controls how the samples at the lowest level are permuted. The default is to form unrestricted permutations via option "type". Options "series" and "grid" form restricted permutations via cyclic or toroidal shifts, respectively. The former is useful for samples that are a time series or line-transect, whilst the latter is used for samples on a regular spatial grid. The final option, "none", will result in the samples at the lowest level not being permuted at all. This option is only of practical use when there are plots within the permutation/experimental design[^As blocks are never permuted, using type = "none" at the within level is also of no practical use.]. constant : this argument only has an effect when there are plots in the design[^Owing to the current implementation, whilst this option could also be useful when blocks to define groups of samples, it will not have any influence over how permutations are generated. As such, only use blocks for simple blocking structures and use plots if you require greater control of the permutations at the group (i.e. plot) level.]. constant = TRUE stipulates that each plot should have the same within-plot permutation. This is useful for example when you have time series of observations from several plots. If all plots were sampled at the same time points, it can be argued that at the plot level, the samples experienced the same time and hence the same permutation should be used within each plot. mirror : when type is "series" or "grid", argument "mirror" controls whether permutations are taken from the mirror image of the observed ordering in space or time. Consider the sequence 1, 2, 3, 4. The relationship between observations is also preserved if we reverse the original ordering to 4, 3, 2, 1 and generate permutations from both these orderings. This is what happens when mirror = TRUE. For time series, the reversed ordering 4, 3, 2, 1 would imply an influence of observation 4 on observation 3, which is implausible. For spatial grids or line transects, however, this is a sensible option, and can significantly increase the number of possible permutations[^Setting mirror = TRUE will double or quadruple the set of permutations for "series" or "grid" permutations, respectively, as long as there are more than two time points or columns in the grid.]. ncol, nrow : define the dimensions of the spatial grid. How Within() is used has already been encountered in earlier sections of this vignette; the function is used to supply a value to the within argument of how(). You may have noticed that all the arguments of Within() have default values? This means that the user need only supply a modified value for the arguments they wish to change. Also, arguments that are not relevant for the type of permutation stated are simply ignored; nrow and ncol, for example, could be set to any value without affecting the permutation design if type != "grid"[^No warnings are currently given if incompatible arguments are specified; they are ignored, but may show up in the printed output. This infelicity will be removed prior to permute version 1.0-0 being released.]. Permuting samples at the Plot level Permutation of samples at the plot level is configured via the Plots() function. As with Within(), Plots() is supplied to the plots argument of how(). Plots() takes many of the same arguments as Within(), the two differences being strata, a factor variable that describes the grouping of samples at the plot level, and the absence of a constant argument. As the majority of arguments are similar between Within() and Plots(), I will not repeat the details again, and only describe the strata argument strata : a factor variable. strata describes the grouping of samples at the plot level, where samples from the same plot are take the same level of the factor. When a plot-level design is specified, samples are never permuted between plots, only within plots if they are permuted at all. Hence, the type of permutation within the plots is controlled by Within(). Note also that with Plots(), the way the individual plots are permuted can be from any one of the four basic permutation types; "none", "free", "series", and "grid", as described above. To permute the plots only (i.e. retain the ordering of the samples within plots), you also need to specify Within(type = "none", ...) as the default in Within() is type = "free". The ability to permute the plots whilst preserving the within-plot ordering is an impotant feature in testing explanatory factors at the whole-plot level in split-plot designs and in multifactorial analysis of variance (Braak and Šmilauer 2012). Specifying blocks; the top of the permute hierarchy In constrast to the within and plots levels, the blocks level is simple to specify; all that is required is an indicator variable the same length as the data. Usually this is a factor, but how() will take anything that can be coerced to a factor via as.factor(). It is worth repeating what the role of the block-level structure is; blocks simply restrict permutation to within, and never between, blocks, and blocks are never permuted. This is reflected in the implementation; the split-apply-combine paradigm is used to split on the blocking factor, the plot- and within-level permutation design is applied separately to each block, and finally the sets of permutations for each block are recombined. Examples To do. Using permute in R functions permute originally started life as a set of functions contained within the vegan package (Oksanen et al. 2013) designed to provide a replacement for the permuted.index() function. From these humble origins, I realised other users and package authors might want to make use of the code I was writing and so Jari oksanen, the maintainer of vegan, and I decided to spin off the code into the permute package. Hence from the very beginning, permute was intended for use both by users, to defining permutation designs, and by package authors, with which to implement permutation tests within their packages. In the previous sections, I described the various user-facing functions that are employed to set up permutation designs and generate permutations from these. Here I will outline how package authors can use functionality in the permute package to implement permutation tests. In Section~@ref{sec:simple} I showed how a permutation test function could be written using the shuffle() function and allowing the user to pass into the test function an object created with how(). As mentioned earlier, it is more efficient to generate a set of permutations via a call to shuffleSet() than to repeatedly call shuffle() and large number of times. Another advantage of using shuffleSet() is that once the set of permutations has been created, parallel processing can be used to break the set of permutations down into smaller chunks, each of which can be worked on simultaneously. As a result, package authors are encouraged to use shuffleSet() instead of the simpler shuffle(). To illustrate how to use permute in R functions, I’ll rework the permutation test I used for the jackal data earlier in Section~@ref{sec:simple}. pt.test <- function(x, group, nperm = 199) { ## mean difference function meanDif <- function(i, x, grp) { grp <- grp[i] mean(x[grp == "Male"]) - mean(x[grp == "Female"]) } ## check x and group are of same length stopifnot(all.equal(length(x), length(group))) ## number of observations N <- nobs(x) ## generate the required set of permutations pset <- shuffleSet(N, nset = nperm) ## iterate over the set of permutations applying meanDif D <- apply(pset, 1, meanDif, x = x, grp = group) ## add on the observed mean difference D <- c(meanDif(seq_len(N), x, group), D) ## compute & return the p-value Ds <- sum(D >= D) # how many >= to the observed diff? Ds / (nperm + 1) # what proportion of perms is this (the pval)? } The commented function should be reasonably self explanatory. I’ve altered the in-line version of the meanDif() function to take a vector of permutation indices i as the first argument, and internally the grp vector is permuted according to i. The other major change is that shuffleSet() is used to generate a set of permutations, which are then iterated over using apply(). In use we see set.seed(42) ## same seed as earlier pval <- with(jackal, pt.test(Length, Sex, nperm = 4999)) pval ``` 0.0024 ``` which nicely agrees with the test we did earlier by hand. Iterating over a set of permutation indices also means that adding parallel processing of the permutations requires only trivial changes to the main function code. As an illustration, below I show a parallel version of pt.test() ppt.test <- function(x, group, nperm = 199, cores = 2) { ## mean difference function meanDif <- function(i, .x, .grp) { .grp <- .grp[i] mean(.x[.grp == "Male"]) - mean(.x[.grp == "Female"]) } ## check x and group are of same length stopifnot(all.equal(length(x), length(group))) ## number of observations N <- nobs(x) ## generate the required set of permutations pset <- shuffleSet(N, nset = nperm) if (cores > 1) { ## initiate a cluster cl <- makeCluster(cores) on.exit(stopCluster(cl = cl)) ## iterate over the set of permutations applying meanDif D <- parRapply(cl, pset, meanDif, .x = x, .grp = group) } else { D <- apply(pset, 1, meanDif, .x = x, .grp = group) } ## add on the observed mean difference D <- c(meanDif(seq_len(N), x, group), D) ## compute & return the p-value Ds <- sum(D >= D) # how many >= to the observed diff? Ds / (nperm + 1) # what proportion of perms is this (the pval)? } In use we observe require("parallel") set.seed(42) system.time(ppval <- ppt.test(jackal$Length, jackal$Sex, nperm = 9999, cores = 2)) ``` user system elapsed 0.049 0.005 0.258 ``` ppval ``` 0.0018 ``` In this case there is little to be gained by splitting the computations over two CPU cores set.seed(42) system.time(ppval2 <- ppt.test(jackal$Length, jackal$Sex, nperm = 9999, cores = 1)) ``` user system elapsed 0.164 0.003 0.168 ``` ppval2 ``` 0.0018 ``` The cost of setting up and managing the parallel processes, and recombining the separate sets of results almost negates the gain in running the permutations in parallel. Here, the computations involved in meanDif() are trivial and we would expect greater efficiencies from running the permutations in parallel for more complex analyses. Accesing and changing permutation designs The object created by how() is a relatively simple list containing the settings for the specified permutation design. As such one could use the standard subsetting and replacement functions in base R to alter components of the list. This is not recommended, however, as the internal structure of the list returned by how() may change in a later version of permute. Furthermore, to facilitate the use of update() at the user-level to alter the permutation design in a user-friendly way, the matched how() call is stored within the list along with the matched calls for any Within() or Plots() components. These matched calls need to be updated too if the list describing the permutation design is altered. To allow function writers to access and alter permutation designs, permute provides a series of extractor and replacement functions that have the forms getFoo() and setFoo<-(), respectively,where Foo is replaced by a particular component to be extracted or replaced. The getFoo() functions provided by permute are getWithin(), getPlots(), getBlocks() : these extract the details of the within-, plots-, and blocks-level components of the design. Given the current design (as of permute version 0.8-0), the first two of these return lists with classes "Within" and "Plots", respectively, whilst getBlocks() returns the block-level factor. getStrata() : returns the factor describing the grouping of samples at the plots or blocks levels, as determined by the value of argument which. getType() : returns the type of permutation of samples at the within or plots levels, as determined by the value of argument which. getMirror() : returns a logical, indicating whether permutations are drawn from the mirror image of the observed ordering at the within or plots levels, as determined by the value of argument which. getConstant() : returns a logical, indicating whether the same permutation of samples, or a different permutation, is used within each of the plots. getRow(), getCol(), getDim() : return dimensions of the spatial grid of samples at the plots or blocks levels, as determined by the value of argument which. getNperm(), getMaxperm(), getMinperm() : return numerics for the stored number of permutations requested plus two triggers used when checking permutation designs via check(). getComplete() : returns a logical, indicating whether complete enumeration of the set of permutations was requested. getMake() : returns a logical, indicating whether the entire set of permutations should be produced or not. getObserved() : returns a logical, which indicates whether the observed permutation (ordering of samples) is included in the entire set of permutation generated by allPerms(). getAllperms() : extracts the complete set of permutations if present. Returns NULL if the set has not been generated. The available setFoo()<- functions are setPlots<-(), setWithin<-() : replaces the details of the within-, and plots-, components of the design. The replacement object must be of class "Plots" or "Within", respectively, and hence is most usefully used in combination with the Plots() or Within() constructor functions. setBlocks<-() : replaces the factor that partitions the observations into blocks. value can be any R object that can be coerced to a factor vector via as.factor(). setStrata<-() : replaces either the blocks or strata components of the design, depending on what class of object setStrata<-() is applied to. When used on an object of class "how", setStrata<-() replaces the blocks component of that object. When used on an object of class "Plots", setStrata<-() replaces the strata component of that object. In both cases a factor variable is required and the replacement object will be coerced to a factor via as.factor() if possible. setType<-() : replaces the type component of an object of class "Plots" or "Within" with a character vector of length one. Must be one of the available types: "none", "free", "series", or "grid". setMirror<-() : replaces the mirror component of an object of class "Plots" or "Within" with a logical vector of length one. setConstant<-() : replaces the constant component of an object of class "Within" with a logical vector of length one. setRow<-(), setCol<-(), setDim<-() : replace one or both of the spatial grid dimensions of an object of class "Plots" or "Within" with am integer vector of length one, or, in the case of setDim<-(), of length 2. setNperm<-(), setMinperm<-(), setMaxperm<-() : update the stored values for the requested number of permutations and the minimum and maximum permutation thresholds that control whether the entire set of permutations is generated instead of nperm permutations. setAllperms<-() : assigns a matrix of permutation indices to the all.perms component of the design list object. setComplete<-() : updates the status of the complete setting. Takes a logical vector of length 1 or any object coercible to such. setMake<-() : sets the indicator controlling whether the entrie set of permutations is generated during checking of the design via check(). Takes a logical vector of length 1 or any object coercible to such. setObserved<-() : updates the indicator of whether the observed ordering is included in the set of all permutations should they be generated. Takes a logical vector of length 1 or any object coercible to such. Examples I illustrate the behaviour of the getFoo() and setFoo<-() functions through a couple of simple examples. Firstly, generate a design object hh <- how()<- how This design is one of complete randomization, so all of the settings in the object take their default values. The default number of permutations is currently 199, and can be extracted using getNperm() getNperm(hh) getNperm ``` 199 ``` The corresponding replacement function can be use to alter the number of permutations after the design has been generated. To illustrate a finer point of the behaviour of these replacement functions, compare the matched call stored in hh before and after the number of permutations is changed getCall(hh) getCall ``` how() ``` setNperm(hh) <- 999 getNperm(hh) ``` 999 ``` getCall(hh) getCall ``` how(nperm = 999) ``` Note how the call component has been altered to include the argument pair nperm = 999, hence if this call were evaluated, the resulting object would be a copy of hh. As a more complex example, consider the following design consisting of 5 blocks, each containing 2 plots of 5 samples each. Hence there are a total of 10 plots. Both the plots and within-plot sample are time series. This design can be created using hh <- how(within = Within(type = "series"), plots = Plots(type = "series", strata = gl(10, 5)), blocks = gl(5, 10)) To alter the design at the plot or within-plot levels, it is convenient to extract the relevant component using getPlots() or getWithin(), update the extracted object, and finally use the updated object to update hh. This process is illustrated below in order to change the plot-level permutation type to "free" pl <- getPlots(hh) setType(pl) <- "free" setPlots(hh) <- pl We can confirm this has been changed by extracting the permutation type for the plot level getType(hh, which = "plots") getTypewhich = "plots" ``` "free" ``` Notice too how the call has been expanded from gl(10, 5) to an integer vector. This expansion is to avoid the obvious problem of locating the objects referred to in the call should the call be re-evaluated later. getCall(getPlots(hh)) getCall getPlots ``` Plots(strata = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L), type = "free") ``` At the top level, a user can update the design using update(). Hence the equivalent of the above update is (this time resetting the original type; type = "series") hh <- update(hh, plots = update(getPlots(hh), type = "series")) getType(hh, which = "plots") ``` "series" ``` However, this approach is not assured of working within a function because we do not guarantee that components of the call used to create hh can be found from the execution frame where update() is called. To be safe, always use the setFoo<-() replacement functions to update design objects from within your functions. Computational details This vignette was built within the following environment: ``` ─ Session info ─────────────────────────────────────────────────────────────── setting value version R version 4.5.0 (2025-04-11) os macOS Sequoia 15.5 system aarch64, darwin20 ui X11 language (EN) collate C ctype en_US.UTF-8 tz Europe/Copenhagen date 2025-06-25 pandoc 3.7.0.2 @ /opt/homebrew/bin/ (via rmarkdown) quarto 1.7.31 @ /usr/local/bin/quarto ─ Packages ─────────────────────────────────────────────────────────────────── package version date (UTC) lib source bookdown 0.43 2025-04-15 CRAN (R 4.5.0) bslib 0.9.0 2025-01-30 CRAN (R 4.5.0) cachem 1.1.0 2024-05-16 CRAN (R 4.5.0) cli 3.6.5 2025-04-23 CRAN (R 4.5.0) digest 0.6.37 2024-08-19 CRAN (R 4.5.0) evaluate 1.0.3 2025-01-10 CRAN (R 4.5.0) fastmap 1.2.0 2024-05-15 CRAN (R 4.5.0) htmltools 0.5.8.1 2024-04-04 CRAN (R 4.5.0) jquerylib 0.1.4 2021-04-26 CRAN (R 4.5.0) jsonlite 2.0.0 2025-03-27 CRAN (R 4.5.0) knitr 1.50 2025-03-16 CRAN (R 4.5.0) lifecycle 1.0.4 2023-11-07 CRAN (R 4.5.0) permute 0.9-8 2025-06-25 local R6 2.6.1 2025-02-15 CRAN (R 4.5.0) rlang 1.1.6 2025-04-11 CRAN (R 4.5.0) rmarkdown 2.29 2024-11-04 CRAN (R 4.5.0) sass 0.4.10 2025-04-11 CRAN (R 4.5.0) sessioninfo 1.2.3 2025-02-05 CRAN (R 4.5.0) xfun 0.52 2025-04-02 CRAN (R 4.5.0) yaml 2.3.10 2024-07-26 CRAN (R 4.5.0) /private/var/folders/yq/gqj7yv7j3n17w__256r9vxn40000gn/T/RtmpPeTY4j/Rinst86431c29d9d9 /private/var/folders/yq/gqj7yv7j3n17w__256r9vxn40000gn/T/Rtmp0ZTPKf/temp_libpath426e566d1c7f /Library/Frameworks/R.framework/Versions/4.5-arm64/Resources/library ── Packages attached to the search path. ────────────────────────────────────────────────────────────────────────────── ``` Besag, J., and P. Clifford. 1989. “Generalized Monte Carlo Significance Tests.” Biometrika 76 (4): 633–42. Braak, C. J. F. ter. 1990. Update Notes: CANOCO Version 3.1. Wageningen: Agricultural Mathematics Group. Braak, C. J. F. ter, and P. Šmilauer. 2012. Canoco Reference Manual and User’s Guide: Software for Ordination (Version 5.0). Microcomputer Power. Higham, C. F. W, A. Kijngam, and B. F. J. Manly. 1980. “An Analysis of Prehistoric Canid Remains from Thailand.” Journal of Archaeological Science 7: 149–65. Manly, B. F. J. 2007. Randomization, Bootstrap and Monte Carlo Methods in Biology. 3rd ed. Boca Raton: Chapman & Hall/CRC. Oksanen, Jari, F. Guillaume Blanchet, Roeland Kindt, Pierre Legendre, Peter R. Minchin, R. B. O’Hara, Gavin L. Simpson, Peter Solymos, M. Henry H. Stevens, and Helene Wagner. 2013. Vegan: Community Ecology Package. The trigger is via the utility function check(), which calls another utility function, allPerms(), to generate the set of permutations for the stated design. The trigger for complete enumeration is set via how() using argument minperm; below this value, by default check() will generate the entire set of permutations.↩︎
189141	https://pobchemteam.weebly.com/uploads/2/5/7/7/25772138/mole_conversion_problems_key.pdf	Date___Name_____ Per___ Mole Conversion Problems Complete the following practice problems for mole conversion. Show your work and units! 1. How many moles are in 72.9 g of HCl? Molar mass HCl = 36.46 72.9g 36.46 = 1.999 mol 2. How many moles are in 79.85 g Fe2O3? Molar mass = 159.7 79.85g 159.7 = 0.5 mol 3. How many molecules are in 720 g of C6H12O6? Molar mass = 180.18 720g 3.996 mol x (6.02 x 1023 ) = 2.406 x 1024 particles 180.18 = 3.996 mol 4. How many grams are in 3.5 mol of Ca3(PO4)2? Molar mass = 310.18 310.18 x 3.5 mol = 1085.63 g 5. How many molecules are in 8550g of SO2? Molar mass = 64.07 8550g 133.45 mol x (6.02 x 1023 ) = 8.03 x 1025 particles 64.07 = 133.45 mol 6. How many grams are in 3.01 × 1024 molecules of (NH4)2SO4? Molar mass = 132.17 3.01× 1024 particles 5 mol x 132.17 = 660.85 g 6.02 x 1023 = 5 mol 7. How many molecules are in 85 g of AgNO3? Molar mass = 169.88 85g 0.5 mol x (6.02 x 1023 ) = 3.01 x 1023 169.88 = 0.5 mol 8. How many grams are in 1.204 × 1024 molecules of CH3COOH? Molar mass = 60.06 1.204 × 1024 particles = 2.0 mol 2.0 mol x 60.06 = 120.12 g 6.02 x 1023 Date___Name_____ Per___ 9. Convert 86.84 g of LiBr to moles: Molar mass = 86.84 86.84 g = 1.0 mol 86.84 10. Convert 8.045 g of H2CO3 to moles: Molar mass = 62.03 8.045 g = 0.1297 mol 62.03 11. How many grams of lithium are there in 3.45 moles? Molar mass = 6.94 6.94 x 3.45 mol = 23.943 g 12. How many moles of nitrogen are there in 4.3 × 1023 molecules? 4.3 × 1023 particles = .714 mol 6.02 x 1023 13. How many cadmium atoms are there in 6.57 × 103 moles? 6.57 × 103 moles x (6.02 x 1023 ) = 3.955 x 1027 particles 14. How many grams of SO2 are 4.5 × 1024 molecules? Molar mass = 64.07 4.5 × 1024 particles = 7.475 mol 7.475 mol x 64.07 = 478.93 g 6.02 x 1023 15. How many copper atoms are in 5.6 mole of Cu2O3? 5.6 mol x (6.02 x 1023 ) = 3.37 x 1024 particles 16. How many grams of sulfur are in 3.45 × 1022 molecules of SO2? Molar mass Sulfur = 32.07 3.45 × 1022 particles = 0.0573 mol 0.0573 mol x 32.07 = 1.838 g 6.02 x 1023
189142	https://amuthan.github.io/tensors-intro/inner_product_spaces/	Skip to content Inner Product Spaces We begin with a discussion of the algebraic properties of vectors, which are defined as elements of a special kind of a set called a vector space. We will then define an additional structure called the inner product that significantly simplifies the mathematical development. We will learn how to represent a vector with respect to a chosen basis, and how this representation changes when the basis changes. Finally, we will study linear maps between vector spaces, and their representation with respect to chosen bases of the vector spaces. To keep the presentation simple, technical proofs for many of the statements given here are omitted. Basic notions from set theory and matrix algebra, reviewed in two appendices, are assumed to be known. Introduction A vector is typically introduced in high school algebra as a quantity with both a magnitude and a direction. A representation of a vector in the familiar three dimensional Euclidean space is shown in the following figure: Representation of a vector in R3 If i,j,k represent the unit vectors along the x,y,z axes, respectively, a vector v can be expressed uniquely as v=vxi+vyj+vzk, where vx,vy,vz are the Cartesian components of the vector v (with respect to the basis vectors i,j and k). There are two core operations associated with vectors: Vector addition: Given two vectors u=uxi+uyj+uzk and v=vxi+vyj+vzk, we can add them to get a new vector u+v, defined as u+v=(ux+vx)i+(uy+vy)j+(uz+vz)k. Geometrically, the new vector u+v is obtained by placing the tail of v at the head of u. The sum of the two vectors is then the vector which shares it’s tail with u and head with v, as shown below: Vector addition in R3 Note that we get the same vector independent of the order of addition: u+v=v+u. For this reason, vector addition is said to be commutative. It can also be shown easily that if u,v,w are three vectors, then (u+v)+w=u+(v+w). This property is called associativity. Scalar multiplication of a vector with a real number: Given any vector v=vxi+vyj+vzk, we can multiply it by some real number a to get the new vector that is a times as long as v: av=avxi+avyj+avzk. This is illustrated in the following figure: Scalar multiplication of a vector in R3 Note that we will need only real vector spaces in what follows. Some vector spaces admit an algebraic operation called the inner product. For instance, in three dimensional space, given two vectors u=uxi+uyj+uzk and v=vxi+vyj+vzk, we can combine them using the dot product to produce a real number u⋅v=uxvx+uyvy+uzvz. Using the dot product, it is customary to define the length or Euclidean norm of a vector v=vxi+vyj+vzk as the (non-negative) real number ‖v‖=(v⋅v)12=(v2x+v2y+v2z)12. For vectors in three dimensional space (only), we also an additional important algebraic operation called the cross product: we can combine two vectors u=uxi+uyj+uzk and v=vxi+vyj+vzk using the cross product to obtain a new vector u×v=(uyvz−uzvy)i+(uzvx−uxvz)j+(uxvy−uyvx)k. In what follows, we will first focus on just vector addition and scalar multiplication. These two operations embody a concept known as linearity, which is fundamental to appreciate what a vector is. We will then study inner product spaces, which are vector spaces with an additional structure known as an inner product, and highlight Euclidean spaces as an important example of inner product spaces. Remark The generalization of the cross product is known as the wedge product. We will not study the wedge product in detail in these notes since it is beyond the scope of these notes. Notice that all the information about the vector v=vxi+vyj+vzk is contained in the ordered set of three real numbers (vx,vy,vz). What this means is that given any vector v in three dimensional space, we can uniquely associate with it a triple of real numbers, and vice versa. The set of all such ordered triples is the set R3, which is the set of all triples of real numbers. If we define addition of two such ordered triples and multiplication of an ordered triple with a real number as, (ux,uy,uz)+(vx,vy,vz)=(ux+vx,uy+vy,uz+vz),c(ux,uy,uz)=(cux,cuy,cuz), then the elements of R3, which are ordered triples of real numbers, behave exactly as the geometric picture of vectors as arrows that we just discussed, as far as the core properties of vector addition and scalar multiplication are concerned. We now have two ways of representing a vector: as an arrow in three dimensional space, and as a set of ordered triple of real numbers. Both of these represent the same object. But the representation in terms of ordered tuples of real numbers immediately admits a generalization to cases where the pictorial representation fails. It is evident from the above discussion that there is nothing special about the number 3 when we considered an ordered triple of real numbers. We can easily generalize this to a set of ordered n-tuple of real numbers (u1,u2,…,un)∈Rn, where n∈N could be any arbitrary positive integer. We can define addition and scalar multiplication in Rn analogous to the case of the ordered triples, (u1,u2,…,un)+(v1,v2,…,vn)=(u1+v1,u2+v2,…,un+vn),a(u1,u2,…,un)=(au1,au2,…,aun), where (u1,u2,…,un) and (v1,v2,…,vn) are two elements of Rn, and a is a real number. We will call this the standard linear structure on Rn, and elements of Rn as vectors in Rn. Note that there is no obvious way to picture an arrow in the n-dimensional space Rn for n>3. Thus, by choosing the right representation, we can extend the elementary notion of vectors as quantities with magnitude and direction to more general objects. Linear structure Let us now generalize the previous discussion to define abstract vector spaces. Suppose that V is a set such that it is possible to define two maps +:V×V→V;+(u,v)↦u+v,⋅:V×V→V;⋅(a,u)↦au, called vector addition and scalar multiplication, respectively, in V, that satisfy the following properties: for any u,v,w∈V and a,b∈R, Associativity of addition: u+(v+w)=(u+v)+w, Commutativity of addition: u+v=v+u, Existence of additive identity: there exists a unique element 0∈V, called the additive identity of V such that u+0=u, Existence of additive inverse: for every u∈V, there exists a unique element −u∈V such that u+(−u)=0, Distributivity of scalar multiplication over vector addition: a(u+v)=au+av, Distributivity of scalar multiplication over scalar addition: (a+b)u=au+bu, Compatibility of scalar multiplication with field multiplication: a(bu)=(ab)u, Scaling property of scalar multiplication: 1u=u. A set V that has two maps that satisfy these axioms is called a (real) vector space, or a (real) linear space. What we have accomplished through these axioms is to endow a set with a notion of addition that allows us to add two elements of the set to get a third element. We have also provided a mechanism to multiply a member of this set by a real number to get another element of this set. The maps + and ⋅ are said to provide a linear structure on V. Elements of V are called vectors. Remark Some textbooks mention additional closure axioms that indicate that if u,v∈V, then u+v∈V, and given any u∈V and a∈R, au∈V. We don’t specify this explicitly since this is already implied by the function definitions +:V×V→V and ⋅:R×V→V. As mentioned in the previous discussion on set theory, we will always insist on mentioning the domain and codomain of every map/function we encounter. Hence, the so-called closure axioms are redundant for our purposes. Remark Following standard convention, we will often use the shorthand notation u−v for u+(−v). Example The simplest example of a real vector space is the set of real numbers R with addition and multiplication defined in the standard manner. More generally, consider the set Rn consisting of all n-tuples of real numbers: Rn={(u1,…,un)\|∀i=1,…,n,ui∈R}. Given (u1,…,un)∈Rn, (v1,…,vn)∈Rn, and a∈R, let us define addition +:Rn×Rn→Rn and scalar multiplication ⋅:R×Rn→Rn as (u1,…,un)+(v1,…,vn)=(u1+v1,…,un+vn),a⋅(u1,…,un)=(au1,…,aun).It is straightforward to verify that with addition and scalar multiplication thus defined, the triple (Rn,+,⋅) is a real vector space. Note that the additive identity in Rn is the zero vector (0,…,0)∈Rn, and the additive inverse of (u1,…,un)∈Rn is (−u1,…,−un)∈Rn. Example Consider the set Rm×n of all m×n matrices with real entries and defined addition of matrices and scalar mutliplication of a matrix with a real number in the usual sense (see Appendix ). It is easily checked that the set of all m×n matrices is a vector space. The zero vector in Rm×n is the matrix with zero in all of its entries, and the additive inverse of a given matrix is just its negative. Example The definition of vector spaces admits more general kinds of objects. As a simple example, consider the set C0(R,R) consisting of all real-valued and continuous functions of one real variable. Given any f,g∈C0(R,R), and any a∈R, we can define addition and scalar multiplication pointwise as follows: for any x∈R, (f+g)(x)=f(x)+g(x),(a⋅f)(x)=af(x). It is not difficult to verify that (C0(R,R),+,⋅) is a real vector space. The additive identity in C0(R,R) is the zero function 0∈C0(R,R) defined as follows: for any x∈R, 0(x)=0. The additive inverse of f∈C0(R,R) is the function −f∈C0(R,R) defined as follows: for any x∈R, (−f)(x)=−f(x). Subspaces and linear independence A subset U⊆V of a vector space V is said to be a linear subspace of V if U is also a vector space. Note that it is implicitly assumed that both U and V share the operations of vector addition and scalar multiplication. It can be easily checked that if a subset U⊆V of a real vector space V has the property that for any u,v∈U, and any a,b∈R, (au+bv)∈U, then U is a linear subspace of V. This property is often used to check if a given subset of a vector space is a linear subspace. An immediate consequence of this is the fact that every linear subspace of a given vector space must contain the additive identity 0∈V. Example Consider the following subsets of R3: S1={(x,y,z)∈R3\|x+y+z=0},S2={(x,y,z)∈R3\|x+y+z=1},S3={(x,y,z)∈R3\|y=z=0}. To see that S1 is a linear subspace of R3, note that if (x1,y1,z1)∈S1, (x2,y2,z2)∈S1, and a,b∈R, then a(x1,y1,z1)+b(x2,y2,z2)=((ax1+bx2),(ay1+by2),(az1+bz2))∈R3. Since (ax1+bx2)+(ay1+by2)+(az1+bz2)=a(x1+y1+z1)+b(x2+y2+z2)=(0,0,0), we see that ((ax1+bx2),(ay1+by2),(az1+bz2))∈S1. This shows that S1 is indeed a linear subspace of R3. It is likewise verified that S3 is also a linear subspace R3, while S2 is not. The intersection of two linear subspaces is also a linear subspace. Moreover, any finite intersection of linear subspaces of a vector space V is also a linear subspace of V, as can be easily checked. Example In the previous example, the intersection of the subspaces S1={(x,y,z)∈R3\|x+y+z=0} and S3={(x,y,z)∈R3\|y=z=0} of R3 is easily seen to be S1∩S3={(0,0,0)}, which is the trivial subspace of R3. Two non-zero vectors u,v∈V in a vector space V are said to be linearly independent iff au+bv=0, where a,b are real numbers, implies that a=0 and b=0. Thus u and v are linearly independent iff the only linear combination of u and v that yields 0 is the trivial linear combination 0u+0v. If this is not true, then u and v are said to be linearly dependent. This definition can be easily extended to a finite set of non-zero vectors v1,v2,…,vn, where n∈N. Thus, the set of vectors v1,v2,…,vn in V is said to be linearly independent iff the only real numbers a1,a2,…,an that satisfy the equation n∑i=1aivi=0, are a1=a2=…=an=0. Example Consider the vectors v1=(1,2)∈R2 and v2=(2,3)∈R2 in the two dimensional Euclidean space R2. For real numbers a,b∈R, note that av1+bv2=0⇒(a+2b,2a+3b)=(0,0). Solving the equations a+2b=0 and 2a+3b=0, we immediately see that a=b=0, which shows that v1 and v2 are linearly independent vectors in R2. If v3=(2,4)∈R2, then v1 and v3 are linearly dependent since if for a,b∈R, av1+bv3=0⇒(a+2b,2a+4b)=(0,0).These equations do not imply that a=b=0. For instance, a=−2,b=1 satisfies the condition. We thus see that v1 and v3 are linearly dependent. Basis of a vector space We will now introduce a very important tool called the basis of a vector space. The basic idea is that once we identify a basis for a vector space, we can use the linear structure inherent in the space to reduce all computations related to the vector space as a whole, to just computations on the basis set. We will need a few definitions first in order to define the basis. The linear span of a set of vectors {v1,v2,…,vm} in V, written as span({v1,…,vm}), is defined as the set of all it’s linear combinations: span({v1,…,vm})={m∑i=1aivi\|∀1≤i≤m,ai∈R}. It is also common to refer to the linear span as just the span. It is straightforward to check that the linear span of any finite collection of vectors in a vector space V is a linear subspace of the vector space V. Example Let us consider the vectors v1=(1,2,3)∈R3 and v2=(2,3,4)∈R3 in R3. The span of these two vectors is the following subset of R3: span({v1,v2})={((a+2b,2a+3b,3a+4b)∈R3\|a,b∈R}. It is left as an easy exercise to verify that this is indeed a linear subspace of R3. An ordered subset S⊆V a vector space V is said to constitute a basis of V if S is linearly independent, and span(S)=V. In the special case when a vector space V is spanned by a finite ordered set of vectors (v1,…,vn)⊆V, for some n∈N, the vector space V is said to be finite dimensional, and the number n is called the dimension of the vector space V (written dim(V)). A vector space that is not finite dimensional is said to be infinite dimensional. In what follows, we will only deal with finite dimensional vector spaces. If V is an n-dimensional vector space, and (v1,…,vn) is a basis for V, then we will often abbreviate the basis as (vi)ni=1, or just (vi), when the dimension n is evident from the context. Example Let us revisit the n-dimensional Euclidean space Rn, and consider the following vectors: for any 1≤i≤n, ei=(0,…,1⏟ith position,…0). It is easy to check that the (e1,…,en) is a basis of Rn. In particular, note that any (u1,…,un)∈Rn can be written as (u1,…,un)=u1(1,0,…,0)+…+un(0,…,0,1)=n∑i=1uiei.The ordered set (ei)ni=1 is called the standard basis of Rn. Note that in the special case of R3, the set of basis vectors (e1,e2,e3) is identical to the basis set (i,j,k) introduced in the beginning of this section. Given a vector space V of dimension n, it is possible to choose an infinite number of bases for V. This non-uniqueness in the choice of the basis can be easily understood as follows. Pick any g1∈V. Choose g2 from the set V∖span(g1), g3 from the set V∖span(g1,g2), and so on. This process will terminate in n steps since the vector space V is of dimension n. The resulting set of vectors (g1,…,gn) is a basis for V. Inner products and norms We will now introduce a special additional structure on an abstract vector space V called the inner product. There are two reasons for introducing the inner product right at the outset: first, the mathematical development becomes significantly simpler, and second, many important applications in science and engineering can be studied in this setting. Remark There is an elegant theory of abstract linear spaces, both in the finite and infinite dimensional cases, where inner products are not defined. We will however not develop this general theory here. Given a vector space V, an inner product on V is defined as a map of the form g:V×V→R such that, for any u,v,w∈V and a,b∈R, Symmetry: g(u,v)=g(v,u), Bilinearity: g(u,(av+bw))=ag(u,v)+bg(u,w), Positive definiteness: g(u,u)≥0, and g(u,u)=0 iff u=0. A vector space V endowed with a map ⋅:V×V→R that satisfies the three properties mentioned above is said to be an inner product space. All vector spaces considered henceforth will be assumed to be inner product spaces, unless stated otherwise. Remark Given u,v∈V, we will often write g(u,v) as just u⋅v, with the understanding that the inner product g is evident from the context. Example The simplest, and also the most important, example of an inner product space is the vector space Rn defined earlier, with the inner product ⋅:V×V→R defined as follows: for any (u1,…,un)∈Rn and (v1,…,vn)∈Rn, (u1,…,un)⋅(v1,…,vn)=n∑i=1uivi. It is easy to check that this is indeed an inner product. The vector space Rn with this inner product is called the Euclidean space of dimension n. We will use the same symbol Rn to denote the n-dimensional Euclidean space. Example Define the set Pn([a,b],R) as the set of all real valued polynomials of degree less than or equal to n on the interval [a,b]⊆R: Pn([a,b],R)={f:[a,b]⊆R→R\|∀x∈[a,b],f(x)=a0+a1x+…+anxn, where a0,a1,…,an∈R}. Define the function ⋅:Pn([a,b],R)×Pn([a,b],R)→R as follows: for any f,g∈Pn([a,b],R), f⋅g=∫baf(x)g(x)dx.It is left as a simple exercise to verify that this function is actually an inner product on the linear space Pn([a,b],R) with addition and scalar multiplication defined pointwise. The inner product on a vector space V can be used to define a norm on V. A norm on a vector space V is a function of the form ‖⋅‖:V→R such that, for any u,v∈V and a∈R, Positive definiteness: ‖u‖≥0, and ‖u‖=0 iff u=0, Homogeneity: ‖au‖=\|a\|‖u‖, Sub-additivity: ‖u+v‖≤‖u‖+‖v‖. Here, \|a\| refers to the absolute value of a∈R. The property of sub-additivity is also referred to as the triangle inequality. A vector space V equipped with a norm ‖⋅‖:V→R that satisfies these properties is called a normed vector space, or a normed linear space. Note that every inner product space is a normed linear space. To see this, note that given an an inner product ⋅:V×V→R, the norm ‖⋅‖:V→R induced by this inner product is defined as follows: for any v∈V, ‖v‖=√v⋅v. The norm induced by the Euclidean inner product on R3 is called the standard Euclidean norm on Rn. Remark In general, a normed vector space is not an inner product space. If, however, the norm ‖⋅‖:V→R on a normed vector space V satisfies the following relation, called the parallelogram identity, 12(‖u+v‖2+‖u−v‖2)=‖u‖2+‖v‖2, for any u,v∈V, then it is possible to define an inner product ⋅:V×V→R using the norm as follows: for any u,v∈V, u⋅v=14(‖u+v‖2−‖u−v‖2).This relation is called the polarization identity. Example Let us consider the n-dimensional Euclidean space Rn with the standard inner product, defined earlier. The norm of a vector (x1,…,xn)∈Rn is easily computed as ‖(x1,…,xn)‖2=n∑i=1x2i. This norm is called the standard Euclidean norm, or the L2-norm on Rn. A variety of other norms can be defined on a given inner product space. For instance, the Lp-norm on Rn can be defined as follows: for any (x1,…,xn)∈Rn and 1≤p<∞, ‖(x1,…,xn)‖p=(n∑i=1xpi)1p. The L∞-norm on Rn is defined as ‖(x1,…,xn)‖∞=max {x1,…,xn}.Note that the standard Euclidean norm corresponds to ‖⋅‖2. Remark There is an important theorem that states that all norms on a finite dimensional vector space are equivalent. This means the following: given norms ‖⋅‖1:V→R and ‖⋅‖2:V→R on a finite dimensional vector space V, there exists constants cL,cU∈R such that, for any v∈V, cL‖v‖2≤‖v‖1≤cU‖v‖2. Without getting into technical details, this roughly means that the conclusions we draw about topological notions in a normed vector space are independent of the specific norm chosen. Cauchy-Schwarz inequality Let V be a real vector space equipped with an inner product ⋅:V×V→R. An important property of inner products that turns out to be quite useful in practice is discussed now. Given any u,v∈V, the Cauchy-Schwarz inequality states that \|u⋅v\|≤‖u‖‖v‖. Furthermore, the equality holds iff u and v are linearly dependent. The proof of the Cauchy-Schwarz inequality is quite easy: for any u,v∈V, and a∈R, ‖u+av‖2≥0⇒‖u‖2+2au⋅v+a2‖v‖2≥0.Substituting a=−u⋅v‖v‖2in this inequality, we get (u⋅v)2≤‖u‖2‖v‖2,which immediately yields the Cauchy-Schwartz inequality. To prove the second part, note that if u and v are linearly dependent, then, without loss of generality, v=au for some a inR. In this case, u⋅v=\|a\|‖u‖2=‖u‖‖v‖. On the other hand, if u⋅v=‖u‖‖v‖, consider the vector w∈V, where, w=u−‖u‖‖v‖v.It is straightforward to show that ‖w‖=0, and hence that w=0. This shows that u and v are linearly dependent. The Cauchy-Schwartz inequality is thus proved. The angle θ:V×V→∈[0,2π)⊆R between two vectors u,v∈V is defined via the relation cosθ(u,v)=u⋅v‖u‖‖v‖. Note how the Cauchy-Schwarz inequality implies that this definition is well-defined. If the angle between two vectors u,v∈V is π/2, they are said to be orthogonal. Equivalently, u,v∈V are said to be orthogonal if u⋅v=0. If u,v∈V are orthogonal, and if it is further true that ‖u‖=‖v‖=1, then u and v are said to be orthonormal. Example The vectors u=(1,√3)∈R2 and v=(−√3,1)∈R2 are orthogonal since u⋅v=0. They are not orthonormal since ‖u‖=‖v‖=2. The vectors ˜u=u/‖u‖=(1/2,√3/2)∈R2 and v=v/‖v‖=(−√3/2,1/2)∈R2 are, however, orthonormal. The following inequality also holds for any u,v∈V in an inner product space V: ‖u+v‖≤‖u‖+‖v‖. Recall that this is the triangle inequality. The triangle inequality is readily proved using the Cauchy-Schwarz inequality: for any u,v∈V: ‖u+v‖2=‖u‖2+‖v‖2+2u⋅v≤‖u‖2+‖v‖2+2‖u‖‖v‖=(‖u‖+‖v‖)2.The triangle inequality follows by taking the square root on both sides. Gram-Schmidt orthogonalization Let us now reconsider the notion of a basis of an n-dimensional vector space V in the special case when the vector space also has an inner product ⋅:V×V→R defined on it. We say that a basis (gi) of V is orthogonal if gi⋅gj=0 whenever i,j∈{1,…,n}, and i≠j. If it is further true that gi⋅gi=1 for every i∈{1,…,n}, we say that the basis (gi) is orthonormal. The fact that a basis (gi) of V is orthonormal can be succinctly expressed by the following equation: for any i,j∈{1,…,n}, gi⋅gj=δij, where δij is the Krönecker delta symbol that is defined as follows: δij={1,i=j,0,i≠j. Example It is straightforward to verify that the standard basis (ei) of Rn is an orthonormal basis, since it follows from the definition of the standard basis that ei⋅ej=δij. The Gram-Schmidt orthogonalization procedure is an algorithm that helps us to transform any given basis (˜gi) of V into an orthonormal basis (gi). The algorithm works as follows: Let g1=˜g1‖˜g1‖. We now define g2 by removing the component of ˜g2 along the direction g1: g2=˜g2−(˜g2⋅g1)e1‖˜g2−(˜g2⋅g1)g1‖. It is easy to check that g2⋅g1=0 and ‖g2‖=1. We then obtain g3 in a similar manner by removing the components of ˜g3 along g1 and g2: g3=˜g3−(˜g3⋅g2)g2−(˜g3⋅g1)e1‖˜g3−(˜g3⋅g2)g2−(˜g3⋅g1)g1‖. It is straightforward to verify that g1⋅g3=0, g2⋅g3=0 and ‖g3‖=1. Continuing this process, we can construct an orthonormal basis (gi)ni=1. Example As a simple illustration of the Gram-Schmidt orthogonalization process, let us consider the vectors ˜g1=(1,2)∈R2 and ˜g2=(2,3)∈R2. We verified earlier that these vectors are linearly independent, and that they form a basis of R2. They are however not orthogonal since ˜g1⋅˜g2=8≠0. Let us orthonormalize this basis using the Gram-Schmidt process. To start with, let us normalize ˜g1: g1=˜g1‖˜g1‖=(1√5,2√5)∈R2. We can now construct g2∈R2 by projecting out the component of ˜g2 along g1: g2=˜g2−(˜g2⋅g1)g1‖˜g2−(˜g2⋅g1)g1‖=(2,3)−((2,3)⋅(1/√5,2/√5))(1/√5,2/√5)‖(2,3)−((2,3)⋅(1/√5,2/√5))(1/√5,2/√5)‖=(2/5,−1/5)‖(2/5,−1/5)‖=(2√5,−1√5)∈R2.It is easily checked that ‖g1‖=1 and ‖g2‖=1, and that g1⋅g2=(1√5,2√5)⋅(2√5,−1√5)=0.We have thus constructed an orthonormal basis (g1,g2) of R2 starting from the general basis (˜g1,˜g2) of R2 by following the Gram-Schmidt algorithm. Note that the orientation of the basis (g1,g2) obtained here is the opposite of the orientation of the standard basis (e1,e2). To see this, embed these vectors in R3 to get the vectors e1=(1,0,0), e2=(0,1,0), g1=(1/√5,2/√5,0), g2=(2/√5,−1/√5,0), and note that e1×e2=e3, whereas g1×g2=−√5ˆe3 - they have opposite signs! This doesn’t really affect the Gram-Schmidt algorithm because if (g1,g2) is a basis of R2, then so is (g2,g1). Remark It turns out that the choice of an orthonormal basis is sufficient for most applications. In the discussion below, we will first study various concepts with respect to the choice of an orthonormal basis, since the calculations are much simpler in this case. The general case of arbitrary bases will be discussed after this to give an idea of how some calculations can be more involved with respect to general bases. Basis representation of vectors We will now study the representation of a vector v∈V in an n-dimensional inner product space V with respect to an orthonormal basis (e1,…,en) of V. It is worth reiterating that we will only deal with orthonormal bases unless otherwise stated. The fact that (e1,…,en) is a basis of V implies that every v∈V can be written as v=n∑i=1viei where vi∈R for every i∈{1,…,n}. This is called the representation of v with respect to the basis (e1,…,en). The real numbers v1,…vn are called the components of v with respect to the basis (e1,…,en). To compute the components vi, we can exploit the fact that (ei) is an orthonormal basis: v⋅ej=vj⇒v=∑(v⋅ei)ei. The components vi thus computed are unique since we have explicitly constructed each component vi as v⋅ei. We can alternatively show the uniqueness of the components based on the fact that the basis vectors are linearly independent by definition. Remark Notice how we have represented the sum on the right without representing the summation index, and the range of summation. We will write ∑i, or just ∑ in place of ∑ni=1, whenever the range under consideration is obvious from the context. If no index is associated with the summation symbol, as in ∑, it will be assumed that the sum is with respect to all repeating indices. In addition, we will assume that the range of the all the indices involved is known from the context. While on this, it is worth noting that many authors employ the Einstein summation convention, according to which, a sum of the form ∑uiei is written simply as uiei, with the summation over i being implicitly understood as long as the indices repeat twice. For pedagogical reasons, we will not follow the Einstein summation convention in these notes. Example As a trivial example of the basis representation of a vector, consider any (x1,…,xn)∈Rn. This vector can be written with respect to the standard basis (ei) of Rn as (x1,…,xn)=∑xiei. Notice that xi=(x1,…,xn)⋅ei. Example As a non-trivial, yet simple, example of basis representation of a vector, consider the orthonormal basis (g1,g2) of R2, where g1=(2/√5,−1/√5) and g2=(1/√5,2/√5) - notice that we have swapped the order of the basis constructed earlier in the context of the Gram-Schmidt procedure to maintain orientation. Consider any (x1,x2)∈R2, we can express this in terms of the basis (g1,g2) as (x1,x2)=∑ˉxigi, for some real constants (ˉxi). To compute this, note that we can write (x1,x2)=∑xiei using the standard basis of R2. The constants (ˉxi) are easily computed by taking the appropriate dot products: ˉxi=(∑xjej)⋅gi=(gi⋅e1)x1+(gi⋅e2)x2.For instance, the vector (1,2)∈R2 can be expressed in terms of the (g1,g2) basis as (1,2)=∑ˉxigi, where ˉx1=2√51−1√52=0,ˉx2=1√51+2√52=√5.We thus see that (1,2)=√5g2, a fact that can be easily checked directly. We will now introduce a useful notion called component maps to collect the components vi of any v∈V with respect to the (not necessarily orthonormal) basis B=(e1,…,en) of V. Define the component map ϕV,B:V→Rn as follows: ϕV,B(v)=[v1,…,vn]T. Notice how we have collected together the components of v as a column vector of size n using the component map. It is useful at this point to introduce the following notation: v=[v1,…,vn]TWhen the choice of basis is evident from the context, we will often write v as just [v]. We can thus alternatively the basis representation of any v∈V with respect to a general basis (ei) of V as follows: v=∑viei=∑(ϕV,B(v))iei=∑[v]iei.We will however use the simpler notation v=∑viei, and use the [v] notation only when we want to refer to the components alone as a column vector. We will see shortly that the component map is an example of an isomorphism between the vector spaces V and Rn. Remark We will often write the component map ϕV,B as ϕV, or just ϕ, when the vector space and its basis are evident from the context. At times, we will omit ϕ altogether and refer to the component map using the following notation: v∈V↦[v]∈Rn. Example As a quick illustration of this, notice that the vector (1,2)∈R2 has the following matrix representation with respect to the orthonormal basis (e1,e2), where e1=(2/√5,−1/√5) and e2=(1/√5,2/√5) of R2 is [0,√5]T. Change of basis rules for vectors Given an n-dimensional inner product space V, let us first consider the case where (ei) and (gi) are two general bases of V, not necessarily orthonormal. Then, any v∈V can be written as v=∑viei=∑˜vigi, where (vi) and (˜vi) are the components of v with respect to the bases (ei) and (gi), respectively. The fact that (ei) is a basis of V implies that gi=∑Ajiej,where Aij∈R for every 1≤i,j≤n. Similarly, we have ei=∑Bjigj,where Bij∈R for every 1≤i,j≤n. Notice how the first index of the transformation coefficients pairs with the corresponding basis vector. The reason for this specific choice will become clear shortly. Combining these two transformation relations, we see that gi=∑AjiBkjgk⇒∑BkjAji=δki, and ei=∑BjiAkjek⇒∑AkjBji=δki.It is convenient to collect together the constants {Aij} as a matrix A whose (i,j)th entry is Aij. We will similarly collect the constants {Bij} in a matrix B. In matrix notation, we can write the foregoing equations succinctly as AB=I,BA=I,⇒A=B−1,where I is the identity matrix of order n. We thus see that matrices A and B are inverses of each other. Given the transformation relations between the two bases, we can use the identity ∑viei=∑˜vigi to see that ∑˜vigi=∑viBjigj⇒˜vi=∑Bijvj=∑A−1ijvj. In matrix notation, we can summarize the foregoing result as follows: gi=∑Ajiej⇒v=A−1v. Remark In a more general treatment of linear algebra, the fact that the components of a vector v∈V transform in a manner contrary to that of the manner in which the basis vectors transform is used to call elements of V as contravariant vectors. We will not develop the general theory here, but make a few elementary remarks occasionally regarding this. Let us now consider the special case when both (ei) and (gi) are orthonormal bases of V. In this case, we can use the fact that ei⋅ej=δij,gi⋅gj=δij, to simplify the calculations. Suppose that gi=∑Qjiej.It follows immediately that Qji=gi⋅ej.Let us now see how the components of any v∈V transform upon this change of basis: v=∑˜vigi=∑viei⇒˜vi=∑ej⋅givj=∑Qjivj.In matrix notation, this is written as v=QTv,where Q is the matrix whose (i,j)th entry is Qij. But, based on the calculation we carried out earlier in the context of general bases, we see that gi=∑Qjiej⇒v=Q−1v.Comparing these two expressions, we are led to the following conclusion: QT=Q−1.Recall that matrices that satisfy this condition are called orthogonal matrices. It is an easy consequence of orthogonality that the determinant of an orthogonal matrix is ±1, as the following calculation shows: if Q is an orthogonal matrix (det(Q))2=det(QTQ)=det(I)=1⇒det(Q)=±1.If det(Q)=1, then the orthogonal matrix Q is called proper orthogonal, or special orthogonal. Remark It is important to note that the foregoing conclusion that the matrix involved in the change of basis is orthogonal is true only in the special case when both bases are orthonormal. Example Consider the orthonormal bases (e1,e2) and (g1,g2) of R2, where (e1,e2) is the standard basis of R2 and g1=(2/√5,−1/√5), g2=(1/√5,2/√5). The transformation matrix Q from (e1,e2) to (g1,g2) is computed using the relation Qij=gj⋅ei as Q=[g1⋅e1g2⋅e1g1⋅e2g2⋅e2]=1√5[21−12]. It is easily checked that Q is orthogonal: QTQ=15[2−112][21−12]=.It can be similarly checked that QQT=I. As a quick check, note that also the determinant of the transformation map Q in the previous example is 1: det(1√5[21−12])=1. This informs us that the transformation matrix Q is in fact special orthogonal. General basis Most of the discussion thus far regarding the representation of a vector in a finite dimensional inner product space has been restricted to the special case of orthonormal bases. Let us briefly consider the general case when a general basis, which is not necessarily orthonormal, is chosen. In what follows, V denotes an inner product space of dimension n, and (gi) is a general basis of V. Representation of vectors Any v∈V can be written in terms of the basis (g1,…,gn) of V as v=∑vigi, where (v1,…,vn) are the components of v with respect to this basis. To compute these components, start with taking the inner product of this equation with the basis vector gi; this yields v⋅gi=∑gi⋅gjvj.This equation can be written in the form of a matrix equation, as follows: [g1⋅g1…g1⋅gn⋮⋱⋮gn⋅g1…gn⋅gn][v1⋮vn]=[v⋅g1⋮v⋅gn].The fact that (gi) is a basis of V implies that the components v1,…,vn exist and are unique. This implies that the matrix introduced above, whose (i,j)th entry is gij=gi⋅gj, is invertible. It is conventional, and convenient, to represent the inverse of this matrix as the matrix with entries gij; thus, [g11…g1n⋮⋱⋮gn1…gnn]−1=[g11…g1n⋮⋱⋮gn1…gnn].A proper justification for this choice of notation will be given shortly when we study reciprocal bases. The fact that these two matrices are inverses of each other can be written succinctly as follows: ∑gikgkj=δij=∑gikgkj.Using this result, a trite calculation yields the following result: for any v∈V, v=∑vigi⇒vi=∑gijgj⋅v.The components of any vector with respect to a general basis can thus be computed explicitly. Example Consider the basis (g1,g2) of R2, where g1=(1,2) and g2=(2,3). Let us now compute the components of v=(2,5)∈R2 with respect to this basis. The first step in to compute the matrix whose entries are gij: [g1⋅g1g1⋅g2g2⋅g1g2⋅g2]=. Notice that this matrix is symmetric, as expected, since gij=gji, in general. The inverse of this matrix gives the scalars (gij) as follows: [g11g12g21g22]=−1=[13−8−85].The components of v with respect to the basis (gi) are now easily computed using the result vi=∑gijgj⋅v as, v1=∑g1jgj⋅v=g11⋅+g12⋅=4,v2=∑g2jgj⋅v=g21⋅+g22⋅=−1.We thus see that v=4g1−g2. As a consistency check, substitute the representations of v,g1,g2 with respect to the standard basis of R2 and verify that this is correct. Reciprocal basis The computations presented in the previous section can be greatly simplified by introducing the reciprocal basis corresponding to a given basis. Given a basis (g1,…,gn) of V, its reciprocal basis is defined as the basis (g1,…,gn) such that gi⋅gj=δij, where 1≤i,j≤n. Based on the preceding development, it can be seen that the reciprocal basis is explicitly given by the following equations: gi=∑gijgj.This can be readily inverted to yield the following equation: gi=∑gijgj.Note that in the special case of the standard basis (ei) of R3, ei=ei. More generally, if (gi) is an orthonormal basis of an inner product space V, then gi=gi. This is one of the reasons why many calculations are much simpler when using orthonormal bases. It follows from the definition of the reciprocal basis (gi) of V that gi⋅gj=(∑gikgk)(∑gjlgl)=∑gikgjlgkl=∑δilgjl=gij. Thus, the following useful formulate are obtained: if (gi) is a general basis of V and (gi) is its reciprocal basis, then gi⋅gj=gij and gi⋅gj=gij. Remark The use of superscripts here is done purely for notational convenience. It is however possible to justify such a notation when considering a more detailed treatment of this subject, as will be briefly noted later. Example Consider the previous example involving the basis (g1,g2) of R2, where g1=(1,2) and g2=(2,3). In this case, the reciprocal basis (g1,g2) is computed as follows: g1=∑g1jgj=13⋅(1,2)−8⋅(2,3)=(−3,2),g2=∑g2jgj=−8⋅(1,2)+5⋅(2,3)=(2,−1). It can be checked with a simple calculation that gi⋅gj=δij, as expected. Further more, the matrix whose (i,j)th entry is gi⋅gj is computed as [g1⋅g1g1⋅g2g2⋅g1g2⋅g2]=[13−8−85]=−1=[g1⋅g1g1⋅g2g2⋅g1g2⋅g2]−1.This confirms that the matrix whose (i,j)th entry is gi⋅gj is indeed the inverse of the matrix whose (i,j)th entry is gi⋅gj. Example The reciprocal basis can be computed easily in the special case of the three dimensional Euclidean space R3 using the cross product. Given any basis (gi) of R3, the corresponding reciprocal basis (gi) of R3 can be computed as g1=g2×g3g1⋅g2×g3,g2=g3×g1g1⋅g2×g3,g3=g1×g2g1⋅g2×g3. It is a simple exercise to check these formulae satisfy the defining condition of the reciprocal basis: gi⋅gj=δij. The expressions for the coefficients of a vector with respect to a given basis simple form when expressed in terms of the reciprocal basis. Given any v∈V and a basis (gi) of V, v=∑˜vigi⇒˜vi=v⋅gi. Thus, any v∈V has the compact representation v=∑(v⋅gi)gi.Compare this with the representation v=∑(v⋅ei)ei of v with respect to an orthonormal basis (ei) of V. Remark Given any v∈V and a basis (gi) of V, the components of v with respect to the (gi) and its reciprocal basis (gi) are written as follows: v=∑vigi=∑v∗igi. The components (vi) and (v∗i) are called the contravariant and covariant components of v, respectively. In many textbooks, the following alternative notation is used: v=∑vigi=∑vigi.The components vi and vi are related as follows: vi=gijvj and vi=gijvj. For this reason, gij and gij are said to raise and lower, respectively, indices. Since we will largely restrict ourselves to the case of orthonormal bases and rarely represent a vector in terms of the reciprocal basis to a given basis, we will not adopt this more nuanced notation here. Change of basis rules The ideas presented so far can be used to express a given vector v∈V with respect to different bases. Suppose that v has the following representations, with respect to two different bases (fi) and (gi) of V: v=∑ˉvifi=∑˜vigi. Taking the inner product of these representations with respect to the appropriate reciprocal basis vectors, it is evident that v=∑ˉvifi=∑˜vigi⇒ˉvi=∑fi⋅gj˜vj,and a similar formula expressing (˜vi) in terms of (ˉvi). Notice how the use of the reciprocal basis significantly simplifies the computations. Example Consider the example considered earlier where the vector (2,5)∈R2 was expressed in terms of the basis (g1,g2) of R2, where g1=(1,2) and g2=(2,3). We saw earlier that v=4g1−g2. Let us now consider another basis (f1,f2) of R2, where f1=(2,1), f2=(1,3). To compute the representation of v with respect to the basis (f1,f2), we first need to compute its reciprocal basis (f1,f2). This is easily accomplished as follows: [f1⋅f1f1⋅f2f2⋅f1f2⋅f2]=⇒[f11f12f21f22]=15[2−1−11]The reciprocal basis is computed using the relations fi=∑fijfj: f1=∑f1jfj=25(2,1)−15(1,3)=15(3,−1),f2=∑f2jfj=−15(2,1)+15(1,3)=15(−1,2).It is left as an easy exercise to verify that fi⋅fj=δij. Using these relations the components (ˉv1,ˉv2) of v with respect to the basis (f1,f2) can be computed using the relations ˉvi=v⋅fi as follows: ˉv1=v⋅f1=(2,5)⋅15(3,−1)=15,ˉv2=v⋅f2=(2,5)⋅15(−1,2)=85.We thus see get the representation of v in the basis (f1,f2) as v=15(f1+8f2).It is left as a simple exercise to verify by direct substitution that this is true. Finally, note that the components (ˉv1,ˉv2) of v with respect to the basis (f1,f2) can be directly obtained from its components (˜v1,˜v2) with respect to the basis (g1,g2) using the relations ˉvi=∑fi⋅gj˜vj as follows: ˉv1=∑f1⋅gj˜vj=45(3,−1)⋅(1,2)−15(3,−1)⋅(2,3)=15,ˉv2=∑f2⋅gj˜vj=45(−1,2)⋅(1,2)−15(−1,2)⋅(2,3)=85. We thus see that all the different ways to compute the components of v with respect to two different choices of bases are consistent with each other.
189143	https://arminstraub.com/downloads/teaching/numbertheory-fall20/lectures-15-17.pdf	Notes for Lecture 15 Tue, 10/13/2020 Example 128. (a) Show that 7 is a primitive root modulo 26. (b) Using the ﬁrst part, make a complete list of all primitive roots modulo 26. Solution. (a) We need to show that 7 has order φ(26) = 12. The order of 7 (or any invertible residue) must divide φ(26) = 12. Hence, the only possibilities for orders are 1; 2; 3; 4; 6; 12. The fact that 74 ≡(−3)2 ≡9 ≡ / 1 (mod 26) and 76 ≡(−3)3 ≡−1 ≡ / 1 (mod 26) is enough (why?!) to conclude that the order of 7 must be 12. (b) Since 7 is a primitive root, all other invertible residues are of the form 7a. Recall that 7a has order 12 gcd (12; a). Thus, 7a is a primitive root if and only if gcd(12; a) = 1. Therefore, a list of all primitive roots modulo 26 is: 7; 75; 77; 711 [These are φ(φ(26)) = φ(12) = 4 many primitive roots.] The same logic applies whenever there is at least one primitive root: Theorem 129. (number of primitive roots) Suppose there is a primitive root modulo n. Then there are φ(φ(n)) primitive roots modulo n. Proof. Let x be a primitive root. It has order φ(n). All other invertible residues are of the form xa. Recall that xa has order φ(n) gcd (φ(n); a). This is φ(n) if and only if gcd (φ(n); a) = 1. There are φ(φ(n)) values a among 1; 2; :::; φ(n), which are coprime to φ(n). In conclusion, there are φ(φ(n)) primitive roots modulo n. □ Comment. Recall that, for instance, there is no primitive root modulo 8. That's why we needed the assumption that there should be a primitive root modulo n (which is the case if and only if n is of the form 1; 2; 4; pk; 2pk for some odd prime p). Corollary 130. There are φ(φ(p)) = φ(p −1) primitive roots modulo a prime p. Example 131. Let p be an odd prime. Show that at most half of the invertible residues modulo p are primitive roots. Solution. In other words, we need to show that φ(p −1) p −1 6 1 2. Let p1; p2; ::: be the primes, in increasing order, dividing p −1. Since p = / 2, p −1 is divisible by 2, so that p1 = 2. Then, φ(p −1) = (p −1) 1 −1 p1 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } =1/2 1 −1 p2 ··· \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } 61 6 1 2(p −1). Consequently, φ(p −1) p −1 6 1 2(p −1) p −1 = 1 2, as claimed. In fact. Note that 1 −1 p2 <1 if there is a second prime. Our proof therefore actually shows that φ(p −1) p −1 = 1 2 if and only if p −1 is of the form 2n (i.e. the only prime dividing p −1 is 2). Equivalently, if p is of the form 2n+1. Comment. Primes of the form 2n+1 are known as Fermat primes. It can be shown that such a prime is, in fact, necessarily of the form Fk = 22k + 1. The ﬁrst ﬁve numbers F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537 are prime, and Fermat conjectured that Fk is prime for all k >0. This was proven wrong by Euler who demonstrated that F5 = 232 + 1 = 641 · 6700417 (this was way before the time, we could ask a computer to factor not-too-large numbers). To this day, it is not known whether any further Fermat primes exist. Armin Straub straub@southalabama.edu 34 Example 132. Recall that, for every prime p, primitive roots exist. The total number of primitive roots is φ(φ(p))= φ(p−1). The following computations in Sage indicate that typically a “decent” proportion (25-50%) of all invertible residues are primitive roots. The exact proportion is, of course φ(p −1) p −1 but to say more about the magnitude, we need the factorization of p −1. Advanced comment. However, the number of primitive roots can (though this is very rare) be an arbitrarily small proportion. In fact, a result of Kátai shows that, for each x 2 [0; 1], there is a proportion P (x) of primes with φ(p −1) p −1 6 x, and that P (x) is a strictly increasing continuous function with P (0) = 0 and P (1/2) = 1. Sage] prime_range(30) [2; 3; 5; 7; 11; 13; 17; 19; 23; 29] Sage] euler_phi(26) 12 Sage] [p^2 for p in prime_range(30)] [4; 9; 25; 49; 121; 169; 289; 361; 529; 841] Sage] [euler_phi(p-1)/(p-1) for p in prime_range(30)] 1; 1 2; 1 2; 1 3; 2 5; 1 3; 1 2; 1 3; 5 11; 3 7 Sage] list_plot([euler_phi(p-1)/(p-1) for p in prime_range(3,10000)]) 0 200 400 600 800 1000 1200 0.25 0.3 0.35 0.4 0.45 0.5 Armin Straub straub@southalabama.edu 35 Notes for Lecture 16 Thu, 10/15/2020 14 Applying the CRT to computing powers If we know the factorization of the modulus, the CRT speeds up the computation of powers: Example 133. Compute 329 (mod 77) using the Chinese remainder theorem. Solution. We determine x = 329 both modulo 7 and 11: • 329 ≡35 ≡3 · 4 ≡−2 (mod7) [Here, we used 29≡5 (mod φ(7)) and 32 ≡2, 34 ≡4 (mod7).] • 329 ≡3−1 ≡4 (mod11) [Here, we proceeded unusually and used 29≡−1 (mod φ(11)).] Therefore, x ≡−2 (mod7) and x ≡4 (mod11). Using the Chinese remainder theorem, x = −2 · 11· 11mod7 −1 2 + 4 · 7 · 7mod11 −1 −3 ≡−128≡26 (mod77). Comment. Alternatively, we can proceed modulo n = 77 directly and use binary exponentiation. However, if we already know the factorization of n (that's a big “if” for large n), then applying the Chinese remainder theorem (followed by binary exponentiation) is a little faster. Example 134. (review) Compute 7100 (mod 60). Solution. φ(60) = φ(22)φ(3)φ(5) = 2 · 2 · 4 = 16. Since gcd (7; 60) = 1, we obtain that 716 ≡1 (mod 60) by Euler's theorem. Since 100≡4 (mod16), we have 7100 ≡74 (mod60). It remains to notice that 72 = 49≡−11 and hence 74 ≡(−11)2 = 121≡1 (mod60). So, 7100 ≡1 (mod60). Comment. The next example shows that we actually have a4 ≡1 (mod60) for all integers a coprime to 60. Euler's theorem doesn't necessarily provide an optimal exponent. For instance: Example 135. Show that a4 ≡1 (mod 60) for all integers a coprime to 60. Note. Since φ(60) = φ(22)φ(3)φ(5) = 2 · 2 · 4 = 16, Euler's theorem shows that a16 ≡1 (mod60). Proof. By the Chinese remainder theorem, a4 ≡1 (mod60) is equivalent to a4 ≡1 (mod4); a4 ≡1 (mod3); a4 ≡1 (mod5): All three of these congruences are true: • a4 ≡1 (mod5) is true by Fermat's little theorem. • a4 ≡1 (mod3) is true, because a2 ≡1 (mod3) by Fermat's little theorem. • a4 ≡1 (mod4) is true, because a2 ≡1 (mod4) by Euler's theorem (φ(4) = 2). (Note that a is coprime to 60 if and only if a is coprime to each of 4, 3, 5.) □ A brute-force veriﬁcation in Sage. The following computation also proves the claim. Even if you have never coded yourself, you can surely ﬁgure out what the following code is doing: Sage] [1..59] [1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; 49; 50; 51; 52; 53; 54; 55; 56; 57; 58; 59] Sage] [ x for x in [1..59] if gcd(x,60)==1 ] [1; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 49; 53; 59] Sage] [ power_mod(x,4,60) for x in [1..59] if gcd(x,60)==1 ] [1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1] Armin Straub straub@southalabama.edu 36 Proceeding as in Example 135, we obtain the following result: Theorem 136. (strengthening of Euler's theorem) Suppose n has the prime factorization n = p1 r1p2 r2···pm rm. If gcd (a; n) = 1, then aλ(n) ≡1 (mod n) where λ(n) = lcm (φ(p1 r1); φ(p2 r2); :::; φ(pm rm)): Advanced. The exponent λ(n) in this result is best possible unless n is divisible by 8; see, for instance: The only improvement that can be made is that, in the computation of λ(n), φ(2r) may be replaced with 1 2φ(2r) if r > 3. This is known as Carmichael's theorem. Example 137. Since 60 = 22 · 3 · 5, we have λ(60) = lcm (φ(22); φ(3); φ(5)) = lcm (2; 2; 4) = 4. By the theorem, a4 ≡1 (mod 60) whenever gcd (a; 60) = 1, as we had observed in Example 135. Example 138. Based on Euler's theorem and the Chinese remainder theorem, ﬁnd the smallest exponent k such that ak ≡1 (mod 42) for all integers a coprime to 42. Solution. Since 42= 2 · 3 · 7, the smallest such exponent is λ(42) = lcm(φ(2); φ(3); φ(7)) = lcm(1; 2; 6) = 6. Note. Since φ(42)= φ(2)φ(3)φ(7)=1·2·6=12, this improves the exponent in Euler's theorem by a factor of 2. Note. In this case, we can easily see that the exponent 6 cannot be further decreased. For instance, take a residue that is a primitive root modulo 7; its order modulo 42 must be at least 6 as well. Example 139. Based on Euler's theorem and the Chinese remainder theorem, ﬁnd the smallest exponent k > 1 such that ak ≡1 (mod 675) for all integers a coprime to 675. Solution. Since 675= 33 · 52, the smallest such exponent is λ(675) = lcm(φ(33); φ(52)) = lcm(18; 20) = 180. Note. Since φ(675)= φ(33)φ(52)=18·20=360, this improves the exponent in Euler's theorem by a factor of 2. The following is a variation of the same idea we used in Example 135. Lemma 140. Suppose that m and n are coprime. If x (mod n) has multiplicative order r and x (mod m) has multiplicative order s, then x (mod mn) has multiplicative order lcm (r; s). Proof. Suppose that xk ≡1 (modmn). Equivalently, by the CRT, xk ≡1 (modm) as well as xk ≡1 (modn). By Lemma 126, this is further equivalent to rjk and sjk. Clearly, the smallest such k is k = lcm(r; s). □ Example 141. Determine the order of 2 (mod 77) using the Chinese remainder theorem. Solution. We ﬁrst determine the orders of 2 (mod7) and 2 (mod11). • modulo 7 (since φ(7) = 6, the possible orders are 2; 3; 6): 22 ≡4, 23 ≡4 · 2 ≡1. Thus, 2 (mod7) has order 3. • modulo 11 (since φ(11) = 10, the possible orders are 2; 5; 10): 22 ≡4, 25 ≡24 · 2 ≡5 · 2 ≡−1 ≡ / 1. Thus, 2 (mod11) has order 10. Taken together, it follows from the lemma that 2 (mod77) has order lcm(3; 10) = 30. Note. Can you see, from these considerations, why there cannot exist a primitive root modulo 77? Armin Straub straub@southalabama.edu 37 Notes for Lecture 17 Tue, 10/20/2020 Example 142. Fermat's little theorem can be stated in the slightly stronger form: n is a prime ( ) an−1 ≡1 (mod n) for all a 2 f1; 2; :::; n −1g Why? Fermat's little theorem covers the “= )” part. The “( =” part is a direct consequence of the fact that, if n is composite with divisor d, then dn−1 ≡ / 1 (modn). (Why?!) Review. In the second part, we used that the contrapositive of A = ) B is the logically equivalent statement :B = ) :A. 15 Primality testing Recall that it is extremely diﬃcult to factor large integers (this is the starting point for many cryptosystems). Surprisingly, it is much simpler to tell if a number is prime. Example 143. The following is the number mentioned earlier, for which RSA Laboratories, until 2007, oﬀered $100,000 to the ﬁrst one to factorize it. To this day, nobody has been able to do so. Has the thought crossed your mind that the challengers might be tricking everybody by choosing M to be a huge prime that cannot be factored further? Well, we'll talk more about primality testing soon. But we can actually quickly convince ourselves that M cannot be a prime. If M was prime then, by Fermat's little theorem, 2M −1 ≡1 (mod M). Below, we compute 2M −1 (mod M) and ﬁnd that 2M −1 ≡ / 1 (mod M). This proves that M is not a prime. It doesn't bring us any closer to factoring it though. Comment. Ponder this for a while. We can tell that a number is composite without ﬁnding its factors. Both sides to this story (ﬁrst, being able to eﬃciently tell whether a number is prime, and second, not being able to factor large numbers) are of vital importance to modern cryptography. Sage] rsa = Integer("135066410865995223349603216278805969938881475605667027524485143851\ 526510604859533833940287150571909441798207282164471551373680419703\ 964191743046496589274256239341020864383202110372958725762358509643\ 110564073501508187510676594629205563685529475213500852879416377328\ 533906109750544334999811150056977236890927563") Sage] power_mod(2, rsa-1, rsa) 12093909443203361586765059535295699686754009846358895123890280836755673393220205933853\ 34853414711666284196812410728851237390407107713940535284883571049840919300313784787895\ 22602961512328487951379812740630047269392550033149751910347995109663412317772521248297\ 950196643140069546889855131459759160570963857373851 Comment. Just for giggles, let us emphasize once more the need to compute 2N −1 (modN) without actually computing 2N −1. Take, for instance, the 1024 bit RSA challenge number N = 135:::563 in this example. The number 2N −1 itself has N −1 ≈21024 ≈10308.3 binary digits. It is often quoted that the number of particles in the visible universe is estimated to be between 1080 and 10100. Whatever these estimates are worth, our number has WAY more digits (!) than that. Good luck writing it out! [Of course, the binary digits are a single 1 followed by all zeros. However, we need to further compute with that!] Comment. There is nothing special about 2 in this discussion. You could just as well use, say, 3. Fermat primality test Input: number n and parameter k indicating the number of tests to run Output: “not prime” or “likely prime” Algorithm: Repeat k times: Pick a random number a from f2; 3; :::; n −2g. If an−1 ≡ / 1 (modn), then stop and output “not prime”. Output “likely prime”. Armin Straub straub@southalabama.edu 38 If an−1 ≡1 (modn) although n is composite, then a is called a Fermat liar modulo n. On the other hand, if an−1 ≡ / 1 (modn), then n is composite and a is called a Fermat witness modulo n. Flaw. There exist certain composite numbers n (see Deﬁnition 145) for which every a is a Fermat liar (or reveals a factor of n). For this reason, the Fermat primality test should not be used as a general test for primality. That being said, for very large random numbers, it is exceedingly unlikely to meet one of these troublesome numbers, and so the Fermat test is indeed used for the purpose of randomly generating huge primes (for instance in PGP). In fact, in that case, we can even always choose a = 2 and k = 1 with virtual certainty of not messing up. There do exist extensions of the Fermat primality test which solve these issues. [For instance, Miller-Rabin, which checks whether an−1 ≡1 (mod n) but also checks whether values like a(n−1)/2 are congruent to ±1.] Advanced comment. If n is composite but not an absolute pseudoprime (see Deﬁnition 145), then at least half of the values for a satisfy an−1 ≡ / 1 (modn) and so reveal that n is not a prime. This is more of a theoretical result: for most large composite n, almost every a (not just half) will be a Fermat witness. Example 144. Suppose we want to determine whether n = 221 is a prime. Simulate the Fermat primality test for the choices a = 38 and a = 24. Solution. • First, maybe we pick a = 38 randomly from f2; 3; :::; 219g. We then calculate that 38220 ≡1 (mod221). So far, 221 is behaving like a prime. • Next, we might pick a = 24 randomly from f2; 3; :::; 219g. We then calculate that 24220 ≡81≡ / 1 (mod221). We stop and conclude that 221 is not a prime. Important comment. We have done so without ﬁnding a factor of n. (To wit, 221= 13· 17.) Comment. Since 38 was giving us a false impression regarding the primality of n, it is called a Fermat liar modulo 221. Similarly, we say that 221 is a pseudoprime to the base 38. On the other hand, we say that 24 was a Fermat witness modulo 221. Comment. In this example, we were actually unlucky that our ﬁrst “random” pick was a Fermat liar: only 14 of the 218 numbers (about 6.4%) are liars. As indicated above, for most large composite numbers, the proportion of liars will be exceedingly small. Somewhat suprisingly, there exist composite numbers n with the following disturbing property: every residue a is a Fermat liar or gcd (a; n) > 1. This means that the Fermat primality test is unable to distinguish n from a prime, unless the randomly picked number a happens to reveal a factor (namely, gcd(a; n)) of n (which is exceedingly unlikely for large numbers). [Recall that, for large numbers, we do not know how to ﬁnd factors even if that was our primary goal.] Such numbers are called absolute pseudoprimes: Deﬁnition 145. A composite positive integer n is an absolute pseudoprime (or Carmichael number) if an−1 ≡1 (mod n) holds for each integer a with gcd (a; n) = 1. The ﬁrst few are 561; 1105; 1729; 2465; ::: (it was only shown in 1994 that there are inﬁnitely many of them). These are very rare, however: there are 43 absolute pseudoprimes less than 106. (Versus 78; 498 primes.) Armin Straub straub@southalabama.edu 39 Example 146. Show that 561 is an absolute pseudoprime. Solution. (using the strengthening of Euler's theorem) We need to show that a560 ≡1 (mod 561) for all invertible residues a modulo 561. Since 561 = 3 · 11 · 17 and lcm(φ(3); φ(11); φ(17)) = lcm (2; 10; 16) = 80, it follows from Theorem 136 that a80 ≡1 (mod 561) for all invertible residues a modulo 561. Since 560 is a multiple of 80, it follows that a560 ≡1 (mod561). Solution. (direct) We need to show that a560 ≡1 (mod561) for all invertible residues a modulo 561. Since 561= 3 · 11· 17, a560 ≡1 (mod561) is eqivalent to a560 ≡1 (mod p) for each of p = 3; 11; 17. By Fermat's little theorem, we have a2 ≡1 (mod3), a10 ≡1 (mod11), a16 ≡1 (mod17). Since 2; 10; 16 each divide 560, it follows that indeed a560 ≡1 (mod p) for p = 3; 11; 17. Comment. Korselt's criterion (1899) states that what we just observed in fact characterizes absolute pseudo-primes. Namely, a composite number n is an absolute pseudoprime if and only if n is squarefree, and for all primes p dividing n, we also have p −1jn −1. Comment. Our argument above shows that, in fact, a80 ≡1 (mod561) for all invertible residues a modulo 561. Note Theorem 147. (Korselt's Criterion) A composite positive integer n is an absolute pseudoprime if and only if n is squarefree and (p −1)j(n −1) for each prime divisor p of n. Proof. Here, we will only consider the “if” part (the “only if” part is also not hard to show but the typical proof requires a little more insight into primitive roots than we currently have). To that end, assume that n is squarefree and that (p −1)j(n −1) for each prime divisor p of n. Let a be any integer with gcd(a; n) = 1. We will show that an−1 ≡1 (modn). n being squarefree means that its prime factorization is of the form n = p1·p2···pd for distinct primes pi (this is equivalent to saying that there is no integer m > 1 such that m2jn). By Fermat's little theorem api−1 ≡1 (mod pi) and, since (pi −1)j(n −1), we have an−1 ≡1 (mod pi) for all pi. It therefore follows from the Chinese remainder theorem that an−1 ≡1 (modn). □ Comment. Modulo a prime p, Fermat's little theorem implies that ap ≡a (mod p) for each integer a. (Why?!) It therefore follows from the above argument that, for an absolute pseudoprime n, we have an ≡a (modn) for each integer a (and this property characterizes absolute pseudoprimes). Example 148. Using Sage, determine all numbers n up to 5000, for which 2 is a Fermat liar. Sage] def is_fermat_liar(x, n): return not is_prime(n) and power_mod(x, n-1, n) == 1 Sage] [ n for n in [1..5000] if is_fermat_liar(2, n) ] [341; 561; 645; 1105; 1387; 1729; 1905; 2047; 2465; 2701; 2821; 3277; 4033; 4369; 4371; 4681] Even if you have never written any code, you can surely ﬁgure out what's going on! Armin Straub straub@southalabama.edu 40
189144	https://artofproblemsolving.com/wiki/index.php/Modular_arithmetic/Introduction?srsltid=AfmBOopzJZQTRYAdtxl035EySHeh-UDHCRhrbiX1-akZH7K6lJG0trnX	Art of Problem Solving Modular arithmetic/Introduction - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Modular arithmetic/Introduction Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Modular arithmetic/Introduction Modular arithmetic is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic. Contents [hide] 1 Introductory Video 2 Understand Modular Arithmetic 3 Residue 4 Congruence 4.1 Examples 4.2 Sample Problem 4.2.1 Solution: 4.2.2 Another Solution: 4.2.3 Another Solution: 5 Making Computation Easier 5.1 Addition 5.1.1 Problem 5.1.2 Solution 5.1.3 Why we only need to use remainders 5.1.4 Solution using modular arithmetic 5.1.5 Addition rule 5.1.6 Proof of the addition rule 5.2 Subtraction 5.2.1 Problem 5.2.2 Solution 5.2.3 Subtraction rule 5.3 Multiplication 5.3.1 Problem 5.3.2 Solution 5.3.3 Solution using modular arithmetic 5.3.4 Multiplication rule 5.4 Exponentiation 5.4.1 Problem #1 5.4.2 Problem #2 5.4.3 Problem #3 6 Summary of Useful Facts 7 Problem Applications 8 Applications of Modular Arithmetic 9 Resources 10 See also Introductory Video Understand Modular Arithmetic Let's use a clock as an example, except let's replace the at the top of the clock with a . This is the way in which we count in modulo 12. When we add to , we arrive back at . The same is true in any other modulus (modular arithmetic system). In modulo , we count We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from to , when written in modulo 5, are where is the same as in modulo 5. Because all integers can be expressed as , , , , or in modulo 5, we give these integers their own name: the residue classes modulo 5. In general, for a natural number that is greater than 1, the modulo residues are the integers that are whole numbers less than : This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily! Residue We say that is the modulo-residue of when , and . Congruence There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol : In other words, this means in base 5, these integers have the same residue modulo 5: The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. In general, two integers and are congruent modulo when is a multiple of . In other words, when is an integer. Otherwise, , which means that and are not congruent modulo . Examples because is a multiple of . because , which is an integer. because , which is not a multiple of . because , which is not an integer. Sample Problem Find the modulo residue of . Solution: Since R , we know that and is the modulo residue of . Another Solution: Since , we know that We can now solve it easily and is the modulo residue of Another Solution: We know is a multiple of since is a multiple of . Thus, and is the modulo residue of . Making Computation Easier We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by , then we can work directly with those remainders in modulo . This can be more easily understood with a few examples. Addition Problem Suppose we want to find the units digit of the following sum: We could find their sum, which is , and note that the units digit is . However, we could find the units digit with far less calculation. Solution We can simply add the units digits of the addends: The units digit of this sum is , which must be the same as the units digit of the four-digit sum we computed earlier. Why we only need to use remainders We can rewrite each of the integers in terms of multiples of and remainders: . When we add all four integers, we get At this point, we already see the units digits grouped apart and added to a multiple of (which will not affect the units digit of the sum): . Solution using modular arithmetic Now let's look back at this solution, using modular arithmetic from the start. Note that Because we only need the modulo residue of the sum, we add just the residues of the summands: so the units digit of the sum is just . Addition rule In general, when , and are integers and is a positive integer such that the following is always true: . And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle. Proof of the addition rule Let , and where and are integers. Adding the two equations we get: Which is equivalent to saying Subtraction The same shortcut that works with addition of remainders works also with subtraction. Problem Find the remainder when the difference between and is divided by . Solution Note that and . So, Thus, so 1 is the remainder when the difference is divided by . (Perform the subtraction yourself, divide by , and see!) Subtraction rule When , and are integers and is a positive integer such that the following is always true: Multiplication Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point. Problem Jerry has boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are cans of soda in each box. Jerry plans to pack the sodas into cases of cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover? Solution First, we note that this word problem is asking us to find the remainder when the product is divided by . Now, we can write each and in terms of multiples of and remainders: This gives us a nice way to view their product: Using FOIL, we get that this equals We can already see that each part of the product is a multiple of , except the product of the remainders when each and are divided by 12. That part of the product is , which leaves a remainder of when divided by . So, Jerry has sodas leftover after making as many cases of as possible. Solution using modular arithmetic First, we note that Thus, meaning there are sodas leftover. Yeah, that was much easier. Multiplication rule When , and are integers and is a positive integer such that The following is always true: . Exponentiation Since exponentiation is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the intermediate modular arithmetic article. Note to everybody: Exponentiation is very useful as in the following problem: Problem #1 What is the last digit of if there are 1000 7s as exponents and only one 7 in the middle? We can solve this problem using mods. This can also be stated as . After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. is simply 1, so therefore , which really is the last digit. Problem #2 What are the tens and units digits of ? We could (in theory) solve this problem by trying to compute , but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by . In other words, all of the information we need can be found using arithmetic mod . We begin by writing down the first few powers of mod : A pattern emerges! We see that So for any positive integer , we have (mod ). In particular, we can write . By the "multiplication" property above, then, it follows that (mod ). Therefore, by the definition of congruence, differs from by a multiple of . Since both integers are positive, this means that they share the same tens and units digits. Those digits are and , respectively. Problem #3 Can you find a number that is both a multiple of but not a multiple of and a perfect square? No, you cannot. Rewriting the question, we see that it asks us to find an integer that satisfies . Taking mod on both sides, we find that . Now, all we are missing is proof that no matter what is, will never be a multiple of plus , so we work with cases: This assures us that it is impossible to find such a number. Summary of Useful Facts Consider four integers and a positive integer such that and . In modular arithmetic, the following identities hold: Addition: . Subtraction: . Multiplication: . Division: , where is a positive integer that divides and . Exponentiation: where is a positive integer. Problem Applications Applications of Modular Arithmetic Modular arithmetic is an extremely flexible problem solving tool. The following topics are just a few applications and extensions of its use: Divisibility rules Linear congruences Resources The AoPS Introduction to Number Theory by Mathew Crawford. The AoPS Introduction to Number Theory Course. Thousands of students have learned more about modular arithmetic and problem solving from this 12 week class. 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189145	https://m.mikeppt.com/sou-ppt/%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%20%20%E5%87%BD%E6%95%B0.html	高一数学函数PPT-高一数学函数PPT模板下载-麦克PPT网麦克PPT网 PPT模板 PPT模板 Word Excel PPT背景 PPT素材 PPT图表 PPT教程首页 PPT模板 PPT课件行业PPT 节日PPT PPT背景 PPT素材 PPT图表求职简历 Word模板 Excel模板当前位置：首页>PPT模板> 高一数学函数PPT 含教案人教A高一数学必修第一册5.4.1正弦函数、余弦函数的图象PPT课件含教案页数：49\|大小：5M 这是一套专为人教A版高一数学必修第一册第五章“三角函数”中“5.4.1正弦函数、余弦函数的图象”设计的PPT课件模板，总页数为49页，内容系统地分为四个主要部分，旨在帮助学生全面而深入地理解和掌握相关知识。在第一部分“正弦函数、余弦函数图象”中，详细介绍了正弦函数和余弦函数图象的基本概念。通过单位圆的直观展示，引导学生逐步掌握如何绘制这两种函数的图象，并深入阐述了函数的周期性特点，为学生后续学习函数的性质和应用奠定了基础。第二部分聚焦于“五点（画图）法”这一实用的作图方法。课件不仅详细讲解了这种方法的具体步骤和关键技巧，还通过典型例题的逐步演示，帮助学生学会如何绘制函数的简图，并引导学生分析图象的特征，使学生能够更加直观地理解正弦函数和余弦函数的图象形态。第三部分“题型强化训练”内容丰富多样，涵盖了用五点法作图、图象变换、解三角方程与不等式等多个重点题型。针对每一类问题，课件都提供了详细的示例解析和解题策略总结，旨在通过多样化的练习，提升学生的综合应用能力，帮助学生更好地掌握和运用所学知识。最后的“小结及随堂练习”部分，对全课的知识要点和方法进行了系统的梳理和归纳。通过多种练习题的设计，为学生提供了自我检测和巩固理解的机会，帮助学生进一步加深对正弦函数和余弦函数图象绘制方法的理解，并能够灵活运用于实际问题的解决中。整个PPT课件结构层次清晰，逻辑严谨，内容丰富实用，非常适合用于课堂教学，能够有效地帮助学生扎实掌握正弦函数与余弦函数图象的绘制方法，并将其灵活运用到实际问题的解决中，从而提升学生的数学素养和解题能力。含教案人教A高一数学必修第一册3.3幂函数PPT课件含教案页数：48\|大小：8M 这套人教A版高一数学必修第一册 3.3《幂函数》的PPT课件共48页，旨在帮助学生深入理解幂函数的定义，掌握其图像和性质，并能够根据这些性质解决简单问题。通过具体实例和自主探究，学生将逐步建立起对幂函数的直观认识和系统理解。课件内容围绕四个板块展开：第一部分：幂函数的概念这一部分首先复习回顾了函数的基本性质，为引入幂函数做好铺垫。接着，通过分析具体实例，如 f(x)=x 2、f(x)=x 3、f(x)=x −1等，帮助学生理解幂函数的定义，即形如 f(x)=x α的函数，其中 α 是常数。为了加深学生对幂函数图像特征及其性质的理解，课件以表格形式详细总结了五种常见幂函数（α=−1,0,1,2,3）的图像和性质，包括定义域、值域、奇偶性、单调性等。通过这种系统化的总结，学生能够清晰地看到不同幂函数之间的相似性和差异性。第二部分：幂函数的图像与性质在这一部分，课件进一步深入探讨幂函数的图像与性质。通过动态演示和图像分析，学生可以直观地看到幂函数在不同指数 α 下的图像变化。例如，当 α0 时，函数图像通过原点且在第一象限单调递增；当 α0 时，函数图像在第一象限单调递减且有垂直渐近线。课件还通过表格形式总结了五种常见幂函数的图像特征和性质，帮助学生系统地掌握这些函数的行为规律。通过具体的图像和表格，学生能够更好地理解幂函数的性质，并能够在实际问题中灵活运用。第三部分：题型强化训练为了巩固学生对幂函数的理解和应用能力，这一部分提供了丰富的练习题。这些题目涵盖了不同类型的幂函数，包括求定义域、值域、判断奇偶性、比较大小等。通过这些练习，学生能够熟练掌握幂函数的性质，并能够运用这些性质解决实际问题。每道题目都配有详细的解题步骤和解析，帮助学生理解每一步的逻辑和方法。通过重复练习，学生能够提升解题速度和准确性，增强对幂函数性质的掌握。第四部分：小结及随堂练习最后，通过思维导图的方式，课件帮助学生系统回顾本节课的关键知识点，包括幂函数的定义、图像特征和性质。随堂练习部分提供了即时反馈的机会，让学生在课堂上就能检验自己的学习效果，及时发现并纠正错误。通过梳理本节课的知识点，学生能够构建完整的知识体系，为后续学习打下坚实的基础。整套课件设计科学，内容丰富，通过从直观到抽象、从定义到应用的逐步引导，帮助学生全面掌握幂函数的概念和性质。通过具体的实例和自主探究，学生不仅能够提升数学思维能力，还能增强解决实际问题的能力，感受到数学在实际生活中的广泛应用。含教案第1课时人教A高一数学必修第一册5.4.2正弦函数、余弦函数的性质PPT课件含教案页数：37\|大小：48M 这是一套专为人教A版高一数学必修第一册第五章“三角函数”中“5.4.2正弦函数、余弦函数的性质第1课时”设计的PPT课件模板，总页数为37页，内容系统地分为四个主要部分，旨在帮助学生全面而深入地理解和掌握正弦函数与余弦函数的性质。在第一部分“正弦函数、余弦函数的周期”中，重点介绍了周期函数的基本概念以及最小正周期的定义。课件通过公式法和定义法，详细讲解了如何求解正弦、余弦函数及其复合函数的周期。通过具体的例子和推导过程，帮助学生理解周期的计算方法，为后续学习函数的性质奠定了基础。第二部分“正弦函数、余弦函数的奇偶性”从函数图象的对称性入手，结合诱导公式，深入分析了正弦函数为奇函数、余弦函数为偶函数的本质。课件通过图象展示和公式推导，帮助学生直观理解奇偶性的定义，并探讨了奇偶性在研究函数性质中的重要作用。通过这部分内容的学习，学生能够更好地理解函数的对称性，从而更全面地掌握函数的性质。第三部分“题型强化训练”通过丰富的例题和练习，涵盖了函数周期性的判断、奇偶性的判别，以及周期性与奇偶性的综合应用等多类问题。课件不仅提供了详细的解题步骤，还对解题策略和方法进行了归纳总结。通过多样化的练习，帮助学生巩固所学知识，提升解题能力，使学生能够灵活运用周期性和奇偶性解决实际问题。最后的“小结及随堂练习”部分，对周期性与奇偶性的核心知识进行了系统的梳理。课件总结了本节课的重点内容，包括周期和奇偶性的定义、求解方法以及它们在函数性质研究中的应用。同时，提供了多种类型的练习题，供学生自我检测和巩固所学内容，帮助学生进一步加深对正弦函数和余弦函数性质的理解。整个PPT课件结构层次清晰，内容丰富实用，非常适合用于课堂教学。通过系统的讲解和多样化的练习，能够有效地帮助学生扎实掌握正弦函数与余弦函数的周期性和奇偶性，并将其灵活运用到实际问题的解决中，从而提升学生的数学素养和解题能力。含教案人教A高一数学必修第一册5.4.2正弦函数、余弦函数的性质第2课时PPT课件含教案页数：52\|大小：5M 这是一套专为人教A版高一数学必修第一册第五章“三角函数”中“5.4.2正弦函数、余弦函数的性质第2课时”设计的PPT课件模板，总页数为52页，内容系统地分为四个主要部分，旨在帮助学生全面而深入地理解和掌握正弦函数与余弦函数的单调性和最值性质。在第一部分“正弦函数、余弦函数的单调性”中，课件从观察函数图像入手，详细分析并归纳了正弦函数和余弦函数的单调递增和递减规律。通过直观的图像展示和详细的推导过程，课件提供了清晰的单调区间结论，并总结了便于学生记忆的方法。这部分内容帮助学生理解函数值随角度变化的规律，为后续学习函数的性质奠定了基础。第二部分“正弦函数、余弦函数的最值”结合图象和函数特性，明确指出了正弦函数和余弦函数取得最大值与最小值的条件及其取值集合。课件通过具体的例题演示了如何求解复合三角函数的最值，帮助学生掌握在不同情境下求解最值的方法。这部分内容不仅加深了学生对函数性质的理解，还提升了学生解决实际问题的能力。第三部分“题型强化训练”通过丰富的例题和练习，涵盖了求正弦型、余弦型函数的单调区间、利用单调性比较函数值大小等多类经典题型。课件不仅提供了详细的解题步骤，还总结了相应的解题策略、步骤和技巧。通过多样化的练习，帮助学生巩固所学知识，提升解题能力，使学生能够灵活运用单调性和最值性质解决实际问题。最后的“小结及随堂练习”部分，对单调性和最值性质的核心知识进行了系统的梳理。课件总结了本节课的重点内容，包括单调性和最值的定义、求解方法以及它们在函数性质研究中的应用。同时，提供了不同层次的练习题，供学生自我检测和巩固所学内容，帮助学生进一步加深对正弦函数和余弦函数性质的理解。整个PPT课件结构层次清晰，内容丰富实用，非常适合用于课堂教学。通过系统的讲解和多样化的练习，能够有效地帮助学生扎实掌握正弦函数与余弦函数的单调性和最值性质，并将其灵活运用到实际问题的解决中，从而提升学生的数学素养和解题能力。含教案人教A高一数学必修第一册3.4函数的应用一PPT课件含教案页数：70\|大小：31M 这套人教A版高一数学必修第一册 3.4《函数的应用（一）》的PPT课件共70页，旨在帮助学生深入理解函数模型在实际问题中的应用，并掌握用函数模型解决实际问题的基本步骤。通过具体实例，引导学生自主探究函数模型的应用，激发学生对学习数学的兴趣，培养学生的数学思维能力和应用能力，让学生感受到数学在实际生活中的广泛应用。课件内容围绕四个板块展开：第一部分：分段函数模型的应用这一部分通过具体实例，帮助学生了解解决实际问题的一般步骤，包括审题、建模、求模、还原。例如，通过分析出租车计费、阶梯电价等实际问题，学生将学习如何将复杂问题分解为多个阶段，并用分段函数进行建模。通过具体的解题步骤，学生能够掌握如何根据实际情境选择合适的函数形式，如何求解函数模型，并将结果还原到实际问题中。这种系统化的解题方法不仅帮助学生理解分段函数的应用，还提升了他们的逻辑思维能力。第二部分：用函数模型解决实际问题在这一部分，课件通过一系列实际问题，展示了如何用函数模型解决实际问题。这些问题涵盖了经济、物理、生物等多个领域，如成本与收益分析、物体运动轨迹、种群增长等。通过具体的函数模型（如一次函数、二次函数、指数函数等），学生将学习如何根据问题的特征选择合适的函数类型，如何通过函数模型进行预测和决策。这些实例不仅帮助学生理解函数模型的多样性，还展示了数学在不同领域的广泛应用。第三部分：题型强化训练为了巩固学生对函数模型的理解和应用能力，这一部分提供了丰富的练习题。这些题目涵盖了不同类型的函数模型，包括分段函数、一次函数、二次函数、指数函数等，帮助学生在多样化的题目中灵活运用所学知识。每道题目都配有详细的解题步骤和解析，帮助学生理解每一步的逻辑和方法。通过重复练习，学生能够熟练掌握解题方法和技巧，提升解题速度和准确性，增强对函数模型应用的掌握。第四部分：小结及随堂练习最后，通过思维导图的方式，课件帮助学生系统回顾本节课的关键知识点，包括分段函数模型的应用、用函数模型解决实际问题的基本步骤等。随堂练习部分提供了即时反馈的机会，让学生在课堂上就能检验自己的学习效果，及时发现并纠正错误。通过梳理本节课的知识点，学生能够构建完整的知识体系，为后续学习打下坚实的基础。整套课件设计科学，内容丰富，通过从具体实例到系统总结、从理论到实践的逐步引导，帮助学生全面掌握函数模型的应用。通过具体的实例和自主探究，学生不仅能够提升数学思维能力，还能增强解决实际问题的能力，感受到数学在实际生活中的广泛应用。含教案人教A高一数学必修第一册4.4.1对数函数的概念PPT课件含教案页数：36\|大小：18M 这套《人教A版必修第一册 4.4.1 对数函数的概念》PPT 课件共 36 张，以“历史溯源—情境建模—符号抽象—迁移应用”为脉络，引领高一学生完成从“幂运算”到“对数运算”的视角转换。课程目标定位于：理解并熟记对数函数 y=log_a x 的严格定义，准确写出其定义域 (0, +∞) 与值域 (-∞, +∞)；能依据定义快速判断给定解析式是否为对数函数，并能处理含参、含根号、含分式等复杂情境下的定义域求解；同时通过“化指数问题为对数问题”的转化实践，发展学生的数学建模素养与数形结合能力，培养以函数视角整体把握变化规律的意识。课件内容分四大板块展开。第一板块“对数函数的概念及应用”从数学史切入：先简介对数创始人纳皮尔的生平与 400 年前“化乘为加”的革命性思想，再通过“地震里氏震级每增 1 级能量增 32 倍”的真实问题，引导学生列出指数方程 32^x = 10^y，进而产生“已知幂值求指数”的强烈需求，自然引出 log_a b 的符号表达；接着用双向箭头直观呈现指数式 a^b = c 与对数式 log_a c = b 的等价互化，帮助学生建立“指数—对数”一一对应的整体框架。第二板块“对数函数模型的应用”设置三道梯度任务：①手机拍照亮度调节遵循 log 模型，让学生用图像直观感受“亮度对数级差 0.3，人眼恰可分辨”；②溶液 pH 值计算，把氢离子浓度指数方程转化为对数函数，体验跨学科价值；③银行复利转连续复利，通过 ln(1+r)≈r 的近似，让学生领悟对数在简化运算中的威力。每例均配有 GeoGebra 动态演示，强化“形”与“数”的同步认知。第三板块“题型强化训练”聚焦两大核心能力：一是“概念辨析”——5 道选择题让学生在给定解析式中快速识别对数函数，并说明底数 a0 且 a≠1、真数 x0 的限定原因；二是“定义域求解”——由易到难呈现 4 道典型题：含根式√(log_2 x)、含分式 1/log_3 (x-1)、含参数 log_a (x-a) 等，教师现场示范“三步法”：列不等式、解不等式、用数轴检验，确保学生学得会、做得对。第四板块“小结与随堂练习”首先由学生独立绘制“对数函数知识速写卡”，涵盖定义、底数限制、定义域、值域、互化公式五要素；教师再补充“函数三看”口诀：看底数、看真数、看定义域。随后推送 6 题分层随堂检测：前 3 题聚焦基础概念，后 3 题融入实际情境，现场扫码提交即时统计，实现精准反馈。整份课件以“历史故事激趣—真实问题驱学—多元训练固能—反思导图提能”的闭环设计，帮助学生在“数”与“形”的往复对话中真正掌握对数函数的本质与力量。含教案人教A高一数学必修第一册3.1.2函数的表示法（第2课时)PPT课件含教案页数：60\|大小：8M 这套 60 页的演示文稿紧扣《人教 A 版数学必修第一册》3.1.2《函数的表示法（第 2 课时）》，是继第 1 课时之后的深化与提升。课堂目标定位于：让学生在“会认”三种表示法的基础上，真正“会用”并“用得好”。教师首先用一道“快递运费”情境题唤醒旧知——同一规则分别用解析式、列表、图像呈现，引导学生讨论“何时解析式最省力、何时列表最精确、何时图像最直观”，在真实任务中体会“选择合适表示方法”的策略意识。随后，针对学生在画图环节常见的“不会分段、不会取空圈、不会标箭头”三大痛点，教师集中展示“水费阶梯计价”“出租车分段计费”“手机流量限速”等生活案例，让学生通过观察、描点、连线、平移，在反复操作中归纳出“分段函数画图三步诀”：一看断点、二判空心、三标趋势，从而把抽象规则内化为可迁移的技能。课件结构同样分为四大板块。第一板块“函数的三种表示法”不再停留于概念罗列，而是用“同题异构”的方式，把一段文字题同时翻译成解析式、数据表和坐标图，让学生直观比较三种语言的优劣；第二板块“函数的图像”以分段函数为核心，先通过动画演示“折线—跳跃—平台”的视觉特征，再总结“左闭右开、空圈实心、箭头延伸”的绘图规范；第三板块“题型强化训练”双线并行：一条线给出“求分段函数值”的四步程序——找区间、代解析、写结果、合表达，另一条线设置“画分段函数图”的五级闯关，从一次—二次—绝对值层层递进，并在每关嵌入即时反馈；第四板块“小结及随堂练习”先让学生用“三句话”总结本节收获，再布置“基础题 + 拓展题”双层作业：基础题侧重巩固分段函数求值与画图，拓展题则引入“自定义分段规则”的微项目，鼓励学生用手机记录家庭用电曲线、设计节能方案，实现课堂知识向生活情境的迁移。整堂课以“问题驱动—操作体验—反思提升”为主线，既突破“画图难”这一现实障碍，又通过多元任务培养学生的数学建模意识与实际应用能力。含教案人教A高一数学必修第一册4.4.3不同函数增长的差异PPT课件含教案页数：47\|大小：17M 《4.4.3 不同函数增长的差异》这套共 47 张幻灯片的课件，立足于人教 A 版高一数学必修第一册，旨在让学生在“一次、二次、指数、对数”四大函数之间搭建一条“看得见的增长赛道”。课程通过数值列表、函数图像与解析式三条路径并驾齐驱，引导学生用量化眼光、图形直觉与代数语言同时发力，比较它们在增速、增量与增长方式上的迥异特征。更重要的是，课堂以“猜想—数值实验—图像验证—归纳结论”的螺旋式探究为主线，让学生在反复验证中体验数学建模的完整周期，在数据驱动中锤炼数据分析的核心素养，最终形成对“指数爆炸”“对数缓增”“线性匀速”“二次加速”等现象的深刻洞察。整套课件的结构围绕四个学习阶段展开：第一阶段“几个函数模型增长差异的比较”，教师创设人口增长、投资收益、病毒传播等真实情境，先让学生凭直觉猜想“谁长得更快”，再用计算器或在线工具生成同步增长的数值表，通过“数据打脸”激活认知冲突，为后续探究埋下悬念；第二阶段“函数增长速度的比较”，借助动态几何软件在同一坐标系中实时绘制四条曲线，并通过“放大镜”功能聚焦局部区间，引导学生观察斜率变化、切线斜率与二阶导数的符号意义，从而把“快慢”的感性认识上升为“凹凸”“爆炸”“饱和”的理性描述；第三阶段“题型强化训练”，选取工程、经济、环境等跨学科案例，分层设置填空、选择、建模三种题型，让学生在独立求解、同伴互评、教师点拨的循环中，学会用恰当函数刻画现实问题并用差异比较指导决策；第四阶段“小结及随堂练习”，先由学生用思维导图自主串联“比较视角—研究方法—典型结论—易错警示”四大关键词，教师再补充完善，并以 3 分钟限时测即时诊断学习成效，确保知识网络牢固、方法迁移到位。整节课在信息技术与数学思维的深度融合中，让学生亲历“用数据说话、用图像讲理、用符号归纳”的全过程，真正实现从“学会”到“会学”、从“解题”到“解决问题”的跨越。含教案人教A高一数学必修第一册4.5.1函数的零点与方程的解PPT课件含教案页数：45\|大小：17M 本套《4.5.1 函数的零点与方程的解》PPT课件共 45 张幻灯片，对应人教 A 版高一数学必修第一册，核心目标是让学生能够用严谨的数学语言刻画“函数零点”的本质，准确理解并灵活运用零点存在性定理的前提与结论；同时熟练掌握图像法、代数法、信息技术计数法三种手段，为超越方程寻求精度可控的近似解。课堂以“问题—探究—应用—反思”为逻辑主线，在层层递进的活动中同步发展学生的数学抽象、逻辑推理与直观想象三大核心素养。课件的整体架构由四大板块铺陈展开：第一板块“函数的零点与方程的解”从“方程的根”与“函数的零点”的双向视角切入，先给出符号化、形式化的定义，再通过二次函数、三次函数等典型示例，示范如何把“求方程 f(x)=0 的根”翻译为“求函数 y=f(x) 的零点”；随后系统梳理代数法（因式分解、求根公式）与几何法（图像交点、对称变换）两条经典路径，为后续综合应用埋下伏笔。第二板块聚焦“零点存在性定理”，利用 GeoGebra 动态演示“连续曲线跨越 x 轴”的微观过程，引导学生归纳定理的“闭区间连续”“端点异号”两大条件，并通过反例辨析“缺一不可”的严谨性，强化逻辑推理。第三板块“题型强化训练”精选物理抛物运动、经济盈亏平衡、生物种群阈值等跨学科情境，设计“判断零点区间—选择合适方法—控制误差范围—给出近似解”四步任务链，让学生在真实问题中体验“数学建模—算法实现—结果解释”的完整流程。第四板块“小结及随堂练习”先由学生用思维导图自主整理“概念—定理—方法—易错点”四位一体知识网络，教师再补充拓展，最后通过分层随堂练习即时检测、即时反馈，确保不同层次学生都能准确迁移本节所学，实现知识、能力、思维品质的同步提升。含教案人教A高一数学必修第一册5.4.3正切函数的性质与图象PPT课件含教案页数：87\|大小：6M 这是一套精心设计的“数学第五章三角函数中正切函数的性质与图像课件 PPT”模板，整套 PPT 共有 87 张幻灯片，内容分为两个主要部分。在演示文稿的开篇部分，通过新课导入环节，迅速将学生的注意力聚焦到正切函数的核心性质上。模板首先展示了正切函数的周期性和奇偶性这两个重要性质，并以清晰的公式推导展示了这些性质的来源，让学生从数学原理层面理解其依据。在讲解完这些基础性质后，模板巧妙地引导学生思考几个与正切函数相关的问题，这些问题设计得富有启发性，旨在激发学生的好奇心和求知欲，通过问题探究的方式自然地过渡到本堂课的深入学习环节。第二部分是学习新知的环节。在这一部分，模板在前面提出的问题基础上，引导学生通过动手画图来探究正切函数的图像和性质。这种由简入深、层层递进的教学方法，符合学生的认知规律，让学生在实践中逐步理解正切函数的复杂性。通过画图探究，学生最终得出了正切函数的另外三个性质。为了进一步加深学生对这些新学知识的印象，模板再次通过直观的图形展示，将抽象的数学概念具象化，帮助学生更好地理解和记忆。整个演示文稿以图形展示为主，这种直观的教学方式简洁易懂，非常适合数学这门注重逻辑和形象思维的课程。在讲解过程中，模板循序渐进，从基础知识入手，逐步引导学生发现新知、学习新知、应用新知，并在最后通过复习和巩固环节，强化学生对所学内容的理解和掌握。这种教学流程符合学生的学习心理，能够有效提高学生的学习效率和兴趣，使学生在轻松愉快的氛围中掌握正切函数的性质与图像。含教案人教A高一数学必修第一册3.1.1函数的概念（第1课时）PPT课件含教案页数：44\|大小：10M 本套课件共44页，围绕人教A版《数学必修第一册》3.1.1节“函数的概念”（第1课时）精心设计，是一堂集知识建构、思维训练与素养提升于一体的新授课。课堂结束后，学生将在以下方面获得显著收获：一是能够准确理解函数的本质内涵，牢固掌握“定义域、对应关系、值域”这三大核心要素；二是具备辨析两个函数是否相同的能力，能够运用集合与对应的观点进行严谨论证；三是通过教师呈现的大量生活化实例与层层递进的对比探究，亲历概念生成的全过程，在“举三反一”中发展抽象概括与逻辑推理等数学思维品质；四是深刻体会函数在刻画变化规律、解决实际问题中的价值，感受数学与现实世界的紧密联系，从而激发持续的学习兴趣。课件结构清晰，由四大板块构成。第一部分“初识概念”从学生已有经验出发，借助“投信与邮箱”“出租车计价”等形象情境，抽象出对应关系，并通过类比、归纳等思维方式回顾初中“变量说”，自然过渡到高中“集合-对应说”的严格定义，实现认知的螺旋上升。第二部分“三要素解读”依次展开：先用通俗语言阐释“定义域是舞台、对应关系是剧本、值域是演出效果”的比喻，帮助学生建立整体图景；再系统梳理解析式、图像、列表、语言描述等多种表征方式，强调“形异质同”的转化思想；最后通过“判断两个函数是否相同”的典型错例，强化“定义域与对应关系完全一致”的判别标准。第三部分“题型强化”精选两类训练：一是“单项选择”快速诊断易错点，如忽视定义域限制、混淆对应顺序等；二是“解决问题”设置“阶梯水费”“疫情传播模型”等真实任务，引导学生用函数观点建模、运算、解释，体验完整的数学应用流程。第四部分“回顾提升”先以时间轴呈现函数概念从莱布尼茨到康托尔的演进史，彰显数学文化；再用“五点说明”——对象、符号、语言、思想、价值——进行课堂总结，配以即时检测与分层作业，确保学生带着问题来、带着方法走、带着兴趣学。整堂课以“情境—问题—探究—应用—反思”为主线，既关注知识的系统性，又突出思维的深刻性，最终实现“教、学、评”一体化的教学目标。含教案人教A高一数学必修第一册3.1.1函数的概念（第2课时）PPT课件含教案页数：39\|大小：8M 本节课所用 PPT 共 39 页，与《人教 A 版数学必修第一册 3.1.1 函数的概念（第 2 课时）》完全匹配。课堂伊始，教师首先带着学生“温故”，通过简洁明快的提问与板书，回顾上节课提炼出的函数定义及其三要素（定义域、对应法则、值域），并顺势抛出两三个贴近生活的实际问题——如气温随时间变化的曲线、出租车计费规则等——让学生在“旧知”与“现实”之间架起桥梁，自然过渡到今天的新内容。接着，教师利用精心设计的四个环节层层推进：第一环节聚焦“求函数的定义域”。PPT 先用生活化的例子解释区间概念，再用集合、区间、数轴三种语言同步呈现，帮助学生在多重表征中灵活切换；随后归纳出求定义域时必须关注的五大注意点，提醒学生“分母不为零、偶次根号下非负、对数真数为正”等易错细节。第二环节以“判断函数相等”为核心，教师给出若干组看似相同却实则不同的对应关系，引导学生从定义域与对应法则两个维度进行辨析，强化“函数相等必须两要素完全一致”的本质认识。第三环节是“题型强化训练”，PPT 先呈现一组梯度分明的填空题，考察学生对概念细节的把握；再给出两道情境化“解决问题”——如根据限速标志写出分段函数、利用几何图形建立面积模型并求值域——让学生在真实任务中体验“从文字到符号、从符号到图像”的完整建模过程。最后一个环节是“小结及随堂练习”，教师先用思维导图回顾本节四大核心要点，再布置“基础作业”与“拓展作业”双层任务：基础作业紧扣课本例题，巩固求定义域、值域的基本套路；拓展作业则引入跨学科情境，如利用指数函数描述药物浓度衰减，要求学生综合运用新旧知识进行探究。整堂课以问题链贯穿始终，既让学生在“回顾—迁移—应用”的循环中不断深化对函数概念的理解，又通过分层训练与实时反馈，确保不同层次的学生都能获得成就感与提升空间。含教案人教A高一数学必修第一册3.1.2函数的表示法（第1课时）PPT课件含教案页数：43\|大小：8M 本套 PPT 共 43 页，对应《人教 A 版数学必修第一册》3.1.2《函数的表示法（第 1 课时）》。课堂伊始，教师并未直接灌输概念，而是把天平、弹簧测力计、温度计等实物带进教室，让学生在“称一称、拉一拉、量一量”的亲身体验中，先感受变量之间的依赖关系；随后，教师用同一组数据依次用解析式、列表、图像三种方式呈现，引导学生对比“哪种方法更直观”“哪种方法更精确”“哪种方法便于预测”，在对比分析中自然生成“各有千秋”的认知。为了点燃学习热情，教师布置“生活寻宝”任务：一周内，每位同学至少找到一个生活里的函数——如公交车票价、手机电量、外卖配送费——并用三种方式加以表示，下节课交流时重点说明各自优缺点，借此训练数学抽象与表达能力。PPT 的第一板块“函数的三种表示方式”依次介绍解析法、列表法和图像法，每介绍一种便配一个“微动画”演示其生成过程，让学生看到“数”如何变“式”、“式”如何变“图”；第二板块“函数的图像”先抛出“作图三大注意”——定义域、关键点、变化趋势，再示范描点法和变换作图法两种常用技巧，现场用几何画板动态演示“平移—伸缩—对称”的魔术效果；第三板块“题型强化训练”分层设计：第一层聚焦“表达方式转换”，让学生把文字情境译成解析式；第二层聚焦“图像识读”，给出折线图、曲线图让学生反推对应法则；第三层聚焦“解析式求解”，将应用题拆分为“建模—求式—验图”三步走；第四板块“小结及随堂练习”先由学生用“思维导图”自主梳理本节三大收获，再完成当堂“闯关题”：基础题巩固描点作图，拓展题则引入分段函数与绝对值函数的图像变换，为下一节埋下伏笔。整节课以“实物—数据—模型—应用”的主线贯穿，既让学生在多元表征中深刻体会函数表示的灵活性与统一性，又通过生活化任务与分层训练，培养其用数学眼光观察世界、用数学语言表达世界的核心素养。含教案人教A高一数学必修第一册4.5.3函数模型的应用PPT课件含教案页数：75\|大小：22M 这套总计 75 张幻灯片的《4.5.3 函数模型的应用》PPT 课件，对应人教 A 版高一数学必修第一册，旨在引领学生综合运用函数图像、方程、不等式及信息技术，从实际问题中抽象变量关系，求出未知参数、最值或预测值，并完整体验“情境—假设—建模—求解—检验—解释”的闭环流程，从而切实提升数学建模能力与数据分析素养。课件以“问题情境驱动、技术深度介入、反思及时跟进”为主线，层层递进地设置四大板块。首板块“已知函数模型解决实际问题”精选人口增长、药物代谢、金融复利等典型案例，引导学生辨析一次、二次、指数、对数及分段模型的适用边界，借助表格、图像与代数运算多维度解析模型参数的现实意义，让学生在“拿来就用”的过程中体会函数语言的精准与高效。第二板块“建立适当的函数模型解决实际问题”以“共享单车投放优化”“温室番茄产量预测”等任务为载体，系统呈现建模六环节：提炼变量、作出假设、选择函数、建立方程（不等式）、技术求解、回归检验；教师示范如何用 GeoGebra 或 Excel 进行数据拟合与残差分析，学生则在拆解步骤中领悟“模型不是越复杂越好，而是越合适越好”的建模哲学。第三板块“题型强化训练”围绕交通流量、电商促销、环境降解等跨学科情境，设计“填空—选择—开放”三级梯度练习，鼓励小组合作完成“数据采集—模型选择—误差评估—结果汇报”的完整链条，在反复迭代中固化技能、拓展思维。第四板块“小结及随堂练习”先让学生用思维导图自主梳理“模型选择—求解技术—结果解释—反思改进”四大关键词，教师再补充“过度拟合、灵敏度分析”等高阶视角，随后通过分层随堂练习即时检测：基础层聚焦模型识别与参数求解，提高层则要求依据误差容忍度反向调整函数形式并给出经济或科学建议，确保不同层次学生都能把本节习得的建模策略迁移至新的现实场景，实现知识、能力与责任意识的同步跃升。含教案人教A高一数学必修第一册4.4.2对数函数的图象和性质第1课时PPT课件含教案页数：47\|大小：18M 这套《人教A版必修第一册 4.4.2 对数函数的图像与性质（第1课时）》PPT 课件共 47 页，以“图像先行—性质聚焦—迁移应用—反思升华”为逻辑主线，引导学生在“看、说、比、用”的完整循环中掌握对数函数的四条核心性质：定义域(0,+∞)、值域(-∞,+∞)、恒过定点(1,0)、当底数a1时单调递增且图像“缓升”，当0a1时单调递减且图像“缓降”。课程旨在使学生不仅能用符号语言准确表述上述性质，还能借助图像直观比较对数值大小，并在解题中灵活转化“数”与“形”，从而同步发展直观想象与逻辑推理素养，树立牢固的数形结合意识。课件内容分四大板块展开。第一板块“对数函数的图像”首先借助 GeoGebra 动态演示，先回顾指数函数 y=a^x 的图像与特征，再在同一坐标系中同步生成其反函数 y=log_a x 的图像，让学生通过“描点—连线—观察”体验互为反函数的对称美；随后以双列表格式梳理指数与对数函数图像的“定义域/值域互换、单调性一致、渐近线位置对调”等关键差异，为性质探究奠定直观基础。第二板块“对数函数的性质”采用“例题驱动”策略：先给出 log_2 x 与 log_{0.5} x 两组具体数值，引导学生猜想单调区间；再通过代数证明“若 a1，x1x2 ⇒ log_a x1log_a x2”，在严谨推理中完成从感性到理性的过渡；最后以对照表形式将指数与对数函数的四条性质并列呈现，突出“反函数视角”下的内在统一，帮助学生构建系统化知识网络。第三板块“题型强化训练”设置三层梯度：A 层“识图说话”——根据给定图像快速写出底数范围及增减性；B 层“比大小”——结合图像与单调性比较 log_3 5 与 log_3 7、log_{0.4} 2 与 log_{0.4} 3；C 层“情境建模”——以“声音分贝与能量对数关系”为例，让学生利用图像估算能量翻 10 倍时分贝增量，体验跨学科应用价值。每题均配“画图—说性质—得结论”三步策略，确保思路可视化、过程可迁移。第四板块“小结与随堂练习”先让学生手绘“对数函数思维导图”，串联定义域、值域、定点、单调性四大关键词；教师再展示优秀范例，补充“看底数、看真数、看图像”三看口诀。随后推送 5 题随堂检测：前 2 题基础巩固，后 3 题拓展拔高，在线实时统计正确率，实现精准反馈。整份课件以“形”启“思”、以“思”促“用”，帮助学生在图像与符号的往复对话中真正吃透对数函数的本质，养成自觉运用数形结合解决问题的思维习惯。含教案人教A高一数学必修第一册4.4.2对数函数的图象和性质第2课时PPT课件含教案页数：53\|大小：36M 本课《4.4.2 对数函数的图像与性质（第 2 课时）》共 53 张幻灯片，定位于人教 A 版高一数学必修第一册。课程以“渐进线”为抓手，引导学生用几何语言精确刻画对数函数曲线的无限逼近特征，在动态演示与静态分析的双重视角中，培养学生的直观想象力和逻辑推理能力；同时借助信息技术平台，让学生亲历数据生成—图像绘制—模型验证的完整过程，体会数学表达的高度简洁与统一，感受数学与信息技术深度融合的时代魅力。整套 PPT 的展开逻辑分为四个板块。第一板块“对数函数性质的综合应用”首先呈现指数函数与对数函数性质的对照一览表，以表格形式唤醒学生对定义域、值域、单调性、对称性、渐近线等要素的记忆，随后精选典型例题，引导学生在复杂情境下灵活调用性质，完成求值、比较大小、解不等式等任务，在“温故”中“知新”。第二板块聚焦“反函数的概念与图像特点”，通过“互为反函数”的对称映射关系，揭示指数函数与对数函数图像关于直线 y＝x 的对称本质，并利用动态几何软件演示点、线、面的实时对应，帮助学生建立“函数—反函数—图像对称”三位一体的认知结构。第三板块“题型强化训练”精选来源于生活、科技、经济等领域的真实问题，以分组探究、即时反馈、错因剖析的方式，强化学生运用对数函数模型解决实际问题的能力，突出数学建模的核心素养。第四板块“小结及随堂练习”先由学生自主梳理本节的知识网络与思想方法，教师再用思维导图进行系统归纳，随后安排分层递进的随堂练习，既巩固基础又拔高思维，确保不同层次的学生都能在课堂内获得成就感与获得感。整节课在问题驱动、技术支撑、素养导向的融合路径中，努力实现知识、能力、情感的三维目标统一。含教案人教A高一数学必修第一册5.6.2函数y＝Asin(ωx＋φ)的图象与性质的应用PPT课件含教案页数：56\|大小：17M 这是一套“数学第五章三角函数中函数 y=Asin(ωx+ψ)的图像第二课时课件 PPT”模板，该 PPT 共有 56 张幻灯片，整个演示文稿分为三个主要部分。在第一部分，模板通过具体的题目讲解和分析，引导学生逐步掌握函数 y=Asin(ωx+ψ)的图像绘制方法。特别地，模板详细展示了如何使用“五点法”来画出该函数的图像。在文字讲解之后，模板还通过图形步骤的展示，使学生能够更加直观地理解每个步骤，确保学生能够清晰明了地掌握图像绘制的全过程。这种图文结合的方式有助于学生更好地理解和记忆图像绘制的方法。第二部分，模板讲解了函数 y=Asin(ωx+ψ)在匀速圆周运动中的应用。这一部分首先通过具体的例题讲解来引入应用背景，帮助学生理解函数在实际问题中的作用。随后，模板展示了几道相关题目，先引导学生自主完成，再进行探究分析。最后，模板引导学生发表自己的感悟，总结所学知识。这种设计不仅帮助学生理解函数的应用，还通过自主探究和总结，提升了学生的自主学习能力和思维能力。第三部分是题型强化训练环节。这一部分主要围绕求三角函数的解析式相关题型展开练习。通过大量的题目训练，学生可以在实践中巩固所学知识，进一步提升解题能力。这些题目不仅涵盖了基础知识，还通过公式的变化引导学生进行发散思维，帮助学生学会举一反三，从而更好地应对各种题型。整个演示文稿包含了大量的题目，这种设计有利于学生通过题目来探究学习新知。在讲解分析题目的过程中，学生不仅能够巩固所学新知，还能通过题型和公式的多样化变化，提升自己的发散思维能力。这种教学设计符合学生的认知规律，能够有效帮助学生系统地学习函数 y=Asin(ωx+ψ)的图像及其应用，为后续的学习打下坚实的基础。人教版数学变量与函数一次函数PPT课件页数：23\|大小：4M 该PPT以一次函数变量与函数为主题，用一些老师，和实际生活示例作为元素呼应主题。内容上，该PPT模板首先抛出学习目标，阐述本章的学习的目标，其一是探索数量关系和变化规律，其二是了解变量，常量。其次用五个示例得出结论，在变化过程中，有些量是变化的，有些是始终不变的。然后是课堂小结，总结这节课的内容，梳理知识结构。最后是课后作业，巩固学习。初中数学说课变量与函数一次函数PPT课件页数：26\|大小：9M PPT模板内容主要通过PowerPoint软件分几个部分来向我们展开介绍有关于八年级变量与函数数学教学课件的相关内容。PPT模板内容第一部分主要向我们详细的讲述了本节数学课的学习目标。第二部分主要带领同学们回顾了上节课所学习的内容。第三部分主要是有关于本节课一次函数重点知识的相关定义。第四部分主要向我们列举出了一些有关于一次函数的习题。最后一部分主要是有关于一次函数相关的解题方法。一次函数变量与函数PPT课件页数：22\|大小：5M 本PPT以数学中一次函数变量与函数为主题，以蓝色为主打色调，搭配书包、笔记本、学生漫画形象等元素，主题突出。PPT在内容上，首先介绍了本节课的学习目标、分析重难点。紧接着，以习题的形式进行新课的导入，让学生了解何为变量、常量等概念，通过跟踪训练和随堂小练对所学知识点进行练习掌握。最后，通过小结，让学生对本堂课知识有了整体感知。 12345678910下一页您是不是想找高一函数数学函数数学函数ppt 高一数学一元二次函数高一数学二次函数与一元高一数学对数函数高一数学二次函数数学一次函数初二数学函数初中数学函数课件数学函数幂函数数学课件相关PPT 数学反比例函数数学函数单调性数学讲课函数幂函数 excel,函数 excel函数 PPT全称是PowerPoint，麦克素材网为你提供高一数学函数PPT模板免费下载资源。让你3分钟学会幻灯片怎么做的诀窍，打造高质量的专业演示文稿模版合集。关于我们成立于2018年5月版权及下载声明供学习研究使用投稿专栏请将您的作品发送至 3114549753@qq.com 联系方式 462325726@qq.com; PPT标签湘ICP备2025126737号-2长沙市互联网违法和不良信息举报中心更多精美模板，可在此搜索，麦克PPT网每日更新高质量模板！
189146	https://endaq.com/pages/shock-testing-analysis?srsltid=AfmBOoqSYtbAjnTM3H5tshB8TLZ_T-pv7VBTp2WmGrcZuaiT5DOrS1GJ	Shock Testing & Analysis \| enDAQ This website uses cookies to ensure you get the best experience on our website. Learn More Got it! Products All Sensors Standard Sensors Large Battery Wi-Fi Enabled Accessories Software Software Overview LAB Configuration Cloud Analysis Open Source Libraries Custom Software Services Re-Calibration Applications All Applications Vibration Shock & Impact Machine Learning Environmental High Value Asset Monitoring Wireless Monitoring Resources All Resources Handbooks Webinars Case Studies Video Tutorials Application Areas Cheat Sheets Blog Help Center Contact Use the up and down arrows to select a result. Press enter to go to the selected search result. Touch device users can use touch and swipe gestures. SECTIONS SECTIONS Shock Isolator Photo Gallery Simple Drop Shock Classical Shock Half-Sine Shock Example Response to Arbitrary Excitation Seismic Shock Pyrotechnic Shock Pyrotechnic Shock Data Water Impact Shock Shock Response Spectrum Synthesis Sections SECTIONS Shock Isolator Photo Gallery Simple Drop Shock Classical Shock Half-Sine Shock Example Response to Arbitrary Excitation Seismic Shock Pyrotechnic Shock Pyrotechnic Shock Data Water Impact Shock Shock Response Spectrum Synthesis Products All Sensors Standard Sensors Large Battery Wi-Fi Enabled Accessories Software Software Overview LAB Configuration Cloud Analysis Open Source Libraries Custom Software Services Re-Calibration Applications All Applications Vibration Shock & Impact Machine Learning Environmental High Value Asset Monitoring Wireless Monitoring Resources All Resources Handbooks Webinars Case Studies Video Tutorials Application Areas Cheat Sheets Blog Help Center Contact × Products All Sensors Standard Sensors Large Battery Wi-Fi Enabled Accessories Software Software Overview LAB Configuration Cloud Analysis Open Source Libraries Custom Software Services Re-Calibration Applications All Applications Vibration Shock & Impact Machine Learning Environmental High Value Asset Monitoring Wireless Monitoring Resources All Resources Handbooks Webinars Case Studies Video Tutorials Application Areas Cheat Sheets Shock Testing & Analysis By Tom Irvine Home Shock & Vibration Resources Center Handbooks Shock Testing & Analysis Shock Isolator Photo Gallery Simple Drop Shock Classical Shock Half-Sine Shock Example Response to Arbitrary Excitation Seismic Shock Pyrotechnic Shock Pyrotechnic Shock Data Water Impact Shock Shock Response Spectrum Synthesis SECTION 1 Shock Isolator Photo Gallery Figure 1.1. Titan II Missile Silo, Launch Control Room The control room is mounted underground via huge isolation springs. A typical spring is shown in the background. The purpose is to isolate the control room from mechanical shock and vibration in the event of a nuclear strike above the launch site. The springs allow 18 inches of relative displacement. The control room could thus carry out a retaliatory strike, as ordered by the U.S. president. This site is located south of Tucson, Arizona. It has been decommissioned and is now a museum. SECTION 2 Simple Drop Shock Figure 2.1. iPhone 6 Accidental Drop The first iPhone 6 was purchased in Perth, Australia on September 19, 2014. The event was covered by a live TV report. The buyer mishandled the phone as he unboxed it. The phone survived the drop onto the ground but may have had some fatigue or fracture damage. Portable electronic devices (PEDs) are expected to survive multiple drops. Most original equipment suppliers specifying between 30 and 50 drops. Recommended test methods are given in Reference . Figure 2.2. Drop Shock Analytical Model The accidental drop shock of a component onto a hard surface is difficult to model accurately. The item may undergo rigid-body translation and rotation in each of its six degrees-of-freedom during freefall. The item may have a nonlinear response with plastic deformation during impact. It may or may not bounce. Furthermore, a box-shaped object may strike the ground on any of its corners, edges or faces. A very simple method, as a first approximation, is to assume that the object is a linear, undamped, single-degree-of-freedom subjected to initial velocity excitation as it strikes the ground and remains attached to it via its spring. The object then undergoes free vibration in his configuration. The initial velocity is calculated using a familiar physics formula where the change in kinetic energy is equal to the change in potential energy due to gravity. Assume that the object is dropped from rest. The initial velocity as it strikes the ground is (2.1) The equation of motion is (2.2) Or equivalently (2.3) The peak displacement is (2.4) The peak velocity is equal to the initial velocity in equation (1.1). The peak acceleration is (2.5) An example is shown in the following table for three natural frequency cases. Table 2.1. Peak Response Values for 36 inch Drop Height\| Natural Freq (Hz) \| Displacement (in) \| Velocity (in/sec) \| Acceleration (G) \| --- --- \| \| 200 \| 0.133 \| 167 \| 543 \| \| 600 \| 0.044 \| 167 \| 1630 \| \| 1000 \| 0.027 \| 167 \| 2710 \| SECTION 3 Classical Shock Figure 3.1. Idealized Classical Pulse Examples Classical pulses are the simplest base excitation pulses. They are deterministic and can be represented by simple mathematical functions. They are typically one-sided. An SDOF system’s response to a classical pulse can be solved for exactly using Laplace transforms. Four classical pulse types are shown in Figure 18.4. Other types include rectangular and trapezoidal pulses. These pulses do not necessarily represent real field environments, but they are still used throughout industry to test equipment ruggedness for convenience. Shock tests are performed on military equipment to: a. Provide a degree of confidence that materiel can physically and functionally withstand the relatively infrequent, non-repetitive shocks encountered in handling, transportation, and service environments. This may include an assessment of the overall materiel system integrity for safety purposes in any one or of the handling, transportation, and service environments. b. Determine the materiel's fragility level, in order that packaging may be designed to protect the materiel's physical and functional integrity. c. Test the strength of devices that attach materiel to platforms that can crash. Potential equipment failure modes due to shock excitation include: a. Materiel failure resulting from increased or decreased friction between parts, or general interference between parts. b. Changes in materiel dielectric strength, loss of insulation resistance, variations in magnetic and electrostatic field strength. c. Materiel electronic circuit card malfunction, electronic circuit card damage, and electronic connector failure. (On occasion, circuit card contaminants having the potential to cause short circuit may be dislodged under materiel response to shock.) d. Permanent mechanical deformation of the materiel resulting from overstress of materiel structural and nonstructural members. e. Collapse of mechanical elements of the materiel resulting from the ultimate strength of the component being exceeded. f. Accelerated fatiguing of materials (low cycle fatigue). g. Potential piezoelectric activity of materials, and materiel failure resulting from cracks in fracturing crystals, ceramics, epoxies, or glass envelopes. Figure 3.2. Drop Shock Test Machine, Initial Velocity Excitation (Courtesy of Lansmont) Classical pulse shock testing has traditionally been performed on a drop tower. The component is mounted on a platform which is raised to a certain height. The platform is then released and travels downward to the base, which has pneumatic pistons to control the impact of the platform against the base. In addition, the platform and base both have cushions for the model shown. The pulse type, amplitude, and duration are determined by the initial height, cushions, and the pressure in the pistons. This is a textbook example of case where the initial potential energy of the raised platform and test item are converted to kinetic energy. The final velocity of the freefall becomes the initial velocity of the shock excitation. Figure 3.3. A 50 G, 11 msec, Terminal Sawtooth Pulse for Shaker Shock Test Classical pulse shock testing can sometimes be performed on shaker tables, but there some constraints. The net velocity and net displacement must each be zero. Also, the acceleration, velocity and displacement peaks must each be within the shaker table stroke limits. Pre and post pulses are added to classical pulses to meet these requirements. A hypothetical terminal sawtooth pulse suitable for shaker shock testing is shown in Figure 1.6. SECTION 4 Half-Sine Shock Example Consider a single-of-freedom is subjected to a 50 G, 11 msec half-sine pulse applied as base excitation per Figure 12.12. Set the amplification factor to Q=10. Allow the natural frequency to be an independent variable. Solve for the absolute response acceleration. The equation of motion is for the relative displacement z is (4.1) The primary response occurs during the half-sine pulse input. The residual response occurs during the quiet period thereafter. The total response is the combination of primary and residual. The exact response for a given time can be calculated via a Laplace transform. Note that the quiet period solution is free vibration with initial velocity and displacement excitation. Figure 4.1. Fourier Transform Half-Sine Shock A common misunderstanding is to regard the half-sine shock pulse as having a discrete frequency which would be the case if it were extended to a full-sine pulse. The Fourier transform in Figure 18.7 shows that the half-sine pulse has a continuum of spectral content beginning at zero and then rolling-off as the frequency increases. There are also certain frequencies where the magnitude drops to zero. The magnitude represents the acceleration, but the absolute magnitude depends on the total duration include the quiet period after the pulse is finished. The post-pulse duration was 10 seconds in the above example, for a total of 10.011 seconds. Figure 4.2. SDOF System Response to Half-Sine Shock, 10 Hz The response in the above figure has an absolute peak value less than the peak input. This is an isolation case. The response positive and negative peaks occur after the base input pulse is over. Figure 4.3. SDOF System Response to Half-Sine Shock, 75 Hz The response in Figure 1.9 has an absolute peak value that is 1.65 times the peak input. This is resonant amplification case. The absolute peak response occurs during the base input pulse. Figure 4.4. SDOF System Response to Half-Sine Shock, 400 Hz The response converges to the base input as the natural frequency becomes increasingly high. This becomes a unity gain case. The system is considered as hard-mounted. The peak results from the three cases are shown in Table 1.2. The calculation can be repeated for a family of natural frequencies. The peak acceleration results can then be plotted as a shock response spectrum (SRS) as shown in Figure 1.11. The peak relative displacement values can likewise be plotted as an SRS as shown in Figure 1.12. Figure 4.5. Acceleration SRS, 50 G, 11 msec Half-Sine Pulse The two curves in Figure 1.11 contain the coordinates in Table 1.2. The initial slope of each SRS curve is 6 dB/octave indicating a constant velocity line. The curves also indicate that the peak response can be lowered by decreasing the natural frequency. A low natural frequency could be achieved for a piece of equipment by mounting it via soft, elastomeric isolator bushings or grommets. But a lower natural frequency leads to a higher relative displacement as shown in Figure 1.12. Table 4.1. Summary of Peak Response, 50 G, 11 msec, Half-Sine Base Input\| Natural Frequency (Hz) \| Peak Positive (G) \| Absolute Value of Peak Negative (G) \| --- \| 10 \| 20.3 \| 17.3 \| \| 75 \| 82.5 \| 65 \| \| 400 \| 51.8 \| 4.38 \| Figure 4.6. Relative Displacement SRS, 50 G, 11 msec Half-Sine Pulse The curves in the above figure could be used for designing isolator mounts for a component. The isolators must be able to take up the relative displacement without bottoming or topping out. There must also be enough clearance and sway space around the component. Figure 4.7. Acceleration SRS, 50 G, 11 msec Terminal Sawtooth Pulse The positive and negative SRS curves are reasonably close for the terminal sawtooth pulse shown in Figure 1.13. In contrast, the positive and negative SRS curves for the half-sine pulse in Figure 1.11 diverge as the natural frequency increases above 80 Hz. The terminal sawtooth pulse is thus usually preferred over the half-sine pulse for classical shock testing. Another means of visualizing the SRS concept is given in Figure 1.14. Figure 4.8. Half-Sine Shock Applied as Base Input to Independent SDOF Systems The systems are arranged in order of ascending natural frequency from left to right and subjected to a common half-sine base input. The Soft-mounted system on the left has high spring relative deflection, but its mass remains nearly stationary. The Hard-mounted system on the right has low spring relative deflection, and its mass tracks the input with near unity gain. The Middle system ultimately has high deflection for both its mass and spring. The peak positive and negative responses of each system are plotted as a function of natural frequency in the shock response spectrum. SECTION 5 Response to Arbitrary Excitation Recall from Section 12.3 that the response of a single-degree-of-freedom system to base excitation can be expressed in terms of a second order, ordinary differential equation for the relative displacement for a base acceleration ÿ. (5.1) Equation (1.7) can be solved via Laplace transforms if the base acceleration is deterministic such as a half-sine pulse. A convolution integral is needed if the excitation varies arbitrary with time. The convolution integral is computationally inefficient, however. An alternative is to use the Smallwood ramp invariant digital recursive filtering relationship , . Again, the recursive filtering algorithm is fast and is the numerical engine used in almost all shock response spectrum software. It is also accurate assuming that the data has a sufficiently high sample rate and is free from aliasing. One limitation is that it requires a constant time step. The equation for the absolute acceleration is (5.2) The damped natural frequency is (5.3) The digital recursive filtering relationship for relative displacement is omitted for brevity but is available in Reference . The relationship in equation (1.8) is recursive because the response at the current time depends on the two previous responses, which are the first two terms on the righthand side of the equation. This is a feedback loop in terms of control theory. The relationship is filtering because the energy at and near the natural frequency is amplified whereas higher frequency energy above √2 times the natural frequency is attenuated. See Figure 12.14. Shock Response Spectrum Test Specification Objective Figure 5.1. Mid-Field Pyrotechnic Shock Time History Consider the measured mid-field shock time history in the above figure as taken from Reference . The time history would be essentially impossible to reproduce in a test lab given that it is a high-frequency, high-amplitude complex, oscillation pulse. The aerospace practice instead is to derive an SRS to represent the damage potential of the shock event. The test conductor may then use an alternate pulse to satisfy the SRS specification within reasonable tolerance bands. This is an indirect method of achieving the test goal. There are some limitations to this approach. One is that the test item is assumed to be linear. Another is that it behaves as a single-degree-of-freedom system. Nevertheless, this method is used in aerospace, military and earthquake engineering fields, for both analysis and testing. Figure 5.2. Mid-Field Shock Response Spectrum & Envelope The customary approach is to draw a conservative envelope over the measured SRS. The ramp-plateau format is the most common, although there are variations. The enveloping process shown in the above vibration is very conservative in the mid frequency domain. An additional dB uncertainty factor may be needed to develop the envelope into a test specification, given that the SRS envelope is derived from a single time history. Industry standards, such as Reference , give guidelines for the dB factor. SECTION 6 Seismic Shock Seismic Waveforms Figure 6.1. Mother Earth The Earth experiences seismic vibration. The fundamental natural frequency of the Earth is 309.286 micro Hertz, equivalent to a period of approximately 54 minutes . The structure of Earth's deep interior cannot be studied directly. But geologists use seismic waves to determine the depths of layers of molten and semi-molten material within Earth. Figure 6.2. P-Wave The primary wave, or P-wave, is a body wave that can propagate through the Earth’s core. This wave can also travel through water. The P-wave is also a sound wave. It thus has longitudinal motion. Note that the P-wave is the fastest of the four waveforms. Figure 6.3. S-Wave The secondary wave, or S-wave, is a shear wave. It is a type of body wave. The S-wave produces an amplitude disturbance that is at right angles to the direction of propagation. Note that water cannot withstand a shear force. S-waves thus do not propagate in water. Figure 6.4. Love Wave Love waves are shearing horizontal waves. The motion of a Love wave is similar to the motion of a secondary wave except that Love wave only travel along the surface of the Earth. Love waves do not propagate in water. Figure 6.5. Rayleigh Wave, Retrograde Rayleigh waves produce retrograde elliptical motion. The ground motion is thus both horizontal and vertical. The motion of Rayleigh waves is similar to the motion of ocean waves in Figure 1.22 except that ocean waves are prograde. Rayleigh waves resulting from airborne acoustical sources may either be prograde or retrograde per Reference . In some cases, the motion may begin as prograde and then switch to retrograde. Airborne acoustical sources include above ground explosions and rocket liftoff events. Figure 6.6. Ocean Surface Wave Particle Motion, Prograde The Love and Rayleigh waves are both surface waves. These are the two seismic waveforms which can cause the most damage to building, bridges and other structures. As an aside, seismic and volcanic activity at the ocean floor generates a water-borne longitudinal wave called a T-wave, or tertiary wave. These waves propagate in the ocean’s SOFAR channel, which is centered on the depth where the cumulative effect of temperature and water pressure combine to create the region of minimum sound speed in the water column. SOFAR is short for “Sound Fixing and Ranging channel.” These T-waves may be converted to ground-borne P or S-waves when they reach the shore. Acoustic waves travel at 1500 m/s in the ocean whereas seismic P and S-waves travel at velocities from 2000 to 7000 m/s in the crust. Seismic Response Spectrum Method Professors Theodore von Kármán and Maurice Biot were very active in the early 1930s in the theoretical dynamics aspects of what would later become known as the response spectrum method in earthquake engineering. Biot proposed that rather than being concerned with the shape of the input time history, engineers should instead use a method describing the response of systems to those shock pulses. The emphasis should instead be on the effect, as represented by the response of a series of single-degree-of-freedom oscillators, similar to that previously shown for the case of a half-sine input in Figure 18.14. Practical use of the response spectrum method had to wait until the 1970s due to the intricacy of the response calculation for complex, oscillating pulses which required digital computers. Time was also needed to establish and publicize databases of strong motion acceleration time histories. The response spectrum method was adopted for pyrotechnic shock in the aerospace industry and renamed as shock response spectrum. El Centro Earthquake Figure 6.7. El Centro, Imperial Valley Earthquake Damage Nine people were killed by the May 1940 Imperial Valley earthquake. At Imperial, 80 percent of the buildings were damaged to some degree. In the business district of Brawley, all structures were damaged, and about 50 percent had to be condemned. The shock caused 40 miles of surface faulting on the Imperial Fault, part of the San Andreas system in southern California. Total damage has been estimated at about $6 million. The magnitude was 7.1. The was the first major earthquake for which strong motion acceleration data was obtained that could be used for engineering purposes. Figure 6.8. El Centro Earthquake, Triaxial Time History The highest excitation is in the North-South axis, parallel to the ground. Figure 6.9. El Centro Earthquake North-South SRS, Three Damping Cases The acceleration levels reached 1.5 G for the 1% damping curve. Recall that large civil engineering structures can have nonlinear damping. The damping values tend to increase as the base excitation levels increase as shown for the Transamerica Title Building in Section 10.7. Figure 6.10. El Centro Earthquake Tripartite SRS Seismic SRS curves are often plotted in tripartite format which displays relative displacement, pseudo velocity and acceleration responses all on the same graph. The pseudo velocity is calculated from the relative displacement as (6.1) Stress can be calculated from pseudo velocity using the methods in Section 19. The acceleration curve might be the most important design metric for equipment mounted inside a building. The relative displacement might be the most important concern for analyzing the foundational strength. The curves also show design tradeoffs. Lowering the building’s natural frequency below 1 Hz reduces the acceleration response but increases the relative displacement. Note that Q=10 is the same as 5% damping. Golden Gate Bridge Figure 6.11. Golden Gate Suspension Bridge, San Francisco, California In addition to traffic loading, the Golden Gate Bridge must withstand the following environments: Earthquakes, primarily originating on the San Andreas and Hayward faults Winds of up to 70 miles per hour Strong ocean currents The Golden Gate Bridge has performed well in all earthquakes to date, including the 1989 Loma Prieta Earthquake. Several phases of seismic retrofitting have been performed since the initial construction. The bridge’s fundamental mode is a transverse mode with a natural frequency of 0.055 Hz, with a period of 18.2 seconds Note that California Department of Transportation (CALTRANS) standards require bridges to withstand an equivalent static earthquake force (EQ) of 2.0 G. This level is plausibly derived as a conservative envelope of the El Centro SRS curves in Figure 1.25. Vandenberg AFB Figure 6.12. Rocket Launch from Vandenberg AFB, California The Vandenberg launch site is near the San Andreas fault system. The vehicle is mounted on the pad as a tall cantilever beam. The vehicle must be analyzed to verify that it can withstand a major seismic event. The vehicle may be mounted to the pad for only two weeks prior to launch. The odds of an earthquake occurring to that time window are miniscule. But the launch vehicle and payload together may cost well over $1 billion. The risk thus necessitates the analysis. Areas of concern are the loads imparted at the launch vehicle’s joints and to the payload. Figure 6.13. NASA SRS Curves for Launch Vehicles at Vandenberg SRS curves are given for three damping cases. The curves are taken from Reference . The vehicle would typically be analyzed as a multi-degree-of-freedom system via a finite element model. Each SRS curve could be applied to the model using a modal combination method. An alternative is to synthesize a time history to satisfy a selected SRS curve. The time history could then be applied to the model via a modal transient analysis. Seismic Testing Figure 6.14. Electrical Power Generator Seismic Shock Test The diesel generator is mounted onto a platform at the top of a shaker table which is located below the ground floor. This could be an emergency power generator for a hospital in an active seismic zone. A video clip of the test is available on YouTube at: SECTION 7 Pyrotechnic Shock Flight Events Figure 7.1. Stage Separation Ground Test, Linear Shaped Charge The plasma jet cuts the metal inducing severe mechanical shock energy, but the smoke and fire would not occur in the near-vacuum of space. Launch vehicle avionics components must be designed and tested to withstand pyrotechnic shock from various stage, fairing and payload separation events that are initiated by explosive devices. Solid rocket motor ignition is another source of pyrotechnic shock. The source shock energy can reach thousands of acceleration Gs at frequencies up to and beyond 100 kHz. The corresponding velocities can reach a few hundred in/sec, well above the severity thresholds in the empirical rules-of-thumb in Section 19.4. Empirical source shock levels for a variety of devices are given in References , , . These levels are intended only as preliminary estimates for new launch vehicle designs. The estimates should be replaced by ground test data once the launch vehicle hardware is built and tested. Pyrotechnic Device Photo Gallery Figure 7.2. Metal Clad Linear Shaped Charge The chevron focuses a pyrotechnic plasma jet at the launch vehicle’s separation plane material. Severe shock levels are generated as a byproduct. Figure 7.3. Frangible Joint A frangible joint may be used for stage or fairing separation. The key components of a frangible joint: Mild Detonating Fuse (MDF) Explosive confinement tube Separable structural element Initiation manifolds Attachment hardware The hot gas pressure from the MDF detonation cause the internal tube to expand and fracture the joint. Figure 7.4. Frangible Nuts, Hold Down Posts The purpose of the nuts was to hold the SRBs in place against wind and ground-borne excitation. The nuts were separated just before liftoff. Figure 7.5. Clamp Band with Pyrotechnic Bolt-Cutters (image courtesy of European space agency) Clamp bands are often used for payload separation from launch vehicle adapters. They are also commonly used for stage separation in suborbital launch vehicles, similar to the one in Figure 4.1. A pyrotechnic bolt-cutter uses an explosive charge to drive a chisel blade to cut the band segments’ connecting bolt. The cutters produce some shock energy, but much of the shock is due to the sudden release of strain energy in the preloaded clamp band. This action can excite a ring mode in the radial axis. Recall Section 7.3. The total clamp band release shock is significantly less than linear shaped charge and frangible joint. Note that an analysis must be performed to verify that no gapping will occur in the between the band and the joint as the vehicle undergoes bending mode vibration during powered flight. SECTION 8 Pyrotechnic Shock Data Initial SRS Slopes Figure 8.1. Expected Pyrotechnic SRS Initial Slope Limits Near-field pyrotechnic shock can be difficult to measure accurately. The accelerometer data may have a baseline shift or spurious low frequency transient. This error could be a result of the accelerometer’s own natural frequency being excited or to some other problem. Aerospace pyrotechnic shock SRS specifications usually begin at 100 Hz due to the difficult in accurately measuring low-amplitude, low-frequency shock, while simultaneously measuring high-amplitude, high-frequency shock. There are several methods for checking whether the data is acceptable. One is to the check the initial slopes of both the positive and negative SRS curves. Each should have an overall trend of 6 to 12 dB/octave. The actual slopes may have local peaks and dips due to low frequency resonances. A 12 dB/octave slope represents constant displacement and zero net velocity change. A 6 dB/octave slope indicates constant velocity. Recall the slope formulas in Section 16.1.2. A second method for checking data accuracy is to verify that the positive and negative SRS curves are within about 3 dB of each other across the entire natural frequency domain. A third method is to integrate the acceleration time history to velocity. The velocity time history should oscillate in a stable manner about the zero baselines. These verification goals are challenging to meet with near-field shock measurements of high-energy source shock, such as that from linear shaped charge and frangible joints. In practice, some high-pass filtering or spurious trend removal may be necessary. There is no one right way to perform this “data surgery.” It is a matter of engineering judgment. Re-entry Vehicle Separation, Flight Data Figure 8.2. Re-Entry Vehicle, Separation Shock, Near-Field Measurement, Time History The source device was linear shaped charge. Figure 8.3. Re-Entry Vehicle, Separation Shock, Near-Field Measurement, SRS The SRS reached 20,040 G at 2400 Hz, which is an extreme level per the rule-of-thumb Section 19.4. SECTION 9 Water Impact Shock Figure 9.1. Solid Rocket Booster Water Impact Shock Each of the two Space Shuttle Boosters was recovered and refurbished after every flight. Each booster contained sensitive avionics components which underwent shock testing for the water impact event. Figure 9.2. Solid Rocket Booster Water Impact Shock, Time History The data is from the STS-6 Mission. The accelerometer was mounted at the forward end of the booster adjacent to a large IEA avionics box. This was the worst-case shock event for this component. Figure 9.3. Solid Rocket Booster Water Impact Shock, Tripartite SRS The maximum acceleration response is 257 G at 85 Hz. The maximum pseudo velocity response is 201 in/sec at 76 Hz, which is severe per the rule-of-thumb Section 1.4. SECTION 10 Shock Response Spectrum Synthesis Synthesis Objectives Consider an SRS specification for a component, subsystem or a large structure. The article should be tested, if possible, to verify that it can withstand the shock environment. Certain shock tests can be performed on a shaker table, like the generator test in Figure 18.30. This requires synthesizing an acceleration time history to satisfy the SRS. The net velocity and net displacement must each be zero for this test. These requirements can be met with a certain type of wavelet series, where the individual wavelets may be nonorthogonal. The resulting wavelet time history should meet the SRS within reasonable tolerance bands, but it may not “resemble” the expected time history which is a limitation of this method. Another requirement is that the peak acceleration, velocity and displacement values must be within the shaker table’s capabilities. An innovative method for meeting the SRS with a wavelet series that resembles one or more measured shock time histories is given in Reference . A synthesized time history can also be used for modal transient analysis. This could be done for articles which are too large or heavy for shaker tables. This analysis can also be done on small components prior to shock testing to determine whether they will pass the test. Or the analysis could be done in support of isolator mounting design. There is again a need for the synthesized time history to have net velocity and net displacement which are each zero, to maintain numerical stability in stress calculations from relative displacement values. Damped-sines can be used for modal transient analysis where the goal is to meet the SRS with a time history that plausibly resembles the expected field shock event. But damped-sines do not meet the desired zero net velocity and displacement goals. The workaround is to first synthesize a damped-sine series to meet the SRS and then reconstruct it via a wavelet series, in Rube Goldberg fashion. Wavelet and damped-sine synthesis are shown in the following examples. Wavelet Synthesis Wavelet Equation A wavelet is a sine function modulated by another sine function. The equation for an individual wavelet is (10.1) where = acceleration of wavelet at time = wavelet acceleration amplitude = wavelet frequency = number of half-sines = wavelet time delay Note that must be an odd integer greater than or equal to 3. This is required so that the net velocity and net displacement will each be zero. The total acceleration at time for a set of wavelets is (10.2) Figure 10.1. Sample Wavelet A sample, individual wavelet is shown in the above figure. This wavelet was a component of a previous analysis for an aerospace project. Wavelet Synthesis Example Consider the specification: MIL-STD-810E, Method 516.4, Crash Hazard for Ground Equipment. Table 10.1. SRS Q=10, Crash Hazard Specification\| Natural Frequency (Hz) \| Peak Accel (G) \| --- \| \| 10 \| 9.4 \| \| 80 \| 75 \| \| 2000 \| 75 \| Synthesize a series of wavelets as a base input time history for a shaker shock test to meet the Crash Hazard SRS. The goals are: Satisfy the SRS specification Minimize the displacement, velocity and acceleration of the base input The synthesis steps are shown in the following table. Table 10.2. Wavelet Synthesis Steps\| Step \| Description \| --- \| \| 1 \| Generate a random amplitude, delay, and half-sine number for each wavelet. Constrain the half-sine number to be odd. These parameters form a wavelet table. \| \| 2 \| Synthesize an acceleration time history from the wavelet table. \| \| 3 \| Calculate the shock response spectrum of the synthesis. \| \| 4 \| Compare the shock response spectrum of the synthesis to the specification. Form a scale factor for each frequency. \| \| 5 \| Scale the wavelet amplitudes. \| \| 6 \| Generate a revised acceleration time history. \| \| 7 \| Repeat steps 3 through 6 until the SRS error is minimized or an iteration limit is reached. \| \| 8 \| Calculate the final shock response spectrum error. Also calculate the peak acceleration values. Integrate the signal to obtain velocity, and then again to obtain displacement. Calculate the peak velocity and displacement values. \| \| 9 \| Repeat steps 1 through 8 many times. \| \| 10 \| Choose the waveform which gives the lowest combination of SRS error, acceleration, velocity and displacement. \| The resulting time history and SRS are shown in Figure 1.43 and Figure 1.44, respectively. Figure 10.2. Crash Hazard Time History Synthesis The acceleration time history has a reverse sine sweep character. It is an efficient and optimized waveform for a shaker shock test, and it satisfies the SRS as shown in the next figure. A drawback is that it does not resemble an actual crash shock time history. Figure 10.3. Crash Hazard SRS The positive and negative curves are from the synthesized waveform. The tolerance bands are set at +3 dB. Damped-Sine Synthesis Damped-Sine Equation The equation for an individual damped sinusoid is (10.3) where = acceleration of damped sinusoid at time = acceleration amplitude = angular frequency = damping value = time delay The total acceleration at time for a set of damped sinusoids is (10.4) Damped-Sine Example Consider the following specification which could represent a stage separation shock level as some location in a launch vehicle. A modal transient finite element analysis is to be performed on a component to verify that the component will pass its eventual shock test. The immediate task is to synthesize a time history to satisfy the SRS. The time history should “resemble” an actual pyrotechnic shock pulse. Note that pyrotechnic SRS specifications typically begin at 100 Hz. The author’s rule-of-thumb is to extrapolate the specification down to 10 Hz in case there are any component modes between 10 and 100 Hz. This guideline is also to approximate the actual shock event which should have an initial ramp somewhere between 6 and 12 dB/octave. Table 10.3. SRS Q=10, Stage Separation Shock\| Natural Frequency (Hz) \| Peak Accel (G) \| --- \| \| 10 \| 10 \| \| 2000 \| 2000 \| \| 10,000 \| 2000 \| The specification has an initial slope of 6 dB/octave. The synthesis steps are shown in the following table. Table 10.4. Damped-Sine Synthesis Steps\| Step \| Description \| --- \| \| 1 \| Generate random values for the following for each damped sinusoid: amplitude, damping ratio and delay. The natural frequencies are taken in one-twelfth octave steps. \| \| 2 \| Synthesize an acceleration time history from the randomly generated parameters. \| \| 3 \| Calculate the shock response spectrum of the synthesis. \| \| 4 \| Compare the shock response spectrum of the synthesis to the specification. Form a scale factor for each frequency. \| \| 5 \| Scale the amplitudes of the damped sine components. \| \| 6 \| Generate a revised acceleration time history. \| \| 7 \| Repeat steps 3 through 6 as the inner loop until the SRS error diverges. \| \| 8 \| Repeat steps 1 through 7 as the outer loop until an iteration limit is reached. \| \| 9 \| Choose the waveform which meets the specified SRS with the least error. \| \| 10 \| Perform wavelet reconstruction of the acceleration time history so that velocity and displacement will each have net values of zero. \| The resulting time history and SRS are shown in Figure 1.45 and Figure 1.46, respectively. Figure 10.4. Stage Separation Time History Synthesis The acceleration time history somewhat resembles a mid or far-field pyrotechnic shock. The velocity and displacement time histories each have a stable oscillation about their respective baselines. Figure 10.5. Stage Separation SRS The positive and negative curves are from the damped-sine synthesis. Modal Transient Finite Element Analysis for Uniform Base Excitation Consider a rectangular plate mounted to a base at each of its four corners. The plate is to be subject to uniform seismic excitation. There are two methods to apply the base excitation in a finite element analysis, as shown in Figure 1.47 and Figure 1.48. Figure 10.6. Direct Enforce Acceleration Method The direct enforcement method is computationally intensive, requiring matrix transformations and a matrix inversion as shown in Reference . Figure 10.7. Seismic Mass Method The seismic mass is chosen to be several orders of magnitude higher in mass than the plate. An equivalent force is calculated and apply to the seismic mass to excite the desired acceleration at each of the plate’s corners. This method adds a degree-of-freedom to the plate system resulting in a rigid-body mode at zero frequency. But the remaining natural frequencies and mode shapes should be the same as if the plate were mounted normally to its joining structure. The author’s experience is that the seismic mass method is faster and more accurate and reliable than the direct enforced method. This method was introduced in Section 12.3.3. Shock Fields Near-Field The near-field environment is dominated by direct stress wave propagation from the source. Peak accelerations in excess of 5000 G occur in the time domain with a frequency content extending beyond 100 kHz. The near-field usually includes structural locations within approximately 15 cm of the source for severe devices such as linear shaped charge and frangible joint. No shock-sensitive hardware should be mounted where it would be exposed to a near-field environment. Mid-Field The mid-field environment is characterized by a combination of wave propagation and structural resonances. Peak accelerations may occur between 1000 and 5000 G, with substantial spectral content above 10 kHz. The mid-field for intense sources usually includes structural locations between approximately 15 and 60 cm of the source, unless there are intervening structural discontinuities. Far-Field The far-field environment is dominated by structural resonances. The peak accelerations tend to fall below 2000 G, and most of the spectral content below 10 kHz. The far-field distances occur outside the mid-field. The typical far-field SRS has a knee frequency corresponding to the dominant frequency response. The knee frequency is the frequency at which the initial ramp reaches the plateau in the log-log SRS plot. Joint & Distance Attenuation The source shock energy is attenuated by intervening material and joints as it propagates from the near-field to the far-field. Empirical distance and joint attenuation factors for the SRS reduction are given in References , , . A typical attenuation curve from is given in Figure 18.49, which assumes an input source shock SRS consisting of a ramp and plateau in log-log format. Such curves should be used with caution given that the attenuation is highly dependent on damping and structural details. The curves can be used for preliminary estimates for new launch vehicle designs, but ground separation tests are still needed for the actual launch vehicle hardware. These tests are needed to measure the source shock as well as the levels at key component mounting locations. Figure 10.8. Shock Response Spectrum Versus Distance from Pyrotechnic Shock Source Shock Mitigation Figure 10.9. Sensor Electronics, Wire Rope Isolators (image courtesy of NASA/JPL) The source shock energy is attenuated as it propagates to avionics mounting locations through the launch vehicle’s material and joints. The input shock to a component can be mitigated by mounting the component as far away from the source device as possible. Another effective attenuation technique is to mount the component via elastomeric bushings or wire rope isolators. The NASA Mars Science Laboratory Sensor Support Electronics unit is mounted on vibration isolators in Figure 1.50. Figure 10.10. SCUD-B Avionics Component Isolation The bushings are made from some type of rubber or elastomeric compound. The bushings provide damping, but their main benefit is to lower the natural frequency of the system. The isolators thus attenuate the shock and vibration energy which flows from the instrument shelf into the avionics component. Pyrotechnic Shock Testing Methods Figure 10.11. Near-Field Shock Simulation using a Plate The test component is mounted on other side of plate. The source device is a textile explosive cord with a core load of 50 gr/ft (PETN explosive). Up to 50 ft of Detonating Cord has been used for some high G tests. The maximum frequency of shock energy is unknown so analog anti-aliasing filters are needed for the accelerometer measurements per the guidelines in Section 14. Note that a component may be mounted in the mid-field shock region of a launch vehicle. The SRS test level derivation for this zone may include a significant dB factor for uncertainty or as a qualification margin. This conservatism may require a near-field-type shock test for a component that is actually located in a mid-field zone. This situation can also occur for components mounted in far-fields. Figure 10.12. NASA JPL Tunable Shock Beam The NASA/JPL Environmental Test Laboratory developed and built a tunable beam shock test bench based on a design from Sandia National Laboratory many years ago. The excitation is provided by a projectile driven by gas pressure. The beam is used to achieve SRS specifications, typically consisting of a ramp and a plateau in log-log format. The intersection between these two lines is referred to as the “knee frequency.” The beam span can be varied to meet a given knee frequency. The high frequency shock response is controlled by damping material. Shock Failure Modes Figure 10.13. Sensitive Electronic Parts Pyrotechnic shock can cause crystal oscillators to shatter. Large components such as DC-DC converters can detached from circuit boards. In addition, mechanical relays can experience chatter or transfer. Figure 10.14. Shock Test Case History, Adhesive & Solder Joint Failure The image shows adhesive failure and rupture of solder joint after a stringent shock test. A large deflection of the PCB resulting from an insufficient support/reinforcement of the PCB combined with high shock loads can lead to these failures. Staking is needed for parts weighing more than 5 grams. Figure 10.15. Shock Test Case History, Lead Wire Failure The image shows a sheared lead between solder joint and winding of coil. 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189147	https://www.sciencedirect.com/science/article/pii/S0092867423005287	Skip to article My account Sign in View PDF ## Cell Volume 186, Issue 13, 22 June 2023, Pages 2897-2910.e19 Article Structures of sperm flagellar doublet microtubules expand the genetic spectrum of male infertility Author links open overlay panel, , , , , , , , , , , , , , , , , rights and content Under an Elsevier user license Open archive Highlights ¢ Cryo-EM structure of the mouse sperm DMT reveals 10 sperm-specific MIPs ¢ Sperm DMTs feature a polymorphic Tektin5 network and multiple FAM166 proteins ¢ The human sperm DMT has fewer MIPs compared with the mouse sperm DMT ¢ MIP variants are associated with a subtype of asthenozoospermia we termed MIVA Summary Sperm motility is crucial for successful fertilization. Highly decorated doublet microtubules (DMTs) form the sperm tail skeleton, which propels the movement of spermatozoa. Using cryo-electron microscopy (cryo-EM) and artificial intelligence (AI)-based modeling, we determined the structures of mouse and human sperm DMTs and built an atomic model of the 48-nm repeat of the mouse sperm DMT. Our analysis revealed 47 DMT-associated proteins, including 45 microtubule inner proteins (MIPs). We identified 10 sperm-specific MIPs, including seven classes of Tektin5 in the lumen of the A tubule and FAM166 family members that bind the intra-tubulin interfaces. Interestingly, the human sperm DMT lacks some MIPs compared with the mouse sperm DMT. We also discovered variants in 10 distinct MIPs associated with a subtype of asthenozoospermia characterized by impaired sperm motility without evident morphological abnormalities. Our study highlights the conservation and tissue/species specificity of DMTs and expands the genetic spectrum of male infertility. Graphical abstract Keywords doublet microtubule sperm flagella sperm motility microtubule inner proteins Tektin5 FAM166 family male infertility asthenozoospermia cryo-EM Data and code availability ¢ The cryo-EM maps of the mouse and human sperm DMTs have been deposited at the Electron Microscopy Data Bank ( The corresponding atomic coordinate data of the mouse sperm DMT has been deposited at the Protein Data Bank ( The map and coordinate data are publicly available as of the date of publication. Accession numbers are listed in the key resources table. ¢ This paper does not report original code. ¢ Any additional information required to reanalyze the data reported in this paper is available from the lead contact upon request. Cited by (0) 10 : These authors contributed equally 11 : Lead contact © 2023 Elsevier Inc.
189148	https://codegolf.stackexchange.com/questions/269180/find-a-fraction-with-the-smallest-denominator	code golf - Find a fraction with the smallest denominator - Code Golf Stack Exchange Join Code Golf By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find a fraction with the smallest denominator Ask Question Asked 1 year, 8 months ago Modified1 year, 8 months ago Viewed 462 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. Input Two non-negative floating point numbers x<y. You can assume they are close enough to each other that there is no integer between x and y. Output A fraction with the smallest possible denomination that lies strictly between x and y. Examples Input: 1 and 2 Output: 3/2 Input: 0 and 0.33 Output: 1/4 Input: 0.4035 and 0.4036 Output: 23/57 code-golf Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Jan 12, 2024 at 22:01 SimdSimd asked Jan 12, 2024 at 21:50 SimdSimd 3,167 1 1 gold badge 9 9 silver badges 55 55 bronze badges 4 1 How do we deal with float imprecision? Say we're given the float representing 1/3 which actually equals 0.3333333333333333148..., is the fraction 1/3 greater than that?xnor –xnor 2024-01-12 22:00:46 +00:00 Commented Jan 12, 2024 at 22:00 @xnor The input is a float so if it just has 3s in its decimal expansion, it is less than 1/3. Is this going to cause a difficulty?Simd –Simd 2024-01-12 22:02:52 +00:00 Commented Jan 12, 2024 at 22:02 1 I'm pretty sure this is a duplicate, but I can't find the original.Wheat Wizard –Wheat Wizard♦ 2024-01-12 23:37:23 +00:00 Commented Jan 12, 2024 at 23:37 5 @WheatWizard Are you thinking of In between fractions? The only difference in that challenge is that x and y are fractions instead of floats.Dingus –Dingus 2024-01-13 02:00:42 +00:00 Commented Jan 13, 2024 at 2:00 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. JavaScript (Node.js), 42 bytes javascript x=>g=(y,d)=>(p=-~(xd))<yd?[p,d]:g(y,-~d) Try it online! If ⌊x d⌋+1<y d, then ⌊x d⌋+1 is an answer Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 12, 2024 at 23:54 answered Jan 12, 2024 at 23:29 l4m2l4m2 32.1k 2 2 gold badges 25 25 silver badges 114 114 bronze badges 1 Which is fortunate since ⌈xd⌉<yd is the wrong test.Neil –Neil 2024-01-12 23:49:45 +00:00 Commented Jan 12, 2024 at 23:49 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. JavaScript (ES6), 47 bytes Expects (x)(y), returns [numerator, denominator]. javascript (x,p=q=1)=>g=y=>p<yq++&&p++>x--q?[--p,q]:g(y) Try it online! Commented javascript ( x, // x = first number p = // p = numerator q = 1 // q = denominator ) => // g = y => // y = second number p < y q++ // we test whether p/q < y // and increment q afterwards in case it's not && // if truthy, p++ > x --q // we restore q, test whether p/q > x // and increment p afterwards in case it's not ? // if truthy again: [--p, q] // we restore p and return [p, q] : // else: g(y) // we try again with either p or q incremented Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 12, 2024 at 22:55 answered Jan 12, 2024 at 22:12 ArnauldArnauld 204k 21 21 gold badges 186 186 silver badges 667 667 bronze badges 2 But how does it work?Simd –Simd 2024-01-12 22:37:03 +00:00 Commented Jan 12, 2024 at 22:37 @Simd I've added a commented version. This is a really simple algorithm, but written in a slightly convoluted way.Arnauld –Arnauld 2024-01-12 22:56:36 +00:00 Commented Jan 12, 2024 at 22:56 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Charcoal, 33 bytes ＮθＮη≔⌕Ｅ⁺²∕¹⁻ηθ⁻⌈×ηι⌊×θι²ζＩ⟦⊖⌈×ηζζ Try it online! Link is to verbose version of code. Explanation: ＮθＮη Input x and y. ≔⌕Ｅ⁺²∕¹⁻ηθ⁻⌈×ηι⌊×θι²ζ Find d such that ⌈yd⌉-⌊xd⌋=2. d is never more than ⌊1+1/(y-x)⌋, but I have to add 2 due to 0-indexing. Ｉ⟦⊖⌈×ηζζ Output ⌈yd⌉-1 and d. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Jan 12, 2024 at 23:46 NeilNeil 184k 12 12 gold badges 76 76 silver badges 287 287 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Jelly, 18 bytes ×ḞĊƭ€r/ḊṖ 1ç1#Ḣṭç¥ Try it online! A pair of links which is called as a monad with a pair of floats and returns a pair of [numerator, denominator]. Explanation ``` ×ḞĊƭ€r/ḊṖ \| Helper link: takes a denominator as the left argument and a pair of floats as the right and returns a valid numerator, if any × \| Multiply (denominator with floats) ḞĊƭ€ \| Floor first, ceiling second r/ \| Reduce using range ḊṖ \| Remove first and last 1ç1#Ḣṭç¥ \| Main link 1ç1# \| Starting with one, find first denominator where there’s a valid numerator (will be wrapped in a list) Ḣ \| Head (to unwrap the list) ṭç¥ \| Tag this onto the result of running the helper link again to find the numerator Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 13, 2024 at 0:44 answered Jan 12, 2024 at 22:26 Nick KennedyNick Kennedy 21.2k 3 3 gold badges 18 18 silver badges 44 44 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Python 2, 95 bytes python (a,b),(c,d)=[input().as_integer_ratio()for m in 0,0] while(md/c+1)a>=mb:m+=1 print m,md/c+1 Try it online! Not a math guy, basically uses the formula of this answer. The input should be in decimal format: 1.0 instead of just 1. -2 bytes: replace // with / -4 bytes: inlining m Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 12, 2024 at 22:59 answered Jan 12, 2024 at 22:36 enzoenzo 2,153 9 9 silver badges 25 25 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. R, 47 bytes r \(x,y){while((p=(xF)%/%1+1)>=yF)F=F+1 c(p,F)} Attempt This Online! Port of @l4m2's JavaScript answer. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Jan 13, 2024 at 15:36 pajonkpajonk 19.1k 4 4 gold badges 29 29 silver badges 76 76 bronze badges Add a comment\| Your Answer If this is an answer to a challenge… …Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead. …Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one. Explanations of your answer make it more interesting to read and are very much encouraged. …Include a short header which indicates the language(s) of your code and its score, as defined by the challenge. 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189149	https://jackson.engr.tamu.edu/wp-content/uploads/sites/229/2021/06/2011-Jackson-ICDERS-shocktube.pdf	23rd ICDERS July 24–29, 2011 Irvine, USA Planar Blast Scaling with Condensed-Phase Explosives in a Shock Tube Scott I. Jackson Shock and Detonation Physics Group, Los Alamos National Laboratory, Los Alamos, NM 87544 1 Introduction Blast waves are strong shock waves that result from large power density deposition into a ﬂuid. The rapid energy release of high-explosive (HE) detonation provides sufﬁciently high power density for blast wave generation. Often it is desirable to quantify the energy released by such an event and to determine that energy relative to other reference explosives to derive an explosive-equivalence value. In this study, we use condensed-phase explosives to drive a blast wave in a shock tube. The explosive material and quantity were varied to produce blast waves of differing strengths. Pressure transducers at varying lengths measured the post-shock pressure, shock-wave arrival time and sidewall impulse associated with each test. Blast-scaling concepts in a one-dimensional geometry were then used to both determine the energy release associated with each test and to verify the scaling of the shock position versus time, overpressure versus distance, and impulse. Most blast scaling measurements to-date have been performed in a three-dimensional geometry such as a blast arena. Testing in a three-dimensional geometry can be challenging, however, as spherical shock-wave symmetry is required for good measurements. Additionally, the spherical wave strength decays rapidly with distance and it can be necessary to utilize larger (several kg) quantities of explosive to prevent signiﬁcant decay from occurring before an idealized blast wave has formed. Such a mode of testing can be expensive, require large quantities of explosive, and be limited by both atmospheric conditions (such as rain) and by noise complaints from the population density near the test arena. Testing is possible in more compact geometries, however. Non-planar blast waves can be formed into a quasi-planar shape by conﬁning the shock diffraction with the walls of a shock tube. Regardless of the initial form, the wave shape will begin to approximate a planar front after successive wave reﬂections from the tube walls. Such a technique has previously been used to obtain blast scaling measurements in the planar geometry with gaseous explosives and the condensed-phase explosive nitroguanidine [1–3]. Recently, there has been much interest in the blast characterization of various non-ideal high explosive (NIHE) materials. With non-ideals, the detonation reaction zone is signiﬁcantly larger (up to several cm for ANFO) than more ideal explosives. Wave curvature, induced by charge-geometry, can signiﬁcantly affect the energy release associated with NIHEs. To measure maximum NIHE energy release accurately, it is desirable to minimize any such curvature and, if possible, to overdrive the detonation shock to ensure Correspondence to: sjackson@lanl.gov 1 Jackson Planar Blast Scaling completion of chemical reactions ahead of the sonic locus associated with the reaction zone. This is achieved in the current study through use of a powerful booster HE and a charge geometry consisting of short cylindrical lengths of NIHE initiated along the charge centerline. 2 Blast Scaling Theory Blast scaling concepts were ﬁrst developed almost a century ago and understanding of the theory has been signiﬁcantly improved since that time [5–7]. In this study, we consider the dimensionless group that characterizes the propagation of a shock front from an intense explosion . Π = r E ρ0 1/(2+ν) t(2/2+ν) (1) The group is self-similar and consists of four variables: E, the source energy; ρ0, the ambient atmo-spheric density; r, the distance of the shock front from the source; and t, the time from energy release. The parameter ν is a dimension-dependent index corresponding to 1, 2, or 3 for blast waves of pla-nar, cylindrical, or spherical symmetry, respectively. The source energy accordingly has dimensions of MT−2, MLT−2, or ML2T−2 where M is mass, L is length, and T is time. Proceeding in the planar geometry approximated by a quasi-planar shock propagating longitudinally inside a tube r = αE ρ0 1/3 t2/3 (2) where the proportionality constant α = f(ν, γ) and γ is the atmospheric ratio of speciﬁc heats. Differentiation with respect to time yields Us = 2 3 s αE ρ0 1 √r (3) where Us is the shock velocity. For a strong shock (M2 s ≫1) in a perfect gas, P P0 → 2γ γ + 1M2 s (4) where Ms is the shock Mach number, P0 is the ambient pressure, and P is the post shock pressure. Equations 3 and 4 combine to form P = 8 9 1 γ + 1 αE0 A 1 r (5) Thus, the postshock pressure of a planar blast wave is inversely proportional to r. The planar energy E has been replaced by the physical energy release E0 (with the standard dimensions of ML2T−2) normalized by the cross-sectional area of the tube A. In this study, α is unity . 23rd ICDERS July 24-29, 2011 Irvine 2 Jackson Planar Blast Scaling Figure 1: The shock tube. Test charge Booster Detonator d L Figure 2: The axisymmetric charge geometry. 3 Experiment and Results Data in the current study were obtained from tests in a shock tube with a 15.2-cm inner diameter and a length of 5.1 m. The working ﬂuid was atmospheric air at Los Alamos atmospheric pressure (nominally 0.777 bar). Special consideration was used in the design to prevent plastic deformation of the facility due to the locally high-pressures near the condensed-phase detonation and to minimize the effect of structural noise on the measurements. Pressure transducers recorded the time-resolved shock overpressure as a function of distance and were located 0.64, 1.64, 2.64, 3.64, 4.64 and 5.04 m from the upstream end of the tube. The shock tube is shown inside of a blast chamber in Fig. 1. Testing was performed for both ideal and non-ideal explosives. Ideal explosives tested included HMX-based PBX 9501 and PBX 9404, as well as RDX-based Composition 4 (C4) and PBX 9407. Non-ideal explosives tested included powdered, stoichiometric mixtures of potassium perchlorate (KClO4) bal-anced with sugar (C12H22O11), sodium perchlorate (NaClO4) balanced with sugar, and ANFO (ammonium-nitrate-fuel-oil). Non-ideals were boosted (and signiﬁcantly overdriven) by the PBX HEs. Explosive quantities ﬁelded ranged from 2.5–51.0 g and at least three different masses of each charge were tested. 1 2 3 4 5 6 0 10 20 30 40 Pressure (bar) Distance (m) Fit to data JWL (cylinder tests) Cheetah Figure 3: Comparison of Eq. 5 ﬁt. E0(m) = 2.05m + 40.5 0 20 40 60 80 100 120 0 5 10 15 20 25 30 Test explosive mass (g) Energy release (kJ) Booster contribution Figure 4: Determination of ∆Hdet for an HE. 23rd ICDERS July 24-29, 2011 Irvine 3 Jackson Planar Blast Scaling As some of the non-ideal mixtures were tested at diameters below their unconﬁned failure diameter limit, charges were formed into right cylinders with length-to-diameter ratios less than unity in order to prevent signiﬁcant wave decay before the charge was consumed. Consoldiated charges were not con-ﬁned and powdered ones were contained with a single layer of copy paper. The charge geometry is shown in Fig. 2. Additional details are given in Ref. 8. For a given condensed-phase detonation in the shock tube, measurement of shock arrival times ts or P versus r along with Eqs. 2 or 5, respectively, allow determination of the released detonation energy E0 that couples to the gas in the shock tube. In this study, E0 values were derived from P measurements. For a given test, Eq. 5 was ﬁt to the peak P values varying only E0. The E0 ﬁt is shown for a 25g C4 shot in Fig. 3 and lies in between a prediction from the thermochemical code Cheetah and cylinder test experiments. For each explosive main charge series, identical boosters were used and linear ﬁtting was then performed over all E0 values to determine the speciﬁc energy or heat of detonation ∆Hdet associated with the known booster and main charge masses (NaClO4 data is shown in Fig. 4). The y-intercept of the ﬁt corresponds to the booster energy, while the line slope was ∆Hdet of the main charge. The experimentally measured ∆Hdet values agreed well with accepted values for the PBX and C4 explosives obtained from calculation and other experiment. Additionally, all energy versus mass data ﬁt well to a straight line (Fig. 4), indicating consistent ∆Hdet measurements from test-to-test. Since the ﬂow is self-similar, experiments of different planar explosion lengths E/P0 are expected to collapse together when nondimensionalized. Poor correlation or deviations from the theoretical curve indicate the failure of theoretical assumptions. Such an approach may seem circular, given that E0 values were determined by ﬁtting to theory. This is not the case, however. Each E0 applies to all experimental P data from a test, consisting of at least four measurements that were collected over a range of normalized distance ¯ r. Thus, it is possible for speciﬁc P values to not follow theory. It is also noted that the ﬁt correlation shown in Fig. 4 is characteristic of all datasets, indicating little variation in the speciﬁc energies measured in each test. 1.0 10.0 100.0 0.001 0.010 0.100 1.000 P r 1273:C4/PBX9501(15.1:6.7) 1272:PBX501(6.7) 1271:PBX9404(6.8) 1270:ANFO/PBX9501(30.0:6.7) 1269:ANFO/PBX9501(10.0:6.7) 1268:ANFO/PBX9501(23.3:6.7) 1267:PBX9501(6.8) 1266:NaClO4/9404(30.0:6.8) 1265:NaClO4/9404(10.1:6.8) 1264:NaClO4/9404(20.2:6.8) 1263:KClO4/9404(10.0:6.8) 1262:KClO4/9404(30.2:6.8) 1261:KClO4/9404(20.7:6.8) 1260:PBX9404(6.8) 1259:PBX9407(2.6) 1258:PBX9407(2.6) 1249:C4/PBX9407(10.4:2.6) 1213:C4/PBX9407(50.0:2.6) 1212:C4/PBX9407(51.3:2.6) 1211:C4/PBX9407(25.1:2.6) 1210:C4/PBX9407(13:2.6) 1209:C4/PBX9407(10:2.6) 1208:C4/PBX9407(9.1:2.6) Theory 1st Transducer Figure 5: Nondimensional pressure versus distance. Signiﬁcant variation from theory will occur near the explosive source due to: (i) the ﬁnite CJ-pressure of the HE (P →∞as r →0 for Eq. 5), (ii) the existence of a signiﬁcantly non-planar front before sufﬁcient wave-wall reﬂections have occurred near r = 0, (iii) the ﬁnite HE source size requiring shock-23rd ICDERS July 24-29, 2011 Irvine 4 Jackson Planar Blast Scaling steepening for a well-formed shock front (planar blast scaling theory assumes energy release along a plane of zero thickness), (iv) the “piston-effect” from the expansion of HE product gas, (v) and the breakdown of the perfect gas assumption at high temperature. These near-ﬁeld effects will not scale with above theory and are best modeled numerically. Deviations in the far-ﬁeld can also occur as the strong shock approximation is not longer met, requiring acoustic models . 1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 0.001 0.010 0.100 ts r !"#$%&'#(%"#)+,-"./+0)1'+2"(%3"#+4'5(6(+7%&'+ 1273:C4/PBX9501(15.1:6.7) 1272:PBX501(6.7) 1271:PBX9404(6.8) 1270:ANFO/PBX9501(30.0:6.7) 1269:ANFO/PBX9501(10.0:6.7) 1268:ANFO/PBX9501(23.3:6.7) 1267:PBX9501(6.8) 1266:NaClO4/9404(30.0:6.8) 1265:NaClO4/9404(10.1:6.8) 1264:NaClO4/9404(20.2:6.8) 1263:KClO4/9404(10.0:6.8) 1262:KClO4/9404(30.2:6.8) 1261:KClO4/9404(20.7:6.8) 1260:PBX9404(6.8) 1259:PBX9407(2.6) 1258:PBX9407(2.6) 1257:PBX9407(2.6) 1256:PBX9407(2.6) 1249:C4/PBX9407(10.4:2.6) 1213:C4/PBX9407(50.0:2.6) 1212:C4/PBX9407(51.3:2.6) 1211:C4/PBX9407(25.1:2.6) 1210:C4/PBX9407(13:2.6) 1209:C4/PBX9407(10:2.6) 1208:C4/PBX9407(9.1:2.6) Theory Figure 6: Nondimensional shock position versus time. 0.0010 0.0100 0.1000 1.0000 10.0000 0.001 0.010 0.100 I r !"#$%&'#(%"#)+,&-.('+/'0(.(+1%&'+ 1273:C4/PBX9501(15.1:6.7) 1272:PBX501(6.7) 1271:PBX9404(6.8) 1270:ANFO/PBX9501(30.0:6.7) 1269:ANFO/PBX9501(10.0:6.7) 1268:ANFO/PBX9501(23.3:6.7) 1267:PBX9501(6.8) 1266:NaClO4/9404(30.0:6.8) 1265:NaClO4/9404(10.1:6.8) 1264:NaClO4/9404(20.2:6.8) 1263:KClO4/9404(10.0:6.8) 1262:KClO4/9404(30.2:6.8) 1261:KClO4/9404(20.7:6.8) 1260:PBX9404(6.8) 1259:PBX9407(2.6) 1258:PBX9407(2.6) 1257:PBX9407(2.6) 1256:PBX9407(2.6) 1249:C4/PBX9407(10.4:2.6) 1213:C4/PBX9407(50.0:2.6) 1212:C4/PBX9407(51.3:2.6) 1211:C4/PBX9407(25.1:2.6) 1210:C4/PBX9407(13:2.6) 1209:C4/PBX9407(10:2.6) 1208:C4/PBX9407(9.1:2.6) 1st transducer Figure 7: Nondimensional impulse versus distance. The pressure correlation was previously demonstrated with a more limited data set . Figure 5 contains additional pressure data and shows the nondimensional pressure ¯ P = ∆P P0 , where ∆P = P −P0, versus nondimensional radius ¯ r = rP0A E0 . The plot legend for each shot is of the form ⟨shot identiﬁcation number⟩:⟨booster type⟩/⟨Main charge type⟩(⟨Booster mass in grams⟩:⟨Main charge mass in grams⟩). 23rd ICDERS July 24-29, 2011 Irvine 5 Jackson Planar Blast Scaling The line is the expected data scaling from Eq. 5. The agreement between theory and experiment over most of the range is good, but for low values of ¯ r, near-ﬁeld effects become increasingly signiﬁcant. Data from the transducer nearest the source (0.64m) are circled in grey to highlight this. Shock position versus time is plotted in Fig. 6. The line corresponds to the expected scaling from Eq. 2. The nondimensional shock arrival time is ¯ ts = tsA E0 q γ P 3 0 ρ0 . The experimental data follows the theory well, but does exhibit slight deviations between ¯ r = 0.005–0.020, where experimental arrival times lag slightly behind theory. Experimental impulse values are shown in Fig. 7 with normalized impulse ¯ I = IA E0 q γ P0 ρ0 . Additional theory (not discussed) is required for an impulse scaling prediction, but the nondimensionalized exper-imental values do cluster to a value slightly below ¯ I = 1. As with the pressure measurements, there is signiﬁcant variation in the near-ﬁeld data for the ﬁrst transducer location (circled). This is expected, since I = R P(t)dt. Additionally, the piezoelectric transducers were not thermally protected from the hot postshock ﬂow, so it is possible that thermally induced case expansion decreased the late time pressure and impulse readings. The above results, their correlation to blast scaling theory, as well as additional experiments and numerical calculations underway will be presented in more detail. References S. Ohyagi, E. Nohira, T. Obara, and P. Cai, “Propagation of pressure waves initiated by a ﬂame and detonation in a tube,” JSME International Journal Series B, vol. 45, no. 1, pp. 192–200, 2002. S. Ohyagi, T. Yoshihashi, and Y. Harigaya, “Direct initiation of planar detonation waves in methane/oxygen/nitrogen mixtures,” in Progress in Astronautics and Aeronautics, pp. 3–22, AIAA, 1984. Presented in 1983 at the 9th International Colloquium on the Dynamics of Explosions and Reactive Systems (ICDERS). P. Thibault, J. Penrose, J. Shepherd, W. Benedick, and D. Ritzel, “Blast waves generated by planar detonations,” in Shock Tubes and Waves: Proceedings of the Sixteenth International Symposium on Shock Tubes and Waves, pp. 765–771, Wiley VCH, 1988. B. Hopkinson, “British ordinance board minutes,” tech. rep., 1915. G. Taylor, “The formation of a blast wave by a very intense explosion. I. theoretical discussion,” Pro-ceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, vol. 201, no. 1065, pp. 159–174, 1950. L. Sedov, Similarity and Dimensional Methods in Mechanics, Tenth Edition. CRC, 1993. R. Sachs, “The dependence of blast on ambient pressure and temperature,” Tech. Rep. BRL Report 466, Aberdeen Proving Ground, Aberdeen, Maryland, 1944. S. Jackson, J. Morris, and L. Hill, “Determination of explosive blast loading equivalencies with an explosively driven shock tube,” in Proceedings of the APS Topical Group on Shock Compression of Condensed Matter, vol. 1195, pp. 323–326, American Institute of Physics, 2009. 23rd ICDERS July 24-29, 2011 Irvine 6
189150	https://louis.pressbooks.pub/trigonometry/chapter/exercises-5-3-trigonometric-identities/	Exercises: 5.3 Trigonometric Identities – Trigonometry Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Book Contents Navigation Contents Introduction Introduction About This Book About the Authors Authors Chapter 1: Triangles and Circles Introduction to Triangles and Circles Introduction Class Activities 1.1 Triangles and Angles Algebra Refresher Triangles and Angles Triangles Angles Section 1.1 Summary 1.2 Similar Triangles Algebra Refresher Congruent Triangles Similar Triangles Using Proportions with Similar Triangles Similar Right Triangles Overlapping Triangles Section 1.2 Summary 1.3 Circles Algebra Refresher The Distance Formula Equation for a Circle Section 1.3 Summary Chapter 1 Summary and Review Exercises: 1.1 Triangles and Angles Exercises: 1.2 Similar Triangles Exercises: 1.3 Circles Exercises: Chapter 1 Review Problems Chapter 2: Trigonometric Ratios Introduction to the Trigonometric Ratios Introduction Class Activities 2.1 Side and Angle Relationships Algebra Refresher Introduction to Triangles The Triangle Inequality Right Triangles: The Pythagorean Theorem Section 2.1 Summary 2.2 Right Triangle Trigonometry Algebra Refresher The Sine of an Angle Using the Sine Ratio to Find an Unknown Side The Cosine and the Tangent The Three Trigonometric Ratios Section 2.2 Summary 2.3 Solving Right Triangles Algebra Refresher Finding an Angle The Special Angles Section 2.3 Summary Chapter 2 Summary and Review Exercises: 2.1 Side and Angle Relationships Exercises: 2.2 Right Triangle Trigonometry Exercises: 2.3 Solving Right Triangles Exercises: Chapter 2 Review Problems Chapter 3: Laws of Sines and Cosines Introduction to Laws of Sines and Cosines Introduction Class Activities 3.1 Obtuse Angles Algebra Refresher Angles in Standard Position Trigonometric Ratios for Obtuse Angles Trigonometric Ratios for Supplementary Angles Supplements of the Special Angles The Area of a Triangle Section 3.1 Summary 3.2 The Law of Sines Algebra Refresher Law of Sines Finding a Side Solving Triangles with the Law of Sines Finding an Angle Applications Measuring Astronomical Distances Small Angles: Minutes and Seconds Section 3.2 Summary 3.3 The Law of Cosines Algebra Refresher Law of Cosines Finding a Side Finding an Angle Using the Law of Cosines for the Ambiguous Case Application: Navigation Which Law to Use Section 3.3 Summary Chapter 3 Summary and Review Exercises: 3.1 Obtuse Angles Exercises: 3.2 The Law of Sines Exercises: 3.3 The Law of Cosines Exercises: Chapter 3 Review Problems Chapter 4: Trig Functions Introduction to Trigonometric Functions Introduction Class Activities 4.1 Angles and Rotation Algebra Refresher Angles in Standard Position Trigonometric Ratios for All Angles Reference Angles Using Reference Angles Coterminal Angles Solving Trigonometric Equations The Special Angles The Unit Circle Section 4.1 Summary 4.2 Graphs of Trigonometric Functions Algebra Refresher Location by Coordinates Bearings Function Notation The Tangent Function Angle of Inclination Section 4.2 Summary 4.3 Periodic Functions Algebra Refresher Period, Midline, and Amplitude Sinusoidal Functions Other Periodic Functions Section 4.3 Summary Chapter 4 Summary and Review Exercises: 4.1 Angles and Rotation Skills Exercises Homework 4.1 Exercises: 4.2 Graphs of Trigonometric Functions Skills Homework 4.2 Exercises: 4.3 Periodic Functions Skills Exercises Homework Exercises: Chapter 4 Review Problems Chapter 5: Equations and Identities Introduction to Equations and Identities Introduction 5.1 Algebra with Trigonometric Ratios Algebra Refresher Evaluating Trigonometric Expressions Simplifying Trigonometric Expressions Powers of Trigonometric Ratios Products Factoring Section 5.1 Summary 5.2 Solving Equations Algebra Refresher Trigonometric Equations Graphical Solutions Equations with Squares of Trig Ratios Snell’s Law Section 5.2 Summary 5.3 Trigonometric Identities Algebra Refresher Using Trigonometric Ratios in Identities Pythagorean Identity Tangent Identity Solving Equations Proving Identities Section 5.3 Summary Chapter 5 Summary and Review Exercises: 5.1 Algebra with Trigonometric Ratios Skills Exercises Homework 5.1 Exercises: 5.2 Solving Equations Skills Exercises Homework 5.2 Exercises: 5.3 Trigonometric Identities Skills Exercises Homework 5.3 Exercises: Chapter 5 Review Problems Exercises for Chapter 5 Review Chapter 6: Radians Introduction to Trig Radians Introduction Class Activities 6.1 Arclength and Radians Algebra Refresher Arclength and Radians Arclength Measuring Angles in Radians Converting between Degrees and Radians Arclength Formula Unit Circle Section 6.1 Summary 6.2 The Circular Functions Algebra Refresher Trigonometric Functions of Angles in Radians Values of Special Angles Reference Angles in Radians Sine and Cosine of Real Numbers Coordinates on a Unit Circle The Tangent Function The Circular Functions Section 6.2 Summary 6.3 Graphs of the Circular Functions Algebra Refresher The Sine and Cosine Functions The Tangent Function Solving Equations Modeling with Circular Functions Domain and Range Section 6.3 Summary Chapter 6 Summary and Review Exercises 6.1 Arclength and Radians Exercises 6.2 The Circular Functions Exercises 6.3 Graphs of the Circular Functions Exercises: Chapter 6 Review Problems Chapter 7: Circular Functions Introduction to Circular Functions Introduction Class Activities 7.1 Transformations of Graphs Algebra Refresher Transformations of Graphs Graphs of Sinusoidal Functions Modeling with Sinusoidal Functions The Tangent Function Section 7.1 Summary 7.2 The General Sinusoidal Function Algebra Refresher The General Sinusoidal Function Combining Transformations Modeling with Sinusoidal Functions Section 7.2 Summary 7.3 Solving Equations Algebra Refresher Multiple Solutions Using a Substitution Applications Section 7.3 Summary Chapter 7 Summary and Review Exercises 7.1 Transformations of Graphs Exercises 7.2 The General Sinusoidal Function Exercises 7.3 Solving Equations Exercises: Chapter 7 Review Problems Chapter 8: More Functions and Identities Introduction to More Functions and Identities Introduction Class Activities 8.1 Sum and Difference Formulas Algebra Refresher The Sum of Angles Identities The Difference of Angles Identities Sum and Difference Identities for Tangent Double Angle Identities Solving Equations Section 8.1 Summary 8.2 Inverse Trigonometric Functions Algebra Refresher Inverse of a Function The Graph of the Inverse The Inverse Cosine and Inverse Tangent Functions Modeling with Inverse Functions Simplifying Inverse Expressions Section 8.2 Summary 8.3 The Reciprocal Functions Algebra Refresher Three More Functions Application to Right Triangles Graphs of the Reciprocal Functions Solving Equations Using Identities Section 8.3 Summary Chapter 8 Summary and Review Exercises: 8.1 Sum and Difference Formulas Exercises: 8.2 Inverse Trigonometric Functions Exercises Exercises: 8.3 The Reciprocal Functions Exercises Exercises: Chapter 8 Review Problems Chapter 9: Vectors Introduction to Vectors Introduction 9.1 Geometric Form Skills Refresher Notation for Vectors Scalar Multiplication of Vectors Addition of Vectors Velocity Components of a Vector Using Components Section 9.1 Summary 9.2 Coordinate Form Skills Refresher Unit Vectors Converting between Geometric and Coordinate Form Scalar Multiples of Vectors in Coordinate Form Adding Vectors in Coordinate Form Force Section 9.2 Summary 9.3 The Dot Product Skills Refresher Components Coordinate Form for Components The Dot Product Geometric Meaning of the Dot Product Section 9.3 Summary Chapter 9 Summary and Review Exercises: 9.1 Geometric Form Exercises: 9.2 Coordinate Form Exercises: 9.3 The Dot Product Exercises: Chapter 9 Review Problems Chapter 10: Polar Coordinates and Complex Numbers Introduction to Polar Coordinates and Complex Numbers Introduction 10.1 Polar Coordinates Algebra Refresher Plotting Points Regions in the Plane Converting between Polar and Cartesian Coordinates Equations in Polar Coordinates Section 10.1 Summary 10.2 Polar Graphs Algebra Refresher Graphing in Polar Coordinates Sketching Familiar Equations Finding Intersection Points Section 10.2 Summary 10.3 Complex Numbers Algebra Refresher Imaginary Numbers Complex Numbers Arithmetic of Complex Numbers Products of Complex Numbers Quotients of Complex Numbers Graphing Complex Numbers Zeros of Polynomials Section 10.3 Summary 10.4 Polar Form for Complex Numbers Algebra Refresher Polar Form Products and Quotients in Polar Form Powers and Roots of Complex Numbers Section 10.4 Summary Chapter 10 Summary and Review Exercises: 10.1 Polar Coordinates Exercises: 10.2 Polar Graphs Exercises: 10.3 Complex Numbers Exercises: 10.4 Polar Form of Complex Numbers Exercises: Chapter 10 Review Problems Glossary Summary of Adaptations Full Resource Adaptations Chapter-Specific Changes Trigonometry Chapter 5: Equations and Identities Exercises: 5.3 Trigonometric Identities Skills Practice each skill in the Homework Problems listed: Recognize identities Verify identities Rewrite expressions using identities Use identities to evaluate expressions Solve trigonometric equations Given one trig ratio, find the others Suggested Problems Problems: #4, 8, 16, 44, 22, 24, 74, 28, 32, 50, 36, 58, 60, 70 Exercises Homework 5.3 Exercise Group For Problems 1–8, decide which of the following equations are identities. Explain your reasoning. 1. (a+b)2=a+b 2. a 2−b 2=a−b 3. 1 a+b=1 a+1 b 4. a+b a=b 5. tan⁡(α+β)=sin⁡(α+β)cos⁡(α+β) 6. 1 tan⁡θ=cos⁡θ sin⁡θ 7. (1+tan⁡θ)2=1+tan 2⁡θ 8. 1−sin 2⁡ϕ=1−sin⁡ϕ Exercise Group For Problems 9–16, use graphs to decide which of the following equations are identities. 9. sin⁡2 t=2 sin⁡t 10. cos⁡θ+sin⁡θ=1 11. sin⁡(30°+β)=1 2+sin⁡β 12. cos⁡(90°−C)=sin⁡C 13. tan⁡(90°−θ)=1 tan⁡θ 14. tan⁡2 θ=2 tan⁡θ 1−tan 2⁡θ 15. tan 2⁡x 1+tan 2⁡x=sin 2⁡x 16. tan⁡x+1 tan⁡x=sin⁡x cos⁡x Exercise Group For Problems 17–26, show that the equation is an identity by transforming the left side into the right side. 17. (1+sin⁡w)(1−sin⁡w)=cos 2⁡w 18. (cos⁡θ−1)(cos⁡θ+1)=−sin 2⁡θ 19. (cos⁡θ−sin⁡θ)2=1−2 sin⁡θ cos⁡θ 20. sin 2⁡x−cos 2⁡x=1−2 cos 2⁡x 21. tan⁡θ cos⁡θ=sin⁡θ 22. sin⁡μ tan⁡μ=cos⁡μ 23. cos 4⁡x−sin 4⁡x=cos 2⁡x−sin 2⁡x 24. 1−2 cos 2⁡v+cos 4⁡v=sin 4⁡v 25. sin⁡u 1+cos⁡u=1−cos⁡u sin⁡u Hint. Multiply numerator and denominator of the left side by 1−cos⁡u. 26. sin⁡v 1−cos⁡v=tan⁡v(1+sin⁡v)cos⁡v Hint. Multiply numerator and denominator of the left side by 1+sin⁡v. Exercise Group For Problems 27–34, simplify, using identities as necessary. 27. 1 cos 2⁡β−sin 2⁡β cos 2⁡β 28. 1 sin 2⁡ϕ−1 tan 2⁡ϕ 29. cos 2⁡α(1+tan 2⁡α) 30. cos 3⁡ϕ+sin 2⁡ϕ cos⁡ϕ 31. tan 2⁡A−tan 2⁡A sin 2⁡A 32. cos 2⁡B tan 2⁡B+cos 2⁡B 33. 1−cos 2⁡z cos 2⁡z 34. sin⁡t cos⁡t tan⁡t Exercise Group For Problems 35–40, evaluate without using a calculator. 35. 3 cos 2⁡1.7°+3 sin 2⁡1.7° 36. 4−cos 2⁡338°−sin 2⁡338° 37. (cos 2⁡20°+sin 2⁡20°)4 38. 18 cos 2⁡17°+sin 2⁡17° 39. 6 cos 2⁡53°−6 tan 2⁡53° 40. 1 sin 2⁡102°−cos 2⁡102°sin 2⁡102° Exercise Group For Problems 41–46, one side of an identity is given. Graph the expression and make a conjecture about the other side of the identity. 41. 2 cos 2⁡θ−1=? 42. 1−2 sin 2⁡(θ 2)=? 43. 1−sin 2⁡x 1+cos⁡x=? 44. sin⁡x 1−sin 2⁡x=? 45. 2 tan⁡t cos 2⁡t=? 46. 2 tan⁡t 1−tan 2⁡t=? Exercise Group For Problems 47–50, use identities to rewrite each expression. 47. 2−cos 2⁡θ+2 sin⁡θ as an expression in sin⁡θ only 48. 3 sin 2⁡B+2 cos⁡B−4 as an expression in cos⁡B only 49. cos 2⁡ϕ−2 sin 2⁡ϕ as an expression in cos⁡ϕ only 50. cos 2⁡ϕ sin 2⁡ϕ as an expression in sin⁡ϕ only Exercise Group For Problems 51–58, solve the equation for 0°≤θ≤360°. Round angles to three decimal places if necessary. 51. cos⁡θ−sin 2⁡θ+1=0 52. 4 sin⁡θ+2 cos 2⁡θ−3=−1 53. 1−sin⁡θ−2 cos 2⁡θ=0 54. 3 cos 2⁡θ−sin 2⁡θ=2 55. 2 cos⁡θ tan⁡θ+1=0 56. cos⁡θ−sin⁡θ=0 57. 1 3 cos⁡θ=sin⁡θ 58. 5 sin⁡C=2 cos⁡C Exercise Group For Problems 59–62, use identities to find exact values for the other two trig ratios. 59. cos⁡A=12 13 and 270°<A<360° 60. sin⁡B=−3 5 and 180°<B<270° 61. sin⁡ϕ=1 7 and 90°<ϕ<180° 62. cos⁡t=−2 3 and 180°<t<270° Exercise Group For Problems 63–66, use the identity below to find the sine and cosine of the angle. 1+tan 2⁡θ=1 cos 2⁡θ 63. tan⁡θ=−1 2 and 270°<θ<360° 64. tan⁡θ=2 and 180°<θ<270° 65. tan⁡θ=3 4 and 180°<θ<270° 66. tan⁡θ=−3 and 90°<θ<180° Exercise Group For Problems 67–72, find exact values for the sine, cosine, and tangent of the angle. 67. 2 cos⁡A+9=8 and 90°<A<180° 68. 25 sin⁡B+8=−12 and 180°<B<270° 69. 8 tan⁡β+5=−11 and 90°<β<180° 70. 6(tan⁡β−4)=−24 and 90°<β<270° 71. tan 2⁡C−1 4=0 and 0°<C<180° 72. 4 cos 2⁡A−cos⁡A=0 and 00°<A<180° Exercise Group For Problems 73–76, prove the identity by rewriting tangents in terms of sines and cosines. (These problems involve simplifying complex fractions. See the Algebra Refresher to review this skill.) 73. tan⁡α 1+tan⁡α=sin⁡α sin⁡α+cos⁡α 74. 1−tan⁡u 1+tan⁡u=cos⁡u−sin⁡u cos⁡u+sin⁡u 75. 1+tan 2⁡β 1−tan 2⁡β=1 cos 2⁡β−sin 2⁡β 76. tan 2⁡v−sin 2⁡v=tan 2⁡v sin 2⁡v 77. Prove the Pythagorean identity cos 2⁡θ+sin 2⁡θ=1 by carrying out the following steps. Sketch an angle θ in standard position and label a point (x,y) on the terminal side, at a distance r from the vertex. Begin with the equation x 2+y 2=r, and square both sides. Divide both sides of your equation from part (a) by r 2. Write the left side of the equation as the sum of the squares of two fractions. Substitute the appropriate trigonometric ratio for each fraction. 78. Prove the tangent identity tan⁡θ=sin⁡θ cos⁡θ by carrying out the following steps. Sketch an angle θ in standard position and label a point (x,y) on the terminal side, at a distance r from the vertex. Write sin⁡θ in terms of y and r, and solve for y. Write cos⁡θ in terms of x and r, and solve for x. Write tan⁡θ in terms of x and y, then substitute your results from parts (a) and (b). Simplify your fraction in part (c). Previous/next navigation Previous: Exercises: 5.2 Solving Equations Next: Exercises: Chapter 5 Review Problems Back to top License Trigonometry Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted. Share This Book Pressbooks Powered by Pressbooks Pressbooks User Guide \|Pressbooks Directory \|Contact Pressbooks on YouTubePressbooks on LinkedIn
189151	https://www.bigideasmath.com/external/state-resources/pdfs/NC_math2_09_02.pdf	9.2 Perpendicular and Angle Bisectors For use with Exploration 9.2 Name ________ Date ___ Essential Question What conjectures can you make about a point on the perpendicular bisector of a segment and a point on the bisector of an angle? Go to BigIdeasMath.com for an interactive tool to investigate this exploration. Work with a partner. Use dynamic geometry software. a. Draw any segment and label it . AB Construct the perpendicular bisector of . AB b. Label a point C that is on the perpendicular bisector of AB but is not on . AB c. Draw CA and CB and find their lengths. Then move point C to other locations on the perpendicular bisector and note the lengths of CA and . CB d. Repeat parts (a)–(c) with other segments. Describe any relationship(s) you notice. Go to BigIdeasMath.com for an interactive tool to investigate this exploration. Work with a partner. Use dynamic geometry software. a. Draw two rays AB ⎯→ and AC ⎯→ to form . BAC ∠ Construct the bisector of . BAC ∠ b. Label a point D on the bisector of . BAC ∠ 1 EXPLORATION: Points on a Perpendicular Bisector 2 EXPLORATION: Points on an Angle Bisector Sample Points A(1, 3) B(2, 1) C(2.95, 2.73) Segments AB = 2.24 CA = ? CB = ? Line 2 2.5 x y − + = 0 1 2 3 0 1 2 A B C 3 4 5 289 Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 Perpendicular and Angle Bisectors (continued) Name ________ Date _ c. Construct and find the lengths of the perpendicular segments from D to the sides of . BAC ∠ Move point D along the angle bisector and note how the lengths change. d. Repeat parts (a)–(c) with other angles. Describe any relationship(s) you notice. Communicate Your Answer 3. What conjectures can you make about a point on the perpendicular bisector of a segment and a point on the bisector of an angle? 4. In Exploration 2, what is the distance from point D to AB ⎯→ when the distance from D to AC ⎯→ is 5 units? Justify your answer. Sample Points A(1, 1) B(2, 2) C(2, 1) D(4, 2.24) Rays 0 AB x y = − + = 1 AC y = = Line 0.38 0.92 0.54 x y − + = 2 EXPLORATION: Points on an Angle Bisector (continued) 0 1 2 3 4 0 1 2 A C F D E B 3 4 5 6 290 Copyright © Big Ideas Learning, LLC All rights reserved. Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 For use after Lesson 9.2 Name ________ Date _ In your own words, write the meaning of each vocabulary term. equidistant Theorems Perpendicular Bisector Theorem In a plane, if a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP ← → is the ⊥ bisector of , AB then . CA CB = Notes: Converse of the Perpendicular Bisector Theorem In a plane, if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment. If , DA DB = then point D lies on the ⊥ bisector of . AB Notes: A B P C A B P D C 9.2 For use after Lesson 9.2 Name ________ Date ___ In your own words, write the meaning of each vocabulary term. equidistant Theorems Perpendicular Bisector Theorem In a plane, if a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP ← → is the ⊥ bisector of , AB then . CA CB = Notes: Converse of the Perpendicular Bisector Theorem In a plane, if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment. If , DA DB = then point D lies on the ⊥ bisector of . AB Notes: A B P C A B P D C 285 Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 Notetaking with Vocabulary For use after Lesson 9.2 In your own words, write the meaning of each vocabulary term. equidistant Theorems Perpendicular Bisector Theorem In a plane, if a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP ← → is the ⊥ bisector of , AB then . CA CB = Notes: Converse of the Perpendicular Bisector Theorem In a plane, if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment. If , DA DB = then point D lies on the ⊥ bisector of . AB Notes: A B P C A B P D C 9.2 Notetaking with Vocabulary (continued) Name ________ Date _ Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the two sides of the angle. If AD ⎯→ bisects BAC ∠ and DB AB ⎯→ ⊥ and , DC AC ⎯→ ⊥ then . DB DC = Notes: from the two sides of the angle, then it lies on the B D C A B 285 Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 Notetaking with Vocabulary For use after Lesson 9.2 equidistant Theorems Perpendicular Bisector Theorem In a plane, if a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP ← → is the ⊥ bisector of , AB then . CA CB = Notes: Converse of the Perpendicular Bisector Theorem In a plane, if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment. If , DA DB = then point D lies on the ⊥ bisector of . AB Notes: A B P C A B P D C Practice 291 Copyright © Big Ideas Learning, LLC All rights reserved. Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 Name ________ Date _ Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the two sides of the angle. If AD ⎯→ bisects BAC ∠ and DB AB ⎯→ ⊥ and , DC AC ⎯→ ⊥ then . DB DC = Notes: Converse of the Angle Bisector Theorem If a point is in the interior of an angle and is equidistant from the two sides of the angle, then it lies on the bisector of the angle. If DB AB ⎯→ ⊥ and DC AC ⎯→ ⊥ and , DB DC = then AD ⎯→ bisects . BAC ∠ Notes: B D C A C 9.2 Name ________ Date _ Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the two sides of the angle. If AD ⎯→ bisects BAC ∠ and DB AB ⎯→ ⊥ and , DC AC ⎯→ ⊥ then . DB DC = Notes: Converse of the Angle Bisector Theorem If a point is in the interior of an angle and is equidistant from the two sides of the angle, then it lies on the bisector of the angle. If DB AB ⎯→ ⊥ and DC AC ⎯→ ⊥ and , DB DC = then AD ⎯→ bisects . BAC ∠ Notes: B D C A C 286 Copyright © Big Ideas Learning, LLC All rights reserved. Notes: Converse of the Angle Bisector Theorem If a point is in the interior of an angle and is equidistant from the two sides of the angle, then it lies on the bisector of the angle. If DB AB ⎯→ ⊥ and DC AC ⎯→ ⊥ and , DB DC = then AD ⎯→ bisects . BAC ∠ Notes: B D C A Copyright © Big Ideas Learning, LLC All i ht d X and Z? B is not collinear with X and Z. Because the two segments containing points X and Z are congruent, B is the same distance from both X and Z , point B is equidistant from X and Z, and point B is in the perpendicular bisector of — XZ . Monitoring Progress and Modeling with Mathematics 3. GH = 4.6; Because GK = KJ and ⃖⃗ HK ⊥ ⃖⃗ GJ , point H is on the perpendicular bisector of — GJ . So, by the Perpendicular Bisector Theorem, GH = HJ = 4.6. 4. QR = 1.3; Because point T is equidistant from Q and S, point T is on the perpendicular bisector of — QS by the Converse of the Perpendicular Bisector Theorem. So, by defi nition of segment bisector, QR = RS = 1.3. 5. AB = 15; Because ⃖⃗ DB ⊥ ⃖⃗ AC and point D is equidistant from A and C, point D is on the perpendicular bisector of — AC by the Converse of the Perpendicular Bisector Theorem. By defi nition of segment bisector, AB = BC. AB = BC 5x = 4x + 3 x = 3 AB = 5 ⋅ 3 = 15 6. UW = 55; Because — VD ≅ — WD and ⃖⃗ UX ⊥ — VW , point U is on the perpendicular bisector of — VW . So, by the Perpendicular Bisector Theorem, VU = WU. VU = UW 9x + 1 = 7x + 13 2x = 12 x = 6 UW = 7 ⋅ 6 + 13 = 55 7. yes; Because point N is equidistant from L and M, point N is on the perpendicular bisector of — LM by the Converse of the Perpendicular Bisector Theorem. Because only one line can be perpendicular to — LM at point K, ⃗ NK must be the perpendicular bisector of — LM , and P is on ⃗ NK . 8. no; You would need to know that either LN = MN or LP = MP. 9. no; You would need to know that ⃖⃗ PN ⊥ ⃖⃗ ML . 10. yes; Because point P is equidistant from L and M, point P is on the perpendicular bisector of — LM by the Converse of the Perpendicular Bisector Theorem. Also, — LN ≅ — MN , so ⃗ PN is a bisector of — LM . Because P can only be on one of the bisectors, ⃗ PN is the perpendicular bisector of — LM . 11. Because D is equidistant from ⃗ BC and ⃗ BA , ⃗ BD bisects ∠ABC by the Converse of the Angle Bisector Theorem. So, m∠ABD = m∠CBD = 20°. ⃗ S is an angle bisector of ∠PQR , — PS ⊥ ⃗ QP , and — SR ⊥ ⃗ QR . So, by the Angle Bisector Theorem, PS = RS = 12. 2. The question that is different is: Is point B collinear with X and Z? B is not collinear with X and Z. Because the two segments containing points X and Z are congruent, B is the same distance from both X and Z, point B is equidistant from X and Z, and point B is in the perpendicular bisector of — XZ . Monitoring Progress and Modeling with Mathematics 3. GH = 4.6; Because GK = KJ and ⃖⃗ HK ⊥ ⃖⃗ GJ , point H is on the perpendicular bisector of — GJ . So, by the Perpendicular Bisector Theorem, GH = HJ = 4.6. 4. QR = 1.3; Because point T is equidistant from Q and S, point T is on the perpendicular bisector of — QS by the Converse of the Perpendicular Bisector Theorem. So, by defi nition of segment bisector, QR = RS = 1.3. 5. AB = 15; Because ⃖⃗ DB ⊥ ⃖⃗ AC and point D is equidistant from A and C, point D is on the perpendicular bisector of — AC by the Converse of the Perpendicular Bisector Theorem. By defi nition of segment bisector, AB = BC. AB = BC 5x = 4x + 3 x = 3 AB = 5 ⋅ 3 = 15 6. UW = 55; Because — VD ≅ — WD and ⃖⃗ UX ⊥ — VW , point U is on the perpendicular bisector of — VW . So, by the Perpendicular Bisector Theorem, VU = WU. VU = UW 9x + 1 = 7x + 13 2x = 12 x = 6 UW = 7 ⋅ 6 + 13 = 55 7. yes; Because point N is equidistant from L and M, point N — LM line can be perpendicular to — LM at point K, ⃗ NK must be the ⃗ Worked-Out Examples Example #1 Find the indicated measure. Explain your reasoning. Example #2 Find the indicated measure. Explain your reasoning. 292 (continued) Practice A C B D 5x 4x + 3 AB B C A D 20° m∠ABD Copyright © Big Ideas Learning, LLC All rights reserved. 9.2 Name ________ Date _ Extra Practice In Exercises 1–3, find the indicated measure. Explain your reasoning. 1. AB 2. EG 3. SU 4. Find the equation of the perpendicular bisector of AB. In Exercises 5–7, find the indicated measure. Explain your reasoning. 5. m CAB ∠ 6. DC 7. BD B A x y 2 −2 2 −2 A B C D 20° B C D A 30° 30° 5 B C D A 3x + 1 5x – 1 B C A D 6 9 6 E H F G 7 7 3 R T S U 10 10 2x + 2 3x Practice A (continued) Practice 293 Copyright © Big Ideas Learning, LLC All rights reserved. 6.2 Practice B 30° 60° B D C A 3(2x − 8) −x + 25 Name _________ Date ___ In Exercises 1–3, tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of RS, or on the angle bisector of DEF. ∠ Explain your reasoning. 1. 2. 3. In Exercises 4–6, find the indicated measure. Explain your reasoning. 4. AC 5. m LNM ∠ 6. m UTW ∠ 7. Write an equation of the perpendicular bisector of the segment with the endpoints ( ) ( ) 3, 7 and 1, 5 . G H − − 8. In the figure, line m is the perpendicular 9. You are installing a fountain in the triangular bisector of . PR Is point Q on line m? Is garden pond shown in the point S on line m? Explain your reasoning. figure. You want to place the fountain the same distance from each side of the pond. Describe a way to determine the location of the fountain using angle bisectors. D P F E (0, 5) x N M y ( , ) 5 2 5 2 (5, 0) K L R P S Q Q D E F P U T W (2x + 3)° (5x − 24)° 9 9 V 10 P Q R S 13 12 10 m Practice B 294 Copyright © Big Ideas Learning, LLC All rights reserved.
189152	https://www.quora.com/What-is-the-easiest-way-of-explaining-how-discounting-impacts-profit-margins-I-have-a-methodology-to-calculate-margin-impact-would-would-love-to-see-how-others-explain-this	Something went wrong. Wait a moment and try again. Margin Line Pricing Strategy Marginal Profit Business Operations Business Finance Margin (economics) 5 What is the easiest way of explaining how discounting impacts profit margins? I have a methodology to calculate margin impact would would love to see how others explain this Garrick Saito former Corp. Controller, large public company · Author has 33.7K answers and 215.5M answer views · 6y I’m positive I’m missing the ball here, but the way my brain works to visually understand and communicate an idea like this is to show various scenarios of how discounts affects margins. So here, we have a very simple P&L scenario showing sales, cost of sales and gross margin. My baseline is a $1,000 widget that costed me $400. Gross margin without discount is $600 or 60%. Then, my model shows how my margin slowly begins to deteriorate as I give discounts off my $1000 widget. I start with one percent, then two, then three and run the model up to 25% (where I arbitrarily stopped). I set up a column I’m positive I’m missing the ball here, but the way my brain works to visually understand and communicate an idea like this is to show various scenarios of how discounts affects margins. So here, we have a very simple P&L scenario showing sales, cost of sales and gross margin. My baseline is a $1,000 widget that costed me $400. Gross margin without discount is $600 or 60%. Then, my model shows how my margin slowly begins to deteriorate as I give discounts off my $1000 widget. I start with one percent, then two, then three and run the model up to 25% (where I arbitrarily stopped). I set up a column for “lost gross margin” so the eyes can easily scan the percentage of discount I’m giving and the related gross margin I’ll be losing at that decision point. The ratio of the lost gross margin : discounts given clearly show me that the more I give, the more margin I’ll lose as a percentage of net sales. If you can put all that into words without the chart, then you are a better man than I. Anyway, if a business owner asked me the question you’re asking, this is how I would show them the effects of discounts and margin, rather than to present them with some complicated formula both of us would roll our eyes at (and probably not understand). Mark Mehling Marketing strategist, author, problem solver for 40+ Years. · Author has 16.3K answers and 8.8M answer views · 6y Very simply: Price is made up of three components: Fixed costs + Variable costs + Profit = Price If you are discounting, there is only one of the three that you control- the Profit. (While there is some small control over the others, the Profit is where you have the most control) When you discount, that is what you are reducing. Profit. It’s possible to get more sales at a reduced price and, overall, get more profit. But in general, most small business owners don’t know how to do that well. Bottom: Never, NEVER, NEVER discount your main product or service. Jeffrey Summers I Build Successful Restaurants · Author has 4.4K answers and 1.2M answer views · 6y How discounting impacts margins? Do you really need a methodology to tell you how charging less for a product creates less profit in both the short term and long term? I too have a methodology. It’s called basic math. Related questions What is a margin, and how is it calculated? What is the purpose of calculating profit margin in a business? Are there alternative methods for determining this value? If so, why is calculating profit margin still commonly used? What are the best are examples for gross profit margin? What's the best way to calculate markup percentage on an item? How do you decide on acceptable profit margins? What is the difference between "markup", "margin", "profit margin", "gross margin" and "net margin"? Assistant Bot · 1y Explaining how discounting impacts profit margins can be done effectively using a simple analogy and straightforward calculations. Here’s a concise way to approach the explanation: Explanation Using a Simple Analogy Basic Concept: Think of profit margin as the "slice of the pie" that a business keeps after selling a product. When you discount a product's price, you're essentially making the pie slice smaller. Analogy: Imagine you bake a pie (your product) and plan to sell it for $10. If it costs you $6 to make, your profit margin is $4, or 40%. Now, if you decide to discount the pie to $8 to attr Explaining how discounting impacts profit margins can be done effectively using a simple analogy and straightforward calculations. Here’s a concise way to approach the explanation: Explanation Using a Simple Analogy Basic Concept: Think of profit margin as the "slice of the pie" that a business keeps after selling a product. When you discount a product's price, you're essentially making the pie slice smaller. Analogy: Imagine you bake a pie (your product) and plan to sell it for $10. If it costs you $6 to make, your profit margin is $4, or 40%. Now, if you decide to discount the pie to $8 to attract more customers, your profit margin shrinks: Before Discount: Selling Price = $10 Cost = $6 Profit = $4 Profit Margin = 410×100=40% After Discount: Selling Price = $8 Cost = $6 Profit = $2 Profit Margin = 28×100=25% Key Points to Highlight Margin Reduction: Discounting reduces the selling price, which directly lowers the profit per unit sold. Volume vs. Margin: While discounts can increase sales volume, businesses must consider whether the increased volume compensates for the reduced margin. If the volume increase doesn’t cover the loss in margin, overall profits may decline. Break-Even Point: It’s essential to calculate how many additional units need to be sold at the discounted price to maintain or increase total profit. Conclusion In summary, discounting can be a useful strategy for boosting sales but must be approached carefully to ensure it doesn’t erode profit margins to the point of harming overall profitability. It’s always beneficial to analyze the potential trade-offs involved. Calculation Methodology If you have a specific methodology for calculating margin impact, consider sharing it alongside these explanations, as it can provide a more detailed understanding of how discounting affects profitability in various scenarios. Would you like to discuss your methodology further? Fu Fei Founder at Gooodbye (2023–present) · Author has 117 answers and 51K answer views · 6y Related How should one determine the appropriate profit margin for their product? Short version: Ask ten possible customers what they would be willing to pay. Long version: There are three SUPER easy ways to get started, and if you are not deliberate about this decision, it is more than likely that you are being told what your products are worth. Your Suppliers If you are in a highly commoditised market, your supplier is likely supplying the same good to multiple retailers. Just ask them what they think is a fair estimate. Cost of good (1 + Markup) Your Competition If you know that your potential customers are extremely price-sensitive and they will only compare you to the exist Short version: Ask ten possible customers what they would be willing to pay. Long version: There are three SUPER easy ways to get started, and if you are not deliberate about this decision, it is more than likely that you are being told what your products are worth. Your Suppliers If you are in a highly commoditised market, your supplier is likely supplying the same good to multiple retailers. Just ask them what they think is a fair estimate. Cost of good (1 + Markup) Your Competition If you know that your potential customers are extremely price-sensitive and they will only compare you to the existing competition, then do some research and markdown Their Price (1 - Markdown) Your Customers If you did the work and researched what customers were willing to pay for something, price accordingly. There should be three questions What would be a fair price for this? What would be too low that you would question the quality? What would be absurdly too expensive to pay for this? Price based on the first question. If you have the time, check out this talk. Good luck! Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Ron Auerbach Degrees in: economics, finance, accounting, human resources · Author has 13.4K answers and 20.3M answer views · 4y Related How do I calculate a sales discount in accounting? This is something I would teach students in my accounting classes. You’d multiply the sale amount (not including freight and taxes) by the cash discount. For example, you buy $1000 worth of merchandise that has a 2% cash discount if you pay within 30 days of the invoice date. You make the payment 25 days after the invoice date. So on the seller’s end, they’d record it as follows: Debit Cash $980 for the amount the customer had paid Debit Sales Discount for the $20 cash discount they had given the customer Credit Accounts Receivable for the $1000 the customer had owed This way, you zero out the c This is something I would teach students in my accounting classes. You’d multiply the sale amount (not including freight and taxes) by the cash discount. For example, you buy $1000 worth of merchandise that has a 2% cash discount if you pay within 30 days of the invoice date. You make the payment 25 days after the invoice date. So on the seller’s end, they’d record it as follows: Debit Cash $980 for the amount the customer had paid Debit Sales Discount for the $20 cash discount they had given the customer Credit Accounts Receivable for the $1000 the customer had owed This way, you zero out the customer’s owing you anything to reflect full payment. And you separate out the cash you actually received from the cash discount because the customer had paid you within the discount period (window). FYI, on the buyer’s end, you’d calculate the discount in the same way. But it would be considered a Purchase Discount rather than a Sales Discount. And exactly how you’d record it as the buyer would depend upon whether you’re using perpetual or periodic inventory. Under perpetual inventory, the purchase discount would be put toward the Inventory account. But with periodic inventory, you’d separate it out by putting it into the Purchases Discount account. So in accounting, sales discount = a specific account used by the seller for the recording of cash discounts taken by the customer. This is different than offering the customer a price reduction right away. That would be a trade or chain discount. For example, the seller gives the customer 20% off the price. So if the price is $10, the customer would only be charged $8. In this case, you would not record the price reduction. Instead, you’d record the actual price to your customer as revenue. So trade and chain discounts are not journalized. This is why terminology is very important! Related questions What is profit margin and how do you calculate it in your business? Can YOY Gross profit margin (GP%) be reconciled with exchange rate impact, sales mix impact and sales price impact? What is the best way to calculate profit margin, gross margin, and net profit margin? How do I calculate the profit percentage if I want a 30% margin and I want to give a 10% discount to customers? What goes into profit margins? Mark Gandy Founder of G3CFO.com · Author has 2.4K answers and 8M answer views · 1y Related What is the best way to calculate profit margin, gross margin, and net profit margin? There is no ‘best’ way. The math is straightforward: profit margin is pre-tax earnings divided by revenue gross margin is gross profit divided by revenue net profit is net income (after taxes) divided by revenue In some industries, through judgment and common sense, we occasionally use gross profit as the denominator because the gross margins are so tiny. The other margins are meaningless when divided by revenue, the top line. Otherwise, always use your profit numbers and revenue as your denominator. I’ve added an example below where you can recompute the margins to test your knowledge. There is no ‘best’ way. The math is straightforward: profit margin is pre-tax earnings divided by revenue gross margin is gross profit divided by revenue net profit is net income (after taxes) divided by revenue In some industries, through judgment and common sense, we occasionally use gross profit as the denominator because the gross margins are so tiny. The other margins are meaningless when divided by revenue, the top line. Otherwise, always use your profit numbers and revenue as your denominator. I’ve added an example below where you can recompute the margins to test your knowledge. Sponsored by Zoho Books Accounting shouldn’t be too hard. Make it easy with Zoho Books! Zoho Books is an all-in-one accounting platform. Migrate to Zoho Books for free today. Avinash More Business Development Executive at Haldiram's (2024–present) · 3y Related How do I calculate my profit margin on any product I sell? Profit margins are one of the most basic and widely used financial ratios in business. On an income statement, a company's profit is calculated at three levels, beginning with the most basic—gross profit—and progressing to the most comprehensive, net profit. Operating profit is the difference between these two. Profit margins for all three are calculated by dividing the profit figure by revenue and multiplying by 100. Gross Profit Margin The simplest profitability metric is gross profit, which defines profit as all income remaining after deducting the cost of goods sold (COGS). COGS only include Profit margins are one of the most basic and widely used financial ratios in business. On an income statement, a company's profit is calculated at three levels, beginning with the most basic—gross profit—and progressing to the most comprehensive, net profit. Operating profit is the difference between these two. Profit margins for all three are calculated by dividing the profit figure by revenue and multiplying by 100. Gross Profit Margin The simplest profitability metric is gross profit, which defines profit as all income remaining after deducting the cost of goods sold (COGS). COGS only includes expenses directly related to the production or manufacture of items for sale, such as raw materials and labour wages required to make or assemble goods. The formula for gross profit margin is: Operating Profit Margin Operating profit is a slightly more complicated metric that includes all overhead, operating, administrative, and sales expenses required to run the business on a daily basis. While this figure still does not include debts, taxes, or other non-operational expenses, it does include asset amortisation and depreciation. This mid-level profitability margin is calculated by dividing operating profit by revenue and reflects the percentage of each dollar that remains after all expenses necessary to keep the business running have been paid. The formula for gross profit margin is: Net Profit Margin The infamous bottom line, net income, represents the total amount of revenue remaining after all expenses and additional income streams have been deducted. This includes, in addition to the COGS and operational expenses mentioned above, debt payments, taxes, one-time expenses or payments, and any income from investments or secondary operations. The net profit margin measures a company's ability to convert income into profit. The formula for gross profit margin is: Manish Pratap Yadav BBA from National PG College, Lucknow (Graduated 2020) · 5y Related How do you calculate profit margins? Profit Margin is calculated with respect to the Selling Price instead of Cost price. For e.g. Assume CP to be 100 and SP to be 125 Here, Profit=125–100 i.e 25 If Profit is calculated on CP its called Profit Percent and is equal to [(Profit/CP)100] percent but If profit is calculated on SP it's termed as Profit Margin and is equal to [(Profit/SP)100] percent In above assumption, Profit Percent=(25/100)100 i.e 25 percent. Profit Margin=(25/125)100 i.e 20 percent. Here you'll notice that margin percent is less than actual Profit percent which is true in every situation. For more details please refer t Profit Margin is calculated with respect to the Selling Price instead of Cost price. For e.g. Assume CP to be 100 and SP to be 125 Here, Profit=125–100 i.e 25 If Profit is calculated on CP its called Profit Percent and is equal to [(Profit/CP)100] percent but If profit is calculated on SP it's termed as Profit Margin and is equal to [(Profit/SP)100] percent In above assumption, Profit Percent=(25/100)100 i.e 25 percent. Profit Margin=(25/125)100 i.e 20 percent. Here you'll notice that margin percent is less than actual Profit percent which is true in every situation. For more details please refer to below picture. Promoted by Savings Pro Mark Bradley Economist · Updated Aug 14 What are the stupidest money mistakes most people make? Where do I start? I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits. Here are the biggest mistakes people are making and how to fix them: Not having a separate high interest savings account Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up. Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one of th Where do I start? 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Pull up Coverage.com, a free site that will compare prices for you, answer the questions on the page, and it will show you how much you could be saving. That’s it. You’ll likely be saving a bunch of money. Here’s a link to give it a try. Consistently being in debt If you’ve got $10K+ in debt (credit cards…medical bills…anything really) you could use a debt relief program and potentially reduce by over 20%. Here’s how to see if you qualify: Head over to this Debt Relief comparison website here, then simply answer the questions to see if you qualify. It’s as simple as that. You’ll likely end up paying less than you owed before and you could be debt free in as little as 2 years. Missing out on free money to invest It’s no secret that millionaires love investing, but for the rest of us, it can seem out of reach. Times have changed. There are a number of investing platforms that will give you a bonus to open an account and get started. All you have to do is open the account and invest at least $25, and you could get up to $1000 in bonus. Pretty sweet deal right? Here is a link to some of the best options. Having bad credit A low credit score can come back to bite you in so many ways in the future. From that next rental application to getting approved for any type of loan or credit card, if you have a bad history with credit, the good news is you can fix it. Head over to BankRate.com and answer a few questions to see if you qualify. It only takes a few minutes and could save you from a major upset down the line. How to get started Hope this helps! Here are the links to get started: Have a separate savings account Stop overpaying for car insurance Finally get out of debt Start investing with a free bonus Fix your credit Marsha Hughes Production Artist · Apr 1 Related How do I determine your profit margin? Part 1 of 2:Calculating Profit Margin 1. Know the difference between gross profit, gross profit margin, and net profit. Gross profit is your total revenue earned from your goods or services, minus the cost of producing or providing those goods or services (COGS). This calculation does not include expenses like payroll, rent, or utilities; it only considers the cost directly related to creating those goods and services. Gross profit margin is the gross profit divided by revenues. Net profit takes all business expenditures into account and is calculated as gross profit minus administrative expense Part 1 of 2:Calculating Profit Margin 1. Know the difference between gross profit, gross profit margin, and net profit. Gross profit is your total revenue earned from your goods or services, minus the cost of producing or providing those goods or services (COGS). This calculation does not include expenses like payroll, rent, or utilities; it only considers the cost directly related to creating those goods and services. Gross profit margin is the gross profit divided by revenues. Net profit takes all business expenditures into account and is calculated as gross profit minus administrative expenses and other relevant expenses. This includes regular operational costs (payroll, rent, etc.) and one-time costs (taxes, contractor invoices, etc.). You must also include any additional earnings, such as investment income. Net profit provides a more complete and detailed rendering of the business health and is generally what is used to manage the business. The steps below detail how to find this number. Net profit is also known as "the bottom line." Determine your calculation period. To calculate your business's profit margin, choose the period of time you want to analyze. Generally, people use either months, quarters, or years to calculate their profit margins. Consider why you want to calculate your margins. If you are applying for loans or looking to attract investors, these people will want to know more than just how your business did over a single month. However, if you're comparing your profit margin between different months for your own purposes, it's fine to use shorter periods of time. Calculate the total revenue generated by your business during the calculation period. Revenue is everything the business brings in through sale of goods, services, or earnings of interest. If your business only sells goods, such a retail shop or restaurant, your total revenue is all the sales you had during the period you've chosen to analyze minus any returns or discounts. If you don't already have this figure on hand, multiply the total number of items you sold by the price of each of those items and then adjust for returns and discounts. Similarly, if your business provides services, such as lawn mowing, your total revenue is all of the amounts you collected for your services during a period. Finally, if the business involves owning securities, you should include the interest and dividend income from those sources in your total revenue calculation. Subtract all your expenses to calculate your net income. Expenses are the opposite of revenue. They're any amounts you have had to pay, or will pay in the future for things you did and/or used during the calculation period. This includes expenses incurred to operate as well as the expense required to carry investments. Common expenses are the cost of labor, rent, electricity, equipment, supplies, inventory, banking, and interest expense on loans. Generally if you run a small business you can just add up everything you paid for during the period. For example, if your business earned $100,000 in revenue during the calculation period, and in order to earn that revenue the business spent $70,000 on rent, supplies, equipment, taxes, and interest payments, you subtract $70,000 from $100,000, your remaining revenue after expenses was $30,000. Divide your net income by the total revenue you already calculated. The resulting percentage is your profit margin, which is the percent of your revenue that you keep as income. In our example above, our difference was $30,000. $30,000 ÷ $100,000 = .3 (30%) As a further example, if your business sells paintings, the profit margin calculation tells you on average, when a person pays for a painting, how much of that money you will keep in profit. Part 2 of 2:Making Sense of Your Profit Margin 1. Assess whether your profit margin meets your business needs. If you plan to live solely off income from your business, consider your profit margin and the amount of sales you generally make in a year. You will want to reinvest some of your income into developing your business, so when you take that amount out, is the remaining profit enough to sustain your lifestyle? For example, like above, your business netted $30,000 in cash after $100,000 in sales. If you use $15,000 of the profit to reinvest in your business (and potentially pay off loans), you have $15,000 left over. Compare your profit margin to other similar businesses. Another useful aspect of knowing your profit margin is comparing it to similar businesses to determine where you stand. If you are applying for a loan, the bank will likely tell you what kind of profit margin they expect for your size and/or business type. If you are a larger company with competitors, you can likely research those companies and learn their profit margins to compare them to yours. Say that Company 1 has revenue of $500,000 and total expenses of $230,000. This would give it a profit margin of 54%. Assume that Company 2 has revenue of $1,000,000 and total expenses of $580,000. This means that Company 2's profit margin is 42%. Company 1 has a better profit margin, even though Company 2 makes double of what Company 1 does and has a higher net profit. Compare apples with apples when comparing profit margins. Companies have widely varied profit margins based on their size and industry. It is best to compare two or more companies in the same industry and with similar revenues in order to make the most of the comparison. For example, the airline industry averages around only 3% profit margins, while technology and software companies average in the 20% margin range. When comparing your company, also consider size to ensure your comparison is meaningful. Adjust your profit margin if necessary. You can change your profit margin percentage by making more revenue (such as by increasing the price of your products or selling more of them), or by reducing the expenses associated with your business. Also, even if your profit margin remains the same, if you increase your total revenue and expenses, your net income will increase in dollar value. Consider your business, competition, and risk tolerance as you experiment with raising prices or cutting costs. Generally you should make small changes and work up to larger ones to prevent a dive in business or angering your customers. Remember that there is a cost to increasing your profit margin, and doing so too aggressively can have the reverse effect by tanking your business. Don't confuse profit margins with markup. Markup is the difference between what something costs to produce and how much it is sold for. Was this worth your time? This helps us sort answers on the page. Absolutely not MaryBeth VanderMeulen Business Owner - 3 Businesses Over 30 Years · Author has 19.8K answers and 8.9M answer views · 2y Related What is the definition of successive discounts? What is the effect of successive discounts on profit margin? ‘My Q Book’ shares this definition: - Sometimes merchants announce more than one discount on some of their items to increase the sale of that item. When two or more discounts are allowed one after another on the list price of an item, they are known as successive discounts. ‘Complete Financial Training’ says - Discounting prices is a common strategy employed with the intention to increase sales and improve the bottom line. While it may seem to make sound financial sense, the impact it has may not always be quite so straightforward. A seemingly insignificant discount can often have a dramatic ef ‘My Q Book’ shares this definition: - Sometimes merchants announce more than one discount on some of their items to increase the sale of that item. When two or more discounts are allowed one after another on the list price of an item, they are known as successive discounts. ‘Complete Financial Training’ says - Discounting prices is a common strategy employed with the intention to increase sales and improve the bottom line. While it may seem to make sound financial sense, the impact it has may not always be quite so straightforward. A seemingly insignificant discount can often have a dramatic effect on your overall profits. It’s important to consider how price wars and discounts can impact your margins. The relationship between selling price, margins and volume helps to anticipate the impact of a discounting strategy. To work this out, you firstly need to understand the margin that you are currently making on the product or service that you are considering discounting. The margin is calculated by subtracting the unit costs of sale, i.e. the extra costs incurred each time you sell one more unit of this product/service, from the unit selling price. This will give you the value of the margin that you currently make per unit. Next, you need to work out the percentage margin that you are making. The final step before assessing the impact of the proposed discount is to identify how much volume of this product/service you are currently selling. This volume can then be multiplied by the margin made per unit to calculate the total amount of margin you are making on this product/service. Mark Mehling Marketing strategist, author, problem solver for 40+ Years. · Author has 16.3K answers and 8.8M answer views · 6y Related How do you calculate your profit margin when you offer a service rather than a product? Profit margin is the difference between what you collect in sales versus what it costs you. There are various models. You can compute profit margin by product when pricing. Such as selling widgets. You ad a specific margin to the widget and that establishes the selling price. The best way for a small business is to measure the difference between what you collected versus your costs. And that number then becomes a baseline for comparison to future years or months. As others have already noted, you need to include your salary in this even if you work for yourself. And all your costs, to give you a Profit margin is the difference between what you collect in sales versus what it costs you. There are various models. You can compute profit margin by product when pricing. Such as selling widgets. You ad a specific margin to the widget and that establishes the selling price. The best way for a small business is to measure the difference between what you collected versus your costs. And that number then becomes a baseline for comparison to future years or months. As others have already noted, you need to include your salary in this even if you work for yourself. And all your costs, to give you a realistic figure, must be included. This means a portion of electricity, phone, administrative costs, etc if you work from home. Let’s say you drive for Uber as your service. The allocated (part associated with Uber) cost of car maintenance, depreciation, gas, insurance, phone, your time, car cleaning, etc would all be part of your costs. Then that is compared to what you brought in. Let me know if that is clear enough. Paul Gobat Level 5 teacher, teaching troubled kids · Author has 23.8K answers and 30.3M answer views · 7y Related How are gross margins calculated? Gross margin is the sales price of the item minus the cost expressed in percentage. Let’s say that you sell an item for $150 and it cost you $90. The GROSS margin is 40% (the amount you keep is 40% of the selling price). NET margin is adding all of your operating costs to the item cost- from salary to insurance to rent to utilities- and doing the same calculation. So, if that same item you sell for $150 and costs $90 and you have combined operating costs of $40 you “keep” $20. That is a 7.5% NET margin. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Related questions What is a margin, and how is it calculated? What is the purpose of calculating profit margin in a business? Are there alternative methods for determining this value? If so, why is calculating profit margin still commonly used? What are the best are examples for gross profit margin? What's the best way to calculate markup percentage on an item? How do you decide on acceptable profit margins? What is the difference between "markup", "margin", "profit margin", "gross margin" and "net margin"? What is profit margin and how do you calculate it in your business? Can YOY Gross profit margin (GP%) be reconciled with exchange rate impact, sales mix impact and sales price impact? What is the best way to calculate profit margin, gross margin, and net profit margin? How do I calculate the profit percentage if I want a 30% margin and I want to give a 10% discount to customers? What goes into profit margins? What is the difference between gross margin and profit margin in business? How does one calculate it? Why would margins matter to individuals who run businesses (not just owners)? How do insurance company profit margins impact healthcare quality? What does "profit margin" mean? What is the difference between gross profit margins and standard margins? Is there a difference between gross margin and profit margin? How do they affect each other? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
189153	https://www.edubull.com/Content/eRepository/SubjectContent/Notes/JEE(M)Math5_Conic_Section.pdf	Conic Section Everything should be made as simple as possible, but not simpler...... Einstein, Albert This chapter focusses on parabolic curves, which constitutes one category of various curves obtained by slicing a cone by a plane, called conic sections. A cone (not necessarily right circular) can be out in various ways by a plane, and thus different types of conic sections are obtained. Let us start with the definition of a conic section and then we will see how are they obtained by slicing a right circular cone. 1. Definition of Conic Sections: A conic section or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line. • The fixed point is called the Focus. • The fixed straight line is called the Directrix. • The constant ratio is called the Eccentricity denoted by e. PS e PM = • The line passing through the focus & perpendicular to the directrix is called the Axis. • A point of intersection of a conic with its axis is called a Vertex. If S is (p, q) & directrix is x + my + n = 0 Then PS = 2 2 (x – ) (y – ) α + β & PM = 2 2 \| x my n \| m + + +   PS PM = e ⇒ (2 + m2) [(x – p)2 + (y – q)2] = e2 (x + my + n)2 Which is of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 1.1 Section of right circular cone by different planes A right circular cone is as shown in the figure – 1 A(vertex) Plane Generator Circular base B Axis α β α is angle between generator and axis. β is angle between plane and axis Section of a right circular cone by a plane passing through its vertex is a pair of straight lines. Section of a right circular cone by a plane not passing through vertex is either circle or parabola or ellipse or hyperbola which is shown in table below : Conic Section Type of conic section 3-D view of section of right circular cone with plane Condition of conic in definition of conic condition of conic in ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 Two distinct real lines A(vertex) Plane Generator Circular base Plane passes through vertex A and 0 ≤ β < α P Q e > 1, focus lies on directrix a h g h b f g f c = 0, h2 > ab Two real same lines A (vertex) Plane Generator Circular base Plane passes through vertex A and β = α e = 1, focus lies on directrix a h g h b f g f c = 0, h2 = ab, (either g2 = ac or f2 = bc) Two imaginary lines/point A(vertex) Plane Generator Circular base Plane passes through vertex A and β > α 0 < e < 1, focus lies on directrix a h g h b f g f c = 0, h2 < ab Parabola A(vertex) Plane Generator Circular base Plane does not passes through vertex A and β = α e = 1, focus does not lies on directrix a h g h b f g f c ≠ 0, h2 = ab Conic Section Ellipse A(vertex) Plane Generator Circular base B Axis α β Plane does not passes through vertex A and α < β < 90 0 < e < 1, focus does not lies on directrix a h g h b f g f c ≠ 0, h2 < ab, (either a ≠ b or h ≠ 0) Circle A(vertex) Plane Generator Circular base B Axis α Plane does not passes through vertex A and β = 90 e = 0, focus does not lies on directrix a h g h b f g f c ≠ 0, a = b, h = 0 Hyperbola A(vertex) Generator Axis Plane does not passes through vertex A and 0 ≤ β < α e > 1, focus does not lies on directrix a h g h b f g f c ≠ 0, h2 > ab Note : (i) Pair of real parallel lines is not the part of conic but it is part of general two degree equation. For it a h g h b f g f c = 0, h2 = ab, (either g2 > ac or f2 > bc) ⇒ General two degree equation can represent real curve other than conic section. (ii) For rectangular hyperbola a h g h b f g f c ≠ 0, h2 > ab, a + b = 0 2. Elementary Concepts of Parabola 2.1 Definition and terminology of parabola Z A M L P' L' N X S P Y y = 4ax 2 Conic Section A parabola is the locus of a point, whose distance from a fixed point (focus) is equal to perpendicular distance from a fixed straight line (directrix). Four standard forms of the parabola are y² = 4ax; y² = − 4ax; x² = 4ay; x² = − 4ay For parabola y² = 4ax. (i) Vertex is (0, 0) (ii) focus is (a, 0) (iii) Axis is y = 0 (iv) Directrix is x + a = 0 Focal Distance: The distance of a point on the parabola from the focus. Focal Chord : A chord of the parabola, which passes through the focus. Double Ordinate: A chord of the parabola perpendicular to the axis of the symmetry. Latus Rectum: A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called the Latus Rectum (L.R.). For y² = 4ax. ⇒ Length of the latus rectum = 4a. ⇒ ends of the latus rectum are L(a, 2a) & L’ (a, − 2a). NOTE : (i) Perpendicular distance from focus on directrix = half the latus rectum. (ii) Vertex is middle point of the focus & the point of intersection of directrix & axis. (iii) Two parabolas are said to be equal if they have the same latus rectum. Example # 1: Find the equation of the parabola whose focus is at (– 1, – 2) and the directrix is x – 2y + 3 = 0. Solution : Let P(x, y) be any point on the parabola whose focus is S(– 1, – 2) and the directrix x – 2y + 3 = 0. Draw PM perpendicular to directrix x – 2y + 3 = 0. Then by definition, SP = PM ⇒ SP2 = PM2 ⇒ (x + 1)2 + (y + 2)2 = 2 x 2y 3 1 4   − +   +   ⇒ 5 [(x + 1)2 + (y + 2)2] = (x – 2y + 3)2 ⇒ 5(x2 + y2 + 2x + 4y + 5) = (x2 + 4y2 + 9 – 4xy + 6x – 12y) ⇒ 4x2 + y2 + 4xy + 4x + 32y + 16 = 0 This is the equation of the required parabola. Example # 2 : Find the vertex, axis, focus, directrix, latusrectum of the parabola, also draw their rough sketches. x2 – 2x + 4y + 9 = 0. Solution : The given equation is x2 – 2x + 4y + 9 = 0 ⇒ (x – 1)2 = – 4(y + 2) which of the form X2 = –4bY Vertex - (X, Y) ≡ (0, 0) (x, y) ≡ (1, –2) Axis X = 0 ⇒ x = 1 Focus- (X, Y) = (0, –b) (x, y) ≡ (1, –1 –2) = (1, –3) Directrix - Y = b ⇒ y + 2 = 1 y = –1 Latusrectum - x = 1 (1, –2) x y Conic Section The length of the latusrectum of the given parabola is 4b = 4. Self Practice Problems : (1) Find the equation of the parabola whose focus is the point (0, 0)and whose directrix is the straight line 4x – 3y – 2 = 0. (2) Find the extremities of latus rectum of the parabola y = x2 – 2x + 3. (3) Find the latus rectum & equation of parabola whose vertex is origin & directrix is x + y = 2. (4) Find the equation of the parabola whose focus is (–1, 1) and whose vertex is (1, 2). Also find its axis and latusrectum. Ans. (1) 9x2 + 16y2 + 24xy + 16x – 12y – 4 = 0 (2) 1 9 , 2 4       3 9 , 2 4       (3) 4 2 , x2 + y2 – 2xy + 8x + 8y = 0 (4) (2y – x – 3)2 = – 20 (y + 2x – 4), Axis 2y – x – 3 = 0. LL′ = 4 5 . 2.2 Parametric representation of parabola The simplest & the best form of representing the co−ordinates of a point on the parabola is (at², 2at) i.e. the equations x = at² & y = 2at together represents the parabola y² = 4ax, t being the parameter. Parametric form for : y2 = – 4ax (–at2, 2at) x2 = 4ay (2at , at2) x2 = – 4ay (2at , – at2) Example # 3: Find the parametric equation of the parabola (x + 1)2 = –6 (y + 2) Solution:  4a = – 6 ⇒ a = 3 2 − , y + 2 = at2 x + 1 = 2 at ⇒ x = –1 – 3t, y = –2 – 3 2 t2 Self Practice Problems: (5) Find the parametric equation of the parabola x2 = 4a(y – 1) Ans. x = 2at, y = 1 + at2 . 2.3 Position of a point relative to a parabola: The point (x1 , y1) lies outside, on or inside the parabola y² = 4ax according as the expression 2 1 y − 4ax1 is positive, zero or negative. Inside P(x , y ) 1 1 Outside S1 : 2 1 y – 4ax1 S1 < 0 → Inside S1 > 0 → Outside Example # 4 : Check whether the point (4, 5) lies inside or outside the parabola y2 = 4x. Solution : y2 – 4x = 0  S1 ≡ y1 2 – 4x1 = 25 – 16 = 9 > 0 ∴ (3, 4) lies outside the parabola. Self Practice Problems : (6) Find the set of value's of α for which (α, – 2 – α) lies inside the parabola y2 + 4x = 0. Ans. α ∈ (– 4 – 2 3 , – 4 + 2 3 ) Conic Section 3. Elementary Concepts of Ellipse 3.1 Definition of Ellipse It is locus of a point which moves in such a way that the ratio of its distance from a fixed point called focus and a fixed line called directrix (not passes through fixed point and all points and line lies in same plane) is constant (e = eccentricity), which is less than one. Example # 5 : Find the equation to the ellipse whose focus is the point (– 1, 1), whose directrix is the straight line x – y + 3 = 0 and eccentricity is 1 2 . Solution : Let P ≡ (h, k) be moving point, e = PS PM = 1 2 ⇒ (h + 1)2 + (k – 1)2 = 1 4 2 h k 3 2   − +     ⇒ locus of P(h, k) is 8 {x2 + y2 + 2x – 2y + 2} = (x2 + y2 – 2xy + 6x – 6y + 9) 7x2 + 7y2 + 2xy + 10x – 10 y + 7 = 0. Self Practice Problems : (7) Find the equation to the ellipse whose focus is (0, 0) directrix is x + y – 1 = 0 and e = 1 2 . Ans. 3x2 + 3y2 – 2xy + 2x + 2y – 1 = 0. 3.2 Standard Equation of Ellipse Standard equation of an ellipse referred to its principal axes along the co−ordinate axes is 2 2 2 2 x y a b + = 1, where a > b & b² = a² (1 − e²). Eccentricity: e = 2 2 b 1 a − , (0 < e < 1) Focii : S ≡ (a e, 0) & S′ ≡ (− a e, 0). Equations of Directrices: x = a e & x = −a e . Major Axis: The line segment A′A in which the focii S′ & S lie is of length 2a & is called the major axis (a > b) of the ellipse. Point of intersection of major axis with directrix is called the foot of the directrix (Z). Minor Axis: The y−axis intersects the ellipse in the points B′ ≡ (0, − b) & B ≡ (0, b). The line segment B′B is of length 2b (b < a) is called the minor axis of the ellipse. Principal Axis : The major & minor axes together are called principal axis of the ellipse. Vertices: Point of intersection of ellipse with major axis. A′ ≡ (− a, 0) & A ≡ (a, 0) . Conic Section Focal Chord: A chord which passes through a focus is called a focal chord. Double Ordinate: A chord perpendicular to the major axis is called a double ordinate. Latus Rectum: The focal chord perpendicular to the major axis is called the latus rectum. Length of latus rectum (LL′) = ( ) ( ) 2 2 2 minor axis 2b 2a 1 e a major axis = = − = 2 e (distance from focus to the corresponding directrix) Centre: The point which bisects every chord of the conic drawn through it, is called the centre of the conic. C ≡ (0, 0) the origin is the centre of the ellipse 2 2 2 2 y x a b + = 1. Note : (i) If the equation of the ellipse is given as 2 2 2 2 y x a b + = 1 and nothing is mentioned, then the rule is to assume that a > b. (ii) If b > a is given, then the y−axis will become major axis and x-axis will become the minor axis and all other points and lines will change accordingly. Equation : 2 2 x a + 2 y b 2 = 1 Foci (0, ± be) Directrices : y = ± b e a2 = b2 (1 – e2), a < b. ⇒ e = 2 2 a 1– b Vertices (0, ± b) ; L.R. y = ± be  (L·R.) = 2 2a b , centre : (0, 0) Example # 6: Find the equation to the ellipse whose centre is origin, axes are the axes of co-ordinate and passes through the points (2, 2) and (3, 1). Solution: Let the equation to the ellipse is 2 2 x a + 2 2 y b = 1 Since it passes through the points (2, 2) and (3, 1) ∴ 2 4 a + 2 4 b = 1 .......... (i) and 2 9 a + 2 1 b = 1 .........(ii) from (i) – 4 (ii), we get 2 4 36 a − = 1– 4 ⇒ a2 = 32 3 from (i), we get 2 1 b = 1 4 – 3 32 = 8 3 32 − ⇒ b2 = 32 5 ∴Ellipse is 3x2 + 5y2 = 32 Example # 7 : Find the equation of the ellipse whose focii are (4, 0) and (– 4, 0) and eccentricity is 1 3 Solution: Since both focus lies on x-axis, therefore x-axis is major axis and mid point of focii is origin which is centre and a line perpendicular to major axis and passes through centre is minor axis which is y-axis. Let equation of ellipse is 2 2 x a + 2 2 y b = 1 Conic Section  ae = 4 and e = 1 3 (Given) ∴ a = 12 and and b2 = a2 (1 – e2) ⇒ b2 = 144 1 1 9   −     ⇒ b2 = 16 × 8 ⇒ b = 8 2 Equation of ellipse is 2 x 144 + 2 y 128 = 1 Example # 8 : In the given figure find the eccentricity of the ellipse if SS’ subtends right angle at B. AA BB SS O Solution: here b = ae --- (i) in ellipse b2 = a2 – a2 e2 ----- (ii) from (i) & (ii) a2e2 = a2–a2 e2 2e2 = 1 ⇒ e = 1 2 Example # 9 : From a point Q on the circle x2 + y2 = a2, perpendicular QM are drawn to x-axis, find the locus of point 'P' dividing QM in ratio 2 : 1. Solution : Let Q ≡ (a cosθ, a sinθ) M ≡ (a cosθ, 0) Let P ≡ (h, k) ∴h = a cosθ, k = asin 3 θ ∴ 2 3k a       + 2 h a       = 1 ⇒ Locus of P is 2 2 x a + 2 2 y (a/3) = 1 Example # 10 : Find the equation of axes, directrix, co-ordinate of focii, centre, vertices, length of latus - rectum and eccentricity of an ellipse 16x2 + 25y2 – 96x – 100 y + 156 = 0. Solution : The given ellipse is 2 (x 3) 25 − + 2 (y 2) 16 − = 1. Let x – 3 = X, y – 2 = Y, so equation of ellipse becomes as 2 2 X 5 + 2 2 Y 4 = 1. equation of major axis is Y = 0 ⇒ y = 2. equation of minor axis is X = 0 ⇒ x = 3. centre (X = 0, Y = 0) ⇒ x = 3, y = 2 C ≡ (3, 2) Length of semi-major axis a = 5 Length of major axis 2a = 10 Length of semi-minor axis b = 4 Length of minor axis = 2b = 8. Let 'e' be eccentricity ∴ b2 = a2 (1 – e2) e = 2 2 2 a b a − = 25 16 25 − = 3 5 . Conic Section Length of latus rectum = LL′ = 2 2b a = 2 16 5 × = 32 5 Co-ordinates focii are X = ± ae, Y = 0 ⇒ S ≡ (X = 3, Y = 0) & S′ ≡ (X = –3, Y = 0) ⇒ S ≡ (6, 2) & S′ ≡ (0, 2) Co-ordinate of vertices Extremities of major axis A ≡ (X = a, Y = 0) & A′ ≡ (X = – a, Y = 0) ⇒ A ≡ (x = 8, y = 2) & A′ = (x = – 2, 2) A ≡ (8, 2) & A′ ≡ (– 2, 2) Extremities of minor axis B ≡ (X = 0, Y = b) & B′ ≡ (X = 0, Y = – b) B ≡ (x = 3, y = 6) & B′ ≡ (x = 3, y = – 2) B ≡ (3, 6) & B′ ≡ (3, – 2) Equation of directrix X = ± a e x – 3 = ± 25 3 ⇒ x = 34 3 & x = – 16 3 Self Practice Problems: (8) Find the equation to the ellipse whose axes are of lengths 6 and 2 and their equations are x – 3y + 3 = 0 and 3x + y – 1 = 0 respectively. (9) Find the co-ordinates of the focii of the ellipse 4x2 + 9y2 = 1. (10) A point moves so that the sum of the squares of its distances from two intersecting lines is constant (given that the lines are neither perpendicular nor they make complimentry angle). Prove that its locus is an ellipse. Hint. : Assume the lines to be y = mx and y = – mx. Ans. (8) 3(x – 3y + 3)2 + 2(3x+ y – 1)2 = 180, 21x2 – 6xy + 29y2 + 6x – 58y – 151 = 0. (9) 5 , 0 6   ±       3.3 Auxiliary Circle / Eccentric Angle of Ellipse A circle described on major axis of ellipse as diameter is called the auxiliary circle. Let Q be a point on the auxiliary circle x² + y² = a² such that line through Q perpendicular to the x − axis on the way intersects the ellipse at P, then P & Q are called as the Corresponding Points on the ellipse & the auxiliary circle respectively. ‘θ’ is called the Eccentric Angle of the point P on the ellipse (− π < θ ≤ π). Q ≡ (a cosθ , a sinθ) P ≡ (a cosθ , b sinθ) Note that : (PN) b Semi minor axis (QN) a Semi major axis = =   NOTE : If from each point of a circle perpendiculars are drawn upon a fixed diameter then the locus of the points dividing these perpendiculars in a given ratio is an ellipse of which the given circle is the auxiliary circle. Conic Section Example # 11 : Find the focal distance of a point P(θ) on the ellipse 2 2 x a + 2 2 y b = 1 (a > b) Solution : Let 'e' be the eccentricity of ellipse. ∴ PS = e . PM = e a acos e   − θ     PS = (a – a e cosθ) and PS′ = e. PM′ = e a acos e   θ +     PS′ = a + ae cosθ ∴ focal distance are (a ± ae cosθ) Note : PS + PS′ = 2a PS + PS′ = AA′ Example # 12 : Find the distance from centre of the point P on the ellipse 2 2 x a + 2 2 y b = 1 whose radius makes angle α with y – axis in clockwise direction. Solution : Let P ≡ (a cosθ, b sinθ) ∴ m(op) = b a tanθ = tan(π/2 – α) ⇒ tanθ = a b tan (π/2 – α) OP = 2 2 2 2 a cos b sin θ + θ = 2 2 2 2 a b tan sec + θ θ = 2 2 2 2 a b tan 1 tan + θ + θ = 2 2 2 2 2 2 2 2 a a b tan ( /2 ) b a 1 tan ( /2 ) b + × π −α + π −α ⇒ OP= 2 2 2 2 ab a cos b sin α + α Self Practice Problems : (11) Find the distance from centre of the point P on the ellipse 2 2 x a + 2 2 y b = 1 whose eccentric angle is α (12) Find the eccentric angle of a point on the ellipse 2 x 16 + 2 y 9 = 1whose distance from the centre is 3. (13) Show that the area of triangle inscribed in an ellipse bears a constant ratio to the area of the triangle formed by joining points on the auxiliary circle corresponding to the vertices of the first triangle. Ans. (11) 2 2 2 2 r a cos b sin = α + α (12) ± 2 π 3.4 Parametric Representation of Ellipse The equations x = a cos θ & y = b sin θ together represent the ellipse 2 2 2 2 y x a b + = 1. Where θ is a parameter. Note that if P(θ) ≡ (a cos θ, b sin θ) is on the ellipse then; Q(θ) ≡ (a cos θ, a sin θ) is on the auxiliary circle. The equation to the chord of the ellipse joining two points with eccentric angles α & β is given by x y cos sin cos a 2 b 2 2 α + β α + β α −β + = Conic Section Example # 13 : Write the equation of chord of an ellipse 2 x 25 + 2 y 16 =1 joining two points P 4 π       and Q 5 4 π       . Solution : Equation of chord is x 5 cos 5 4 4 2 π π   +    + y 4 . sin 5 4 4 2 π π   +    = cos 5 4 4 2 π π   −     x 5 . cos 3 4 π       + y 4 . sin 3 4 π       = 0 ⇒ – x 5 + y 4 = 0 ⇒ 4x = 5y Example # 14 : If P(α) and P(β) are extremities of a chord of ellipse which passes through the mid-point of the line segment joining focus & centre then prove that its eccentricity e = 2. cos 2 cos 2 α −β       α + β       Solution : Let the equation of ellipse is 2 2 x a + 2 2 y b = 1 ∴ equation of chord is x a cos 2 α + β       + y b sin 2 α + β       = cos 2 α −β       above chord passes through (ae/2, 0) or (– ae/2, 0) ∴ ± e cos 2 α + β       = 2cos 2 α −β       ∴ e = 2 cos 2 cos 2 α −β       α + β       Ans. Self Practice Problems : (14) Find the locus of the foot of the perpendicular from the centre of the ellipse 2 2 x a + 2 2 y b = 1 on the chord joining two points whose eccentric angles differ by 2 π . Ans. (14) 2(x2 + y2)2 = a2 x2 + b2 y2. 3.5 Position of a Point w.r.t. an Ellipse : The point P(x1, y1) lies outside, inside or on the ellipse according as S1 > 0, S1 < 0 or S1 = 0 where S1 = 2 2 1 1 2 2 x y 1 a b + − . Example # 15 : Check whether the point P(1, –1) lies inside or outside of the ellipse 2 x 25 + 2 y 16 = 1. Solution : S1 ≡ 1 25 + 1 16 – 1 < 0 ∴ Point P ≡ (1, –1) lies inside the ellipse. Example # 16 : Find the set of value(s) of 'α' for which the point P(2α, – 3α) lies inside the ellipse 2 x 16 + 2 y 9 = 1. Solution : If P(2α, – 3α) lies inside the ellipse ∴ S1 < 0 ⇒ 2 4 α + 2 1 α – 1 < 0 ⇒ – 2 5 < α < 2 5 ∴ α∈ 2 2 , 5 5   −     . Conic Section 4. Elementary Concepts of Hyperbola Hyperbolic curves are of special importance in the field of science and technology especially astronomy and space studies. In this chapter we are going to study the characteristics of such curves. 4.1 Definition of Hyperbola A hyperbola is defined as the locus of a point moving in a plane in such a way that the ratio of its distance from a fixed point to that from a fixed line (the point does not lie on the line) is a fixed constant greater than 1. PS PM = e > 1 , e – eccentricity 4.2 Standard equation of Hyperbola Standard equation of hyperbola is − = 2 2 x y 1 2 2 a b , where b2 = a2 (e2 − 1). • Eccentricity (e) : e2 = 1 + 2 2 b a • Foci : S ≡ (ae, 0) & S′ ≡ (− ae, 0). • Equations of directrices : x = a e & x = −a e . • Transverse axis : The line segment A′A of length 2a in which the foci S′ & S both lie is called the transverse axis of the hyperbola. • Conjugate axis : The line segment B′B of length 2b between the two points B′ ≡ (0, − b) & B ≡ (0, b) is called as the conjugate axis of the hyperbola. • Principal axes : The transverse & conjugate axis together are called principal axes of the hyperbola. • Vertices : A ≡ (a, 0) & A′ ≡ (− a, 0) • Focal chord : A chord which passes through a focus is called a focal chord. • Double ordinate : A chord perpendicular to the transverse axis is called a double ordinate. • Latus rectum : Focal chord perpendicular to the transverse axis is called latus rectum. Its length () is given by  = ( ) 2 2 C.A. 2b a T.A. = = 2a (e2 − 1). Conic Section Note : (i) Length of latus rectum = 2 e × (distance of focus from corresponding directrix) (ii) End points of latus rectum are L ≡ 2 b ae, a         , L′ ≡ 2 b ae, a   −       , M ≡ 2 b ae, a   −       , M′ ≡ 2 b ae, a   − −       • Centre: The point which bisects every chord of the conic, drawn through it, is called the centre of the conic. C ≡ (0,0) the origin is the centre of the hyperbola 2 2 2 2 x y 1 a b − = . General note : Since the fundamental equation to hyperbola only differs from that to ellipse in having −b2 instead of b2 it will be found that many propositions for hyperbola are derived from those for ellipse by simply changing the sign of b2. Example #17: Find the equation of the hyperbola whose directrix is x + 2y = 1, focus (2,1) and eccentricity 3 . Solution: Let P(x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix. Then by definition SP = e PM ⇒ (SP)2 = e2 (PM)2 ⇒ (x – 2)2 + (y – 1)2 = 3 2 x 2y 1 4 1 + −     +   ⇒ 2x2 – 7y2 – 12xy – 14x + 2y + 22 = 0 Which is the required hyperbola. Example # 18: Find the eccentricity of the hyperbola whose latus rectum is half of its transverse axis. Solution: Let the equation of hyperbola be 2 2 x a – 2 2 y b = 1. Then transverse axis = 2a and latus–rectum = 2 2b a . According to question 2 2b a = 1 2 (2a) ⇒ 2b2 = a2 ( b2 = a2 (e2 – 1)) ⇒ 2a2 (e2 – 1) = a2 ⇒ 2e2 – 2 = 1 ⇒ e2 = 3 2 ∴ e = 3 2 Hence the required eccentricity is 3 2 . 4.3 Conjugate hyperbola : Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & the transverse axes of the other are called conjugate hyperbolas of each other. eg. 2 2 2 2 x y 1 a b − = & 2 2 2 2 x y 1 a b − + = are conjugate hyperbolas of each other. Equation : 2 2 y b – 2 2 x a = 1 a2 = b2 (e2 – 1) ⇒ e = 2 2 a 1 b + Vertices(0, ± b) ;  (L.R.) = 2 2a b Conic Section Note : (a) If e1 & e2 are the eccentrcities of the hyperbola & its conjugate then e1 −2 + e2 −2 = 1. (b) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. (c) Two hyperbolas are said to be similar if they have the same eccentricity. (d) Two similar hyperbolas are said to be equal if they have same latus rectum. (e) If a hyperbola is equilateral then the conjugate hyperbola is also equilateral. Example # 19 : Find the lengths of transverse axis and conjugate axis, eccentricity, the co-ordinates of foci, vertices, length of the latus-rectum and equations of the directrices of the following hyperbola 16x2 – 9y2 = – 144. Solution : The equation 16x2 – 9y2 = –144 can be written as 2 x 9 – 2 y 16 = – 1 This is of the form 2 2 x a – 2 2 y b = – 1 ∴ a2 = 9, b2 = 16 ⇒ a = 3, b = 4 Length of transverse axis : The length of transverse axis = 2b = 8 Length of conjugate axis : The length of conjugate axis = 2a = 6 Eccentricity : e = 2 2 a 1 b   +       = 9 1 16   +     = 5 4 Foci : The co-ordinates of the foci are (0, + be) i.e., (0, + 5) Vertices : The co–ordinates of the vertices are (0, + b) i.e., (0, + 4) Length of latus–rectum : The length of latus–rectum = 2 2a b = 2 2(3) 4 = 9 2 Equation of directrices : The equation of directrices are y = + b e ⇒ y = + 4 (5/ 4) ⇒ y = + 16 5 Self Practice Problems : (15) Find the equation of the hyperbola whose foci are (6, 4) and (– 4, 4) and eccentricity is 2. (16) Obtain the equation of a hyperbola with coordinates axes as principal axes given that the distances of one of its vertices from the foci are 9 and 1 units. (17) The foci of a hyperbola coincide with the foci of the ellipse 2 x 25 + 2 y 9 = 1. Find the equation of the hyperbola if its eccentricity is 2. Ans. (15) 12x2 – 4y2 – 24x + 32y – 127 = 0 (16) 2 x 16 – 2 y 9 = 1, 2 y 16 – 2 x 9 = 1 (17) 3x2 – y2 – 12 = 0. 4.4 Auxiliary Circle of Hyperbola A circle drawn with centre C and transverse axis as a diameter is called the auxiliary circle of the hyperbola. Equation of the auxiliary circle is x2 + y2 = a2. Note from the following figure that P & Q are called the "corresponding points" of the hyperbola & the auxiliary circle. Conic Section 4.5 Parametric representation of Hyperbola The equations x = a sec θ & y = b tan θ together represent the hyperbola 2 2 2 2 x y 1 a b − = where θ is a parameter. Note that if P(θ) ≡ (a sec θ, b tan θ) is on the hyperbola then, Q(θ) ≡ (a cos θ, a sin θ) is on the auxiliary circle. The equation to the chord of the hyperbola joining the two points P(α) & Q(β) is given by . x y cos sin cos a 2 b 2 2 α −β α + β α + β − = 4.6 Position of a point 'P' w.r.t. a hyperbola : The quantity S1 ≡ 2 2 1 1 2 2 x y 1 a b − − is positive, zero or negative according as the point (x1, y1) lies inside, on or outside the curve. Example # 20 : Find the position of the point (5, – 4) relative to the hyperbola 9x2 – y2 = 1. Solution : Since 9 (5)2 – (–4)2 – 1 = 225 – 16 – 1 = 208 > 0, So the point (5,–4) lies inside the hyperbola 9x2 – y2 = 1.. 5. Rectangular hyperbola (equilateral hyperbola) : The particular kind of hyperbola in which the lengths of the transverse & conjugate axis are equal is called an Equilateral Hyperbola. Note that the eccentricity of the rectangular hyperbola is. Since a = b equation becomes x2 – y2 = a2 whose asymptotes are y = ± x. e = 2 2 b 1 a + = 1 1 + = 2 Rotation of this system through an angle of 45° in clockwise direction gives another form to the equation of rectangular hyperbola. which is xy = c2 where c2 = 2 a 2 . It is referred to its asymptotes as axes of co−ordinates. Vertices : (c, c) & (− c, − c); Foci : ( ) 2 c , 2 c & ( ) 2 c , 2 c − − , Directrices : x + y = ± 2 c Latus Rectum (l ) :  = 2 2 c = T.A. = C.A. Parametric equation x = ct, y = c/t, t ∈ R – {0} Example # 21 : A triangle has its vertices on a rectangular hyperbola. Prove that the orthocentre of the triangle also lies on the same hyperbola. Conic Section Solution : Let "t1", "t2" and "t3" are the vertices of the triangle ABC, described on the rectangular hyperbola xy = c2. ∴ Co–ordinates of A,B and C are 1 1 c ct , t       , 2 2 c ct , t       and 3 3 c ct , t       respectively Now slope of BC is 3 2 2 3 2 3 c(t t ) c(t t )t t − − = – 2 3 1 t t ∴ Slope of AD is t2t3 Equation of Altitude AD is y – 1 c t = t2t3(x – ct1) or t1y – c = x t1t2t3 – ct1 2t2t3 .....(1) Similarly equation of altitude BE is t2y – c = x t1t2t3 – ct1t2 2t3 ......(2) Solving (1) and (2), we get the orthocentre 1 2 3 1 2 3 c , ct t t t t t   − −     Which lies on xy = c2. 6. Line & a parabola : The line y = mx + c meets the parabola y² = 4ax in two points real, coincident or imaginary according as a > cm, a = cm, a < cm respectively. ⇒ condition of tangency is, c = a/m. Tangent Secant A B Length of the chord intercepted by the parabola on the line y = m x + c is : 2 4 m       2 a(1 m )(a mc) + − NOTE : 1. The equation of a chord joining t1 & t2 is 2x − (t1 + t2) y + 2 at1 t2 = 0. 2. If t1 & t2 are the ends of a focal chord of the parabola y² = 4ax then t1t2 = −1.Hence the co−ordinates at the extremities of a focal chord can be taken as (at², 2at) & 2 a 2a , t t   −     A S (focus) B Focal chord 3. Length of the focal chord making an angle α with the x− axis is 4acosec² α. Example # 22 : Discuss the position of line y = x + 3 with respect to parabola y2 = 4(x + 2). Solution : Solving we get (x + 3)2 = 4(x + 2) ⇒ (x – 1)2 = 0 so y = x + 3 is tangent to the parabola. Example # 23 : Prove that focal distance of a point P(at2, 2at) on parabola y2 = 4ax (a > 0) is a(1 + t2). Solution : Conic Section  PS = PM = a + at2 PS = a (1 + t2). Example # 24 : If t1, t2 are end points of a focal chord then show that t1 t2 = –1. Solution : Let parabola is y2 = 4ax P (at , 2at ) 1 1 2 S(a, 0) Q (at , 2at ) 2 2 2 since P, S & Q are collinear ∴ mPQ = mPS ⇒ 1 2 2 t t + = 1 2 1 2t t 1 − ⇒ t1 2 – 1 = t1 2 + t1t2 ⇒ t1t2 = – 1 Example # 25 : If the endpoint t1, t2 of a chord satisfy the relation t1 t2 = –3, then prove that the chord of y2 = 4x always passes through a fixed point. Find the point ? Solution : Equation of chord joining (at1 2, 2at1) and (at2 2, 2at2) is y – 2at1 = 1 2 2 t t + (x – at1 2) ⇒ (t1 + t2) y – 2at1 2 – 2at1t2 = 2x – 2at1 2 y = 1 2 2 t t + (x – 3) ( t1t2 = –3) ∴ This line passes through a fixed point (3, 0). Self Practice Problems : (18) If the line y = 3x + λ intersect the parabola y2 = 4x at two distinct point's then set of value's of 'λ' is (19) Find the midpoint of the chord x + y = 2 of the parabola y2 = 4x. (20) If one end of focal chord of parabola y2 = 16x is (16, 16) then coordinate of other end is. (21) If PSQ is focal chord of parabola y2 = 4ax (a > 0), where S is focus then prove that 1 PS + 1 SQ = 1 a (22) Find the length of focal chord whose one end point is (ap2, 2ap) Ans. (18) (– ∞, 1/3) (19) (4, – 2) (20) (1, – 4) (22) 2 1 a p p   +     6.1 Tangents to the parabola y² = 4ax : Equation of tangent at a point on the parabola can be obtained by replacement method or using derivatives. In replacement method, following changes are made to the second degree equation to obtain T. x2 → x x1, y2 → y y1, 2xy → xy1 + x1y, 2x → x + x1, 2y → y + y1 So, it follows that the targents are : (i) y y1 = 2 a (x + x1) at the point (x1, y1) ; (ii) y = mx + a m (m ≠ 0) at 2 a 2a , m m       (iii) t y = x + a t² at (at², 2at). (iv) Point of intersection of the tangents at the point t1 & t2 is { at1 t2 , a(t1 + t2) }. Example # 26 : Prove that the straight line y = mx + c touches the parabola y2 = 4a (x + a) if c = ma + a m Solution: Equation of tangent of slope ‘m’ to the parabola y2 = 4a(x + a) is y = m(x + a) + a m ⇒ y = mx + a 1 m m   +     But the given tangent is y = mx + c ∴ c = am + a m Example # 27 : A tangent to the parabola y2 = 8x makes an angle of 45° with the straight line y = 3x + 5. Find its equation and its point of contact. Solution : Slope of required tangent’s are m = 3 1 1 3 ±  ⇒ m1 = – 2, m2 = 1 2 Conic Section  Equation of tangent of slope m to the parabola y2 = 4ax is y = mx + a m . ∴ tangent’s y = – 2x – 1 at 1, 2 2   −     ⇒ y = 1 2 x + 4 at (8, 8) Example # 28 : Find the equation to the tangents to the parabola y2 = 9x which goes through the point (4, 10). Solution : Equation of tangent to parabola y2 = 9x is y = mx + 9 4m Since it passes through (4, 10) ∴ 10 = 4m + 9 4m ⇒ 16 m2 – 40 m + 9 = 0 m = 1 4 , 9 4 ∴ equation of tangent’s are y = x 4 + 9 & y = 9 4 x + 1. Example # 29 : Find the equations to the common tangents of the parabolas (y – 1)2 = 4ax and x2 = 4b(y – 1). Solution : Equation of tangent to (y – 1)2 = 4ax is (y – 1) = mx + a m ........(i) Equation of tangent to x2 = 4b(y – 1) is x = m1(y – 1) + 1 b m ⇒ (y – 1) = 1 1 m x – 2 1 b (m ) ........(ii) for common tangent, (i) & (ii) must represent same line. ∴ 1 1 m = m & a m = – 2 1 b m ⇒ a m = – bm2 ⇒ m = 1/ 3 a b   −     ∴equation of common tangent is y = 1/ 3 a b   −     x + a 1/ 3 a b   −     + 1. Self Practice Problems: (23) Find equation tangent to parabola y2 = 4x whose intercept on y–axis is 2. (24) Prove that perpendicular drawn from focus upon any tangent of a parabola lies on the tangent at the vertex. (25) Prove that image of focus in any tangent to parabola lies on its directrix. (26) Prove that the area of triangle formed by three tangents to the parabola y2 = 4ax is half the area of triangle formed by their points of contacts.. Ans. (23) x y 2 2 = + 7. Line and an Ellipse : The line y = mx + c meets the ellipse 2 2 x a + 2 2 y b = 1 in two points real, coincident or imaginary according as c² < a²m² + b², c2 = a²m² + b² or c2 > a²m² + b² Hence y = mx + c is tangent to the ellipse 2 2 2 2 y x a b + = 1 if c² = a²m² + b². NOTE: The equation to the chord of the ellipse 2 2 x a + 2 2 y b = 1 joining two points with eccentric angles α & β is given by x y cos sin cos a 2 b 2 2 α + β α + β α −β + = Example # 30 : Find the set of value(s) of 'λ' for which the line x + y + λ = 0 intersect the ellipse 2 x 16 + 2 y 9 = 1 at two distinct points. Conic Section Solution : Solving given line with ellipse, we get ( ) 2 2 x x 16 9 + λ + = 1 25x2 + 39λ x + 16λ2 – 144 = 0 Since, line intersect the parabola at two distinct points, ∴ roots of above equation are real & distinct ∴ D > 0 ∴ (32λ)2 – 4.25(16λ2 – 144) > 0 ⇒ λ∈ (–5, 5) Self Practice Problems : (27) Find the value of 'λ' for which 2x – y + 109 λ = 0 touches the ellipse 2 x 25 + 2 y 9 = 1 Ans. (27) λ = ± 1 7.1 Tangents to ellipse 2 2 x a + 2 2 y b = 1 (a) Slope form: y = mx ± 2 2 2 a m b + is tangent to the ellipse 2 2 x a + 2 2 y b =1 for all values of m. (b) Point form : 1 1 2 2 x x y y 1 a b + = is tangent to the ellipse 2 2 x a + 2 2 y b = 1 at (x1, y1). (c) Parametric form: xcos ysin 1 a b θ θ + = is tangent to the ellipse 2 2 x a + 2 2 y b = 1 at the point (a cos θ, b sin θ). Note : (i) There are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.These tangents touches the ellipse at extremities of a diameter. (ii) Point of intersection of the tangents at the point α & β is, 2 2 2 2 cos sin a , b cos cos α + β α + β α −β α −β         (iii) The eccentric angles of the points of contact of two parallel tangents differ by π. Example # 31 : Find the equations of the tangents to the ellipse 3x2 + 4y2 = 12 which are parallel to the line x – 2y + 7 = 0 Solution: Slope of tangent = m = 1 2 Given ellipse is 2 x 4 + 2 y 3 = 1 Equation of tangent whose slope is 'm' is y = mx ± 2 4m 3 +  m = 1 2 ∴ y = 1 2 x ± 1 3 + ⇒ 2y = x ± 4 Example # 32 : A tangent to the ellipse 9x2 + 16y2 – 144 = 0 touches at the point P on it in the first quadrant and meets the co-ordinate axes in A and B respectively. If P divides AB in the ratio 3 : 1, find the equation of the tangent. Solution: The given ellipse is 2 2 2 2 x y 4 3 + = 1 ⇒ a = 4, b = 3 Let P ≡ ( a cosθ, b sinθ) ∴ equation of tangent is x a cosθ + y b sinθ = 1 A ≡ (a secθ, 0) B ≡ (0, b cosecθ)  P divide AB internally in the ratio 3 : 1 ∴ a cosθ = asec 4 θ ⇒ cos2θ = 1 4 ⇒ cosθ = 1 2 Conic Section and b sin θ = 3bcosec 4 θ ⇒ sinθ = 3 2 ∴ tangent is x 2a + 3y 2b = 1 ⇒ bx + 3 ay = 2ab ⇒ 3x + 4 3 y = 24 Example # 33 : Prove that the locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by 3 π is an ellipse having the same eccentricity. Solution : Let P (h, k) be the point of intersection of tangents at A(θ) and B(β) to the ellipse. ∴ h = acos 2 cos 2 θ + β       θ −β       & k = bsin 2 cos 2 θ + β       θ −β       ⇒ 2 h a       + 2 k b       = sec2 2 θ −β       but given that θ – β = 3 π ∴ locus is 2 2 2 x a sec 6 π       + 2 2 2 y b sec 6 π       = 1 which is ellipse having same eccentricity. Conic Section Example # 34 : If the locus of foot of perpendicular drawn from centre to any tangent to the ellipse 3x2 + 4y2 = 12 is (x2 + y2)2 = ax2 + by2, then find a + b. Solution : Let P(h, k) be the foot of perpendicular to a tangent y = mx + 2 4m 3 + .......(i) from centre ∴ k h . m = – 1 ⇒ m = – h k .......(ii)  P(h, k) lies on tangent ∴ k = mh + 2 4m 3 + .......(iii) from equation (ii) & (iii), we get 2 2 h k k   +       = 2 2 4h k + 3 ⇒ locus is (x2 + y2)2 = 4x2 + 3y2 Self Practice Problems : (28) Show that the locus of the point of intersection of the tangents at the extremities of any focal chord of an ellipse is the directrix corresponding to the focus. (29) Show that the locus of the foot of the perpendicular on a varying tangent to an ellipse from either of its foci is a concentric circle. (30) Prove that the portion of the tangent to an ellipse intercepted between the ellipse and the directrix subtends a right angle at the corresponding focus. (31) Find the area of parallelogram formed by tangents at the extremities of latera recta of the ellipse 2 2 2 2 x y 1 a b + = . (32) If y1 is ordinate of a point P on the ellipse then show that the angle between its focal radius and tangent at it, is tan–1 2 1 b aey         . (33) Find the eccentric angle of the point P on the ellipse 2 2 x a + 2 2 y b =1 tangent at which, is equally inclined to the axes. Ans. (31) 3 2 2 2a a b − (33) θ = ± tan–1 b a       , π – tan–1 b a       , – π + tan–1 b a       8. Line and a hyperbola : The straight line y = mx + c is a secant, a tangent or passes outside the hyperbola 2 2 2 2 x y 1 a b − = according as : c2 > a2 m2 − b2 or c2 = a2 m2 − b2 or c2 < a2 m2 − b2, respectively. NOTE: The equation to the chord of the hyperbola 2 2 2 2 x y 1 a b − = joining the two points P(α) & Q(β) is given by x y cos sin cos a 2 b 2 2 α −β α + β α + β − = Conic Section 8.1 Tangents to hyperbola 2 2 2 2 x y -= 1 a b : (i) Slope form : y = m x ± − 2 2 2 a m b can be taken as the tangent to the hyperbola 2 2 2 2 x y 1 a b − = , having slope 'm'. (ii) Point form: Equation of tangent to the hyperbola 2 2 2 2 x y 1 a b − = at the point (x1, y1) is 1 1 2 2 xx yy 1 a b − = (iii) Parametric form: Equation of the tangent to the hyperbola 2 2 2 2 x y 1 a b − = at the point. (a sec θ, b tan θ) is xsec y tan 1 a b θ θ − = Note : (i) Point of intersection of the tangents at P(θ1) & Q(θ2) is 1 2 1 2 1 2 cos 2 a , btan 2 cos 2 θ −θ     θ + θ       θ + θ         (ii) If \|θ1 + θ2\| = π, then tangents at these points (θ1 & θ2) are parallel. (iii) There are two parallel tangents having the same slope m. These tangent touches the hyperbola at the extremities of a diameter. Example # 35 : Find c, if x + y = c touch the hyperbola 2 x 4 – y2 = 1. Solution: Solving line and hyperbola we get x2 – 4 (c–x)2 = 4 3x2 + 8cx + 4c2 + 4 = 0 D = 0 64c2 – 4.3.4 (c2–1) = 0 c2 – 3 = 0 c = ± 3 Example # 36 : Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the line 3 x + y + 5 = 0 Solution : y = mx ± 2 36m – 9 , where m = 1 3 ∴ equation of tangents are y = x 3 ± 3 ⇒ 3 y = x ± 3 Example # 37 : Find the point of contact if 3x – 7 y – 9 = 0 is tangent to 2 2 x y 1 16 9 − = . Solution : Let the point of contact is (x1, y1). The equation of tangent is 1 1 xx yy 16 9 − – 1 = 0...........(i) The given equation of tangent is 3x – 7 y – 9 = 0...........(ii) From Equ (i) & (ii) 1 1 x y 1 16 3 9 9 7 = = × ⇒ (x1, y1) = 16 , 7 3       9. Line and Rectangular hyperbola : Equation of a chord joining the points P (t1) & Q(t2) is x + t1 t2 y = c (t1 + t2). Equation of the tangent at P (x1, y1) is 1 1 x y x y + = 2 & at P (t) is x t + t y = 2 c. Conic Section Example # 38 : A, B, C are three points on the rectangular hyperbola xy = c2, find (i) The area of the triangle ABC (ii) The area of the triangle formed by the tangents at A, B and C. Solution : Let co–ordinates of A,B and C on the hyperbola xy = c2 are 1 1 c ct , t       , 2 2 c ct , t       , and 3 3 c ct , t       respectively. (i) ∴ Area of triangle ABC = 1 2 [ 1 1 2 2 c ct t c ct t + 2 2 3 3 c ct t c ct t + 3 3 1 1 c ct t c ct t ] = 2 c 2 3 3 1 2 2 1 2 1 3 2 1 3 t t t t t t t t t t t t − + − + − = 2 1 2 3 c 2t t t 2 2 2 2 2 2 1 3 2 3 1 2 3 1 2 3 1 2 t t t t t t t t t t t t − + − + − = 2 1 2 3 c 2t t t \| (t1 – t2) (t2 – t3) (t3 – t1) \| (ii) Equations of tangents at A,B,C are x + yt1 2 – 2ct1 = 0 x + yt2 2 – 2ct2 = 0 and x + yt3 2 – 2ct3 = 0 ∴ Required Area = 1 2 3 1 2 \| C C C \| 2 2 1 1 2 2 2 2 3 3 1 t 2ct 1 t 2ct 1 t 2ct − − − .........(1) where C1 = 2 2 2 3 1 t 1 t , C2 = – 2 1 2 3 1 t 1 t and C3 = 2 1 2 2 1 t 1 t ∴ C1 = t3 2 – t2 2, C2 = t1 2 – t3 2 and C3 = t2 2 – t1 2 From (1) 2 2 2 2 2 2 3 2 1 3 2 1 1 2 (t t ) (t t )(t t ) − − − = 4c2.(t1 – t2)2 (t2 – t3)2 (t3 – t1)2 = 2c2 1 2 2 3 3 1 1 2 2 3 3 1) (t t )(t t )(t t ) (t t ) (t t ) (t t − − − + + + ∴ Required area is, 2c2 1 2 2 3 3 1 1 2 2 3 3 1 t t ) (t t )(t t ) (t t ) (t t ) (t t ) ( − − − + + + Example # 39 : Prove that the perpendicular focal chords of a rectangular hyperbola are equal. Solution : Let rectangular hyperbola is x2 – y2 = a2 Let equations of PQ and DE are y = mx + c ......(1) and y = m1x + c1 ......(2) respectively. Be any two focal chords of any rectangular hyperbola x2 – y2 = a2 through its focus. We have to prove PQ = DE. Since PQ ⊥ DE. ∴ mm1 = –1 ......(3) Also PQ passes through S (a 2 , 0) then from (1), 0 = ma 2 + c or c2 = 2a2m2 ......(4) Let (x1,y1) and (x2,y2) be the co–ordinates of P and Q then (PQ)2 = (x1 – x2)2 + (y1 – y2)2 ......(5) Conic Section Since (x1,y1) and (x2,y2) lie on (1) ∴ y1 = mx1 + c and y2 = mx2 + c ∴ (y1 – y2) = m (x1 – x2) .......(6) From (5) and (6) (PQ)2 = (x1 – x2)2 (1 + m2) .......(7) Now solving y = mx + c and x2 – y2 = a2 then x2 – (mx + c)2 = a2 or (m2 – 1) x2 + 2mcx + (a2 + c2) = 0 ∴ x1 + x2 = – 2 2mc m 1 − and x1x2 = = 2 2 2 a c m 1 + − ⇒ (x1 – x2)2 = (x1 + x2)2 – 4x1x2 = 2 2 2 2 4m c (m 1) − – 2 2 2 4(a c ) (m 1) + − = 2 2 2 2 2 2 4{a c a m } (m 1) + − − = 2 2 2 2 4a (m 1) (m 1) + − { c2 = 2a2m2} From (7), (PQ)2 = 4a2 2 2 2 m 1 m 1   +     −   Similarly, (DE)2 = 4a2 2 2 1 2 1 m 1 m 1   +     −   = 4a2 2 2 2 1 1 m 1 1 m       − +             − −         = 4a2 2 2 2 m 1 m 1   +     −   = (PQ)2 ( mm1 = – 1) Thus (PQ)2 = (DE)2 ⇒ PQ = DE. Hence perpendicular focal chords of a rectangular hyperbola are equal. Self Practice Problems : (34) Show that the line x cos α + y sin α = p touches the hyperbola 2 2 x a – 2 2 y b = 1 if a2 cos2 α – b2 sin2 α = p2. (35) For what value of λ does the line y = 2x + λ touches the hyperbola 16x2 – 9y2 = 144 ? (36) Find the equation of the tangent to the hyperbola x2 – y2 = 1which is parallel to the line 4y = 5x + 7. Ans. (35) λ = ± 2 5 (36) 4y = 5x ± 3 10 Pair of tangents : The equation to the pair of tangents which can be drawn from any point (x1, y1) to the curve S = 0 is SS1 = T² Curve(S=0) T for point (x1, y1) & S = 0 S1 for point (x1, y1) & S = 0 Combined equation of tangents from external point (x1, y1) to S=0 Parabola (y2 – 4ax = 0) T ≡ y y1 − 2a(x + x1) S1 = y1² − 4ax1 SS1 = T² Ellipse 2 2 2 2 x y 1 0 a b   + − =       T ≡ 1 2 xx a + 1 2 yy b – 1 S1 = 2 2 1 1 2 2 x y a b + – 1 SS1 = T² Conic Section Hyperbola 2 2 2 2 x y 1 0 a b   − − =       T ≡ 1 2 xx a – 1 2 yy b – 1 S1 = 2 2 1 1 2 2 x y a b − – 1 SS1 = T² Example # 40 : Write the equation of pair of tangents to the parabola y2 = 4x drawn from a point P(–1, 2) Solution : We know the equation of pair of tangents are given by SS1 = T² ∴ (y2 – 4x) (4 + 4) = (2y – 2 (x – 1))2 ⇒ 8y2 – 32x = 4y2 + 4x2 + 4 – 8xy + 8y – 8x ⇒ y2 – x2 + 2xy – 6x – 2y = 1 Example # 41 : Find the locus of the point P from which tangents are drawn to parabola y2 = 4ax having slopes m1, m2 such that (i) \|m1 – m2\| = 2 (ii) θ1 + θ2 = π/3 Solution : Equation of tangent to y2 = 4ax, is y = mx + a m Let it passes through P(h, k) ∴ m2h – mk + a = 0 (i) m1 + m2 = k h and m1 . m2= a h ⇒ \|m1 – m2\| = 2 ⇒ (m1 + m2)2 – 4 m1m2 = 4 2 2 k h – 4 a h = 4 ⇒ 4ax = 4x2 (ii) tan π/3 = 1 2 1 2 m m 1 m m + − = k /h 1 a/h − ⇒ y = (x – a) 3 Example # 42 : How many real tangents can be drawn from the point (–2, –5) to the ellipse 2 x 4 + 2 y 25 = 1. Find the equation of these tangents & angle between them. Solution : x = –2 y = –5 (–2, – 5) By direct observation x = –2, y = –5 are tangents. Example # 43 : Find the locus of point of intersection of perpendicular tangents to the ellipse 2 2 2 2 x y a b + = 1 Solution : Let P(h, k) be the point of intersection of two perpendicular tangents equation of pair of tangents is SS1 = T2 ⇒ 2 2 2 2 x y 1 a b   + −       2 2 2 2 h k 1 a b   + −       = 2 2 2 hx ky 1 a b   + −     ⇒ 2 2 x a 2 2 k 1 b   −       + 2 2 y b 2 2 h 1 a   −       + ........ = 0 .........(i) Since equation (i) represents two perpendicular lines ∴ 2 1 a 2 2 k 1 b   −       + 2 1 b 2 2 h 1 a   −       = 0 ⇒ k2 – b2 + h2 – a2 = 0 ⇒ locus is x2 + y2 = a2 + b2 Conic Section Example # 44 : How many real tangents can be drawn from the point (2, 1) to the hyperbola 2 x 16 – 2 y 9 =1. Find the equation of these tangents. Solution : Given point P ≡ (2, 1) Hyperbola S ≡ 2 x 16 – 2 y 9 – 1 = 0  S1 ≡ 4 16 – 1 9 – 1 = 31 36 − < 0 ⇒ Point P ≡ (2, 1) lies outside the hyperbola. ∴ Two tangents can be drawn from the point P(2, 1). Equation of pair of tangents is SS1 = T2 ⇒ 2 2 x y 1 16 9   − −       1 1 1 4 9   − −     = 2 2x y 1 16 9   − −     ⇒ 144 (9x2 – 16y2 – 144) + (9x – 9y – 72)2 = 0 Example # 45 : Find the locus of point of intersection of perpendicular tangents to the hyperbola 2 2 2 2 x y a b − = 1 Solution : Let P(h, k) be the point of intersection of two perpendicular tangents. Equation of pair of tangents is SS1 = T2 ⇒ 2 2 2 2 x y 1 a b   − −       2 2 2 2 h k 1 a b   − −       = 2 2 2 hx ky 1 a b   − −     ⇒ 2 2 x a 2 2 k 1 b   − −       – 2 2 y b 2 2 h 1 a   −       + ........ = 0 .........(i) Since equation (i) represents two perpendicular lines ∴ 2 1 a 2 2 k 1 b   − −       – 2 1 b 2 2 h 1 a   −       = 0 ⇒ – k2 – b2 – h2 + a2 = 0 ⇒ locus is x2 + y2 = a2 – b2 Self Practice Problems : (37) If two tangents to the parabola y2 = 4ax from a point P make angles θ1 and θ2 with the axis of the parabola, then find the locus of P in each of the following cases. (i) tan2θ1 + tan2θ2 = λ (a constant) (ii) cos θ1 cos θ2 = λ (a constant) (38) Find the locus of point of intersection of the tangents drawn at the extremities of a focal chord of the ellipse 2 2 x a + 2 2 y b = 1. Ans. (37) (i) y2 – 2ax = λx2 , (ii) x2 = λ2 {(x – a)2 + y2} (38) x = ± a e 11. Director circle: Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle. Curve(S=0) Locus of Director Circle of (S=0) Figure Parabola (y2 – 4ax = 0) x + a = 0 x=0 Director Circle (x=–a) y2=4ax y=0 Conic Section Ellipse 2 2 2 2 x y 1 0 a b   + − =       x2 + y2 = a2 + b2 Director circle (x2+y2 = a2+b2) 2 2 2 2 x y 1 a b + = y= 0 x= 0 Hyperbola 2 2 2 2 x y 1 0 a b   − − =       x2 + y2 = a2 – b2 2 2 2 2 x y 1 a b − = 2 2 2 2 Director Circle (x y a b ) + = − y=0 x=0 Note: For hyperbola, if b2 < a2 , then the director circle is real. If b2 = a2 (i.e. rectangular hyperbola), then the radius of the director circle is zero and it reduces to a point circle at the origin. In this case centre is the only point from which two perpendicular tangents can be drawn on the curve. If b2 > a2, then the radius of the director circle is imaginary, so that there is no such circle and so no pair of tangents at right angle can be drawn to the curve. Example # 46 : Find the point of the line x – y = 0 for from where perpendicular tangent can be drawn to 2 x 9 + y2 = 1 Solution : Solving director circle x2 + y2 = 10 & x – y = 0 ⇒ ( 5 , 5 ), (– 5 , – 5 ) Self Practice Problems : (39) Find the angle between the tangent drawn from (–2, 1) to x2 + 4y2 = 4 Ans. (39) 2 π 12. Chord of contact: Equation to the chord of contact of tangents drawn from a point P(x1, y1) to the curve S = 0 is T = 0 Curve(S=0) T for point (x1, y1) & S = 0 equation of chord of contact from external point (x1, y1) to S=0 is T = 0 Parabola (y2 – 4ax = 0) T ≡ y y1 − 2a(x + x1) yy1 – 2a (x + x1) = 0 Ellipse 2 2 2 2 x y 1 0 a b   + − =       T ≡ 1 2 xx a + 1 2 yy b – 1 1 2 xx a + 1 2 yy b – 1= 0 Hyperbola 2 2 2 2 x y 1 0 a b   − − =       T ≡ 1 2 xx a – 1 2 yy b – 1 1 2 xx a – 1 2 yy b – 1 = 0 Conic Section Rectangular Hyperbola (xy – c2 = 0) T = 1 1 xy yx 2 + – c2 1 1 xy yx 2 + – c2 = 0 NOTE : The area of the triangle formed by the tangents from the point (x1, y1) & the chord of contact is 1 2a (y1² − 4ax1)3/2 Example # 47 : Find the length of chord of contact of the tangents drawn from point (–2, 3) to the parabola y2 = 8x. Solution : Let tangent at P(t1) & Q(t2) meet at (–2, 3) ∴ 2t1t2 = –2 & 2(t1 + t2) = 3  PQ = 2 2 2 2 1 2 1 2 (2t 2t ) (4(t t )) − + − = 2 2 2 1 2 1 2 1 2 ((t t ) 4t t )((t t ) 4) + − + + = 2 2 2 2 (3 4.2(–2))(3 4.2 ) 2 − + = 25/2 Example # 48 : If the line x – y – 1 = 0 intersect the parabola y2 = 8x at P & Q, then find the point of intersection of tangents at P & Q. Solution : Let (h, k) be point of intersection of tangents then chord of contact is yk = 4(x + h) 4x – yk + 4h = 0 .....(i) But given is x – y – 1 = 0 ∴ 4 1 = k 1 − − = 4h 1 − ⇒ h = – 1, k = 4 ∴ point ≡ (–1, 4) Example # 49 : Find the locus of point whose chord of contact w.r.t to the parabola y2 = 4bx is the tangents of the circle x2 + y2 = a2. Solution : Let it is chord of contact for parabola y2 = 4bx w.r.t. the point P(h, k) ∴ Equation of chord of contact is yk = 2b(x + h) y = 2b k x + 2bh k .....(i) (i) is tangents to x2 + y2 = a2 ⇒ 2 2 2bh k a 4b 1 k = + ⇒ 4b2x2 = a2 (y2 + 4b2) Example # 50 : If tangents to the circle x2 + y2 = b2 intersect the ellipse 2 2 x a + 2 2 y b = 1 at A and B, then find the locus of point of intersection of tangents at A and B. Solution : Let P ≡ (h, k) be the point of intersection of tangents at A & B ∴equation of chord of contact AB is 2 xh a + 2 yk b = 1 .............(i) which touches the circle x2 + y2 = b2 ∴ 2 2 4 4 1 b h k a b = + ⇒ required locus is 2 2 4 4 2 x y 1 a b b + = Example # 51: If tangents to the parabola y2 = 4ax intersect the hyperbola 2 2 x a – 2 2 y b = 1 at A and B, then find the locus of point of intersection of tangents at A and B. Conic Section Solution: Let P ≡ (h, k) be the point of intersection of tangents at A & B ∴ Equation of chord of contact AB is 2 xh a – 2 yk b = 1 ......(i) Which touches the parabola Equation of tangent to parabola y2 = 4ax y = mx + a m ⇒ mx – y = – a m ......(ii) equation (i) & (ii) as must be same ∴ 2 m h a       = 2 1 k b −   −     = a m 1 − ⇒ m = h k 2 2 b a & m = – 2 ak b ∴ 2 2 hb ka = – 2 ak b ⇒ locus of P is y2 = – 4 3 b a . x Self Practice Problems : (40) Prove that locus of a point whose chord of contact w.r.t. parabola passes through focus is directrix (41) If from a variable point ‘P’ on the line 2x – y – 1 = 0 pair of tangent’s are drawn to the parabola x2 = 8y then prove that chord of contact passes through a fixed point, also find that point. (42) Find the locus of point of intersection of tangents at the extremities of normal chords of the ellipse 2 2 x a + 2 2 y b = 1. (43) Find the locus of point of intersection of tangents at the extremities of the chords of the ellipse 2 2 x a + 2 2 y b = 1 subtending a right angle at its centre. Ans. (41) (8,1) (42) 6 2 a x + 6 2 b y = (a2 – b2) 2 (43) 2 4 x a + 2 4 y b = 2 1 a + 2 1 b Conic Section 13. Chord with a given middle point: Equation of the chord of the curve S = 0 whose middle point is (x1, y1) is T = S1. Curve(S=0) T for point (x1, y1) & S = 0 S1 for point (x1, y1) & S = 0 Chord with middle point (x1, y1) for S=0 is T = S1 Parabola (y2 – 4ax = 0) T = y y1 − 2a(x + x1) S1 = y1² − 4ax1 y y1 − 2a(x + x1) = y1² − 4ax1 Ellipse 2 2 2 2 x y 1 0 a b   + − =       T = 1 2 xx a + 1 2 yy b – 1 S1 = 2 2 1 1 2 2 x y a b + – 1 1 2 xx a + 1 2 yy b = 1 2 xx a + 1 2 yy b Hyperbola 2 2 2 2 x y 1 0 a b   − − =       T = 1 2 xx a – 1 2 yy b – 1 S1 = 2 2 1 1 2 2 x y a b − – 1 1 2 xx a – 1 2 yy b = 2 2 1 1 2 2 x y a b − Rectangular Hyperbola (xy – c2 = 0) T = 1 1 xy yx 2 + – c2 S1 = x1y1 – c2 xy1 + yx1 = 2x1y1 Example # 52 : Find the locus of middle point of the chord of the parabola y2 = 16x which pass through a given point (7, –2). Solution : Let P(h, k) be the mid point of chord of parabola y2 = 16x so equation of chord is yk – 8(x + h) = k2 – 16h. Since it passes through (7, –2) –2k – 8 (7 + h) = k2 – 16h ∴ Required locus is y2 + 2y – 8x + 56 = 0 Example # 53 : Find the locus of middle point of the chord of the parabola y2 = 4ax which is parallel to line y = mx + c Solution : Let P(h, k) be the mid point of chord of parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah. but slope = 2a k = m ∴ locus is y = 2a m Example # 54 : Find the locus of the mid - point of chords of the ellipse 2 2 x a + 2 2 y b = 1 Which are focal chords of y2 = 4ax Solution : Let P ≡ (h, k) be the mid-point ∴ equation of chord whose mid-point is given 2 xh a + 2 yk b – 1 = 2 2 h a + 2 2 k b – 1 since it is a focal chord, ∴ it passes through focus (a, 0) ⇒ 2 2 2 2 h h k a a b = + ⇒ required locus is 2 2 2 2 x y x a a b + = Example # 55 : Find the mid point of chord x + 2y = 4 of ellipse 9x2 + 36y2 = 324 Solution : Let (h,k) be mid point of chord . So T = S1 9xh + 36yk = 9h2 + 36k2 ------ (i) x + 2y = 4 ------ (ii) From (i) and (ii) 2 2 9h 36k 9h 36k 1 2 4 + = = ⇒ (h, k) = (2,1) Example # 56 : Find the locus of the mid - point of focal chords of the hyperbola 2 2 x a – 2 2 y b = 1. Conic Section Solution : Let P ≡ (h, k) be the mid-point ∴ equation of chord whose mid-point is given is 2 xh a – 2 yk b – 1 = 2 2 h a – 2 2 k b – 1 since it is a focal chord, ∴ it passes through focus, either (ae, 0) or (–ae, 0) If it passes through (ae, 0) ∴ locus is ex a = 2 2 x a – 2 2 y b If it passes through (–ae, 0) ∴ locus is – ex a = 2 2 x a – 2 2 y b Example # 57 : Find the condition on 'a' and 'b' for which two distinct chords of the hyperbola 2 2 x 2a – 2 2 y 2b = 1 passing through (a, b) are bisected by the line x + y = b. Solution : Let the line x + y = b bisect the chord at P(α, b – α) ∴ equation of chord whose mid-point is P(α, b – α) 2 x 2a α – 2 y(b ) 2b −α = 2 2 2a α – 2 2 (b ) 2b −α Since it passes through (a, b) ∴ 2a α – (b ) 2b −α = 2 2 2a α – 2 2 (b ) 2b −α α2 2 2 1 1 a b   −     + α 1 1 b a   −     = 0 α = 0, α = 1 1 1 a b + ∴ a ≠ ± b Example # 58 : Find the locus of the mid point of the chords of the hyperbola 2 2 x a – 2 2 y b = 1 which subtend a right angle at the origin. Solution : let (h,k) be the mid–point of the chord of the hyperbola. Then its equation is 2 hx a – 2 ky b – 1 = 2 2 h b – 2 2 k b – 1 or 2 hx a – 2 ky b = 2 2 h a – 2 2 k b ........(1) The equation of the lines joining the origin to the points of intersection of the hyperbola and the chord (1) is obtained by making homogeneous hyperbola with the help of (1) ∴ 2 2 x a – 2 2 y b = 2 2 2 2 2 2 2 2 hx ky a b h k a b   −       −       ⇒ 2 1 a 2 2 2 2 2 h k a b   −       x2 – 2 1 b 2 2 2 2 2 h k a b   −       y2 = 2 4 h a x2 + 2 4 k b y2 – 2 2 2hk a b xy .......(2) The lines represented by (2) will be at right angle if coefficient of x2 + coefficient of y2 = 0 ⇒ 2 1 a 2 2 2 2 2 h k a b   −       – 2 4 h a – 2 1 b 2 2 2 2 2 h k a b   −       – 2 4 k b = 0 ⇒ 2 2 2 2 2 h k a b   −       2 2 1 1 a b   −     = 2 4 h a + 2 4 h a hence, the locus of (h,k) is 2 2 2 2 2 x y a b   −       2 2 1 1 a b   −     = 2 4 x a + 2 4 y b Conic Section Self Practice Problems : (44) Find the mid point of chord x – y – 2 = 0 of parabola y2 = 4x. (45) Find the locus of mid - point of chord of parabola y2 = 4ax which touches the parabola x2 = 4by. (46) Find the equation of the chord 2 x 36 + 2 y 9 = 1 which is bisected at (2, 1). (47) Find the locus of the mid-points of normal chords of the ellipse 2 2 x a + 2 2 y b = 1. (48) Find the equation of the chord 2 x 36 – 2 y 9 = 1 which is bisected at (2, 1). (49) Find the point 'P' from which pair of tangents PA & PB are drawn to the hyperbola 2 x 25 – 2 y 16 = 1 in such a way that (5, 2) bisect AB (50) From the points on the circle x2 + y2 = a2, tangent are drawn to the hyperbola x2 – y2 = a2, prove that the locus of the middle points of the chords of contact is the curve (x2 – y2)2 = a2 (x2 + y2). Ans. (44) (4, 2) (45) y (2ax – y2) = 4a2b (46) x + 2y = 4 (47) 2 2 2 2 2 x y a b   +       6 6 2 2 a b x y   +       = (a2 – b2)2 (48) x = 2y (49) 20 8 , 3 3       14 NORMAL 14.1 Normal to the parabola : Normal is obtained using the slope of tangent. Normal P Slope of tangent at (x1 , y1) = 1 2a y ⇒ Slope of normal = – 1 y 2a (i) y − y1 = – 1 y 2a (x − x1) at (x1, y1) ; (ii) y = mx − 2am − am3 at (am2, − 2am) (iii) y + tx = 2at + at3 at (at2, 2at). NOTE : (i) Point of intersection of normals at t1 & t2 is (a (t + t + t1t2 + 2), − a t1 t2 (t1 + t2)). (ii) If the normals to the parabola y² = 4ax at the point t1, meets the parabola again at the point P(t ) 1 Q(t ) 2 t2, then t2 = – 1 1 2 t t   +     . (iii) If the normals to the parabola y² = 4ax at the points t1 & t2 intersect again on the parabola at the point 't3' then t1 t2 = 2; t3 = − (t1 + t2) and the line joining t1 & t2 passes through a fixed point (−2a, 0) Example # 59 : If the normal at point ‘t1’ intersects the parabola again at ‘t2’ then find value of \|t1.t2 + t1 2\|. Solution : Slope of normal at P = – t1 and slope of chord PQ = 1 2 2 t t + ∴ – t1 = 1 2 2 t t + ⇒ t2 = – t1 – 1 2 t ⇒ t2 = – t1 – 1 2 t ⇒ t2.t1 = –t1 2 – 2 ⇒ \|t1.t2 + t1 2\| = 2 Conic Section Example # 60 : If the normals at points t1, t2 meet at the point t3 on the parabola then find value of (t1 + t2 + t3)2 + (t1 . t2)2 Solution : Since normal at t1 & t2 meet the curve at t3 ∴ t3 = – t1 – – 1 2 t .....(i) t3 = – t2 – 2 2 t .....(ii) ⇒ (t1 2 + 2) t2 = t1 (t2 2 + 2) ⇒ t1t2 (t1 – t2) + 2 (t2 – t1) = 0  t1 ≠ t2 , t1t2 = 2 ......(iii) Hence (i) t1 t2 = 2 from equation (i) & (iii), we get t3 = – t1 – t2 Hence (ii) t1 + t2 + t3 = 0 ....(iv) from (iii) & (iv) (t1 + t2 + t3)2 + (t1 . t2)2 = 4 Example # 61 : Find the locus of the point N from which 3 normals are drawn to the parabola y2 = 4ax are such that (i) Two of them are equally inclined to y-axis (ii) Two of them have product of their slops is equal to 2. Solution : Equation of normal to y2 = 4ax is y = mx – 2am – am3 Let the normal passes through N(h, k) ∴ k = mh – 2am – am3 ⇒ am3 + (2a – h) m + k = 0 For given value’s of (h, k) it is cubic in ‘m’. Let m1, m2 & m3 are root’s of above equation ∴ m1 + m2 + m3 = 0 ......(i) m1m2 + m2m3 + m3m1 = 2a h a − ......(ii) m1m2m3 = – k a ......(iii) (i) If two normal are equally inclined to x-axis, then m1 + m2 = 0 ∴ m3 = 0 ⇒ y = 0 (ii) If two normal’s have product of their slops = 2 ∴ m1 m2 = 2 from (3) m3 = – k 2a .....(iv) from (2) 2 – k 2a (m1 + m2) = 2a h a − .....(v) from (1) m1 + m2 = k 2a .....(vi) from (5) & (6), we get k2 = 4ax Conic Section Self Practice Problems: (51) Find the points of the parabola y2 = 4ax at which the normal is inclined at 45° to the axis. (52) If chord drawn from point P(9, –6) on the parabola y2 = 4x is normal at point Q then Q = ? (53) Find the length of normal chord at point ‘t’ to the parabola y2 = 4ax. (54) If the normals at 3 points P, Q & R on the parabola (x – 3)2 = y + 2 are concurrent, then show that (i) The sum of slopes of normals is zero, (ii) The locus of centroid of ∆PQR is x – 3 = 0. Ans. (51) (a, –2a), (a, 2a) (52) (9, –6) (53) 3 2 2 2 4a(t 1) t + =  14.2 Normal to Ellipse (i) Equation of the normal at (x1, y1) to the ellipse 2 2 x a + 2 2 y b = 1 is 2 2 1 1 a x b y x y − = a² − b². (ii) Equation of the normal at the point (acos θ, bsin θ) to the ellipse 2 2 x a + 2 2 y b = 1 is; ax. sec θ − by. cosec θ = (a² − b²). (iii) Equation of a normal in terms of its slope 'm' is y = mx −( ) 2 2 2 2 2 a b m a b m − + . Example # 62 : A and B are corresponding points on the ellipse 2 2 x a + 2 2 y b =1 and the auxiliary circles respectively. The normal at A to the ellipse meets CB in R, where C is the centre of the ellipse. Prove that locus of R is a circle of radius a + b. Solution : Let A ≡ (acos θ, b sinθ) ∴ B ≡ (a cosθ, a sinθ) A B Equation of normal at A is (a secθ) x – (b cosec θ) y = a2 – b2 ..........(i) equation of CB is y = tanθ . x .........(ii) Solving equation (i) & (ii), we get (a – b) x = (a2 – b2) cosθ x = (a + b) cosθ, & y = (a + b) sinθ ∴ R ≡ ((a + b) cosθ, (a + b) sinθ) = (h, k) h2 + k2 = (a + b)2 ⇒ x2 + y2 = (a + b)2 Example # 63 : Find the shortest distance between the line 3x + 4y = 12 and the ellipse 2 2 x y 16 9 + = 1 Solution : Shortest distance occurs between two non-intersecting curve always along common normal. Let 'P' be a point on ellipse and Q is a point on given line for which PQ is common normal. ∴ Tangent at 'P' is parallel to given line ∴ Equation of tangent parallel to given line is (y = mx ± 2 2 2 a m b + ) y = 3x 4 ±3 2 Conic Section ∴ minimum distance = distance between 3x + 4y = 12 & 3x + 4y = 3 2 ⇒ shortest distance = 12 3 2 5 − Example # 64 : Prove that, in an ellipse, the distance between the centre and any normal does not exceed the difference between the semi-axes of the ellipse. Solution : Let the equation of ellipse is 2 2 2 2 x y 1 a b + = Equation of normal at P (θ) is (a secθ)x – (bcosec θ)y – a2 + b2 = 0 distance of normal from centre = OR = 2 2 2 2 2 2 \| a b \| a b (atan ) (bcot ) − + + θ + θ = 2 2 2 2 \| a b \| (a b) (atan bcot ) − + + θ − θ  (a + b)2 + (a tanθ – b cotθ)2 ≥ (a + b)2 or ≤ 2 2 2 \| a b \| (a b) − + ⇒ \|OR\| ≤ (a – b) Self Practice Problems : (55) Find the value(s) of 'a' for which the line x + y = a is a normal to the ellipse 3x2 + 4y2 = 12 (56) If the normal at the point P(θ) to the ellipse 2 2 x y 14 5 + = 1 intersects it again at the point Q(2θ) then find the value of cosθ Ans. (55) a = 1 7 ± (56) – 2 3 14.3 Normal to Hyperbola (a) The equation of the normal to the hyperbola 2 2 2 2 x y 1 a b − = at the point P (x1, y1) on it is 2 2 1 1 a x b y x y + = a2 + b2 = a2 e2. (b) The equation of the normal at the point P (a sec θ, b tan θ) on the hyperbola 2 2 2 2 x y 1 a b − = is ax by sec tan + θ θ = a2 + b2 = a2 e2. (c) Equation of normals in terms of its slope 'm' are y = mx ± 2 2 2 2 2 (a b )m a b m + − . 14.4 Normal to Rectangular hyperbola Equation of the normal at P (t) is x t3 − y t = c (t4 − 1) Conic Section Example # 65 : A normal to the hyperbola 2 2 x a – 2 2 y b = 1 meets the axes in M and N. find a locus of point R on segment MN such that NR : RM = 2 :1. Solution : The equation of normal at the point Q (a sec φ, b tan φ) to the hyperbola 2 2 x a – 2 2 y b = 1 is ax cos φ + by cot φ = a2 + b2 ........(1) The normal (1) meets the x–axis in M 2 2 a b sec , 0 a   + φ       and y–axis in N 2 2 a b 0, tan b   + φ       Let R (h, k) is point whose locus we have to find. as NR : RM = 2 :1. ⇒ h = 2 2 2 (a b ) sec 3 a + φ , k = 2 2 (a b ) tan 3b + φ we know that sec2φ – tan2φ = 1 ⇒ 2 2 2 2 2 2 2 2 2 2 9a 9b x y 1 4(a b ) (a b ) − = + + ⇒ 2 2 2 2 2 2 2 a x (a b ) b y 4 9 + − = Self Practice Problems : (57) Prove that the line lx + my – n = 0 will be a normal to the hyperbola 2 2 x a – 2 2 y b = 1 if 2 2 a  – 2 2 b m = 2 2 2 2 (a b ) n + . 15. Important Highlights of Parabola : (i) If the tangent & normal at any point ‘P’ of the parabola intersect the axis at T & G then ST = SG = SP where ‘S’ is the focus. In other words the tangent and the normal at a point P on the parabola are the bisectors of the angle between the focal radius SP & the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to the axis of theparabola after reflection. (ii) The portion of a tangent to a parabola cut off between the directrix & the curve subtends a right angle at the focus. See figure above. (iii) The tangents at the extremities of a focal chord intersect at right angles on the directrix, and hence a circle on any focal chord as diameter touches the directrix. Also a circle on any focal radii of a point P (at2, 2at) as diameter touches the tangent at the vertex and intercepts a chord of length a 2 1 t + on a normal at the point P. (iv) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent at the vertex. Conic Section (v) If the tangents at P and Q meet in T, then: ⇒ TP and TQ subtend equal angles at the focus S. ⇒ ST2 = SP. SQ & ⇒ The triangles SPT and STQ are similar. S Q P ∠ ∠ TSP = TSQ T (vi) Semi latus rectum of the parabola y² = 4ax, is the harmonic mean between segments of any focal chord of the parabola. P S Q 2(PS)(SQ) PS SQ + = 2a (vii) The area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. (viii) If normal are drawn from a point P(h, k) to the parabola y2 = 4ax then k = mh − 2am − am3 i.e. am3 + m(2a − h) + k = 0. m1 + m2 + m3 = 0 ; m1m2 + m2m3 + m3m1 = 2a h a − ; m1 m2 m3 = k a − . Where m1, m2, & m3 are the slopes of the three concurrent normals. Note that A B C P(h, k) A, B, C → Conormal points ⇒ algebraic sum of the slopes of the three concurrent normals is zero. ⇒ algebraic sum of the ordinates of the three conormal points on the parabola is zero ⇒ Centroid of the ∆ formed by three co−normal points lies on the x−axis. ⇒ Condition for three real and distinct normals to be drawn froma point P (h, k) is h > 2a & k2 < 4 27a (h – 2a)3. (ix) Length of subtangent at any point P(x, y) on the parabola y² = 4ax equals twice the abscissa of the point P. Note that the subtangent is bisected at the vertex.. x D N P O T y TD = 2(OD) , DN = 2a (x) Length of subnormal is constant for all points on the parabola & is equal to the semi latus rectum. See figure above. Conic Section 16. Important Highlights of Ellipse : Refering to the ellipse 2 2 2 2 x y a b + = 1 (i) If P be any point on the ellipse with S & S′ as its foci then  (SP) +  (S′P) = 2a. (ii) The tangent & normal at a point P on the ellipse bisect the external & internal angles between the focal distances of P. This refers to the well known reflection property of the ellipse which states that rays from one focus are reflected through other focus & vice−versa. Hence we can deduce that the straight lines joining each focus to the foot of the perpendicular from the other focus upon the tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis. (iii) The product of the length’s of the perpendicular segments from the foci on any tangent to the ellipse is b² and the feet of these perpendiculars lie on its auxiliary circle and the tangents at these feet to the auxiliary circle meet on the ordinate of P and that the locus of their point of Intersection is a similiar ellipse as that of the original one. (iv) The portion of the tangent to an ellipse between the point of contact & the directrix subtends a right angle at the corresponding focus. (v) If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively, & if CF be Perpendicular upon this normal, then (i) PF. PG = b² (ii) PF. Pg = a² (iii) PG. Pg = SP. S′ P (iv) CG. CT = CS2 (v) Locus of the mid point of Gg is another ellipse having the same eccentricity as that of the original ellipse. [Where S and S′ are the focii of the ellipse and T is the point where tangent at P meet the major axis] (vi) The circle on any focal distance as diameter touches the auxiliary circle. Perpendiculars from the centre upon all chords which join the ends of any perpendicular diameters of the ellipse are of constant length. (vii) If the tangent at the point P of a standard ellipse meets the axis in T and t and CY is the perpendicular on it from the centre then, Conic Section (i) T t. PY = a2 − b2 and (ii) least value of T t is a + b. 17. Important Highlights of Hyperbola: (i) Difference of focal distances is a constant, i.e. \|PS – PS′\| = 2a (ii) Locus of the feet of the perpendicular drawn from focus of the hyperbola 2 2 2 2 x y 1 a b − = upon any tangent is its auxiliary circle i.e. x2 + y2 = a2 & the product of these perpendiculars is b2. S′ T P S T′ (iii) The portion of the tangent between the point of contact & the directrix subtends a right angle at the corresponding focus. P T S (iv) The tangent & normal at any point of a hyperbola bisect the angle between the focal radii. This explains the reflection property of the hyperbola as "An incoming light ray " aimed towards one focus is reflected from the outer surface of the hyperbola towards the other focus. It follows that if an ellipse and a hyperbola have the same foci, they cut at right angles at any of their common point. Note that the ellipse 2 2 2 2 x y 1 a b + = & the hyperbola 2 2 2 2 2 2 x y a k k b − − − = 1 (a > k > b > 0) are confocal and therefore orthogonal. (v) The foci of the hyperbola and the points P and Q in which any tangent meets the tangents at the vertices are concyclic with PQ as diameter of the circle. A S P Q S′ Conic Section (vi) A rectangular hyperbola circumscribing a triangle also passes through the orthocentre of this triangle. If i i c t , c t       i = 1, 2, 3 be the angular points P, Q, R then orthocentre is 1 2 3 1 2 3 , c t t t c t t t   −     − . (vii) If a circle and the rectangular hyperbola xy = c2 meet in the four points t1, t2, t3 & t4,, then (a) t1 t2 t3 t4 = 1 (b) the centre of the mean position of the four points bisects the distance between the centres of the two curves. (c) the centre of the circle through the points t1, t2 & t3 is : 1 2 3 1 2 3 1 2 3 1 2 3 t t t , t t t c c 1 1 1 1 t t t t t t 2 2       + + + + + +             Example # 66 : A ray originating from the point (5, 0) is incident on the hyperbola 9x2 – 16y2 = 144 at the point P with abscissa 8. Find the equation of the reflected ray after first reflection and point P lying in first quadrant. Solution : Given hyperbola is 9x2 – 16y2 = 144. This equation can be rewritten as 2 x 16 – 2 y 9 = 1 ....(1) Since x co–ordinate of P is 8. Let y co–ordinate of P is α.  (8,α) lies on (1) ∴ 64 16 – 2 9 α = 1 ⇒ α2 = 27 ⇒ α = 3 3 ( P lies in first quadrant) Hence co-ordinate of point P is (8,3 3 ).  Equation of reflected ray passes through P (8,3 3 ) and S′(–5,0) ∴ Its equation is y – 3 3 = 0 3 3 5 8 − −− (x – 8) or 13y – 39 3 = 3 3 x – 24 3 or 3 3 x – 13y + 15 3 = 0.
189154	https://math.stackexchange.com/questions/1255277/finding-equation-of-hyperbola-with-only-foci-and-asymptote	graphing functions - Finding equation of hyperbola with only foci and asymptote - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding equation of hyperbola with only foci and asymptote Ask Question Asked 10 years, 5 months ago Modified10 years, 5 months ago Viewed 20k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This is a concept we learned in class today, which I still can't seem to grasp. I have no specific question that necessarily has to be done, so I will use one of the examples my book gives me: Given the foci are: (6√5, 10) and (-6√5,10) The asymptote is y = 1/2x + 10 Find the equation of the hyperbola I know that... (y-k)²/a² - (x-h)²/b² = 1 (Hyperbola equation) Center (h,k) foci (h, k±c), c² = a²+b² Asmyptote y = -(a/b)x + k+(a/b)h (The -/+ are interchangeable) I am just confused on how to find the equation with the givens above. Thanks! graphing-functions hyperbolic-functions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Apr 28, 2015 at 2:44 TEEBQNETEEBQNE 221 1 1 gold badge 3 3 silver badges 10 10 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. It looks like you know all of the equations you need to solve this problem. I also see that you know that the slope of the asymptote line of a hyperbola is the ratio b a b a for a simple hyperbola of the form x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1 So, given y=1 2 x+10 y=1 2 x+10, we can see that the ratio b a=1 2 b a=1 2. Furthermore, we can say that b=n b=n and a=2 n a=2 n since the n′s n′s would cancel in the ratio. Now that we have our a a and b b depending on only one variable n n, we can use the relationship you mentioned, c 2=a 2+b 2 c 2=a 2+b 2, to determine the value of n n. a 2+b 2(2 n)2+n 2 n 2 n=c 2=(6 5–√)2=36=6 a 2+b 2=c 2(2 n)2+n 2=(6 5)2 n 2=36 n=6 Therefore, we have a=12 a=12 and b=6 b=6, so then our equation for this hyperbola can be expressed with the following: x 2 144−(y−10)2 36=1 x 2 144−(y−10)2 36=1 Where the 10 came from shifting the hyperbola up 10 units to match the y y value of our foci. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 28, 2015 at 5:21 answered Apr 28, 2015 at 4:43 michael10000michael10000 66 3 3 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graphing-functions hyperbolic-functions See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2How to find the equation of a hyperbola knowing its asymptotes and one focus? Related 1how to find the foci, directrix, center of a polar conic section. 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189155	https://www.etymonline.com/word/opulence	ABCDEFGHIJKLMNOPQRSTUVWXYZ Advertisement Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Origin and history of opulence opulence(n.) "wealth, riches, affluence," c. 1500, from French opulence (16c.), from Latin opulentia "riches, wealth," from opulentus "wealthy," a dissimilation of op-en-ent-, which is related to ops "wealth, power, ability, resources," and to opus "work, labor, exertion" (from PIE root op- "to work, produce in abundance"). Opulence is a dignified and strong word for wealth. Wealth and riches may mean the property possessed, and riches generally does mean it; the others do not. Affluence suggests the flow of wealth to one, and resulting free expenditure for objects of desire. There is little difference in the strength of the words. [Century Dictionary, 1895] also from c. 1500 Entries linking to opulence opulent(adj.) "wealthy, rich, affluent," c. 1600, from French opulent and directly from Latin opulentem (nominative opulens) "wealthy, rich; splendid, noble," from opulentus (see opulence). op- Proto-Indo-European root meaning "to work, produce in abundance." It might form all or part of: cooperate; cooperation; copious; copy; cornucopia; hors d'oeuvre; inure; maneuver; manure; oeuvre; office; official; officinal; omni-; omnibus; omnium gatherum; op. cit.; opera; operate; operation; operose; optimism; optimum; opulence; opulent; opus; Oscan. It might also be the source of: Sanskrit apas- "work, religious act," apnas- "possession, property;" Hittite happina- "rich;" Avestan huapah- "doing good work, masterly;" Latin opus "a work, labor, exertion;" Greek ompne "food, corn;" Old High German uoben "to start work, to practice, to honor;" German üben "to exercise, practice;" Dutch oefenen, Old Norse æfa, Danish øve "to exercise, practice;" Old English æfnan "to perform, work, do," afol "power." Advertisement Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Trends of opulence adapted from books.google.com/ngrams/ with a 7-year moving average; ngrams are probably unreliable. More to explore luxury c. 1300, "sexual intercourse;" mid-14c., "lasciviousness, sinful self-indulgence;" late 14c., "sensual pleasure," from Old French luxurie "debauchery, dissoluteness, lust" (12c., Modern French luxure), from Latin luxuria "excess, extravagant living, profusion; delicacy" (source a riches modified from richesse (c. 1200), a singular form misunderstood as a plural, from Old French richesse, richece "wealth, opulence... quadrillion each case a millionaire (we shall soon have billionaires, trillionaires, quadrillionaires) fettered, imprisoned, by abject opulence... opus "a work, composition," especially a musical one, 1809, from Latin opus "a work, labor, exertion" (source of Italian opera, French oeuvre, Spanish obra), from Proto-Italic opes- "work," from PIE root op- "to work, produce in abundance." The plural, seldom used as such, is opera. vernacular c. 1600, "native to a country," from Latin vernaculus "domestic, native, indigenous; pertaining to home-born slaves," from verna "home-born slave, native," a word of Etruscan origin. Used in English in the sense of Latin vernacula vocabula, in reference to language. As a noun, "n pandemonium 1667, Pandæmonium, in "Paradise Lost" the name of the palace built in the middle of Hell, "the high capital of Satan and all his peers," and the abode of all the demons; coined by John Milton (1608-1674) from Greek pan- "all" (see pan-) + Late Latin daemonium "evil spirit," from protest c. 1400, "avowal, pledge, solemn declaration," from Old French protest, from protester, from Latin protestari "declare publicly, testify, protest," from pro- "forth, before" (from PIE root per- (1) "forward," hence "in front of, before") + testari "testify," from testis "witness endure late 14c., "to undergo or suffer" (especially without breaking); also "to continue in existence," from Old French endurer (12c.) "make hard, harden; bear, tolerate; keep up, maintain," from Latin indurare "make hard," in Late Latin "harden (the heart) against," from in- (from PIE flamboyant 1832, originally in reference to a 15c.-16c. architectural style with wavy, flame-like curves, from French flamboyant "flaming, wavy," present participle of flamboyer "to flame," from Old French flamboiier "to flame, flare, blaze, glow, shine" (12c.), from flambe "a flame, flame cybernetics "theory or study of communication and control," coined 1948 by U.S. mathematician Norbert Wiener (1894-1964), with -ics + Latinized form of Greek kybernetes "steersman" (metaphorically "guide, governor"), from kybernan "to steer or pilot a ship, direct as a pilot," figuratively " Share opulence Page URL: HTML Link: APA Style: Chicago Style: MLA Style: IEEE Style: Advertisement Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Dictionary entries near opulence optimum option optional optometrist optometry opulence opulent opus or -or orach Advertisement Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads.
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NSWVIC Part 3: Vector Projections \| Beginner’s Guide to Year 12 Maths Ext 1 Ace your next vector projections test now! Share: Facebook Twitter Linkedin Email Guide Chapters Guide Chapters 0. Overview 1. Introduction to vectors 2. Operations with vector and dot product 3. Vector projections 4. Projectile motion 5. Vector proofs 6. Integrating squares of sine and cosine 7. Integration by substitution 8. Differential equations 9. Mathematical induction Still confused about vector projections? Well, you don’t need to be! In this article, we’ll go through everything you need to know about vector projections. In this article we discuss: The NESA Syllabus Outcomes Vectors Projections Applications of Projection Concept Check Questions Ace Vector Projections with these practice questions! Free Y12 Maths Ext1 Vector Projections Worksheet A worksheet to test your knowledge. Download now Free Y12 Maths Ext1 Vector Projections Worksheet Fill out your details below to get this resource emailed to you. "" indicates required fields First Name Last Name Email Postcode Grade in 2025 This field is hidden when viewing the form Resource (DO NOT DELETE OR CHANGE) By submitting this form, you agree to receive promotional messages from Matrix about its products and services. You can unsubscribe at any time by clicking on the link at the bottom of our emails. Year 12 Mathematics Extension 1: Vector Projections Within the topic of vectors there is a specific requirement for students to be familiar with the concept of the projection of one vector onto another. This syllabus dot point allows for students to develop their understanding of vector operations and can be used in applications to both harder exam questions and real world problems. NESA Syllabus Outcomes Define and use the projection of one vector onto another Assumed Knowledge Students should have a basic understanding of vectors, as well as some fundamental properties and operations including the dot product. Students should also feel comfortable with concepts of trigonometry and coordinate geometry. This content can be found in our previous mathematics guides should students want to solidify their understanding. Vector Projection The idea of a vector projection, in its simplest form is just the question of how much one vector goes in the direction of another. This idea is geometrically represented by the figure below, with vector a being projected onto vector b. The projection in this case would be the vector O P. We can see that O P is parallel to vector b. We can also see from the triangle that the following relationship should hold: c o s(θ)=O P a Hence, O P\|=\|a\|×\|c o s(θ)\| Using our definition of the dot product as a⋅b=\|a\|×\|b\|×c o s θ, we can rewrite the above expression: \|O P\|=\|a⋅b\|\|b\| From here, we can use the idea that the unit vector b\|b\| is simply a vector with unit length pointing in the direction of vector b to formally define a formula for the vector projection in terms of the dot product. For vectors a,b∈R n and b≠0, the projection of a onto b is given by: p r o j b a=(a⋅b\|b\|2)b Scalar Projection The scalar projection is essentially derived from the concept of a vector projection. The only difference between the two is that the scalar projection, as implied in its name is simply a scalar measure of the length of the projection created from one vector onto another. For vectors a,b∈R n and b≠0, the scalar projection of a onto b is given by: \|p r o j b a\|=(\|a⋅b\|\|b\|) We can observe that the scalar projection is simply the expression for the vector projection with the removal of the unit vector b\|b\|. This can be attributed to the scalar projection being a scalar value, hence the removal of the directional vector component of the vector projection should simply leave us with a measurement of the projection’s length. Applications to Problems Now that we understand the concept of projections, let’s try applying this idea to some examples: Example 1 Find the distance from the point (2,−1,3) to the line through the points (0,1,4) and (4,2,9) Solution 1 At first glance, it might not be obvious that the idea of vector projection can be used in solving this question. However, recall that the distance between a point and a line is simply the perpendicular distance taken from the base of the line to the point, similar to the formula taught in year 10. A good place to start for this question would be to label each point. Hence let A,B and C be the points (0,1,4),(2,−1,3) and (4,2,9) respectively. This subsequently implies the vectors A B and A C are represented by (2,−2,−1) and (4,1,5) respectively. We can see in the previous diagram that a vector projection essentially creates a perpendicular from the end of the projecting vector to the base of the vector being projected upon. If we let P denote the base of said perpendicular, we can see that the perpendicular distance between the point and the line is simply the distance B P. By taking the scalar projection of the vector A B onto the vector A C : \|p r o j A C A B\|=\|A B⋅A C\|\|A C\|=\|8−2−5\|16+1+25−−−−−−−−−√=1 42−−√ We have now found the length of the vector A P. With reference to our diagram, we can see that to find the length of the vector B P, we can just apply Pythagoras theorem as it’s relatively simple to determine the length of the vector A B: \|A B\|=4+4+1−−−−−−−−√=3 Hence, the distance from the point to the line is given by: \|B P\|=3 2–1 42−−−−−−√=377−−−√42−−√ Example 2 Let u and v be non-zero vectors such that \|u\|=\|v\|. Show that \|p r o j v u\|=\|p r o j u v\| Solution 2 For proof questions such as this one, it can be a good idea to first convert functions or operations into simpler terms. In this case, let’s begin the question by writing out the formula for vector projection: L H S=∣∣∣(u⋅v\|v\|2)v∣∣∣R H S=∣∣∣(v⋅u\|u\|2)u∣∣∣ Now that we’ve done that, it can be beneficial to also extract all the information provided to us in the question. In this case, the only condition we’re given is that \|u\|=\|v\|. Let’s try to manipulate the LHS with this condition to try and make it resemble the RHS a bit more: L H S=∣∣∣(u⋅v\|v\|2)v∣∣∣ Since \|u\|=\|v\|, it should follow that \|u\|2=\|v\|2. Hence we can rewrite the LHS: L H S=∣∣∣(u⋅v\|u\|2)v∣∣∣ We can also use the property that u⋅v=v⋅u to further manipulate the expression: L H S=∣∣∣(v⋅u\|u\|2)v∣∣∣ From here, it can be a little difficult to see that we’ve essentially finished the proof already. It’s important for students to make the observation that (v⋅u\|u\|2) is simply a scalar value, meaning that it can be separated from the vector component of the expression. L H S=∣∣∣(v⋅u\|u\|2)∣∣∣×\|v\| Now using the given condition that\|u\|=\|v\|: L H S∴\|p r o j v u\|=∣∣∣(v⋅u\|u\|2)∣∣∣×\|u\|=∣∣∣(v⋅u\|u\|2)u∣∣∣=R H S=\|p r o j u v\| The proof is complete. Concept Questions 1. Find the dot product of vectors a=(−4,1) and b=(1,2) 2. Sketch the vectors a=(−4,1) and b=(1,2) 3. Find the vector resulting from the projection of a=(−4,1) onto b=(1,2) 4. Find the length of the projection of (−4,1) onto (1,2) Concept Question Answers 1.a⋅b=−2 2. Diagram 3.(8 17,–2 17) 4.2 17√ Save time and boost your marks Our HSC experts will break down Maths concepts through our engaging, structured lessons. And our comprehensive resources will give you plenty of practice to ace your exams. Learn more now! Ace Maths Ext 1 with Matrix+ Expert HSC teachers and Band 6 resources! Learn at your own pace and join 8000+ students getting ahead. Start free trial ### Part 4: Projectile Motion \| Beginner’s Guide to Year 12 Ext 1 Maths Read next chapter © Matrix Education and www.matrix.edu.au, 2025. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content. 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189157	https://www.chegg.com/homework-help/questions-and-answers/8-vi-hypobromous-acid-hobr-pka-864-calculate-ph-0300-m-solution-acid-percent-dissociation--q31899106	Solved (8) VI. Hypobromous acid, HOBr, has a pKa of 8.64. \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Chemistry Chemistry questions and answers (8) VI. Hypobromous acid, HOBr, has a pKa of 8.64. Calculate the pH of a 0.300 M solution of this acid. What is the percent dissociation? (7) VII. Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Ki-2.5 × 10-4,K2-5.6 × 10-8, and Ks = 3 × 10-13. What is the pH of a 0.500 M solution Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: (8) VI. Hypobromous acid, HOBr, has a pKa of 8.64. Calculate the pH of a 0.300 M solution of this acid. What is the percent dissociation? (7) VII. Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Ki-2.5 × 10-4,K2-5.6 × 10-8, and Ks = 3 × 10-13. What is the pH of a 0.500 M solution Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link View the full answer Previous questionNext question Transcribed image text: (8) VI. Hypobromous acid, HOBr, has a pKa of 8.64. Calculate the pH of a 0.300 M solution of this acid. What is the percent dissociation? (7) VII. Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Ki-2.5 × 10-4,K2-5.6 × 10-8, and Ks = 3 × 10-13. What is the pH of a 0.500 M solution of arsenic acid? Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. 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189158	https://en.wikipedia.org/wiki/Class_(biology)	Jump to content Search Contents (Top) 1 History 2 See also 3 Explanatory notes 4 References Class (biology) Afrikaans Alemannisch العربية Aragonés অসমীয়া Asturianu Avañe'ẽ Azərbaycanca Basa Bali বাংলা 閩南語 / Bn-lm-gí Башҡортса Беларуская Беларуская (тарашкевіца) भोजपुरी Български Boarisch Bosanski Brezhoneg Català Чӑвашла Čeština Cymraeg Dansk الدارجة Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Frysk Gaeilge Galego ГӀалгӀай 한국어 Հայերեն हिन्दी Hornjoserbsce Hrvatski Ilokano Bahasa Indonesia Interlingua IsiZulu Íslenska Italiano עברית Jawa Kapampangan ქართული کٲشُر Қазақша Kiswahili Kreyòl ayisyen Kurdî Кыргызча Latina Latviešu Lëtzebuergesch Lietuvių Limburgs Magyar Македонски മലയാളം Malti مازِرونی Bahasa Melayu မြန်မာဘာသာ Nederlands 日本語 Napulitano Norsk bokmål Norsk nynorsk Occitan Oromoo Oʻzbekcha / ўзбекча Plattdüütsch Polski Português Romnă Runa Simi Русиньскый Русский Саха тыла Shqip Sicilianu සිංහල Simple English سنڌي Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் Taqbaylit Татарча / tatarça తెలుగు ไทย Türkçe Українська اردو Vèneto Tiếng Việt West-Vlams Winaray 吴语粵語 Zeêuws 中文 Betawi ရခိုင် Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Taxonomic rank between phylum and order For other uses, see Class (disambiguation). This article is about the taxonomic grouping of related organisms and is not to be confused with the ecological grouping of unrelated plant taxa in phytosociology. In biological classification, class (Latin: classis) is a taxonomic rank, as well as a taxonomic unit, a taxon, in that rank. It is a group of related taxonomic orders.[a] Other well-known ranks in descending order of size are domain, kingdom, phylum, order, family, genus, and species, with class ranking between phylum and order. History [edit] The class as a distinct rank of biological classification having its own distinctive name – and not just called a top-level genus (genus summum) – was first introduced by French botanist Joseph Pitton de Tournefort in the classification of plants that appeared in his Eléments de botanique of 1694. Insofar as a general definition of a class is available, it has historically been conceived as embracing taxa that combine a distinct grade of organization—i.e. a 'level of complexity', measured in terms of how differentiated their organ systems are into distinct regions or sub-organs—with a distinct type of construction, which is to say a particular layout of organ systems. This said, the composition of each class is ultimately determined by the subjective judgment of taxonomists. In the first edition of his Systema Naturae (1735), Carl Linnaeus divided all three of his kingdoms of nature (minerals, plants, and animals) into classes. Only in the animal kingdom are Linnaeus's classes similar to the classes used today; his classes and orders of plants were never intended to represent natural groups, but rather to provide a convenient "artificial key" according to his Systema Sexuale, largely based on the arrangement of flowers. In botany, classes are now rarely discussed. Since the first publication of the APG system in 1998, which proposed a taxonomy of the flowering plants up to the level of orders, many sources have preferred to treat ranks higher than orders as informal clades. Where formal ranks have been assigned, the ranks have been reduced to a very much lower level, e.g. class Equisitopsida for the land plants, with the major divisions within the class assigned to subclasses and superorders. The class was considered the highest level of the taxonomic hierarchy until George Cuvier's embranchements, first called Phyla by Ernst Haeckel, were introduced in the early nineteenth century. See also [edit] Biology portal Cladistics List of animal classes Phylogenetics Systematics Taxonomy Explanatory notes [edit] ^ When the term denotes taxonomic units, the plural is classes (Latin classes). References [edit] ^ "Class". Biology Articles, Tutorials & Dictionary Online. 23 July 2021. ^ Huxley, Thomas Henry (1853). Henfrey, Arthur (ed.). Scientific memoirs, selected from the transactions of foreign academies of science, and from foreign journals. Natural history. Taylor and Francis. doi:10.5962/bhl.title.28029. ^ Mayr E. (1982). The Growth of Biological Thought. Cambridge: The Belknap Press of Harvard University Press. ISBN 0-674-36446-5 ^ Chase, Mark W. & Reveal, James L. (2009), "A phylogenetic classification of the land plants to accompany APG III", Botanical Journal of the Linnean Society, 161 (2): 122–127, doi:10.1111/j.1095-8339.2009.01002.x ^ Collins, A.G., Valentine, J.W. (2001). "Defining phyla: evolutionary pathways to metazoan body plans" Archived 2020-04-27 at the Wayback Machine. Evol. Dev. 3: 432–442. \| v t e Taxonomic ranks \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Realm (vir.) Subrealm (vir.) Domain/Superkingdom Kingdom Subkingdom Infrakingdom/Branch \| Superphylum/Superdivision (bot.) Phylum/Division (bot.) Subphylum/Subdivision (bot.) Infraphylum Parvphylum \| Superclass Class Subclass Infraclass Subterclass Parvclass \| Division (zoo.) Legion Cohort \| Magnorder Superorder Order Suborder Infraorder Parvorder \| Section (zoo.) Superfamily Family Subfamily Infrafamily \| Supertribe Tribe Subtribe Infratribe \| Genus Subgenus Section (bot.) Series (bot.) \| "Species complex" Species Subspecies Variety (bot.) Form (bot.) \| \| \| Authority control databases \| \| International \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Classes (biology) Bacterial nomenclature Zoological nomenclature Botanical nomenclature Plant taxonomy Hidden categories: Webarchive template wayback links Articles with short description Short description is different from Wikidata Articles containing Latin-language text Class (biology) Add topic
189159	https://math.stackexchange.com/questions/3487539/proof-that-the-product-of-3-sequential-numbers-is-divisible-by-3	induction - Proof that the product of 3 sequential numbers is divisible by 3 - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Proof that the product of 3 sequential numbers is divisible by 3 [duplicate] Ask Question Asked 5 years, 9 months ago Modified5 years, 9 months ago Viewed 4k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. This question already has answers here: Among any three consecutive positive integers one is a multiple of 3 (6 answers) n n divides a product of n n consecutive integers (by n n divides one of them) (4 answers) Prove that one of n n consecutive integers must be divisible by n n (5 answers) Closed 5 years ago. I am starting to learn about proofs and I ran into this problem: Prove that the product of 3 sequential numbers is divisible by 3 I am going to present my thoughts on how to prove that and any feedback about whether it is wrong or not would be very appreciated. Thank you in advance Lets fisrt prove that n³-n is divisible by 3 using Induction: Constraints n ∈ Z - {-1, 0, 1} m ∈ Z When n = 2, n³ - n = 6 ->Proved for the base case n = k Then k³ - k = 3m Lets proof for (k + 1) (k + 1)³ - (k + 1) = k³ - k + 3k² + 3k = 3m + 3k² + 3k = 3(k² + k + m) -> Proved Now, lets prove the conjecture of this question by deduction, which is: Prove that the product of 3 sequential numbers is divisible by 3 (n - 1)(n)(n + 1) = (n² - n)(n + 1) = n³ - n Because It was previously proved that n³ - n is in fact divisible by 3, then it is proved as well elementary-number-theory induction solution-verification Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Dec 25, 2019 at 22:16 Matheus Simões MinghiniMatheus Simões Minghini 157 3 3 silver badges 13 13 bronze badges 5 n 3≡n(mod 3)n 3≡n(mod 3) by Fermat’s little theoremJ. W. Tanner –J. W. Tanner 2019-12-25 22:20:41 +00:00 Commented Dec 25, 2019 at 22:20 @J.W.Tanner, thanks for your reply, I am not familiar with modular arithmetic notation, would you be so kind as to explain to me what the sentence above means?Matheus Simões Minghini –Matheus Simões Minghini 2019-12-25 22:29:17 +00:00 Commented Dec 25, 2019 at 22:29 It means 3 3 divides n 3−n n 3−n; according to Fermat’s little theorem, p p divides n p−n n p−n for p p prime J. W. Tanner –J. W. Tanner 2019-12-25 22:30:51 +00:00 Commented Dec 25, 2019 at 22:30 Awesome, I did not know that, massive thank you!Matheus Simões Minghini –Matheus Simões Minghini 2019-12-25 22:32:05 +00:00 Commented Dec 25, 2019 at 22:32 You’re welcome; it’s good to know J. W. Tanner –J. W. Tanner 2019-12-25 22:32:22 +00:00 Commented Dec 25, 2019 at 22:32 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Yeah, it is ok, but you can cut it down drastically. Notice that among 3 3 consecutive numbers, exactly one is divisible by 3 3 (look at their remainder when divided by 3 3). So their product is divisible by 3 3, and you are done. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 25, 2019 at 22:21 ViHdzP 4,884 2 2 gold badges 20 20 silver badges 51 51 bronze badges answered Dec 25, 2019 at 22:19 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges 2 Aqua, thank you for your reply! I noticed that I was kinda overworking the problem. But for the sake of someone who is learning those things, would you say I am on the right path? any advices?Matheus Simões Minghini –Matheus Simões Minghini 2019-12-25 22:34:03 +00:00 Commented Dec 25, 2019 at 22:34 Thank you, I really appreciate it :D Matheus Simões Minghini –Matheus Simões Minghini 2019-12-25 22:40:16 +00:00 Commented Dec 25, 2019 at 22:40 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Since the OP is "starting to learn about proofs" it is worthwhile to expand Aqua's answer with more detail. Let m m be any integer. Using Euclidean_division, we can write m=3 q+r where r∈{0,1,2}(1)(1)m=3 q+r where r∈{0,1,2} We want to show that n=m(m+1)(m+2)(2)(2)n=m(m+1)(m+2) is divisible by 3 3. If any one of the three factors on the rhs of (2)(2) is divisible by 3 3, then n n is divisible by 3 3. Case 1: r = 0. Then m=3 q m=3 q is divisible by 3 3. Case 2: r = 1. Then m+2=(3 q+1)+2=3(q+1)m+2=(3 q+1)+2=3(q+1) is divisible by 3 3. Case 3: r = 2. Then m+1=(3 q+2)+1=3(q+1)m+1=(3 q+2)+1=3(q+1) is divisible by 3 3. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 26, 2019 at 0:36 CopyPasteItCopyPasteIt 11.9k 1 1 gold badge 26 26 silver badges 47 47 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let 3 sequential integer numbers be k-1,k&k+1 Now, k(k-1)(k+1)=k^3-k We know for any integer k={0,1,2}(mod 3) Case 1: if, k=0(mod3) Then, k^3-k=0(mod 3) Case 2:if, k=1(mod 3) ,let k=3n+1 Then, k^3-k=(27n^3+27n^2+9n+1)-(3n+1)=0(mod 3) Case 3:if, k=2(mod 3) ,let k=3n+2 Then,k^3-k=(27n^3+54n^2+36n+8)-(3n+2)=0(mod 3) We see in each case the product is divisible by 3. So,we're done and hence proved. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 25, 2019 at 22:53 answered Dec 25, 2019 at 22:42 pankaj kumarpankaj kumar 139 1 1 silver badge 9 9 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory induction solution-verification See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 7Prove that one of n n consecutive integers must be divisible by n n 1Among any three consecutive positive integers one is a multiple of 3 3n n divides a product of n n consecutive integers (by n n divides one of them) Related 2Complete induction proof that every n>1 n>1 can be written as a product of primes 2What is the differences between proving by Deduction and Induction 0Proof ∀n∈N∀n∈N, that 9\|10 n−1 9\|10 n−1 by mathematical induction 0How much Base Proof is required? 3Induction Proof for the Sum of the First n Fibonacci Numbers Hot Network Questions Passengers on a flight vote on the destination, "It's democracy!" 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189160	https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2019-20/06c_fin_ab_Fourier.pdf	(September 21, 2019) Fourier analysis on ﬁnite abelian groups Paul Garrett garrett@math.umn.edu http: / /www.math.umn.edu/egarrett/ [This document is http: / /www.math.umn.edu/egarrett/m/mfms/notes 2019-20/06c ﬁn ab Fourier.pdf] 1. Fourier analysis on ﬁnite abelian groups 2. Appendix: spectral theorem for unitary operators 3. Appendix: cancellation lemma There are Fourier expansions on ﬁnite abelian groups essentially identical in form to Fourier expansions of periodic functions on the real line. This follows from the spectral theory of unitary operators on ﬁnite-dimensional complex vector spaces. 1. Fourier analysis on ﬁnite abelian groups The main conclusion, existence of Fourer expansions on ﬁnite abelian groups, can be given independently of most of the ideas of the proofs. Let G be a ﬁnite abelian group, and L2(G) the complex vectorspace of complex-valued functions on G, with inner product ⟨f, ϕ⟩= X x∈G f(x) ϕ(x) A character χ on a group is a group homomorphism χ : G − →C× Let b G be the collection of characters χ : G →C×. For f a complex-valued function on G, the Fourier transform b f of f is the function on b G deﬁned by b f(χ) = ⟨f, χ⟩ (for χ ∈b G) The Fourier expansion or Fourier series of f is f ∼ 1 \|G\| X χ∈b G b f(χ) χ In general, for a space X with some sort of integral on it, the notation L2(X) means functions f so that R X \|f\|2 < ∞. On ﬁnite sets integrals become sums, possibly weighted, and this ﬁniteness condition becomes vacuous. Nevertheless, it is good to use this notation as a reminder of the larger context. The notation L2(G) is meant to suggest the presence of the inner product on this space of functions. On a general space X with an integral, the iner product is ⟨f1, f2⟩= R X f1f 2. The term character has diﬀerent meanings in diﬀerent contexts. The simplest sense is a group homomorphism to C×. However, an equally important use is for the trace of a group homomorphism ρ : G →GLn(k) from G to invertible n-by-n matrices with entries in a ﬁeld k. In the latter sense, (character of ρ)(g) = trace ρ(g) For inﬁnite-dimensional representations, further complications appear. Except from context, there is no way to know which sense is intended. 1 Paul Garrett: Fourier analysis on ﬁnite abelian groups (September 21, 2019) [1.1] Theorem: On a ﬁnite abelian group, the Fourier expansion of a complex-valued function f represents f, in the sense that, for every g ∈G, f(g) = 1 \|G\| X χ∈b G b f(χ) χ(g) The elements of b G form an orthogonal basis for L2(G). In particular, the Fourier coeﬃcients are unique. [1.2] Remark: What are we not doing? The theorem asserts nothing directly about the collection b G of characters of G. Its proof uses no information about these characters. Its proof uses nothing about the structure theorem for ﬁnite abelian groups. All that is used is a spectral theorem. The proof is in the following paragraphs. [1.3] Translation action on functions The distinguishing feature of functions on a group is that the group acts on itself by right or left multiplication (or whatever the group operation is called), thereby moving around the functions on it. The group operation in G will be written multiplicatively, not additively, to ﬁt better with other notational conventions. The group G acts on the vector space L2(G) of functions on itself by translation: for g ∈G, the translate Tgf of a function f by g is the function on G deﬁned by (Tgf)(x) = f(xg) (for function f, and x, g ∈G) The maps-on-function Tg are vectorspace endomorphisms of the vectorspace of functions on G:    Tg(f1 + f2)(x) = (f1 + f2)(xg) = f1(xg) + f2(xg) = Tgf1(x) + Tgf2(x) (additivity) Tg(c · f)(x) = (c · f)(gx) = c f(gx) = c · (Tgf) (x) (scalar c) To reduce clutter, the action of g ∈G on functions f may be written simply gf or g · f. The associativity property (gh)f = g(hf) (for g, h ∈G, function f) comes from the associativity of the group operation itself: (gh)f (x) = f(x(gh)) = f((xg)h) = (hf)(xg) = g(hf) (x) For non-abelian groups G, there are two translation actions, namely, left and right    T right g f(x) = f(xg) T left g f(x) = f(g−1x) The inverse in the left translation is for associativity T left gh f = Tg Thf For abelian groups, the two translation actions become essentially the same thing, insofar as either alone gives all the information that the two together could give. Also, for abelian groups, the inverse in the deﬁnition of the left translation action loses some of its signiﬁcance, since for abelian groups g →g−1 is a group automorphism. 2 Paul Garrett: Fourier analysis on ﬁnite abelian groups (September 21, 2019) The associativity property is equivalent to the assertion that the map g →Tg is a group homomorphism from G to C-linear automorphisms of L2(G) (and that the identity element of g acts trivially). Since g →Tg is a group homomorphism, the abelian-ness of G implies that the linear maps Tg, Th commute: since gh = hg, Tg ◦Th = Tgh = Thg = Th ◦Tg (for all g, h ∈G) Since G is ﬁnite, there is a positive integer N such that, for all g ∈G, gN = e ∈G. Thus, χ(g)N = χ(gN) = χ(e) = 1 (for any χ ∈b G) That is, the values of χ lie on the unit circle in C×, so \|χ(g)\| = 1. In particular, χ is unitary in the sense that χ(g)−1 = χ(g) We claim that the linear operators Tg are also unitary, in the sense that ⟨Tgf, TgF⟩= ⟨f, F⟩ (for g ∈G, functions f, F) To prove this, compute directly: ⟨Tgf, TgF⟩= X h∈G (Tgf)(h) (TgF)(h) = X h∈G f(hg) F(hg) Change variables in the sum, by replacing h by hg−1. Here the fact that G is a group is used: g−1 exists, and is closed under the group law: X h∈G f(hg) F(hg) = X h∈G f(h) F(h) = ⟨f, F⟩ proving the unitarity. For a single linear operator T on a complex vector space V , and for a complex number λ, the λ-eigenspace Vλ of T on V is Vλ = {v ∈V : Tv = λ · v} The Spectral Theorem for a single unitary operator T on a ﬁnite-dimensional complex vector space with inner product ⟨, ⟩asserts that V decomposes as an orthogonal direct sum of eigenspaces of T: V = M λ Vλ (orthogonal direct sum) We claim that another unitary operator S commuting with T stabilizes the T-eigenspaces Vλ. To see this, take v ∈Vλ: T(Sv) = (TS)v = (ST)v = S(Tv) = S(λv) = λ · Sv since the linearity of S implies that S commutes with scalar multiplication. This sets up an induction, as follows. We want to prove that a group H of commuting unitary operators on a ﬁnite-dimensional complex vectorspace V with hermitian inner product ⟨, ⟩has simultaneous eigenspaces Vλ on V giving an orthogonal direct sum decomposition V = L λ Vλ. In this situation, the notion of eigenvalue must be a little more complicated than individual numbers: for each T ∈H, there must be a number λT ∈C. That is, an eigenvalue is really a map T →λT from H to C. 3 Paul Garrett: Fourier analysis on ﬁnite abelian groups (September 21, 2019) In this context, for two eigenvalues λ, µ to be distinct means that λT ̸= µT for some T ∈H (not necessarily for all T ∈H). Now we do the induction. Suppose we have the conclusion for vector spaces of dimension < n. Let V be of dimension n. First, a silly case: if all operators T ∈H are scalar, then every vector is a simultaneous eigenvector for all the operators in H, and we are done. So now consider the (serious) case that not all operators in H are scalar. Let T ∈H be a non-scalar operator. By the spectral theorem for unitary operators, V has an orthogonal decomposition into eigenspaces for T, implicitly with diﬀerent eigenvalues. Since T is non-scalar, every one of these eigenspaces has dimension < n. By induction, and by the fact that the operators all commute, each such eigenspace decomposes as an orthogonal direct sum of simultaneous eigenspaces for H. Thus, the whole space V is an orthogonal direct sum of simultaneous eigenspaces. This completes the induction. Thus, in the case that H is the group of automorphisms Tg with g ∈G and V = L2(G), L2(G) = M λ L2(G)λ (with simultaneous eigenvalues λ : G →C×) In fact, for H an abelian group of unitary automorphisms, we claim that the eigenvalues T →λT for T ∈H are group homomorphisms H →C×: for S, T ∈H, and for v in the λ-eigenspace, λST · v = (ST)(v) = S(T(v)) = S(λT · v) = λT · S(v) = λT · λS · v Thus, λST = λT · λS (for all S, T ∈H, simultaneous eigenvalue λ : H →C×) That is, for groups of automorphisms, eigenvalues are characters. We’ll write Vχ instead of Vλ to emphasize this information, and for L2(G) write L2(G) = M χ∈b G L2(G)χ We will see that each L2(G)χ is one-dimensional, spanned by χ itself. On one hand, every χ ∈b G is a complex-valued function on G, so is in L2(G). In fact, we claim that χ ∈Vχ: (Tgχ)(h) = χ(hg) = χ(h) χ(g) = χ(g) χ(h) (since C× is abelian) On the other hand, we claim that Vχ is exactly scalar multiples C · χ of χ. To see this, let f ∈Vχ. With e the identity element of G, f(g) = f(e · g) = (Tgf)(e) = χ(g) · f(e) = f(e) · χ(g) That is, f = f(e) · χ (for f ∈Vχ) By the orthogonality of Vχ and Vτ for distinct χ, τ, the characters are an orthogonal basis for L2(G). Their lengths are readily determined, using the earlier-noted unitariness χ = χ−1: ⟨χ, χ⟩= X g∈G χ(g) · χ(g) = X g∈G χ(g) · χ(g)−1 = X g∈G 1 = \|G\| Any f ∈L2(G) can be written as a linear combination of orthogogonal basis elements ei f = X i ⟨f, ei⟩· ei ⟨ei, ei⟩ 4 Paul Garrett: Fourier analysis on ﬁnite abelian groups (September 21, 2019) Using the orthogonal basis χ ∈b G, f = X χ∈b G ⟨f, χ⟩· χ ⟨χ, χ⟩ = 1 \|G\| X χ∈b G ⟨f, χ⟩· χ This is an equality of functions on the ﬁnite set G, and b f(χ) is deﬁned to be ⟨f, χ⟩, so f(g) = 1 \|G\| X χ∈b G ⟨f, χ⟩· χ(g) = 1 \|G\| X χ∈b G b f(χ) · χ(g) (for all g ∈G) This proves the representability of functions on ﬁnite abelian groups by their Fourier series. / / / 2. Appendix: spectral theorem for unitary operators Let V be a ﬁnite-dimensional complex vector space with a hermitian inner product ⟨, ⟩. A linear map T : V →V is unitary if it preserves the inner product, in the sense that ⟨Tv, Tw⟩= ⟨v, w⟩ (for all v, w ∈V ) Thus, the adjoint T ∗of a unitary operator T has the property ⟨v, w⟩= ⟨Tv, Tw⟩= ⟨v, T ∗Tw⟩ Subtracting, ⟨v, T ∗Tw −w⟩= 0 for all v, so T ∗Tw = w for all w ∈V . That is, unitary T is invertible, and T ∗= T −1. This also shows that T ∗T = TT ∗. The inverse of a unitary operator is unitary, since ⟨T −1v, T −1w⟩= ⟨T ∗v, T −1w⟩= ⟨v, TT −1w⟩= ⟨v, w⟩ Eigenvalues λ of a unitary operator T are of absolute value 1, since for a λ-eigenvector v λλ⟨v, v⟩= ⟨λv, λv⟩= ⟨Tv, Tv⟩= ⟨v, v⟩ In particular, eigenvalues λ are non-zero, and λ−1 = λ. Given λ ∈C, let λ-eigenspace of T = Vλ = {v ∈V : Tv = λ · v} [2.1] Theorem: The vectorspace is an orthogonal direct sum V = M λ Vλ (eigenspaces of unitary T) Proof: We grant ourselves the more elementary fact that, because V is ﬁnite-dimensional and C is algebraically closed, there is at least one one eigenvalue λ and non-zero eigenvector v for T. Thus, the λ-eigenspace Vλ is not {0}. Now the unitariness is used, to set up an induction on dimension. We claim that T stabilizes the orthogonal complement V ⊥ λ = {w ∈V : ⟨w, v⟩= 0 for all v ∈Vλ} 5 Paul Garrett: Fourier analysis on ﬁnite abelian groups (September 21, 2019) Indeed, for w in that orthogonal complement and v ∈Vλ, ⟨Tw, v⟩= ⟨w, T ∗v⟩= ⟨w, T −1v⟩= ⟨w, λ−1v⟩= λ⟨w, v⟩= 0 (for all v ∈Vλ) The restriction of a unitary operator T to a T-stable subspace is obviously still unitary. By induction on the dimension of the vectorspace, V ⊥ λ is an orthogonal direct sum of T-eigenspaces: V ⊥ λ = L µ V ′ µ. Then V = Vλ ⊕ M µ V ′ µ is the orthogonal direct sum decomposition of the whole space. / / / 3. Appendix: cancellation lemma The orthogonality of distinct characters can be proven directly. Let G be a ﬁnite group, not necessarily abelian. First, we have the cancellation lemma: [3.1] Lemma: For a non-trivial group homomorphism σ : G →C×, X g∈G σ(g) = 0 Proof: Let go ∈G be such that σ(go) ̸= 1. Then X g∈G σ(g) = X g∈G σ(gog) = X g∈G σ(go) σ(g) = σ(go) X g∈G σ(g) by replacing g by gog in the sum, using the fact that left multiplication by go is a bijection of G to itself. Subtracting, (1 −σ(go) · X g∈G σ(g) = 0 Since σ(go) ̸= 1, necessarily the sum is 0. / / / [3.2] Corollary: Let σ ̸= τ be group homomorphisms G →C×. Then X g∈G σ(g) τ(g) = 0 Proof: Since G is ﬁnite, there is N such that gN = e for every g ∈G. Thus, τ(g)N = τ(gN) = τ(e) = 1 Thus, τ(g) is a root of unity, and \|τ(g)\| = 1. In particular, τ(g) = τ(g)−1. Then στ = στ −1 is a character of G, and is not the trivial character. The previous lemma gives the vanishing. / / / 6
189161	https://www.usna.edu/Users/physics/mungan/_files/documents/Publications/TPT57.pdf	tions is no! If the applied pulling force (i.e., the tension in the unrolled portion of the ribbon) is constant for all positive times t, then v = at, where a is the acceleration of the center of mass of the spool, because the spool starts from rest at t = 0. In that case, questions 3 and 5 could equally well be referring to the direction of a rather than v. The geometry in Fig. 2 shows that there is a critical angle θc = cos–1(Ri/Ro) (1) at which the spool cannot roll without slipping, because there is no torque on the spool about the point P at which it contacts the table. Following Hewitt,4 the spool is oriented so that the horizontal component of the pulling force Fx is always right-ward. (In other words, forward is defined to be rightward, and backward is defined to be leftward.) Accordingly, the spool is oriented as shown in Figs. 2–4 for pulling angles in the range 0° < 90°, but the spool is turned around if the angle is instead in the range 90° < 180° as in Fig. 5. (The special case of 90° is excluded because then the pulling force has zero horizontal component.) A simple model for the kinetic frictional force on an ob-ject sliding across a surface is adopted in introductory physics. Its magnitude is given by a constant coeffi-cient of kinetic friction multiplied by the normal force exerted on the object by the surface, and the force points opposite the direction that the object slides relative to the surface. In con-trast, the standard model for the static frictional force is more complicated. There is no formula for its magnitude, but instead it can vary from zero up to some maximum value, as required to keep the point of contact of the object at rest relative to the surface it is on. Furthermore, it points in the direction opposite the way in which the object would slip at its point of contact if there were zero friction, but it is not always obvious which way that is. In an attempt to bypass these complications, students can make erroneous assumptions about the magnitude and direction of the static frictional force.1 To counter them, it is helpful to present some example that shows students that those assumptions need not hold. Such a system is a pulled spool that rolls without slipping on a horizontal table. Since the point of contact between the spool and the table does not slip, any fric-tional force between them must be static.2 Introducing the spool in class Consider a cylindrically symmetric spool that is placed at rest on a table with a ribbon3 wrapped around its inner axle of radius Ri that is smaller than the radius Ro of the outer rim with which the spool contacts the table. The ribbon is gently pulled (so that the spool does not slip) at a variety of different angles relative to the table, where = 0° means the ribbon is unwinding from the bottom of the inner axle, = 90° is when the ribbon is being pulled straight up into the air, and = 180° implies the ribbon is unwinding from the top of the inner ax-le. (The pulling angle is restricted to values between 0° and 180°, because other angles would require that there be a slot in the table to pass the ribbon.) For example, Fig. 1 is a photo-graph of such a spool being pulled at an angle near = 70°. Denote the static frictional force and the normal force exerted on the spool by the table as, respectively, f and N, the coefficient of static friction between the spool and the table as s, and the velocity of the center of mass (coincident with the geometric center O) of the spool as v. The following questions are now presented to students for discussion in small groups, assuming the ribbon is pulled such that the spool rolls without slipping starting from rest, where the horizontal component of the pulling force defines the forward direction: 1. Must f be nonzero? 2. Must f be equal to sN? 3. Must v point in the forward direction? 4. Must f point in the backward direction? 5. Must f point in the direction opposite to v? As will be shown, the answer to all five of these ques-Pulling a Spool Carl E. Mungan, U.S. Naval Academy, Annapolis, MD Fig. 1. Spool being pulled by its ribbon at an angle rela-tive to the tabletop of about q = 70° . Fig. 2. Critical pulling angle qc for a spool of inner radius Ri and outer radius Ro. The static friction f is equal and opposite to Fx up until slipping occurs. In all figures in this article, O denotes the center of the uniform spool, P is the point of contact between the spool and the table, and S is the point at which the ribbon unwinds from the inner axle. 178 THE PHYSICS TEACHER t Vol. 61, March 2023 DOI: 10.1119/5.0042450 Directions of rolling and of the frictional force for gentle pulling Which way does the spool roll if the ribbon is gently pulled at angle ? To put it another way, what is the sign of the accel-eration ax of the center of mass of the spool when the pulling force has a positive horizontal component Fx? The answer is that ax is positive (so that the spool rolls in the direction it is pulled) if either 0° < c as in Fig. 3 or if 90° < 180° as in Fig. 5. However, if c < < 90°, then ax is negative (so that the spool rolls opposite to the direction the ribbon is pulled) as in Fig. 4. Those results can be quickly demonstrated to students using a spool like the one shown in Fig. 1. Another interesting feature of a pulled spool (which can also be demonstrated5 but not as easily) is that, referring to Fig. 5, the frictional force f is usually not opposite in direc-tion to the horizontal component Fx of the applied force when > 90°. These assertions are proven as follows. Consider the free-body diagrams (all of which omit the vertical normal and gravitational forces) first for the case of < 90°as in Figs. 2–4. Newton’s second law for horizontal translations of the center of mass becomes F cos – f = Max , (2) where M is the mass of the spool. Likewise, the angular form of Newton’s second law for clockwise rotations about the cen-ter of mass (without slipping relative to the table) is Rof – RiF = Iax/ Ro, (3) where I is the moment of inertia of the spool (about its center O). Eliminating f between Eqs. (2) and (3) leads to (4) using Eq. (1). Equation (4) implies that • ax > 0, so the spool rolls rightward in Fig. 3 for 0° < c, • ax = 0, so the spool does not roll in Fig. 2 when = c, and • ax < 0, so the spool rolls leftward in Fig. 4 for c < < 90°. Second, if > 90°, then because the spool in Fig. 5 was turned around, Eq. (2) becomes –F cos + f = Max , (5) since cos < 0. Likewise, Eq. (3) has both signs flipped on the left side and is now –Rof + RiF = Iax/Ro. (6) Eliminating f between Eqs. (5) and (6) implies that Eq. (4) changes sign to (7) which shows that ax > 0, so the spool rolls rightward in Fig. 5 for 90° < 180°. On the other hand, eliminating ax between either Eqs. (2) and (3) or Eqs. (5) and (6) results in Fig. 3. Spool pulled with force F directed at any small angle in the range 0° £ q < qc. The static friction f causes the spool to rotate clockwise around O. Fig. 4. Spool pulled with force F directed at any interme-diate angle in the range qc < q < 90°. The static friction f causes the center of the spool to translate leftward. Fig. 5. Spool pulled with force F directed at any large angle in the range 90° < q £ 180° (i.e., when point S is vertically above point O). This diagram assumes the static friction f points rightward, so that its torque about point O opposes that due to F. A formal analy-sis shows, however, that f is instead directed leftward for large enough values of the moment of inertia of the spool, but only for pulling angles in the range qm < q £ 180°. THE PHYSICS TEACHER t Vol. 61, March 2023 179 180 THE PHYSICS TEACHER t Vol. 61, March 2023 mass O of the spool or the rotations of the spool about O as follows: • at small angles 0° < c, the friction must point back-ward because there must be a clockwise net torque to cause the spool to roll rightward starting from rest; • at = c, the frictional force must point backward (equal and opposite to the horizontal component of the pulling force) so that the spool does not move; • at intermediate angles c < < 90°, the friction must point backward because there must be a backward net force to cause the spool to start rolling leftward; • at large pulling angles 90° < 180°, if the spool’s mo-ment of inertia I is smaller than MRiRo (as would typi-cally be the case), then the friction must point forward to keep the net torque about point O small; and, finally, • if I > MRiRo (such as for a specially designed spool with most of its mass near its outer rim), then the frictional force points forward for pulling angles in the range 90° < < m but f points backward for angles in the range m < 180° (and f = 0 when the pulling angle is m). Conclusion To recap the answers to the five set-off questions asked near the beginning of this article for gentle pulling on the string: 1. By adding enough mass near the outer circumference of a spool, one can ensure that m exists and lies between 90° and 180° in value.5 When pulled at that angle, the spool will roll without slipping even though the frictional force on the spool is zero! The spool could be on an arbitrarily slippery table or levitated on an air table and it would roll such that the hori-zontal component of the instantaneous velocity of its bottom point P is zero relative to the table. 2. The frictional force is given by Eq. (8). It is assumed that the pulling force F is small enough that \|f \| sN, but they are not equal to each other except in the limiting case when the spool is on the verge of slipping. 3. The acceleration and hence the velocity of the center of mass of the spool points in the backward direction for pulling angles in the range c < < 90°, as illustrated in Fig. 4. That is, a forward pulling force results in backward translational ac-celeration of the spool! 4. As Fig. 5 shows, the frictional force is directed forward for pulling angles in the range 90° < < m if m exists, or in the range 90° < 180° if m is undefined (as is the case for most ordinary spools). The frictional force need not oppose the pulling force!8 5. The frictional force points in the same direction as the acceleration and hence the velocity of the center of mass for pulling angles in the range c < < 90° as sketched in Fig. 4. In (8) for all pulling angles. Since MRiRo/I < MR2 o /I and \|cos \| 1, the magnitude of the frictional force is always smaller than the magnitude of the pulling force. (However, the magnitude of the frictional force must be larger than the x-component of the ap-plied force for pulling angles in the range c < < 90°, because the spool rolls leftward in Fig. 4.) Equation (8) implies that f must be positive for < 90°, and thus, the direction of the fric-tional force f is necessarily leftward in Figs. 2–4. On the other hand, cos < 0 in Fig. 5, so the frictional force can point in either direction, depending on the relative magnitudes of the two terms in the numerator of Eq. (8). However, the second term can be smaller than \|cos \| only if I > MRiRo. Although that inequality will not be satisfied for most typical spools (which have I MRi 2 ), the moment of inertia can be as large as MRo 2 (or even larger if the spool has “outriggers” and rolls on a special track6). For unusual spools having I > MRiRo, the frictional force is zero at the angle m > 90° such that cos m = –MRiRo/I. In particular, if the radius of gyration of the spool is equal to the geometric average of its inner and outer radii, then m = 180°, and the spool with the ribbon unwinding from the top point can roll without slipping regardless of how small the coeffi-cient of static friction is! If such a spool is pulled at 180° start-ing from rest, it will roll at any velocity and acceleration;7 the friction will be zero because the tension alone provides the re-quired ratio of force to torque. For example, take a thin-walled pipe so that I = MR2 with Ri = Ro R , wrap a strip of paper around it, and pull on the strip’s end unwinding from the top of the pipe. More generally, with reference to Fig. 5, if I > MRiRo, then f points rightward for 90° < m but left-ward for m < 180° when f is negative in Eq. (8). On the other hand, if I < MRiRo so that m is undefined, then the fric-tion force f points rightward for all pulling angles in the range 90° < 180°. To summarize, define the x-axis to point in the direction of the horizontal component of the applied pulling force F, which is called the forward direction. Then the direction of rolling can be found by considering the torque due to F about the point of contact P between the spool and table, because the frictional force f has zero torque about that point. Conse-quently, one finds that • the spool rolls forward for pulling angles in the range 0° < c , • the spool cannot roll without slipping when the pulling angle equals the critical angle = c because then F also has zero torque about point P, • the spool rolls backward for angles in the range c < < 90°, and • the spool rolls forward for angles in the range 90° < 180°. On the other hand, to find the direction of the static fric-tional force, one can examine the translations of the center of THE PHYSICS TEACHER t Vol. 61, March 2023 181 addition, since the spool in Fig. 5 always rolls forward, f points in the same direction as a and v for the angular range 90° < < m if m exists, or 90° < 180° if m is undefined. The frictional force need not oppose the translational motion either! The online Appendix9 explains what happens if the string is pulled hard rather than gently. Acknowledgments Thanks to Greg Veteto for taking the photograph in Fig. 1, and to the Chesapeake Section of AAPT for feedback on this content at the Spring 2022 meeting at Radford University. References 1. P. S. Carvalho and A. S. e Sousa, “Rotation in secondary school: Teaching the effects of frictional force,” Phys. Educ. 40, 257–265 (2005). 2. Both the spool and the table are assumed not to deform nor to collide inelastically at their points of contact, and there is no vertical adhesion between the outer rim of the spool and the surface of the table. In that case, the spool is not subject to roll-ing friction as discussed in R. Cross, “Origins of rolling friction,” Phys. Educ. 52, 055001:1–4 (2017). Air drag on the spool is also neglected. 3. A ribbon prevents the spool from twisting laterally out of a one-dimensional line of motion along the tabletop. 4. P. G. Hewitt, “December Figuring Physics answer,” Phys. Teach. 59, 67 (2021). 5. C. E. Mungan, “Acceleration of a pulled spool,” Phys. Teach. 39, 481–485 (2001). 6. M. Mears, “The spiderwheel: A new apparatus to demonstrate energy conservation and moment of inertia,” Am. J. Phys. 83, 817–820 (2015). 7. A spool pulled at 180° cannot lift off the table because the nor-mal force always balances the weight. 8. A. Pinto and M. Fiolhais, “Rolling cylinder on a horizontal plane,” Phys. Educ. 36, 250–254 (2001). 9. Readers can access the Appendix at TPT Online at org/10.1119/5.0042450, under the Supplemental tab. Carl Mungan is visiting the U.S. Air Force Academy and teaching laboratory techniques during the spring 2023 semester. mungan@usna.edu 1 Online Appendix for “Pulling a Spool” The main text discusses what happens if the spool is pulled gently, assuming the coefficient of static friction µs between the spool and table is nonzero. Namely, the spool rolls without slipping for any pulling angle other than θc; at that critical angle the spool cannot do so and instead it remains at rest (neither rotating nor translating) if the tension in the string is small. Now consider what happens if the applied force is gradually increased in strength until it is no longer “gentle.” The first change in behavior is that the spool will begin to slip1 for any pulling angle other than θm; if θm is defined and is smaller than 180° then at that pulling angle the first change in behavior will instead be that the spool will lift off the table (since the static friction force is zero at θm and thus it cannot attain its slipping value prior to lift off). If the magnitude of F continues to increase, then the final change in behavior is that the spool will lift off the table for any pulling angle other than 0° or 180°; at those two angles the applied force has no vertical component and so the spool cannot be lifted upward. Add the two missing forces to any of Figs. 2−5 in the main text: the normal force N vertically upward and the gravitational force Mg vertically downward. Then the vertical force balance becomes sin sin N F Mg N Mg F θ θ + = ⇒ = − . (A1) This result shows that N must monotonically decrease as F is increased starting from zero for any pulling angle other than 0° or 180°, and at the latter two angles N will be constantly equal to Mg for any value of F. On the other hand, we see from Eq. (8) that the absolute value of the static friction f must monotonically increase as F is increased starting from zero for any pulling angle other than θm. Consequently, f must eventually become equal to sN µ as the applied force is increased starting from zero, for any pulling angle other than θm. Thus, the first thing to happen is that the spool will begin to slip at any angle other than that, as claimed above. The second thing to happen is that if the value of F contines to increase, the spool will lift off the table when the normal force becomes zero,2 which according to Eq. (A1) occurs when sin Mg F θ = . (A2) This lifting value of F is well defined for all pulling angles other than 0° or 180°, as also claimed above. As an illustration of these ideas, consider the specific example of a spool having an inner radius of i 5 cm R = , an outer radius of o 10 cm R = , a mass of 1 kg M = , and coefficients of static and kinetic friction with the table of s 1 µ = and k 0.9 µ = , respectively. Suppose the string is pulled at the critical angle of c 60 θ = °. Then a plot of the applied versus the frictional force starting from zero is shown in Fig. A1. Equation (7) implies that 0 x a = , so that Eq. (5) or (6) gives c sec F f θ = , which means that section A of the plot is a straight line that ends when the static friction attains its maximum s max f whose value from Eq. (A1) is 2 s s max s s max c s max s c ( tan ) 3.59 N 1 tan Mg f Mg f f µ µ θ µ θ = − ⇒ = ≈ + (A3) using 9.8 N/kg g = . To rebalance the horizontal force components once slipping starts, one must immediately decrease the pulling force slightly by backing up along section B of the plot because the maximum kinetic friction is likewise equal to k k max k c 3.45 N 1 tan Mg f µ µ θ = ≈ + (A4) when 0 x a = . Equation (A4) gives the maximum value of the kinetic friction, because as one thereafter increases the pulling force, the kinetic friction decreases according to k k c ( sin ) f Mg F µ θ = − (A5) from Eq. (A1). Equation (A5) describes section C of the plot in Fig. A1 beginning at an applied force of k max c sec 6.89 N F f θ = ≈ and ending at the lifting value of the tension, which Eq. (A2) gives as lift c csc 11.32 N F Mg θ = ≈ (A6) at which point fk has decreased in value to zero because N is then zero. Endnotes 1. At a pulling angle of θc, the spool can slip in place. At any other pulling angle, if the spool slips it must do so while continuing to translate which is often described as “rolling with slipping.” 2. The normal force N has to decrease to the value 1 s \| \| f µ− for the spool to start slipping; for it to lift off, N must further decrease to zero in value. 3 Figure A1. Pulling force versus the frictional force at the critical angle. Starting from the origin, the static friction prevents the spool from moving and we rise up along line A until the maximum value of the static friction fs max is attained. Since the coefficient of kinetic friction is a bit smaller than the coefficient of static friction, we must immediately reduce the applied force by backing up along line B once slipping starts. At the maximum value of the kinetic friction fk max the spool slips in place. From that point, if we now increase the pulling force, we will progressively reduce the normal force and hence the kinetic friction (on the now translating spool) along line C until we lift the spool off the table as the friction force falls to zero. 0 3 6 9 12 0 1 2 3 tension F (N) friction f (N) A B C Flift fs max fk max
189162	https://www.youtube.com/watch?v=sOOzpxBF7lk	Rotated Ellipse. Find Equation Of Tangent Line At Given Point Using Implicit Differentiation Ms Shaws Math Class 51100 subscribers 10 likes Description 1149 views Posted: 30 May 2017 2 comments Transcript: Hi everyone. We're going to find the equation of the tangent line at a given point using implicit differentiation. This is a rotated ellipse. So we're going to just go ahead and get started. This would be 14x. And then we have a product thing going on here. So going to write down um the first product, the first part function times the derivative of the y And then we're going to have another minus here. You write y times dative of the x. So that's going to be a square whoops 6 3 times your y. Then we take the derivative of this piece. That's going to be 26 y times the derivative of the y - 0 = 0. Now, what I'm going to do, we can just get rid of that. Um, is group my derivative parts with the y primes in them together. So, that's going to be 26 y y derivative of y - 6 3x times the derivative of y. And I'm going to move um these two over. And so that's going to be um 6 3 y - uh 14x. Now I'm going to factor out the derivative of y. We're just factoring it out. We're not dividing it. So 26 y - 6 3x. Now we divide both sides by um this quantity here. So our derivative = 6 3 y - 14x over 26 y - 6 3x. Um if I want to simplify that, I could, but all I'm going to do is plug in my x's and y's right in here. So, we're going to have six. If I plug all that in, I get 6 3 y -4 the square of 3. And then I have 26 - 6 3 3. So this uh the numerator is going to be -8 3 these two equ= 3. So 26 - 8 is just 8. So my slope is going to be 3. Now we're going to go ahead and find the equation of the tangent line. So I'm going write y = 3x + b because that's my slope. My point again was u 3, 1. And just plug those in. So you get 1 = 3 the 3 um + b. So this is going to be 1 = 3 + b. So b = 4. So the equation of the tangent line is y = um 3 x 3x + 4. Now don't laugh at my graph, but this is the rotated ellipse. This is the point. Um my y intercept is four. So if I choose this would be three. This would be four. And so the tangent line would be right in there on the rotated ellipse. And that's all. Thank you. Have a nice day. Bye bye.
189163	http://philsci-archive.pitt.edu/16104/1/A%20Case%20Study%20in%20Polyhedra%20-%20Synthese%20preprint.pdf	Synthese manuscript No. (will be inserted by the editor) Word Choice in Mathematical Practice: A Case Study in Polyhedra Lowell Abrams,⋆Landon D. C. Elkind⋆⋆ Accepted 2019-06-09, forthcoming Abstract We examine the inﬂuence of word choices on mathematical practice, i.e. in developing def-initions, theorems, and proofs. As a case study, we consider Euclid’s and Euler’s word choices in their inﬂuential development and, in particular, their use of the term ‘polyhedron’. Then, jumping to the 20th century, we look at word choices surrounding the use of the term ‘polyhedron’ in the work of Coxeter and of Gr¨ unbaum. We also consider a recent and explicit conﬂict of approach between Gr¨ unbaum and Shephard on the one hand and that of Hilton and Pedersen on the other, elucidating that the conﬂict was engendered by disagreement over the proper conceptualization, and so also the appropriate word choices, in the study of polyhedra. Keywords mathematical practice, word choice, polyhedron, history of mathematics, philosophy of mathematics 1 Introduction Mathematics is typically developed and conveyed in language. Most mathematicians know they must choose terms and formulate both deﬁnitions and theorems carefully so as to accurately communicate their conceptualization of a mathematical subject-matter. What is less clear is the extent to which this reliance on language inﬂuences the actual development of a subject-matter in mathematical practice. As a case study in the inﬂuence of language on mathematical practice, we examine the term ‘polyhe-dron’ and consider how the word choices one uses in developing a mathematical subject-matter can (and do) embody distinct (and sometimes divergent) conceptualizations. We take the case of word choices used in the history and development of ‘polyhedron’ as a fascinating exemplar of the non-trivial role language plays in the development of mathematical subject-matter – a role that we note it necessarily plays so long as language is the primary means of one’s mathematical practice. Indeed, drawing the conclusions of his study of a programmatic eﬀort to express mathematics in Maori, Barton tells us that mathematics arises after, not before, human activity. The development of mathematical language is consistent with the idea that mathematical concepts, objects, and relationships arise through language, and within particular sociocultural environments, in response to human thinking about quantity, relationships, and space. [1, p. 88] Our analysis of word choice in the context of the development of the notion of polyhedron will in fact reveal diﬀerent mathematicians emphasizing the perspectives of quantity, relationships, and space diﬀerently. ⋆George Washington University, University Writing Program and Department of Mathematics, Ames Hall, Room 207, 2100 Foxhall Road NW, Washington, DC 20007, labrams@gwu.edu ⋆⋆University of Iowa, Department of Philosophy, English-Philosophy Building, Room 570, 251 West Iowa Avenue, Iowa City, IA 52242, ORCid 0000-0003-0513-2937, landon-elkind@uiowa.edu Address(es) of author(s) should be given We focus on word choice as one aspect of language’s non-trivial inﬂuence on mathematical practice. We do not think, and do not claim, that this is the only way in which language non-trivially inﬂuences mathematical practice. Lakoﬀand Nu˜ nez, for instance, have developed an extensive theory explaining the ways in which metaphor has played, and continues to play, a critical role in the development of mathematics by providing the means to concretize, at least linguistically, the abstract. Of course, nonlinguistic factors also inﬂuence mathematical cognition and thought processes. Mani-festly, mathematics, like other institutions, has a history and culture that powerfully shape the direction of mathematical research. Determining to what extent each of these inﬂuences mathematical develop-ments, and even disentangling the distinct features of mathematical practice, is not our present concern. As such, we make no conjecture as to if and how word choice determines mathematical developments. Our case study is presented as neutral between competing global accounts of mathematical practice: it is even consistent with the data examined in our case study of word choice that other aspects of mathematical practice explain the inﬂuence of word choices in developing a mathematical subject-matter. That said, even without embracing here a taxonomy of the means through which mathematical prac-tice shapes mathematics, we claim what is supported by our case study of ‘polyhedra’, that word choice plays a detectable and signiﬁcant role in mathematical developments. The denial that word choice is inﬂuential in mathematics is inconsistent with the data examined in this case study. With this exception, various accounts of what inﬂuences word choice, and even divergent global accounts of mathematical practice, are consistent with the data presented here: we show merely that word choice inﬂuences math-ematics. For all that, word choice as a critical feature of mathematical practice is understudied.1 Hence the need for our case study of ‘polyhedra’ supporting that word choice inﬂuences mathematical practice. This claim, however, needs some clariﬁcation before its truth-conditions are comprehensible. To that end, we ﬁrst explain (although do not deﬁne) some vital terms in our case study – word choice, conceptualization, encoding, mathematical subject-matter, and inﬂuencing.2 By word choice, we mean the (written) words used to develop a mathematical subject-matter.3 For example, in presenting set theory one might describe sets as ‘containing’ their elements, or one might use ‘including’ them. These are two diﬀerent word choices that one might use to develop set theory. Word choices have the virtue that they are observable: they occur as concrete objects on a page. Authorial intent, in contrast, is not observable and word choices may be a misleading guide to authorial intent. Indeed, word choices can misrepresent authorial intent. For example, a mathematician may not intend to communicate about sets as spatial in any way. But a reader may (mis)understand their talk of sets ‘containing’ elements as a spatial metaphor. Because authorial intent can be imperfectly reﬂected in word choice, here we avoid consideration or reconstruction of (unobservable) authorial intent, we prefer to speak instead of conceptualizations of mathematical subject-matter encoded by word choice. By a conceptualization of a mathematical subject-matter, we mean an understanding of some math-ematics. The nature of mathematical understanding is controversial. For example, one might view un-derstanding as a mental content one masters and has ‘in their head’, or in contrast as a social, cultural, and linguistic practice in which one trains and engages. We do not propose a deﬁnition of mathematical understanding here: we keep our case study neutral as between divergent accounts of it. This much is assumed about mathematical understanding in our case study: an understanding of mathematics is, at minimum, the kind of thing an author, whether intentionally or accidentally, imparts to or trains in their reader by communicating a mathematical subject-matter in (usually written) words. Methodologically, we assume that words can communicate a conceptualization of a mathematical subject-matter with varying degrees of success. If words are transparent, so that the conceptualization is clear and distinct, then this is the highest degree of success. But words might communicate a conceptual-ization of a mathematical subject-matter less well, just as ‘containment’ may misleadingly communicate a spatial conceptualization of sets. Assuming that words can communicate a conceptualization more or less well, we can speak of word choices encoding a conceptualization of a mathematical subject-matter. By word choices encoding a conceptualization of a mathematical subject-matter, we mean that word choices communicate some (perhaps more than one) conceptualization of the subject-matter (which 1 A search of “word choice” and “mathematical practice” returns only six articles, all of which are in education studies. A search for “word choice” and “mathematics” returns only 29 works, mostly in education studies and some in linguistics. 2 We thank an anonymous reviewer for inviting us to state and clarify our methodological assumptions. 3 We do not discuss here spoken mathematics or visual mathematics. But we conjecture that our remarks about word choices in written mathematics also apply to those in spoken mathematics and to diagram choices in visual mathematics. 2 can be independent of, in harmony with, or contrary to, what the author intends to communicate). The evidence that word choices do this at all is to be found by considering how the conceptualization encoded by word choice inﬂuences mathematical practice, i.e. the development of a mathematical subject-matter. By a mathematical subject-matter, we mean to include its objects of study, theorems, constructions, examples, proofs, methods, and informal framing remarks.4 It is controversial whether a subject-matter also needs some kind of foundation or internal unity; it is also controversial whether mathematical subject-matters are amenable to sharp delimitation.5 The precise nature of these things – e.g. the ontology of mathematical objects, what makes a string of symbols a proof, and so on – is also controversial. Our case study leaves these issues open. But what we mean by ‘the development of these things’ in a text (e.g. an article or a book) is comprehensible even though the nature of the ‘things’ is controversial. The development of a mathematical subject-matter is what we call mathematical practice. By something’s inﬂuence on mathematical practice, we mean to include how it expands or restricts the range of objects in the domain of inquiry, how it facilitates or hinders the statement or discovery of theorems, and how it makes available or eases constructions or proofs. ‘X inﬂuences mathematical practice’ means X inﬂuences (in the speciﬁed ways or others) the actual development of a mathematical subject-matter. In this case study of ‘polyhedra’, we examine and show how word choice does just this. In Section 2, we review the treatment of polyhedra by Euclid and Euler to illustrate the impact of word choice on the development of mathematical results. In Section 3 we turn to two book-length texts, Regular Polytopes by H. S. M. Coxeter and Convex Polytopes by B. Gr¨ unbaum, to examine the role of language choices in mathematical exposition. We conclude Section 3 on a connection with D. Reed’s work, which highlights an aspect of the connection between mathematics and meta-mathematics, namely how a text delimits a mathematical investigation. In Section 4, we discuss some thoughts from I. Lakatos on formalizing mathematical ideas, then we describe a recent disagreement regarding how to extend Euler’s famous counting formula V −E +F = 2, between Gr¨ unbaum and Shephard on the one hand and Hilton and Pedersen on the other. We show that this conﬂict partly rests on the issues of language and word choice raised in Sections 2 and 3. In Section 5, we draw out some of the lessons of our study. 2 Word Choice and Conceptions of Polyhedra We examine Euclid and Euler (in that order) to show that word choice conveys diﬀerent conceptions of solids and also that diﬀerent notions of an object of study can lead to diﬀerent mathematical results.6 We summarize our ﬁndings at the end of this section. 2.1 Euclid’s Elements We note ﬁrst that in our treatment of Euclid, distinguishing between a motion and an action is important. A motion is a speciﬁed operation upon some object or objects resulting in some change. An action is some change enacted upon an object or objects without detailing explicitly how that change occurs. Both motion and action may involve motion-language, but in Euclid motions necessarily do while actions need not. Falling is an instance of motion. Separating is an example of an action. The former speciﬁes how the end result is achieved – passing through a spatial region – whereas the latter does not. Euclid invokes both motion and action, but as we will see, Euclid’s word choice encodes a conception of polyhedra on which motion-language is critical to his construction of polyhedra.7 Euclid employs motion-language to communicate his mathematics. Consider, for example, his deﬁni-tion of a circle: “A circle is a plane ﬁgure contained by one line such that all the straight lines falling upon it from one point...” [4, Book 1, Deﬁnition 15]8 His deﬁnition of a circle utilizes the motion-language of falling. The conception of geometric objects thereby communicated is one of moving objects about an 4 We do not explain what makes a subject-matter mathematical on the assumption that any plausible account of this would have as a consequence that solid geometry, including the study of polyhedra, is a mathematical subject-matter. 5 Confer [14, Introduction, esp. §2, and Chs 6 and 7]. 6 Diﬀerent notions may also result in the same theorems while developing a subject by distinct paths. 7 We thank Michael Blaustein of St. John’s College for pointing out to Abrams the importance of motion in Euclid’s work. 8 When we quote Euclid, we use the standard Thomas L. Heath translation. Emphasis and brackets come from the original text. We do rely on translations alone in our discussions of Greek and Latin texts. 3 ambient space. Thus, even in examples such as “A circle does not cut a circle at more points than two” [4, Book 3, Proposition 10], the action implicitly relies on motion-language of cutting, so that one circle is said to “cut” another. The conceptualization of geometric objects as given through an ambient space is further evidenced by the deﬁnition of circle invoking containment by a line, as part of a plane ﬁgure, thus conceiving of a circle as what has been spatially surrounded. In contrast, talk about set-theoretic containment, meaning set-membership, does not critically rely on motion-language. Euclid’s ﬁrst postulate similarly utilizes motion-language: “To draw a straight line from any point to any point.” [4, Book 1, Postulate 1] This postulate involves the motion of drawing a line through space between two part-less occupants of that space. Propositions of the Elements also incorporate motion-language, as here in the antecedent of a theorem mid-way through Book 1: “If a straight line falling on the two straight lines...” [4, Book 1, Proposition 27] This ubiquitous conceptualization of geometry as a spatial endeavor grounds the development of Euclid’s geometry, and is manifested in Euclid’s mathematical practice in his word choices. As our guiding interest here is in polyhedra, we skip to the later sections of the Elements to estab-lish that Euclid communicates his solid geometry using spatial metaphors, including motion-language. Consider Euclid’s construction of a tetrahedron: If then, KL remaining ﬁxed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, . . . [4, Book 13, Proposition 13] Motion language cannot be removed from Euclid’s proof because Euclid cannot refer to a single semicircle being carried around if that very semicircle is not being moved. So Euclid’s choice of words, including “carrying” and “passing through,” communicates conceptual and mathematical content necessary for his argument to go through. Without it, his words communicate a gap-ridden proof at best, for without such motion-language, we cannot readily describe the construction that Euclid means to communicate, despite our being able to oﬀer a wholly diﬀerent construction that does not appeal to motion-language. Some of Euclid’s deﬁnitions for his solid geometry also utilize motion-language: The inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up. [4, Book 11, Deﬁnition 5] The completed motions here – “so drawn” and “[so] standing up” – communicate the concept demarcated in Deﬁnition 5 of Book 11. Euclid is not just augmenting the reader’s informal understanding of his deﬁnition; drawing and standing, completed motions, are constitutive of his deﬁnition of the notion of the inclination of a straight line to a plane.9 Euclid again employs motion-language in Proposition 18 of Book 13, “To set out the sides of the ﬁve ﬁgures and to compare them with one another.” The construction begins: Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB. From CD, let CE, DF, be drawn at right angles to AB, and let AF, FB, EB be joined. [4, Book 13, Proposition 18] It is on the basis of motion-language, like “joined”, “set out”, “cut”, and “drawn” that Euclid com-municates his notions. It is through motion-language word choices that Euclid compares the sides of solid ﬁgures. Compare this conceptualization with the modern notion of a metric, which tends not to be communicated in motion-language. Motion-language is also the bridge between Euclid’s two-dimensional geometry and his three-dimensional one; for Euclid deﬁnes and constructs three-dimensional objects as completed motions of two-dimensional objects. Consider Euclid’s deﬁnition of a sphere: When, the diameter of a semicircle remaining ﬁxed, the semicircle is carried round and restored again to the same position from which it began to be moved, the ﬁgure so comprehended is a sphere. [4, Book 11, Deﬁnition 14] 9 As logicians might put it, Euclid’s meta-theoretic language is the language of motion. 4 Completed motions of a semicircle constitute a sphere: for Euclid, a sphere is a ﬁgure enclosed by the completed motion of a semicircle.10 As another example, take Euclid’s construction of a tetrahedron: To construct a pyramid, to comprehend it in a given sphere...Let the diameter AB of a given sphere be set out... from the point H let HK be set up at right angles to the plane...therefore the semicircle described on KL will pass through E also. If then, KL remaining ﬁxed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G . . . [4, Book 13, Proposition 13] All these steps either involve or result from completed motions, as in laying out the diameter or drawing on a line. The construction begins by enclosing a space within a sphere, and the study of the tetrahedron relies on this spatial enclosure. Euclid studies the completed tetrahedron by encompassing it spatially; for him, this mathematical task is a spatial one. In the section that immediately follows Proposition 18’s proof, Euclid proves that only ﬁve regular solids can be constructed.11 For Euclid, this assertion involves the spatial notion of containment: “I say next that no other ﬁgure, besides the said ﬁve ﬁgures, can be constructed which is contained by equilateral and equiangular ﬁgures equal to one another.” [4, Book 13] Euclid’s proof exhausts the possible arrange-ments of two-dimensional components of solid angles. But motion-language occurs here as elsewhere: “...but a solid angle cannot be formed by six equilateral triangles placed together at one point...” [4, Book 13]12 Here the claim is that we cannot “place together” equilateral triangles at a point to form a solid angle. An inability to construct a solid ﬁgure is equated with the impossibility of placing together two-dimensional ﬁgures, and so the proof of Euclid’s capstone result utilizes motion-language. We close this section by noting that Euclid’s motion-language is part-and-parcel of a conception of polyhedra on which a polyhedron is a region of a pre-given (unconstructed) ambient space. Euclid’s motion-language relies on there being such an ambient space to move objects through. For example, his deﬁnition of solid angle presumes that there is a space such that multiple lines drawn through it can consequently contain or bound regions, which are then deﬁned to be solids in the Elements: A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines. Otherwise: A solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point. [4, Book 11, Deﬁnition 11] A solid is that which has length, breadth, and depth... Similar solid ﬁgures are those contained by similar planes equal in multitude... When, the diameter of a semicircle remaining ﬁxed, the semicircle is carried round and restored again to the same position from which it began to be moved, the ﬁgure so comprehended is a sphere. [4, Book 11, Deﬁnitions 1, 9, 14] Euclid deﬁnes the notion of similarity in terms of the ﬁgures that bound the spatial regions that solids are said to be. The solid itself, the spatial region, is constructed (bounded) by a motion that cordons oﬀpart of the ambient space. That is how he deﬁnes the pyramid and the sphere. Euclid’s deﬁnitions of the Platonic solids are likewise given in terms of ﬁgures that, when they are moved about the ambient space, consequently bound a spatial region: A pyramid is a solid ﬁgure, contained by planes, which is constructed from one plane to one point... A cube is a ﬁgure contained by six equal squares. An octahedron is a solid ﬁgure contained by eight equal and equilateral triangles. An icosahedron is a solid ﬁgure contained by twenty equal and equilateral triangles. A dodecahedron is a solid ﬁgure contained by twelve equal, equilateral, and equiangular pentagons [4, Book 11, Deﬁnitions 12 and 25-28]. To summarize, Euclid deﬁnes, studies, and categorizes polyhedra using motion-language. A polyhedron, as Euclid conceives of it, is a region cordoned from the ambient space by the trace of two-dimensional 10 Note that a ﬁgure is any thing “contained by any boundary or boundaries,” another use of spatial containment to communicate Euclid’s geometry [4, Book 1, Deﬁnition 14]. 11 Heath assigns no number to the proposition that only ﬁve regular solids can be constructed, preventing us from using the briefer ‘Proposition 19’. 12 In fact, Euclid proves that only ﬁve regular solids exist by showing that only ﬁve combinations of triangles or pentagons result in a solid angle: “For a solid angle cannot be constructed with two triangles, or indeed planes. With three triangles the angle of the pyramid is constructed... but a solid angle cannot be formed by six equilateral triangles placed together at one point...” [4, 1, p. 480] 5 ﬁgures moved through that space. Note, therefore, that Euclid’s mathematical language and word choice are nonsensical if we distinguish a region of space from the polyhedron itself as he conceives of it. Euclid’s word choice ties his conception of polyhedra into his mathematical practice – his deﬁnitions, theorems, constructions, and proofs. Let us consider other examples of this phenomenon. 2.2 Euler’s “Proof of Some Notable Properties with which Solids Enclosed by Plane Faces are Endowed” In contrast to Euclid, Euler’s conception of polyhedra requires treating polyhedra as bodies with struc-tural features that are separable from the underlying ambient space. Euler writes, “If the solid angle is formed from ﬁve plane angles...” [5, p. 6] Since Euler has chosen words for polyhedra that communicate a distinct relationship between the part and the whole, polyhedra are no longer identiﬁed with a spatial region. Polyhedra, and their parts, can now be considered apart from the underlying ambient space. Euler’s conception of polyhedra can thus accommodate word choices that communicate operations on polyhedra that are nonsensical in Euclid’s framework. In particular, Euler’s polyhedra can be dissected into pieces. More generally, it is now feasible to discuss a polyhedron’s solid angle apart from the ambient space and motion-based manner of construction, unlike on Euclid’s conception. Indeed, Euler explicitly instructs the reader to consider a speciﬁc feature of polyedra apart from the underlying space and the way in which the polyhedron was formed – namely, convex polyhedra with planar faces:13 In Solid Geometry those bodies which are bounded on all sides by plane faces rightly merit ﬁrst consideration, just as rectilinear ﬁgures do in Planar Geometry, or what is properly called Geometry. [5, p. 1] After restricting his inquiry to planar-bounded solids, Euler states his purpose: to found solid geometry on sound ﬁrst principles. Among Euler’s ﬁrst principles is his famous formula for polyhedra: “In every solid enclosed by plane faces the number of solid angles, along with the number of faces, exceeds the number of edges by two.” [5, p. 2] Euler, despite being unable to prove his formula, maintains his “perception” of its truth for all solids [5, p. 2]. As Francese and Richeson note, Euler’s formula in part arises from his coining a word for a poly-hedron’s edge, ‘acies’: “Remarkably, before Euler introduced it, no one had a name for an edge of a polyhedron.” [6, p. 288] Echoing the remarks of Francese-Richeson, we note that this novel word choice foregrounded the features of a polyhedron that are related in Euler’s formula. It was chieﬂy this new and (in Euler’s time) non-standard word choice that spurred a conceptualization of polyhedron that led to a new and landmark development: “The importance of Euler’s use of these three quantities should not be understated. Descartes came very close to stating Euler’s polyhedral formula but was unable to do so because he had not singled out the edge as an important entity.” [6, 288]14 Euler analogizes two dimensions and three dimensions in his proofs, leading him to propose the following method for studying three-dimensional ﬁgures: In Geometry any rectilinear ﬁgure can be ultimately reduced to a triangle by successive division of angles. Likewise, given any solid enclosed by plane faces, I observed that the solid angles can be continuously divided so that, ﬁnally, a triangular pyramid remains. Since a triangular pyramid is the most simple ﬁgure among solids, I perceived that on the basis of its known properties one could generalize to the properties of all solids. [5, p. 2] The idea of dividing – “dissecting” in Euler’s vocabulary – a solid is introduced as an extension to three dimensions of the operation of dividing a planar ﬁgure [5, p. 3]. Note that dividing a polyhedron cannot be done so long as one conceives of polyhedra as Euclid did, in terms of a spatial region bounded by the trace of planar ﬁgures. For Euclid, one cannot divide the spatial region that constitutes the polyhedron, or separate the polyhedron’s structure from its underlying and constitutive cordoned spatial region, without thereby annihilating the polyhedron. In contrast, Euler’s conception of polyhedra, as communicated in his word choice, allows him to manipulate the solid and its pieces apart from the region of the ambient space that Euclid said constituted the polyhedron, thereby allowing Euler to study a polyhedron’s structural features without regard for its manner of construction. 13 Francese and Richeson claim Euler assumed polyhedra to be convex [6, p. 289] [5, p. 1]. 14 In contrast, Francese and Richeson note that there were old and familiar terms for face (hedra) and solid angle (angulus solidus) [6, 288]. 6 Euler rather describes solids as ‘enclosed by plane faces’ and not as constituted by being so-enclosed. To illustrate the distinction, consider cattle in a holding pen: the holding pen is the boundary, but it does not constitute the cattle in any sense of the word. For Euler, the polyhedron is conceived of as independent from its boundary, just as a cow is conceived of as independent from its enclosure.15 The phrase “cut a given solid angle from [a solid]...” lacks meaning unless the polyhedron has been conceived apart from its boundary and solid angles; similarly for the phrase “a part must be separated from the solid in such a way that solid angle O is completely removed but everything else remains...” [5, p. 4] It is Euler’s conception of polyhedra as apart from the underlying space that allows him to conceive of them with parts as distinct from the region of the ambient space that they occupy. Euler does occasionally invoke motion language. In the piece we are discussing, Euler refers to a previous article16 in which he uses “drawing” in his proofs, arguing that since one can draw lines in two dimensions, one can draw planes in three: If straight lines are drawn from any point within a solid to each solid angle, then the solid will be divided into as many pyramids as there are faces, in as much as each face will form the base of a pyramid, while their vertices meet at the point. [5, p. 3] This process of drawing planes divides the polyhedron into multiple solid parts. Euler has a diﬀerent method of decomposition in the article we are presently considering, a method “by which any solid is reduced by successive cutting of its solid angles to triangular pyramids.” [5, p. 3] In the ﬁrst article, Euler draws from a single interior point of a polyhedron to all of its vertices. In the article we treat in depth, Euler traces planes between solid angles [5, pp. 2-3]. Euler claims that, in either case, one arrives at triangular pyramids; the diﬀerence, Euler claims, consists merely in how one eﬀects the decomposition.17 Euler describes the two-dimensional analogue of his three-dimensional method thusly: This operation is similar to that by which any rectilinear ﬁgure is customarily reduced to a triangle by successive cuttings of its angles. For if we have a plane ﬁgure with sides ABCDEFGA, and if the triangle CDE is cut from it by the line CE, the ﬁgure that remains is ABCEFGA, whose number of angles will be less by one. Now if, again, the triangle CFE is cut by the line CF the ﬁgure ABCFGA will remain. If, from this, we next remove the triangle BCF and then the triangle BGF, ﬁnally the triangle ABG will remain [5, p. 3]. Euler’s choice of the word ‘cut’ indicates a speciﬁc operation applied to the ﬁgure. This is reinforced by his repeating the phrase “successive cutting” to communicate his technique: “By successive cuttings of solid angles, I shall reduce all solids enclosed by plane faces, ﬁnally, to triangular pyramids.” [5, p. 4] Euler describes this process as “applying the knife” to particular line segments. [5, p. 4] As indicated by his use of resulting pyramidal pieces to calculate the polyhedron’s volume, Euler’s cutting operation is a decomposition [5, p. 13]. But not all methods of decomposition are logically equivalent. If we interpret Euler’s proof method as allowing the decomposition of the interior to be along arbitrary lines, then his proof contains a gap because his method of decomposing polyhedra into pyramids can generate non-convex solids from convex ones [6, p. 293]. But if Euler only permits planar decompositions of a polyhedron into pyramids, as his word choice of “cuts” and “applying the knife” strongly suggests, then Euler’s proof does arguably preserve convexity as he claims; such a procedure applied to a convex solid always results in another convex solid. His word choice, that is, can close the gap in his proof discussed in [6, p. 292]. So depending on how closely we adhere to the conceptual framework which Euler encodes by his word choices (‘cut’, ‘applying the knife’), we have either a gap-ridden proof or a gap-free proof. So we have a conception of polyhedron, communicated by Euler’s word choice, that is foreign to Euclid and allows for speaking of a polyhedron’s structure as distinct from the underlying space and a motion-based means, or any other means, of constructing it. Not only that, but we have two readings of Euler’s argument based on what his word choice conveys. If Euler allows any division of a polyhedron into pyramids, then his proof is faulty. If he only allows planar cuts that decompose a polyhedron into pyramids, then his proof is arguably correct. Word choice impacts his mathematics, enabling him to 15 For Euclid, it is rather that the object of study is the ambient space marked oﬀby the holding pen. 16 Leonhard Euler, Elementa Doctrinae Solidorum” (“Elements of the Doctrine of Solids”), Novi Commentarii Academiae Scientiarum Petropolitanae 4: pp. 72-93, 1758. 17 As noted in [6, pp. 292-293] and [16, Chapter 3], Euler’s claim is mistaken, at least on a natural reading. 7 conduct studies of solid geometry that Euclid could not. Such consequences of word choice in mathematics indicates how word choice is critical in understanding and interpreting mathematics. To summarize, word choices encode speciﬁc conceptualizations of the underlying objects of math-ematical study, and thus distinct mathematical proofs and theorems arise from distinct word choices. Corresponding methods may even be rendered admissible or inadmissible by one’s word choices. Diﬀerent interpretations of Euler’s formula and Euler’s proof, along with Euler’s conceptual shifts, led to arguments over the legacy of Euler and his polyhedral formula. Some of these arguments revolve around word choice. We discuss these issues further in Section 3; presently, we conclude that Euler conceives of polyhedra as solids with component parts that may be removed or manipulated without destroying the solid, and that Euler’s coining of the Latin word ‘acies’ for ‘edge’ partly spurred the discovery of his famous formula. 3 Word Choice in Mathematical Texts We examine two book-length mathematical texts that illustrate vastly diﬀerent developments of solid geometry and, correspondingly, exemplify vastly diﬀerent modes of interaction between word choice and mathematical exposition. In the same text, one may adhere to a speciﬁc word choice, thereby employing a unifying conceptualization, or one may adroitly shift between word choices and the conceptualiza-tions that they encode. In diﬀerent contexts, with diﬀerent aims or audiences, either approach may be hazardous. Some word choices may snugly suit the desired end of the exposition, or leave the audience bewildered and lost. So there is a balance to be struck between shifting word choices, on the one hand, and correspondingly shifting conceptualizations and adhering to a unifying word choice and developing mathematics within that one corresponding framework on the other. We study both ends on the spectrum to see how a text can advance a conceptualization of polyhedra via word choice. 3.1 Coxeter’s Regular Polytopes H. S. M. Coxeter’s Regular Polytopes is a text in which there is dynamic shifting between word choices and, correspondingly, in conceptualizations of polyhedra; he shifts his word choice to adopt a conceptual orientation that eases the mathematical development at hand, ﬂuidly moving between word choices that embody distinct conceptualizations. Coxeter oﬀers two basic deﬁnitions of ’polyhedron’, one of which we term the “complex-deﬁnition” (here, ‘complex’ is a noun, in the sense of a conglomerate of interconnected parts) and the other we term the “ﬁgure deﬁnition.” His ﬁgure deﬁnition appears in the preface: “A polytope is a geometrical ﬁgure bounded by portions of lines, planes, or hyperplanes; e.g., in... three [dimensions it is] a polyhedron.” [2, p. vi] At ﬁrst glance, this harks back to Euclid: “A solid is that which has length, breadth, and depth. An extremity of a solid is a surface.” [4, Book 11, Deﬁnitions 1 and 2] Coxeter’s ﬁgure deﬁnition of polyhedron deﬁnes the ﬁgure in terms of a bounded region of space; the boundary in turn is composed of previously deﬁned component objects of lower dimension, namely lines and planes. Coxeter’s complex-deﬁnition appears in Chapter 1: “A polyhedron may be deﬁned as a ﬁnite, con-nected set of plane polygons, such that every side of each polygon belongs also to just one other polygon, with the proviso that the polygons surrounding each vertex form a single circuit...” [2, p. 4] Coxeter’s deﬁnitions of polyhedra shift between a bounded region of space and a set of oriented polygons. We will see that this makes a diﬀerence. Coxeter invokes his complex-deﬁnition to deﬁne regular polyhedra: “A convex polyhedron is said to be regular if its faces are all regular and equal, and its vertices are all surrounded alike.” [2, p. 5] Conceived as in the ﬁgure deﬁnition, as regions of space, Coxeter’s polyhedra do not have vertices or other structural features that are separable from the underlying space. Conceived as in the complex-deﬁnition, as sets of oriented polygons, their structural features, such as their number of vertices, are brought to the fore. It is not accidental that Coxeter’s deﬁnition of regular polyhedra draws on his complex-deﬁnition of polyhedra. Coxeter’s construction of the Platonic solids also relies on his complex-deﬁnition: By placing two equal pyramids base to base, we obtain a dipyramid bounded by 2p triangles... Similarly, by adjusting the altitude of an antiprism, we may take its 2p lateral triangles to be 8 equilateral... If p = 4 or 5, we can place pyramids on the two bases, making 4p equilateral triangles altogether... There is no such simple way to construct the ﬁfth Platonic solid. But if we ﬁt six pentagons together so that one is entirely surrounded by the other ﬁve, making a kind of bowl, we observe that the free edges are the sides of a skew decagon. Two such bowls can then be ﬁtted together, decagon to decagon, to form the dodecahedron... [2, pp. 5-6] Coxeter’s complex-deﬁnition leads his proof to proceed upon the arrangement of a polyhedron’s polygons. Note that Coxeter’s proof uses motion-language reminiscent of Euclid’s. Coxeter’s treatment of symmetry also utilizes his complex-deﬁnition of polyhedra. “When we say that a ﬁgure is “symmetrical”, we mean that there is a congruent transformation which leaves it unchanged as a whole, merely permuting the component elements.” [2, p. 44] This deﬁnition requires that a polyhedron consist of structured components distinct from the ambient space. Coxeter also departs from motion-language here; a transformation is not a motion, but an action. Referring to points An which “may be regarded as the vertices of a generalized regular polygon,” Coxeter observes “...there is a symmetry operation interchanging An and A−n for all values of n (simultaneously).” [2, p. 45] A simultaneous interchange of distinct objects is not a motion. Here the complex-deﬁnition is utilized. Coxeter switches to his ﬁgure deﬁnition of polyhedron to discuss kaleidoscopes and reﬂections: If the group [generated by reﬂections] is discrete, the whole set of planes eﬀects a partition of space into a ﬁnite or inﬁnite number of congruent convex regions, and the group is generated by reﬂections in the bounding planes of any one of the regions. Let these bounding planes or walls be denoted by w1, w2, ...The dihedral angle between two adjacent walls, wi and wj, is π/pij... [2, pp. 80-81] In Coxeter’s treatment of symmetry in general, his words describe actions on a polyhedron’s component pieces. Here, in discussing reﬂections, Coxeter’s object of inquiry is the underlying space and its regions; Coxeter uses words invoking a polyhedron’s boundary. “In particular, two trihedra (or trihedral solid angles) are congruent if the three face-angles of one are equal to respective face-angles of the other.” [2, p. 33] As Coxeter moves to name regions of reﬂection, he invokes the polyhedron’s boundary, describing how one may “pass through” a region’s wall [2, p. 81]. The aptness of this procedure depends on a conceptual shift from the complex-deﬁnition to the ﬁgure deﬁnition, which Coxeter readily undertakes. Once this deﬁnition shift is made, motion language occurs. We mentioned “passing through”; we also have: “Whenever the path goes from one region into another and then immediately returns... when the path momentarily crosses an edge...” [2, pp. 80-81] Coxeter uses Euclidean language: we have “standing”, “joined”, and “inclined” in, respectively, [2, pp. 82, 84, 86].18 Motion language also appears in Coxeter’s treatment of star polyhedra: The regular polyhedron p is traced out by a moving point which continuously describes equal chords of a ﬁxed circle and returns to its original position... The general regular polyhedron p can be derived from the convex polyhedron np by either of two reciprocal processes: stellating and faceting. In the former process, we retain the positions of the sides of np, and produce them at both ends, all to the same extent, until they meet to form new vertices. In the latter, we retain the vertices of np and insert a fresh set of sides, so that each new side subtends the same central angle... [2, pp. 94-95] In addition to the use of motion-language, Coxeter opens his chapter on star polyhedra by invoking rota-tions to generate star polygons [2, p. 93]. He intermixes his two conceptions of polyhedra, discussing both the surface and an interior region in the same sentence: “By removing one zone from the triacontahe-dron, and bringing together the two remaining pieces of the surface, we obtain the rhombic icosahedron, which has a decidedly “oblate” appearance.” [2, p. 29] Elsewhere, he writes, “[A] second rhombic do-decahedron... can be derived from the same rhombic icosahedron by removing one zone and bringing together the two remaining pieces of the surface.” [2, p. 31] While describing the stellation of Platonic solids, he says, “In order to stellate a polyhedron, we have to extend its faces symmetrically until they again form a polyhedron.” [2, p. 96] Here, Coxeter draws upon his polygonal notion of polyhedra; three pages later, he utilizes his ﬁgure deﬁnition: “Thus stellating involves the addition of solid pieces, while faceting involves the removal of solid pieces.” [2, p. 99] Thus, Coxeter switches between deﬁnitions and their corresponding conceptions – and not fallaciously – throughout his text. 18 Euclid uses “passing through” and “joined” in [4, Book 13, Proposition 13]. Euclid uses both “standing” and “inclined” in, respectively, [4, Book 11, Deﬁnitions 5 and 7]. 9 We take Coxeter’s dynamism in moving among conceptions as evidence of his geometric ability: Coxeter mastered multiple approaches to his objects of study. But we remark that Coxeter never explicitly mentions his shifts between conceptions,19 and only indirectly touches on this issue in the preface: “In fact, this book might have been subtitled “A sequel to Euclid’s Elements”.” [2, p. vi] He recognizes his geometric heritage from Euclid, leaving unmentioned his motion-language word choices – an inheritance from that same author. 3.2 Gr¨ unbaum’s Convex Polytopes Gr¨ unbaum, in contrast to Coxeter, adheres to one conception on polyhedra throughout his Convex Polytopes. As we will see, a polyhedron for Gr¨ unbaum is conceived with the language of sets: it is a set of points in the Euclidean space Rd with certain properties. Gr¨ unbaum begins his book by limiting the scope of the inquiry: “With few exceptions, we shall be concerned with convexity in Rd, the d-dimensional real Euclidean space.” [7, p. 1] Gr¨ unbaum develops a conception of polyhedra as sets of d-tuples in Rd. He deﬁnes geometric entities as sets of such points.20 For example, Gr¨ unbaum writes, “A hyperplane H is a set which may be deﬁned as H = {x ∈Rd \| ⟨x, y⟩= α} for a suitable y ∈Rd, y ̸= 0, and α.” [7, p. 2] Similarly, Gr¨ unbaum writes, “Let K be a convex subset of Rd. A set F ⊂K is a face of K if either F = ∅or F = K, or if there exists a supporting hyperplane H of K such that F = K∩H.” [7, p. 17] His adoption of the convention dimension(∅) = −1 shows that Gr¨ unbaum aims to explicitly channel his conception of polytopes as sets of d-tuples in his technical development and even his notation [7, p. 3].21 Another example of his strict adherence to a single conception of polyhedra involves his deﬁnition of ‘section’: “A section of a polytope P is the [set] intersection of P with some ﬂat...” [7, p. 71] Furthermore, Gr¨ unbaum couches spatial relationships in set-language: “A set A ⊂X is bounded if there exists δ > 0 and x ∈X such that ρ(a, x) < δ for all a ∈A.” [7, p. 5]22 Gr¨ unbaum also deﬁnes convexity in the language that utilizes operations on sets: A set K ⊂Rd is convex if and only if for each pair of distinct points a, b ∈K the closed segment with endpoints a and b is contained in K. [So] K is convex if its intersection with every straight line is either empty, or a connected set. [7, p. 8] Gr¨ unbaum is indiﬀerent as to whether we rely on subset or intersection to ground convexity, but either way, we consider convex ﬁgures as sets of d-tuples. Gr¨ unbaum also conceptualizes the relationships ‘beyond’ and ‘beneath’ in set-language [7, p. 78]. Similarly, Gr¨ unbaum deﬁnes historically spatial operations in set-language. For Euclid, motion lan-guage grounds words like ‘cut’; for Gr¨ unbaum, set-language explains the meaning of ‘cut’: Let A be a subset of Rd. We shall say that a hyperplane H = {x ∈Rd \| ⟨x, u⟩= α} cuts A provided both open halfspaces determined by H contain points of A. In other words, H cuts A provided there exist x1, x2 ∈A such that ⟨x1, u⟩< α and ⟨x2, u⟩> α. [7, p. 10] So Gr¨ unbaum, rather than conceiving of a ‘cut’ as a spatial action, conceived of a ‘cut’ as a relationship between sets. Gr¨ unbaum proceeds with a similarly set-theoretic conception in constructing objects of study: “To ‘construct’ means to draw appropriate lines in the plane, i.e. to specify a suitable conﬁguration.” [7, p. 94] Here Gr¨ unbaum unpacks ‘to draw’ as specifying a conﬁguration of points in a coordinate space. And Gr¨ unbaum deﬁnes a conﬁguration as “a ﬁnite set of points and lines in a projective plane, with 19 This is diﬀerent from generalizing a deﬁnition within one conception, as here: “the deﬁnition of polyhedron can be generalized by allowing non-adjacent faces to intersect...” [2, p. 96] We remark that it is important to explicitly recognize when conceptual shifts occur so as to facilitate understanding. If, say, a student-reader might miss such conceptual shifts, explicitly marking them in one’s treatment seems warranted. 20 Part of our point is that one can approach the same entities from other vantage points; labeling polytopes as ‘geometric’ may bias one perspective over other possibilities, as Gr¨ unbaum’s text shows. 21 Instead of, say, regions of space or polygonal arrangements. Coxeter does not adopt this convention. 22 The departure from Euclid here is striking; containment is no longer conveyed by a spatial metaphor. 10 prescribed incidence relations.” [7, p. 93] Similarly, terms like ‘separates’ and ‘supports’ each receive similar set-theoretic casting [7, p. 10], as do both ‘pasting’ [7, p. 59] and ‘pulling’ [7, p. 80]. Gr¨ unbaum meticulously keeps to his set-language in presenting his geometry. Gr¨ unbaum’s theorems and proofs also conceptualize historically geometric notions in the language of sets. Consider the following two theorems and some excerpts from the proof of the ﬁrst theorem: The following results are of fundamental importance. 1. If A and A′ are convex subsets of Rd such that A′ is bounded and cl A ∩cl A′ = ∅, then A and A′ may be strictly separated by a hyperplane. 2. If A and A′ are convex subsets of Rd such that aﬀ(A ∪A′) = Rd then A and A′ may be separated by a hyperplane if and only if relint A ∩relint A′ = ∅. PROOF OF THEOREM 1 Since the distance between two sets... Therefore T ϵ>0 (A′ ∩B(ϵ)) = A′ ∩B(0) ̸= 0. In other words, there exists a point y = y(x) ∈A′ such that δ(x, A′) = ρ(x, y(x))... Each point of the ﬁrst interval, suﬃciently near to x0, is at a distance less than δ from y(x0) and thus can not belong to A; similarly, each point of the second interval... Since both A and A′ are convex, it follows that z belongs to neither of them... [7, p. 11] Rather than talking about spatial position or using motion language, Gr¨ unbaum uses set language,23 reasoning about a historically spatial concept (convexity) using the language of a set-theoretic notion (membership). Another example of conceptualizing in the language of sets what has historically been a motion occurs in [7, pp. 83-84], a typical modern application of the notion of convergence. Set-theoretic language inﬂuences Gr¨ unbaum’s vocabulary and proof methods. We conclude with a look at Gr¨ unbaum’s treatment of polytopes and Euler’s formula. Gr¨ unbaum deﬁnes, “A compact convex set K ⊂Rd is a polytope provided ext K is a ﬁnite set.” [7, p. 31] This is a set-theoretic conceptualization of the notion of a polytope. He even uses set-language to motivate Euler’s formula: “What d-tuples of numbers can occur as the numbers of vertices, edges, ..., (d−1)-faces of convex d-polytopes?” [7, p. 130] He answers: We formulate the result known as Euler’s theorem: 1. The aﬃne hull of the f-vectors of all members of the family Pd of all d-polytopes is given by aﬀf(Pd) = ( f = (f0, . . . , fd−1) d X i=0 (−1)ifi = 1 −(−1)d ) . The relation Pd i=0(−1)ifi = 1 −(−1)d, which by the theorem holds for every d-polytope P, is known as Euler’s equation. [7, p. 131] Gr¨ unbaum’s conceptualization of V +F = E +2 ties Euler’s formula to set-talk. Gr¨ unbaum is well-aware of this! He remarks: Some remarks of a methodological nature seem indicated in view of the proofs given in the present Section [Polytopes]. It is hoped that readers who worked their way through the proofs are by now ready to accept the validity of the results proved. The author doubts, however, that the above formal proofs give a good idea of why the proofs work. In a subject as elementary and intuitively comprehensible as the theory of polytopes, it seems a pity to obscure the simple idea of a proof by the – almost necessarily – involved and complicated notation and symbolism... In this context, as in many other cases, the idea of the proof becomes clearly comprehensible with the help of a graphic presentation of the two- or three-dimensional case... [7, p. 33] 23 A non-exhaustive list of examples include the deﬁnition of convex hull and Carath´ eodory’s Theorem on [7, p. 14-15], poonem (Hebrew for ‘face’) [7, p. 20], unboundedness [7, p. 23], the notion of a polytope’s dual [7, p. 46], combinatorial scheme [7, p. 90], and the Euler characteristic of a proper face [7, p. 138]. 11 Gr¨ unbaum acknowledges that the reader may have diﬃculty visualizing the proven facts due to his “notation and symbolism,” which we have noted is heavy with set-theoretic word choices. While adherence to a particular framework in approaching an object of study can preserve purity of perspective, it can also result in confusion at certain steps of the treatment, at least for those readers who conceive of polyhedra through diﬀerent, e.g. spatial, word choices. The words a mathematical writer chooses bear signiﬁcantly on his or her success in communicating to the reader a solid connection between the underlying intuitions and the objects of study. 3.3 Word Choices as Codifying Conceptualizations David Reed’s Figures of Thought studies mathematical texts “as texts”, i.e. as encoding a conceptual stance towards their objects [15, p. x]. He writes: Although all mathematics is set forth in texts which argue, explicitly or implicitly, for their own version of ‘doing mathematics’... the general view remains that mathematical facts and mathe-matical subject matters exist somehow independently of the texts in which they are expressed. [15, p. xi] Reed aims to refute this “general view”. Using Euclid as his case study, Reed studies Euclid’s approach to mathematics and argues that critiques of Euclid’s ‘imprecision’ or ‘gap-ridden’ proofs have ignored the conceptual setting of Euclid’s arguments [15, p. xii-xiii]. Reed advances the claim that Euclid’s work presents a systematized whole, one that moves from a study of parts in Books I-X to wholes in Book XI-XIII [15, p. 164]. On his account, Euclid’s Elements has been anachronistically misrepresented as containing bad proofs and other ﬂaws by the axiom-driven mathematicians of the modern era, who had goals and standards for rigor rather diﬀerent from Euclid’s own [15, p. 162-163]. They pursued a form of completeness in geometry that Euclid did not espouse: ...from Euclid’s point of view... [t]he study of plane geometry does not constitute a whole and cannot achieve completion; the study of solid ﬁgures in which plane ﬁgures function as boundaries and parts is a study of wholes and arrives at a natural conclusion. [15, p. 145] Reed argues that Euclid’s work, from start to ﬁnish, follows a careful plan and explores the limits of deﬁnability, and that contemporary authors ignore the philosophical elements underlying Euclid’s reasoning [15, p. 162]. We mention Reed’s thesis because he highlights a phenomenon of mathematical practice: the con-ceptualization(s) of a mathematical subject-matter embodied by an entire text. We here focus on the phenomenon of word choice as encoding a conceptualization of the mathematical objects of study. This phenomenon resurfaces in a recent dispute over the appropriate word choice for and conceptualization of polyhedra, as we will see in Section 4. Reed also discusses the vital role of conceptualization. His remarks equally apply to whole texts or individual word choices: Mathematical understanding is achieved by ﬁnding appropriate mathematical terms in which problems can be formulated and theorems enunciated. It might be said that the mathematics in a mathematical text is precisely that in it which can be translated into other (mathematical) terms. The text and its argument is what cannot be translated or so transformed. [15, p. 161] As with translating languages, moving from one conceptualization to another may result in some loss of information or rigor, or gains in the same. As we have seen in the above discussion, word choice is a factor in conceptualizing an object of mathematical study. Gr¨ unbaum is alert to this fact. He notes that a text encodes a conceptualization, or conceptualizations, of polyhedra: he writes, “...‘polyhedron’ can mean either a solid... or a surface...” [8, p. 461] Gr¨ unbaum brings this up in discussing past attempts at complete classiﬁcations of regular polyhedra and persistent counterexamples to them. He argues: ...all the results mentioned are completely valid; what changed is the meaning in which the word “polyhedron” is used. As long as diﬀerent people interpret the concept in diﬀerent ways there is always the possibility that results true under one interpretation are false with other under-standings... even slight variations in the deﬁnitions of concepts often entail signiﬁcant changes in results. [8, p. 462] 12 A similar sentiment occurs in Lakatos’ Proofs and Refutations; informal proofs that we ﬁnd acceptable are indexed to a conceptualization [12, p. 56]. We can switch conceptual frameworks, changing, say, the accompanying word choices, and this can invalidate a given result. Gr¨ unbaum observes: If [Kepler’s star polyhedra] are interpreted as solids, their boundary is a topological sphere bounded by sixty triangles in each case. Under this interpretation, neither of the two polyhe-dra is regular. However, they are regular if they are interpreted the way Kepler intended, as being constituted by 12 pentagrams (“star pentagons”) each. [9, p. 451] Moreover, by their eﬀects on the reader or researchers’ conceptualization of the object of study, word choice can foreclose possibilities and discoveries to the mathematician: “But the impossibility of Steinitz having used this formulation is obvious from the fact that in 1916 nobody used terms such as ‘3-connected graph’.” [9, p. 448] Switching to a diﬀerent conceptualization can clarify a proof or bring out a feature of the object being studied, but such switches in conceptualization can also obstruct a proof: However, the ideas of Steinitz’s original proofs can be reworked in a graph-theoretic setting, gaining in clarity and simplicity. [9, p. 447] Eberhard’s proof is long and messy, since he has to accommodate the operations he performs on the convex polyhedra. A simpler proof, utilizing graphs and Steinitz’s theorem, appears in [18, Section 13.3]. [9, p. 450] Researchers and teachers can choose from conceptual frameworks available: one may select more than one framework, but at least one framework must be chosen. This selection may result in varying degrees of successful communication and, importantly, this selection impacts the mathematics developed – proofs, techniques, and theorems. Success depends partly on the target audience. On the one hand, the dynamism of Coxeter may be easier to follow with a high degree of exposure to the subject. On the other hand, the new student may ﬁnd the interconnections bridged in language compelling, which could inspire greater interest in the subject. Again, on the one hand, a stricter presen-tation like that of Gr¨ unbaum may provide greater intelligibility of the argument in the sense that each step can be followed more easily. On the other hand, the global perspective may be lost in the process. The ‘right’ approach(es), if any, can only be decided on a case-by-case basis: the preferred conceptual background, and so the preferred choice of words, may change with the author’s conceptualization of the subject-matter, the intended goal(s), and the audience(s). Facilitating mathematical communication and understanding manifestly requires sensitivity to the conceptualization(s) encoded in one’s word choices. 4 Euler’s Formula We next discuss a recent dispute as to what Euler meant by his formula for polyhedra. To introduce this dispute, we ﬁrst revisit the language with which Euler conveys his formula in “Proof of Some Notable Properties”, and then we discuss Imre Lakatos’ Proofs and Refutations, which treats the formula in detail. We will then consider ‘conﬂicting’ interpretations of Euler’s formula oﬀered by two pairs of modern authors. We draw attention to the fact that they are partly disputing word choice for polyhedra. Euler makes his formula part of the basis for his solid geometry: ...the general properties of solids... depend on one property [Euler’s formula] in such a way that if it were possible to prove this property, then all of the ﬁrst principles of Solid Geometry which I have proposed would be equally as ﬁrm as the ﬁrst principles of Geometry. [5, p. 1] Recall that Euler’s formula, in his (translated) words, is this: “In every solid enclosed by plane faces the number of solid angles, along with the number of faces, exceeds the number of edges by two.” [5, p. 2] In notation, this reads V + F = E + 2. Then there is the matter of proving it. Euler describes his proof method as follows: I arrived at proofs similar to the one customarily used for the analogous proposition from Geometry regarding the sum of the angles of any rectilinear ﬁgure. In Geometry any rectilinear ﬁgure can be ultimately reduced to a triangle by a successive division of angles. Likewise, given any solid enclosed by plane faces, I observed that the solid angles can be continuously divided so that ﬁnally, a triangular pyramid remains... Here I will explain a... method by which any solid is reduced by successive cutting of its solid angles to triangular pyramids. [5, pp. 2-3] 13 Euler describes his proof as shearing oﬀsolid angles from the solid one at a time and bases his argument on the equation V +F = E+2 holding after each removal [5, p. 12]. From these excerpts, we know Euler’s proof requires that a polyhedron, as a region of space bounded by planes, be amenable to decomposition into components of space, e.g., into pyramids. If the solid is conceived as constituted by its boundary, as in Euclid, then this process is incoherent. Euler’s language does not disambiguate his notion of ‘counting’ vertices, faces, and edges; the counting procedure occurs vicariously via triangular pyramids. His statement only requires that the number of vertices, edges, and faces satisfy his formula, but how one determines the number of vertices, edges and faces remains open. Not surprisingly, given this ambiguity, the meaning of Euler’s formula has long been a matter of dispute, as Lakatos’ book exploring the history of Euler’s formula shows. Lakatos’ Proofs and Refutations: The Logic of Mathematical Discovery explores relationships be-tween theorem, proof, and counterexample using Euler’s formula as the case study. He tracks attempts of a hypothetical class of students to prove Euler’s formula deﬁnitively, i.e., without gaps appearing or counterexamples. Lakatos’ ﬁctional students explore the content, truth, applicability, and meaning of Euler’s theorem: they catalog, in their oﬀered guesses and counterexamples, historical instances of mathematicians undergoing the same process. A ﬁctional teacher leads the students through speculation as to the relationship between the number of vertices, edges, and faces of a polyhedron, asking them to consider whether some relation, analogous to V = E for polygons, holds for polyhedra [12, p. 6]. Most of this discussion paraphrases quotes, proofs, theorems, and counterexamples from the historical development of the mathematics of polyhedra.24 The teacher oﬀers a proof lifted from Cauchy [12, pp. 7-8], which the students subsequently consider and refute with a counterexample [12, p. 10]. The students then engage in oﬀering improved deﬁnitions, including those oﬀered by Legendre, Jonqui eres, M¨ obius, and Baltzer, and new counterexamples, some of which may be found in the work of Lhuilier, Hessel, and Kepler.25 Lakatos’ ﬁctional setting distorts little of the troubled history of Euler’s formula: counterexamples – solids satisfying a given deﬁnition of a polyhedron for which V + F = E + 2 does not hold – have repeatedly been oﬀered in response to proposed proofs of the result; in turn, these counterexamples (or ‘monsters’) have been ‘barred’ by improvements to the proof or the incorporation of lemmas; labeled ‘exceptions’ and so excluded from the theorem’s domain of application; ignored altogether; or admitted into a revised ‘Euler’s formula’ or deﬁnition of ‘polyhedron’ that masks changes to the proof. Save for the ﬁctional pupils and dialogue, the result is an accurate depiction of the history of Euler’s formula. Lakatos’ imaginary pupil Alpha describes this procedure: “...the identity of the linguistic expressions of the naive conjecture and the monster-barring theorem hides, behind surreptitious changes in the meaning of the terms, an essential improvement.” [12, p. 41] The phenomenon Alpha is alluding to is how changes in word choice invite reconceptualization of objects of mathematical study and thus mathematical changes. We next consider a modern instance of this phenomenon: In relatively recent issues of the American Mathematical Monthly, two pairs of authors debated the meaning of Euler’s formula. Both pairs use the term ‘polyhedron’, but this hides their diﬀering interpretations of Euler’s theorems. One pair emphasizes Euler’s word “counting”; the other pair emphasizes Euler’s word “division” (of space). We shall see that these word choices demarcate distinct conceptualizations of polyhedra. 4.1 Gr¨ unbaum’s and Shephard’s “A New Look at Euler’s Theorem for Polyhedra” Branko Gr¨ unbaum and Geoﬀrey Colin Shephard aim to incorporate all counterexamples to Euler’s for-mula under one conceptual apparatus, “polyhedral sets.” Gr¨ unbaum and Shephard intend polyhedral sets to rigorously capture the act of counting vertices, edges, and faces of a polyhedron [11, Response from Gr¨ unbaum and Shephard, p. 962]. They hold that any polyhedron will fall under the framework of polyhedral sets: “polyhedral sets... generalize familiar polyhedra; they may be closed, open, or neither, connected or not, bounded or not, and their parts may have diﬀerent dimensions.” [10, p. 110] 24 Much of this historical background receives a masterful treatment in Peter R. Cromwell’s Polyhedra; we recommend [3, pp. 181-218], especially [3, pp. 198-210]. 25 [12, p. Legendre, 14, Footnote 1], [12, p. Jonqui` eres, 14, Footnote 2], [12, p. M¨ obius, 15, Footnote 2], [12, p. Baltzer, 16, Footnote 1], [12, p. Lhuilier, 13], [12, p. Hessel, 15], and [12, p. Kepler, 16, Footnote 2]. 14 For Gr¨ unbaum and Shephard, the domain of objects caught under the umbrella word ‘polyhedra’ includes any geometric realization of a set of vertices and edges, so long as these can be counted by the methods they describe. The word “counting” sets the limits of inquiry [11, Response, p. 962]. What justiﬁes their use of ‘polyhedron’ to refer to such a wide variety of objects? They argue that their view is “simple and elementary,” relating “an integer χ(P) to the geometrical features of P,” where P is a polyhedral set [10, pp. 110-111]. Gr¨ unbaum and Shephard intend χ(P) to be a generalization of Euler’s formula. Gr¨ unbaum and Shephard deﬁne a “closed convex polyhedron” as “the intersection of a ﬁnite number of closed half-spaces.” [10, p. 114] They allude to the inclusion of traditional examples of polyhedron while stressing their notion’s broader applicability: Familiar examples of closed convex polyhedra are (closed) cubes, (closed) squares, (closed) line segments, and single points... But our deﬁnition includes also unbounded convex polyhedra... their edges can be rays (halﬂines) or straightlines, and faces and cells can be unbounded as well. [10, p. 114] We arrive at Gr¨ unbaum and Shephard’s notion of polyhedral set: If a set P is the union of members a ﬁnite family C = {C1, C2, . . . , Cn} of pairwise disjoint sets {C1, C2, . . . , Cn}, we say that C is a dissection of P and we write P = S C, where the dot in the union symbol indicates that the sets Ci are pairwise disjoint. If P is a set that admits a dissection C all elements of which are relatively open convex polyhedra we shall say that P is a polyhedral set, and express this by writing P = [ c C where the subscript c is to remind us that each element of C is a relatively open convex polyhedron. Gr¨ unbaum and Shephard remark once again that their deﬁnition is quite general: It will be observed that the deﬁnition is very general in that it does not require P to be open or closed, connected, simply connected or even homogenous in the sense that neighborhoods of all points of P are of the same dimension. [10, p. 116] By only demanding that a polyhedron’s pieces be separable and requiring nothing of the polyhedron’s underlying space, e.g., that all pieces have the same dimension, Gr¨ unbaum and Shephard depart from the language of space of Euler’s original proof, the language that enables a polyhedron to be cut into pyramidal pieces. The upshot of their departure from Euler’s language of space is new examples for consideration. “All the sets shown in Figures 1 to 9 are polyhedral sets...” [10, p. 116] Ever-further generalization drives Gr¨ unbaum and Shephard’s extension of Euler’s formula: Thus the deﬁnition of χ(P) given here may be regarded as an extension of the traditional approach to sets which need not be closed. On the other hand, our deﬁnition is restricted to polyhedral sets, hence not applicable to non-polyhedral sets even if they are not convex. [10, p. 117] Their deﬁnition of χ(P) for a polyhedral set P is as follows [10, p. 117]: 1. χ(∅) = 0 2. If P is a relatively open convex polyhedron of dimension d, then χ(P) = (−1)d. 3. If P = S c C is a relatively open convex dissection of P, then χ(P) = P C∈C χ(C). How does χ(P) capture V + F = E + 2? Gr¨ unbaum and Shephard introduce four k-scaﬀolds of a polyhedral set P. These sets are deﬁned as follows [10, p. 118]: 1. scaf 3(P) = int(P) (interior points) 2. scaf 2(P) = int(P\scaf3(P)) (faces) 3. scaf 1(P) = int((P\scaf3(P))\scaf2(P)) (edges) 4. scaf 0(P) = int(((P\scaf3(P))\scaf2(P))\scaf1(P)) (vertices) 15 Gr¨ unbaum and Shephard do not care, for example, what dimension of space or region of space a polyhe-dron’s parts occupy; their scaﬀolding will capture vertices, edges, and faces because the structure of the parts, how one “divides” a polyhedron, does not impact the tabulation of a polyhedron’s components: “the deﬁnition of scafk(P) is independent of the relatively open convex dissection of P into polyhedral sets.” [10, p. 118] We have here an interpretation of the Euler characteristic such that the polyhedron’s structure can be as ‘bizarre’ as one pleases. A polyhedron’s structure is given and not adjusted to ﬁt the Euler characteristic; rather, the Euler characteristic accommodates whatever the polyhedron’s structure may be. We see a very diﬀerent perspective in the next section. We need not speculate as to Gr¨ unbaum and Shephard’s purpose, as they explicitly state their dis-satisfaction with the history of Euler’s theorem: “At least two books have been dedicated to the early history and attempts at clariﬁcation of Euler’s theorem.” [10, p. 122] They spend about six pages of the article detailing past confusions, and then give the following diagnosis: All the discussions of [Euler’s theorem] were dogged by two diﬃculties... On the one hand, no precise deﬁnitions were given for the polyhedra under consideration or for their faces, edges, and vertices. It was more or less generally assumed that one is dealing with solids and considering features on their surfaces – but how to determine faces (or edges) was illustrated by examples rather than deﬁned by unambiguous rules... On the other hand, in its original formulation and in the minds of many early workers, it was a hallmark of Euler’s theorem that it involved only the numbers of vertices, edges, and faces – without any consideration of the nature of the faces. [10, p. 123] Gr¨ unbaum and Shephard claim to clarify Euler’s theorem and preserve Euler’s insight. They believe they have provided deﬁnitive rules for counting edges, faces, and vertices which are guided by Euler’s language, especially his choice of the word ‘counting’. On the other hand, Gr¨ unbaum and Shephard ﬁnd that focusing on space obscures the idea of Euler’s formula: Starting with Poincar´ e... these concepts acquired precise meaning for simplicial complexes and more general objects... However, this development led to a loss of the connection to the origins of Euler’s theorem as a relation involving the vertices, edges and faces of a polyhedral objects. [10, p. 125] They similarly discount Betti numbers as a precise interpretation of Euler’s original insight. [10, p. 123] This evaluation is contrary to Hilton and Pedersen’s conceptualization of polyhedra, which we consider in the next section. 4.2 Hilton’s and Pedersen’s “Euler’s Theorem for Polyhedra: A Topologist and Geometer Respond” Peter Hilton and Jean Pedersen approach polyhedra through the perspective of underlying space. They dispute Gr¨ unbaum’s and Shephard’s version of the history of Euler’s theorem: [Gr¨ unbaum’s and Shephard’s] paper appears, in contrast to its obvious positive contribution, to both neglect and distort the contribution which algebraic and combinatorial topologists have made to the development of the Euler characteristic. We claim that the ‘problem’ referred to on page 110 of ... is no real problem at all, and that topologists have completely and satisfactorily formalized the Euler characteristic... Moreover, we categorically deny that ‘this development led to a loss of the connection to the origins of Euler’s theorem as a relation involving the vertices, edges and faces of a polyhedral object.’ [11, p. 959] Hilton and Pedersen believe that topologists solved all conceptual diﬃculties relating to the Euler char-acteristic and that no controversy obtains over the present state of the theorems. Note that they do not contest the mathematical merit of Gr¨ unbaum and Shephard’s article: ...resemblance of their Theorem 4 to the classical theorem of Euler might seem at ﬁrst to owe more to a judicious choice of notation than to a conceptual closeness; however, careful inspection of their deﬁnition of a k-scaﬀold and contemplation of their examples show that their notation may, indeed, be justiﬁed. [11, p. 959] 16 Hilton and Pedersen do not dispute Gr¨ unbaum and Shephard’s mathematical innovation, yet they argue that Gr¨ unbaum and Shephard’s rendition of the history of Euler’s formula wrongly identiﬁes the content of Euler’s theorem as concerned with the act of counting; Hilton and Pedersen believe the topological development of Euler’s formula remains true to Euler’s development of his formula as a description of spatial properties: We regard the theorem (referred to somewhat disparagingly in ) that the Euler-Poincar´ e char-acteristic equals the alternating sum of Betti numbers as fundamental. It not only explains Euler’s original result (since orientable closed surfaces bounding convex regions are homeomorphic to the sphere), but establishes the homotopy invariance (not merely topological invariance) of χ. Of course, it also explains formula (2) of , that is v −e + f = 2 −2g, for an orientable closed surface of genus g, since the Betti numbers of such a surface are given by p0 = 1, p1 = 2g, p2 = 1. Thus, formula (2) is scarcely a stride in extending Euler’s Theorem and so should not be seen as ‘dealing with a discrepancy’. [11, p. 960] Hilton and Pedersen think that the topological development of Euler’s theorem exposes Euler’s true insight, and that Euler’s theorem gives an invariant between surfaces, i.e., that solids with diﬀerent Euler characteristics are (topologically-speaking) diﬀerent. The notion of counting is entirely absent from their discussion; Hilton and Pedersen emphasize Euler’s language of dividing or decomposing a space over his language of counting. Hilton and Pedersen claim that formula (2), i.e., v −e+f = 2−2g, is a case of Euler’s formula rather than an extension, and argue against Gr¨ unbaum and Shephard’s expansion of the term ‘polyhedron’: Indeed, one might legitimately comment that it is unreasonable to expect that one can count the number of holes by counting subsets which themselves are allowed to have holes! Moreover, if one allows strange (2-dimensional) faces... then why not regard the entire surface as a face and thus obtain v −e + f = 1 for any surface! Finally, we remark that it should never happen that scratching a few extra lines on a surface... should alter a really signiﬁcant invariant quantity. ...Nor should Figure 2 of create a problem... Certainly χ(P) = 1 for this polyhedron, so formula (2) does not apply. But P is not a closed orientable surface, so how could formula (2) apply? [11, p. 960] Hilton and Pedersen deny Gr¨ unbaum and Shephard’s application of Euler’s theorem (to them, a fact about space) to some objects with bizarre (to Hilton and Pedersen) underlying spaces, such as topolog-ically nontrivial two-dimensional faces [11, p. 960]. Their conceptualization of polyhedra as intimately bound to the underlying (topological) space – encoded in their spatial-language word choices – leads them to such conclusions. They see, e.g., an “error in permitting Figure 1(b) as a legitimate dissection of the torus...” [11, p. 960] Hilton and Pedersen specify what they consider the appropriate domain of inquiry26 corresponding to the term ‘polyhedron’ by oﬀering a characterization of the objects of study: We claim that a polyhedron is made of cells and a closed ( n)-cell is the homeomorph of the n-dimensional ball x2 1 + x2 2 + · · · + x2 n ≤1 in Rn. In particular, a two dimensional face must be a polygonal region [11, p. 959-960]. In other words, the substructure of a surface does not matter; a polyhedron’s regions must merely be amenable to deformation into cells for the solid to count as a polyhedron. Gr¨ unbaum and Shephard, on the other hand, deny in their response the ‘spatial divisibility’ limits Hilton and Pedersen would apply to the polyhedra: In particular, we reject the implication that the geometrical approach we propose is redundant. The purpose of our paper... is to show how the “traditional” approach of counting vertices, edges, and faces, can, with care and appropriate modiﬁcations, lead to results which are valid and far-reaching. In particular, there is no need, as topologists generally do, to restrict attention to bounded polyhedra which can be expressed as ﬁnite unions of topological balls all of which are 26 Lakatos calls this the “domain of proof,” but here we use the term in a slightly broader sense that corresponds to an area of mathematical inquiry rather than a single proof from such an area [12, p. 64]. 17 open or all of which are closed... it seems unreasonable to us to reject a polyhedron... by stating dogmatically that every “2-dimensional face must be a polygonal region” and then to deﬁne a polygonal region in such a way as to eliminate multiply-connected regions. Euler himself, if he were alive today, would, we believe, feel far more comfortable with our approach to his theorem than that advocated in the “Response”... Our purpose... was to show that a simple geometrical treatment of a classical topic can lead to results that are more general, and more intelligible to the majority of mathematicians, than the recondite and much more abstract topological approach. [11, Response, p. 962] Gr¨ unbaum and Shephard make it explicit: their conceptualization of Euler’s formula ties it to counting. They ﬁnd Hilton’s and Pedersen’s spatial restrictions unappealingly ad hoc. Again, the mathematical results are not issue; the issue is what one means by ‘polyhedron’, and what Euler meant 240 years ago in formulating his formula. The same word, ‘polyhedron’, conveys two diﬀerent concepts based on Euler’s choice of two other words, ‘sum’ and ‘divides’, to describe his fundamental result. We see this by noticing how one pair of authors uses counting language while the other uses spatial language. (Interestingly, each side of the dispute traces their word choices and conceptualization back to Euler’s words explicitly without ever citing Euler’s words.) This argument between the two pairs of authors over Euler’s formula turns chieﬂy on their divergent word choices for their respective conceptualizations of polyhedra. The diﬀerent conceptualizations lead each pair of authors to use distinct word choices – ‘count’ and ‘spatial divisibility’ – to direct their mathematical discussion of Euler’s formula. The pairs of authors disagree about the set of objects to which the word ‘polyhedron’ rightfully applies and the meaning of Euler’s formula: is it the division of space or the act of counting that Euler emphasized? Euler himself uses both the language of counting and the language of dividing space, and his text does not seemingly decide which choice of words is the right rendering of his meaning some 240 years ago. The result of these divergent word choices is a conceptual dispute between mathematicians over the appropriateness, motivation, and meaning of ostensible developments of Euler’s formula and its modern interpretation.27 For Hilton and Pedersen, a (three-dimensional) polyhedron must be decomposable into closed balls homeomorphic to a sphere and the faces must be homeomorphic to a circle; moreover, certain alterations to the polyhedron must not change the Euler characteristic of a polyhedron. For Gr¨ unbaum and Shephard, the polyhedron need not be decomposable into ‘nice’ pieces and Euler’s characteristic does not track topological invariance; Euler’s formula applies to any polyhedron if one employs precise notions of vertex, edge, and face. The crucial point is that the argument revolves around the range of objects to which the concept expressed by the word ‘polyhedron’ applies. This disagreement is manifested in the divergent word choices for the objects of study. Whether we use the language of counting or of dividing space alters which conceptualization of Euler’s formula, and of polyhedra, we study. The mathematical realization of the concept, and thus the direction of one’s inquiry, will be correspondingly diﬀerent. It will not thereby be ‘wrong’. 5 Conclusions As we stated in Section 1, our case study of ‘polyhedra’ evidences the inﬂuence of word choice on mathematical practice. This relates to other meta-mathematical issues relevant to mathematical practice. For example, Lakatos’ work highlights the importance of guesswork and counterexamples in producing theorems with rigorous proofs [12, pp. 73-74]. He also makes insightful remarks about language inﬂuencing mathematics. Pi: ...I certainly wouldn’t call a whale a ﬁsh, a radio a noisy box... and I am not upset when a physicist refers to glass as a liquid. Progress indeed replaces naive classiﬁcation by theoretical classiﬁcation, that is, by theory-generated (proof-generated, or if you like, explanation-generated) 27 One might prefer to say that the conceptual disagreement leads to divergent word choice. Here we stake no claim as to the priority of explanation – though there is no doubt about the close tie between word choice and conceptualization – since our purpose is to examine the vital phenomenon of word choice in mathematical practice, particularly as encoding a conceptualization of mathematical subject-matter. 18 classiﬁcation... Naive conjectures and naive concepts are superseded by improved conjectures (the-orems) and concepts (proof-generated or theoretical concepts) growing out of the method of proofs and refutations. And as theoretical ideas and concepts supersede naive ideas and concepts, theo-retical language supersedes naive language. [12, p. 91] The disagreement of Gr¨ unbaum and Shephard with Hilton and Pedersen centers on how to resolve the issues of ‘naive’ classiﬁcation in our theoretical language. More precisely, they dispute which theo-retical language’s vocabulary more appropriately captures ‘the’ conceptualization of polyhedra lurking behind Euler’s original insight. But we need only to acknowledge that no one such language captures all conceptualizations of the object of inquiry to retreat from the dispute and acknowledge that both conceptualizations as reﬂected in the diﬀerent choices of words bear mathematical fruit. There is then no need to legislate against alternative conceptualizations by ﬁxing word choices in our theoretical language. Lakatos’ imaginary pupil Omega illustrates the sort of mindset engendering such disputes: Omega: In the end we shall arrive from naive, accidental, merely nominal classiﬁcation to the ﬁnal true, real, classiﬁcation, to perfect language! [12, p. 92] The dispute in the American Mathematical Monthly may have been better framed by acknowledging that realizing Omega’s dream is both improbable and potentially stiﬂes new research needlessly, and by focus-ing on the key issue underlying the debate: how change in our choice of words alters our interpretation of results and ﬁndings. As we have seen, refutation by counterexamples depends on the meaning of the terms in question. If a counterexample is to be an objective criticism, we have to agree on the meaning of our terms. [12, p. 16] As knowledge grows, languages change. [12, p. 93] Word choices are our looking-glass for mathematical inquiry, but they are not a transparent medium for conceptualizing mathematical subject-matter: our language non-trivially alters our mathematical practices – how we deﬁne objects, develop theorems and proofs, and what mathematical facts are noticed (or noticeable). We no longer use words as Euclid or Euler did; it makes little sense to bind our word ‘polyhedron’ to their historical uses of it, which are in any case quite divergent. Yet tracing the diﬀerent uses of ‘polyhedron’ back to Euclid and Euler as we have done here shows how shifting word choices can indicate or inspire shifts in mathematical practices. These shifts in word choice, as we saw in Section 2, can produce new mathematical results about the objects of inquiry. As Lambda says: Alpha: I propose one single rule instead: Construct rigorous (crystal clear) proofs. Lambda: ...Cannot ‘crystal-clear’ thought-experiments lead to paradoxical or even contradictory results? Alpha: Language is vague, but thought can achieve absolute rigour. [12, p. 52] We rely on language to show our concepts to others, but language can (and frequently does) imper-fectly perform that task for us. We think the argument between Gr¨ unbaum and Shephard and Hilton and Pedersen provides an instance of this phenomenon: the word ‘polyhedron’ historically has given rise to many diﬀerent conceptualizations of a mathematical subject-matter with a common core of objects, but with divergence in the range of objects falling under that term. Our two pairs of authors have shown that these conceptualizations can be considered as ‘rival’ ones, even though nothing about them leading to diﬀerent results makes them ‘rivals’. Moreover, these pairs of authors demonstrated that diﬀerent word choices (embodying diﬀerent conceptions) lead to diﬀerent mathematics, showing that mathematicians’ word choices in practice demand non-trivial selections. In short, the encoding of mathematical concepts in language, including our choice of words, is a key component of mathematical practice and meta-mathematical inquiry. It is just as much a part of mathematics as theorem discovery and proof writing. And in the present authors’ view, diﬀerent word choices encoding diﬀerent mathematics is a boon to mathematical activity – at least so long as our communal understanding of the subject-matter of mathematics remains incomplete.28 28 We thank our anonymous reviewers for numerous helpful suggestions and criticisms. This work was undertaken with the support of a Luther Rice Fellowship from George Washington University, and we are grateful for that support. 19 References 1. Bill Barton, The Language of Mathematics, Mathematics Education Library 46, Springer, 2008. 2. H. S. M. Coxeter, Regular Polytopes, Dover Publications, Inc., 1973. 3. Peter R. Cromwell, Polyhedra, Cambridge University Press, 1997. 4. Euclid, Thomas L. Heath (translator), Euclid’s Elements, Green Lion Press, 2010. 5. Leonhard Euler, Christopher Francese and David Richeson (translators), “Proof of Some Notable Properties with Which Solids Encased by Plane Faces are Endowed,” Novi Commentarii Academiae Scientiarum Petropolitanae, 4: 72-93, 1758. 6. Christopher Francese and David Richeson, “The Flaw in Euler’s Proof of His Polyhedral Formula,” American Mathe-matical Monthly, Vol. 114, No. 4: 286-296, Apr. 2007. 7. Branko Gr¨ unbaum, Convex Polytopes, John Wiley & Sons, Ltd., 1967. 8. Branko Gr¨ unbaum, “Are Your Polyhedra the Same as My Polyhedra?” pp. 461-488 in Discrete and Computational Geometry: The Goodman-Pollock Festschrift, B. Aronov, S. Basu, J. Pach, and Sharir, M. (editors), Springer, 2003. 9. Branko Gr¨ unbaum, “Graphs of polyhedra; polyhedra as graphs,” Discrete Mathematics Volume 307, Issues 3-5: pp. 445-463, 2007. 10. Branko Gr¨ unbaum and G. C. Shephard, “A New Look at Euler’s Theorem for Polyhedra,” American Mathematical Monthly, Vol. 101, Issue 2: 109-128, 1994. 11. Peter Hilton and Jean Pedersen, “Euler’s Theorem for Polyhedra: A Topologist and Geometer Respond,” American Mathematical Monthly, Vol. 101, Issue 10: 959-962, 1994. 12. Imre Lakatos, John Worrall and Elie Zahar (editors), Proofs and Refutations: The Logic of Mathematical Discovery, Cambridge University Press, 1976. 13. George Lakoﬀand Rafael Nu˜ nez, Where Mathematics Come From: How The Embodied Mind Brings Mathematics Into Being, Basic Books, 2001. 14. Paolo Mancosu (editor), The Philosophy of Mathematical Practice, Oxford University Press, 2008. 15. David Reed, Figures of Thought, Routledge, 1995. 16. C. Edward Sandifer, How Euler Did It, The Mathematical Association of America, 2007. 20
189164	https://www.youtube.com/watch?v=KvGsNISgcdg	Solving a pre-algebra word problem with fractions Wyzant 5950 subscribers Description 47 views Posted: 13 Jun 2022 View full question and answer details: Question: Math Question . Ruben baked 63 muffins. Of these, 2/7 are bananas, 3/7 are blueberries, and the rest are apples. How many apple muffins did he bake? Answered By: Richard C. Confidence-building Geometry tutor with 18 years experience More information: See full answer: About: Wyzant Ask an Expert offers free answers to your toughest academic and professional questions from over 65,000 verified experts. It’s trusted by millions of students each month with the majority of questions receiving an answer within 1 hour of being asked. If you ever need more than just an answer, Wyzant also offers personalized 1-on-1 sessions with experts that will work with you to help you understand whatever you’re trying to learn. Ask your own question for free: Find a tutor for a 1-on-1 session: Subscribe to Wyzant on YouTube: Transcript: [Music] [Applause] [Music] hi i'm richard c i'm a math tutor at wiseant here and answering a question submitted by a student i'm including this question i'm answering this question because it's a nice easy question and as from early algebra pre-algebra type thing you might see it in middle school but sometimes you'll see something like this is kind of a warm-up question and a heart in harder uh math classes and you might even see something like this as an opening question an early question on an s.a.t or a ct so i think it's worth talking about and it's kind of just a good way to kind of get your juices flowing when you're working on math um so here we have a reuben who's baking muffins and uh two sevenths two sevenths of the muffins are banana muffins which happens to be my favorite and then three sevenths are the blueberry and the rest are apple and we want to know how many apple muffins he's going to bake so the question is what fraction is the apple so we could say that 2 7 plus 3 7 plus x over 7 has to equal 1 because these all have to add up to one now one is also seven over seven and so you could see this is two plus three is five so x is obviously two two sevenths instead of a question mark there i should just put an x so let me put an x there okay so now we know that the apples two sevens but we we're actually want to know the actual we want to know the actual number of muffins so what we do is we take two sevenths of 63. and i'm going to do this by hand because a problem like this can be done by hand easily and 7 goes into 63 9 times so this is going to be 2 times 9 which is going to be 18 apple muffins and that's it that's the whole problem but it's a nice little pre-algebra question um definitely this is something that could be handled in middle school and it's an introductory style question for anything coming after that but it's just kind of a nice little one to do and um i'm just like i'm adding it to our inventory of problems because i think we should have some variety not just answer the harder ones okay so that's it for today i'll see you again next time bye [Music]
189165	https://math.stackexchange.com/questions/3121219/does-ssa-congruence-criterion-work-if-the-non-included-angle-is-obtuse	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Does SSA congruence criterion work if the non-included angle is obtuse? Ask Question Asked Modified 1 year, 1 month ago Viewed 2k times 1 $\begingroup$ I was trying to make some triangles when $2$ sides and the non-included angle was given. When the angle given was below $90$ degrees, there were two possible triangles which could be made. But, when it went above $90$ degrees, only one possibility came. I was using a compass to determine the third side's possibilities. I checked the internet if there were questions like this, but there were none. Also, no site said specifically that it works for $90+$ angles. Does this mean that the SSA congruence can work for angles greater than $90$ degrees? If so, I was taught very wrong in school. geometry triangles Share edited Feb 21, 2019 at 11:19 Vinyl_cape_jawa 4,93477 gold badges3030 silver badges4949 bronze badges asked Feb 21, 2019 at 10:55 archmundadaarchmundada 16988 bronze badges $\endgroup$ 2 $\begingroup$ See SSA : "If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent. The opposite side is sometimes longer when the corresponding angles are acute, but it is always longer when the corresponding angles are right or obtuse." $\endgroup$ Mauro ALLEGRANZA – Mauro ALLEGRANZA 2019-02-21 11:01:20 +00:00 Commented Feb 21, 2019 at 11:01 $\begingroup$ What I call "Side-Side-non-acute-Angle" (SSnaA) does work, as you have determined. That said, it's not fair to say that you were taught "very wrong" in school. "Side-Side-Angle" is not a valid congruence pattern, due to the ambiguity in the acute case; the nature of the angle is additional information that sometimes resolves any ambiguity, but that information isn't always available. (Compare SAS or ASA, which work without additional information.) Note: If you were taught the "Hypotenuse-Leg" congruence pattern, then you were taught "Side-Side-right-Angle", so there's that. :) $\endgroup$ Blue – Blue 2019-02-21 13:51:23 +00:00 Commented Feb 21, 2019 at 13:51 Add a comment \| 4 Answers 4 Reset to default 2 $\begingroup$ Here is a trigonometric proof. Say $\triangle ABC$ has an angle at $A$ with a measure of $x$, and we know side $AC$ has length $b$ and side $BC$ has length $a$. We now want to find the side length of $AB$, or $c$. Using the law of cosines, $$b^2 + c^2 -2bc(\cos(x)) = a^2,$$ and rearranging gives us a quadratic in $c$: $$c^2 - 2b(\cos(x))(c) + (b^2- a^2) = 0.$$ Using the quadratic formula, we get c = 2acos(x) +/- sqrt(4a^2cos^2(x)-4(b^2-a^2))/2. Taking the 4s out of the square root and dividing, we get c = acos(x) +/- sqrt(a^2cos^2(x) - (b^2-a^2)) We know that if x > 90, or triangle ABC is obtuse, then cos (x) must be negative. Therefore, when triangle ABC is obtuse, we have that c is a negative number plus or minus a positive number. (Square roots are always positive. ) This means that only the positive sign works, since the negative sign means a negative minus a positive, or a negative, and side lengths can't be negative, so there's only one solution. You're right! It does work. Following this logic, we know that there is only one solution when sqrt(a^2cos^2(x) - (b^2-a^2)) > acos(x), since then the negative sign is invalid. Doing a bunch of algebra yields b < a, which is the general formula for when SSA works. Share edited Jan 22, 2020 at 23:30 Xander Henderson♦ 32.8k2525 gold badges7373 silver badges122122 bronze badges answered Jan 22, 2020 at 22:57 AnonymousPerson.1xAnonymousPerson.1x 3633 bronze badges $\endgroup$ 1 1 $\begingroup$ Your answer is quite hard to read, as it lacks proper mathematical formatting and whitespace. I have made a few edits to demonstrate how you can format your post. Can you please finish the job? You can find additional help in the MathJax tutorial. $\endgroup$ Xander Henderson – Xander Henderson ♦ 2020-01-22 23:31:23 +00:00 Commented Jan 22, 2020 at 23:31 Add a comment \| 0 $\begingroup$ There is the following theorem. Let $AB=PQ$, $BC=QR$, $\measuredangle A=\measuredangle P$ and $BC>AB$. Thus, $\Delta ABC\cong\Delta PQR.$ If $\measuredangle A\geq 90^{\circ}$ we can use this theorem immediately, but it's true also for $\measuredangle A<90^{\circ}.$ Share edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Feb 21, 2019 at 11:03 Michael RozenbergMichael Rozenberg 208k3131 gold badges171171 silver badges294294 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ If another triangle has parts congruent to angle A and sides AB and BC of our triangle, and angle A is not acute, then the two triangles are congruent. If angle A is acute, and BC > or = to AB, then they are congruent. Also congruent if either angle B or C is a right angle. Otherwise (i.e., angle A is acute, ABC is not a right triangle, and BC < AB) there are two triangles, the smaller of which has an obtuse angle at C which is supplementary to its cousin of the larger triangle. For this latter case, there is only congruence if we also know both original triangles (ours and the other one) are obtuse or both acute. Share answered Mar 19, 2021 at 18:27 Harold HannettHarold Hannett 1 $\endgroup$ Add a comment \| 0 $\begingroup$ Make a triangle with marking the following: given sides a, b, unknown side c and non- included angle d (take it near a and not b). Drop an altitude on c. Take one part of c as x and so the other as c - x; and apply Pythagoras Theorem on both triangles and then simultaneously solve the two equations by elimination method. You will get a quadratic equation like this: c^2 - 2xc - (a^2 - b^2) = 0 After solving the equation, you will get the formula to calculate the third side, if two sides and non-included angle are given: c = a cos d +- {a^2 (1 + (cos d)^2) - b^2} Which can be changed to: c = a cos d +- {a^2 (2 - (sin d)^2) - b^2} c = third side, a and b = given two sides and d = the non-included angle Since, cos d is negative for obtuse angles, the square root should only be added and not subtracted to get positive answer as sides of triangle are positive. For right angles, cos d = cos 90 = 0 and sin d = sin 90 = 1 c = a 0 +- {a^2 (2 - 1) - b^2} = +- (a^2 - b^2) To get positive answer, we only take the positive square root and not the negative one. So, there is only one value of third side, if the two other sides and the non-included angle are given, for right and obtuse angles. So, congruence theorems (SSA) holds for right angles (RHS) and obtuse angles (SSO). Share edited Aug 26, 2024 at 13:43 answered May 15, 2024 at 10:28 Syed Owais AliSyed Owais Ali 4655 bronze badges $\endgroup$ 1 $\begingroup$ Go to lostmathlessons.blogspot.com for good explanation. $\endgroup$ Syed Owais Ali – Syed Owais Ali 2024-05-31 09:55:45 +00:00 Commented May 31, 2024 at 9:55 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry triangles See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related geometry triangles side-side-side \| prove my teacher she is wrong? 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189166	https://thirdspacelearning.com/gcse-maths/geometry-and-measure/perimeter-of-compound-shapes/	What is perimeter of compound shapes? How to calculate the perimeter of compound shapes Perimeter of compound shapes worksheet Perimeter of compound shapes examples Example 1: rectilinear shapes Example 2: rectilinear shapes Example 3: finding the missing side given the perimeter Common misconceptions Practice perimeter of compound shapes questions Perimeter of compound shapes GCSE questions Learning checklist Next lessons Still stuck? GCSE Tutoring Programme Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring. Teacher-trusted tutoring In order to access this I need to be confident with: Arithmetic Fractions, decimals and percentages 2D shapes How to work out perimeter Area of a rectangle Circumference of a circle This topic is relevant for: Introduction What is perimeter of compound shapes? How to calculate the perimeter of compound shapes Perimeter of compound shapes worksheet Perimeter of compound shapes examples ↓ Example 1: rectilinear shapes Example 2: rectilinear shapes Example 3: finding the missing side given the perimeter Common misconceptions Practice perimeter of compound shapes questions Perimeter of compound shapes GCSE questions Learning checklist Next lessons Still stuck? GCSE Maths Geometry and Measure 2D Shape Perimeter Compound Shape Perimeter Of Compound Shapes Here we will learn about the perimeter of compound shapes, including how to find the perimeter of compound shapes. There are also perimeter of compound shapesworksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck. What is perimeter of compound shapes? Perimeter of compound shapes is the distance around a compound shape. The perimeter of a shape is finding the total distance around the outside of a 2D shape. A compound shape (or composite shape) is any shape which is made up of two or more shapes. They are often rectilinear shapes which have straight sides and right angles. For example, Find the perimeter of this compound shape. To find the perimeter we add all the lengths of the sides. 3+2+4+3+7+5=24 The perimeter is 24 \ cm. What is perimeter of compound shapes? How to calculate the perimeter of compound shapes In order to calculate the perimeter of compound shape: Work out any missing lengths. Add all the side lengths. Write the final answer with the correct units. Explain how to calculate the perimeter of compound shapes Perimeter of compound shapes worksheet Get your free perimeter of compound shapes worksheet of 20+ questions and answers. Includes reasoning and applied questions. DOWNLOAD FREE x Perimeter of compound shapes worksheet Get your free perimeter of compound shapes worksheet of 20+ questions and answers. Includes reasoning and applied questions. DOWNLOAD FREE Related lessons on how to work out perimeter Perimeter of compound shapes is part of our series of lessons to support revision on 2D shapes and how to work out perimeter. You may find it helpful to start with the main 2D shapes lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include: 2D shapes How to work out perimeter Polygons Symmetry Perimeter of compound shapes examples Example 1: rectilinear shapes Calculate perimeter. Work out any missing lengths. There are some missing lengths we need to work out first. We can look at the vertical lengths. The missing vertical length is 7 \ cm because 11-4=7. We can then look at the missing horizontal length. The missing side is 5 \ cm because 7-2=5. 2Add all the side lengths. Starting in the top left corner and going clockwise around the shape, we can add the sides together. 2+7+5+4+7+11=36 3Write the final answer with the correct units. The perimeter of the compound shape is 36 \ cm. Example 2: rectilinear shapes Calculate perimeter. Work out any missing lengths. There are some missing lengths we need to work out first. We can look at the vertical lengths. The missing vertical lengths are 6 \ cm and 13 \ cm. We can then look at the missing horizontal length. The missing side is 16 \ cm because 7+5+4=16. Add all the side lengths. Starting in the top left corner and going clockwise around the shape, we can add the sides together. 16+13+4+6+5+6+7+13=70 Write the final answer with the correct units. The perimeter of the compound shape is 70 \ cm. Example 3: finding the missing side given the perimeter The perimeter is 58.6 \ m, find x. Work out any missing lengths. We can work out the missing horizontal length. The missing side is 11.9 \ m because 4.1+7.8=11.9. We still don’t know the missing vertical length. We can write it as x+10.3. Add all the side lengths. Starting in the bottom left corner and going clockwise around the shape, we can add the sides we have together. x+7.8+10.3+4.1+(x+10.3)+11.9 This algebraic expression simplifies to 2x+44.4. We can then use the information that the perimeter is 58.6 \ m to form an equation and solve it to find x. \begin{aligned} 2x+44.4 &= 58.6 \\ 2x &= 14.2 \\ x&=7.1 \end{aligned} Write the final answer with the correct units. The perimeter of the compound shape is 58.6 \ m. The missing side is 7.1 \ m, so x=7.1. Common misconceptions Order of adding the sides Addition is commutative. This means it doesn’t matter what order the sides of the shape are added in. Check the units Check which units are involved in the question as there may be a mixture. Practice perimeter of compound shapes questions Find the perimeter of this 2D shape. 31 \ cm 30 \ cm 39 \ cm 38 \ cm Add all the sides together. The perimeter is 38 \ cm because 4+2+7+8+11+6=38. Find the perimeter of this 2D shape. 23 \ cm 25 \ cm 28 \ cm 20 \ cm Add all the sides together. The perimeter is 28 \ cm because 5+1+4+3+7+8=28. Find the perimeter of this 2D shape. 33 \ m 44 \ m 78 \ m 148 \ m The missing sides are 5 \ m and 6 \ m. The perimeter is 44 \ m because 5+10+12+4+7+6=44. Find the perimeter of this 2D shape. 31 \ m 114 \ m 90 \ m 62 \ m The missing sides are 9 \ m and 22 \ m. The perimeter is 62 \ m because 9+22+3+14+6+8=62. Find the perimeter of this 2D shape. 49 \ cm 270 \ cm 66 \ cm 80 \ cm The missing sides are 6 \ cm , 7 \ cm and 18 \ cm. The perimeter is 80 \ cm because 6+7+6+7+3+18+15+18=80. The perimeter of this 2D shape is 48 \ m. Find the value of x. 12 14 16 18 The missing horizontal side is 10 \ cm. The two vertical missing sides will add together to make x \ cm. The perimeter of the shape can be used to form an equation which can be solved to find x. \begin{aligned} 2x+20 &= 48 \\ 2x &= 28 \\ x&=14 \end{aligned} Perimeter of compound shapes GCSE questions Calculate the perimeter of the shape below. (2 marks) Show answer For finding the missing sides. (1) 42 \ m (1) Calculate the perimeter of the shape below. (3 marks) Show answer For finding the missing base, 10 \ cm. (1) (9 \times 2)+10+(2 \times 9) (1) 46 \ cm (1) (a) Calculate the perimeter of the shape below. (b) A regular pentagon has the same perimeter as the shape in part a. Find the length of one side of the regular pentagon. (5 marks) Show answer (a) 45 \ mm = 4.5 \ cm \ (or 6.8 \ cm = 68 \ mm) (1) 1.8 \ cm \ ( or 18 \ mm) and 3.1 \ cm ( or 31 \ mm). (1) Perimeter is 22.6 \ cm \ ( or 226 \ mm) . (1) (b) 22.6 \div 5 \ ( or 226 \div 5) (1) 4.52 \ cm \ ( or 45.2 \ mm) (1) Learning checklist You have now learned how to: ");--ub-list-item-fa-li-top:3px;--ub-list-item-spacing:0px;"> ");"> Work out the perimeter of compound shapes ");"> Find the missing side of a compound shape, given its perimeter ");"> Solve problems involving the perimeter of compound shapes (or composite shapes) The next lessons are Area Circles, sectors and arcs Angles Still stuck? Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors. Find out more about our GCSE maths tuition programme. Introduction What is perimeter of compound shapes? How to calculate the perimeter of compound shapes Perimeter of compound shapes worksheet Perimeter of compound shapes examples ↓ Example 1: rectilinear shapes Example 2: rectilinear shapes Example 3: finding the missing side given the perimeter Common misconceptions Practice perimeter of compound shapes questions Perimeter of compound shapes GCSE questions Learning checklist Next lessons Still stuck? We use essential and non-essential cookies to improve the experience on our website. Please read our Cookies Policy for information on how we use cookies and how to manage or change your cookie settings.Accept Privacy & Cookies Policy Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. 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This ensures that the cookie consent box will not be presented again upon re-entry. \| Session \| HTTP Cookie \| \| 1.gif \| Cookiebot \| Used to count the number of sessions to the website, necessary for optimizing CMP product delivery. \| Session \| Pixel Tracker \| \| cu \| Adelphic \| Used to detect if the visitor has accepted the marketing category in the cookie banner. This cookie is necessary for GDPR-compliance of the website. \| 1 year \| HTTP Cookie \| \| bcookie \| LinkedIn \| Used in order to detect spam and improve the website's security. \| 1 year \| HTTP Cookie \| \| li_gc \| LinkedIn \| Stores the user's cookie consent state for the current domain \| 180 days \| HTTP Cookie \| \| m \| Stripe \| Determines the device used to access the website. This allows the website to be formatted accordingly. \| 400 days \| HTTP Cookie \| \| _ab \| m.stripe.com \| This cookie is necessary for making credit card transactions on the website. The service is provided by Stripe.com which allows online transactions without storing any credit card information. \| Session \| HTML Local Storage \| \| _mf \| m.stripe.com \| This cookie is necessary for making credit card transactions on the website. The service is provided by Stripe.com which allows online transactions without storing any credit card information. \| Session \| HTML Local Storage \| \| id \| m.stripe.com \| Pending \| Session \| HTML Local Storage \| \| mw \| Zeotap \| Determines whether the user should have access to restricted areas on the website. \| Session \| Pixel Tracker \| \| [SessionID#27] \| PayPal \| This cookie is provided by Paypal. The cookie is used in context with transactions on the website - The cookie is necessary for secure transactions. \| 1 year \| HTTP Cookie \| \| ts_c \| PayPal \| Used in context with the PayPal payment-function on the website. The cookie is necessary for making a safe transaction through PayPal. \| 1 year \| HTTP Cookie \| \| opt_out \| Nativo \| Used to detect if the visitor has accepted the marketing category in the cookie banner. This cookie is necessary for GDPR-compliance of the website. \| 1 year \| HTTP Cookie \| \| audit \| Rubicon Project \| Determines whether the visitor has accepted the cookie consent box. This ensures that the cookie consent box will not be presented again upon re-entry. \| 1 year \| HTTP Cookie \| \| U[x4] \| storygize.net \| Necessary for the sign-up function on the website. \| Session \| HTTP Cookie \| \| cookietest \| Vidoomy \| This cookie is used to determine if the visitor has accepted the cookie consent box. \| Session \| HTTP Cookie \| \| AWSALB \| widgets.guidestar.org \| Registers which server-cluster is serving the visitor. This is used in context with load balancing, in order to optimize user experience. \| 7 days \| HTTP Cookie \| \| AWSALBCORS \| widgets.guidestar.org \| Registers which server-cluster is serving the visitor. This is used in context with load balancing, in order to optimize user experience. \| 7 days \| HTTP Cookie \| [x] Preferences (9) Show details (Preferences ) Preference cookies enable a website to remember information that changes the way the website behaves or looks, like your preferred language or the region that you are in. \| Name \| Provider \| Purpose \| Maximum Storage Duration \| Type \| --- --- \| __lc_cid \| LiveChat \| Necessary for the functionality of the website's chat-box function. \| 400 days \| HTTP Cookie \| \| __lc_cst \| LiveChat \| Necessary for the functionality of the website's chat-box function. \| 400 days \| HTTP Cookie \| \| __oauth_redirect_detector \| LiveChat \| Allows the website to recoqnise the visitor, in order to optimize the chat-box functionality. \| 1 day \| HTTP Cookie \| \| dc \| Between Digital \| Necessary for the functionality of the website's chat-box function. \| 1 year \| HTTP Cookie \| \| ss \| Between Digital \| Necessary for the functionality of the website's chat-box function. \| 1 year \| HTTP Cookie \| \| g \| RTB House \| Used in context with the language setting on the website. Facilitates the translation into the preferred language of the visitor. \| 1 year \| HTTP Cookie \| \| lidc \| LinkedIn \| Registers which server-cluster is serving the visitor. This is used in context with load balancing, in order to optimize user experience. \| 1 day \| HTTP Cookie \| \| 1 \| m.stripe.com \| This cookie is used in conjunction with the payment window - The cookie is necessary for making secure transactions on the website. \| Session \| HTML Local Storage \| \| drip_client# \| Drip \| Pending \| 2 years \| HTTP Cookie \| [x] Statistics (7) Show details (Statistics ) Statistic cookies help website owners to understand how visitors interact with websites by collecting and reporting information anonymously. \| Name \| Provider \| Purpose \| Maximum Storage Duration \| Type \| --- --- \| c.gif \| Microsoft \| Collects data on the user’s navigation and behavior on the website. This is used to compile statistical reports and heatmaps for the website owner. \| Session \| Pixel Tracker \| \| _livechat_has_visited \| Livechat \| Identifies the visitor across devices and visits, in order to optimize the chat-box function on the website. \| Persistent \| HTML Local Storage \| \| log \| Media.net \| Used to distinguish between internal and external visitors to the website, in order to obtain more concise statistical data regarding the use of the website. \| Session \| Pixel Tracker \| \| g.gif \| safecatch.com \| Registers statistical data on users' behaviour on the website. Used for internal analytics by the website operator. \| Session \| Pixel Tracker \| \| browser_id \| Nazwa.pl \| Used to recognise the visitor's browser upon reentry on the website. \| 1 year \| HTTP Cookie \| \| tk_rl \| WordPress.com \| Registers data on visitors' website-behaviour. This is used for internal analysis and website optimization. \| Session \| HTTP Cookie \| \| tk_ro \| WordPress.com \| Registers data on visitors' website-behaviour. This is used for internal analysis and website optimization. \| Session \| HTTP Cookie \| [x] Marketing (314) Show details (Marketing ) Marketing cookies are used to track visitors across websites. The intention is to display ads that are relevant and engaging for the individual user and thereby more valuable for publishers and third party advertisers. \| Name \| Provider \| Purpose \| Maximum Storage Duration \| Type \| --- --- \| rxuuid[x2] \| RhythmOne Unruly \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 1 year \| HTTP Cookie \| \| dsp_match/275 \| Improve Digital \| Pending \| Session \| Pixel Tracker \| \| ebda \| Triplelift \| Pending \| Session \| Pixel Tracker \| \| getuid \| Triplelift \| This cookie is set by the audience manager of the website to determine the time and frequencies of visitor data synchronization - cookie data synchronization is used to synchronize and gather visitor data from several websites. \| Session \| Pixel Tracker \| \| sa-user-id[x7] \| Triplelift Beachfront Gumgum Zeotap Seedtag Smart StackAdapt \| Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. \| Session \| HTTP Cookie \| \| sa-user-id-v2[x7] \| Triplelift Beachfront Gumgum Zeotap Seedtag Smart StackAdapt \| Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. \| Session \| HTTP Cookie \| \| sa-user-id-v3[x7] \| Triplelift Beachfront Gumgum Zeotap Seedtag Smart StackAdapt \| Collects data on the user across websites - This data is used to make advertisement more relevant. \| Session \| HTTP Cookie \| \| sync \| Triplelift \| Detects whether partner data synchronization is functioning and currently running - This function sends user data between third-party advertisement companies for the purpose of targeted advertisements. \| 30 days \| HTTP Cookie \| \| sync3 \| Triplelift \| Pending \| 30 days \| HTTP Cookie \| \| tluid \| Triplelift \| This cookie is used to identify the visitor and optimize ad-relevance by collecting visitor data from multiple websites – this exchange of visitor data is normally provided by a third-party data-center or ad-exchange. \| 3 months \| HTTP Cookie \| \| tluidp \| Triplelift \| Pending \| 3 months \| HTTP Cookie \| \| xuid \| Triplelift \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| Pixel Tracker \| \| sync[x3] \| Sportradar ads.servenobid.com Bidswitch \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| Pixel Tracker \| \| C \| Adform \| Used to check if the user's browser supports cookies. \| 1 month \| HTTP Cookie \| \| uid[x9] \| Adform admatic.de Adotmob adtarget.biz Criteo server.smartytech.io static.cdn.admatic.de sync.contextualadv.com Amobee \| Registers a unique user ID that recognises the user's browser when visiting websites that use the same ad network. The purpose is to optimise display of ads based on the user's movements and various ad providers' bids for displaying user ads. \| 2 months \| HTTP Cookie \| \| UserID1 \| Adition Technologies \| This cookie is used to collect information on a visitor. This information will become an ID string with information on a specific visitor – ID information strings can be used to target groups with similar preferences, or can be used by third-party domains or ad-exchanges. \| 3 months \| HTTP Cookie \| \| ac_r \| Adman Media \| Pending \| 7 days \| HTTP Cookie \| \| admtr \| Adman Media \| Used in context with video-advertisement. The cookie limits the number of times a user is shown the same advertisement. The cookie is also used to ensure relevance of the video-advertisement to the specific user. \| 7 days \| HTTP Cookie \| \| uids[x3] \| static.cdn.admatic.de pixfuture.com prebidserver.pixfuture.com \| Registers user behaviour and navigation on the website, and any interaction with active campaigns. This is used for optimizing advertisement and for efficient retargeting. \| 3 months \| HTTP Cookie \| \| tuuid[x6] \| Admedo Between Digital Bidswitch DemandBase KPN Gumgum \| Collects visitor data related to the user's visits to the website, such as the number of visits, average time spent on the website and what pages have been loaded, with the purpose of displaying targeted ads. \| 1 year \| HTTP Cookie \| \| tuuid_lu[x5] \| Admedo Bidswitch DemandBase KPN Gumgum \| Contains a unique visitor ID, which allows Bidswitch.com to track the visitor across multiple websites. This allows Bidswitch to optimize advertisement relevance and ensure that the visitor does not see the same ads multiple times. \| 1 year \| HTTP Cookie \| \| am-uid[x3] \| Admixer E-planning.net \| This cookie is used to identify the visitor and optimize ad-relevance by collecting visitor data from multiple websites – this exchange of visitor data is normally provided by a third-party data-center or ad-exchange. \| 3 months \| HTTP Cookie \| \| uuid2 \| Appnexus \| Registers a unique ID that identifies a returning user's device. The ID is used for targeted ads. \| 3 months \| HTTP Cookie \| \| XANDR_PANID \| Appnexus \| This cookie registers data on the visitor. The information is used to optimize advertisement relevance. \| 3 months \| HTTP Cookie \| \| partners \| Adotmob \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 180 days \| HTTP Cookie \| \| uuid[x2] \| Adotmob MediMath \| This cookie is used to optimize ad relevance by collecting visitor data from multiple websites – this exchange of visitor data is normally provided by a third-party data-center or ad-exchange. \| 180 days \| HTTP Cookie \| \| av-mid \| AntVoice \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 400 days \| HTTP Cookie \| \| av-tp-bsw \| AntVoice \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 2 days \| HTTP Cookie \| \| track/cmf/generic \| The Trade Desk \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| Pixel Tracker \| \| track/cmf/rubicon \| The Trade Desk \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| Pixel Tracker \| \| vmuid \| Adtelligent \| This cookie allows visitors to leave comments on their use of the website - This information is used to optimize the website. The cookie also assigns a specific session ID on the current visit. This allows the website’s operator to make reports on the specific visit. \| 3 months \| HTTP Cookie \| \| UID[x2] \| adx.opera.com Freewheel \| Unique user ID that recognizes the user on returning visits \| 1 year \| HTTP Cookie \| \| ad-id \| Amazon \| Used by Amazon Advertising to register user actions and target content on the website based on ad clicks on a different website. \| 7 months \| HTTP Cookie \| \| ad-privacy \| Amazon \| Used by Amazon Advertising to register user actions and target content on the website based on ad clicks on a different website. \| 400 days \| HTTP Cookie \| \| _ga \| Google \| Used to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels. \| 2 years \| HTTP Cookie \| \| _ga# \| Google \| Used to send data to Google Analytics about the visitor's device and behavior. Tracks the visitor across devices and marketing channels. \| 2 years \| HTTP Cookie \| \| gcl_au[x2] \| Google \| Used by Google AdSense for experimenting with advertisement efficiency across websites using their services. \| 3 months \| HTTP Cookie \| \| aniC[x3] \| Aniview omnitagjs.com \| Used in context with video-advertisement. The cookie limits the number of times a user is shown the same advertisement. The cookie is also used to ensure relevance of the video-advertisement to the specific user. \| 30 days \| HTTP Cookie \| \| ut \| Between Digital \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| 1 year \| HTTP Cookie \| \| __#_cid \| Beachfront \| Pending \| 1 year \| HTTP Cookie \| \| __141_exp \| Beachfront \| Pending \| 7 days \| HTTP Cookie \| \| __bfio_sync \| Beachfront \| Pending \| 3 days \| HTTP Cookie \| \| __io_cid \| Beachfront \| Used in conjunction with video-advertisement - Detects where on the website the video-advertisements should be displayed. \| 1 year \| HTTP Cookie \| \| bito \| Beeswax \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 13 months \| HTTP Cookie \| \| bitoIsSecure \| Beeswax \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 13 months \| HTTP Cookie \| \| bsw_origin_init \| Bidswitch \| Collects visitor data related to the user's visits to the website, such as the number of visits, average time spent on the website and what pages have been loaded, with the purpose of displaying targeted ads. \| Session \| HTTP Cookie \| \| custom_data \| Bidswitch \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| 1 day \| HTTP Cookie \| \| syncd \| Bidswitch \| Pending \| Session \| Pixel Tracker \| \| MR \| Microsoft \| Pending \| 7 days \| HTTP Cookie \| \| MUID \| Microsoft \| Pending \| 1 year \| HTTP Cookie \| \| pixel/cookiesync \| Bidtellect \| Detects whether partner data synchronization is functioning and currently running - This function sends user data between third-party advertisement companies for the purpose of targeted advertisements. \| Session \| Pixel Tracker \| \| pixel/cookiesyncredir \| Bidtellect \| Used for data-synchronization with advertisement networks. \| Session \| Pixel Tracker \| \| CMID \| Casale Media \| Collects visitor data related to the user's visits to the website, such as the number of visits, average time spent on the website and what pages have been loaded, with the purpose of displaying targeted ads. \| 1 year \| HTTP Cookie \| \| CMPRO \| Casale Media \| Collects data on visitor behaviour from multiple websites, in order to present more relevant advertisement - This also allows the website to limit the number of times that they are shown the same advertisement. \| 3 months \| HTTP Cookie \| \| CMPS \| Casale Media \| Collects visitor data related to the user's visits to the website, such as the number of visits, average time spent on the website and what pages have been loaded, with the purpose of displaying targeted ads. \| 3 months \| HTTP Cookie \| \| pb_rtb_ev_part \| Pulsepoint \| Pending \| 1 year \| HTTP Cookie \| \| VP \| Pulsepoint \| Pending \| 1 year \| HTTP Cookie \| \| cto_bundle \| Criteo \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 1 year \| HTTP Cookie \| \| demdex \| Adobe Inc. \| Via a unique ID that is used for semantic content analysis, the user's navigation on the website is registered and linked to offline data from surveys and similar registrations to display targeted ads. \| 180 days \| HTTP Cookie \| \| dpm \| Adobe Inc. \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 180 days \| HTTP Cookie \| \| IDE \| Google \| Used by Google DoubleClick to register and report the website user's actions after viewing or clicking one of the advertiser's ads with the purpose of measuring the efficacy of an ad and to present targeted ads to the user. \| 400 days \| HTTP Cookie \| \| __eDId \| eskimi.com \| Collects data on visitors' behaviour and interaction - This is used to optimize the website and make advertisement on the website more relevant. \| 30 days \| HTTP Cookie \| \| ct/upi/pid/gjIEMT18 \| Adobe Inc. \| Pending \| Session \| Pixel Tracker \| \| everest_g_v2 \| Adobe Inc. \| Used for targeted ads and to document efficacy of each individual ad. \| 1 year \| HTTP Cookie \| \| _uid \| Freewheel \| Enables the visitor to share content from the website onto social media platforms or websites. \| 180 days \| HTTP Cookie \| \| NID \| Google \| Registers a unique ID that identifies a returning user's device. The ID is used for targeted ads. \| 6 months \| HTTP Cookie \| \| pagead/1p-user-list/# \| Google \| Tracks if the user has shown interest in specific products or events across multiple websites and detects how the user navigates between sites. This is used for measurement of advertisement efforts and facilitates payment of referral-fees between websites. \| Session \| Pixel Tracker \| \| usersync[x2] \| Gumgum SpringServe \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| Session \| Pixel Tracker \| \| vst \| Gumgum \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 1 year \| HTTP Cookie \| \| callback \| ID5 \| Collects data on visitor behaviour from multiple websites, in order to present more relevant advertisement - This also allows the website to limit the number of times that they are shown the same advertisement. \| Session \| HTTP Cookie \| \| car \| ID5 \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| HTTP Cookie \| \| cf \| ID5 \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| Session \| HTTP Cookie \| \| cip \| ID5 \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| HTTP Cookie \| \| cnac \| ID5 \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| HTTP Cookie \| \| gpp \| ID5 \| Pending \| Session \| HTTP Cookie \| \| s/846/9.gif \| ID5 \| Pending \| Session \| Pixel Tracker \| \| _ljtrtb# \| Sovrn \| Collects data related to reader interests, context, demographics and other information on behalf of the Lijit platform with the purpose of finding interested users on websites with related content. \| 1 year \| HTTP Cookie \| \| ljt_reader \| Sovrn \| Collects data related to reader interests, context, demographics and other information on behalf of the Lijit platform with the purpose of finding interested users on websites with related content. \| 1 year \| HTTP Cookie \| \| ljtrtb \| Sovrn \| Collects data related to reader interests, context, demographics and other information on behalf of the Lijit platform with the purpose of finding interested users on websites with related content. \| 1 year \| HTTP Cookie \| \| merge \| Sovrn \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Session \| Pixel Tracker \| \| viewer_token \| LoopMe \| Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. \| 3 months \| HTTP Cookie \| \| [empty name] \| Media.net \| Pending \| Session \| HTTP Cookie \| \| bping.php \| Media.net \| This cookie is used to register information on the ad-delivery-system – this optimizes the ad relevance towards the visitor. \| Session \| Pixel Tracker \| \| receive-cookie-deprecation[x2] \| Media.net Rubicon Project \| Collects information on user behaviour on multiple websites. This information is used in order to optimize the relevance of advertisement on the website. \| 180 days \| HTTP Cookie \| \| t/v2/sync \| Xaxis \| Used for data-synchronization with advertisement networks. \| Session \| Pixel Tracker \| \| nauid \| N.Rich \| Collects data on visitor behaviour from multiple websites, in order to present more relevant advertisement - This also allows the website to limit the number of times that they are shown the same advertisement. \| 400 days \| HTTP Cookie \| \| ayl_visitor \| omnitagjs.com \| This cookie registers data on the visitor. The information is used to optimize advertisement relevance. \| 30 days \| HTTP Cookie \| \| visitor/sync \| omnitagjs.com \| This cookie is used to assign specific visitors into segments, this segmentation is based on visitor behavior on the website - the segmentation can be used to target larger groups. \| Session \| Pixel Tracker \| \| sync/prebid \| Openx \| Pending \| Session \| Pixel Tracker \| \| obuid \| Outbrain \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 3 months \| HTTP Cookie \| \| suid/101967 \| Nativo \| Pending \| Session \| Pixel Tracker \| \| setuid \| prebidserver.pixfuture.com \| Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. \| Session \| Pixel Tracker \| \| csuuid \| primis.tech \| Collects information on user behaviour on multiple websites. This information is used in order to optimize the relevance of advertisement on the website. \| 25 days \| HTTP Cookie \| \| KCCH \| Pubmatic \| Registers if the PubMatic partner-cookie has been set in the user's browser. \| 1 day \| HTTP Cookie \| \| KTPCACOOKIE \| Pubmatic \| Registers a unique ID that identifies the user's device during return visits across websites that use the same ad network. The ID is used to allow targeted ads. \| 3 months \| HTTP Cookie \| \| PubMatic_USP \| PubMatic \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Persistent \| HTML Local Storage \| \| iqm.retarget.uid \| IQM \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 15 days \| HTTP Cookie \| \| pixel/p-EtBqU4Lj3YbAv.gif \| Quantcast \| Pending \| Session \| Pixel Tracker \| \| pixel/p-zLwwakwy-hZw3.gif \| Quantcast \| Pending \| Session \| Pixel Tracker \| \| eud \| Zeta Global \| Registers user data, such as IP address, geographical location, visited websites, and what ads the user has clicked, with the purpose of optimising ad display based on the user's movement on websites that use the same ad network. \| 1 year \| HTTP Cookie \| \| rud \| Zeta Global \| Registers user data, such as IP address, geographical location, visited websites, and what ads the user has clicked, with the purpose of optimising ad display based on the user's movement on websites that use the same ad network. \| 1 year \| HTTP Cookie \| \| ruds \| Zeta Global \| Registers user data, such as IP address, geographical location, visited websites, and what ads the user has clicked, with the purpose of optimising ad display based on the user's movement on websites that use the same ad network. \| Session \| HTTP Cookie \| \| [empty name] \| Nazwa.pl \| Pending \| Session \| Pixel Tracker \| \| rc \| Nazwa.pl \| Pending \| 1 day \| HTTP Cookie \| \| audit_p \| Rubicon Project \| Pending \| 1 year \| HTTP Cookie \| \| fcap \| Rubicon Project \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| Persistent \| HTML Local Storage \| \| khaos \| Rubicon Project \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Persistent \| HTML Local Storage \| \| khaos \| Rubicon Project \| Registers user data, such as IP address, geographical location, visited websites, and what ads the user has clicked, with the purpose of optimising ad display based on the user's movement on websites that use the same ad network. \| 1 year \| HTTP Cookie \| \| khaos_p \| Rubicon Project \| Pending \| 1 year \| HTTP Cookie \| \| pux \| Rubicon Project \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Persistent \| HTML Local Storage \| \| tap.php \| Rubicon Project \| Registers data on visitor such as IP addresses, geographical location and advertisement interaction. This information is used optimize the advertisement on websites that make use of Rubicon Project-services. \| Session \| Pixel Tracker \| \| _gcl_ls[x2] \| safecatch.com Google \| Tracks the conversion rate between the user and the advertisement banners on the website - This serves to optimise the relevance of the advertisements on the website. \| Persistent \| HTML Local Storage \| \| sbjs_current \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| sbjs_current_add \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| sbjs_first \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| sbjs_first_add \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| sbjs_migrations \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| sbjs_session \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| 1 day \| HTTP Cookie \| \| sbjs_udata \| WordPress.com \| Collects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. \| Session \| HTTP Cookie \| \| tk_lr \| WordPress.com \| Collects data on visitors' preferences and behaviour on the website - This information is used make content and advertisement more relevant to the specific visitor. \| 1 year \| HTTP Cookie \| \| tk_or \| WordPress.com \| Collects data on visitors' preferences and behaviour on the website - This information is used make content and advertisement more relevant to the specific visitor. \| 5 years \| HTTP Cookie \| \| tk_r3d \| WordPress.com \| Collects data on visitors' preferences and behaviour on the website - This information is used make content and advertisement more relevant to the specific visitor. \| 3 days \| HTTP Cookie \| \| tk_tc \| WordPress.com \| Collects data on visitors' preferences and behaviour on the website - This information is used make content and advertisement more relevant to the specific visitor. \| Session \| HTTP Cookie \| \| st_cs \| Seedtag \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 1 year \| HTTP Cookie \| \| st_csd \| Seedtag \| This cookie is set by the audience manager of the website to determine the time and frequencies of visitor data synchronization - cookie data synchronization is used to synchronize and gather visitor data from several websites. \| 1 year \| HTTP Cookie \| \| pid# \| servenobid.com \| Presents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers. \| 7 days \| HTTP Cookie \| \| suid \| Simpli.fi \| Collects information on user preferences and/or interaction with web-campaign content - This is used on CRM-campaign-platform used by website owners for promoting events or products. \| 1 year \| HTTP Cookie \| \| suid_legacy \| Simpli.fi \| Collects information on user preferences and/or interaction with web-campaign content - This is used on CRM-campaign-platform used by website owners for promoting events or products. \| 1 year \| HTTP Cookie \| \| csync \| Smart \| Optimises ad display based on the user's movement combined and various advertiser bids for displaying user ads. \| 30 days \| HTTP Cookie \| \| pid \| Smart \| Registers a unique ID that identifies a returning user's device. The ID is used for targeted ads. \| 3 months \| HTTP Cookie \| \| HAPLB8G \| Sonobi \| Pending \| Session \| HTTP Cookie \| \| us.gif \| Sonobi \| Used to track the visitor's usage of GIFs - This serves for analytical and marketing purposes. \| Session \| Pixel Tracker \| \| idsync/ex/receive/check \| Tapad \| This cookie is set by the audience manager of the website to determine the time and frequencies of visitor data synchronization - cookie data synchronization is used to synchronize and gather visitor data from several websites. \| Session \| Pixel Tracker \| \| TapAd_3WAY_SYNCS \| Tapad \| Used for data-synchronization with advertisement networks. \| 2 months \| HTTP Cookie \| \| TapAd_DID \| Tapad \| Used to determine what type of devices (smartphones, tablets, computers, TVs etc.) is used by a user. \| 2 months \| HTTP Cookie \| \| TapAd_TS \| Tapad \| Used to determine what type of devices (smartphones, tablets, computers, TVs etc.) is used by a user. \| 2 months \| HTTP Cookie \| \| SONATA_ID \| TapTap \| Used to track the visitor across multiple devices including TV. This is done in order to re-target the visitor through multiple channels. \| 19 days \| HTTP Cookie \| \| vidoomy-uids \| Vidoomy \| Used in context with video-advertisement. The cookie limits the number of times a user is shown the same advertisement. The cookie is also used to ensure relevance of the video-advertisement to the specific user. \| 1 year \| HTTP Cookie \| \| statid \| WP \| Sets a unique ID for the visitor, that allows third party advertisers to target the visitor with relevant advertisement. This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 400 days \| HTTP Cookie \| \| sync/gumgum \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/iponweb \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/rubicon/# \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1050195632187828380838 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1146416960883353224063 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1166557178287137406470 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1177844984753799941369 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1240382808428054108099 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1248681216559195866360 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1495518049130577464008 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1558017364139613955081 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1565981029497596791254 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1637956745329411197203 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1646836138013195897713 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1660262809648298776583 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/166491263189247617460 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1853206943953760322297 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1880635045170591233339 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1893458310858221416161 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1921743851012202810548 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/1951211668184741430153 \| Yahoo \| Pending \| 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Pending \| Session \| Pixel Tracker \| \| sync/triplelift/782385594243771149584 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/82785051910396541137 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/901769686252726739860 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| sync/triplelift/905074871482278128601 \| Yahoo \| Pending \| Session \| Pixel Tracker \| \| #-# \| YouTube \| Used to track user’s interaction with embedded content. \| Session \| HTML Local Storage \| \| __Secure-ROLLOUT_TOKEN \| YouTube \| Used to track user’s interaction with embedded content. \| 180 days \| HTTP Cookie \| \| __Secure-YEC \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTTP Cookie \| \| iU5q-!O9@$ \| YouTube \| Registers a unique ID to keep statistics of what videos from YouTube the user has seen. \| Session \| HTML Local Storage \| \| LAST_RESULT_ENTRY_KEY \| YouTube \| Used to track user’s interaction with embedded content. \| Session \| HTTP Cookie \| \| LogsDatabaseV2:V#\|\|LogsRequestsStore \| YouTube \| Used to track user’s interaction with embedded content. \| Persistent \| IndexedDB \| \| remote_sid \| YouTube \| Necessary for the implementation and functionality of YouTube video-content on the website. \| Session \| HTTP Cookie \| \| TESTCOOKIESENABLED \| YouTube \| Used to track user’s interaction with embedded content. \| 1 day \| HTTP Cookie \| \| VISITOR_INFO1_LIVE \| YouTube \| Tries to estimate the users' bandwidth on pages with integrated YouTube videos. \| 180 days \| HTTP Cookie \| \| YSC \| YouTube \| Registers a unique ID to keep statistics of what videos from YouTube the user has seen. \| Session \| HTTP Cookie \| \| YtIdbMeta#databases \| YouTube \| Used to track user’s interaction with embedded content. \| Persistent \| IndexedDB \| \| yt-remote-cast-available \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTML Local Storage \| \| yt-remote-cast-installed \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTML Local Storage \| \| yt-remote-connected-devices \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Persistent \| HTML Local Storage \| \| yt-remote-device-id \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Persistent \| HTML Local Storage \| \| yt-remote-fast-check-period \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTML Local Storage \| \| yt-remote-session-app \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTML Local Storage \| \| yt-remote-session-name \| YouTube \| Stores the user's video player preferences using embedded YouTube video \| Session \| HTML Local Storage \| \| usersync/adyoulike/ \| Zemanta \| Pending \| Session \| Pixel Tracker \| 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This pairing service is provided by third party advertisement hubs, which facilitates real-time bidding for advertisers. \| 3 months \| HTTP Cookie \| \| zc \| Zeotap \| Registers data on visitors from multiple visits and on multiple websites. This information is used to measure the efficiency of advertisement on websites. \| 1 year \| HTTP Cookie \| \| zsc \| Zeotap \| Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. \| 1 day \| HTTP Cookie \| Allow selection Español Ovulation Calendar Due Date Calculator Pregnancy Questions Live Help: 1-800-672-2296 Click to Donate Your gift Makes a Difference Unplanned Pregnancy Unplanned Pregnancy Am I Pregnant? Pregnancy Symptoms How Pregnancy Happens Pregnancy Tests Can I Get Pregnant If… ? 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See all results Top Pages ### Pregnancy Questions Center ### Blog ### Free Live Chat Top Tools Ovulation Calendar Due Date Calculator Pregnancy Questions Pregnancy Provider Search Baby Name Generator Search Donate Triple Screen Test Also Known as Triple Test, Multiple Marker Screening and AFP Plus The triple screen test is a maternal blood screening test that looks for three specific substances: AFP, hCG, and Estriol. AFP:alpha-fetoprotein is a protein that is produced by the fetus. hCG:human chorionic gonadotropin is a hormone produced within the placenta Estriol:estriol is an estrogen produced by both the fetus and the placenta It is a non-invasive procedure done through a blood test with little to no known risk to the mother or developing baby. What is a screening test? It is very important to remember what a screening test is before getting one performed. This will help alleviate some of the anxiety that can accompany test results. Screening tests do not look only at results from the blood test. They compare a number of different factors (including age, ethnicity, results from blood tests, etc…) and then estimate what a person’s chances are of having an abnormality. These tests DO NOT diagnose a problem; they only signal that further testing should be done. How is the triple screen test performed? The triple screen test involves drawing blood from the mother which takes about 5 to 10 minutes. The blood sample is then sent to the laboratory for testing. The results usually take a few days to receive. What are the risks and side effects to the mother or baby? Except for the discomfort of drawing blood, there are no known risks or side effects associated with the triple screen test. When is the triple screen test performed? The triple screen test is performed between the 15th and 20th week of pregnancy although results obtained in the 16th -18th week are said to be the most accurate. All pregnant women should be offered the triple screen, but it is recommendedfor women who: Have a family history of birth defects Are 35 years or older Used possible harmful medications or drugs during pregnancy Have diabetes and use insulin Had a viral infection during pregnancy Have been exposed to high levels of radiation What does the triple screen test look for? The triple screen is measuring high and low levels of AFP and abnormal levels of hCG and estriol. The results are combined with the mother’s age, weight, ethnicity, and gestation of pregnancy in order to assess probabilities of potential genetic disorders. High levels of AFP may suggest the developing baby has a neural tube defect such as spina bifidaor anencephaly. However, the most common reason for elevated AFP levels is inaccurate dating of the pregnancy. Low levels of AFP and abnormal levels of hCG and estriol may indicate that the developing baby has Trisomy 21 (Down Syndrome), Trisomy 18 (Edwards Syndrome) or another type of chromosome abnormality. Although the primary reason for conducting the test is toscreen for genetic disorders, the results of the triple screen can also be used toidentify: A multiples pregnancy Pregnancies that are more or less advanced than thought What do the triple test results mean? It is important to remember the triple test is a screening test and not a diagnostic test. This test only notes that a mother is at possible risk of carrying a baby with a genetic disorder. The triple screen test is known to have a high percentage of false-positive results. Abnormal test results warrant additional testing for making a diagnosis. A more conservative approach involves performing a second triple screen followed by a high definition ultrasound. If the testing still maintains abnormal results, a more invasive procedure like amniocentesis may be performed. Invasive testing procedures should be discussed thoroughly with your healthcare provider and between you and your partner. Additional counseling and discussions with a counselor, social worker or minister may prove helpful. What are the reasons for further testing? The triple screen is a routine screening that is not an invasive procedure and poses no risks to the mother or baby. The abnormal triple screen results often warrant additional testing. The reasons to pursue further testing or not vary from person to person and couple to couple. Performingfurther testing allows you to confirm a diagnosis and then providesyou with certain opportunities: Pursue potential interventions that may exist (i.e. fetal surgery for spina bifida) Begin planning for a child with special needs Start addressing anticipated lifestyle changes Identify support groups and resources Make a decision about carrying the child to term Some individuals or couples may elect not to pursue testing or additionaltesting for various reasons: They are comfortable with the results no matter what the outcome is Because of personal, moral, or religious reasons, making a decision about carrying the child to term is not an option Some parents choose not to allow any testing that poses any risk of harming the developing baby It is important to discuss the risks and benefits of testing thoroughly with your healthcare provider. Your healthcare provider will help you evaluate if the benefits from the results could outweigh any risks from the procedure. Compiled using information from the following sources: Mayo Clinic Complete Book of Pregnancy and Babys First Year Johnson, Robert V., et al, Ch. 6 American Family Physician American Academy of Family Physicians Search BLOG CATEGORIES Can I get pregnant if… ?3 Pregnancy Symptoms 5 Paternity Tests 2 Healthy Pregnancy 18 Multiple Births 11 Pregnancy Complications 68 Pregnancy Concerns 62 Cord Blood 4 Pregnancy Supplements & Medications 14 Pregnancy Products & Tests 8 Your Developing Baby 16 Changes In Your Body 5 general 2 Health & Nutrition 2 Pregnancy Health and Wellness 149 Genetic Disorders & Birth Defects 17 The First Year 40 Is it Safe While Pregnant 55 Week by Week Newsletter 40 Breastfeeding 29 Planning and Preparing 23 Labor and Birth 65 Options for Unplanned Pregnancy 18 Uncategorized 7 Unplanned Pregnancy 21 Birth Control Pills, Patches & Devices 21 Abstinence 3 Prenatal Testing 16 Women's Health 34 Child Adoption 19 Getting Pregnant 59 Pregnancy Loss 11 Fertility 53 NAVIGATION Thank You for Your Donation Unplanned Pregnancy Getting Pregnant Healthy Pregnancy Resources About Privacy Policy Contact Pregnancy Questions Center FAQs Share this post: Facebook Twitter LinkedIn Email Similar Post Is it Safe While Pregnant Pregnancy and Ice Cream Is eating ice cream during pregnancy safe? YES, it can be, when eaten in moderation. So what is the best… Getting Pregnant Preconception Wellness - Prepare for the Unexpected Planning to get pregnant can be a powerful motivating factor to help improve your preconception health. The rate of smoking… Pregnancy Health and Wellness Leg Cramps During Pregnancy Nearly half of all pregnant women suffer from leg cramps – those painful involuntary muscle spasms that strike your calf, foot or both. It is not uncommon to experience pregnancy leg cramps particularly in the second and third trimesters. What Causes Leg Cramps During Pregnancy? While it isn’t clear exactly what causes these muscle spasms, leg cramps may be caused by pregnancy weight gainand changes in your circulation. Pressure from the growing baby may pitch the nerves and blood vessels that go to your legs. How to avoid and treat pregnancy leg cramps Exercise regularly, and include a stretching routine. For a simple calf stretch – stand at arm’s length from a wall, place your hands on the wall in front of you and move your right foot behind your left foot with toes facing the wall. Slowly bend your left leg forward, keeping your right knee straight and your right heel on the floor. Hold the stretch for about 30 seconds, being careful to keep your back straight and your hips forward. Breath deeply through the stretch. Repeat with the other leg. When spasms hit, gently perform the stretch on the affected side and then rest with your legs elevated. A warm bath, ice massage or muscle massage may help too. Drink lots of water. Your urine will be clear or light yellow when you’re adequately hydrated. Wear support socks or stockings that provide some compression on your calves. Increase the calcium and magnesium in your diet by eating food such as whole grains, beans, dried fruit, nuts and seeds. Contact your health care provider if your leg cramps are severe or persistent. Your health care provider may prescribe additional supplements or medication. Want to Know More? Treating Muscle Cramps Naturally During Pregnancy Exercise During Pregnancy Role of Vitamin B in Pregnancy Compiled using information from the following sources: Mayo Clinic Guide To A Healthy Pregnancy Harms, Roger W., M.D., et al, Part 3. March of Dimes, Danforth’s Obstetrics and Gynecology Ninth Ed. Scott, James R., et al, Ch. 1. COVID-19 Vaccines and Pregnancy Between news stories and multiple opinions about COVID-19 variants and vaccines, many pregnant and breastfeeding moms have concerns and questions about how to keep themselves... Spotting During Pregnancy Spotting is when you see a light or trace amount of pink, red, or dark brown blood. It will be lighter than your menstrual period... Preeclampsia Preeclampsia is a condition that occurs only during pregnancy. Some symptoms may include high blood pressure and protein in the urine, usually occurring after week... 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189168	https://www.sciencedirect.com/topics/medicine-and-dentistry/omphalocele	Skip to Main content My account Sign in Omphalocele In subject area:Medicine and Dentistry Omphalocele is defined as a midline abdominal wall defect where herniated viscera are covered by a membrane consisting of peritoneum, amnion, and Wharton's jelly, with umbilical vessels inserting into the membrane rather than the body wall. The herniated organs can include intestines and, in some cases, part of the liver, depending on the defect's size and location. AI generated definition based on: Surgical Clinics of North America, 2012 How useful is this definition? Add to Mendeley Also in subject areas: Nursing and Health Professions Pharmacology, Toxicology and Pharmaceutical Science Veterinary Science and Veterinary Medicine Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Diseases of the Pediatric Abdominal Wall, Peritoneum, and Mesentery 2015, Textbook of Gastrointestinal Radiology, 2-Volume Set (Fourth Edition)Kevin P. Boyd, ... George S. BissettIII Omphalocele An omphalocele is herniation of abdominal viscera into the intact umbilical cord, with a membranous covering over the eviscerated organs unless there has been membrane rupture (Fig. 123-1). The spectrum of severity ranges from a small umbilical hernia to a large defect, resulting in evisceration of all abdominal organs.6 The incidence of omphalocele is 1 case in 4000 to 7000 live births.2 Although an omphalocele can be an isolated abnormality, associated anomalies are more common with an omphalocele than with gastroschisis.7 Associated genetic abnormalities are seen in up to 54% of patients; trisomy 18 is most common, but trisomy 13, Beckwith-Wiedemann syndrome, and rarely uniparental disomy (i.e., inheritance of both copies of a chromosome pair from one parent and none from the other) also occur. Studies suggest that associated chromosomal abnormalities are more likely in patients with a small omphalocele that does not include herniated liver.2,7 Visceral abnormalities are seen in up to 70%, including neural tube defects and cardiac, renal, facial, and skeletal abnormalities and pentalogy of Cantrell.2,7 Gastrointestinal anomalies can include imperforate anus, colonic atresia, and Hirschsprung's disease. Physiologic herniation of bowel is normally seen between the 8th and 12th weeks of gestation; omphalocele cannot be reliably diagnosed until after this period. Omphalocele usually is diagnosed on prenatal ultrasound; patients with omphalocele demonstrate abdominal viscera within a mass at the umbilical cord insertion site6 with a covering membrane (Fig. 123-2). The abdominal wall defect of an omphalocele tends to be large.7 The herniated viscera usually include liver with various amounts of bowel.6 When the omphalocele contains herniated liver and biliary structures, it is referred to as a giant omphalocele (Fig. 123-3).2 Ascites can be seen within the abdomen or omphalocele, and polyhydramnios may be present.6 Prenatal diagnosis should also include a thorough search for associated anomalies.7 Fetal MRI is increasingly performed in patients with omphalocele to characterize the size and location of the anterior wall defect, to characterize the sac and its contents, and to identify associated anomalies (see Fig. 123-3).8 In addition, fetal MRI can be used to calculate lung volumes in patients with giant omphalocele to predict the degree of pulmonary hypoplasia, which is predictive of postnatal morbidity.9 Postnatal imaging of infants with omphalocele is generally reserved for identification of associated anomalies and should include echocardiography and renal ultrasound. Studies of the route of delivery for infants with omphalocele have demonstrated no significant improvement in outcome with cesarean delivery compared with vaginal delivery. Cesarean delivery usually is reserved for patients with obstetric indications. Exceptions are patients who have a giant omphalocele, who are at increased risk of fatal hemorrhage from liver injury.2 Unlike in gastroschisis, the herniated bowel and liver in an omphalocele are morphologically and functionally normal.7 The membrane that covers the herniated viscera in an omphalocele decreases fluid loss and protects against the metabolic abnormalities commonly seen in patients with gastroschisis.2 The treatment of omphalocele is dictated by the presence of comorbidities (cardiovascular compromise, pulmonary insufficiency), the coexisting congenital syndromes or anomalies, and the size of the defect. Surgical options include primary closure for smaller defects and delayed closure, by the “paint and wait” technique, allowing time for the patient to grow enough to accommodate larger omphaloceles (see Fig. 123-1).10 Functionally, patients with omphalocele are typically spared the problems of bowel motility associated with gastroschisis. The protective covering membrane over the omphalocele affords a protective barrier to amniotic fluid exposure. The prognosis of patients with omphalocele predominantly depends on associated chromosomal or structural anomalies.2,7 Infants with an omphalocele and a cardiac anomaly have a mortality rate approaching 80%.2 Prognosis is poorer when the covering membrane has ruptured because these patients tend to have significantly smaller birth weight than those with an intact membrane, regardless of omphalocele size. Patients with giant omphaloceles also tend to have pulmonary hypoplasia, which can complicate primary repair.6,11 View chapterExplore book Read full chapter URL: Book2015, Textbook of Gastrointestinal Radiology, 2-Volume Set (Fourth Edition)Kevin P. Boyd, ... George S. BissettIII Chapter Congenital Defects of the Abdominal Wall 2006, Pediatric Surgery (Sixth Edition)Michael D. Klein Omphalocele Omphalocele is a central abdominal wall defect, usually larger than 4 cm in diameter. It is always covered by a translucent sac from which the umbilical cord extends. The sac may be torn during delivery, but this is unusual. The muscles of the abdominal wall are normal; however, the rectus abdominis muscles insert laterally on the costal margins, thereby creating a depression defect superiorly. The sac usually contains the liver, midgut, and frequently other organs such as the spleen or gonad. Babies with omphalocele are usually full term. Incidence Before 1970, omphalocele was the most common of the abdominal wall defects; it is now the second after gastroschisis. The overall incidence is 1 to 2.5 per 5000 live births,107 with a male preponderance.28 Associated Conditions Conditions associated with omphalocele are listed in Table 73-2. As many as 45% of patients with omphalocele have been reported to have a cardiac abnormality, including ventricular septal defect, atrial septal defect, ectopia cordis, tricuspid atresia, coarctation of the aorta, and persistent pulmonary hypertension of the newborn.51 Chromosomal abnormalities can be found in up to 40%,118 and an association with Down's syndrome has also been reported.131 Patients with omphalocele are more likely to be large for gestational age (macrosomia or greater than 4 kg in birth weight).168 Musculoskeletal and neural tube defects are also reported in greater than expected incidence.7,98 Gastroesophageal reflux is more likely, with 43% being affected in one study,81 and at 1 year of age 33% may have cryptorchidism.71 Cephalic Fold Omphalocele (Pentalogy of Cantrell) The five elements of this pentalogy are remarkably constant.29 There is a supraumbilical omphalocele, which may contain only the liver or mainly the liver. Through the translucent sac one can often see the heart or a left ventricular diverticulum that has extruded through the diaphragmatic defect. Also present are a lower sternal cleft, a defect in the central tendon of the diaphragm, a defect in the pericardium, and an intracardiac abnormality. View chapterExplore book Read full chapter URL: Book2006, Pediatric Surgery (Sixth Edition)Michael D. Klein Chapter Abdominal Wall Defects 2024, Avery's Diseases of the Newborn (Eleventh Edition)SHILPI CHABRA, ... PATRICK J. JAVID Omphalocele An omphalocele (known as exomphalos in the United Kingdom) is a midline abdominal wall defect with herniation of bowel and possibly liver and other organs outside the abdomen. Although there is no consensus definition of giant omphalocele, it is often described as omphaloceles with fascial defects greater than 5 cm in diameter or those containing large amounts of liver.89 The herniated contents are covered with a membrane consisting of peritoneum on the inside, amnion on the outside, and Wharton’s jelly between those two layers. The umbilical cord inserts into the membrane rather than the abdominal wall (Fig. 62.7). The omphalocele membrane is usually intact, but it occasionally ruptures, resulting in exposure of the herniated visceral contents. A ruptured omphalocele can be difficult to differentiate from gastroschisis, especially during the prenatal period. Until recently, both omphalocele and gastroschisis shared the same International Classification of Diseases (ICD) code; however, they have since been separated when we submitted a proposal to change the two ICD codes.90 Ultrasound findings that favor the diagnosis of gastroschisis include a relatively small abdominal wall defect (usually less than 4 cm in diameter), the absence of covering, no liver protruding outside the body wall, and an umbilical cord that is normally inserted into the body wall just to the left of the defect. Some authors make a distinction between an omphalocele and a hernia of the umbilical cord. In a hernia of the umbilical cord, the umbilical ring is reportedly normal. Since there is only an open umbilical ring and no deficiency of abdominal wall, the surgical repair is much easier. More importantly, unlike newborns with omphaloceles, newborns with a hernia of the umbilical cord reportedly have a low risk of associated anomalies. The clinical differentiation can be difficult, and some physicians refer to a hernia of the umbilical cord as a small omphalocele.91 Epidemiology The incidence of omphalocele is estimated to be 1.5 to 3 per 10,000 births.4,5,92 The incidence of omphalocele has been stable over time, unlike the incidence of gastroschisis, which is increasing.92,93 Most cases of omphalocele are sporadic and often associated with chromosomal anomalies, but there are rare familial occurrences.94 Risk factors for omphalocele include a maternal age less than 20 years or greater than 40 years, and maternal obesity.4,95 In contrast to the relatively low risk of associated anomalies in babies with gastroschisis, babies with omphalocele are twice as likely as those with gastroschisis to present with other birth defects. These associated anomalies play a major role in how patients are managed and in their eventual outcome.96 Chromosomal anomalies such as trisomy 13 and 18 are common, especially in fetuses diagnosed early in gestation.92,97 In addition, other congenital malformations are frequent, especially congenital heart disease, which may occur in up to 50% of newborns with omphalocele.93 Finally, omphalocele is a component of several syndromes of congenital malformations, including Beckwith-Wiedemann syndrome, which occurs in 6% of newborns with an omphalocele. Up to 10% to 20% of newborns thought to have an isolated omphalocele have Beckwith-Wiedemann syndrome on prenatal evaluation.98 Pathophysiology Normal development of the abdomen involves the flat, three-layered embryonic disk folding from top, bottom, and both sides to meet at the umbilical ring to form the cylindrical torso during the fourth to fifth week of gestation. The gastrointestinal tract begins to grow very rapidly early in the first trimester, and due to lack of room in the abdominal cavity there is a physiologic bowel herniation through the umbilical ring into the umbilical cord. By 10 to 12 weeks, the bowel returns back into the abdomen in a precise pattern of rotation and fixation to its final position. The embryology of omphalocele is still not clear, and this malformation is thought to occur due to incomplete abdominal wall folding and failure of the intestinal tract to return from the umbilical cord.99,100 Currently, omphalocele is thought to occur due to the embryonic dysplasia combined with malfunction of the ectodermal placodes.101 Clinical Presentation The prenatal diagnosis of omphalocele may be suspected when maternal serum α-fetoprotein levels are elevated or when there is fetal aneuploidy. The definitive diagnosis is made by prenatal ultrasound. The prenatal ultrasound findings of omphalocele are abdominal organs herniated outside the abdominal cavity and covered with a membrane, and an abnormal insertion of the umbilical cord into the membrane rather than into the abdominal wall.97 Prenatal ultrasound diagnosis of omphalocele can be reliably made after the first trimester when the bowel returns to the abdominal cavity. The finding of liver outside the abdomen potentially allows an earlier and more accurate diagnosis of omphalocele.102 It is also important to note the presence or absence of liver within the omphalocele since the presence of liver is associated with a higher risk of other anomalies.97 Management The prenatal management of omphalocele includes evaluation for associated anomalies with imaging for structural anomalies and monitoring of fetal growth because of the risk of IUGR. Because of the high risk of congenital heart disease, fetal echocardiography is indicated. Other specific evaluations for associated pulmonary hypoplasia and the size of the defect, such as fetal MRI, are recommended to provide improved prenatal counseling about the expected hospital course and the long-term prognosis.103,104 Obstetric care including the timing and method of delivery is usually determined by traditional maternal and fetal factors rather than by the presence of the omphalocele. Specifically, there is usually no benefit (and there is potential harm) with preterm delivery105 due to lung immaturity in the setting of possible pulmonary hypoplasia. Cesarean section is indicated only for most fetuses with large defects containing liver (i.e., giant omphaloceles) and is commonly performed in this scenario55,106 to avoid inadvertent rupture of the sac. After delivery, the initial evaluation and resuscitation of a baby with an omphalocele follows the same priorities as for all newborns. During the initial resuscitation, a newborn with an omphalocele should be handled carefully to prevent the omphalocele membrane from tearing. After the initial resuscitation, the omphalocele should be inspected to confirm that it is intact and then covered with a nonadherent dressing or bowel bag to protect the sac. Newborns with omphaloceles, especially large omphaloceles, may have respiratory insufficiency caused by pulmonary hypoplasia and pulmonary hypertension,107 and they may require respiratory support.108 In addition, those babies with associated Beckwith-Wiedemann syndrome may have hypoglycemia and require supplemental glucose. An early evaluation for possible associated anomalies, especially congenital heart disease, is required to diagnose conditions that may need further treatment. The goal of operative treatment of omphalocele is to reduce the herniated organs back into abdomen and close the abdominal wall. Surgical closure of an omphalocele is not an emergency if the omphalocele membrane remains intact. This allows time for the evaluation of associated anomalies and supportive treatment of co-morbidities. If the omphalocele membrane ruptures, it can sometimes be repaired using suture at the bedside.109 When a ruptured omphalocele cannot be repaired, the care of the patient and the abdominal wall defect follows a pathway more similar to the care of newborns with a large gastroschisis, although surgical closure is usually more complicated, and both mortality and morbidity are high.110 Decisions regarding initial operative versus nonoperative management of the omphalocele depend on whether the membrane is ruptured, the size of the omphalocele including the eviscerated contents, and co-morbidities. When the omphalocele is relatively small and the baby is otherwise stable, an early operation with excision of the omphalocele membrane, reduction of the herniated organs, closure of the abdominal wall muscles, and umbilicoplasty is performed. When the defect is too large to safely reduce the viscera and close the abdominal wall, or if the baby has significant comorbidities, the repair can be staged using a “paint-and-wait” approach if the membrane is intact, or silo placement with delayed surgical repair if the membrane is ruptured. The “paint-and-wait” delayed approach is accomplished by treating the intact omphalocele membrane with daily application of topical agents that allow the sac to epithelialize, followed by delayed surgical repair months later.111 A variety of topical agents have been used to treat the omphalocele sac. The agents most commonly used in the United States are silver sulfadiazine cream and povidone iodine solution.89,112–114 As the sac opacifies and eventually epithelializes, the baby continues to grow with adequate caloric intake. Caloric goals in neonates with omphalocele can generally be reached with enteral nutrition and oral feeds. Prolonged need for parenteral nutrition is rare with uncomplicated omphalocele. The growth of the abdominal cavity allows the herniated abdominal organs to reduce back into the abdomen with gravity alone or with mild external compression. Lung growth over time will also improve the baby’s ability to tolerate an operation and an abdominal closure. The eventual repair of the abdominal wall can be done several months or even years later.115 The “paint-and-wait” approach is especially useful in the management of giant omphaloceles when there is no reasonable chance for early reduction, and in the management of smaller omphalocele in patients with significant congenital cardiac or other disease that precludes an upfront surgical repair. If the membrane is ruptured, unable to be repaired, and the defect is very large, a staged surgical repair can be performed. The staged surgical repair involves initial silo placement followed by serial reduction of the silo, and finally closure of the abdominal wall. This strategy is almost identical to the staged approach often used for gastroschisis.111 The early, staged approach is not always successful if the abdominal defect is large and a significant amount of herniated contents need to be reduced into a small abdominal cavity. In addition, reducing the abdominal contents may impair ventilation which can be a significant problem in the setting of underlying pulmonary insufficiency.116 Outcomes The survival and long-term outcomes of newborns with omphaloceles are mainly determined by the severity of associated anomalies.92,117 After the repair of omphalocele, volvulus occurs in up to 4% of omphalocele cases.88 Long-term morbidity includes feeding difficulties, failure to thrive, gastroesophageal reflux disease, inguinal hernias, chronic lung disease, and neurodevelopmental and motor delays.118,119 Babies with giant omphaloceles have increased mortality and morbidity because of the large abdominal wall defect and associated pulmonary hypoplasia and pulmonary hypertension.103,106,120 Overall survival for liveborn omphalocele infants has improved over the last few decades, and 1-year survival rates have been reported as high as 90% in cases of isolated omphalocele.92 View chapterExplore book Read full chapter URL: Book2024, Avery's Diseases of the Newborn (Eleventh Edition)SHILPI CHABRA, ... PATRICK J. JAVID Chapter Surgical Assessment of the Abdomen 2011, Pediatric Clinical Skills (Fourth Edition)Michael Giacomantonio Omphalocele An omphalocele results from a failure of complete closure of the abdominal wall around the umbilical structures. Small omphaloceles represent incomplete closure of the abdominal wall so that an amniotic membrane remains at the umbilical site from which the umbilical vessels extend. Surgical repair is easily accomplished. Larger (giant) omphaloceles also represent incomplete closure of the abdominal wall, with the central area having an amniotic covering from which the umbilical vessels emanate. In giant omphaloceles, the amniotic membrane covers a significant portion of the central abdominal cavity. In infants with a giant omphalocele, it is believed that the abdominal wall, although seemingly fully developed, is inadequate to contain the otherwise normal abdominal viscera. Management often requires a staged repair. Here, too, it is important to look for associated congenital abnormalities, particularly in the cardiovascular system. Outcome depends on the size of the omphalocele and/or associated anomalies. Infants with giant omphaloceles may require multiple interventions, and prognosis is more guarded. View chapterExplore book Read full chapter URL: Book2011, Pediatric Clinical Skills (Fourth Edition)Michael Giacomantonio Chapter Congenital Defects of the Abdominal Wall 2006, Pediatric Surgery (Sixth Edition)Michael D. Klein CLINICAL FEATURES Omphalocele Omphalocele is a central abdominal wall defect, usually larger than 4 cm in diameter. It is always covered by a translucent sac from which the umbilical cord extends. The sac may be torn during delivery, but this is unusual. The muscles of the abdominal wall are normal; however, the rectus abdominis muscles insert laterally on the costal margins, thereby creating a depression defect superiorly. The sac usually contains the liver, midgut, and frequently other organs such as the spleen or gonad. Babies with omphalocele are usually full term. Incidence Before 1970, omphalocele was the most common of the abdominal wall defects; it is now the second after gastroschisis. The overall incidence is 1 to 2.5 per 5000 live births,107 with a male preponderance.28 Associated Conditions Conditions associated with omphalocele are listed in Table 73-2. As many as 45% of patients with omphalocele have been reported to have a cardiac abnormality, including ventricular septal defect, atrial septal defect, ectopia cordis, tricuspid atresia, coarctation of the aorta, and persistent pulmonary hypertension of the newborn.51 Chromosomal abnormalities can be found in up to 40%,118 and an association with Down's syndrome has also been reported.131 Patients with omphalocele are more likely to be large for gestational age (macrosomia or greater than 4 kg in birth weight).168 Musculoskeletal and neural tube defects are also reported in greater than expected incidence.7,98 Gastroesophageal reflux is more likely, with 43% being affected in one study,81 and at 1 year of age 33% may have cryptorchidism.71 Cephalic Fold Omphalocele (Pentalogy of Cantrell) The five elements of this pentalogy are remarkably constant.29 There is a supraumbilical omphalocele, which may contain only the liver or mainly the liver. Through the translucent sac one can often see the heart or a left ventricular diverticulum that has extruded through the diaphragmatic defect. Also present are a lower sternal cleft, a defect in the central tendon of the diaphragm, a defect in the pericardium, and an intracardiac abnormality. Ectopia Cordis Thoracis In this rare anomaly the heart has no pericardium and protrudes from between the two halves of the split sternum with the apex pointing up toward the chin. There are usually associated intracardiac anomalies and occasionally other anomalies such as an associated abdominal wall defect that is not part of the pentalogy of Cantrell. Caudal Fold Omphalocele (Cloacal Exstrophy, Vesicointestinal Fissure) Unlike the cephalic fold defect, the elements of this anomaly are extremely variable. There is usually an infraumbilical omphalocele that seldom contains the liver, exstrophy of the bladder with associated epispadias, diastasis of the pubic rami, and an imperforate anus. The ileum prolapses between the two halves of the exstrophied bladder. The clinical issues are discussed in detail in Chapter 118. Gastroschisis In contrast to omphalocele, gastroschisis is a small defect, usually less than 4 cm. It occurs adjacent to the umbilicus just to the right of the cord in all but a few rare instances. Occasionally, a skin bridge may be present between the cord and the defect, but the abdominal wall and its muscles are normal. There is no sac or remnant of a sac. The midgut is herniated through the defect and occasionally a gonad as well. At birth the bowel can appear perfectly normal, but more than 20 minutes after birth, the extruded intestine may be thickened and covered with a fibrinous exudate matted together so that individual loops cannot be distinguished. There have been several reports of gastroschisis with a small remnant of midgut appearing above a defect that has essentially closed, most likely caused by antenatal volvulus.21,36 Gastroschisis in the fetus is probably associated with intrauterine distress. Neonates with gastroschisis are more frequently premature and commonly have respiratory problems. Even term babies with gastroschisis are more likely to be small for gestational age24,25,50 and to have younger mothers.33 Incidence Gastroschisis has become the most common of the abdominal wall defects over the last 30 years.12,72,108,136 This may be related to the increased incidence of prematurity and the increased survival of premature infants in general or to the fact that it was not until the 1970s that distinction between gastroschisis and omphalocele was regularly made.76 The incidence is about 2 to 4.9 per 10,000 live births,12,88,108,132 with a male preponderance.28 Associated Conditions The anomalies associated with gastroschisis are usually related to the midgut, with the most common being intestinal atresia (see Table 73-2).149 In the first year of life infants with gastroschisis are very likely to have gastroesophageal reflux (16%)81 and undescended testicle (15%), although the latter often corrects spontaneously.71,89 Umbilical Cord Hernia This uncommon defect is generally small and occurs at the umbilicus, with the umbilicus extending from it. It is covered with a sac and often confused with omphalocele. The differences are that it contains only midgut, never liver, and the abdominal wall above the defect is normal, with the rectus muscles meeting in the midline at the xiphoid. Few associated anomalies are reported with this defect. Like all abdominal defects in which the midgut has not returned to the abdominal cavity before birth to allow for rotation and fixation, these patients have malrotation, although it is not usually a cause of intestinal obstruction. View chapterExplore book Read full chapter URL: Book2006, Pediatric Surgery (Sixth Edition)Michael D. Klein Review article Cesarean Delivery: Its Impact on the Mother and Newborn-Part I 2008, Clinics in PerinatologyShannon E.G. Hamrick MD Omphalocele Omphalocele, an abdominal wall defect characterized by an amnioperitoneal membrane covering herniated intra-abdominal contents, is believed to result from a malformation as opposed to a deformation, such as gastroschisis. It represents a persistence of the primitive stalk with failure of the bowel to return to the abdomen and failure to complete lateral body wall closure. The diagnoses often is complicated by the presence of other visceral (50%–70%) and chromosomal abnormalities (7%–50%) [37,54,55]. Prognosis is more closely dependent on associated anomalies and gestational age at delivery rather than mode of delivery [37,56]. Although definitive data are lacking, cesarean delivery is recommended for cases of extracorporeal liver and cases of “giant” omphalocele (>5 cm) . Similar to gastroschisis, the available data suggest no benefit for cesarean delivery except as noted for extracorporeal liver and giant omphalocele [33,36,37,39,45,47]. View article Read full article URL: Journal2008, Clinics in PerinatologyShannon E.G. Hamrick MD Review article Route of delivery of fetuses with structural anomalies 2003, European Journal of Obstetrics & Gynecology and Reproductive BiologyEyal Y Anteby, Simcha Yagel Omphalocele represents a persistence of the body stalk in an area normally occupied by the differentiated abdominal wall. A membranous sac usually covers the herniated viscera. It occurs in one in 4000–5000 and is associated with major chromosomal abnormalities (up to 40%), and with an increased incidence of other major structural defects (up to 70%). Because prognosis is mainly related to the presence of associated anomalies or extracorporeal liver, assessment of independent effect of route of delivery is difficult [23–25]. Moretti et al. compared the neonatal and long term outcome of 54 fetuses delivered by the vaginal route with 15 fetuses delivered by cesarean section. Comparable rates of associated anomalies and extracorporeal liver were noted. No difference in neonatal morbidity, mortality and incidence of visceral trauma was observed. A slightly increased rate of ruptured sacs occurred in the vaginal delivery group (15 versus 7%, statistically non-significant). Yet, this increase could be partially attributed to pre-labor rupture of the sac. Sipes et al. compared the outcome of eight infants after vaginal delivery with eight fetuses delivered by cesarean section. No difference in neonatal hospital days was observed between the groups, and no evidence of visceral injury was noted in either group. View article Read full article URL: Journal2003, European Journal of Obstetrics & Gynecology and Reproductive BiologyEyal Y Anteby, Simcha Yagel Chapter Paediatric gastrointestinal disorders 2013, Grainger & Allison's Diagnostic Radiology EssentialsNyree Griffin MD FRCR, Lee Alexander Grant BA (Oxon) FRCR OMPHALOCELE (EXOMPHALOS) DEFINITION • : Incomplete formation of the embryonic ventral abdominal wall leads to a congenital midline anterior abdominal wall defect around the umbilicus (the umbilical cord inserts at the tip of the defect) ▪ : Larger omphaloceles (containing liver tissue): due to failure of lateral body fold fusion ▪ : Smaller omphaloceles (containing bowel only): due to persistence of physiological gut herniation • : Associated chromosomal abnormalities are common (50%): trisomy 13 and 18 ▪ : Beckwith–Wiedemann syndrome: omphalocele (exomphalos) + macroglossia + gigantism (the ‘EMG’ syndrome) ▸ visceral abnormalities are seen in up to 70% of cases • : Diagnosis: antenatal US View chapterExplore book Read full chapter URL: Book2013, Grainger & Allison's Diagnostic Radiology EssentialsNyree Griffin MD FRCR, Lee Alexander Grant BA (Oxon) FRCR Chapter Gastroschisis 2017, Diagnostic Imaging: Pediatrics (Third Edition) Omphalocele • : Viscera herniated into base of umbilical cord with covering membranous sac • : Always contains bowel; often contains liver, other organs • : Ruptured omphalocele may mimic gastroschisis • : > 50% with associated malformations, chromosomal abnormalities View chapterExplore book Read full chapter URL: Book2017, Diagnostic Imaging: Pediatrics (Third Edition) Chapter Congenital Defects of the Abdominal Wall 2012, Pediatric Surgery (Seventh Edition)Michael D. Klein Omphalocele Before 1970, omphalocele was the most common of the abdominal wall defects; it is now the second after gastroschisis. The overall incidence is 1 to 2.5 per 5000 live births8 with a male preponderance.115 Conditions associated with omphalocele are listed in Table 75-3. Up to 45% of patients with omphalocele have been reported to have a cardiac abnormality including ventricular septal defect, atrial septal defect, ectopia cordis, tricuspid atresia, coarctation of the aorta, and persistent pulmonary hypertension of the newborn.116 Chromosomal abnormalities can be found in up to 20%, and an association with Down syndrome has also been reported.117 Patients with omphalocele are more likely to be large for gestational age (macrosomia or > 4 kg in birth weight).118 Musculoskeletal and neural tube defects are also reported in greater than expected incidence.119,120 Gastroesophageal reflux is more likely, with 43% being affected in one study.121 View chapterExplore book Read full chapter URL: Book2012, Pediatric Surgery (Seventh Edition)Michael D. Klein Related terms: Propylthiouracil Imperforate Anus Neonate Abdominal Wall Bladder Gastroschisis Hernia Abdominal Wall Defect Beckwith Wiedemann Syndrome Congenital Malformation View all Topics
189169	https://www.khanacademy.org/science/biology/plant-biology/plant-responses-to-light-cues/v/photoperiodism	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
189170	https://arxiv.org/pdf/2408.07766	arXiv:2408.07766v1 [math.PR] 14 Aug 2024 COLLISION LOCATION FOR HARD SPHERES IN STATIONARY REGIME KRZYSZTOF BURDZY AND SHUNTAO CHEN Abstract. Consider two balls with radius r > 0 whose centers are at a distance 2, positioned symmetrically with respect to the origin in Rd.Suppose that the ini-tial velocities are independent standard normal vectors. When r→0, the collision probability goes to 0 as rd−1, and the asymptotic collision location distribution is a (defective) t-distribution. This distribution is rotationally symmetric about the origin for no apparent reason. Introduction Physics principles known as “equidistribution of energy” and “microcanonical en-semble formula” (see [ 3, Chaps. 2-3]) suggest that velocities of gas molecules in the air are i.i.d. rotationally invariant normal vectors. Given this assumption, we will find the collision location for two “randomly” chosen molecules. If the gas is dilute, i.e., the distance between a molecule and its nearest neighbor is large relative to its diameter, the probability of a collision of two given molecules is very small. Nevertheless, the probability is non-zero, assuming strictly positive radii r of the molecules. We will find asymptotic formulas for the probability and location of a collision of two molecules, assuming a fixed initial distance, as r → 0. We will assume that two molecules are represented by balls initially positioned sym-metrically with respect to the origin in Rd. We will suppose that their initial velocities are independent standard normal vectors. The cross-section of a ball has a volume of order rd−1, so heuristically, the collision probability should be of order rd−1. Not surprisingly, this is the case; we will give a rigorous and exact formula for the collision probability. To our surprise, we discovered that as r → 0, the asymptotic collision location distribution is a multi-dimensional t-distribution for d ≥ 2. This distribution is rotation invariant, i.e., invariant under the action of the orthogonal group O(d), for no apparent reason. The model is not invariant under O(d). It turns out that the collision location is rotation invariant if the moving objects are strictly convex, the union of the two objects is centrally symmetric with respect to the origin at the initial time, and the initial velocities are independent standard normal vectors. This paper is based on a chapter in the second author’s Ph.D. thesis [ 2]. Research supported in part by Simons Foundation Grant 928958. 12KRZYSZTOF BURDZY AND SHUNTAO CHEN Model and main results Suppose that d ≥ 1, M1, M2 ⊂ Rd are closed, M1 ∩ M2 = ∅, M1 ∪ M2 is centrally symmetric, i.e. x ∈ M1 ∪ M2 if an only if −x ∈ M1 ∪ M2, and M1 and M2 are strictly convex. We do not assume that M1 and M2 are bounded. Hence, for d = 2, we could take M1 = {(x, y ) : y ≥ x2 + 1 }, and make M2 the symmetric set. Suppose that M1 and M2 are moving with random velocities V1 and V2. Let M1(s) = M1 + sV1 and M2(s) = M2 + sV2. Let the collision time and location be defined by T = inf {t ≥ 0 : M1(t) ∩ M2(t) 6 = ∅} and {C} = M1(T ) ∩ M2(T ). Note that M1(T ) ∩ M2(T ) contains only one point by the assumption of strict convexity of M1 and M2. We will write T < ∞ to indicate that there is a collision. Theorem 2.1. If the distribution of (V1, V2)/\|(V1, V2)\| is uniform on the unit sphere in R2d, then the conditional distribution of C given T < ∞ is rotation invariant, i.e., invariant under the action of the orthogonal group O(d). This gives us immediately the following corollary. Corollary 2.2. If V1 and V2 are independent standard d-dimensional normal N (0, I d) random vectors then the conditional distribution of C given T < ∞ is rotation invariant. Remark 2.3. Unfortunately, Corollary 2.2 seems to contain the only interesting explicit example to which Theorem 2.1 applies. More precisely, if V1 and V2 are independent and the distribution of ( V1, V2) is rotation invariant then V1 and V2 are independent standard d-dimensional normal N (0, I d), or a constant multiple of such random vectors. We sketch a proof of this claim kindly provided by Wlodek Bryc. By rotation invariance, (V1 + V2)/√2 and ( V1 − V2)/√2 are independent and have the same distributions as V1. The Bernstein Theorem (see [ 1, Thm. 5.1.1]) implies that for any linear functional A, random variables A(V1) and A(V1) are normal. It follows that V1 and V2 are multidimensional normal. By rotation invariance, all components of V1 and V2 have the same variance, and all pairs of components have the same covariance. The off-diagonal covariances are all zero because V1 and V2 are independent. This completes the proof. In the rest of this section, we will assume that (i) M1 and M2 are “molecules” in the shape of balls with radii r ∈ (0 , 1) and centers x1 = ( −1, 0, · · · , 0) and x2 = (1 , 0, · · · , 0), and (ii) V1 and V2 are independent standard d-dimensional normal N (0, I d) random vectors. Our first result is a simple explicit criterion for a collision. Proposition 2.4. If d ≥ 1 and the initial velocities of M1 and M2 are non-random vectors v1 and v2 then the balls will collide if and only if cos β ≤ − √1 − r2, where β is the angle between the vectors x1 − x2 and v1 − v2. Let pr,d denote the probability that M1 and M2 collide. We have pr, 1 = 1 /2 because, in d = 1, there will be a collision if and only if V1 > V2.COLLISION LOCATION 3 Theorem 2.5. For d ≥ 2, pr,d = 1 2F ( 1 d − 1 · r2 1 − r2 ; d − 1, 1 ) (2.1) = 1 2 ∫ 1 d−1·r2 1−r2 0 Γ (d 2 ) Γ (d−1 2 ) √π (d − 1) d−1 2 xd−3 2 (1 + ( d − 1) x)− d 2 dx, (2.2) where F ( · ; d−1, 1) is the CDF of the F -distribution with d−1 and 1 degrees of freedom. It follows that pr,d = 1 d − 1 · Γ (d 2 ) Γ ( d−1 2 ) √π rd−1 + o(rd−1).(2.3) Remark 2.6. In this and the next remark, we will discuss the special cases d = 2 , 3because these dimensions are physically relevant, and the corresponding formulas are simple. Substituting d = 2 , 3 into ( 2.3 ), we get pr, 2 = 1 π r + o(r) and pr, 3 = 1 4 r2 + o(r2). Actually, we can compute exact values of pr, 2 and pr, 3. If we substitute d = 2 , 3 in ( 2.2 ), we get pr, 2 = 1 2 ∫ r2 1−r2 0 1 π · 1 √x (1 + x)dx = 1 π arctan √x ∣∣∣∣ r2 1−r2 0 (2.4) = 1 π arctan √ r2 1 − r2 = r π + O(r2),(2.5) pr, 3 = 1 2 ∫ r2 2(1 −r2) 0 (1 + 2 x)− 3 2 dx = −1 2 (1 + 2 x)− 1 2 ∣∣∣∣ r2 2(1 −r2) 0 (2.6) = 1 2 − 1 2 ( 1 + r2 1 − r2 )− 1 2 = 1 − √1 − r2 2 = 1 4 r2 + O(r4).(2.7) Next, we turn to the collision location. Theorem 2.7. For d = 1 , (2.8) P(C ≤ x) = ∫ x −∞ 1 − r 2π (u2 + (1 − r)2) du. Theorem 2.8. Suppose d ≥ 2 and A ⊂ Rd is a bounded Borel set with a positive distance from the origin. Then (2.9) lim r→0 P(C ∈ A)r1−d = π−(d+1) /2Γ( d) 2Γ( d/ 2 + 1 /2) ∫ A 1 (1 + \|x\|2)d dx. Remark 2.9. (i) The distribution in ( 2.9 ) is a special case of the multidimensional t-distribution; more precisely, it is a defective version of the multidimensional t-distribution. The formula in ( 2.9 ) agrees with ( 2.8 ), except that the latter is an exact formula, unlike the asymptotic formula in ( 2.9 ). 4 KRZYSZTOF BURDZY AND SHUNTAO CHEN (ii) The values of the coefficient (1 /2) π−(d+1) /2Γ( d)/Γ( d/ 2 + 1 /2) in ( 2.9 ) for d =2, . . . , 11 are 1 π2 , 1 π2 , 4 π3 , 6 π3 , 32 π4 , 60 π4 , 384 π5 , 840 π5 , 6144 π6 , 15120 π6 .(2.10) (iii) We will show that the formula ( 2.9 ) agrees with ( 2.3 ) and ( 2.4 )-( 2.7 ) for d = 2 , 3. Note that according to ( 2.10 ), the value of the coefficient (1 /2) π−(d+1) /2Γ( d)/Γ( d/ 2 + 1/2) in ( 2.9 ) is 1 /π 2 for d = 2 , 3. We compute the integral in ( 2.9 ) for d = 2 using polar coordinates, r π2 ∫ ∫ R2 1 (1 + x2 + y2)2 dxdy = r π2 ∫ ∞ 0 ∫ 2π 0 ρ (1 + ρ2)2 dθdρ = r π . In the case d = 3, we apply the following polar coordinates in R3, x = ρ cos θ1, y = ρ sin θ1 cos θ2, z = ρ sin θ1 sin θ2, ρ > 0, θ 1 ∈ [0 , π ), θ 2 ∈ [0 , 2π), and substitution u = 1 1+ ρ2 to obtain r2 π2 ∫ ∫ R3 1 (1 + x2 + y2 + z2)3 dxdydz = r2 π2 ∫ ∞ 0 ρ2 (1 + ρ2)2 dρ ∫ π 0 sin θ1dθ 1 ∫ 2π 0 dθ 2 = r2 π2 · 2 · 2π · ∫ 10 1 2 √u√1 − udu = r2 π2 · 2 · 2π · B (3 2 , 3 2 ) = r2 4 , where B ( 3 2 , 3 2 ) denotes the Beta function. As expected, ( 2.9 ) agrees with ( 2.3 ) and (2.4 )-( 2.7 ) in the cases d = 2 , 3. 3. General moving sets A collision will occur if and only if the trajectory of t → (V1, V2)t passes through the set A := {(y + z, y − z) ∈ R2d : y ∈ Rd, z ∈ M2}.(3.1) If ( y + z, y − z) is the first point in A encountered by ( V1, V2)t, then C = y. Let A+ = A ∪ { 0}.For an element g of the orthogonal group O(d), let Tg : A+ → A+ be defined by Tg(0) = 0, and for ( y + z, y − z) ∈ A, Tg (y + z, y − z) = ( g(y) + z, g (y) − z). Lemma 3.1. For every g ∈ O(d), Tg is an isometry on A+.Proof. If y1, y 2 ∈ Rd and z1, z 2 ∈ M2 then \|(y1 + z1, y 1 − z1) − (y2 + z2, y 2 − z2)\|2 = \|(( y1 − y2) + ( z1 − z2), (y1 − y2) − (z1 − z2)) \|2 = \|y1 − y2\|2 + 2( y1 − y2) · (z1 − z2) + \|z1 − z2\|2 \|y1 − y2\|2 − 2( y1 − y2) · (z1 − z2) + \|z1 − z2\|2 = 2 \|y1 − y2\|2 + 2 \|z1 − z2\|2.COLLISION LOCATION 5 Every g ∈ O(d) is an isometry, so \|g(y1) − g(y2)\| = \|y1 − y2\| and, therefore, \|Tg (y1 + z1, y 1 − z1) − Tg (y2 + z2, y 2 − z2)\|2 = \|(g(y1) + z1, g (y1) − z1) − (g(y2) + z2, g (y2) − z2)\|2 = 2 \|g(y1) − g(y2)\|2 + 2 \|z1 − z2\|2 = 2 \|y1 − y2\|2 + 2 \|z1 − z2\|2 = \|(y1 + z1, y 1 − z1) − (y2 + z2, y 2 − z2)\|2. This shows that Tg is an isometry on A.If y ∈ Rd and z ∈ M2 then \|(y + z, y − z) − 0\|2 = \|(y + z, y − z)\|2 = 2 \|y\|2 + 2 \|z\|2. Since \|g(y)\| = \|y\|, \|Tg(y + rz, y − rz ) − 0\|2 = 2 \|g(y)\|2 + 2 \|z\|2 = \|(y + rz, y − rz ) − 0\|2. This proves that Tg is an isometry on A+. Proof of Theorem 2.1 . Suppose B ⊂ Rd and let AB = {(y + z, y − z) ∈ R2d : y ∈ B, z ∈ M2}. The collision point C is in B if and only if ( V1, V2)T ∈ AB . The distribution of (V1, V2)/\|(V1, V2)\| is uniform on the unit sphere. By Lemma 3.1 , for any g ∈ O(d), the radial projection of AB on the unit sphere has the same surface area as the projection of Tg (AB ). Hence, the probability that ( V1, V2)t will intersect AB is the same as the probability that it will intersect Tg(AB ). It follows that P(C ∈ B) = P(C ∈ g(B)). Remark 3.2. For B1 ⊂ Rd and B2 ⊂ M2 let AB1,B 2 = {(y + z, y − z) ∈ R2d : y ∈ B1, z ∈ B2}. The argument used in the proof of Theorem 2.1 also shows that P(( V1, V2)T ∈ AB1,B 2 ) = P(( V1, V2)T ∈ Ag(B1),B 2 ), for any B1 ⊂ Rd, B2 ⊂ M2, and g ∈ O(d). 4. Collision probability Proof of Proposition 2.4 . We will use \| · \| to denote the usual norm in the Euclidean space. At time s, the centers of M1 and M2 will be x1 + sv1 and x2 + sv2, so to have a collision, we need \| (x1 + sv1) − (x2 + sv2) \| = 2 r, for some s > 0. This is equivalent to [( x1 − x2) + s (v1 − v2)] T [( x1 − x2) + s (v1 − v2)] = 4 r2, \|v1 − v2\|2s2 + 2 ( v1 − v2)T (x1 − x2) s + \|x1 − x2\|2 − 4r2 = 0 .(4.1) To make the above quadratic equation have a root in (0 , +∞), we need ∆ = 4 [ (v1 − v2)T (x1 − x2) ]2 − 4\|v1 − v2\|2 (\|x1 − x2\|2 − 4r2) ≥ 0,6 KRZYSZTOF BURDZY AND SHUNTAO CHEN which is equivalent to \|v1 − v2\|2\|x1 − x2\|2 cos 2 β − \| v1 − v2\|2 (\|x1 − x2\|2 − 4r2) ≥ 0, cos 2 β − 1 + 4r2 \|x1 − x2\|2 ≥ 0. Since \|x1 − x2\| = 2, the above condition is the same as (4.2) cos 2 β ≥ 1 − r2. If s1, s 2 are the two roots of ( 4.1 ), Vieta’s formulas give {s1 + s2 = −2( v1−v2)T (x1−x2) \|v1−v2\|2 = −2\|x1−x2\| \|v1−v2\| cos β, s1s2 = \|x1−x2\|2−4r2 \|v1−v2\|2 0. Hence, for ( 4.1 ) to have a positive solution, we need cos β ≤ 0. This and ( 4.2 ) yield the condition cos β ≤ − √1 − r2. Remark 4.1. When r is close to 0, to have a collision, cos β should be close to −1. This means β should be close to π, i.e., x1 − x2 and v1 − v2 must have almost the opposite directions. Hence, if M1 and M2 were to collide, v1 − v2 should have almost the same direction as the positive half of the first axis. Proof of Theorem 2.5 . Let (4.3) V1 = ( v1,1, v 1,2, · · · , v 1,d ) , V2 = ( v2,1, v 2,2, · · · , v 2,d ) . Let β be the angle between the vectors x1 − x2 and V1 − V2. Given Proposition 2.4 ,we only need to show P ( cos β ≤ − √1 − r2 ) = 1 2 F ( 1 d − 1 · r2 1 − r2 ; d − 1, 1 ) .COLLISION LOCATION 7 Recall that x1 = ( −1, 0, · · · , 0), x2 = (1 , 0, · · · , 0), and F ( · ; d − 1, 1) denotes the CDF of the F distribution with d − 1 and 1 degrees of freedom. We have P ( cos β ≤ − √1 − r2 ) = P (−(v1,1 − v2,1) \|V1 − V2\| ≤ − √1 − r2 ) = P ( v1,1 − v2,1 ≥ √1 − r2\|V1 − V2\| ) = P ( (v1,1 − v2,1)2 ≥ (1 − r2) d ∑ i=1 (v1,i − v2,i )2 , v 1,1 − v2,1 ≥ 0 ) = P ( r2 (v1,1 − v2,1)2 ≥ (1 − r2) d ∑ i=2 (v1,i − v2,i )2 , v 1,1 − v2,1 ≥ 0 ) = 1 2 P ( 1 d−1 ∑di=2 (v1,i − v2,i )2 (v1,1 − v2,1)2 ≤ 1 d − 1 · r2 1 − r2 ) = 1 2 F ( 1 d − 1 · r2 1 − r2 ; d − 1, 1 ) . (4.4) To justify the last equality, we note that {v1,i − v2,i }di=1 are i.i.d. N(0 , 2), so the random variables ∑di=2 (v1,i − v2,i )2 and ( v1,1 − v2,1)2 are independent. Moreover, the distribu-tion of 1 4 ∑di=2 (v1,i − v2,i )2 is χd−12 , while the distribution of 1 4 (v1,1 − v2,1)2 is χ12. Hence, the last equality follows from the definition of the F distribution. This proves ( 2.1 ). The PDF of F distribution with d − 1 and 1 degrees of freedom is: Γ (d 2 ) Γ (d−1 2 ) √π (d − 1) d−1 2 xd−3 2 (1 + ( d − 1) x)− d 2 , x > 0. This and ( 4.4 ) yield ( 2.2 ), i.e., pr,d = 1 2 ∫ 1 d−1·r2 1−r2 0 Γ (d 2 ) Γ (d−1 2 ) √π (d − 1) d−1 2 xd−3 2 (1 + ( d − 1) x)− d 2 dx. By L’Hˆ opital’s Rule, lim r→0 pr,d rd−1 = lim r→0 1 2 · Γ (d 2 ) Γ (d−1 2 ) √π (d − 1) d−1 2 ( 1 d − 1 · r2 1 − r2 ) d−3 2 × ( 1 + r2 1 − r2 )− d 2 · 1 d − 1 · 2r (1 − r2)2 · 1 (d − 1) rd−2 = lim r→0 1 d − 1 · Γ (d 2 ) Γ (d−1 2 ) √π (1 − r2)− 1 2 = 1 d − 1 · Γ (d 2 ) Γ (d−1 2 ) √π . This proves ( 2.3 ). 8 KRZYSZTOF BURDZY AND SHUNTAO CHEN Collision location in dimension 1The 1-dimensional case is very easy due to the following well-known representation of the Cauchy distribution. Lemma 5.1. Suppose the endpoint of a half-line is a fixed point (a, b ) with b > 0, and its angle relative to the first axis is uniform in (−π, 0) . Then the first coordinate of the intersection point with the first axis has the Cauchy distribution with the probability density function f (x) = 1 π · b (x − a)2 + b2 . Proof of Theorem 2.7 . Let W(s) = ( W1(s), W2(s)). The halfline L := {W(s), s > 0} is the trajectory of the centers of the balls. The initial location is W(0) = ( −1, 1). Note that the ball M1 is always to the left side of ball M2, so if the collision time T is finite then W2(T ) − W1(T ) = ( x2 + T V2) − (x1 + T V1) = 2 r, which is equivalent to L intersecting the line ℓ := {(x, y ) : y − x = 2 r}.Since V1, V2 are distributed as i.i.d. standard normal, the direction of W(s) is uniform on the circle. The transformation T : ( x, y ) 7 → (y + x, y − x) takes ℓ to the line ℓ1 := {(x, y ) : y = 2 r} and W(0) to the point (0 , 2). The distance between (0 , 2) and ℓ1 is 2 − 2r.Applying Lemma 5.1 , the fact that trajectory hits the line ℓ with probability 1 /2 and W2(T ) = W1(T ) + 2 r, we get P (W1(T ) ≤ x) = P (W1(T ) + W2(T ) ≤ 2x + 2 r) = 1 2 ∫ 2x+2 r −∞ 1 π · 2 − 2r u2 + (2 − 2r)2 du = ∫ x −∞ 1 2π · 1 − r (u + r)2 + (1 − r)2 du. This implies ( 2.8 ) because C = W1(T ) + r if d = 1. Collision location in dimensions d ≥ 2 Proof of Theorem 2.8 . We will decompose the motion into the motion of the center of mass and the motion relative to the center of mass. Recall ( 4.3 ) and for k = 1 , . . . , d let vk = ( v1,k + v2,k )/2,vck = ( v1,k − v2,k )/2, V = ( v1, . . . , vd), Vc = ( vc 1 , . . . , v cd). Since the random variables v1,1, . . . , v 1,d , v 2,1, . . . , v 2,d are i.i.d. standard normal, the vectors V and Vc are independent mean zero normal. The covariance matrix for each of these vectors is the identity matrix times 1 /2. COLLISION LOCATION 9 The collision time T is determined by Vc because T = inf {t > 0 : 0 ∈ M1 + Vct} = inf {t > 0 : 0 ∈ M2 − Vct}. Then C = VT . Conditionally on T , the distribution of C is rotationally invariant by independence of V and Vc and rotation invariance of V. It follows that the (uncondi-tional) distribution of C is rotationally invariant. For x ∈ Rn and a ≥ 0, let Sn(x, a ) = {y ∈ Rn : \|y − x\| = a}. For z ∈ Sd(0 , 1), let ρ(z) = inf {b > 0 : 0 ∈ M1 + zb }. Then T = ρ(Vc/\|Vc\|) \|Vc\| . Random variables Vc/\|Vc\| and \|Vc\| are independent. Thus, the conditional distri-bution of C given {ρ(Vc/\|Vc\|) = u} is that of uV/\|Vc\|. Therefore, the conditional distribution of \|C\|2 given {ρ(Vc/\|Vc\|) = u} is that of u2\|V\|2/\|Vc\|2. The random vari-ables √2\|V\|2 and √2\|Vc\|2 are χ2 d so \|V\|2/\|Vc\|2 has the F -distribution with d and d degrees of freedom. We have P(\|C\| < a ) = P(\|C\|2 < a 2) = ∫ P(\|C\|2 < a 2 \| ρ(Vc/\|Vc\|) = u)P(ρ(Vc/\|Vc\|) ∈ du )= ∫ P(u2\|V\|2/\|Vc\|2 < a 2 \| ρ(Vc/\|Vc\|) = u)P(ρ(Vc/\|Vc\|) ∈ du )= ∫ P(\|V\|2/\|Vc\|2 < a 2/u 2)P(ρ(Vc/\|Vc\|) ∈ du ). The density of the F -distribution with d and d degrees of freedom is f (x) = xd/ 2−1 B(d/ 2, d/ 2)(1 + x)d so P(\|V\|2/\|Vc\|2 < a 2/u 2) = ∫ a2 /u 2 0 xd/ 2−1 B(d/ 2, d/ 2)(1 + x)d dx, P(\|C\| < a ) = ∫ ∫ a2/u 2 0 xd/ 2−1 B(d/ 2, d/ 2)(1 + x)d dx P(ρ(Vc/\|Vc\|) ∈ du ). When r → 0, ρ(z) converges uniformly to 1 on the set where it is finite. Hence, ρ(Vc/\|Vc\|) → 1 on the event T < ∞. It follows that lim r→0 P(\|C\| < a )/P(T < ∞) = ∫ a2 0 xd/ 2−1 B(d/ 2, d/ 2)(1 + x)d dx. We have d da ∫ a2 0 xd/ 2−1 B(d/ 2, d/ 2)(1 + x)d dx = 2 a ad−2 B(d/ 2, d/ 2)(1 + a2)d = 2ad−1 B(d/ 2, d/ 2)(1 + a2)d . The above formula gives us the density of the asymptotic distribution of \|C\| given T < ∞. To transform this formula into the density of C, we have to divide it by 10 KRZYSZTOF BURDZY AND SHUNTAO CHEN the surface area of Sd(0 , a ), i.e., 2 πd/ 2ad−1/Γ( d/ 2). The density of the asymptotic distribution of C given T < ∞ is 2\|x\|d−1 B(d/ 2, d/ 2)(1 + \|x\|2)d · Γ( d/ 2) 2πd/ 2\|x\|d−1 = Γ( d/ 2) πd/ 2B(d/ 2, d/ 2) · 1 (1 + \|x\|2)d = Γ( d) πd/ 2Γ( d/ 2) · 1 (1 + \|x\|2)d . at x ∈ Rd. Suppose d ≥ 2 and A ⊂ Rd is a bounded Borel set with a positive distance from the origin. Then lim r→0 P(C ∈ A)/P(T < ∞) = Γ( d) πd/ 2Γ( d/ 2) ∫ A 1 (1 + \|x\|2)d dx. Elementary calculations show that this and ( 2.3 ) yield ( 2.9 ). Acknowledgments We thank W. Bryc, J. Wellner, S. Steinerberger and J. Wesolowski for the most useful advice. References Wlodzimierz Bryc. The normal distribution , volume 100 of Lecture Notes in Statistics . Springer-Verlag, New York, 1995. Characterizations with applications. Shuntao Chen. Pinned Balls, Foldings and Particle Collisions . PhD thesis, University of Washing-ton, 2024. R. K. Pathria and Paul D. Beale. Statistical Mechanics . Academic Press, Amsterdam, 2011. Department of Mathematics, Box 354350, University of Washington, Seattle, WA 98195 Email address : burdzy@uw.edu Email address : shuntao1994@gmail.com
189171	https://arxiv.org/html/2401.10821v2	Back to arXiv License: arXiv.org perpetual non-exclusive license arXiv:2401.10821v2 [math.NT] 07 Mar 2024 On integer distance sets Rachel Greenfeld School of Mathematics, Institute for Advanced Study, 1 Einstein Drive, Princeton, NJ 08540, USA rgreenfeld@ias.edu , Marina Iliopoulou Department of Mathematics, National and Kapodistrian University of Athens, Panepistimiopolis 157 84, Athens, Greece miliopoulou@math.uoa.gr and Sarah Peluse Department of Mathematics, University of Michigan, East Hall, 530 Church Street, Ann Arbor, MI 48109, USA speluse@umich.edu Abstract. We develop a new approach to address some classical questions concerning the size and structure of integer distance sets. Our main result is that any integer distance set in the Euclidean plane has all but a very small number of points lying on a single line or circle. From this, we deduce a near-optimal lower bound on the diameter of any non-collinear integer distance set of size and a strong upper bound on the size of any integer distance set in with no three points on a line and no four points on a circle. 1. Introduction A subset of the plane is an integer distance set if the Euclidean distance between every pair of points in is an integer. In 1945, Anning and Erdős [1, 11] showed that any infinite integer distance set must be contained in a line, but also that there exist non-collinear integer distance sets of arbitrarily large finite size. They constructed two infinite families of arbitrarily large non-collinear integer distance sets: one of concyclic sets and one of sets in which all but one point are collinear. It turns out that all so-far-known integer distance sets, such as those in [16, 20, 26], are of a similar special form: they have all but up to four of their points lying on a single line or circle. In this paper, we develop a new approach to the study of integer distance sets that enables us to prove a structure theorem partially explaining this phenomenon, showing that any integer distance set in has all but at most polylogarithmically many points lying on a single line or circle. Theorem 1.1 (Structure theorem). Let be an integer distance set. Then, there exists a line or circle such that \| \| \| \| --- \| \| \| \| Remark 1.2. This result can be extended to subsets of with pairwise rational distances of height at most . We leave the details to the interested reader. Theorem 1.1 is the first-ever general structure theorem for integer distance sets, and we use it to address two classical questions about integer distance sets. The first is the question of how large an integer distance set can be if it has no three points on a line and no four points on a circle (see, for example, [10, Chapter 13]). This question is often attributed to Erdős, though he says in that it is an “old problem whose origin I cannot trace”. Erdős still posed it on numerous occasions (for instance, in ), and, in particular, asked whether such a set consisting of seven points exists in the plane. In 2008, Kreisel and Kurz indeed found seven points in the plane with no three of them on a line, no four of them on a circle, and with pairwise integer distances. No larger such set has been found since. It is an immediate consequence of Theorem 1.1 that integer distance sets with no three points on a line and no four points on a circle must be very sparse. Corollary 1.3 (New upper bound in Erdős’s integer distance set problem). Let be an integer distance set with no three points on a line and no four points on a circle. Then, \| \| \| \| --- \| \| \| \| The previous best known upper bound for the size of integer distance sets in with no three points on a line and no four points on a circle was , which follows from work of Solymosi . It could even be the case that there is a uniform upper bound on the size of any integer distance set with no three points on a line and no four points on a circle. Indeed, Ascher, Braune, and Turchet proved this conditionally on Lang’s conjecture. The second question we address is that of estimating the minimum diameter of a non-collinear integer distance set of size (see, for example, Problem 8 of [5, Section 5.11]). It is known, thanks to work of Solymosi and Harborth, Kemnitz, and Möller , that this minimum diameter has order of magnitude at least and at most . Theorem 1.1 reduces the problem of proving a lower bound for the diameter to the problem of proving an upper bound for the maximum number of collinear and concyclic points a non-collinear integer distance set in can contain. The paper of Kurz and Wassermann implicitly contains such a bound for the number of collinear points. Proposition 1.4 (). Let be a non-collinear integer distance set. Then, for every line , we have \| \| \| \| --- \| \| \| \| Bat-Ochir proved an upper bound of the same strength for circles contained in with special radii, which we extend to all circles. Proposition 1.5. Let be a circle and be an integer distance set. Then, \| \| \| \| --- \| \| \| \| Combining Theorem 1.1 with Propositions 1.4 and 1.5, we improve Solymosi’s lower bound to obtain almost tight bounds on the smallest possible diameter of a non-collinear integer distance set of size . Corollary 1.6 (New diameter lower bound). Any non-collinear integer distance set of size has diameter at least \| \| \| \| --- \| \| \| \| It is possible to make all implied constants in the proofs of Propositions 1.4 and 1.5 completely explicit, and thus to compute an explicit such that the minimum diameter of a non-collinear integer distance set of size has order of magnitude at least . On the other hand, one can also compute, using the constructions in , an explicit constant such that the minimum diameter of a non-collinear integer distance set of size has order of magnitude at most . We can improve these constructions a bit to obtain a slightly smaller than the best known one currently in the literature, but there is still a gap between this and the largest possible value of produced by our arguments. It is, therefore, still an open problem to determine the exact asymptotics of the minimum diameter of a non-collinear integer distance set of size . Kurz and Wassermann conjectured that, for sufficiently large, an integer distance set of size with minimum diameter has all but one point collinear. This problem is also still open. Finally, the famous question of Erdős and Ulam of whether there exists a dense rational distance set in the plane is still unresolved, though the answer is known to be negative conditionally on the conjecture or on Lang’s conjecture [29, 31]. 1.1. Contrast with prior approaches. Shortly after his original paper with Anning proving that any non-collinear integer distance set is finite, Erdős gave a very short and elegant second proof of this result , which we now reproduce in its entirety: If , , are three distinct non-collinear points in , then the fact that every point has integer distance from each of , , and implies that \| \| \| \| --- \| \| \| \| for both and . In other words, \| \| \| \| --- \| \| \| \| where, for each , is a family of hyperbolas with foci and . Since the points are not collinear, a hyperbola from can intersect a hyperbola from in at most four points. Therefore, must be finite. All previous progress towards Corollaries 1.3 and 1.6 relied heavily on this argument of Erdős, which is one of many applications of “low degree algebraic geometry” to extremal combinatorics and discrete geometry. It appears that the utility of low degree algebraic geometry in understanding the size and structure of integer distance sets is limited, however, as can be seen from the lack of progress beyond Solymosi’s work . Our new innovation is to use “intermediate degree algebraic geometry” to study integer distance sets; we expect this general method to have further applications in combinatorial and discrete geometry. 1.2. Outline of our method. Our technique may be viewed as a new polynomial method to count points that, after appropriate transformations, are rational points of controlled height satisfying additional conditions of an algebraic nature. To motivate our method, let be a set of rational points of height at most . Then, one strategy to control from above would be to show that lies on an irreducible algebraic curve , defined over , of some intermediate degree (or, equivalently, on a small number of such curves that are irreducible). Roughly speaking, if is too low or too high, then may contain many rational points: it is either not curved enough (e.g., it is a line), or it is too curved. There is, however, a small range of intermediate degrees (depending on ) for which there can be very few rational points of height at most on . The exact bounds follow from the analysis in , as recorded in Theorem 2.3 below. In fact, the degrees that would give optimal bounds on would be of magnitude , implying that . Inspired by this, our approach can be outlined as follows: (1) We show that any integer distance set in can be transformed into a set of rational points of height via an appropriate invertible linear transformation of (see Lemma 2.1 below). 2. (2) According to the above analysis, we would ideally like to show that if is not almost completely contained in a single line or circle, then lies in the union of irreducible algebraic curves, defined over , each of degree roughly (for some constant ). 3. (3) To achieve this, we need to search for algebraic conditions that satisfies. In particular, we fix points (if, of course, points exist in ; note that these points are considerably more than the three points fixed in Erdős’s argument). To each (which corresponds to some unique ), we associate the -tuple of the distances of from each of the . More precisely, we do this on the whole of to form a complex algebraic variety: to each , we associate (generically such) -tuples with for all , where for . 4. (4) We prove that the set of all such (as ranges over ) is an irreducible algebraic surface, defined over , of degree roughly . Since is an integer distance set, the set of all lifted points of on consists of rational points of height . Therefore, we are close to the aim in (2): we have transformed our points into rational points of the correct height inside an irreducible algebraic variety of the correct degree. However, this variety is a surface, rather than a curve. 5. (5) To tackle this issue, we begin by covering by irreducible algebraic curves , defined over , of degree , contained in the surface . The existence of such curves is ensured by the results of (see Theorem 2.2). Projecting the first two coordinates to , we are now closer to the aim in (2), but not quite there: we have covered by an acceptable number of irreducible algebraic curves defined over . However, these curves may have very low degree relative to the desired . 6. (6) We resolve this issue by noticing that we have a lot of freedom to choose the points we fix at the start of the argument. Each choice leads to a different , with properties as in (4). We prove that, if is not a line or a circle, or if it is but there are enough points in outside it, then there is a choice of initial points such that the lift of on the corresponding is an irreducible curve of the correct degree . This completes the proof. We will give a more detailed outline of our argument in Section 2. 1.3. Comparison with the polynomial method in incidence geometry. In 2008, Dvir solved the finite field Kakeya problem , introducing, for the first time, the polynomial method into incidence geometry. This revealed a deep algebraic nature underlying incidence geometric problems. A refinement of Dvir’s method, called polynomial partitioning, was developed soon after by Guth and Katz in their solution of the Erdős distinct distances problem in the plane . Polynomial partitioning has since induced a revolution in incidence geometry and, subsequently, in harmonic analysis, leading to groundbreaking progress on some of the longest-standing conjectures in the fields. Polynomial partitioning relies on finding a low-degree variety containing the points we wish to count. This reveals hidden structure and allows computations. Indeed, in an incidence geometric problem, the points in question generally arise as intersections of certain algabraic objects (e.g., lines), and the study of the interaction of these objects with the variety leads to tighter bounds the lower the degree of the variety is (e.g., via Bézout’s theorem). In stark contrast to this, our method relies on finding high-degree varieties containing our points. This allows us to take advantage of the (at first, hidden) grid-like structure of our point set in the absence of any incidence-geometric structure. 1.4. Notation and conventions We will frequently use Vinogradov’s asymptotic notation throughout this paper: we write to mean that , to mean that (i.e., ), and to mean that and . We will write to represent a quantity that is and to represent a quantity that is . We use to denote the cardinality of a set , to denote the usual Euclidean norm on , to denote -dimensional complex projective space, to denote the rational points of , to denote the divisor function, and to denote the Legendre symbol. Finally, we will always use to denote a prime number, and, for any natural number , will write with to mean that divides but does not divide (so that is the highest power of dividing ). Acknowledgments This work was initiated at the Hausdorff Research Institute for Mathematics (HIM) during the trimester program Harmonic Analysis and Analytic Number Theory; we are grateful to HIM for facilitating this collaboration. We also gratefully acknowledge the hospitality and support of the Institute for Advanced Study during the Special Year on Dynamics, Additive Number Theory and Algebraic Geometry, where a significant portion of this research was conducted. We thank Tony Carbery, Polona Durcik, Ben Green, Leo Hochfilzer, Karen Smith, and Terence Tao for helpful discussions, József Solymosi for help with references and informing us about the history of the study of integer distance sets, as well as Ben Green, Alex Iosevich, and Noah Kravitz for useful comments on an earlier version of this paper. We would particularly like to thank Terence Tao for suggesting a strengthening of a prior version of our structure theorem. RG was supported by the Association of Members of the Institute for Advanced Study (AMIAS) and by NSF grant DMS-2242871. SP was supported by the NSF Mathematical Sciences Postdoctoral Research Fellowship Program under Grant No. DMS-1903038 while most of this research was carried out. 2. Setup and overview of the argument We begin by proving the following lemma, which allows us to assume that the elements of an integer distance set are contained in a lattice and whose proof is a quantification of a well-known argument of Kemnitz . Lemma 2.1. Let be an integer distance set containing the origin and at least one other point. Set to be the smallest distance from the origin to another (distinct) point in . Then there exist a rotation and a squarefree such that \| \| \| \| --- \| \| \| \| Proof. Observe first that if and are any points in an integer distance set, then, since \| \| \| \| \| --- --- \| \| \| \| \| (2.1) \| by the law of cosines, the dot product is in . There exists a rotation such that , so if , then is forced to lie in . It therefore follows that all points in that lie on the -axis are contained in \| \| \| \| --- \| \| \| \| So, suppose that , with , is a point in not on the -axis. Then, since is the square of the distance from to the origin, we certainly have \| \| \| \| --- \| \| \| \| so that there exists an integer such that \| \| \| \| --- \| \| \| \| It follows that there exist a squarefree and an integer such that . Observe that this is the same for all points in not on the -axis. Indeed, if \| \| \| \| --- \| \| \| \| for with squarefree, then, by (2.1), the dot product \| \| \| \| --- \| \| \| \| is a rational number, which forces , and since and are both squarefree, they have to be equal. We therefore conclude that is contained in a set of the form \| \| \| \| --- \| \| \| \| as stated. ∎ 2.1. Encoding the points of as rational points on a surface We will use basic definitions and results from algebraic geometry throughout this paper; a good reference for this is [18, Chapter I]. Observe that it suffices to prove Theorem 1.1 in the case that contains the origin and at least one other point. Indeed, either contains at least two points and in , or the result is trivial; moreover, if there exists a line or circle such that all but points of are contained in , then certainly all but points of are contained in a single line or circle. Thus, it follows from Lemma 2.1 that we may further assume, without loss of generality, that there exists a squarefree such that \| \| \| \| --- \| \| \| \| where is the shortest distance from the origin to another point in , which is necessarily at most . Let be an integer parameter increasing slowly with (which we will, eventually, take to be ). If , then we are done. Otherwise, pick distinct points (so that ) for , none of which is the origin, and set for each . For any point , we define the polynomial by \| \| \| \| --- \| \| \| \| Consider the affine variety \| \| \| \| --- \| \| \| \| which is defined over . Note that implies that \| \| \| \| --- \| \| \| \| Each such point corresponds to a point \| \| \| \| --- \| \| \| \| with all components integers in the projective closure of . Thus, the points in correspond to a certain subset of rational points in of height at most . In Section 3, we will prove that is an irreducible surface of degree . 2.2. Covering rational points of small height by few low degree curves Let be a projective variety defined over . For , we set \| \| \| \| --- \| \| \| \| where for with coprime denotes the height of a rational point . It is known, originally thanks to work of Heath-Brown , that almost all rational points of small height on an irreducible projective surface defined over lie on a small number of low degree curves. Salberger refined this result in the course of his proof of the Dimension Growth Conjecture, and then Salberger’s result was further refined by Walsh to remove all factors of and appearing in the bounds. We will apply a subsequent refinement of Walsh’s result (Theorem 2.2 below) that follows from work of Castryk, Cluckers, Dittmann, and Nguyen , which makes explicit the dependence of various implied constants on the degree of the surface and the dimension of the ambient projective space. This will allow us to deduce, after an appropriate projection, that all points in lie on the union of irreducible affine curves of degree . Theorem 2.2. For any irreducible surface of degree defined over , there exists a homogeneous polynomial of degree at most \| \| \| \| \| --- --- \| \| \| \| \| (2.2) \| that vanishes at all rational points of of height at most but does not vanish on the whole of . Proof. By tracing through the proof of Lemma 5.1 of , the implied constant in the lemma statement satisfies and the constant in the proof satisfies . Thus, by this lemma and the definition of in its proof, there exists a linear map defined over the rationals such that is an irreducible surface of degree and such that, if is a rational point, then \| \| \| \| --- \| \| \| \| Applying Theorem 3.1.1 of to produces a homogeneous polynomial of degree \| \| \| \| --- \| \| \| \| (using that the quantity appearing in the theorem statement in is bounded) that vanishes on all rational points of of height at most but does not vanish on all of . This means that the homogeneous polynomial has degree at most , vanishes on every rational point of of height at most , and does not vanish on all of . ∎ 2.3. Counting rational points of small height on curves As mentioned above, by applying Theorem 2.2 to and then an appropriate projection (see Lemma 6.1), we deduce that is covered by irreducible affine curves of degree . Taking reveals that is covered by irreducible affine curves of degree . Therefore, to prove Theorem 1.1, it suffices to bound, for each irreducible curve that either (1) is not a line or a circle or (2) is a line or circle for which , the maximum possible size of an integer distance set contained in by . Indeed, this immediately gives that all but at most points of are contained in a single line or circle. We show in Section 6 that either contains few points in , or is defined over . In the latter case, we proceed as before, selecting distinct, well-chosen points depending on , none of which is the origin, with and for each , and consider the affine variety \| \| \| \| --- \| \| \| \| Then, the points in correspond to a certain subset of rational points in the projective closure of height at most . In Section 4, we will prove that is an irreducible curve of degree between and , provided that are chosen appropriately. We emphasise that these points will be chosen on when is not a line or a circle, and outside of otherwise. We can then apply a refinement of the Bombieri–Pila bound due to Castryk, Cluckers, Dittmann, and Nguyen , who proved an upper bound on , when is an irreducible algebraic curve defined over , with optimal dependence on and with an explicit polynomial dependence on the degree of . By tracing through the proof of [6, Theorem 2] (and the results used to deduce it from the case [6, Proposition 4.3.2 and Lemma 5.1]), one can see that the dependence of their implied constant on is . We record this in the following theorem. Theorem 2.3 (Castryk, Cluckers, Dittmann, Nguyen ). For any irreducible curve of degree defined over , we have \| \| \| \| \| --- --- \| \| \| \| \| (2.3) \| Thus, as long as either is not a line or circle or is a line or circle for which , we can use Theorem 2.3 to bound the number of points in by , which, for , is . Putting everything together yields Theorem 1.1. We will carry out the argument just outlined in Sections 3, 4, and 6, and prove Propositions 1.4 and 1.5 in Section 5. Remark 2.4. Some polynomial dependence on is necessary in the bounds in Theorems 2.2 and 2.3 (see [6, Proposition 5]). As a consequence, no bound better than in Theorem 1.1 seems attainable via our methods. The determinant method produces bounds that apply to any irreducible surface (respectively, curve) in of degree , but significantly better bounds could hold for individual surfaces (respectively, curves) in this collection. Thus, a further analysis of the specific varieties arising in our argument would be needed to improve Theorem 1.1. 3. Degree and irreducibility of The goal of this section is to prove the following result: Lemma 3.1. Let , be squarefree, and be any distinct points, none of which is the origin. Define the affine variety by \| \| \| \| \| --- --- \| \| \| \| \| (3.1) \| Then is an irreducible surface in of degree defined over . For the remainder of this section, will refer to as in the above lemma. There are several paths to showing that is a surface of degree ; we will take one of the longer routes, since some of the intermediate lemmas we prove along the way will be useful later in this section and in Section 4. 3.1. An explicit description of We begin by explicitly describing . Lemma 3.2. Let be as in Lemma 3.1. We have \| \| \| \| --- \| \| \| \| where \| \| \| \| --- \| \| \| \| for each . Let denote the set of monomials in the variables . Any polynomial can be written as \| \| \| \| --- \| \| \| \| where for all but finitely many . We say that a monomial appears in if in the above expression, and denote the homogenization of with respect to by , where is the maximal total degree of any monomial appearing in . To prove Lemma 3.2, it suffices, by standard results (e.g., combine Theorems 4 and 8 of [7, Chapter 8 §4]), to show that is a Gröbner basis111See Definition 5 in [7, Chaper 2 §5] for the definition of a Gröbner basis. for with respect to some graded monomial order. Recall that a monomial order is a total order on satisfying (1) for all and 2. (2) implies that for all . A monomial order on is said to be graded if \| \| \| \| --- \| \| \| \| whenever . Once a monomial order has been fixed, one can write any as \| \| \| \| --- \| \| \| \| where and for all , and where . Then the leading term of is defined to be \| \| \| \| --- \| \| \| \| and the leading monomial of is defined to be \| \| \| \| --- \| \| \| \| Finally, recall that the multivariable polynomial division algorithm (see, for example, [7, Chapter 2 §3]) says that the remainder of under division by a set of polynomials is zero if and only if we can write \| \| \| \| --- \| \| \| \| where each and whenever . To prove Lemma 3.2, we will use the graded lexicographic (grlex) order with , which is defined by declaring that if and only if one of the following two conditions is met: (1) or 2. (2) and if is the smallest index for which . We will also require Buchberger’s criterion (see [7, Chapter 2]), which gives a convenient way to check whether a generating set of an ideal is a Gröbner basis. Lemma 3.3 (Buchberger’s criterion). Let be an ideal of and be a basis for . Then is a Gröbner basis with respect to a monomial order if and only if each -polynomial \| \| \| \| --- \| \| \| \| has remainder zero on division by . Lemma 3.4. With respect to the grlex ordering with on , is a Gröbner basis for . Proof. By Buchberger’s criterion, it suffices to check that each -polynomial with has remainder zero on division by \| \| \| \| --- \| \| \| \| So let , and note that, since and , we have \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| From the second line above, we see that , so since \| \| \| \| --- \| \| \| \| and \| \| \| \| --- \| \| \| \| we certainly have that has remainder zero on division by . ∎ By the discussion above, Lemma 3.2 now follows. 3.2. Dimension and degree of Next, we will confirm that is a surface of degree . There are a few ways to see that is a projective surface, but perhaps the fastest is to note that , and to use the general fact (see [7, Chapter 9 §3]) that if is an ideal and is a Gröbner basis for with respect to a graded monomial order, then equals the maximal size of a subset of the variables such that, for all , the leading term of contains a variable not in . The following lemma is then an immediate consequence of Lemma 3.4 (by taking ). Lemma 3.5. Let be as in Lemma 3.1. Then (i.e., is a surface). Now, we will show that has degree . This will follow easily from two standard facts, which we record here for later use. Both are immediate consequences of Theorem 7.7 from [18, Chapter I]. Lemma 3.6 (Bézout’s inequality). Let be a projective variety. We have \| \| \| \| --- \| \| \| \| Lemma 3.7. Let be a projective variety of dimension . If is any linear subvariety of codimension such that is finite, then is at least . Now we can finish this subsection. Lemma 3.8. Let be as in Lemma 3.1. Then . Proof. By Lemmas 3.2 and 3.6, we certainly have . For the lower bound, consider the linear variety of codimension defined by . Then \| \| \| \| --- \| \| \| \| which has elements, since we stipulated that none of are the origin. Thus, by Lemma 3.7, as well, so that . ∎ 3.3. Irreducibility of Finally, we will show that is irreducible. Lemma 3.9. Let be as in Lemma 3.1. Then is irreducible. Note that it suffices to check that is an irreducible affine variety. We will begin this section by computing the set of singular points of . Lemma 3.10. Let be as in Lemma 3.1, and denote the set of singular points of by . We have \| \| \| \| --- \| \| \| \| Thus, consists of distinct points. Proof. Observe that factors as \| \| \| \| --- \| \| \| \| Thus, if and only if the pair lies on at least one of the lines \| \| \| \| --- \| \| \| \| or \| \| \| \| --- \| \| \| \| which intersect at the point . Since are all distinct, we must have whenever . A short calculation shows that and intersect at the point when . The Jacobian matrix of at a point is the matrix \| \| \| \| --- \| \| \| \| If with for some , then is clearly a singular point of , as the -th row of the Jacobian matrix is zero and thus the rank cannot be . Conversely, can only possibly be a singular point of if for at least one . If for exactly one index and , then and the Jacobian matrix still has rank . Since are all distinct, it follows that and whenever and are pairwise distinct indices. Thus, the only remaining possibility is that for exactly two distinct indices and , in which case is forced to be the intersection point of either and or of and . Without loss of generality, we may assume that we are in the first case, so that \| \| \| \| --- \| \| \| \| This matrix has determinant , and so is nonsingular. Thus, in this case too, the Jacobian matrix still has rank . We conclude that for some is necessary and sufficient for to be a singular point of . ∎ Set to be the set of nonsingular points of . Since is open and dense in , to show that is irreducible, it suffices to show that is irreducible. Further, since is a smooth quasiaffine variety (as it is a smooth, open subvariety of ), it is irreducible if it is connected (see, for example, Remark 7.9.1 of [18, Chapter III]). Set \| \| \| \| --- \| \| \| \| so that, in the notation of the proof of Lemma 3.10, \| \| \| \| \| --- --- \| \| \| \| \| (3.2) \| where is the standard projection map onto the first two coordinates. Note that by, for example, Theorem 1.22 of , since neither of nor vanish on the whole of . Indeed, we proved in Lemma 3.4 that is a Gröbner basis for with respect to the grlex ordering with . If were to vanish on , then we would have , so that there would exist an such that . But, since is a Gröbner basis for , if and only if has remainder zero on division by . Note that if \| \| \| \| --- \| \| \| \| for some , then we must, necessarily, have for some (in particular, for any for which ), since all monomials appearing in are divisible by only the variables and , while if , then is divisible by , and for all . It follows that cannot possibly have remainder zero on division by , so does not vanish on the whole of . An identical argument shows that also does not vanish on the whole of . It follows that must be dense in , since all irreducible components of are two-dimensional (see, for example, Proposition 7.1 of [18, Chapter I]). The proof of Lemma 3.9 thus boils down to showing that is connected. As any set that is connected in the standard (Euclidean) topology is Zariski-connected, it thus suffices to show that is connected in the standard topology on , which we will work in for the remainder of this section. We will next show that is, in fact, path-connected. It will then follow that is connected, and thus Zariski-connected, as desired. Lemma 3.11. The set is path-connected. Proof. Set \| \| \| \| --- \| \| \| \| First, note that itself is path connected. Indeed, is homeomorphic to minus planes, and minus any finite number of planes can be seen to be path connected by taking any two points , noting that there must exist a plane containing both and such that consists of finitely many points, and then using that a plane minus any finite number of points is path connected. Observe that is a -fold covering map. It therefore suffices to show that there exists such that all points in the fiber are in the same path-component in . Indeed, for any two points , there exist paths and in from to and from to , respectively. By (3.2), these paths lift to paths and in from to an element of and from an element of to , respectively. If there exists a path from to , then concatenating , , and shows that and lie in the same path-component. We will select to be the explicit point \| \| \| \| --- \| \| \| \| so that equals \| \| \| \| --- \| \| \| \| Consider the (complex) line , and note that, since for each , we have \| \| \| \| --- \| \| \| \| where the intersection consists of the single point for each . Moreover, since for we have \| \| \| \| --- \| \| \| \| the set consists of distinct elements. Since is homeomorphic to a (real) plane minus points, for each there certainly exists a path contained in starting and ending at with winding number around each of the points in and with winding number around each of the points in . We will show that each of these paths lifts to a path in that starts from and ends at a different point in ; this will prove the desired statement. To study lifts of such paths , we observe that the intersection is parametrized by where ranges over . For all , any element of is of the form , with \| \| \| \| --- \| \| \| \| and , for all . Crucially, is a linear function of that vanishes exactly at , so that the order of vanishing at is exactly . Thus, let be a closed path, beginning and ending at . The path lifts to a path in , that starts at and ends at some point in . By the above, the -th coordinate of is a continuous square root of . • If has winding number around the point , then, viewing the argument of as a continuous function of , it follows (by the formula for described above) that the arguments at and differ by . Therefore, any continuous square root of will have arguments that differ by at and . Since (as this is the -th coordinate of ), we deduce that . • If has winding number around the point , then, viewing the argument of as a continuous function of , we see that the arguments at and at are the same. Therefore, . In other words, if, for some , has winding number 1 around each of the points in , and winding number 0 around each of the points in , then lifts to a path in connecting to the point \| \| \| \| --- \| \| \| \| where denotes the indicator function of . Since such exists for arbitrary , it follows that all elements of are in the same path-component, completing the proof of the lemma. ∎ This completes the proof of Lemma 3.9, and thus the proof of Lemma 3.1. 4. Degree and irreducibility of In this section, we prove that is an irreducible curve of large degree, treating the case that is not a line or circle and the case that is a line or circle separately. Before stating our main results, we will require a preparatory technical lemma. Lemma 4.1. Let and be a polynomial of degree , and write \| \| \| \| --- \| \| \| \| where is the degree homogeneous component of . Then, \| \| \| \| --- \| \| \| \| where each is a polynomial of degree at most and, for some , at least one of or is not identically zero. Proof. A short computation yields \| \| \| \| --- \| \| \| \| for each , which shows that . If for all , then we must have \| \| \| \| --- \| \| \| \| for all and . When , this implies that and are both roots of of multiplicity at least . Thus, . But, since , so we must have , which is impossible when . We conclude that or for some . ∎ Lemma 4.2. Let , be squarefree, and be an irreducible curve that contains the origin and is not a line or circle such that the curve \| \| \| \| --- \| \| \| \| is defined over and the origin is not a singular point of . Let be such that and for be as in Lemma 4.1. Denote the set of singular points of by and define by for . Let be distinct points, none of which is the origin, a singular point of , or, when , a point in , such that for every the point \| \| \| \| --- \| \| \| \| is not on and such that \| \| \| \| \| --- --- \| \| \| \| \| (4.1) \| Define the affine variety by \| \| \| \| --- \| \| \| \| Then is an irreducible curve in of degree at least and at most , defined over . Lemma 4.3. Let , be squarefree, and be a line or circle that contains the origin such that the curve \| \| \| \| --- \| \| \| \| is defined over . Let be distinct points such that, for every , the point \| \| \| \| --- \| \| \| \| is not on and such that, if is a circle, none of is the center of . Define the affine variety by \| \| \| \| --- \| \| \| \| Then is an irreducible curve in of degree at least and at most , defined over . 4.1. Dimension and degree of Given the results of the previous section, it is now easy to deduce the following. Lemma 4.4. Let and be as in Lemma 4.2 or Lemma 4.3. Then is a curve of degree at least and at most . Proof. Let denote the standard projection map onto the first two coordinates. First of all, we certainly have that , since and . To see that , note that is a subvariety of the irreducible affine surface (where is defined as in Lemma 3.1), so that and would imply that (by the irreducibility of ), which is impossible since . We conclude that is a curve. The upper bound \| \| \| \| --- \| \| \| \| is an immediate consequence of Bézout’s inequality (Lemma 3.6), since . Showing that requires essentially the same argument as in the proof of Lemma 3.8, since we assumed that (and thus ) contains the origin. Let be a line that is not equal to . Since the irreducible curve is not , is finite. Consider the subspace defined by . Then \| \| \| \| --- \| \| \| \| where the set on the right-hand side has elements, since none of is the origin and thus \| \| \| \| --- \| \| \| \| Moreover, is finite. Indeed, since is finite, the definition of gives \| \| \| \| --- \| \| \| \| for some finite number of affine hyperplanes , each of the form \| \| \| \| --- \| \| \| \| for some fixed (a point in the finite set ). Since, for each , the fiber \| \| \| \| --- \| \| \| \| in above is finite, it follows that each is finite, hence is finite. Thus, by Lemma 3.7, . ∎ 4.2. Irreducibility of Now, we will show the following. Lemma 4.5. Let be as in Lemma 4.2 or Lemma 4.3. Then is irreducible. As in the proof of Lemma 3.9, it suffices to check that is an irreducible affine variety. We will, similarly, begin this section by roughly describing the set of singular points of . Lemma 4.6. Let and be as in Lemma 4.2. Denote the set of singular points of by . Then is finite, \| \| \| \| \| --- --- \| \| \| \| \| (4.2) \| and, if , then \| \| \| \| \| --- --- \| \| \| \| \| (4.3) \| Proof. First of all, it is a standard fact that the set of singular points of an algebraic curve is finite (see, for example, Theorem 5.3 of [18, Chapter I]). Writing for an appropriate polynomial , we have that the Jacobian matrix of at a point is \| \| \| \| \| --- --- \| \| \| \| \| (4.4) \| If with , then some row of will be the zero vector. This means that can have rank at most , so that must be a singular point of . This proves (4.2). Finally, if , then, by definition, has rank at most . Since , by the definition (3.1) of , any element of \| \| \| \| --- \| \| \| \| has the form for some . Then for the diagonal matrix that has entries along the diagonal. Since is nonsingular, must have the same rank as . It follows that as well, proving (4.3). ∎ Arguing in almost the exact same manner (the only differences being that is smooth and ) yields the analogous lemma for when is a line or circle. Lemma 4.7. Let and be as in Lemma 4.3. Denote the set of singular points of by . Then is finite and, if , then \| \| \| \| \| --- --- \| \| \| \| \| (4.5) \| Next, we will analyze the intersection of with the lines defined in the proof of Lemma 3.10. Lemma 4.8. Let and be as in Lemma 4.2 and, for each , define \| \| \| \| --- \| \| \| \| and \| \| \| \| --- \| \| \| \| Then, for each we have (1) , 2. (2) , and 3. (3) there exists a point \| \| \| \| --- \| \| \| \| not equal to or any of the singular points of . Proof. We certainly have for each , since is an irreducible curve that is not a line. In the proof of Lemma 3.10, we saw that for each , which gives (2) since . To prove (3), first note that . By Bézout’s theorem, the number of intersections, counted with multiplicity, of with and of with are both . Letting and denote the intersection multiplicity of with at and at a point at infinity and with at and at a point at infinity, respectively, it thus suffices to show that or . Let us first consider the intersection multiplicities and of with and of with , respectively, at infinity when . The point at infinity lying on is and the point at infinity lying on is . On , and can be parametrized as \| \| \| \| --- \| \| \| \| and \| \| \| \| --- \| \| \| \| respectively. Thus, is the multiplicity of as a root of and is the multiplicity of as a root of . It then follows from Lemma 4.1 and the assumption that that there exists a such that the coefficient of in or is nonzero. Thus, or is at most when . Now, note that (respectively, ) is the intersection multiplicity of the line (respectively, ) with the curve defined by the vanishing of the polynomial at the point . Thus, (respectively, ) is the multiplicity of the root of the polynomial (respectively, ). If (respectively, ) were greater than , we would have that if (respectively, ), then (respectively, ). But, observe that, as a function of , the linear term of equals \| \| \| \| --- \| \| \| \| so that if and only if both \| \| \| \| \| --- --- \| \| \| \| \| (4.6) \| and \| \| \| \| \| --- --- \| \| \| \| \| (4.7) \| But, since has rational coefficients, neither (4.6) nor (4.7) can hold since, by hypothesis, is not a singular point of . We therefore conclude that . Thus, if , then or is less than , from which (3) follows. Since by the assumption that is not a line, it only remains to rule out that when . In the case , would imply that . Write where is the degree homogeneous component of for . Then implies that , which means that and, thus, that the degree homogeneous component of must equal . This forces to either be a circle or the union of two lines, both of which contradict our assumptions on in Lemma 4.2. We have thus shown that there exists a point not equal to for each . That is, additionally, not a singular point of follows immediately from the assumption (4.1) in Lemma 4.2, which says that none of the singular points of lie in . This completes the proof of the lemma. ∎ Lemma 4.9. Let and be as in Lemma 4.3 and, for each , define \| \| \| \| --- \| \| \| \| and \| \| \| \| --- \| \| \| \| Then, for each we have (1) , 2. (2) , and 3. (3) there exists a point \| \| \| \| --- \| \| \| \| Proof. Since for all , we certainly always have (2), i.e., . For the other two claims, we begin with the case that , and thus , is a line, so that where for . Then no two of are parallel, so that , yielding (1) and (3). Now we deal with the case that is a circle, so that where with . As in the proof of the previous lemma, we certainly have since is irreducible and not a line. Also as in the proof of the previous lemma, we have that the number of intersections (counted with multiplicity) of with and of with are both equal to . So, to conclude (3) in this case, we just need to show that it is impossible for the intersection multiplicities of with at (the point at infinity lying on ) and of with at (the point at infinity lying on ) to both equal . On , and can be parametrized as \| \| \| \| --- \| \| \| \| and \| \| \| \| --- \| \| \| \| respectively. Thus, is the multiplicity of as a root of and is the multiplicity of as a root of . If (respectively, ) were equal to , then (respectively, ) would be equal to a multiple of . But, expanding out the definition of (respectively, ) yields \| \| \| \| --- \| \| \| \| (respectively, \| \| \| \| --- \| \| \| \| which is of the form for some nonzero by the assumption that is not the center of (which implies that ). Thus, we must have . So, in fact, for all , which implies (3) in this case as well. ∎ Let be as in Lemma 4.2 or Lemma 4.3, and set to be the set of nonsingular points of . Analogously to our argument222See the discussion between Lemma 3.10 and Lemma 3.11. in Section 3, to show that is irreducible, it suffices to show that is irreducible, and since is a smooth quasiaffine variety, it further suffices to show that is connected. As in Section 3.3, we define \| \| \| \| --- \| \| \| \| so that \| \| \| \| --- \| \| \| \| Since is dense in (as all irreducible components of are one-dimensional), it suffices to show that is connected, and, analogously to the argument in Section 3, it further suffices to show that is connected in the Euclidean topology on , which we will work in for the remainder of this section. Lemma 4.10. The set is path connected. Proof. We argue along the lines of the proof of Lemma 3.11. Observe that \| \| \| \| --- \| \| \| \| is a -fold covering map by claim (4.3) of Lemma 4.6 or claim (4.5) of Lemma 4.7. Since, by Lemmas 4.6 and 4.8 or by Lemmas 4.7 and 4.9, the set \| \| \| \| --- \| \| \| \| is finite, by applying the normalization theorem for irreducible algebraic curves to and using that any compact Riemann surface is homeomorphic to either the -sphere or a finite connected sum of -tori (see, respectively, and , for example), it follows that is homeomorphic to either a -sphere minus a finite number of points or a finite connected sum of -tori minus a finite number of points, where there is a bijective correspondence between the finitely many points removed from the Riemann surface and the finitely many points in . One consequence of this observation is that is path-connected. So, analogously to the proof of Lemma 3.11, it thus suffices to show that the fiber is path-connected in , where \| \| \| \| --- \| \| \| \| Note that \| \| \| \| --- \| \| \| \| since none of is the origin, is nonsingular at the origin, and none of the lines pass through the origin. By one of the assumptions of Lemma 4.2 and Lemma 4.3, the set \| \| \| \| --- \| \| \| \| is disjoint from for every , so the points \| \| \| \| --- \| \| \| \| or \| \| \| \| --- \| \| \| \| defined as in Lemma 4.8 (3) or Lemma 4.9 (3), respectively, are all distinct. A second consequence of our observation about the topology of is that, for any choice of , there exists a path contained in starting and ending at with winding number around each of the points in and with winding number around each of the points in . Lifting the closed path then gives rise to a path in from to the point in with the property that, for , its -th coordinate is the opposite of the -th coordinate of , i.e., equals \| \| \| \| --- \| \| \| \| while, for , its -th coordinate agrees with the -th coordinate of , i.e., equals \| \| \| \| --- \| \| \| \| In other words, lifts from to a path in connecting to the point \| \| \| \| --- \| \| \| \| Since was arbitrary, it follows that all elements of are in the same path-component, completing the proof of the lemma. ∎ This completes the proof of Lemma 4.5. Combining Lemmas 4.4 and 4.5 now yields Lemmas 4.2 and 4.3. 5. The intersections of integer distance sets with lines and circles 5.1. Lines Proposition 1.4 was essentially proved in [24, Section 5]. However, since some of the terms used in are nonstandard and [24, Theorem 4] is phrased in a manner that is not equivalent to Proposition 1.4, we will present a detailed proof of Proposition 1.4 here. Proof of Proposition 1.4. Let be the maximal number of points of that are all contained in a line . Without loss of generality, we may assume that and for , where . Let be such that the triangle with vertices , , and has maximal area (note that by our assumption that not all the points of are collinear, this maximal area must be positive). Then, is the height of the triangle from to on the base of . For , set and , so that . For , let be the distance between and , so that . See Figure 5.1. Then, by Pythagoras’ theorem, we have \| \| \| \| --- \| \| \| \| which, on subtracting one equation from the other, gives \| \| \| \| \| --- --- \| \| \| \| \| (5.1) \| Letting , then, by (5.1), is also an integer; hence, for , since is in , we have that \| \| \| \| --- \| \| \| \| is an integer too. A further application of Pythagoras’ theorem gives for . Thus, \| \| \| \| \| --- --- \| \| \| \| \| (5.2) \| for . Note that we either have or . If the former holds, then for every we have and ; if the latter holds, then for every we have and . We conclude that, in either case, \| \| \| \| --- \| \| \| \| So, since and , are all integers, (5.2) implies that the natural number has different divisors, hence . Since , we thus have \| \| \| \| --- \| \| \| \| using the standard estimate (see, for example, [17, Theorem 315]), which gives the claim. ∎ 5.2. Circles Recall that Proposition 1.5 controls the size of an integer distance set in that lives on an arbitrary circle. It turns out that the bulk of the argument lies in tackling the case where the circle is fully contained in ; the general case easily follows. In particular, for any positive real number , define to be the maximum size of an integer distance set contained in the circle of radius centered at the origin. Then, this special case of Proposition 1.5 is equivalent to the following. Proposition 5.1. We have \| \| \| \| --- \| \| \| \| We can deduce Proposition 1.5 from Proposition 5.1 as follows. Proof of Proposition 1.5. Let , be a circle, and be an integer distance set. Our goal is to show that \| \| \| \| --- \| \| \| \| Let be an absolute constant to be fixed shortly. If has radius at most , then by Proposition 5.1. Thus, the proof is complete in this case. If has radius , then we claim that , provided that is chosen to be sufficiently large. Indeed, letting denote the minimal angle between any three (necessarily non-collinear) points of , i.e., \| \| \| \| --- \| \| \| \| then, by [30, Observation 1], we have \| \| \| \| \| --- --- \| \| \| \| \| (5.3) \| As the diameter of is , the intersection of with must be contained in an arc of of length . Therefore, the central angle of in satisfies , which gives \| \| \| \| \| --- --- \| \| \| \| \| (5.4) \| Fixing sufficiently large, it follows from (5.3) that is strictly smaller than , and thus, by (5.4), that \| \| \| \| \| --- --- \| \| \| \| \| (5.5) \| Now, suppose to the contrary that intersects in a set containing three distinct points ; reordering the points if needed, we can assume that . Thus, from the definition of , we must have , contradicting (5.5). We conclude that can intersect in at most two points, completing the proof in this case as well. ∎ It remains to prove Proposition 5.1. We will begin by showing that one can assume the radius of is of a certain special form. Lemma 5.2. There exists an absolute constant such that \| \| \| \| \| --- --- \| \| \| \| \| (5.6) \| Proof. Suppose that is the circle of radius centered at the origin. If the largest integer distance set contained in has size , then the desired inequality automatically holds by taking (since one can inscribe an equilateral triangle with side length inside a circle of radius ). So, suppose that contains three distinct points and such that \| \| \| \| --- \| \| \| \| are all integers. Then, by Heron’s formula and the circumradius formula, we have \| \| \| \| --- \| \| \| \| Thus, there exist and a squarefree such that . Since , we certainly have , , and . By dilating by , we see that \| \| \| \| --- \| \| \| \| for the positive integer . Note that the proof of (5.6) is not yet complete, as and are not necessarily coprime. Since is squarefree, we can write \| \| \| \| --- \| \| \| \| for with and both and being squarefree, where we have and (specifically, is the product of those prime factors of that divide ). Next, by adapting the proof of Proposition 1 of , we will show that \| \| \| \| \| --- --- \| \| \| \| \| (5.7) \| whenever are such that , and are squarefree, and . The conclusion of the lemma will then follow by repeated applications of (5.7), which eventually gives that \| \| \| \| --- \| \| \| \| where \| \| \| \| --- \| \| \| \| It therefore remains to prove (5.7). For with , let be any distinct points such that \| \| \| \| --- \| \| \| \| and let denote the origin. We will show that all are divisible by (which will then complete the proof by rescaling). Let us focus on first. By rotating and reflecting , we can assume that lies on the positive -axis, and that \| \| \| \| --- \| \| \| \| see Figure 5.2. If , then , which means that and there is nothing to prove. So, suppose that . Then , and hence, by the law of cosines, since all of the side lengths of the triangle with vertices are positive integers, we deduce that is rational. In addition, from the law of sines in the same triangle, we have that \| \| \| \| --- \| \| \| \| Now, as \| \| \| \| --- \| \| \| \| is rational and , we deduce that there must exist with such that \| \| \| \| --- \| \| \| \| Thus, , which implies that divides . Since is squarefree, it follows that , and so . Therefore, is divisible by , since is squarefree and . Now by rotating and reflecting again, we can assume that (respectively, ) lies on the positive -axis and that (respectively, ). By applying the same argument as above we obtain that (respectively, ) must also be divisible by . Therefore, the distances between the points on the circle are also integers. This gives (5.7), as desired. ∎ We will now prove Proposition 1.5. When the radius of has the special form with , squarefree, and having class number , this result already follows from work of Bat-Ochir . A minor modification of Bat-Ochir’s argument, using a classical result on binary quadratic forms of negative discriminant in place of unique factorization, proves the proposition in general. Lemma 5.3. There exists an absolute constant such that, for all with and squarefree, we have \| \| \| \| --- \| \| \| \| Proof. Using that , we factor , where \| \| \| \| --- \| \| \| \| Let denote the circle of radius centered at the origin , and suppose that is an integer distance set of maximal size , where we may assume, without loss of generality, that . We may also assume, without loss of generality, that contains a point, , that lies on the -axis. Let be any two other distinct points, and set and . By reflecting about the -axis, if needed, we may assume that . Then, as in the proof of Lemma 5.2, must be rational by the law of cosines, since and the side lengths of the triangle with vertices are all integers. On the other hand, by the law of sines, , so that \| \| \| \| --- \| \| \| \| is rational. Thus, must be rational, and thus an integer, say . Squaring and rearranging to get \| \| \| \| --- \| \| \| \| it follows that both and are divisible by , for otherwise would be a quadratic residue modulo some prime dividing , contradicting the definition of . Setting and , so that , we thus get \| \| \| \| --- \| \| \| \| from which it follows that, since the point is completely determined by the angle and the radius , we must have \| \| \| \| --- \| \| \| \| The right-hand side above is well-known to be (see, for example, [21, Chapter 11]). Since \| \| \| \| --- \| \| \| \| this completes the proof of the lemma. ∎ Combining Lemmas 5.2 and 5.3 now proves Proposition 5.1, and thus Proposition 1.5. 6. Conclusion of the argument We will need two final preparatory lemmas to prove Theorem 1.1. Lemma 6.1. Let be an affine curve of degree , and let denote the projection map . Then is contained in the union of at most irreducible affine curves of degree at most . Proof. By Chevalley’s theorem (see, for example, Exercise 3.19 of [18, Chapter I]), there exist a finite number of affine varieties such that \| \| \| \| --- \| \| \| \| Note that we must have for all , for otherwise we could find a line that intersects in infinitely many points for which the hyperplane does not contain , which yields a contradiction since then would consist of finitely many points while its projection under would consist of infinitely many points. Thus, can be written as the union of a finite number of irreducible curves and a finite number of points , minus a finite number of points that are distinct from . A generic line avoids and is distinct from any of the curves that happen to be a line, and thus intersects in at most points. Therefore, , from which the conclusion of the lemma follows. ∎ Lemma 6.2. Let be an irreducible affine curve of degree and be squarefree. Suppose that contains at least points in . Then \| \| \| \| --- \| \| \| \| is an irreducible affine curve of degree defined over . Proof. Since the curve is irreducible of degree , it is the zero set of some irreducible polynomial of degree . Then, is the zero set of the polynomial , which is also irreducible of degree . We will show that there exists a non-zero such that . It will then follow that is an irreducible affine curve of degree defined over , since is also irreducible, has degree , and has the same zero set as in . So, fix a set of rational points on . Consider the general polynomial \| \| \| \| --- \| \| \| \| of degree at most subject to the conditions \| \| \| \| \| --- --- \| \| \| \| \| (6.1) \| Note that (6.1) is a linear system of equations and unknowns, and thus of the form for a certain matrix with rational entries. Since the non-zero polynomial has degree at most and vanishes on , the system (6.1) has a non-trivial solution in . Equivalently, the linear map with has . Since has rational entries, the map with is well-defined, and also has . Indeed, the rank of is the largest order of a non-vanishing minor of , and is thus independent of whether is viewed over or over . By the above, the linear system in has a non-trivial solution in , i.e. there exists a non-zero polynomial , of degree at most , vanishing on . Both and vanish on , and thus on at least points of . By Bézout’s theorem, and have a common factor. Since is irreducible, divides ; and since , we conclude that is a non-zero, constant multiple of . ∎ Now we are ready to prove Theorem 1.1, from which Corollaries 1.3 and 1.6 immediately follow (the latter by invoking Propositions 1.4 and 1.5). In fact, we will prove an even stronger structure theorem that implies Theorem 1.1. Theorem 6.3. Let be an integer distance set. Then either \| \| \| \| --- \| \| \| \| or there exists a line or circle such that \| \| \| \| --- \| \| \| \| Proof. We follow the outline in Section 2. Let be an integer distance set and let , with as and , be a parameter to be chosen later. We may assume, without loss of generality, that contains the origin and at least other points, and, further, that there exists an and a squarefree such that \| \| \| \| --- \| \| \| \| where . Pick distinct points , none of which is the origin, so that each with . Set for . Using the notation of Section 3, consider the affine variety \| \| \| \| --- \| \| \| \| in . Each point corresponds to the rational point \| \| \| \| --- \| \| \| \| of height . By Lemma 3.1, is an irreducible surface of degree . Thus, it follows from Theorem 2.2 and Lemma 6.1 that is contained in the union of irreducible affine curves , each of degree at most . For each , either contains at least points in , in which case is defined over by Lemma 6.2, or else contains at most points. It follows that all but points of are contained in the union of irreducible affine curves of degree at most , where each \| \| \| \| --- \| \| \| \| is defined over . Let . Assume first that is not a line or circle. We will now bound . We may assume, without loss of generality (by applying a translation), that contains the origin and that the origin is not a singular point of . Recall that an irreducible affine curve in of degree can have at most singular points (see, for example, Exercise 3 of [28, Chapter 3 §2]). Denoting the set of singular points of by , either contains at most \| \| \| \| \| --- --- \| \| \| \| \| (6.2) \| nonsingular points of , or else there exist distinct points \| \| \| \| --- \| \| \| \| none of which is the origin or a singular point of , such that \| \| \| \| --- \| \| \| \| and \| \| \| \| \| --- --- \| \| \| \| \| (6.3) \| and, if , \| \| \| \| \| --- --- \| \| \| \| \| (6.4) \| where the are as in Lemma 4.2. Indeed, the condition (6.3) forbids at most \| \| \| \| --- \| \| \| \| points, and, if , the condition (6.4) excludes at most points. The reason for the latter is that is contained in a curve of degree at most (by Lemma 4.1), which thus has no irreducible components that agree with the irreducible curve ; therefore, the intersection of with can contain at most points by Bézout’s inequality. Moreover, if , with , have already been chosen so that \| \| \| \| --- \| \| \| \| then there are at most points for which \| \| \| \| --- \| \| \| \| This is because, for all such points, lies on one of the intersections \| \| \| \| --- \| \| \| \| each of which consists exactly of the points with , where is a linear transformation that maps to the line and . For each choice of and sign or , there are at most points in by Bézout’s inequality (Lemma 3.6), and, for each such , there are at most points with , again by Bézout’s inequality (since gives the equation of a line in the variables and , while is not a line). Therefore, each of the above intersections can contain at most points; by summing \| \| \| \| --- \| \| \| \| we conclude that as long as there are at least (6.2) nonsingular points of in , we can find a suitable choice of . We then set for and consider the affine variety \| \| \| \| --- \| \| \| \| in . Every point in corresponds to a rational point on of height . By Lemma 4.2, is an irreducible curve of degree at least and at most , so that, by Theorem 2.3, whenever contains at least (6.2) nonsingular points of . If contains fewer than (6.2) nonsingular points of , then . Thus, either way, we have that \| \| \| \| --- \| \| \| \| whenever is not a line or circle. Now assume that is a line or circle. To bound , we proceed as in the previous case, with a few modifications. We may again assume, without loss of generality, that contains the origin. Either contains at most points outside of , or else there exist distinct points \| \| \| \| --- \| \| \| \| none of which is the center of in the case that is a circle, such that \| \| \| \| --- \| \| \| \| Indeed, if with have already been chosen so that \| \| \| \| --- \| \| \| \| then there are at most points for which \| \| \| \| --- \| \| \| \| since each of the sets \| \| \| \| --- \| \| \| \| is the union of at most two lines of the form for some ; each such set intersects in at most points (by Bézout’s inequality, as ), while at the same time there can only be one point for which on each of the lines (as each such line contains at most one point in ). By summing \| \| \| \| --- \| \| \| \| we conclude that as long as there are at least points outside of in , we can find a suitable choice of . 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189172	https://suli.pppl.gov/2019/course/Collins_SULISingleParticleMotion_2019.pdf	1 Single Particle Motion Presented by Cami Collins at the SULI Introductory Course in Plasma Physics Princeton Plasma Physics Laboratory June 10, 2019 2 2 My path in plasma physics  3 plasma physics turbulence transport I want fusion energy to work. But what does that mean? stability 4 plasma physics computer science steady state ELM controlled disruption mitigation turbulence transport high confinement I want fusion energy to work. But what does that mean? stability efficient current drive control high bootstrap current shaping high beta dissipative divertor engineering 5 engineering plasma physics computer science materials science tritium breeding steady state ELM controlled disruption mitigation turbulence transport high confinement alpha heating I want fusion energy to work. But what does that mean? stability efficient current drive control power handling high bootstrap current shaping high beta dissipative divertor 6 engineering plasma physics computer science materials science tritium breeding steady state ELM controlled disruption mitigation turbulence transport nuclear materials net electricity metal walls, liquid walls? heat extraction high confinement alpha heating I want fusion energy to work. But what does that mean? licensing stability efficient current drive control power handling high bootstrap current shaping high beta dissipative divertor 7 engineering plasma physics computer science materials science political science? tritium breeding steady state ELM controlled disruption mitigation turbulence transport nuclear materials net electricity metal walls, liquid walls? heat extraction high confinement alpha heating I want fusion energy to work. But what does that mean? licensing stability efficient current drive control power handling high bootstrap current shaping high beta dissipative divertor social science? $ $ $ $ $ 8 Deuterium ~0.01 MeV Helium 3.5 MeV Neutron 14.1 MeV Tritium ~0.01 MeV • Sustained fusion reactions require enough particles (density) that are energetic enough (temperature) and collide often enough (confinement time). Plasma Physics is the Basis for Fusion Research 9 Deuterium ~0.01 MeV Helium 3.5 MeV Neutron 14.1 MeV Tritium ~0.01 MeV • Sustained fusion reactions require enough particles (density) that are energetic enough (temperature) and collide often enough (confinement time). • The fusion triple product is the figure of merit: T~100-200 million K n~2-3x1020 ions/m3 τ ~1-2 s D & T is a plasma at these temperatures Plasma Physics is the Basis for Fusion Research 10 Typical velocity of a 100 million K ion: Even with ~1020 ions/m3 , the ion would travel ~10 km before colliding with another It would be crazy to build a fusion reactor that big! We can understand a lot about how fusion devices confine plasma by studying single particle motion. The trick: use magnetic fields charged particle magnetic field line with magnetic field no magnetic field 11 • Gyromotion about a guiding center • Forces can cause guiding center drift • Real life consequences: – Why do tokamaks have helical B fields? – What is a banana orbit? – Why are instabilities like Alfvén Eigenmodes bad for fusion? Outline • NRL Plasma Formulary www.nrl.navy.mil/ppd/content/nrl-plasma-formulary • Introduction to Plasma Physics and Controlled Fusion by F. Chen References 12 • A particle with charge (q) moving with velocity (v) in the presence of electric and magnetic fields will experience a force: Charged Particles Feel The (Lorentz) Force We know from Newton’s second law of motion that force causes acceleration: A charged particle moving perpendicular to the magnetic field feels a force 13 • Consider the motion of a particle in a constant, uniform B field How Does a Charged Particle Move in a Magnetic Field? y x Then So we can write 14 y x Let’s break this into components: Goal: Solve the Equations of Motion for a Charged Particle In A Magnetic Field The ‘dot’ represents Particles move freely along the field line Matching components: 15 Take Another Time Derivative & Substitute to Obtain Differential Equations For Each Spatial Coordinate Rewriting, we get These may remind you of the equations for a simple harmonic oscillator 16 Solve the Differential Equations These differential equations can be solved using sines and cosines: the magnitude of the initial velocity perpendicular to B an arbitrary phase to match the initial velocity conditions account for positive or negative q 17 Integrating, we obtain Larmor radius Cyclotron frequency • Charged particles undergo circular orbits about a guiding center The Result: Circular Motion About A Guiding Center 18 Let’s take and Gyromotion of a Charged Particle In A Magnetic Field y x For a positively charged particle: 1. At , 19 Let’s take and Gyromotion of a Charged Particle In A Magnetic Field For a positively charged particle: 1. At , 2. At , y x 20 Let’s take and Gyromotion of a Charged Particle In A Magnetic Field For a positively charged particle: 1. At , 2. At , y x 21 Let’s take and y x Gyromotion of a Charged Particle In A Magnetic Field For a negatively charged particle: 1. At , - 22 Let’s take and y x Gyromotion of a Charged Particle In A Magnetic Field For a negatively charged particle: 1. At , 2. At , - 23 Let’s take and y x Gyromotion of a Charged Particle In A Magnetic Field For a negatively charged particle: 1. At , 2. At , - 24 y x Gyromotion of Ions vs. Electrons • Ions generally have a much larger Larmor radius than electrons • The direction of gyromotion depends on the sign of the charge - In ITER, a typical deuterium ion with Ti=10 keV and B=5 Tesla would have An electron with Te=10 keV and B=5 Tesla has (60 times smaller) 25 Magnetic Confinement Devices Should Be Much Larger Than the Larmor Radius • Particles are confined perpendicular to the applied magnetic field Single Magnetic Field Line Image credit: • Tokamak approach: parallel confinement is achieved through toroidal geometry 26 Magnetic Mirrors Result: gyromotion + mirror force in the direction The Br ends up causing additional acceleration in the z direction: The magnetic moment is mirror force 27 Magnetic Moment Is Conserved The magnetic moment is a constant of motion s is the coordinate along the field line We can write this is Then We also have conservation of energy: this is 28 1. As the particle moves to stronger B, must increase. 2. Since energy is conserved, must decrease. 3. If B is strong enough, and the particle is reflected. More Insight Into Magnetic Mirrors The particle is reflected when 29 Magnetic Mirror Confinement In Action Charged particles can be trapped by Earth’s magnetic field Multicusp Confinement Devices Early Fusion Experiments Ex: Tandem Mirror Experiment (LLNL,1980’s) and other variants (Polywell devices) Particles with enough v\|\| can still escape 30 Time-varying Electric and Magnetic Fields Can Be Used To Accelerate & Heat Particles • A high frequency electro-magnetic field can be used to accelerate electrons or ions. y x - 31 y x - • A high frequency electro-magnetic field can be used to accelerate electrons or ions. • Particle gains energy as the applied electric field component oscillates at the cyclotron frequency (“in-phase” with the gyro-orbit). Cyclotron frequency Time-varying Electric and Magnetic Fields Can Be Used To Accelerate & Heat Particles The fluctuating accelerates the electron in the +x direction 32 y x - • A high frequency electro-magnetic field can be used to accelerate electrons or ions. • Particle gains energy as the applied electric field component oscillates at the cyclotron frequency (“in-phase” with the gyro-orbit). Cyclotron frequency Time-varying Electric and Magnetic Fields Can Be Used To Accelerate & Heat Particles Now accelerates the electron in the -x direction 33 y x - • A high frequency electro-magnetic field can be used to accelerate electrons or ions. • Particle gains energy as the applied electric field component oscillates at the cyclotron frequency (“in-phase” with the gyro-orbit). Cyclotron frequency Time-varying Electric and Magnetic Fields Can Be Used To Accelerate & Heat Particles 34 y x - • A high frequency electro-magnetic field can be used to accelerate electrons or ions. • Particle gains energy as the applied electric field component oscillates at the cyclotron frequency (“in-phase” with the gyro-orbit). Ex: For an electron, what B corresponds to 2.45 Ghz (microwave oven frequency)? The Cyclotron Frequency is Important for Cyclotron Resonance Heating Cyclotron frequency 35 Electron Cyclotron Heating In A Plasma Experiment at UW-Madison (2.45 GHz, B=875 Gauss) Example of Cyclotron Heating in Action In the DIII-D tokamak, use 110 GHz second harmonic heating (B~2 Tesla) 36 Other Practical Applications: EM Emission from Charged Particle Acceleration • Electron cyclotron emission (measure Te profiles) Produced by acceleration of gyrating charged particle EM radiation emitted at discrete frequencies: - B Detected radiated power is proportional to Te: - • Bremsstrahlung emission Produced by deceleration of deflected charged particle Radiated power depends on ne, Te, charge state Zeff (can be used to measure Zeff) 37 y x Next Simplest Case to Analyze: Constant, Uniform Electric Field Perpendicular to Magnetic Field 38 y x Accelerates due to E Faster velocity increases vxB Next Simplest Case to Analyze: Constant, Uniform Electric Field Perpendicular to Magnetic Field 39 y x Faster velocity increases vxB Decelerates Next Simplest Case to Analyze: Constant, Uniform Electric Field Perpendicular to Magnetic Field Accelerates due to E 40 y x Next Simplest Case to Analyze: Constant, Uniform Electric Field Perpendicular to Magnetic Field Faster velocity increases vxB Decelerates Accelerates due to E 41 Next Simplest Case to Analyze: Constant, Uniform Electric Field Perpendicular to Magnetic Field y x Ion guiding center drifts in the direction 42 Guiding Center Drift Due to E x B y x - The ExB drift can be written more generally as • ExB drift is independent of charge and mass • Both electrons and ions move together Electron guiding center also drifts in the direction 43 Other Forces Can Cause Guiding Center Drift • Any force perpendicular to B can cause particles to drift Drift due to force: Examples of forces: gravity centrifugal Rc • Bend the magnetic field into a donut shape • No end losses because the field lines go around and close on themselves • BUT a particle following a toroidal magnetic field would experience Fcf 44 A particle moving along a curved field line will drift up or down, depending on the sign of the charge The outward centrifugal force causes curvature drift z Btoroidal Rc Curvature Drift Due to Bending Field Lines 45 Spatially Varying Magnetic Field Strength Also Causes Drift 46 • The gyro-radius will be larger where the field is weaker and smaller where the field is stronger Spatially Varying Magnetic Field Strength Also Causes Drift 47 - • The gyro-radius will be larger where the field is weaker and smaller where the field is stronger • The resulting drift velocity is described by: Spatially Varying Magnetic Field Strength Also Causes Drift 48 What Happens To Charged Particles In A Purely Toroidal Magnetic Field? Ion drift + + + + Electron drift - - - - Btoroidal • Charged particles in a curved magnetic field will experience both ∇B and curvature drift: these effects add z ɸ 49 Charged Particles Will Drift Outward Ion drift Electron drift E + + + - - - + - Btoroidal • This means that no matter what, particles in a torus with a purely toroidal field will drift radially out and hit the walls. • Charged particles in a curved magnetic field will experience both ∇B and curvature drift z ɸ 50 Tokamak Solution: Add Poloidal Magnetic Field Iplasma Bpoloidal Toroidal: long way around Poloidal: short way around 1. Use external coils to apply a toroidal magnetic field 2. Drive toroidal current in the plasma to generate a poloidal magnetic field Btoroidal z ɸ • The resulting helical magnetic field is much better at confining charged particles. • The challenge: how to drive current in plasma in steady state while keeping the plasma stable and free of disruptions? 51 z ɸ R There Are Two Main Classes of Particle Orbits In Tokamaks Passing Bp Bt Particles with sufficient v\|\|will follow the helical magnetic field around the torus z ɸ R Trapped Bp Bt Particles with lower v\|\| are reflected as they encounter stronger B and therefore execute “banana” orbits as they precess around the torus B 52 Image credit: euro-fusion.org Banana Orbits Particles that don’t have enough v\|\| are reflected by the mirror force at the high field side of the tokamak Trapped particles won’t hit the wall if the banana orbit width Δr is small enough 53 Image credit: Pace et. al.,Physics Today (2015) Classifying Particle Orbits In Tokamaks Is Important for Understanding Basic Physics Mechanisms Like Wave-Particle Interactions 54 • Occurs when the wave and particle orbit phases match after many cycles: ω = lωci + pωθ + nωζ Time to complete poloidal orbit Time to complete cyclotron orbit Time to complete toroidal orbit Particle Resonance With Alfvén Eigenmode (AE) Instabilities [W.W. Heidbrink, Phys. Plasmas 15 (2008)] AE mode frequency 55 • Occurs when the wave and particle orbit phases match after many cycles: Time to complete poloidal orbit ω = lωci + pωθ + nωζ Time to complete cyclotron orbit Time to complete toroidal orbit Particle Resonance With Alfvén Eigenmode (AE) Instabilities [W.W. Heidbrink, Phys. Plasmas 15 (2008)] AE mode frequency x Ion Trajectory t = 4.32 μs t = 2.16 μs t = 0.00 μ z R φ z R φ • Power transfer can occur as the ion stays in phase with the wave as it traverses the mode [Pace, Physics Today (Oct. 2015)] 56 -2.0 -1.5 -1.0 -0.5 0.0 0.5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 P / φ ψwall μB/E Magnetic Moment Toroidal Canonical Angular Momentum STAGNATION POTATO COUNTER-PASSING CO-PASSING LOST TRAPPED Projection of 80 keV D+ orbit in the DIII-D tokamak [R. B. White, Theory of toroidally confined plasmas, Imperial College Press (2001)] The Fast Ion Distribution Function Is “Most Simply” Described Using Constants of Motion 57 -2.0 -1.5 -1.0 -0.5 0.0 0.5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 P / φ ψwall μB/E Magnetic Moment Toroidal Canonical Angular Momentum STAGNATION POTATO COUNTER-PASSING CO-PASSING LOST TRAPPED Projection of 80 keV D+ orbit in the DIII-D tokamak [R. B. White, Theory of toroidally confined plasmas, Imperial College Press (2001)] Small Changes in Particle Energy Can Cause Large Changes in Orbit Topology 58 • Example of distribution function F(E,µ,Pϕ) for neutral beam injection, which is anisotropic and non-Maxwellian -2.0 -1.5 -1.0 -0.5 0.0 0.5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 P / φ ψwall μB/E Magnetic Moment Toroidal Canonical Angular Momentum PASSING TRAPPED LOST neutral beam injection [R. B. White, Theory of toroidally confined plasmas, Imperial College Press (2001)] There Can Be Several Different Populations of Fast Ions In Fusion Devices • The distribution function for fusion products (alpha particles) is isotropic 59 -2.0 -1.5 -1.0 -0.5 0.0 0.5 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 P / φ ψwall μB/E Magnetic Moment Toroidal Canonical Angular Momentum n=4, f=79.1010 PASSING TRAPPED LOST diagnostic Beam Deposition Resonance (Many Modes) Active Research: Calculating the Energetic Particle Distribution Function After Transport by Instabilities • Transport can occur if fast ions intersect AE resonances in this “phase space” plot • In certain conditions, AEs can cause significant transport of fast ions and significantly degrade fusion performance à We are working on controlling/avoiding AEs 60 • Charged particles undergo gyromotion about magnetic fields, and are free to move along the magnetic field line. • Depending on magnetic field geometry or the presence of other forces like electric fields, particles can drift across field lines (and even leave the systemàhit the walls). • (One) challenge for fusion energy research is to confine enough charged particles that are energetic enough for long enough lengths of time to achieve sustained fusion. Conclusions
189173	https://en.wikipedia.org/wiki/Centrifugal_force	Centrifugal force - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 History 2 Introduction 3 ExamplesToggle Examples subsection 3.1 Vehicle driving round a curve 3.2 Stone on a string 3.3 Earth 3.3.1 Weight of an object at the poles and on the equator 4 FormulationToggle Formulation subsection 4.1 Time derivatives in a rotating frame 4.2 Acceleration 4.3 Force 4.4 Potential 5 Absolute rotation 6 Applications 7 Other uses of the termToggle Other uses of the term subsection 7.1 In Lagrangian mechanics 7.2 As a reactive force 8 See also 9 Notes 10 References 11 External links [x] Toggle the table of contents Centrifugal force [x] 72 languages Afrikaans العربية Asturianu Azərbaycanca বাংলা Беларуская Български Català Чӑвашла Čeština ChiShona Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Íslenska Italiano עברית ქართული Қазақша Kiswahili Kreyòl ayisyen ລາວ Latina Latviešu Lietuvių Magyar Македонски മലയാളം Bahasa Melayu မြန်မာဘာသာ Nederlands 日本語 Norsk bokmål Norsk nynorsk Occitan Polski Português Română Русский Sicilianu Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska தமிழ் Татарча / tatarça తెలుగు ไทย Türkçe Українська اردو Tiếng Việt 吴语粵語中文 ရခိုင် Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Type of inertial force For the effect in politics, see center squeeze. Not to be confused with Centripetal force. For the real frame-independent Newtonian force that exists as a reaction to a centripetal force in some scenarios, see Reactive centrifugal force. Riders on a swing carousel interpret the cessation of upward motion as a balancing of the force of gravity, the force of the tension of the chains, and a centrifugal force pushing them away from the center of rotation. A stationary observer on the ground observes uniform circular motion, which requires a net centripetal force that is the combination of the force of gravity and the force of the tension of the chains. Part of a series on Classical mechanics F=d p d t{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}} Second law of motion History Timeline Textbooks Branches Applied Celestial Continuum Dynamics Field theory Kinematics Kinetics Statics Statistical mechanics Fundamentals Acceleration Angular momentum Couple D'Alembert's principle Energy kinetic potential Force Frame of reference Inertial frame of reference Impulse Inertia/ Moment of inertia Mass Mechanical power Mechanical work Moment Momentum Space Speed Time Torque Velocity Virtual work Formulations Newton's laws of motion Analytical mechanics Lagrangian mechanics Hamiltonian mechanics Routhian mechanics Hamilton–Jacobi equation Appell's equation of motion Koopman–von Neumann mechanics Core topics Damping Displacement Equations of motion Euler's laws of motion Fictitious force Friction Harmonic oscillator Inertial/ Non-inertial reference frame Motion(linear) Newton's law of universal gravitation Newton's laws of motion Relative velocity Rigid body dynamics Euler's equations Simple harmonic motion Vibration Rotation Circular motion Rotating reference frame Centripetal force Centrifugal force reactive Coriolis force Pendulum Tangential speed Rotational frequency Angular acceleration/ displacement/ frequency/ velocity Scientists Kepler Galileo Huygens Newton Horrocks Halley Maupertuis Daniel Bernoulli Johann Bernoulli Euler d'Alembert Clairaut Lagrange Laplace Poisson Hamilton Jacobi Cauchy Routh Liouville Appell Gibbs Koopman von Neumann Physics portal Category v t e In Newtonian mechanics, a centrifugal force is a kind of fictitious force (or inertial force) that appears to act on all objects when viewed in a rotating frame of reference. It appears to be directed perpendicularly from the axis of rotation of the frame. The magnitude of the centrifugal force F on an object of massm at the perpendicular distanceρ from the axis of a rotating frame of reference with angular velocityω is F=m ω 2 ρ{\textstyle F=m\omega ^{2}\rho }. The concept of centrifugal force simplifies the analysis of rotating devices by adopting a co-rotating frame of reference, such as in centrifuges, centrifugal pumps, centrifugal governors, and centrifugal clutches, and in centrifugal railways, planetary orbits and banked curves. The same centrifugal effect observed on rotating devices can be analyzed in an inertial reference frame as a consequence of inertia and the physical forces without invoking a centrifugal force. History [edit] Main article: History of centrifugal and centripetal forces From 1659, the Neo-Latin term vi centrifuga ("centrifugal force") is attested in Christiaan Huygens' notes and letters. Note, that in Latin centrum means "center" and ‑fugus (from fugiō) means "fleeing, avoiding". Thus, centrifugus means "fleeing from the center" in a literal translation. In 1673, in Horologium Oscillatorium, Huygens writes (as translated by Richard J. Blackwell): There is another kind of oscillation in addition to the one we have examined up to this point; namely, a motion in which a suspended weight is moved around through the circumference of a circle. From this we were led to the construction of another clock at about the same time we invented the first one. [...] I originally intended to publish here a lengthy description of these clocks, along with matters pertaining to circular motion and centrifugal force[a], as it might be called, a subject about which I have more to say than I am able to do at present. But, in order that those interested in these things can sooner enjoy these new and not useless speculations, and in order that their publication not be prevented by some accident, I have decided, contrary to my plan, to add this fifth part [...]. The same year, Isaac Newton received Huygens's work via Henry Oldenburg and replied "I pray you return [Mr. Huygens] my humble thanks [...] I am glad we can expect another discourse of the vis centrifuga, which speculation may prove of good use in natural philosophy and astronomy, as well as mechanics". In 1687, in Principia, Newton further develops vis centrifuga ("centrifugal force"). Around this time, the concept is also further evolved by Newton, Gottfried Wilhelm Leibniz, and Robert Hooke. In the late 18th century, the modern conception of the centrifugal force evolved as a "fictitious force" arising in a rotating reference.[citation needed] Centrifugal force has also played a role in debates in classical mechanics about detection of absolute motion. Newton suggested two arguments to answer the question of whether absolute rotation can be detected: the rotating bucket argument, and the rotating spheres argument. According to Newton, in each scenario the centrifugal force would be observed in the object's local frame (the frame where the object is stationary) only if the frame was rotating with respect to absolute space. Around 1883, Mach's principle was proposed where, instead of absolute rotation, the motion of the distant stars relative to the local inertial frame gives rise through some (hypothetical) physical law to the centrifugal force and other inertia effects. The modern view is based upon the idea of an inertial frame of reference, which privileges observers for which the laws of physics take on their simplest form, and in particular, frames that do not use centrifugal forces in their equations of motion in order to describe motions correctly. Around 1914, the analogy between centrifugal force (sometimes used to create artificial gravity) and gravitational forces led to the equivalence principle of general relativity. Introduction [edit] In the inertial frame of reference (upper part of the picture), the black ball moves in a straight line. However, the observer (brown dot) who is standing in the rotating/non-inertial frame of reference (lower part of the picture) sees the object as following a curved path due to the Coriolis and centrifugal forces present in this frame. Centrifugal force is an outward force apparent in a rotating reference frame. It does not exist when a system is described relative to an inertial frame of reference. All measurements of position and velocity must be made relative to some frame of reference. For example, an analysis of the motion of an object in an airliner in flight could be made relative to the airliner, to the surface of the Earth, or even to the Sun. A reference frame that is at rest (or one that moves with no rotation and at constant velocity) relative to the "fixed stars" is generally taken to be an inertial frame. Any system can be analyzed in an inertial frame (and so with no centrifugal force). However, it is often more convenient to describe a rotating system by using a rotating frame—the calculations are simpler, and descriptions more intuitive. When this choice is made, fictitious forces, including the centrifugal force, arise. In a reference frame rotating about an axis through its origin, all objects, regardless of their state of motion, appear to be under the influence of a radially (from the axis of rotation) outward force that is proportional to their mass, to the distance from the axis of rotation of the frame, and to the square of the angular velocity of the frame. This is the centrifugal force. As humans usually experience centrifugal force from within the rotating reference frame, e.g. on a merry-go-round or vehicle, this is much more well-known than centripetal force. Motion relative to a rotating frame results in another fictitious force: the Coriolis force. If the rate of rotation of the frame changes, a third fictitious force (the Euler force) is required. These fictitious forces are necessary for the formulation of correct equations of motion in a rotating reference frame and allow Newton's laws to be used in their normal form in such a frame (with one exception: the fictitious forces do not obey Newton's third law: they have no equal and opposite counterparts). Newton's third law requires the counterparts to exist within the same frame of reference, hence centrifugal and centripetal force, which do not, are not action and reaction (as is sometimes erroneously contended). Examples [edit] Vehicle driving round a curve [edit] A common experience that gives rise to the idea of a centrifugal force is encountered by passengers riding in a vehicle, such as a car, that is changing direction. If a car is traveling at a constant speed along a straight road, then a passenger inside is not accelerating and, according to Newton's second law of motion, the net force acting on them is therefore zero (all forces acting on them cancel each other out). If the car enters a curve that bends to the left, the passenger experiences an apparent force that seems to be pulling them towards the right. This is the fictitious centrifugal force. It is needed within the passengers' local frame of reference to explain their sudden tendency to start accelerating to the right relative to the car, a tendency which they must resist by applying a rightward force to the car (for instance, a frictional force against the seat) in order to remain in a fixed position inside. Since they push the seat toward the right, Newton's third law says that the seat pushes them towards the left. The centrifugal force must be included in the passenger's reference frame (in which the passenger remains at rest): it counteracts the leftward force applied to the passenger by the seat, and explains why this otherwise unbalanced force does not cause them to accelerate. However, it would be apparent to a stationary observer watching from an overpass above that the frictional force exerted on the passenger by the seat is not being balanced; it constitutes a net force to the left, causing the passenger to accelerate toward the inside of the curve, as they must in order to keep moving with the car rather than proceeding in a straight line as they otherwise would. Thus the "centrifugal force" they feel is the result of a "centrifugal tendency" caused by inertia. Similar effects are encountered in aeroplanes and roller coasters where the magnitude of the apparent force is often reported in "G's". Stone on a string [edit] If a stone is whirled round on a string, in a horizontal plane, the only real force acting on the stone in the horizontal plane is applied by the string (gravity acts vertically). There is a net force on the stone in the horizontal plane which acts toward the center. In an inertial frame of reference, were it not for this net force acting on the stone, the stone would travel in a straight line, according to Newton's first law of motion. In order to keep the stone moving in a circular path, a centripetal force, in this case provided by the string, must be continuously applied to the stone. As soon as it is removed (for example if the string breaks) the stone moves in a straight line, as viewed from above. In this inertial frame, the concept of centrifugal force is not required as all motion can be properly described using only real forces and Newton's laws of motion. In a frame of reference rotating with the stone around the same axis as the stone, the stone is stationary. However, the force applied by the string is still acting on the stone. If one were to apply Newton's laws in their usual (inertial frame) form, one would conclude that the stone should accelerate in the direction of the net applied force—towards the axis of rotation—which it does not do. The centrifugal force and other fictitious forces must be included along with the real forces in order to apply Newton's laws of motion in the rotating frame. Earth [edit] The Earth constitutes a rotating reference frame because it rotates once every 23 hours and 56 minutes around its axis. Because the rotation is slow, the fictitious forces it produces are often small, and in everyday situations can generally be neglected. Even in calculations requiring high precision, the centrifugal force is generally not explicitly included, but rather lumped in with the gravitational force: the strength and direction of the local "gravity" at any point on the Earth's surface is actually a combination of gravitational and centrifugal forces. However, the fictitious forces can be of arbitrary size. For example, in an Earth-bound reference system (where the earth is represented as stationary), the fictitious force (the net of Coriolis and centrifugal forces) is enormous and is responsible for the Sun orbiting around the Earth. This is due to the large mass and velocity of the Sun (relative to the Earth). Weight of an object at the poles and on the equator [edit] If an object is weighed with a simple spring balance at one of the Earth's poles, there are two forces acting on the object: the Earth's gravity, which acts in a downward direction, and the equal and opposite restoring force in the spring, acting upward. Since the object is stationary and not accelerating, there is no net force acting on the object and the force from the spring is equal in magnitude to the force of gravity on the object. In this case, the balance shows the value of the force of gravity on the object. When the same object is weighed on the equator, the same two real forces act upon the object. However, the object is moving in a circular path as the Earth rotates and therefore experiencing a centripetal acceleration. When considered in an inertial frame (that is to say, one that is not rotating with the Earth), the non-zero acceleration means that force of gravity will not balance with the force from the spring. In order to have a net centripetal force, the magnitude of the restoring force of the spring must be less than the magnitude of force of gravity. This reduced restoring force in the spring is reflected on the scale as less weight — about 0.3% less at the equator than at the poles. In the Earth reference frame (in which the object being weighed is at rest), the object does not appear to be accelerating; however, the two real forces, gravity and the force from the spring, are the same magnitude and do not balance. The centrifugal force must be included to make the sum of the forces be zero to match the apparent lack of acceleration. Note:In fact, the observed weight difference is more — about 0.53%. Earth's gravity is a bit stronger at the poles than at the equator, because the Earth is not a perfect sphere, so an object at the poles is slightly closer to the center of the Earth than one at the equator; this effect combines with the centrifugal force to produce the observed weight difference. Formulation [edit] Main article: Rotating reference frame See also: Fictitious force For the following formalism, the rotating frame of reference is regarded as a special case of a non-inertial reference frame that is rotating relative to an inertial reference frame denoted the stationary frame. Time derivatives in a rotating frame [edit] In a rotating frame of reference, the time derivatives of any vector function P of time—such as the velocity and acceleration vectors of an object—will differ from its time derivatives in the stationary frame. If P 1, P 2, P 3 are the components of P with respect to unit vectors i, j, k directed along the axes of the rotating frame (i.e. P = P 1 i + P 2 j +P 3 k), then the first time derivative [⁠d P/d t⁠] of P with respect to the rotating frame is, by definition, ⁠d P 1/d t⁠i + ⁠d P 2/d t⁠j + ⁠d P 3/d t⁠k. If the absolute angular velocity of the rotating frame is ω then the derivative ⁠d P/d t⁠ of P with respect to the stationary frame is related to [⁠d P/d t⁠] by the equation: d P d t=[d P d t]+ω×P,{\displaystyle {\frac {\mathrm {d} {\boldsymbol {P}}}{\mathrm {d} t}}=\left[{\frac {\mathrm {d} {\boldsymbol {P}}}{\mathrm {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {P}}\ ,} where × denotes the vector cross product. In other words, the rate of change of P in the stationary frame is the sum of its apparent rate of change in the rotating frame and a rate of rotation ω × P attributable to the motion of the rotating frame. The vector ω has magnitude ω equal to the rate of rotation and is directed along the axis of rotation according to the right-hand rule. Acceleration [edit] Newton's law of motion for a particle of mass m written in vector form is: F=m a,{\displaystyle {\boldsymbol {F}}=m{\boldsymbol {a}}\ ,} where F is the vector sum of the physical forces applied to the particle and a is the absolute acceleration (that is, acceleration in an inertial frame) of the particle, given by: a=d 2 r d t 2,{\displaystyle {\boldsymbol {a}}={\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\ ,} where r is the position vector of the particle (not to be confused with radius, as used above.) By applying the transformation above from the stationary to the rotating frame three times (twice to d r d t{\textstyle {\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}} and once to d d t[d r d t]{\textstyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]}), the absolute acceleration of the particle can be written as: a=d 2 r d t 2=d d t d r d t=d d t([d r d t]+ω×r)=[d 2 r d t 2]+ω×[d r d t]+d ω d t×r+ω×d r d t=[d 2 r d t 2]+ω×[d r d t]+d ω d t×r+ω×([d r d t]+ω×r)=[d 2 r d t 2]+d ω d t×r+2 ω×[d r d t]+ω×(ω×r).{\displaystyle {\begin{aligned}{\boldsymbol {a}}&={\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}={\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}={\frac {\mathrm {d} }{\mathrm {d} t}}\left(\left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\&=\left[{\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]+{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times {\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\&=\left[{\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]+{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times \left(\left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\&=\left[{\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\right]+{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}\times {\boldsymbol {r}}+2{\boldsymbol {\omega }}\times \left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]+{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})\ .\end{aligned}}} Force [edit] The apparent acceleration in the rotating frame is [d 2 r d t 2]{\displaystyle \left[{\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\right]}. An observer unaware of the rotation would expect this to be zero in the absence of outside forces. However, Newton's laws of motion apply only in the inertial frame and describe dynamics in terms of the absolute acceleration d 2 r d t 2{\displaystyle {\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}}. Therefore, the observer perceives the extra terms as contributions due to fictitious forces. These terms in the apparent acceleration are independent of mass; so it appears that each of these fictitious forces, like gravity, pulls on an object in proportion to its mass. When these forces are added, the equation of motion has the form: F+(−m d ω d t×r)⏟Euler+(−2 m ω×[d r d t])⏟Coriolis+(−m ω×(ω×r))⏟centrifugal=m[d 2 r d t 2].{\displaystyle {\boldsymbol {F}}+\underbrace {\left(-m{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}\times {\boldsymbol {r}}\right)} {\text{Euler}}+\underbrace {\left(-2m{\boldsymbol {\omega }}\times \left[{\frac {\mathrm {d} {\boldsymbol {r}}}{\mathrm {d} t}}\right]\right)} {\text{Coriolis}}+\underbrace {\left(-m{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})\right)} _{\text{centrifugal}}=m\left[{\frac {\mathrm {d} ^{2}{\boldsymbol {r}}}{\mathrm {d} t^{2}}}\right]\ .} From the perspective of the rotating frame, the additional force terms are experienced just like the real external forces and contribute to the apparent acceleration. The additional terms on the force side of the equation can be recognized as, reading from left to right, the Euler force−m d ω/d t×r{\displaystyle -m\mathrm {d} {\boldsymbol {\omega }}/\mathrm {d} t\times {\boldsymbol {r}}}, the Coriolis force−2 m ω×[d r/d t]{\displaystyle -2m{\boldsymbol {\omega }}\times \left[\mathrm {d} {\boldsymbol {r}}/\mathrm {d} t\right]}, and the centrifugal force −m ω×(ω×r){\displaystyle -m{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})}, respectively. Unlike the other two fictitious forces, the centrifugal force always points radially outward from the axis of rotation of the rotating frame, with magnitude m ω 2 r⊥{\displaystyle m\omega ^{2}r_{\perp }}, where r⊥{\displaystyle r_{\perp }} is the component of the position vector perpendicular to ω{\displaystyle {\boldsymbol {\omega }}}, and unlike the Coriolis force in particular, it is independent of the motion of the particle in the rotating frame. As expected, for a non-rotating inertial frame of reference (ω=0){\displaystyle ({\boldsymbol {\omega }}=0)} the centrifugal force and all other fictitious forces disappear. Similarly, as the centrifugal force is proportional to the distance from object to the axis of rotation of the frame, the centrifugal force vanishes for objects that lie upon the axis. Potential [edit] The centrifugal force per unit mass can also be derived as the gradient of a centrifugal potential. For example, the centrifugal potential at the perpendicular distance ρ from the axis of a rotating frame of reference with angular velocity ω is 0.5 ω 2 ρ 2{\textstyle 0.5\omega ^{2}\rho ^{2}} (see also: Geopotential#Centrifugal potential.) Absolute rotation [edit] Main article: Absolute rotation The interface of two immiscible liquids rotating around a vertical axis is an upward-opening circular paraboloid. When analysed in a rotating reference frame of the planet, centrifugal force causes rotating planets to assume the shape of an oblate spheroid; when analysed in an inertial frame, planetary flattening is a result of inertia and real forces. Three scenarios were suggested by Newton to answer the question of whether the absolute rotation of a local frame can be detected; that is, if an observer can decide whether an observed object is rotating or if the observer is rotating. The shape of the surface of water rotating in a bucket. The shape of the surface becomes concave to balance the centrifugal force against the other forces upon the liquid. The tension in a string joining two spheres rotating about their center of mass. The tension in the string will be proportional to the centrifugal force on each sphere as it rotates around the common center of mass. In these scenarios, the effects attributed to centrifugal force are only observed in the local frame (the frame in which the object is stationary) if the object is undergoing absolute rotation relative to an inertial frame. By contrast, in an inertial frame, the observed effects arise as a consequence of the inertia and the known forces without the need to introduce a centrifugal force. Based on this argument, the privileged frame, wherein the laws of physics take on the simplest form, is a stationary frame in which no fictitious forces need to be invoked. Within this view of physics, any other phenomenon that is usually attributed to centrifugal force can be used to identify absolute rotation. For example, the oblateness of a sphere of freely flowing material is often explained in terms of centrifugal force. The oblate spheroid shape reflects, following Clairaut's theorem, the balance between containment by gravitational attraction and dispersal by centrifugal force. That the Earth is itself an oblate spheroid, bulging at the equator where the radial distance and hence the centrifugal force is larger, is taken as one of the evidences for its absolute rotation. Applications [edit] The operations of numerous common rotating mechanical systems are most easily conceptualized in terms of centrifugal force. For example: A centrifugal governor regulates the speed of an engine by using spinning masses that move radially, adjusting the throttle, as the engine changes speed. In the reference frame of the spinning masses, centrifugal force causes the radial movement. A centrifugal clutch is used in small engine-powered devices such as chain saws, go-karts and model helicopters. It allows the engine to start and idle without driving the device but automatically and smoothly engages the drive as the engine speed rises. Inertial drum brake ascenders used in rock climbing and the inertia reels used in many automobile seat belts operate on the same principle. Centrifugal forces can be used to generate artificial gravity, as in proposed designs for rotating space stations. The Mars Gravity Biosatellite would have studied the effects of Mars-level gravity on mice with gravity simulated in this way. Spin casting and centrifugal casting are production methods that use centrifugal force to disperse liquid metal or plastic throughout the negative space of a mold. Centrifuges are used in science and industry to separate substances. In the reference frame spinning with the centrifuge, the centrifugal force induces a hydrostatic pressure gradient in fluid-filled tubes oriented perpendicular to the axis of rotation, giving rise to large buoyant forces which push low-density particles inward. Elements or particles denser than the fluid move outward under the influence of the centrifugal force. This is effectively Archimedes' principle as generated by centrifugal force as opposed to being generated by gravity. Some amusement rides make use of centrifugal forces. For instance, a Gravitron's spin forces riders against a wall and allows riders to be elevated above the machine's floor in apparent defiance of Earth's gravity. Nevertheless, all of these systems can also be described without requiring the concept of centrifugal force, in terms of motions and forces in a stationary frame, at the cost of taking somewhat more care in the consideration of forces and motions within the system. Other uses of the term [edit] While the majority of the scientific literature uses the term centrifugal force to refer to the particular fictitious force that arises in rotating frames, there are a few limited instances in the literature of the term applied to other distinct physical concepts. In Lagrangian mechanics [edit] One of these instances occurs in Lagrangian mechanics. Lagrangian mechanics formulates mechanics in terms of generalized coordinates {q k}, which can be as simple as the usual polar coordinates (r,θ){\displaystyle (r,\ \theta )} or a much more extensive list of variables. Within this formulation the motion is described in terms of generalized forces, using in place of Newton's laws the Euler–Lagrange equations. Among the generalized forces, those involving the square of the time derivatives {(d q k ⁄ d t )2} are sometimes called centrifugal forces. In the case of motion in a central potential the Lagrangian centrifugal force has the same form as the fictitious centrifugal force derived in a co-rotating frame. However, the Lagrangian use of "centrifugal force" in other, more general cases has only a limited connection to the Newtonian definition. As a reactive force [edit] In another instance the term refers to the reactionforce to a centripetal force, or reactive centrifugal force. A body undergoing curved motion, such as circular motion, is accelerating toward a center at any particular point in time. This centripetal acceleration is provided by a centripetal force, which is exerted on the body in curved motion by some other body. In accordance with Newton's third law of motion, the body in curved motion exerts an equal and opposite force on the other body. This reactive force is exerted by the body in curved motion on the other body that provides the centripetal force and its direction is from that other body toward the body in curved motion. This reaction force is sometimes described as a centrifugal inertial reaction, that is, a force that is centrifugally directed, which is a reactive force equal and opposite to the centripetal force that is curving the path of the mass. The concept of the reactive centrifugal force is sometimes used in mechanics and engineering. It is sometimes referred to as just centrifugal force rather than as reactive centrifugal force although this usage is deprecated in elementary mechanics. See also [edit] Physics portal Engineering portal Balancing of rotating masses Centrifugal mechanism of acceleration Equivalence principle Folk physics Lagrangian point Lamm equation Notes [edit] ^In Latin: vim centrifugam. References [edit] ^Gülich, Johann Friedrich (2010-05-25). Centrifugal Pumps. Springer Science & Business Media. ISBN978-3-642-12824-0. Retrieved 2025-09-11. ^ abYoder, Joella (1991). "Christiaan Huygens' Great Treasure"(PDF). Tractrix. 3: 1–13. Archived(PDF) from the original on 13 April 2018. Retrieved 12 April 2018. ^Yoder, Joella (17 May 2013). A Catalogue of the Manuscripts of Christiaan Huygens including a concordance with his Oeuvres Complètes. BRILL. ISBN9789004235656. Archived from the original on 16 March 2020. Retrieved 12 April 2018. ^Blackwell, Richard J. (1986). Christiaan Huygens' the pendulum clock, or, Geometrical demonstrations concerning the motion of pendula as applied to clocks. Ames: Iowa State University Press. p.173. ISBN978-0-8138-0933-5. ^Œuvres complètes de Christiaan Huygens (in French). Vol.7. The Hague: M. Nijhoff. 1897. p.325. Archived from the original on 2023-11-06. Retrieved 2023-01-14. ^An English translation is found at Isaac Newton (1934). Philosophiae naturalis principia mathematica (Andrew Motte translation of 1729, revised by Florian Cajori ed.). University of California Press. pp.10–12. ISBN9780520009271.{{cite book}}: ISBN / Date incompatibility (help) ^Julian B. Barbour; Herbert Pfister, eds. (1995). Mach's principle: from Newton's bucket to quantum gravity. Boston: Birkhäuser. p.69. ISBN0-8176-3823-7. OCLC32664808. ^Science education in the 21st century. Ingrid V. Eriksson. New York: Nova Science Publishers. 2008. ISBN978-1-60021-951-1. OCLC165958146.{{cite book}}: CS1 maint: others (link) ^Richard T. Weidner and Robert L. Sells (1973). Mechanics, mechanical waves, kinetic theory, thermodynamics (2 ed.). Allyn and Bacon. p.123. ^Restuccia, S.; Toroš, M.; Gibson, G. M.; Ulbricht, H.; Faccio, D.; Padgett, M. J. (2019). "Photon Bunching in a Rotating Reference Frame". Physical Review Letters. 123 (11): 110401. arXiv:1906.03400. Bibcode:2019PhRvL.123k0401R. doi:10.1103/physrevlett.123.110401. PMID31573252. S2CID182952610.{{cite journal}}: CS1 maint: article number as page number (link) ^John Robert Taylor (2004). Classical Mechanics. Sausalito CA: University Science Books. Chapter 9, pp. 344 ff. ISBN978-1-891389-22-1. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Kobayashi, Yukio (2008). "Remarks on viewing situation in a rotating frame". European Journal of Physics. 29 (3): 599–606. Bibcode:2008EJPh...29..599K. doi:10.1088/0143-0807/29/3/019. S2CID120947179. ^David P. Stern (2006). "Frames of Reference: The Basics". From Stargazers to Starships. Goddard Space Flight Center Space Physics Data Facility. Archived from the original on 6 April 2020. Retrieved 20 April 2017. ^"Centrifuge". Encyclopædia Britannica. April 30, 2015. Archived from the original on October 7, 2024. Retrieved June 2, 2022. ^"The Feynman Lectures on Physics Vol. I Ch. 12: Characteristics of Force". Archived from the original on 2024-10-07. Retrieved 2022-05-07. ^ abAlexander L. Fetter; John Dirk Walecka (2003). Theoretical Mechanics of Particles and Continua. Courier Dover Publications. pp.38–39. ISBN978-0-486-43261-8. ^Jerrold E. Marsden; Tudor S. Ratiu (1999). Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems. Springer. p.251. ISBN978-0-387-98643-2. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^"Centrifugal force". Encyclopædia Britannica. 17 August 2016. Archived from the original on 21 April 2017. Retrieved 20 April 2017. ^Knight, Judson (2016). "Centripetal Force". In Schlager, Neil (ed.). Science of Everyday Things, Volume 2: Real-Life Physics. Thomson Learning. p.47. Retrieved 19 April 2017. ^"Curious About Astronomy?"Archived January 17, 2015, at the Wayback Machine, Cornell University, retrieved June 2007 ^Boynton, Richard (2001). "Precise Measurement of Mass"(PDF). Sawe Paper No. 3147. Arlington, Texas: S.A.W.E., Inc. Archived from the original(PDF) on 2007-02-27. Retrieved 2007-01-21. ^John L. Synge; Byron A. Griffith (2007). Principles of Mechanics (Reprint of Second Edition of 1942 ed.). Read Books. p.347. ISBN978-1-4067-4670-9. ^Taylor (2005). p. 342. ^LD Landau; LM Lifshitz (1976). Mechanics (Third ed.). Oxford: Butterworth-Heinemann. p.128. ISBN978-0-7506-2896-9. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Louis N. Hand; Janet D. Finch (1998). Analytical Mechanics. Cambridge University Press. p.267. ISBN978-0-521-57572-0. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Mark P Silverman (2002). A universe of atoms, an atom in the universe (2 ed.). Springer. p.249. ISBN978-0-387-95437-0. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Taylor (2005). p. 329. ^Cornelius Lanczos (1986). The Variational Principles of Mechanics (Reprint of Fourth Edition of 1970 ed.). Dover Publications. Chapter 4, §5. ISBN978-0-486-65067-8. ^Morton Tavel (2002). Contemporary Physics and the Limits of Knowledge. Rutgers University Press. p.93. ISBN978-0-8135-3077-2. Archived from the original on 2024-10-07. Retrieved 2020-11-09. Noninertial forces, like centrifugal and Coriolis forces, can be eliminated by jumping into a reference frame that moves with constant velocity, the frame that Newton called inertial. ^Louis N. Hand; Janet D. Finch (1998). Analytical Mechanics. Cambridge University Press. p.324. ISBN978-0-521-57572-0. ^I. Bernard Cohen; George Edwin Smith (2002). The Cambridge companion to Newton. Cambridge University Press. p.43. ISBN978-0-521-65696-2. ^Simon Newcomb (1878). Popular astronomy. Harper & Brothers. pp.86–88. ^Myers, Rusty L. (2006). The basics of physics. Greenwood Publishing Group. p.57. ISBN978-0-313-32857-2. ^For an introduction, see for example Cornelius Lanczos (1986). The variational principles of mechanics (Reprint of 1970 University of Toronto ed.). Dover. p.1. ISBN978-0-486-65067-8. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^For a description of generalized coordinates, see Ahmed A. Shabana (2003). "Generalized coordinates and kinematic constraints". Dynamics of Multibody Systems (2 ed.). Cambridge University Press. p.90 ff. ISBN978-0-521-54411-5. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Christian Ott (2008). Cartesian Impedance Control of Redundant and Flexible-Joint Robots. Springer. p.23. ISBN978-3-540-69253-9. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Shuzhi S. Ge; Tong Heng Lee; Christopher John Harris (1998). Adaptive Neural Network Control of Robotic Manipulators. World Scientific. pp.47–48. ISBN978-981-02-3452-2. In the above Euler–Lagrange equations, there are three types of terms. The first involves the second derivative of the generalized co-ordinates. The second is quadratic in q˙{\displaystyle {\boldsymbol {\dot {q}}}} where the coefficients may depend on q{\displaystyle {\boldsymbol {q}}}. These are further classified into two types. Terms involving a product of the type q˙i 2{\displaystyle {{\dot {q}}{i}}^{2}} are called _centrifugal forces while those involving a product of the type q˙i q˙j{\displaystyle {\dot {q}}{i}{\dot {q}}{j}} for i ≠ j are called Coriolis forces. The third type is functions of q{\displaystyle {\boldsymbol {q}}} only and are called gravitational forces. ^R. K. Mittal; I. J. Nagrath (2003). Robotics and Control. Tata McGraw-Hill. p.202. ISBN978-0-07-048293-7. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^T Yanao; K Takatsuka (2005). "Effects of an intrinsic metric of molecular internal space". In Mikito Toda; Tamiki Komatsuzaki; Stuart A. Rice; Tetsuro Konishi; R. Stephen Berry (eds.). Geometrical Structures Of Phase Space In Multi-dimensional Chaos: Applications to chemical reaction dynamics in complex systems. Wiley. p.98. ISBN978-0-471-71157-5. Archived from the original on 2024-10-07. Retrieved 2020-11-09. As is evident from the first terms ..., which are proportional to the square of ϕ˙{\displaystyle {\dot {\phi }}}, a kind of "centrifugal force" arises ... We call this force "democratic centrifugal force". Of course, DCF is different from the ordinary centrifugal force, and it arises even in a system of zero angular momentum. ^See p. 5 in Donato Bini; Paolo Carini; Robert T Jantzen (1997). "The intrinsic derivative and centrifugal forces in general relativity: I. Theoretical foundations". International Journal of Modern Physics D (Submitted manuscript). 6 (1): 143–198. arXiv:gr-qc/0106014v1. Bibcode:1997IJMPD...6..143B. doi:10.1142/S021827189700011X. S2CID10652293.. The companion paper is Donato Bini; Paolo Carini; Robert T Jantzen (1997). "The intrinsic derivative and centrifugal forces in general relativity: II. Applications to circular orbits in some stationary axisymmetric spacetimes". International Journal of Modern Physics D (Submitted manuscript). 6 (1): 143–198. arXiv:gr-qc/0106014v1. Bibcode:1997IJMPD...6..143B. doi:10.1142/S021827189700011X. S2CID10652293. Archived from the original on 2021-04-29. Retrieved 2023-06-21. ^Mook, Delo E.; Thomas Vargish (1987). Inside relativity. Princeton, N.J.: Princeton University Press. p.47. ISBN0-691-08472-6. OCLC16089285. Archived from the original on 2024-10-07. Retrieved 2016-03-11. ^G. David Scott (1957). "Centrifugal Forces and Newton's Laws of Motion". Vol.25. American Journal of Physics. p.325. ^Signell, Peter (2002). "Acceleration and force in circular motion"Archived 2024-10-07 at the Wayback MachinePhysnet. Michigan State University, "Acceleration and force in circular motion", §5b, p. 7. ^Mohanty, A. K. (1994). Fluid mechanics (2nd ed.). New Delhi: Prentice-Hall of India. p.121. ISBN81-203-0894-8. OCLC44020947. Archived from the original on 2024-10-07. Retrieved 2016-03-11. ^Roche, John (September 2001). "Introducing motion in a circle"(PDF). Physics Education. 43 (5): 399–405. Bibcode:2001PhyEd..36..399R. doi:10.1088/0031-9120/36/5/305. S2CID250827660. ^Lloyd William Taylor (1959). "Physics, the pioneer science". American Journal of Physics. 1 (8): 173. Bibcode:1961AmJPh..29..563T. doi:10.1119/1.1937847. ^Edward Albert Bowser (1920). An elementary treatise on analytic mechanics: with numerous examples (25th ed.). D. Van Nostrand Company. p.357. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Joseph A. Angelo (2007). Robotics: a reference guide to the new technology. Greenwood Press. p.267. ISBN978-1-57356-337-6. Archived from the original on 2024-10-07. Retrieved 2020-11-09. ^Eric M Rogers (1960). Physics for the Inquiring Mind. Princeton University Press. p.302. External links [edit] Media related to Centrifugal force at Wikimedia Commons \| Authority control databases \| \| International \| GND \| \| National \| United States Spain Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Acceleration Fictitious forces Force Mechanics Rotation Hidden categories: CS1 French-language sources (fr) CS1 errors: ISBN date CS1 maint: others CS1 maint: article number as page number Webarchive template wayback links Articles with short description Short description matches Wikidata All articles with unsourced statements Articles with unsourced statements from December 2012 Commons category link is on Wikidata This page was last edited on 19 September 2025, at 04:59(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. 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189174	https://www.benzoinfo.com/mechanism-of-action/	Skip to content Mechanism of Action Benzodiazepines, like alprazolam (Xanax), lorazepam (Ativan), clonazepam (Klonopin) and clonazepam) act on the central nervous system (CNS) and brain. They are known pharmacologically as GABAergic agents, sedative-hypnotics, or minor tranquilizers. Benzodiazepines work by enhancing a very important neurotransmitter called GABA (gamma-aminobutyric acid) at the GABA A receptor. This results in the sedative, hypnotic (sleep-inducing), anxiolytic (anti-anxiety), anticonvulsant, and muscle relaxant properties for which the drugs are prescribed. GABA is the chief inhibitory neurotransmitter in the mammalian central nervous system. Its role is in reducing neuronal excitability and, in humans, it is also responsible for the regulation of muscle tone. If your nervous system was a car, GABA functions much like the “brakes”. When the “car” takes off speeding down the road (excitability of the nervous system), GABA functions as the “brakes” to calm and slow it down. Simply put, GABA sends its inhibitory message by binding at special sites called GABA-A receptors on the outside of the receiving neuron. Once GABA is bound to the GABA-A receptor, the neuron opens a channel which allows chloride ions to pass inside of the neuron. These negative chloride ions make the neuron less responsive to other neurotransmitters (norepinephrine [noradrenaline], serotonin, acetylcholine and dopamine) which would normally excite it. Benzodiazepines also bind to their own receptors (benzodiazepine receptors) that are situated on the GABA-A receptor. Combination of a benzodiazepine at this site acts as a booster to the actions of GABA, allowing more chloride ions to enter the neuron, making it even more resistant to excitation. This short video explains this concept visually: A simplified visual of how benzodiazepines (and barbiturates) work is illustrated in this simple animation video: How Do Benzodiazepines Impact the Brain? Long-term benzo usage can cause what is known as ‘uncoupling’ of the GABA-A receptor. Uncoupling results in a decrease in the ability of BZs to potentiate the action of GABA on GABA-A receptors and in a decrease in the ability of GABA to potentiate BZ binding. This may be due to changes in GABA-A receptor gene expression where the neurons swap out GABA-A receptors that contain subunits benzos bind to with ones that don’t, to combat the action of the drug. FDA information for Ativan states withdrawal symptoms can be experienced by some after as little as one week of use, suggesting uncoupling occurs even with shorter-term use. In depth information on how the GABA-A receptors work with benzodiazepines can be found here. When the brain’s output of excitatory neurons is reduced, a consequence of the enhancement of GABA’s inhibitory activity caused by benzodiazepines, there may be impairment of certain functions, as the excitatory neurotransmitters are necessary for normal alertness, memory, muscle tone and coordination, emotional responses, endocrine gland secretions, heart rate and blood pressure control and a host of other functions. Other benzodiazepine receptors not linked to GABA are present in the kidney, colon, blood cells and adrenal cortex and these may also be affected by some benzodiazepines. These direct and indirect actions are responsible for the well-known adverse effects of dosage with benzodiazepines. There are also various subtypes of benzodiazepine receptors, all of which have slightly different actions. The alpha 1 subtype is responsible for sedative effects, the alpha 2 for anti-anxiety effects, and both alpha 1 and alpha 2 (as well as alpha 5) for anticonvulsant effects. All benzodiazepines combine, to a greater or lesser extent, with all these subtypes and all enhance GABA activity in the brain. What are Z Drugs and How Are They Similar to Benzodiazepines? Non-benzodiazepines, sometimes referred to as ‘Z-drugs’ or hypnotics, are also a class of psychoactive drugs that are very similar to the benzodiazepines. Most Z-drugs, like Zolpidem (Ambien), zaleplon (Sonata) and eszopiclone (Lunesta), are approved and prescribed for insomnia or sleep disorders. For this reason, they have very short half-lives ranging from 2-6 hours (in the non-elderly). The pharmacodynamics (biochemical and physiological effects) of Z-drugs are almost identical to the benzodiazepine drugs. For this reason, the Z-drugs have similar effects and also risks to the benzodiazepines. Z-drugs differ from benzodiazepines in their chemical structure, which makes them unrelated molecularly to benzodiazepines. The nonbenzodiazepines are activators of the GABA-A receptor. Like the benzodiazepines, they exert their effects by binding to and activating the benzodiazepine site of the receptor complex. Many of the Z-drugs are subtype selective (see subtypes discussed above) and are therefore novel in that they can provide specific effects (e.g., hypnotics with no anxiolytic effects). A review of the literature regarding hypnotics, including the nonbenzodiazepine Z-drugs, concluded that these drugs cause an unjustifiable risk to the individual and to public health and lack evidence of long-term effectiveness due to tolerance. The risks include dependence, accidents, and other adverse effects. Gradual discontinuation of hypnotics leads to improved health without worsening of sleep. If difficult interdose withdrawal symptoms arise between doses of these short-acting Z-drugs, sometimes a diazepam substitution taper may be necessary. It is preferred that the Z-drugs should be prescribed for only a few days at the lowest effective dose and avoided altogether whenever possible in the elderly. Both benzodiazepines and Z-drugs are recommended for short-term use only and both classes of drugs can cause tolerance, interdose withdrawal, physical dependence, and withdrawal syndromes upon discontinuation. They both require slow and gradual tapers to discontinue them safely.
189175	https://dummit.cos.northeastern.edu/teaching_sp20_3527/3527_lecture_29_polynomial_congruences.pdf	Math 3527 (Number Theory 1) Lecture #29 Polynomial Congruences: Polynomial Congruences Modulo m Polynomial Congruences Modulo pn and Hensel’s Lemma This material represents §5.1 from the course notes. Overview The goal of this last segment of the course is to discuss quadratic residues (which are simply squares modulo m) and the law of quadratic reciprocity, which is a stunning and unexpected relation involving quadratic residues modulo primes. We begin with some general tools for solving polynomial congruences modulo prime powers, which essentially reduce matters to studying congruences modulo primes. Then we study the quadratic residues (and quadratic nonresidues) modulo p, which leads to the Legendre symbol, a tool that provides a convenient way of determining when a residue class a modulo p is a square. We then discuss quadratic reciprocity and some of its applications. Polynomial Congruences, I In an earlier chapter, we analyzed the problem of solving linear congruences of the form ax ≡b (mod m). We now study the solutions of congruences of higher degree. As a ﬁrst observation, we note that the Chinese Remainder Theorem reduces the problem of solving any polynomial congruence q(x) ≡0 (mod m) to solving the individual congruences q(x) ≡0 (mod pd), where the pd are the prime-power divisors of m. Polynomial Congruences, II Example: Solve the equation x3 + x + 2 ≡0 (mod 36). Polynomial Congruences, II Example: Solve the equation x3 + x + 2 ≡0 (mod 36). By the Chinese remainder theorem, it suﬃces to solve the two separate equations x3 + x + 2 ≡0 (mod 4) and x3 + x + 2 ≡0 (mod 9). We can just test all possible residues to see that the only solutions are x ≡2 (mod 4) and x ≡8 (mod 9). Therefore, by the Chinese remainder theorem, there is a unique solution; namely, the solution to those simultaneous congruences, which is x ≡26 (mod 36). Polynomial Congruences, III Example: Solve the equation x2 ≡0 (mod 12). Polynomial Congruences, III Example: Solve the equation x2 ≡0 (mod 12). By the Chinese remainder theorem, it suﬃces to solve the two separate equations x2 ≡0 (mod 4) and x2 ≡0 (mod 3), and then put the results back together. The ﬁrst equation visibly has the solutions x ≡0, 2 (mod 4) while the second equation has the solution x ≡0 (mod 3). Then applying the Chinese remainder theorem to the 2 possible pairs of congruences x ≡0 (mod 4), x ≡0 (mod 3), and x ≡0 (mod 4), x ≡2 (mod 3), yields the solutions x ≡0, 6 (mod 12) to the original equation. Polynomial Congruences, IV Example: Solve the equation x2 ≡1 (mod 30). Polynomial Congruences, IV Example: Solve the equation x2 ≡1 (mod 30). By the Chinese remainder theorem, it suﬃces to solve the three separate equations x2 ≡1 (mod 2), x2 ≡1 (mod 3), x2 ≡1 (mod 5). We can just test all possible residues to see that the solutions are x ≡1 (mod 2), x ≡1, 2 (mod 3), and x ≡1, 4 (mod 5). Therefore, by applying the Chinese remainder theorem to all 1 · 2 · 2 = 4 ways to pick a solution from each congruence, we see that there are 4 solutions modulo 30, and they are x ≡1, 11, 19, 29 (mod 30). Polynomial Congruences, V We are therefore reduced to solving a polynomial congruence of the form q(x) ≡0 (mod pd). Observe that any solution modulo pd “descends” to a solution modulo p, simply by considering it modulo p. For example, any solution to x3 + x + 3 ≡0 (mod 25), such as x = 6, is also a solution to x3 + x + 3 ≡0 (mod 5). Our basic idea is that this procedure can also be run in reverse, by ﬁrst ﬁnding all the solutions modulo p and then using them to compute the solutions modulo pd. More explicitly, if we ﬁrst solve the equation modulo p, we can then try to “lift” each of these solutions to get all of the solutions modulo p2, then “lift” these to obtain all solutions modulo p3, and so forth, until we have obtained a full list of solutions modulo pd. Polynomial Congruences, VI Example: Solve the congruence x3 + x + 3 ≡0 (mod 25). Polynomial Congruences, VI Example: Solve the congruence x3 + x + 3 ≡0 (mod 25). Since 25 = 52, we ﬁrst solve the congruence modulo 5. If q(x) = x3 + x + 3, we can just try all residues to see the only solution is x ≡1 (mod 5). Now we “lift” to ﬁnd the solutions to the original congruence, as follows: if x3 + x + 3 ≡0 (mod 25) then we must have x ≡1 (mod 5). Now write x = 1 + 5a: plugging in yields (1 + 5a)3 + (1 + 5a) + 3 ≡0 (mod 25), which, upon expanding and reducing, simpliﬁes to 5 + 20a ≡0 (mod 25). Cancelling the factor of 5 yields 4a ≡4 (mod 5), which has the single solution a ≡1 (mod 5). This yields the single solution x ≡6 (mod 25) to our original congruence. Polynomial Congruences, VII Example: Solve the congruence x3 + 4x ≡4 (mod 343). Polynomial Congruences, VII Example: Solve the congruence x3 + 4x ≡4 (mod 343). Since 343 = 73, we ﬁrst solve the congruence modulo 7, then modulo 72, and then ﬁnally modulo 73. By trying all the residue classes, we see that x3 + 4x ≡4 (mod 7) has the single solution x ≡3 (mod 7). Next we lift to ﬁnd the solutions modulo 72: any solution must be of the form x = 3 + 7a for some a. Plugging in yields (3 + 7a)3 + 4(3 + 7a) ≡4 (mod 72), which eventually simpliﬁes to 21a ≡14 (mod 72). Cancelling the factor of 7 yields 3a ≡2 (mod 7), which has the single solution a ≡3 (mod 7). This tells us that x ≡24 (mod 49). Polynomial Congruences, VIII Example (continued): Now that we know that we must have x ≡24 (mod 49), we can lift to ﬁnd the solutions modulo 73 in the same way. Explicitly, any solution must be of the form x = 24 + 49b for some b. Plugging in yields (24 + 72b)3 + 4(24 + 72b) ≡4 (mod 73), which eventually simpliﬁes to 147b ≡147 (mod 73). Cancelling the factor of 72 yields 3b ≡3 (mod 7), which has the single solution b ≡1 (mod 7). Hence we obtain the unique solution x ≡24 + 49b ≡73 (mod 73). Polynomial Congruences, IX Example: Solve the congruence x3 + 4x ≡12 (mod 73). Polynomial Congruences, IX Example: Solve the congruence x3 + 4x ≡12 (mod 73). We ﬁrst solve the congruence modulo 7. By trying all the residue classes, we see that x3 + 4x ≡5 (mod 7) has two solutions, x ≡1 (mod 7) and x ≡5 (mod 7). Next we lift to ﬁnd the solutions modulo 72: any solution must be of the form x = 1 + 7k or x = 5 + 7k for some k. If x = 1 + 7k, then we get (1 + 7k)3 + 4(1 + 7k) ≡12 (mod 72), which simpliﬁes to 0 ≡7 (mod 72). This is contradictory so there are no solutions in this case. If x = 5 + 7k, then we get (5 + 7k)3 + 4(5 + 7k) ≡12 (mod 72), which simpliﬁes to 14k ≡14 (mod 72). Solving this linear congruence produces k ≡1 (mod 7), so we obtain x ≡12 (mod 49). Polynomial Congruences, X Example (continued): Now we lift to ﬁnd the solutions modulo 73: from the previous slide, any solution must be of the form x = 12 + 49k. In the same way as before, plugging in yields (12 + 72k)3 + 4(12 + 72k) ≡4 (mod 73), which after expanding and reducing, simpliﬁes to 98k ≡294 (mod 73). Solving in the same way as before yields k ≡5 (mod 7), whence x ≡12 + 49k ≡257 (mod 73). Hence, there is a unique solution: x ≡257 (mod 73). Polynomial Congruences, XI Example: Solve the congruence x2 ≡9 (mod 16). Polynomial Congruences, XI Example: Solve the congruence x2 ≡9 (mod 16). Since 16 = 24, we ﬁnd the solutions mod 2, then work upward. It is easy to see that there is a unique solution to x2 ≡9 (mod 2), namely, x ≡1 (mod 2). Next we lift to ﬁnd the solutions modulo 22: any solution must be of the form x = 1 + 2k, so we get (1 + 2k)2 ≡9 (mod 22), which simpliﬁes to 1 ≡9 (mod 22). This is always true, so we get two possible solutions, x ≡1, 3 (mod 4). If x = 1 + 4k, then we get (1 + 4k)2 ≡9 (mod 23), which simpliﬁes to 1 ≡9 (mod 23), which is again always true. If x = 3 + 4k, then we get (3 + 4k)2 ≡9 (mod 23), which simpliﬁes to 9 ≡9 (mod 23), which is also always true. Thus we get the four solutions x ≡1, 3, 5, 7 (mod 23). Polynomial Congruences, XII Example (continued): Finally, we must lift each solution x ≡1, 3, 5, 7 (mod 23) to the modulus 24. If x = 1 + 8k then we get (1 + 8k)2 ≡9 (mod 24), which simpliﬁes to 1 ≡9 (mod 24), which is contradictory. If x = 3 + 8k then we get (3 + 8k)2 ≡9 (mod 24), which simpliﬁes to 9 ≡9 (mod 24), which is always true, so we get two solutions x ≡3, 11 (mod 24). If x = 5 + 8k then we get (5 + 8k)2 ≡9 (mod 24), which simpliﬁes to 25 ≡9 (mod 24), which is always true, so we get two solutions x ≡5, 13 (mod 24). If x = 7 + 8k then we get (7 + 8k)2 ≡9 (mod 24), which simpliﬁes to 49 ≡9 (mod 24), which is contradictory. Thus, we get four solutions in total: x ≡3, 5, 11, 13 (mod 24). Polynomial Congruences, XIII: Lucky! The general procedure will work the same way for any prime power modulus pn: We ﬁrst solve the congruence modulo p. For each solution we obtain, we then try to lift it to a solution mod p2, then lift each of those to a solution mod p3, and so forth, until we get the full list of solutions mod pn. In the last few examples we just worked through, we saw a variety of diﬀerent behaviors. Sometimes, when we lift a solution, we obtain exactly one lifted solution. Other times, the lifting might fail, or it might yield more than one possible lifted solution. We would like to understand what determines when each of these behaviors will occur. Hensel’s Lemma, I Rather than building the motivation, we will simply state the result: Theorem (Hensel’s Lemma) Suppose q(x) is a polynomial with integer coeﬃcients. If q(a) ≡0 (mod pd) and q′(a) ̸≡0 (mod p), then there is a unique k (modulo p) such that q(a + kpd) ≡0 (mod qd+1). Explicitly, if u is the inverse of q′(a) modulo p, then k = −u · q(a) pd . This result (and a number of variations) is traditionally called Hensel’s lemma, although for us it is really more of a theorem since the proof is fairly technical. (The full proof is in the notes, but it is just a formalized version of the procedure we were using earlier.) Hensel’s Lemma, II Example: Show that there is a unique solution to the congruence x3 −2x + 7 ≡0 (mod 32020). Hensel’s Lemma, II Example: Show that there is a unique solution to the congruence x3 −2x + 7 ≡0 (mod 32020). The idea is to use Hensel’s lemma to show that the lifting will always yield a unique solution starting from the bottom level. First, we solve the congruence modulo 3: testing all 3 possible residues shows that the only solution is x ≡1 (mod 3). Now we just compute the derivative: if q(x) = x3 −2x + 7, then q′(x) = 3x2 −2 ≡1 (mod 3), no matter what x is. Therefore, Hensel’s lemma guarantees that we will always have a unique solution to this congruence modulo 3d for any d ≥1. In particular, the solution is unique modulo 32020. Hensel’s Lemma, III Example (continued): Solutions of x3 −2x + 7 ≡0 (mod 3d). Hensel’s Lemma, III Example (continued): Solutions of x3 −2x + 7 ≡0 (mod 3d). We can even calculate the various lifts using the formula given in Hensel’s lemma. (Our direct technique will yield the same result, since ultimately it is how Hensel’s lemma is proven.) For example, mod 32, since q′(a) ≡1 (mod 3) has inverse u ≡1 (mod 3), we will obtain the solution x = 1 + 3k where k = −u · q(a) pd = −1 · 6 3 = −2: thus, x ≡−5 ≡4 (mod 9), which indeed works. Lifting again yields x = 4 + 9k where k = −u · q(a) pd = −1 · 63 9 = −7, yielding x ≡4 + 9k ≡22 (mod 27). We can continue in this way and compute the lifts as high as we desire. Summary We discussed how to solve polynomial congruences modulo m and modulo prime powers. We discussed how to use Hensel’s lemma to calculate solutions to congruences modulo pd explicitly in many cases. Next lecture: Quadratic Residues and Legendre Symbols
189176	https://www.chemlin.org/chemical-elements/uranium-isotopes.php	About \| More \| Search Isotopes of Uranium List, data and properties of all known isotopes of Uranium. Atomic properties Decay properties NMR active nuclides Radiation protection References The chemical element uranium has no stable isotopes and occurs in nature as a mixture of three different radioactive nuclides: Naturally Occurring Uranium Isotopes \| \| \| \| \| \| --- --- \| \| Atomic Mass ma \| Quantity \| Half-life \| Spin \| \| UraniumIsotopic mixture \| 238,02891 u \| 100 % \| \| \| \| Isotope 234U \| 234,04095(2) u \| 0,0054(5) % \| 2.455(6) × 105 a \| 0+ \| \| Isotope 235U \| 235,04393(2) u \| 0,7204(6) % \| 7.04(1) × 108 a \| 7/2- \| \| Isotope 238U \| 238,050787618(15) u \| 99,2742(10) % \| 4.468(6) × 109 a \| 0+ \| In addition, extremely small traces of the nuclide uranium-236 can be detected in nature. The relative atomic mass of the natural uranium isotope mixture is given as 238.02891(3) u. Pure uranium isotopes - in particular uranium-235 and uranium-238 - play in all areas of nuclear technology, nuclear weapon technology and as a starting material an important role for the generation of nuclides of other elements. Isotope Table: Uranium The two following tables list the most important data and properties of the Uranium isotopes. Further information on the individual Uranium isotopes is listed on separate pages and can be accessed via the link in column 1. Atomic Properties \| IsotopeNuclide \| E \| N \| Atomic Mass[Nuclear Mass]{Mass Excess} \| Spin I(h/2π) \| Parent \| --- --- --- \| \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| \| Uranium-214 \| 21492U \| 122 \| 214 u[213.9495373 u]{0 MeV} \| 0+ \| \| Uranium-215 \| 21592U \| 123 \| 215.02676(9) u[214.9762973 u]{24.92678 MeV} \| 5/2- \| \| Uranium-216 \| 21692U \| 124 \| 216.02476(3) u[215.9742973 u]{23.06379 MeV} \| 0+ \| \| Uranium-216m \| 216m92U \| 124 \| 216.02476(3) u[215.9742973 u]{23.06379 MeV} \| (8+) \| \| Uranium-217 \| 21792U \| 125 \| 217.02466(8) u[216.9741973 u]{22.97064 MeV} \| \| Uranium-218 \| 21892U \| 126 \| 218.023505(15) u[217.9730423 u]{21.89477 MeV} \| 0+ \| \| Uranium-218m \| 218m92U \| 126 \| 218.023505(15) u[217.9730423 u]{21.89477 MeV} \| (8+) \| \| Uranium-219 \| 21992U \| 127 \| 219.02500(5) u[218.9745373 u]{23.28735 MeV} \| \| Uranium-220 \| 22092U \| 128 \| 220 u[219.9495373 u]{0 MeV} \| \| Uranium-221 \| 22192U \| 129 \| 221.026320(80) u[220.9758573 u]{24.51692 MeV} \| (9/2+) \| \| Uranium-222 \| 22292U \| 130 \| 222.02606(6) u[221.9755973 u]{24.27473 MeV} \| 0+ \| \| Uranium-223 \| 22392U \| 131 \| 223.02774(8) u[222.9772773 u]{25.83964 MeV} \| \| Uranium-224 \| 22492U \| 132 \| 224.027614(25) u[223.9771513 u]{25.72228 MeV} \| 0+ \| 228Pu \| \| Uranium-225 \| 22592U \| 133 \| 225.029385(11) u[224.9789223 u]{27.37195 MeV} \| 229Pu \| \| Uranium-226 \| 22692U \| 134 \| 226.029339(14) u[225.9788763 u]{27.3291 MeV} \| 0+ \| 230Pu \| \| Uranium-227 \| 22792U \| 135 \| 227.031182(10) u[226.9807193 u]{29.04585 MeV} \| (3/2+) \| 231Pu \| \| Uranium-228 \| 22892U \| 136 \| 228.031371(15) u[227.9809083 u]{29.2219 MeV} \| 0+ \| 232Pu \| \| Uranium-229 \| 22992U \| 137 \| 229.033506(6) u[228.9830433 u]{31.21064 MeV} \| (3/2+) \| 233Pu \| \| Uranium-230 \| 23092U \| 138 \| 230.033940(5) u[229.9834773 u]{31.61491 MeV} \| 0+ \| 234Pu230Pa230Np \| \| Uranium-231 \| 23192U \| 139 \| 231.0362923(29) u[230.9858296 u]{33.80606 MeV} \| (5/2-) \| 235Pu231Np \| \| Uranium-232 \| 23292U \| 140 \| 232.0371549(1) u[231.9866922 u]{34.60957 MeV} \| 0+ \| 236Pu232Np232Pa \| \| Uranium-233 \| 23392U \| 141 \| 233.0396343(24) u[232.9891716 u]{36.91911 MeV} \| 5/2+ \| 237Pu233Np233Pa \| \| Uranium-234 \| 23492U \| 142 \| 234.04095(2) u[233.9904873 u]{38.14468 MeV} \| 0+ \| 238Pu234Np234Pa234mPa \| \| Uranium-234m \| 234m92U \| 142 \| 234.04095(2) u[233.9904873 u]{38.14468 MeV} \| 6- \| \| Uranium-235 \| 23592U \| 143 \| 235.04393(2) u[234.9934673 u]{40.92053 MeV} \| 7/2- \| 239Pu235Np235Pa \| \| Uranium-235m \| 235m92U \| 143 \| 235.04393(2) u[234.9934673 u]{40.92053 MeV} \| 1/2+ \| \| Uranium-235m2 \| 235m292U \| 143 \| 235.04393(2) u[234.9934673 u]{40.92053 MeV} \| \| Uranium-236 \| 23692U \| 144 \| 236.0455661(12) u[235.9951034 u]{42.44455 MeV} \| 0+ \| 240Pu236Np236Pa \| \| Uranium-236m1 \| 236m192U \| 144 \| 236.0455662(12) u[235.9951035 u]{42.44464 MeV} \| (4)- \| \| Uranium-236m2 \| 236m292U \| 144 \| 236.0455662(12) u[235.9951035 u]{42.44464 MeV} \| (0+) \| \| Uranium-237 \| 23792U \| 145 \| 237.0487284(13) u[236.9982657 u]{45.39021 MeV} \| 1/2+ \| 241Pu237Pa \| \| Uranium-237m \| 237m92U \| 145 \| 237.0487284(13) u[236.9982657 u]{45.39021 MeV} \| (7/2)- \| \| Uranium-238 \| 23892U \| 146 \| 238.050787618(15) u[238.0003249 u]{47.30836 MeV} \| 0+ \| 242Pu238Pa \| \| Uranium-238m \| 238m92U \| 146 \| 238.05079(2) u[238.0003273 u]{47.31058 MeV} \| 0+ \| \| Uranium-239 \| 23992U \| 147 \| 239.0542920(16) u[239.0038293 u]{50.57267 MeV} \| 5/2+ \| 239Pa \| \| Uranium-239m1 \| 239m192U \| 147 \| 239.0542920(16) u[239.0038293 u]{50.57267 MeV} \| 1/2+ \| \| Uranium-239m2 \| 239m292U \| 147 \| 239.0542920(16) u[239.0038293 u]{50.57267 MeV} \| (5/2+) \| \| Uranium-240 \| 24092U \| 148 \| 240.0565924(27) u[240.0061297 u]{52.71548 MeV} \| 0+ \| 244Pu \| \| Uranium-241 \| 24192U \| 149 \| 241.06033(21) u[241.0098673 u]{56.19703 MeV} \| 7/2+ \| \| Uranium-242 \| 24292U \| 150 \| 242.06293(22) u[242.0124673 u]{58.61892 MeV} \| 0+ \| \| Uranium-243 \| 24392U \| 151 \| 243.06708(32) u[243.0166173 u]{62.48462 MeV} \| Radioactive Decay Properties \| Isotope \| Radioactive Decay \| Extern \| --- \| Half-life \| Decay Mode \| Probability \| Energy \| \| \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| \| U-214 \| 0.52 ms \| α → 210Th \| 100 % \| 8.692(18) MeV \| AL \| \| U-215 \| 0.7(+13-3) ms \| α → 211ThEC/β+ → 214Pa \| > 0 %? \| 8.59(5) MeV7.06(11) MeV \| \| \| U-216 \| 4.5 ms \| α → 212Th \| 8.531(26) MeV \| \| \| U-216m \| 0.7 ms \| α → 212ThIso → 216U \| > 0 %? \| \| \| U-217 \| 16 ms \| α → 213ThEC/β+ → 216Pa \| 100 %? \| 8.43(7) MeV \| AL \| \| U-218 \| 0.51 ms \| α → 214Th \| 100 % \| 8.775(9) MeV \| AL \| \| U-218m \| 0.56 ms \| α → 214Th \| 100 % \| AL \| \| U-219 \| ca. 42 μ s \| α → 215Th \| 100 % \| 9.94(5) MeV \| AL \| \| U-220 \| - ungelistet - \| AL \| \| U-221 \| 0.66(14) μs \| α → 217Th \| ~100 % \| 9.889(71) MeV \| AL \| \| U-222 \| 4.7(7) mu;s \| α → 218Th \| 100 % \| 9.48(5) MeV \| AL \| \| U-223 \| ca. 18 μs \| α → 219ThEC/β+ → 223Pa \| ca. 100 %ca. 0.2 % \| 8.94(5) MeV3.52(10) MeV \| AL \| \| U-224 \| 0.84 ms \| α → 220Th \| 100 % \| 8.628(7) MeV \| AL \| \| U-225 \| 69(15) ms \| α → 221Th \| 100 % \| 8.007(6) MeV \| AL \| \| U-226 \| 268(9) ms \| α → 222Th \| 100 % \| 7.701(4) MeV \| AL \| \| U-227 \| 1.1(1) min \| α → 223Th \| 100 % \| 7.235(3) MeV \| AL \| \| U-228 \| 9.1(2) min \| α → 224ThEC → 228Pa \| > 95 %< 5 % \| 6.804(10) MeV0.299(15) MeV \| AL \| \| U-229 \| 58(3) min \| α → 225ThEC → 229Pa \| ca. 20 %ca. 80 % \| 6.476(3) MeV1.314(7) MeV \| AL \| \| U-230 \| 20.23(2) d \| α → 226ThCD → 22Ne \| 100 %selten \| 5.9925(5) MeV \| AL \| \| U-231 \| 4.2(1) d \| α → 227ThEC → 231Pa \| 0.004(1) %ca. 100 % \| 5.5763(17) MeV0.382(3) MeV \| AL \| \| U-232 \| 68.9(4) a \| α → 228ThSF div \| ca. 100 %Spuren \| 5.41363(9) MeV \| AL \| \| U-233 \| 1.5919(15) × 105 a \| α → 229ThSF div \| 100 %Spuren \| 4.9087(12) MeV \| AL \| \| U-234 \| 2.455(6) × 105 a \| α → 230ThSF div \| ca. 100 %Spuren \| 4.8575(7) MeV \| AL \| \| U-234m \| 33.5(20) μs \| \| \| U-235 \| 7.04(1) × 108 a \| α → 231ThSF div \| ca. 100 %Spuren \| 4.6780(7) MeV \| AL \| \| U-235m \| ca. 26 min \| Iso → 235U \| 100 % \| AL \| \| U-235m2 \| 3.6(19) ms \| SF div \| ? \| \| \| U-236 \| 2.342(4) × 107 a \| α → 232ThSF div \| ca. 100 %selten \| 4.5730(9) MeV \| AL \| \| U-236m1 \| 100(4) ns \| \| \| U-236m2 \| 120(2) ns \| Iso → 236USF α → 232Th \| 87(6) %13(6) % \| \| \| U-237 \| 6.752(2) d \| β- → 237Np \| 100 % \| 0.5185(5) MeV \| AL \| \| U-237m \| 155(6) ns \| \| \| U-238 \| 4.468(6) × 109 a \| α → 234ThSF div \| ca. 100 %10-5 % \| 4.2699(21) MeV \| AL \| \| U-238m \| 280(6) ns \| Iso → 238USF div \| 97.4(4) %2.6(4) % \| AL \| \| U-239 \| 23.45(2) min \| β- → 239Np \| 100 % \| 1.2617(15) MeV \| AL \| \| U-239m1 \| 0.78(4) μs \| \| \| U-239m2 \| 0.25 μs \| \| \| U-240 \| 14.1(1) h \| β- → 240Np \| 100 % \| 4.035(14) MeV \| AL \| \| U-241 \| ca. 40 min \| β- → 241Np \| 1.88(22) MeV \| AL \| \| U-242 \| 16.8(5) min \| β- → 242Np \| 100 % \| 1.20(28) MeV \| AL \| \| U-243 \| 10 s \| β- → 243Np \| ? \| 2.67 MeV \| \| Notes (related to the columns): 1 - name of the nuclide, isotope.2 - E: isotope symbol with mass number (superscript; number of nucleons) and Atomic number (subscript; number of protons). 3 - N: number of neutrons. 4 - relative atomic mass of the Uranium isotope (isotopic mass including electrons) and the mass of the atomic nucleus in square brackets (nuclear mass, nuclide mass without electrons), each related to 12C = 12.00000 . In addition, the mass excess is given in MeV. 5 - nuclear spin I, unit: h/2π. 6 - source nuclides: Possible, assumed or actual source nuclides (mother nuclides, parent nuclides). If applicable, the corresponding decay modes can be found in the data for the respective starting nuclide. 7 - isotope notation in short form. 8 - decay: half-live of the Uranium isotope (a = years; ; d = days; h = hours; min = minutes; s = seconds). 9 - decay mode: type of decay into the respective daughter nuclides with n = neutron emission; p = proton emission; α = alpha decay; β- = beta minus decay with electron emission; EC = electron capture; β+ = positron emission; ε = β+ and/or EC; Iso = isomeric transition; CD = cluster decay; SF = spontaneous decay. 10 - decay probability in percent (%). 11 - decay energy; Particle energy related to decay type. 12 - other information and notes: AL = Adopted Levels (link to external data ). Miscellaneous: ()- Numbers in brackets: uncertainty to represent the spread of the reported value.~ - Theoretical values or systematic trends. - unlisted-: Nuclides that have already been mentioned in the literature but for some reason can no longer be found in the current nuclide tables because their discovery e.g. has not confirmed. NMR active Uranium nuclides \| Nuclidequantity 1)spin \| Nuclearmagneticmomentμ/μN \| Gyromagnetic ratio{Quadrupole moment} \| Resonantfrequencyv0 bei 1 T \| RelativesensitivityH0 = const.v0 = const. 3) \| --- --- \| 235U0,7204(6) %7/2- \| - 0,38(3) \| 0,4926{+ 4,936(6)} \| 0,83 \| 0,000150,4082 \| 1) Quantity Percentage of natural occurrence. 2) Gyromagnetic ratio: 107 rad T-1 s-1 Quadrupole moment: Q [barn] = [100 fm2] 3) Related to 1H = 1,000. Radiation Protection According to the Radiation Protection Ordinance (StrlSchV 2018, Germany), the following values (columns 1 to 7) apply to the handling of Uranium radionuclides: \| Nuclide \| Limit Value \| HASS limit \| SC \| Daughter Nuclides \| Half-life \| --- --- --- \| \| U-230+ \| 105 Bq \| 10 Bq/g \| 0,04 TBq \| 0,1 Bq cm-2 \| Th-226, Ra-222, Rn-218,Po-214 \| 20.8 d \| \| U-231 \| 107 Bq \| 100 Bq/g \| 10 Bq cm-2 \| 4.2 d \| \| U-232+ \| 103 Bq \| 0,1 Bq/g \| 0,06 TBq \| 0,1 Bq cm-2 \| Th-228, Ra-224, Rn-220,Po-216, Pb-212, Bi-212,Tl-208, Po-212 \| 68.9 a \| \| U-233 \| 104 Bq \| 1 Bq/g \| 0,07 TBq \| 1 Bq cm-2 \| 160000 a \| \| U-234 \| 104 Bq \| 1 Bq/g \| 1 Bq cm-2 \| 250000 a \| \| U-235+ \| 104 Bq \| 1 Bq/g \| 0,00008 TBq \| 1 Bq cm-2 \| Th-231 \| 7.0 × 108 a \| \| U-236 \| 104 Bq \| 10 Bq/g \| 0,2 TBq \| 1 Bq cm-2 \| 2.4 × 107 a \| \| U-237 \| 106 Bq \| 100 Bq/g \| 10 Bq cm-2 \| 6.8 d \| \| U-238+ \| 104 Bq \| 1 Bq/g \| Unbegrenzt (UL) \| 1 Bq cm-2 \| Th-234, Pa-234m, Pa-234 \| 4.5 × 109 a \| \| U-238sec \| 103 Bq \| 1 Bq/g \| 1 Bq/cm2 \| 4.4 × 109 a \| \| U-239 \| 106 Bq \| 100 Bq/g \| 100 Bq cm-2 \| 23.5 min \| \| U-240+ \| 106 Bq \| 100 Bq/g \| 10 Bq cm-2 \| 14.1 h \| (HASS = High-Activity Sealed Radioactive Sources; SC = surface contamination) Literature Sources and References Properties of the Uranium nucleides - NuDat: National Nuclear Data Center, Brookhaven National Laboratory, based on ENSDF and the Nuclear Wallet Cards. - G. Audi et. al.: The NUBASE evaluation of nuclear and decay properties. Nuclear Physics, (2003), DOI 10.1016/j.nuclphysa.2003.11.001. - Live Chart of Nuclides. Nuclear structure and decay data. Uranium: NMR properties - 235U-NMR - N. J. Stone: Table of nuclear magnetic dipole and electric quadrupole moments. Atomic Data and Nuclear Data Tables, (2005), DOI 10.1016/j.adt.2005.04.001. - Pekka Pyykkö: Year-2008 nuclear quadrupole moments. Molecular Physics, (2008), DOI 10.1080/00268970802018367. - Pekka Pyykkö: Year-2017 nuclear quadrupole moments. Molecular Physics, (2018), DOI 10.1080/00268976.2018.1426131. - N. J. Stone: Table of recommended nuclear magnetic dipole moments. IAEA, (2019). More sources: - Isotopic abundances, atomic weights and isotopic masses: see respective keyword. - C. Fry, M. Thoennessen:Discovery of actinium, thorium, protactinium, and uranium isotopes.In: Atomic Data and Nuclear Data Tables, (2013), DOI 10.1016/j.adt.2012.03.002. 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189177	https://coefficient.io/excel-tutorials/stdev-s	Skip to content Blog/ STDEV.S Function in Excel: Quick Guide + Tutorial Last Modified: October 17, 2024 - 10 min read Julian Alvarado Join our community Calculating statistical measures is a crucial part of data analysis, and the STDEV.S function in Excel is a powerful tool for determining the sample standard deviation of a dataset. Whether you’re a data analyst, a researcher, or simply someone who needs to analyze numerical information, understanding how to use the STDEV.S function can provide valuable insights and help you make more informed decisions. In this comprehensive guide, we’ll explore the basics of the STDEV.S function, how to use it in Excel, and the key differences between it and its counterpart, STDEV.P. Did you know? Coefficient eliminates the need to export data manually and rebuild stale dashboards. Get started by pulling live data into pre-built Sheets dashboards. STDEV.S Function 101: The Basics Click to copy link The STDEV.S function in Excel is used to calculate the sample standard deviation of a dataset. Standard deviation is a statistical measure that quantifies the amount of variation or dispersion of a set of values from the mean. In other words, it tells us how much the individual data points in a sample tend to deviate from the sample’s mean. The formula for the STDEV.S function is: =STDEV.S(number1, [number2], …) Where number1, number2, and so on, are the values you want to calculate the sample standard deviation for. The key difference between the STDEV.S and STDEV.P functions is that STDEV.S calculates the sample standard deviation, while STDEV.P calculates the population standard deviation. We’ll explore the differences between these two functions in more detail later in this guide. How to Use the STDEV.S Function in Excel Click to copy link Using the STDEV.S function in Excel is a straightforward process. Here’s a step-by-step guide with screenshots to help you get started: Select the cell where you want the STDEV.S result to appear. Type the formula =STDEV.S( and then select the range of cells containing the values you want to calculate the sample standard deviation for. Close the formula with a closing parenthesis ). Press Enter, and the STDEV.S result will be displayed in the selected cell. Here’s an example: In this example, we’re calculating the sample standard deviation for the values in the range C2:C10. The STDEV.S function returns the result of 951.5409, which represents the sample standard deviation of the dataset. Video Tutorial Click to copy link STDEV.S vs. STDEV.P: Key Differences Click to copy link While the STDEV.S and STDEV.P functions may seem similar, there are important differences between them. Understanding these differences is crucial for choosing the right function for your specific data analysis needs. The main distinction between STDEV.S and STDEV.P is the type of standard deviation they calculate: STDEV.S calculates the sample standard deviation, which is used when you have a sample of data and want to estimate the standard deviation of the population. STDEV.P calculates the population standard deviation, which is used when you have access to the entire population of data. The formula for STDEV.S is: =SQRT(SUM((x – mean(x))^2) / (n-1)) Where x is the individual data points, mean(x) is the average of the data points, and n is the number of data points. The formula for STDEV.P is: =SQRT(SUM((x – mean(x))^2) / n) The key difference is the divisor: STDEV.S uses n-1, while STDEV.P uses n. This means that STDEV.S will always return a slightly larger value than STDEV.P, as it takes into account the uncertainty of estimating the population standard deviation from a sample. Here’s an example to illustrate the difference: Suppose we have a dataset of 10 values: 5, 7, 8, 6, 9, 8, 7, 6, 8, 7. The STDEV.S of this dataset is 1.14. The STDEV.P of this dataset is 1.12. As you can see, the STDEV.S value is slightly higher than the STDEV.P value, reflecting the fact that we’re working with a sample rather than the entire population. In general, you should use STDEV.S when you have a sample of data and want to estimate the standard deviation of the population. Use STDEV.P when you have access to the entire population of data. Practical Applications of the STDEV.S Function Click to copy link The STDEV.S function in Excel is a powerful tool that has a wide range of practical applications across various industries and scenarios. Let’s explore some real-world examples of how this function can be utilized: Analyzing Sales Performance Click to copy link One of the most common use cases for STDEV.S is in the analysis of sales data. By calculating the standard deviation of sales figures, you can gain valuable insights into the consistency and variability of your sales performance. This information can help you identify outliers, detect patterns, and make more informed decisions about your sales strategies. Evaluating Investment Portfolios Click to copy link In the financial sector, the STDEV.S function is essential for assessing the risk associated with investment portfolios. By calculating the standard deviation of returns, you can determine the volatility of your investments and make more informed decisions about asset allocation and risk management. Monitoring Quality Control Click to copy link In manufacturing and production environments, the STDEV.S function can be used to monitor quality control. By tracking the standard deviation of key metrics, such as product dimensions or defect rates, you can identify process variations and take corrective actions to maintain consistent quality. Analyzing Survey Data Click to copy link When conducting market research or customer surveys, the STDEV.S function can help you understand the distribution and variability of responses. This information can be valuable for identifying trends, segmenting your target audience, and making data-driven decisions. Forecasting and Budgeting Click to copy link In the realm of financial planning and budgeting, the STDEV.S function can be used to analyze historical data and make more accurate forecasts. By understanding the standard deviation of expenses, revenue, or other financial metrics, you can better anticipate and plan for potential fluctuations. Calculating STDEV.S Manually Click to copy link While the STDEV.S function in Excel provides a convenient way to calculate the standard deviation of a dataset, it’s important to understand the underlying mathematical formula and the manual calculation process. This knowledge can be particularly useful when working with datasets that are too large to fit in a spreadsheet or when you need to perform the calculation in a programming environment. The formula for calculating the standard deviation using the STDEV.S function is: STDEV.S = √(Σ(x – μ)^2 / (n-1)) Where: x represents each data point in the dataset μ is the mean or average of the dataset n is the total number of data points To calculate the STDEV.S manually, follow these steps: Calculate the mean of the dataset. For each data point, subtract the mean and square the result. Sum up all the squared differences. Divide the sum by the number of data points minus 1. Take the square root of the result to get the standard deviation. Let’s illustrate this with an example. Suppose we have the following dataset: [10, 15, 12, 18, 14]. Calculate the mean: (10 + 15 + 12 + 18 + 14) / 5 = 13.8 Subtract the mean and square the results: (10 – 13.8)^2 = 9.64 (15 – 13.8)^2 = 1.44 (12 – 13.8)^2 = 4.84 (18 – 13.8)^2 = 16.84 (14 – 13.8)^2 = 0.04 Sum the squared differences: 9.64 + 1.44 + 4.84 + 16.84 + 0.04 = 32.8 Divide by the number of data points minus 1: 32.8 / (5 – 1) = 8.2 Take the square root: √8.2 = 2.86 Therefore, the manual calculation of the STDEV.S for this dataset is 2.86. Try the Free Spreadsheet Extension Over 500,000 Pros Are Raving About Stop exporting data manually. Sync data from your business systems into Google Sheets or Excel with Coefficient and set it on a refresh schedule. Get Started Using STDEV.S in Google Sheets and VBA Click to copy link While the STDEV.S function is primarily associated with Microsoft Excel, it’s also available in other spreadsheet applications, such as Google Sheets. The usage and syntax are very similar, making it easy to transition between the two. In Google Sheets, you can use the STDEV.S() function to calculate the standard deviation of a dataset. The syntax is the same as in Excel: =STDEV.S(range_of_values) For example, to calculate the standard deviation of the values in cells A1 to A10, you would use the following formula: =STDEV.S(A1:A10) In addition to using the function directly in the spreadsheet, you can also automate the STDEV.S calculation using VBA (Visual Basic for Applications) in Excel. Here’s a simple example of a VBA subroutine that calculates the standard deviation of a range of values: Sub CalculateSTDEV() Dim data_range As Range Dim result As Double ‘ Specify the range of values to calculate the STDEV.S Set data_range = Range(“A1:A10”) ‘ Calculate the standard deviation result = Application.WorksheetFunction.StDevP(data_range) ‘ Display the result MsgBox “The standard deviation is: ” & result End Sub This subroutine first defines the range of values to be used in the calculation, then calls the StDevP() function (the population standard deviation function) to compute the standard deviation. Finally, it displays the result in a message box. By understanding how to use the STDEV.S function in both Google Sheets and VBA, you can expand your data analysis capabilities and apply the standard deviation calculation in a wider range of scenarios. Common Pitfalls and Troubleshooting Tips for STDEV.S Click to copy link While the STDEV.S function is relatively straightforward to use, there are a few common pitfalls and potential issues to be aware of: Incorrect Data Range: Ensure that you have selected the correct range of cells for the STDEV.S calculation. Accidentally including or excluding data points can significantly impact the result. Handling Blank or Non-Numeric Cells: The STDEV.S function will ignore blank cells or non-numeric values in the dataset. If your data contains these types of cells, make sure to either remove them or replace them with appropriate values before performing the calculation. Interpreting the Result: Remember that the standard deviation is a measure of the spread or variability of the data. A higher standard deviation indicates greater dispersion, while a lower standard deviation suggests the data points are more tightly clustered around the mean. Dealing with Outliers: Outliers, or data points that are significantly different from the rest of the dataset, can skew the standard deviation calculation. Consider identifying and addressing outliers before performing the STDEV.S analysis. Selecting the Appropriate Function: Depending on your specific needs, you may need to use the STDEV.P function instead of STDEV.S. The STDEV.P function calculates the population standard deviation, while STDEV.S is used for sample standard deviation. To troubleshoot any issues with the STDEV.S function, consider the following tips: Verify the Data: Double-check your dataset to ensure that it contains only valid, numeric values and that there are no blank or non-numeric cells. Check the Function Syntax: Ensure that you are using the correct function name and that the syntax is correct, including the appropriate range of cells. Perform Manual Calculations: As demonstrated earlier, you can calculate the standard deviation manually to verify the results from the STDEV.S function. Consult Excel Help or Online Resources: If you’re still encountering issues, refer to the Excel help documentation or search for relevant online resources and tutorials for further guidance. By being aware of these common pitfalls and following the troubleshooting tips, you can ensure that the STDEV.S function functions effectively and obtains accurate results for your data analysis needs. Use STDEV.S to Uncover Data Patterns Click to copy link STDEV.S helps you measure data spread, useful for sales analysis, investment evaluation, or quality control. However, manually updating data for these calculations can be tedious. Coefficient pulls live data from Salesforce, QuickBooks, and other systems directly into Excel. Update your spreadsheets automatically today! Connect Live Data to Excel Instantly Automatically sync data from any source into Excel and keep it on a refresh schedule with Coefficient. Get Started Free Learn More 5.0 Stars - Top Rated Try the Spreadsheet Automation Tool Over 500,000 Professionals are Raving About Tired of spending endless hours manually pushing and pulling data into Google Sheets? Say goodbye to repetitive tasks and hello to efficiency with Coefficient, the leading spreadsheet automation tool trusted by over 350,000 professionals worldwide. Sync data from your CRM, database, ads platforms, and more into Google Sheets in just a few clicks. Set it on a refresh schedule. And, use AI to write formulas and SQL, or build charts and pivots. Julian Alvarado AI Content Engineering Julian is a dynamic B2B marketer with 8+ years of experience creating full-funnel marketing journeys, leveraging an analytical background in biological sciences to examine customer needs. 500,000+ happy users Wait, there's more! 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189178	https://www.varsitytutors.com/practice/subjects/sat-math/help/sequences/integers/arithmetic/consecutive-integers	Consecutive Integers Help Questions SAT Math › Consecutive Integers The sum of four consecutive odd integers is equal to 96. How many of the integers are prime? 0 1 2 3 4 Explanation Let x be the smallest of the four integers. We are told that the integers are consecutive odd integers. Because odd integers are separated by two, each consecutive odd integer is two larger than the one before it. Thus, we can let x + 2 represent the second integer, x + 4 represent the third, and x + 6 represent the fourth. The sum of the four integers equals 96, so we can write the following equation: x + (x + 2) + (x + 4) + (x + 6) = 96 Combine x terms. 4_x_ + 2 + 4 + 6 = 96 Combine constants on the left side. 4_x_ + 12 = 96 Subtract 12 from both sides. 4_x_ = 84 Divide both sides by 4. x = 21 This means the smallest integer is 21. The other integers are therefore 23, 25, and 27. The question asks us how many of the four integers are prime. A prime number is divisible only by itself and one. Among the four integers, only 23 is prime. The number 21 is divisible by 3 and 7; the number 25 is divisible by 5; and 27 is divisible by 3 and 9. Thus, 23 is the only number from the integers that is prime. There is only one prime integer. The answer is 1. The sum of four consecutive odd integers is equal to 96. How many of the integers are prime? 0 1 2 3 4 Explanation Let x be the smallest of the four integers. We are told that the integers are consecutive odd integers. Because odd integers are separated by two, each consecutive odd integer is two larger than the one before it. Thus, we can let x + 2 represent the second integer, x + 4 represent the third, and x + 6 represent the fourth. The sum of the four integers equals 96, so we can write the following equation: x + (x + 2) + (x + 4) + (x + 6) = 96 Combine x terms. 4_x_ + 2 + 4 + 6 = 96 Combine constants on the left side. 4_x_ + 12 = 96 Subtract 12 from both sides. 4_x_ = 84 Divide both sides by 4. x = 21 This means the smallest integer is 21. The other integers are therefore 23, 25, and 27. The question asks us how many of the four integers are prime. A prime number is divisible only by itself and one. Among the four integers, only 23 is prime. The number 21 is divisible by 3 and 7; the number 25 is divisible by 5; and 27 is divisible by 3 and 9. Thus, 23 is the only number from the integers that is prime. There is only one prime integer. The answer is 1. Explanation Explanation In the repeating pattern 9,5,6,2,1,9,5,6,2,1......What is the 457th number in the sequence? 5 9 1 2 1 Explanation There are 5 numbers in the sequnce. How many numbers are left over if you divide 5 into 457? There would be 2 numbers! The second number in the sequence is 9,5,6,2,1 In the repeating pattern 9,5,6,2,1,9,5,6,2,1......What is the 457th number in the sequence? 5 9 1 2 1 Explanation There are 5 numbers in the sequnce. How many numbers are left over if you divide 5 into 457? There would be 2 numbers! The second number in the sequence is 9,5,6,2,1 Four consecutive integers have a mean of 9.5. What is the largest of these integers? 12 13 9 8 11 Explanation Four consecutive integers could be represented as n, n+1, n+2, n+3 Therefore, by saying that they have a mean of 9.5, we mean to say: (n + n+1 + n+2 + n+ 3)/4 = 9.5 (4n + 6)/4 = 9.5 → 4n + 6 = 38 → 4n = 32 → n = 8 Therefore, the largest value is n + 3, or 11. Four consecutive integers have a mean of 9.5. What is the largest of these integers? 12 13 9 8 11 Explanation Four consecutive integers could be represented as n, n+1, n+2, n+3 Therefore, by saying that they have a mean of 9.5, we mean to say: (n + n+1 + n+2 + n+ 3)/4 = 9.5 (4n + 6)/4 = 9.5 → 4n + 6 = 38 → 4n = 32 → n = 8 Therefore, the largest value is n + 3, or 11. Four consecutive odd integers have a sum of 32. What are the integers? Explanation Consecutive odd integers can be represented as x, x+2, x+4, and x+6. We know that the sum of these integers is 32. We can add the terms together and set it equal to 32: x + (x+2) + (x+4) + (x+6) = 32 4x + 12 = 32 4x = 20 x = 5; x+2=7; x+4 = 9; x+6 = 11 Our integers are 5, 7, 9, and 11. Four consecutive odd integers have a sum of 32. What are the integers? Explanation Consecutive odd integers can be represented as x, x+2, x+4, and x+6. We know that the sum of these integers is 32. We can add the terms together and set it equal to 32: x + (x+2) + (x+4) + (x+6) = 32 4x + 12 = 32 4x = 20 x = 5; x+2=7; x+4 = 9; x+6 = 11 Our integers are 5, 7, 9, and 11. Powered by Varsity Tutors ⋅ © 2025 All Rights Reserved
189179	https://web.mit.edu/nnf/education/wettability/summerreading-2005short.pdf	June 29, 2005 1 Surface Tension of Polymers Abhinandan Agrawal Hatsopoulos Microfluids Laboratory Department of Mechanical Engineering Massachusetts Institute of Technology June 29, 2005 2 Physical origin of surface tension/surface energy • “Unhappy” molecules at the surface: they are missing half their attractive interactions • Unbalanced forces for the molecules at the surface lead to additional energy • The additional free energy at the surface is known as surface energy • This is the fundamental reason behind liquids adjusting their shapes to expose the smallest possible area Examples of “minimal surfaces” spherical shape of bubbles www.fairfied.edu full dry hair vs. sticky wet hair Jose Bico, Nature, 2004 June 29, 2005 3 Mechanical definition: as a surface energy Supply of energy is necessary to create surfaces dW dA γ = ⋅ (mJ/m2) specific free energy or surface tension γ = Specific surface energy of a material is the excess energy per unit area due to the existence of the free surface June 29, 2005 4 Estimation of surface tension based on intermolecular forces U U/2 Surface tension is then of order For van der Waals type interactions, At a temperature of 25 oC, kT is equal to 1/40 eV, which gives γ = 20 mJ /m2 2 2 U a γ ≅ U kT ≅ U is the cohesion energy per molecule inside the liquid Energy shortfall for a molecule sitting at the surface ~ U/2 a is the size of a molecule; a2 is the exposed area of a molecule (close to actual value for oils and alcohols) Helium (4 K) Ethanol Acetone Glycerol Water Water (100oC) Molten glass Mercury 0.1 23 24 63 73 58 ~300 485 Surface tension of a few common liquids in mJ/m2 strongly cohesive liquid hydrogen bonding 1eV U ≈ June 29, 2005 5 Mechanical definition: as a capillary force Surface tension (γ) can also be viewed as a force per unit length (mN/m or N/m) The term “surface tension” is tied to the concept that the surface stays under a tension Examples where surface tension manifests itself as force 3) Objects on water 2) Capillary adhesion F 1) Slider 2 F dx l dx γ = end view of the leg W F1 F2 (F1+F2) cosθ θ F1=F2=γl June 29, 2005 6 Contents Relationship of surface tension with other material properties Derivation of surface tension theoretically from knowledge of intermolecular forces 1) Surface energy of polymer liquids and melts 2) Surface energy of solid polymers 3) Interfacial tension between a solid and a liquid June 29, 2005 7 1. Surface energy of liquids and melts Existing methods for determining the surface tension of liquids 1) Wilhelmy’s method, in which one dips a thin plate or ring in the liquid and measures the capillary force acting on the plate 2) The rise of a liquid in a small capillary tube 3) The method of drops, in which one characterizes the shape of drops F cos( ) pγ θ 2cos h gr ρ γ θ = is the perimeter of the plate =2(length + width) p θ r h water washing-liquid θ C gz γ ρ = ( ) ( ) 3/ 2 1/ 2 2 2 1 1 1 zz z z r C r r r = − + + + r z June 29, 2005 8 Estimation of surface tension from related properties Since the surface tension is a manifestation of intermolecular forces, it may be expected to be related to other properties derived from intermolecular forces, such as work of cohesion internal pressure, and compressibility. 1) Relationship between work of cohesion and surface tension (Grunberg 1949) molecular vol.; A V N = 2/3 molecules/unit surf. area; A N V ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ molar vol.; V = 1/3 2/3 2 coh A W N V γ = 2/3 2/3 / surface energy per molecule; A V N γ = surf. energy per unit area; γ = 1/3 2/3 molar surf. energy; A N V γ = 1/3 2/3 2 work of cohesion; A N V γ = 2) Relationship between surface tension and solubility parameter (Hildebrand 1950) 1/3 2 / molar surf. energy/ / ; v V V E V γ δ = ∆ = ∼ 0.43 1/3 4.1 (cgsunits) V γ δ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ is the energy of vaporization; is the solubility parameter; v E δ ∆ 3) Relationship between compressibility and surface tension (McGowan 1967) κ γ ↑→ ↓ 3/ 2 8 1.33 10 (cgsunits) κγ − = × June 29, 2005 9 Introduction to parachor Macleod et al. (Trans. Faraday So., 1923, 19, 38) γ is the surface tension, D and d are the density of liquid and vapor at same temperature. C is a a characteristic constant for a given liquid. ( ) 4 C D d γ = − valid for different compounds and at different temperatures 1/ 4 1/ 4 s M P C M D d γ = = − at low temperature, d becomes small Sugden S. J. (Chem. Soc., 1924, 125, 1177) molecular volume V Unit of Ps (cm3/mol)×(erg/cm2)1/4 or (m3/mol)×(mJ/m2)1/4 Ps for different substances is a comparison of molecular volumes at temperatures at which liquids have the same surface tension Ps bears an constant ratio to the critical volume, which suggests that it is a true measure of the molecular volume 0.78 s c P V = Ps is called parachor (from παρα = by the side of, and χορος = space to signify comparative volumes) June 29, 2005 10 Useful properties of parachor Ps is only a function of chemical composition and it is an additive function From experimental data for Ps, it is found that • Ps can be reproduced by adding together two sets of constants, one for the atoms in the molecule, the other for the unsaturation or ring closure. • The values for a particular atom is independent of the manner in which it is situated • The values for individual element are same from compound to compound. X R C C The molecular parachor is a useful means of estimating surface tensions M is the molecular weight, V is the molar volume, ρ is the density. If the group contributions of Ps and V are known, γ results from the above expression 4 s P V γ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ H R1 O June 29, 2005 11 Atomic and structural contributions to the parachor Unit Value CH2 40.0 C 9.0 H 15.5 O 19.8 O2 (in esters) 54.8 N 17.5 S 49.1 F 26.1 Cl 55.2 Unit Value Br 68.0 I 90.3 Double bond 16.3–19.1 Triple bond 40.6 Three-membered ring 12.5 Four-membered ring 6.0 Five-membered ring 3.0 Six-membered ring 0.8 Seven-membered ring 4.0 Deduction of atomic constants: Difference for CH2 in paraffins, esters, ethers, and ketones has a mean value of 40.0 Subtracting nCH2 from a series of values for the paraffins CnH2n+2, values for 2H can be calculated Quayle O.R. Chem. Revs. 53 (1953) 439 June 29, 2005 12 Example: ethanol O H H Unit Contribution to parachor O 19.8 C 9.0 H 15.5 H H C C H H ( ) 3 2 1/4 3 3 4 4 4 2 2 1 19.8 6 15.5 2 9 130.8(cm /mol)×(erg/cm ) 0.789g/cm ; 46g/mol 58.3 cm /mol 130.8 = 2.24 =25.3 erg/cm or25.3mJ/m 58.3 s s P M M V P V ρ ρ γ = × + × + × = = = = = ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ close to the actual value of 24 mJ/m2 June 29, 2005 13 Example: poly (ethylene oxide) O H Unit Contribution to parachor O 19.8 C 9.0 H 15.5 H C C H H ( ) 3 2 1/4 3 3 4 4 4 2 2 1 19.8 4 15.5 2 9 99.8(cm /mol)×(erg/cm ) 1.12g/cm ; 44g/mol cm /mol 99.8 = 2.54 =41.6 erg/cm or41.6mJ/m 39.28 s s P M M V P V ρ ρ γ = × + × + × = = = = = ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ close to the actual value of 43 mJ/m2 June 29, 2005 14 Contact between solids and liquids Solid/liquid pair partially wetting non-wetting wetting dA cos θ dA θ θ' γs γsl γl air liquid Change in Gibbs free energy dG when the drop spreads an infinitesimal amount cos sl s l dG dA dA dA γ γ γ θ = − + At equilibrium / 0 cos 0 sl s l dG dA γ γ γ θ = → − + = γl and θ are directly measurable; γs and γsl are not wetting behavior dependent on interfacial and solid surface energy as well cos s sl l γ γ θ γ − = June 29, 2005 15 2. Surface tension of solid surfaces (γs) Methods for determining the surface tension of solids a) By measuring the contact angle between the solid and liquid b) By determining critical surface tension ( ) according to Zisman (1964), with the assumption that c) By extrapolating surface tension data of polymer melts to room temperature (Roe, 1965; Wu, 1969–71) cr γ cr s γ γ ≈ No direct method available for measurement of surface tension of solid polymers June 29, 2005 16 2a. Estimation of solid surface energy from contact angle ( ) 1/3 2 1/3 1/3 4( ) s l s l VV V V Φ = + 1/2 2 ( ) sl s l s l γ γ γ γ γ = + −Φ cos 0 sl s l γ γ γ θ − + = Girifalco and Good (1956) ( ) 2 2 1 cos 4 s l θ γ γ ⎛ ⎞ + ≈ ⎜ ⎟ ⎜ ⎟ Φ ⎝ ⎠ Example: PMMA ( ) 1/3 2 1/3 1/3 4(18 86.5) 0.93 18 86.5 × Φ = = + 3 1.17g/cm ; 100.1g/mol PMMA PMMA M ρ = = 3 86.5 cm /mol PMMA PMMA PMMA M V ρ = = 2 2 3 1g/cm ; 18g/mol H O H O M ρ = = 68–72o 2 2 3 2 18 cm /mol; 72mJ/m H O H O M V γ ρ = = = ( ) 2 2 2 1 cos(69 ) 72.8 38 mJ/m 4 0.93 s γ ⎛ ⎞ + ⎜ ⎟ ≈ ≈ ⎜ ⎟ × ⎝ ⎠ D By means of the boxed equation, surface tension of solid surface can be calculated from measurements of contact angle and surface tension of liquid June 29, 2005 17 2b. Determining solid surface tension using Zisman plot 30 40 50 60 70 80 0.4 0.6 0.8 1.0 Propylene carbonate Dimethyl sulfoxide Ethylene glycol Glycerol Water cos(θ ) Surface Tension (mN/m) Toluene 2 38 mJ/m cr γ = Zisman found that cosθ is usually a monotonic function of γl ( ) cos 1 l l cr a b θ γ β γ γ = − = − − γcr is called the “critical surface tension” of a solid and is a characteristic property of any given solid Any liquid with γl <γcr will wet the surface It is found that critical surface tension is close to the solid surface tension of polymer s cr γ γ ≈ Zisman plot for PMMA using various testing liquids June 29, 2005 18 2c. Estimation of γ from extrapolating data of polymer melt Extrapolation of surface tensions of melts to room temperature leads to reliable values for solid polymers Hence, surface tension of solid polymers may be calculated using 4 s P V γ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Example: PMMA C CH2 CH3 C O O CH3 8 H 5 C 2 O 1 double bond s P = × + × + × + × 8 17.1 5 4.8 2 20 1 23.2 s P = × + × + × + × Unit Contribution to parachor O 20 C 9 H 17.1 Double bond 23.2 3 2 1/4 220.8(cm /mol)×(erg/cm ) = 3 1.17g/cm ; 100.1g/mol M ρ = = ( ) 4 4 4 220.8 = 2.55 86.5 s P V γ ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3 86.5 cm /mol M V ρ = = 2 2 =42.5 erg/cm or42.5mJ/m June 29, 2005 19 Example: contact angle of water on PMMA 1/2 cos 2 1 s l γ θ γ ⎛ ⎞ ≈Φ − ⎜ ⎟ ⎝ ⎠ ( ) 1/3 2 1/3 1/3 4( ) s l s l VV V V Φ = + 2 2 2 2 3 3 2 3 2 1g/cm ; 18g/mol 18 cm /mol; 72mJ/m 86.5 cm /mol; 42.5 mJ/m H O H O H O H O PMMA PMMA M M V V ρ γ ρ γ = = = = = = = ( ) 1/3 2 1/3 1/3 4(18 86.5) 0.93 18 86.5 × Φ = = + 1/2 o 42.5 cos 2 0.93 1 0.43 72 65 θ θ ⎛ ⎞ ≈ × −= ⎜ ⎟ ⎝ ⎠ = contact angle of water on PMMA = 68–72 o June 29, 2005 20 Temperature dependence of surface tension 1) Since the parachor Ps is independent of the temperature 4 s P V γ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 4 ( ) (298) (298) T ρ γ γ ρ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 1 (0) 1 r cr T T γ γ + ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 2) Guggenheim (1945) relationship Tcr is the imaginary critical temperature of the polymer 2/9 11 (0) 1 9 cr cr d T dT T T γ γ ⎛ ⎞ − = − ⎜ ⎟ ⎝ ⎠ --surface tension of a liquid decreases with temperature --surface tension vanishes at the critical point r=2/9 for low values of T/Tcr , dγ/dT will be constant June 29, 2005 21 Molecular mass dependence of surface tension The surface tension of homologous series tends to increase with increasing molecular weight The bulk properties (e.g. heat capacity, tensile strength, specific volume etc.) vary linearly with the reciprocal of molecular weight The surface tension, instead, varies linearly with Mn -2/3 At infinite molecular weight, surface tension is finite, γ∞ 2/3 e n k M γ γ ∞ = − Polymer γ∞ ke 36 30 21 43 polyethylene 386 polystyrene 373 Polydimethylsiloxane 166 polyethylene oxide 343 ke is a constant “Polymer Interface and Adhesion” Wu S. June 29, 2005 22 3. Estimation of interfacial surface energy The additional free energy at the interface between two condensed phases is known as interfacial energy ( ) 1/3 2 1/3 1/3 4( ) s l s l VV V V Φ = + cos 0 sl s l γ γ γ θ − + = 1/2 2 ( ) sl s l s l γ γ γ γ γ = + −Φ eliminating γs ( ) 2 2 1 cos cos 4 sl l θ γ γ θ ⎛ ⎞ + = − ⎜ ⎟ ⎜ ⎟ Φ ⎝ ⎠ By means of this equation, interfacial tension can be calculated from measurements of contact angle and surface tension of liquid June 29, 2005 23 Estimation of interfacial tension Lorentz-Berthelot Mixing Rules interaction energy between unlike molecules ab A = , interaction energy between like molecules aa bb A A = ( ) 1/ 2 1 ab aa bb A A A = Girifalco and Good (1956) ( ) 1/3 2 1/3 1/3 4( ) s l s l VV V V Φ = + ( ) 1/ 2 ab aa bb F F F ∆ − = Φ ∆ ∆ free energy of adhesion ab F ∆ = free energy of cohesion aa F ∆ = interaction parameter Φ = γa γb γab γa γa ab ab b a F γ γ γ ∆ = − − 2 aa a F γ ∆ = − adhesion cohesion 1/2 2 ( ) ab a b a b γ γ γ γ γ = + −Φ June 29, 2005 24 General expression for the interfacial tension ( ) 1/ 2 2 1/ 2 1/ 2 2 ( ) since 1for most polymers sl s l s l sl l s γ γ γ γ γ γ γ γ = + −Φ ≈ − Φ ≈ Equation is only valid for substances interacting with additive dispersive forces and without hydrogen bonds. (Fowkes, 1964) Fowkes has suggested that total free energy at a surface is the sum of contributions from the different intermolecular forces at the surface. d and a refer to the dispersion forces and a-scalar forces (the combined polar interactions: dipole, induction, and hydrogen bonding d a γ γ γ = + ( ) ( ) 1/2 1/2 2 2 d d a a sl s l l s l s γ γ γ γ γ γ γ = + − − ( ) 1/2 2 d d sl s l l s γ γ γ γ γ = + − for apolar liquids/solids Owens and Wendt extended the formulation June 29, 2005 25 General expression for the interfacial tension ( ) ( ) ( ) ( ) 2 2 1/2 1/2 1/2 1/2 12 1 2 1 2 d d a a γ γ γ γ γ ⎡ ⎤ ⎡ ⎤ = − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ Substances 1 and 2 may either be liquids, or solids, or they may be a combination of a solid and a liquid Liquid n-hexane 18.4 18.4 0 cyclohexane 25.5 25.5 0 glycol 48 33.8 14.2 methylene iodide 50.8 49.5 1.3 glycerol 63.4 37.0±4 26 water 72.8 21.8±0.7 51 1 γ 1 d γ 1 a γ ( ) ( ) ( ) ( ) 2 2 1/ 2 1/ 2 1/ 2 1/ 2 12 1 2 1 2 d d a a γ γ γ γ γ ⎡ ⎤ ⎡ ⎤ = − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ( ) ( ) ( ) ( ) 2 2 1/ 2 1/ 2 1/ 2 1/ 2 12 1 2 1 2 d d a a γ γ γ γ γ ⎡ ⎤ ⎡ ⎤ = − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ( ) ( ) ( ) ( ) 2 2 1/2 1/2 1/2 1/ 2 12 1 2 1 2 d d a a γ γ γ γ γ ⎡ ⎤ ⎡ ⎤ = − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ June 29, 2005 26 Calculation of force components of surface tension Liquid 1 (e.g. water): 1 1 , a d γ γ ( ) ( ) ( ) ( ) 2 2 1/2 1/2 1/2 1/2 12 1 2 1 2 d d a a γ γ γ γ γ ⎡ ⎤ ⎡ ⎤ = − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ • Measure the surface tension of this liquid • Take another liquid (e.g. cyclohexane) which is immiscible with Liquid 1 and is apolar • Measure the interfacial tension by one of the available methods • Solve the following the two equations simultaneously 2 2 2 2 2 ( 0; 25mJ/m ; 25mJ/m ) a d γ γ γ = = → = 1 1 1 d a γ γ γ = + 2 1 ( 72.3 mJ/m ) γ = 2 12 ( 50.2 mJ/m ) γ = ( ) 2 2 1/ 2 , 50.2 25 72 2 25 d H O ch H O γ γ = = + − 2 2 22.7, 72 22.7 49.3 d a H O H O γ γ = = − = June 29, 2005 27 Conclusions • The specific energy of a polymer can be estimated by means of an additive quantity, the parachor • It may be alternatively calculated from different related material properties e.g. cohesive energy, compressibility, solubility parameter • Rules are given for the estimation of the interfacial tension and the contact angle of a liquid on a solid June 29, 2005 28 Extra slides: estimation of Φ ( ) 1/3 2 1/3 1/3 4( ) s l s l VV V V Φ = + ( ) 2 ab ab b j a z j f F z n d f df n dr r ε π ∞ ∞ ∞ ∂ = ∂ ∫ ∫ ∫ ( ) 2 ab ab b j a z x j f F z dx n d f df n dr r ε π ∞ ∞ ∞ ∞ ∂ −∆ = ∂ ∫ ∫ ∫ ∫ 6 ab ab ab m A B r r ε = − + Lennnard-Jones potential 7 1 6 ab ab ab m A mB r r r ε + ∂ = − ∂ Interaction force Carrying out the integrations and equating the net force to 0 when z=dAB (equilibrium distance) 2 1 1 6 2 4 a b ab ab ab n n A F d m ⎛ ⎞ ∆ = − ⎜ ⎟ − ⎝ ⎠ 2 2 1 1 6 2 4 a aa aa aa n A F d m ⎛ ⎞ ∆ = − ⎜ ⎟ − ⎝ ⎠ ( ) 1/ 2 ab aa bb F F F ∆ − = Φ ∆ ∆ ( ) 2 2 aa bb ab ab aa bb d d A d A A Φ = × Girifalco and Good (1956) June 29, 2005 29 1) Floating needles l end view of needle W F1 F2 (F1+F2) cosθ θ F1=F2=γl
189180	https://www.khanacademy.org/science/generations-old-and-new/x74478e8d979438e2:reproduction/x74478e8d979438e2:events-of-sexual-reproduction/v/germ-cells-gametes-sexual-reproduction-how-do-organisms-reproduce-biology-khan-academy	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
189181	https://www.maths.tcd.ie/~dwilkins/Courses/MA1S11Calc/MA1S11Calc_Mich2016_CourseNotes/MA1S11Calc_Mich2016_Section06.pdf	Module MA1S11 (Calculus) Michaelmas Term 2016 Section 6: Trigonometric Functions and their Derivatives D. R. Wilkins Copyright c ⃝David R. Wilkins 2016 Contents 6 Trigonometrical Functions and their Derivatives 133 6.1 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 133 6.2 Periodicity of the Trigonometrical Functions . . . . . . . . . . 135 6.3 Values of Trigonometric Functions at Particular Angles . . . . 136 6.4 Addition Formulae satisﬁed by the Sine and Cosine Functions 137 6.5 Derivatives of Trigonometrical Functions . . . . . . . . . . . . 140 6.6 The Inverse Tangent Function . . . . . . . . . . . . . . . . . . 146 6.7 The Inverse Sine and Cosine Functions . . . . . . . . . . . . . 147 i 6 Trigonometrical Functions and their Deriva-tives 6.1 Trigonometric Functions There are six standard trigonometric functions. They are the sine function (sin), the cosine function (cos), the tangent function (tan), the cotangent function (cot), the secant function (sec) and the cosecant function (csc). Angles will always be represented in the following discussion using radian measure. If one travels a distance s around a circle of radius r, then the angle subtended by the starting and ﬁnishing positions at the centre of the circle is s/r radians. The standard trigonometrical functions represent ratios of sides of right-angled triangles, as indicated in the following diagrams. A B C θ \|AC\| \|AC\| cos θ \|AC\| sin θ In the above triangle ABC, in which the angle at the vertex B is a right angle, the lengths \|AC\|, \|AB\| and \|BC\| satisfy the identities \|AB\| = \|AC\| cos θ, \|AB\| = \|AC\| sin θ, where θ denotes the angle of the triangle at the vertex A. A B C θ \|AB\| \|AB\| sec θ \|AB\| tan θ 133 In the above triangle ABC, in which the angle at the vertex B is a right angle, the lengths \|AC\|, \|AB\| and \|BC\| satisfy the identities \|BC\| = \|AB\| tan θ, \|AC\| = \|AB\| sec θ, where θ denotes the angle of the triangle at the vertex A. A B C θ \|BC\| \|BC\| cot θ \|BC\| csc θ In the above triangle ABC, in which the angle at the vertex B is a right angle, the lengths \|AC\|, \|AB\| and \|BC\| satisfy the identities \|AB\| = \|BC\| cot θ, \|AC\| = \|BC\| csc θ, where θ denotes the angle of the triangle at the vertex A. The identities described above that determine the ratios of the sides of a right angled triangle are summarized in the following proposition. Proposition 6.1 Let ABC be a triangle in which the angle at B is a right angle, and let θ denote the angle at A. Then the lengths \|AB\|, \|BC\| and \|AC\| of the sides AB, BC and AC respectively satisfy the following identities:— \|AB\| = \|AC\| cos θ, \|BC\| = \|AC\| sin θ; \|BC\| = \|AB\| tan θ, \|AC\| = \|AB\| sec θ; \|AB\| = \|BC\| cot θ, \|AC\| = \|BC\| csc θ. The following trigonometrical formulae follow directly from the results stated in Proposition 6.1. Proposition 6.2 The tangent, cotangent, secant and cosecant functions are determined by the sine and cosine functions in accordance with the following identities:— tan θ = sin θ cos θ; cot θ = cos θ sin θ ; sec θ = 1 cos θ; csc θ = 1 sin θ. 134 Proposition 6.3 The sine and cosine functions are related by the following relationship, when angles are speciﬁed using radian measure:— sin θ = cos( 1 2π −θ); cos θ = sin( 1 2π −θ). Proof The trigonometrical functions are determined by ratios of edges of a right angled triangle ABC in which the angle B is a right angle and the angle A is θ radians. The angles of a triangle add up to two right angles, and two right angles are equal to π in radian measure. Thus if ∠B denotes the angle of the right-angled triangle ABC then ∠A + ∠B + ∠C = π, and thus θ + 1 2π + ∠C = π, and therefore C = 1 2π −θ. The result then follows from the deﬁnitions of the sine and cosine functions. The nth powers of trigonometric functions are usually presented using the following traditional notation, in instances where n is a positive integer:— sinn θ = (sin θ)n, cosn θ = (cos θ)n, tann θ = (tan θ)n, etc. Proposition 6.4 The trigonometric functions satisfy the following identities:— sin2 θ + cos2 θ = 1; 1 + tan2 θ = sec2 θ; 1 + cot2 θ = csc2 θ; Proof These identities follow from the deﬁnitions of the trigonometric func-tions on applying Pythagoras’ Theorem. 6.2 Periodicity of the Trigonometrical Functions Suppose that a particle moves with speed v around the circumference of a circle of radius r, where that circle is represented in Cartesian coordinates by the equation x2 + y2 = r2. The centre of the circle is thus at the origin of the Cartesian coordinate system. We suppose that the particle travels in an anticlockwise direction 135 and passes through the point (r, 0) when t = 0. Then the particle will be at the point r cos vt r , r sin vt r . at time t. The quantities sin θ and cos θ are deﬁned for all real numbers θ so that the above formula for the position of the particle moving around the circumference of the circle at a constant speed remains valid for all times. Now the particle moving round the circumference of the circle of radius r with speed v will complete each revolution in time 2πr v . Thus cos(θ + 2π) = cos θ and sin(θ + 2π) = sin θ for all real numbers θ. It follows that cos(θ + 2nπ) = cos θ and sin(θ + 2nπ) = sin θ for all real numbers θ and for all integers n. These equations express the periodicity of the sine and cosine functions. −1 1 x y = sin x 1 2π π 3 2π 2π y −1 1 x y = cos x 1 2π π 3 2π 2π y 6.3 Values of Trigonometric Functions at Particular Angles The following table sets out the values of sin θ and cos θ for some angles θ that are multiples of 1 2π:— 136 θ −π −1 2π 0 1 2π π 3 2π 2π 5 2π sin θ 0 −1 0 1 0 −1 0 1 cos θ −1 0 1 0 −1 0 1 0 The following values of the sine and cosine functions can be derived using geometric arguments involving the use of Pythagoras’ Theorem:— θ 0 1 6π 1 4π 1 3π 1 2π sin θ 0 1 2 1 √ 2 √ 3 2 1 cos θ 1 √ 3 2 1 √ 2 1 2 0 6.4 Addition Formulae satisﬁed by the Sine and Cosine Functions We derive the standard addition formulae for trigonometric functions by considering the formulae that implement rotations of the plane about some origin chosen within that plane. An anticlockwise rotation about the origin through an angle of θ radians sends a point (x, y) of the plane to the point (x′, y′), where x′ = x cos θ −y sin θ y′ = x sin θ + y cos θ (This follows easily from the fact that such a rotation takes the point (1, 0) to the point (cos θ, sin θ) and takes the point (0, 1) to the point (−sin θ, cos θ).) An anticlockwise rotation about the origin through an angle of ϕ radians then sends the point (x′, y′) of the plane to the point (x′′, y′′), where x′′ = x′ cos ϕ −y′ sin ϕ y′′ = x′ sin ϕ + y′ cos ϕ Now an anticlockwise rotation about the origin through an angle of θ + ϕ radians sends the point (x, y), of the plane to the point (x′′, y′′), and thus x′′ = x cos(θ + ϕ) −y sin(θ + ϕ) y′′ = x sin(θ + ϕ) + y cos(θ + ϕ) But if we substitute the expressions for x′ and y′ in terms of x, y and θ obtained previously into the above equation, we ﬁnd that x′′ = x(cos θ cos ϕ −sin θ sin ϕ) −y(sin θ cos ϕ + cos θ sin ϕ) y′′ = x(sin θ cos ϕ + cos θ sin ϕ) + y(cos θ cos ϕ −sin θ sin ϕ) 137 On comparing equations, we see that cos(θ + ϕ) = cos θ cos ϕ −sin θ sin ϕ, and sin(θ + ϕ) = sin θ cos ϕ + cos θ sin ϕ. On replacing ϕ by −ϕ, and noting that cos(−ϕ) = cos ϕ and sin(−ϕ) = −sin ϕ, we ﬁnd that cos(θ −ϕ) = cos θ cos ϕ + sin θ sin ϕ, and sin(θ −ϕ) = sin θ cos ϕ −cos θ sin ϕ. We have therefore established the addition formulae for the sine and co-sine functions stated in the following proposition. Proposition 6.5 The sine and cosine functions satisfy the following identi-ties for all real numbers θ and ϕ:— sin(θ + ϕ) = sin θ cos ϕ + cos θ sin ϕ; sin(θ −ϕ) = sin θ cos ϕ −cos θ sin ϕ; cos(θ + ϕ) = cos θ cos ϕ −sin θ sin ϕ; cos(θ −ϕ) = cos θ cos ϕ + sin θ sin ϕ. Remark The equations describing how Cartesian coordinates of points of the plane transform under rotations about the origin may be written in ma-trix form as follows: x′ y′ = cos θ −sin θ sin θ cos θ x y , x′′ y′′ = cos ϕ −sin ϕ sin ϕ cos ϕ x′ y′ . Also equation (6.4) may be written x′′ y′′ = cos(θ + ϕ) −sin(θ + ϕ) sin(θ + ϕ) cos(θ + ϕ) x y . It follows from basic properties of matrix multiplication that cos(θ + ϕ) −sin(θ + ϕ) sin(θ + ϕ) cos(θ + ϕ) = cos ϕ −sin ϕ sin ϕ cos ϕ cos θ −sin θ sin θ cos θ . 138 Therefore cos(θ + ϕ) = cos θ cos ϕ −sin θ sin ϕ sin(θ + ϕ) = sin θ cos ϕ + cos θ sin ϕ. This provides an alternative derivation of the addition formulae stated in Proposition 6.5. Corollary 6.6 The sine and cosine functions satisfy the following identities for all real numbers θ:— sin(θ + 1 2π) = cos θ, cos(θ + 1 2π) = −sin θ, sin(θ + π) = −sin θ, cos(θ + π) = −cos θ, Proof These results follow directly on applying Proposition 6.5 in view of the identities sin 1 2π = 1, cos 1 2π = 0, sin π = 0 and cos π = −1. The formulae stated in the following corollary follow directly from the addition formulae stated in Proposition 6.5 on adding and subtracting those addition formulae. Corollary 6.7 The sine and cosine functions satisfy the following identities for all real numbers θ and ϕ:— sin θ sin ϕ = 1 2(cos(θ −ϕ) −cos(θ + ϕ)); cos θ cos ϕ = 1 2(cos(θ + ϕ) + cos(θ −ϕ)); sin θ cos ϕ = 1 2(sin(θ + ϕ) + sin(θ −ϕ)). Corollary 6.8 The sine and cosine functions satisfy the following identities for all real numbers θ:— sin 2θ = 2 sin θ cos θ; cos 2θ = cos2 θ −sin2 θ = 2 cos2 θ −1 = 1 −2 sin2 θ. 139 Proof The formula for sin 2θ and the ﬁrst formula for cos 2θ follow from the identities stated in Proposition 6.5 on setting ϕ = θ in the formulae for sin(θ + ϕ) and cos(θ + ϕ). The second and third formulae for cos 2θ then follow on making use of the identity sin2 θ + cos2θ = 1. The following formulae then follow directly from those stated in Corol-lary 6.8. Corollary 6.9 The sine and cosine functions satisfy the following identities for all real numbers θ:— sin2 θ = 1 2(1 −cos 2θ); cos2 θ = 1 2(1 + cos 2θ). 6.5 Derivatives of Trigonometrical Functions Lemma 6.10 Let ε be a positive real number. Then there exists some pos-itive real number δ satisfying 0 < δ < 1 2π with the property that 1 −ε < cos θ < 1 whenever 0 < θ < δ. Proof Choose a real number u satisfying 0 < u < 1 for which 1−ε ≤u. Let a right-angled triangle OFG be constructed so that the angle at F is a right angle, \|OF\| = u and \|FG\| = √ 1 −u2, and let δ be the angle of this triangle at the vertex O. Then \|OG\|2 = \|OF\|2 + \|FG\|2 = 1, and therefore u = cos δ. It follows that if θ is a positive real number satisfying 0 < θ < δ then 1 −ε ≤cos δ < cos θ < 1. The result follows. O F G δ \|OG\| = 1 \|OF\| = u 140 Proposition 6.11 Let sin: R →R be the sine function whose value sin θ, for a given real number θ is the sine of an angle of θ radians. Then lim θ→0 sin θ θ = 1. Proof Let a circle of unit radius pass through points A and B, so that the angle θ in radians between the line segements OA and OB at the centre O of the circle satisﬁes the inequalities 0 < θ < 1 2π. Let C be the point on the line segment OA for which the angle OCB is a right angle, and let the line OB be produced to the point D determined so that the angle OAD is a right angle. O A B C D θ The sector OAB of the unit circle is by deﬁnition the region bounded by the arc AB of the circle and the radii OA and OB. Now the area of a sector of a circle subtending at the centre an angle of θ radians is equal to the area of the circle multiplied by θ 2π. But the area of a circle of unit radius is π. It follows that a sector of the unit circle subtending at the centre an angle of θ radians has area 1 2θ. O A B D θ The the area of a triangle is half the base of the triangle multiplied by the height of the triangle. The base \|OA\| and height \|BC\| of the triangle 141 AOB satisfy \|OA\| = 1, \|BC\| = sin θ. It follows that area of triangle OAB = 1 2 × \|OA\| × \|BC\| = 1 2 sin θ. O A B C D θ \|OA\| = 1 \|BC\| = sin θ Also the base \|OA\| and height \|AD\| of the triangle AOD satisfy \|OA\| = 1, \|AD\| = sin θ cos θ. It follows that area of triangle OAD = 1 2 × \|OA\| × \|AD\| = sin θ 2 cos θ. O A B D θ \|OA\| = 1 \|AD\| = sin θ cos θ The results concerning areas just obtained can be summarized as follows:— area of triangle OAB = 1 2 × \|OA\| × \|BC\| = 1 2 sin θ, 142 area of sector OAB = θ 2π × π = 1 2θ, area of triangle OAD = 1 2 × \|OA\| × \|AD\| = 1 2 tan θ = sin θ 2 cos θ. Moreover the triangle OAB is strictly contained in the sector OAB, which in turn is strictly contained in the triangle OAD. It follows that area(△OAB) < area(sector OAB) < area(△OAD), and thus 1 2 sin θ < 1 2θ < sin θ 2 cos θ for all real numbers θ satisfying 0 < θ < 1 2π. O A B C D θ Multiplying by 2, and then taking reciprocals, we ﬁnd that 1 sin θ > 1 θ > cos θ sin θ for all real numbers θ satisfying 0 < θ < 1 2π. If we then multiply by sin θ, we obtain the inequalities cos θ < sin θ θ < 1, for all real numbers θ satisfying 0 < θ < 1 2π. Now, given any positive real number ε, there exists some positive real number δ satisfying 0 < δ < 1 2π such that 1 −ε < cos θ < 1 whenever 0 < θ < δ (see Lemma 6.10). But then 1 −ε < sin θ θ < 1 143 whenever 0 < θ < δ. These inequalities also hold when −δ < θ < 0, because the value of sin θ θ is unchanged on replacing θ by −θ. It follows that lim θ→0 sin θ θ = 1, as required. Corollary 6.12 Let cos: R →R be the cosine function whose value cos θ, for a given real number θ is the cosine of an angle of θ radians. Then lim θ→0 1 −cos θ θ = 0. Proof Basic trigonometrical identities ensure that 1 −cos θ = 2 sin2 1 2θ and sin θ = 2 sin 1 2θ cos 1 2θ for all real numbers θ (see Corollary 6.8 and Corollary 6.9). Therefore 1 −cos θ sin θ = sin 1 2θ cos 1 2θ = tan 1 2θ for all real numbers θ. It follows that lim θ→0 1 −cos θ sin θ = lim θ→0 tan 1 2θ = 0, and therefore lim θ→0 1 −cos θ θ = lim θ→0 1 −cos θ sin θ × lim θ→0 sin θ θ = 0 × 1 = 0, as required. Proposition 6.13 The derivatives of the sine and cosine functions satisfy d dx (sin x) = cos x, and d dx (cos x) = −sin x. Proof Limits of sums, diﬀerences and products of functions are the cor-responding sums, diﬀerences and products of the limits of those functions, provided that those limits exist (see Proposition 4.17). Also sin(x + h) = sin x cos h + cos x sin h and cos(x + h) = cos x cos h −sin x sin h 144 for all real numbers h (see Proposition 6.5). Applying these results, together with those of Proposition 6.11 and Corollary 6.12, we see that d dx (sin x) = lim h→0 sin(x + h) −sin x h = lim h→0 sin x cos h + cos x sin h −sin x h = cos x lim h→0 sin h h −sin x lim h→0 1 −cos h h = cos x. Similarly d dx (cos x) = lim h→0 cos(x + h) −cos x h = lim h→0 cos x cos h −sin x sin h −cos x h = −sin x lim h→0 sin h h −cos x lim h→0 1 −cos h h = −sin x, as required. Corollary 6.14 The derivative of the tangent function satisﬁes d dx (tan x) = 1 cos2 x = sec2 x. Proof Using the formulae for the derivatives of the sine and cosine func-tions (Proposition 6.13), together with the Quotient Rule for diﬀerentiation (Proposition 5.4) we ﬁnd that d dx (tan x) = d dx sin x cos x = 1 cos2 x d dx (sin x) cos x −d dx (cos x) sin x = cos2 x + sin2 x cos2 x = 1 cos2 x = sec2 x as required. 145 6.6 The Inverse Tangent Function Deﬁnition The inverse tangent function arctan: R →(−1 2π, 1 2π) is deﬁned so that, given any real number x, the quantity arctan x is the unique angle (speciﬁed in radian measure) for which tan(arctan x)) = x. Thus the inverse tangent function is the unique function mapping the set R of real numbers into the interval (−1 2π, 1 2π) with the property that tan(arctan x)) = x for all real numbers x. Lemma 6.15 The inverse tangent function arctan: R →(−1 2π, 1 2π) is continuous. Proof Let s be a real number, and let β = arctan s. Let some positive real number ε be given. Then real numbers α and γ can be chosen so that −1 2π < α < 1 2π, −1 2π < γ < 1 2π, and β −ε < α < β < γ < β + ε. Let u = tan α and v = tan γ. The tangent function is increasing. It follows that the inverse tangent function is also increasing. Thus if x is a real number satisfying u < x < v then α < arctan x < γ. Let δ be the smaller of the positive numbers v −s and s −u. if x is a real number satisfying s −δ < x < s + δ then u < x < v. But then arctan s −ε < α < arctan x < γ < arctan s + ε. Thus the inverse tangent function arctan is continuous at s, as required. Proposition 6.16 The inverse tangent function arctan is diﬀerentiable, and d dx (arctan x) = 1 1 + x2 for all real numbers x. 146 Proof Let s be a real number. Then there exists a real number β satisfying −1 2π < β < 1 2π for which tan β = s. Let Q: (−1 2π, 1 2π) →R be deﬁned such that Q(θ) =      tan θ −tan β θ −β if θ ̸= β; 1 cos2 β if θ = β. Now the continuity of the tangent function, together with standard theorems on continuity, ensures that Q(θ) is a continuous function of θ when θ ̸= β. The function Q is also continuous at β because lim θ→β Q(θ) = lim θ→β tan θ −tan β θ −β = d dθ(tan θ) θ=β = 1 cos2 β = Q(β) (see Corollary 6.14). It follows that the function Q(θ) is continuous at θ for all values of θ satisfying −1 2π < θ < 1 2π. Now the inverse tangent function is continuous (Lemma 6.15), and com-positions of continuous functions are continuous (Proposition 4.26). It follows that Q(arctan x) is a continuous function of x. Also tan β = s. Therefore lim x→s Q(arctan x) = Q(arctan s) = Q(β) = 1 cos2 β = 1 + tan2 β = 1 + s2. (see Proposition 6.4). But lim x→s Q(arctan x) = lim x→s x −s arctan x −arctan s. It follows that d dx (arctan x) x=s = lim x→s arctan x −arctan s x −s = lim x→s 1 Q(arctan x) = 1 1 + s2. Thus the inverse tangent function is diﬀerentiable, and moreover its deriva-tive at any real number x is equal to 1 1 + x2, as required. 6.7 The Inverse Sine and Cosine Functions Deﬁnition The inverse sine function arcsin: [−1, 1] →[−1 2π, 1 2π] 147 is deﬁned so that, given any real number x satisfying −1 ≤x ≤1, the quantity arcsin x is the unique angle (speciﬁed in radian measure) satisfying the inequalities −1 2π ≤arcsin x ≤1 2π for which sin(arcsin x)) = x. Deﬁnition The inverse cosine function arccos: [−1, 1] →[0, π] is deﬁned so that, given any real number x satisfying −1 ≤x ≤1, the quantity arccos x is the unique angle (speciﬁed in radian measure) satisfying the inequalities 0 ≤arcsin x ≤π for which cos(arccos x)) = x. The inverse sine and cosine functions are related by the identity arccos x = 1 2π −arcsin x for all real numbers x satisfying −1 ≤x ≤1. We explore the relationship between the inverse sine and inverse tangent functions. Lemma 6.17 The inverse sine function arcsin is a diﬀerentiable function of x on the interval (−1, 1) which satisﬁes the identity arcsin x = arctan x √ 1 −x2 . when −1 < x < 1. Proof Let x be a real number satisfying −1 < x < 1 and let θ be the unique real number in the range −1 2π < θ < 1 2π that satisﬁes sin θ = x. Then cos θ = √ 1 −x2, and therefore tan θ = x √ 1 −x2. It follows that arcsin x = θ = arctan x √ 1 −x2 . Now it follows from the Chain Rule (Proposition 5.5) that any composition of diﬀerentiable functions is diﬀerentiable. Therefore the inverse sine func-tion is diﬀerentiable at x for all real numbers x satisfying −1 < x < 1, as required. 148 Proposition 6.18 The inverse sine function arcsin satisﬁes d dx (arcsin x) = 1 √ 1 −x2 for all real numbers x satisfying −1 < x < 1. Proof Diﬀerentiating the identity sin(arcsin x) = x with respect to x using the Chain Rule (Proposition 5.5), we ﬁnd that cos(arcsin x) d dx (arcsin x) = 1. Let θ = arcsin x. Then x = sin θ. It follows that cos(arcsin x) = cos θ = p 1 −sin2 θ = √ 1 −x2, and therefore d dx (arcsin x) = 1 cos(arcsin x) = 1 √ 1 −x2, as required. Remark The inverse sine and tangent functions are related by the identity arcsin x = arctan x √ 1 −x2 . when −1 < x < 1. Diﬀerentiating the right hand side of this identity using the Chain Rule, and using the result that d dx (arctan x) = 1 1 + x2 (Proposition 6.16), we ﬁnd that d dx arctan x √ 1 −x2 = 1 1 + x2 1 −x2 d dx x √ 1 −x2 = 1 −x2 (1 −x2) + x2 × (1 −x2) −1 2(−2x)x (1 −x2) 3 2 = (1 −x2) × 1 (1 −x2) 3 2 = 1 √ 1 −x2. 149 This agrees with the formulae already found for the derivative of the inverse sine function (Proposition 6.18). The inverse cosine function satisﬁes the identity arccos x = 1 2π −arcsin x. It therefore follows from Proposition 6.18 that d dx (arccos x) = −d dx (arcsin x) = − 1 √ 1 −x2 for all real numbers x satisfying −1 < x < 1. 150
189182	https://www.youtube.com/watch?v=EwOUmojITss	Arithmetic Operations (Math Tutorial. Arithmetic #2- part 1) Open Math Camp 3130 subscribers 160 likes Description 11637 views Posted: 29 Jan 2022 Full Course of Arithmetic (Playlist): Arithmetic Problems (Playlist): Description: This Tutorial covers next topics: Basic Arithmetic Operations: - Addition and Subtraction, - Multiplication and Division, - Raising Number to a Power and Root of a Number. Like us and Subscribe! Best Regards, Dr.RFZ ARITHMETIC PLAYLIST CONTENT: 1.Numbers (part 1) 1.Numbers (part 2) 2.Arithmetic Operations (part 1) 2.Arithmetic Operations (part 2) 3.Order of Arithmetic Operations 4.Decimal Positional Numeral System 5.Divisibility Criteria for 2,3,4,5,6,7,8 and 9 (part 1) 5.Divisibility Criteria for 2,3,4,5,6,7,8 and 9 (part 2) 5.Divisibility Criteria for 2,3,4,5,6,7,8 and 9 (part 3) 6.Prime and Composite Numbers. Sieve of Eratosthenes 7.Prime number Factorization 8.Greatest Common Divisor, co-prime numbers 9.Least Common Multiple 10.Fractions. Proper and Improper Fractions (part 1) 10.Fractions. Proper and Improper Fractions (part 2) 11.Reducing Fractions 12.Comparing Fractions 13.Adding and Subtracting Fractions (part 1) 13.Adding and Subtracting Fractions (part 2) 14.Multiplication and Division of Fractions 15.Operations Involving Zero 16.Decimal Fractions and its Properties 17.Addition and Subtraction of Decimal Fractions 18.Division of Decimal Fraction 19.Multiplication of Decimal Fractions 20.Percentage 21.Rounding Off. Absolute and Relative Errors 2 comments Transcript: Intro hello my friends welcome to openmathcamp we continue our series of tutorials on arithmetic and today we are going to talk about basic arithmetic operations arithmetic operations such as addition subtraction multiplication and division also we will learn arithmetic operations as raising a number to a power and finding the root of the number let's get started like us and subscribe to see new videos on openmathcamp Arithmetic Operations arithmetic operations addition subtraction multiplication and division are main operations in arithmetic and these operations let's denote our operation as a star so this operation can be any of these four arithmetic operations in this case assume we have two numbers a and b where a and b is are an element of real line or a and b are real numbers once again this symbol in mathematics stands for element off so assume a and b are real numbers and when we perform arithmetic operation with this two number in this case the result will be also some number c also real number and only exception is division by zero because in mathematics if we divide the number by zero this operation is meaningless assume we have a number a which is not 0 and a is an element of it this is real any number which is not zero and we write a over zero and this operation is meaningless because if we assume this is equal to b then we have if we multiply the both side of this to zero we have a equal to b multiplied to zero but we have left part not equal to zero but the right part is zero because b is multiplied to zero and the result will be zero so that's why this is the meaningless and division by zero is also meaningless operation let's talk about first talk about addition and subtraction these two operations addition and subtraction are opposite operation and we will talk about this later let's first about let's first talk about definition actually addition and subtraction have no definition and we can describe addition as the process of combining two num two numbers say nine plus two equal to eleven and these two elements nine and two two numbers called add-ins and eleven is called sum another example is 7.5 plus 3 equal to equals to 10 10 and 10.5 and this operation can be described using the real line assuming we have a real line we discussed real line in the previous session as soon we have a point seven and five then if we add number three to this point then we have to move to the right side of the real line by the amount of tree and result will be ten 0.5 so um plus is moving to the right side of the of the real line uh when we perform subtraction and we write 7 and 5 minus 3 in this case this is this can be described as moving to the left side of the real line minus three and it will be obviously four and five four point five so this is a description of this operation and we cannot define this operation it's very hard to define this next example assume we have two minus five in this case if we have a two and here we have a zero then minus five it can be described as moving to the left side by the amount of five one two three four five one two five and this will be minus three so this is obvious operation and as we already said the result of this operation is also real number it's important to understand that addition addition and subtraction are these two operations are opposite operations what we mean by this assume we have two then if we write plus three we will have five and opposite operation to the plus is minus and we can go back writing minus 3. so 2 plus 3 is 5 and 5 minus 3 is again initial point which is 2. so plus and minus are opposite operation and further we will talk about multiplication and division and they are also opposite opposite operation operations to each other Multiplication let's further describe multiplication and division these two operations multiplication first let's talk about the multiplication multiplication is i will give a definition which is quite easy to understand and to remember this is like several several addition several additions and multiplication is like if we add the number to itself several times for example if we write if we write two times three this means two three times two plus two plus two we have to add 2 to itself and this is done 3 times so 2 plus 2 plus 2. three twos we have two two two three times and this will be six this will be six along the same line if we write five times two then this is five plus five and this is done two times which is equal to ten along the same line four times five is equal to four plus four plus four plus four and plus four and this is done five times which is equal to twenty in this case four is called multiplicand multi applicant 5 is called multiplier and 20 is called product and product as i said is also an element of the real numbers set of real numbers next let's talk about division
189183	https://math.stackexchange.com/questions/725386/rewriting-quadratic-form-using-coordinate-transformation	linear algebra - Rewriting quadratic form using coordinate transformation - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Rewriting quadratic form using coordinate transformation Ask Question Asked 11 years, 6 months ago Modified11 years, 6 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I just finished a linear algebra exam and I could answer most of it except for the bonus question. From memory, it looked something like this: x 2 1−x 2 3−4 x 1 x 2+4 x 2 x 3 x 1 2−x 3 2−4 x 1 x 2+4 x 2 x 3 can be written as −3 y 2 2+3 y 2 3−3 y 2 2+3 y 3 2 using a well chosen coordinate transformation. Show it without actually doing that transformation, and afterwards with the transformation. I've tried doing a transformation, but you're supposed to be able to do this without doing one. I am lost. Any pointers? linear-algebra Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 24, 2014 at 22:54 Cameron Buie 105k 10 10 gold badges 106 106 silver badges 245 245 bronze badges asked Mar 24, 2014 at 22:34 ChronicleChronicle 103 4 4 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. The matrix for the quadratic form is ⎡⎣⎢1−2 0−2 0 2 0 2−1⎤⎦⎥[1−2 0−2 0 2 0 2−1] Its characteristic equation is (1−λ)(−λ)(−1−λ)−4(1−λ)+4(1+λ)=λ−λ 3−4+4 λ+4+4 λ=−(λ 3−9 λ)(1−λ)(−λ)(−1−λ)−4(1−λ)+4(1+λ)=λ−λ 3−4+4 λ+4+4 λ=−(λ 3−9 λ) so the eigenvalues are 0, +3 and -3. Under the orthogonal change of basis (which will exist because the matrix is symmetric), you would get the diagonal matrix ⎡⎣⎢0 0 0 0 3 0 0 0−3⎤⎦⎥[0 0 0 0 3 0 0 0−3] which corresponds to the quadratic form 3 y 2 2−3 y 2 3 3 y 2 2−3 y 3 2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 24, 2014 at 22:49 user137500user137500 655 3 3 silver badges 4 4 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Perhaps the intent of the question was to complete the square: x 2 1−x 2 3−4 x 1 x 2+4 x 2 x 3=(x 1−2 x 2)2−(x 3−2 x 2)2 x 1 2−x 3 2−4 x 1 x 2+4 x 2 x 3=(x 1−2 x 2)2−(x 3−2 x 2)2 and so on. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 24, 2014 at 22:40 DavidDavid 84.9k 9 9 gold badges 96 96 silver badges 166 166 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Just show that the eigenvalues match. Because of symmetry, the eigenvectors are always orthogonal. You can find eigenvalues without actually finding the coordinate frame y 2 y 2 and y 3 y 3. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 24, 2014 at 22:41 orionorion 16.2k 1 1 gold badge 34 34 silver badges 45 45 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra See similar questions with these tags. 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189184	https://math.stackexchange.com/questions/2893388/perpendicular-lines-and-vectors	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Perpendicular lines and vectors Ask Question Asked Modified 7 years, 1 month ago Viewed 352 times 2 $\begingroup$ My textbook is confusing me a little. Here is a worked example from my textbook: Line $l$ has the equation $\begin{pmatrix}3\ -1\ 0\end{pmatrix}+\lambda \begin{pmatrix}1\ -1\ 1\end{pmatrix}$ and point $A$ has co-ordinates $(3, 9, -2)$. Find the coordinates of point $B$ on $l$ so that $AB$ is perpendicular to $l$. $\vec{AB\cdot }\begin{pmatrix}1\ -1\ 1\end{pmatrix}=0$ $\vec{OB}=r=\begin{pmatrix}3+\lambda \ -1-\lambda \ \lambda \end{pmatrix}$ $\vec{AB}=\begin{pmatrix}\lambda \ -10-\lambda \ \lambda +2\end{pmatrix}$ $\begin{pmatrix}\lambda \ -10-\lambda \ \lambda +2\end{pmatrix}\cdot \begin{pmatrix}1\ -1\ 1\end{pmatrix}=0$ $3\lambda= -12, \lambda = -4$ Coordinates of $B$: $(-1, 3, -4)$ The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, $a\cdot b = 0 $ and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line $AB$ and then just the direction vector of line l? Or am I missing something as usual? geometry vectors Share edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Aug 24, 2018 at 18:05 ChxChx 40155 silver badges1616 bronze badges $\endgroup$ Add a comment \| 2 Answers 2 Reset to default 0 $\begingroup$ There is nothing special here the dot product $AB\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line. Note also that vector $AB$ is a direction vector for the line orthogonal to the given line in $B$ and passing through $A$. Share answered Aug 24, 2018 at 18:10 useruser 164k1414 gold badges8484 silver badges157157 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ So to get the coordinates of $B$ given the parametric equation of line $\ell$ and coordinates of $A$. You need $\vec{AB}$ and the direction vector of $\ell$, namely $\vec{u_{\ell}}$ to be orthogonal, i.e. $$\vec{AB}.\vec{u_{\ell}}=0$$ Share answered Aug 24, 2018 at 18:11 Ahmad BazziAhmad Bazzi 12.3k22 gold badges1616 silver badges3232 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry vectors See similar questions with these tags. 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189185	https://www.cuemath.com/algebra/pascals-triangle/	LearnPracticeDownload Pascal's Triangle Pascals triangle or Pascal's triangle is an arrangement of binomial coefficients in triangular form. It is named after the French mathematician Blaise Pascal. The numbers in Pascal's triangle are placed in such a way that each number is the sum of two numbers just above the number. Pascals triangle is used widely in probability theory, combinatorics, and algebra. Generally, we can use Pascal's triangle to find the coefficients of binomial expansion, to find the probability of heads and tails in a toss, in combinations of certain things, etc. Let us discuss Pascals triangle in detail in the following section. \| \| \| --- \| \| 1. \| What is Pascal's Triangle? \| \| 2. \| Pascals Triangle Explained \| \| 3. \| Pascal's Triangle Formula \| \| 4. \| Pascal's Triangle Binomial Expansion \| \| 5. \| Pascals Triangle Probability \| \| 6. \| Pascals Triangle Pattern \| \| 7. \| FAQs on Pascals Triangle \| What is Pascal's Triangle? A pascal's triangle is an arrangement of numbers in a triangular array such that the numbers at the end of each row are 1 and the remaining numbers are the sum of the nearest two numbers in the above row. This concept is used widely in probability, combinatorics, and algebra. Pascal's triangle is used to find the likelihood of the outcome of the toss of a coin, coefficients of binomial expansions in probability, etc. Pascals Triangle Explained Pascals triangle or Pascal's triangle is a special triangle that is named after Blaise Pascal, in this triangle, we start with 1 at the top, then 1s at both sides of the triangle until the end. The middle numbers are so filled that each number is the sum of the two numbers just above it. The number of elements in the nth row is equal to (n + 1) elements. Pascal's triangle can be constructed by writing 1 as the first and the last element of a row and the other elements of the row are obtained from the sum of the two consecutive elements of the previous row. Pascal's triangle can be constructed easily by just adding the pair of successive numbers in the preceding lines and writing them in the new line. Pascals triangle or Pascal's triangle is shown in the image below. Here, we can see that any number is the sum of the two numbers just above that number. Pascal's Triangle Formula The formula to fill the number in the nth column and mth row of Pascal's triangle we use the Pascals triangle formula. The formula requires the knowledge of the elements in the (n-1)th row, and (m-1)th and nth columns. The elements of the nth row of Pascal's triangle are given by, nC0, nC1, nC2, ..., nCn. The formula for Pascal's triangle is: nCm = n-1Cm-1 + n-1Cm where nCm represents the (m+1)th element in the nth row. n is a non-negative integer, and 0 ≤ m ≤ n. Let us understand this with an example. If we want to find the 3rd element in the 4th row, this means we want to calculate 4C2. Then according to the formula, we get 4C2 = 4-1C2-1 + 4-1C2 ⇒ 4C2 = 3C1 + 3C2 So, this means we need to add the 2nd element in the 3rd row (i.e. 3) with the 3rd element in the 3rd row (i.e. 3.). So our answer will be 4C2 = 3 + 3 = 6 Pascal's Triangle Binomial Expansion Pascals triangle can also be used to find the coefficient of the terms in the binomial expansion. Pascal's triangle is a handy tool to quickly verify if the binomial expansion of the given polynomial is done correctly or not. Let us understand this with an example We know the expansion of (x+y)2 is x2 + 2xy + y2. If we write all the hidden terms and coefficient of this expansion, we can write that x2 + 2xy + y2 = 1x2 y0+ 2x1y1 + 1x0y2. And now if we check the elements in the second row of the Pascals triangle, we will find the numbers 1 2 1. This is the exact match of the coefficients of the terms in the expansion of (x+y)2. Now if we take the binomial expansion of the polynomial (x+y)n, we have the following expression. (x+y)n = a0 xn y0+ a1 xn-1 y1 + a2 xn-2 y2 + ... + an x0 yn We can get the am by using the binomial formula nCm = n-1Cm-1 + n-1Cm Pascals Triangle Probability Pascal's triangle can be used in various places in the field of mathematics. Pascal's triangle is used in probability, can be used to find the number of combinations, etc. It gives us the number of combinations of heads or tails that are possible from the number of tosses. For example, if we toss a coin two times, we get 1 time HH, 2 times HT or TH, and 1 time TT, which is the exact match of the elements in the second row of the Pascals triangle. Similarly, we get the following results in the various number of tosses: \| Number of Tosses or Row of Pascals Triangle \| Outcomes in Combinations \| Elements in Pascals Triangle \| --- \| 1 \| {H} {T} \| 1, 1 \| \| 2 \| {HH} {HT TH} {TT} \| 1, 2, 1 \| \| 3 \| {HHH} {HHT, HTH, THH} {HTT, THT, TTH} {TTT} \| 1, 3, 3, 1 \| \| 4 \| {HHHH} {HHHT, HHTH, HTHH, THHH} {HHTT, HTHT, HTTH, THHT, THTH, TTHH} {HTTT, THTT, TTHT, TTTH} {TTTT} \| 1, 4, 6, 4, 1 \| \| ...etc... \| ... etc ... \| etc... \| Pascal's Triangle Pattern Pascal's triangle has various patterns within the triangle which were found and explained by Pascal himself or were known way before him. A few of the Pascal triangle patterns are: The sum of values in the nth row is 2n. For example, in the 4th row 1 4 6 4 1, sum of the elements is 1 + 4 + 6 + 4 + 1 = 16 = 24. If a row has the second element a prime number, then all the following elements in the row are divisible by that prime number (not considering 1s). ex. 1 5 10 10 5 1. By adding the different diagonal elements of Pascal's triangle, we get the Fibonacci series. Few more patterns on the diagonal of a Pascal's triangle can be seen in the following image: Important Notes on Pascal's Triangle: The elements in the Pascals triangle can find out by finding the sum of the two adjoint elements in the preceding row. The sum of values in the nth row is 2n. Pascal's triangle is used to determine the coefficients of binomial expansions. Related Topics: Binomial Expansion Formula Algebraic Identities Variable Expressions Algebraic Formulas Read More Pascal's Triangle Examples Example 1: A coin is tossed three times, find the probability of getting exactly 2 tails. Solution: Using the Pascal triangle formula, the total number of outcomes will be 23 = 8 (1 + 3 + 3 + 1 = 8 ) Where 3 of them give exactly two tails. So the probability of getting exactly two tails is 3/8, or 37.5% Answer: The probability of getting exactly two tails is 37.5% 2. Example 2: Find all the coefficients of the expansion of the polynomial (x + y)6 using the Pascals triangle. Solution: According to the Pascals triangle, the coefficients of the expansion of the (x + y)6 will be the elements in row 6 of the Pascals triangle. Elements in the 6th row of the Pascals triangle are 1, 6, 15, 20, 15, 6, 1. Answer: The coefficients of the expansion of (x + y)6 are 1, 6, 15, 20, 15, 6, 1. 3. Example 3: Find the sum of the elements in the 20th row of the Pascals triangle. Solution: Using the Pascals triangle formula for the sum of the elements in the nth row of the Pascals triangle: Sum = 2n where n is the number of the row. Hence Sum = 220 Sum = 1048576 Answer: The sum of the elements in the 20th row is 1048576. View More > Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class Pascals Triangle Questions Try These! > FAQs on Pascal's Triangle What is Pascal's Triangle? A pascal's triangle is an arrangement of numbers in a triangular array such that the numbers at the end of each row are 1 and the remaining numbers are the sum of the nearest two numbers in the above row. What is Pascal's Triangle Used For? Pascals triangle can be used for various purposes in mathematics. It is used in the binomial expansion of a polynomial, in probability, to find the number of combinations, and can be used to find the Fibonacci series. Pascal's triangle is a very useful tool and has various properties that can be useful in various aspects of mathematics. What is The Rule for Pascal's Triangle? The rule that Pascal's triangle has is that we start with 1 at the top, then 1s at both sides of the triangle until the end. The middle numbers, each is the sum of the two consecutive numbers just above it. Hence to construct a Pascal's triangle we just need to add the two numbers just above the number. What are 3 Patterns In Pascal's Triangle? The three patterns in the Pascals triangle are: The sum of values in the nth row is 2n. For. example we have the values in the 4th row 1 4 6 4 1, and the sum of these elements is 1 + 4 + 6 + 4 + 1 = 16 = 24. If a row has the second element which is a prime number, then all the following elements in the row are divisible by that prime number(not considering 1s), for. ex. 1 5 10 10 5 1. By adding the different diagonals elements of a Pascals triangle, we get the Fibonacci series. How Do You Use Pascal's Triangle in Binomial Expansion? Pascals triangle can also be used to find the coefficient of the terms in the binomial expansion. Let us understand this with an example. We know the expansion of (x+y)2 is x2 + 2xy + y2 If we write all the hidden terms and coefficient of this expansion, we can write that x2 + 2xy + y2 = 1x2 y0+ 2x1y1 + 1x0y2, and now if we check the elements in the second row of the Pascals triangle we will find the numbers 1 2 1. Hence if we want to find the coefficients in the binomial expansion, we use Pascals triangle. What is Pascal's Triangle Formula? Pascal's formula is used to find the element in the Pascal triangle. The formula for Pascal's triangle is nCm = n-1Cm-1 + n-1Cm where nCm represents the (m+1)th element in the nth row. n is a non-negative integer, and 0 ≤ m ≤ n. What is the 5th Row of Pascal's Triangle? There are 6 elements in the 5th row of the pascal triangle. The 5th row in Pascal's triangle is 1 5 10 10 5 1. The sum of the elements in the 5th row of the Pascals triangle is 32 which can also be verified by 25 = 32. What is the First Element in Each Row of Pascal's Triangle? The first element in each row of Pascal's triangle is 1. This term is referred to as the 0th term of each row. Q1: What is the middle element of the 6th row of Pascal's triangle? Q2: What is the sum of the nth row of a Pascal's triangle? Q3: What does the 2nd downward diagonal on each side of a Pascal's triangle consist of? Q4: For (x+y)¹⁰, what is the coefficient of the term x⁴y⁶? Q5: Pascal's triangle is a triangular array constructed by summing adjacent elements in _____. 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189186	https://math.libretexts.org/Courses/Santa_Ana_College/Mathematics_Concepts_and_Skills_for_Elementary_School_Teachers/07%3A_From_Parts_to_Percents-_Decimals_Ratios_Proportions_Percentages/7.03%3A_Ratios_and_Proportions	7.3: Ratios and Proportions - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: From Parts to Percents- Decimals, Ratios, Proportions, Percentages Mathematics Concepts and Skills for Elementary School Teachers { "7.3.01:_Written_Retrieval_and_Problem_Solving_Practice" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.3.02:_Online_Retrieval_and_Problem_Solving_Practice" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "7.01:_Decimals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Understanding_Decimal_Operations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Ratios_and_Proportions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Percentages" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_Chapter_7_Resources" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Mathematical_Thinking_Problem_Solving_and_Math_as_a_Language" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Sets_Counting_and_Numeration-_Connecting_Early_Math_to_Advanced_Ideas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Building_Number_Sense-_Understanding_Whole_Number_Operations_and_Their_Properties" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Strengthening_Number_Sense-_Strategies_Algorithms_and_Estimation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_The_Foundations_of_Number_Theory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_The_Story_of_Fractions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_From_Parts_to_Percents-_Decimals_Ratios_Proportions_Percentages" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Integers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Tue, 05 Aug 2025 11:55:49 GMT 7.3: Ratios and Proportions 186589 186589 Martin Romero { } Anonymous Anonymous 2 false false [ "article:topic", "rate", "unit rate", "licenseversion:40", "ratios", "mathematics education", "proportional reasoning", "middle school", "part-to-part relationship", "part-to-whole relationship", "whole-to-part relationship", "fraction notation" ] [ "article:topic", "rate", "unit rate", "licenseversion:40", "ratios", "mathematics education", "proportional reasoning", "middle school", "part-to-part relationship", "part-to-whole relationship", "whole-to-part relationship", "fraction notation" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Santa Ana College 4. Mathematics Concepts and Skills for Elementary School Teachers 5. 7: From Parts to Percents- Decimals, Ratios, Proportions, Percentages 6. 7.3: Ratios and Proportions Expand/collapse global location 7.3: Ratios and Proportions Last updated Aug 5, 2025 Save as PDF 7.2.2: Online Retrieval and Problem Solving Practice 7.3.1: Written Retrieval and Problem Solving Practice Page ID 186589 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Ratios 1. 1. Ratio 2. Example 7.3.1 3. Example 7.3.2 2. Rates 1. Rate 2. Example 7.3.3) 3. Unit Rates 1. Unit Rate 2. Example 7.3.4) 3. Definition: Equality of Ratios 4. Example 7.3.5) Use the Definition of Proportion Definition: Proportion Example 7.3.6 Example 7.3.7 Answer Example 7.3.8 Scaling Wonder, Play, Grow Advanced Connections The formal introduction and detailed study of ratios and proportional reasoning primarily occur later than the early elementary grades, specifically in middle school. However, early mathematics learning plays a crucial foundational role in preparing students for later understanding of ratios (CDE, 2013). Ratios and proportions are essential concepts in the middle school mathematics curriculum, offering a powerful way to understand relationships between quantities. In everyday life, ratios appear all around us. For instance, a community might have 2 parks for every 5000 residents. In a classroom, the student-to-laptop ratio might be 3 to 2, meaning there are three students for every two laptops available. Further,Learning about ratios and proportions develops relational reasoning and thinking, the ability to compare two quantities and recognize invariant relationships. This form of mathematical thinking is foundational for advanced topics like rates of change, which are central to calculus. Ratios A ratio is a comparison of two quantities using division. Some ratios show a part-to-part relationship. For example, the ratio of left-handed students to right-handed students in a class could be 1 to 4. Other ratios represent part-to-whole or whole-to-part relationships. If there are 10 students who have dogs in the class of 25, then the ratio of students who have a dog (part) to all students (whole) is 10 to 25. Conversely, the ratio of all students (whole) to those who have a dog (part) would be 25 to 10. We will also see that this ratio can be stated as 2 to 5 or 5 to 2. Ratio A ratio is the quotient of two numerical quantities or two quantities with the same physical units written as an ordered pair of numbers. There are a number of equivalent ways of expressing ratios, three of which we will use in this text: fraction notation, “to” notation, and “colon” notation. 3/4 is a ratio, read as “the ratio of 3 to 4.” 3 to 4 is a ratio, read as “the ratio of 3 to 4.” 3:4 is a ratio, read as “the ratio of 3 to 4.” In general, a/b, a to b, or a:b. NOTE: Unlike fractions, there are instances of ratios in which b could be zero. Example 7.3.1 Express each of the following ratios as a fraction reduced to lowest terms: a) 36 to 24, and b) 0.12 : 0.18. Answer a) To express the ratio “36 to 24” as a fraction, place 36 over 24 and reduce. 36 24=3⋅12 2⋅12 Factor.=3⋅12 2⋅12 Cancel common factor.=3 2 Thus, the ratio 36 to 24 equals 3/2.(b) To express the ratio “0.12:0.18” as a fraction, place 0.12 over 0.18 and reduce. 0.12 0.18=(0.18)⁢(100)(0.18)⁢(100)Multiply numerator and denominator by 100.=12 18 Move each decimal 2 places right.=2⋅6 3⋅6 Factor.=2⋅6 3⋅6 Cancel.2 3 Thus, the ratio 0.12:0.18 equals 2/3. Example 7.3.2 For the rectangle that follows, express the ratio of length to width as a fraction reduced to lowest terms. Answer The ratio length to width can be expressed as a fraction and reduced as follows. length width=3 1 4 ft 2 1 2 ft Length to width as a fraction.=3 1 4⁢ft 2 1 2⁢ft Cancel common units.=13 4 5 2 Mixed to improper fractions. Invert and multiply, factor, and cancel common factors. =13 4⋅2 5 Invert and multiply.=26 20 Multiply numerators and denominators.=13⋅2 10⋅2 Factor numerator and denominator.=13⋅2 10⋅2 Cancel common factors.=13 10 Hence, the ratio length to width is 13/10. Rates We now introduce the concept of rate, a special type of ratio. Rate A rate is a quotient of two measurements with different units. Example 7.3.3 An automobile travels 224 miles on 12 gallons of gasoline. Express the ratio distance traveled to gas consumption as a fraction reduced to lowest terms. Write a short sentence explaining the physical significance of your solution. Include units in your description. Answer Place miles traveled over gallons of gasoline consumed and reduce. 224 mi 12 gal=56⋅4 mi 3⋅4 gal Factor.=56⋅4 mi 3⋅4 gal Cancel common factor.=56 mi 3 gal Thus, the rate is 56 miles to 3 gallons of gasoline. In plain-speak, this means that the automobile travels 56 miles on 3 gallons of gasoline. Unit Rates When making comparisons, it is helpful to have a rate in a form where the denominator is 1. Such rates are given a special name. Unit Rate A unit rate is a rate whose denominator is 1. Example 7.3.4 a) Herman drives 120 miles in 4 hours. Find his average rate of speed. b) Aditya works 8.5 hours and receives $95 for his efforts. What is his hourly salary rate? Answer a) Place the distance traveled over the time it takes to drive that distance. 120 miles 4 hours=30 miles 1 hour Divide:120/4=30.=30 miles/hour Hence, Herman’s average rate of speed is 30 miles per hour. b) Let’s place money earned over hours worked to get the following rate: 95 dollars 8.5 hours We will get a much better idea of Aditya’s salary rate if we express the rate with a denominator of 1. To do so, divide. 95 dollars 8.5 hours=11.18 dollars 1 hour Divide:95/8.5≈11.18.=11.18 dollars/hour. That is, his salary rate is 11.18 dollars per hour. Definition: Equality of Ratios For any proportion of the form a b=c d,a b=c d, where b≠0,⁢d≠0,b≠0,d≠0, its cross products are equal. Example 7.3.5 One automobile travels 422 miles on 15 gallons of gasoline. A second automobile travels 354 miles on 13 gallons of gasoline. Which automobile gets the better gas mileage? Answer Decimal division (rounded to the nearest tenth) reveals the better gas mileage. In the case of the first automobile, we get the following rate: 422 mi 15⁢gal In the case of the second autombile, we get the following rate: 354 mi 13 gal In the case of the first automobile, the mileage rate is 28.1 mi/1 gal, which can be read “28.1 miles per gallon.” In the case of the second automobile, the mileage rate is 27.2 mi/1 gal, which can be read “27.2 miles per gallon.” Therefore, the first automobile gets the better gas mileage. Use the Definition of Proportion When two ratios or rates are equal, the equation relating them is called a proportion. Definition: Proportion A proportion is an equation of the form a b=c d,a b=c d, where b≠0,⁢d≠0.b≠0,d≠0. The proportion states two ratios or rates are equal. The proportion is read “a“a is to b,b, as c c is to d⁢”. The equation 1 2=4 8 1 2=4 8 is a proportion because the two fractions are equal. The proportion 1 2=4 8 1 2=4 8 is read “1“1 is to 2 2 as 4 4 is to 8⁢”.8”. If we compare quantities with units, we have to be sure we are comparing them in the right order. To determine if a proportion is true, we find the cross products of each proportion. To find the cross products, we multiply each denominator with the opposite numerator (diagonally across the equal sign). The results are called a cross product because of the cross formed. If, and only if, the given proportion is true, that is, the two sides are equal, then the cross products of a proportion will be equal. Example 7.3.6 Determine whether each relationship is a proportion: a) 4:9 and 12:28 b) 1.50⁢f⁡o⁢r⁢6⁢o⁢u⁢n⁢c⁢e⁢s⁢i⁢s⁢e⁢q⁢u⁢i⁢v⁢a⁢l⁢e⁢n⁢t⁢t⁢o 2.25 for 9 ounces. Answer a) To determine if the equation is a proportion, we find the cross products. If they are equal, the equation is a proportion. 28 ⋅ 4 = 112, 9 ⋅ 12 = 108 Since the cross products are not equal, 28 · 4 ≠ 9 · 12 , 28 · 4 ≠ 9 · 12 , the relatiobship is not a proportion. b) $o⁢u⁢n⁢c⁢e⁢s = $o⁢u⁢n⁢c⁢e⁢s, 1.50 6 = 2.25 9 9 ⋅ 1.50 = 6 ⋅ 2.25, 1.50⁢f⁡o⁢r⁢6⁢o⁢u⁢n⁢c⁢e⁢s⁢i⁢s⁢e⁢q⁢u⁢i⁢v⁢a⁢l⁢e⁢n⁢t⁢t⁢o 2.25 for 9 ounces. The strategy for solving applications that we have used earlier in this chapter, also works for proportions, since proportions are equations. When we set up the proportion, we must make sure the units are correct—the units in the numerators match and the units in the denominators match. Example 7.3.7 Josiah went to Mexico for spring break and changed 325 325 dollars into Mexican pesos. At that time, the exchange rate had 1 1 U.S. is equal to 12.54 12.54 Mexican pesos. How many Mexican pesos did he get for his trip? Answer Identify what you are asked to find.How many Mexican pesos did Josiah get? Choose a variable to represent it.Let p=p= number of pesos. Write a sentence that gives the information to find it.If 1⁢U.S.i⁢s⁢e⁢q⁢u⁢a⁢l⁢t⁢o⁢12.54⁢M⁢e⁢x⁢i⁢c⁢a⁢n⁢p⁢e⁢s⁢o⁢s,t⁢h⁡e⁢n 325 is how many pesos? Translate into a proportion. Substitute given values. The variable is in the denominator, so find the cross products and set them equal. Simplify. Check if the answer is reasonable. Yes, 100⁢w⁢o⁢u⁢l⁢d⁢b⁢e 1,254 pesos. $325 is a little more than 3 times this amount. Write a complete sentence.Josiah has 4075.5 pesos for his spring break trip. Example 7.3.8 In a scale drawing, 0.5 centimeter represents 35 miles. a. How many miles will 4 centimeters represent? b. How many centimeters will represent 420 miles? Answer a) 0.5 35=4 x.Solving, we obtain x=35⋅4 0.5,or x=280. b) 0.5 35=y 420,or 0.5×420 35=y.Therefore,y=210 35=6 centimeters. NOTE: We could have been solved mentally by using a technique called scaling up/scaling down, that is, by multiplying/dividing each number in a ratio by the same number. 0.5 centimeter:35 miles=1 centimeter:70 miles=2 centimeters:140 miles=4 centimeters:280 miles. See if you can do part b) using this technique. Scaling is when one quantity changes in a consistent ratio with another. For fun, they some problems below. Scaling Wonder, Play, Grow Solve the following problem in as many different ways as possible. Compare and contrast your methods with those of your peers. Lena has a new smoothie recipe that uses a 117-ounce container of almond milk every 6.5 weeks. If she keeps making smoothies at the same rate, how many ounces of almond milk will she need for a full year? Advanced Connections Proportional reasoning prepares you to understand rates of change, such as velocity (change in distance over time) or slope of a function. In calculus, these ideas are formalized with derivatives. For instance, the concept of unit rate — "miles per hour" — becomes the instantaneous rate of change in calculus. Even related rates problems in calculus (e.g., how fast a shadow lengthens as someone walks) are grounded in the ability to set up ratios and proportions between changing quantities.Understanding ratios and proportions isn't just about solving everyday problems — it's about developing the habits of mind that support mathematical modeling and reasoning about change. These skills are foundational as you move toward algebra, functions, and eventually calculus, where the relationships between changing quantities are central. 7.3: Ratios and Proportions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 7.2.2: Online Retrieval and Problem Solving Practice 7.3.1: Written Retrieval and Problem Solving Practice Was this article helpful? Yes No Recommended articles 8.1: RatiosWe use ratios to compare two numeric quantities or quantities with the same units. 6.1: RatiosWe use ratios to compare two numeric quantities or quantities with the same units. 4.2: Introduction to Ratios and RatesWe use ratios to compare two numeric quantities or quantities with the same units. 4.2: Introduction to Ratios and RatesWe use ratios to compare two numeric quantities or quantities with the same units. 6.2: Introduction to Ratios and RatesWe use ratios to compare two numeric quantities or quantities with the same units. Article typeSection or PageLicense Version4.0 Tags fraction notation mathematics education middle school part-to-part relationship part-to-whole relationship proportional reasoning rate ratios unit rate whole-to-part relationship © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status×
189187	https://www.khanacademy.org/math/revision-term-1-ka-math-class-12/x8b87e674238aecfc:week-3/x8b87e674238aecfc:integrals/v/partial-fraction-expansion-to-integrate	Partial fraction decomposition to evaluate integral (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Revision Term 1 KA Math Class 12 Course: Revision Term 1 KA Math Class 12>Unit 3 Lesson 1: Integrals Antiderivatives and indefinite integrals Antiderivatives and indefinite integrals Indefinite integrals of sin(x), cos(x), and eˣ Indefinite integrals: sin & cos Indefinite integral of 1/x Indefinite integrals: eˣ & 1/x 𝘶-substitution: rational function 𝘶-substitution: logarithmic function 𝘶-substitution: indefinite integrals 𝘶-substitution: double substitution u-substitution challenge Integration using trigonometric identities Integration using long division Partial fraction decomposition to evaluate integral Integration with partial fractions Integration by parts: ∫𝑒ˣ⋅cos(x)dx Integration by parts Finding derivative with fundamental theorem of calculus Finding derivative with fundamental theorem of calculus Finding definite integrals using area formulas Area between a curve and the x-axis: negative area Area using definite integrals Definite integral of trig function Definite integrals: common functions 𝘶-substitution: definite integral of exponential function 𝘶-substitution: definite integrals Finding definite integrals using algebraic properties Math> Revision Term 1 KA Math Class 12> Week 3> Integrals © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Partial fraction decomposition to evaluate integral Google Classroom Microsoft Teams About About this video Transcript When you are integrating a function in the form of a fraction, it helps to find a way to break apart the expression. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Q 10 years ago Posted 10 years ago. Direct link to Q's post “In this video, Sal distri...” more In this video, Sal distributed the A and B into (x-1) and (x+1) respectively, then he factored out the (A+B) from x and used the fact that that equals the coefficient of the x-term and the other expression (B-A) equals the constant term. In the algebra partial fraction videos, however, Sal would set (x-4) = A(x-1) + B(x+1) then plug in arbitrary values of x so A or B would be multiplied by zero, and he would solve for A and B that way. My question is this: Which way is better? Or does it really matter which way you go since both ways give the same answer? Personally, I like plugging in x-values and solving that way, but is the way Sal did in this video better than the way I like? Thanks! Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer robshowsides 10 years ago Posted 10 years ago. Direct link to robshowsides's post “Great question. The meth...” more Great question. The method of equating coefficients is technically more correct, but the method of plugging in x-values so that either A or B is multiplied by zero is usually much faster, so I always use it. The reason it is technically wrong is that in order to get (x-4) = A(x-1) + B(x+1), you multiply both sides by (x-1)(x+1), right? But when you do that, you should be very nervous about ever setting x = 1 or x = -1 at any point in the future, because that means that when you multiplied both sides by (x-1)(x+1), you were multiplying both sides by 0, and cancelling 0/0 = 1! That is always a dangerous thing to do and often leads to extraneous or just plain false solutions in many situations. However, as far as the method of partial fractions is concerned, that quick method of multiplying by the denominator and then plugging in "clever" values for x is safe, and you can use it to save time. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Johnmonkeys3 8 years ago Posted 8 years ago. Direct link to Johnmonkeys3's post “Did anyone else hear the ...” more Did anyone else hear the ringtone at 4:48 :) Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Connor Hodge 11 years ago Posted 11 years ago. Direct link to Connor Hodge's post “5:45 why must A+B always ...” more 5:45 why must A+B always be equal to that coefficient and B-A always equal to the constant? I see that for any example I do it's true, but I don't understand why. Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer vlahosandrew 11 years ago Posted 11 years ago. Direct link to vlahosandrew's post “Is there a shorter way to...” more Is there a shorter way to find the partial fractions? For example, a simple function of the denominators to find the numerators instead of calculating out the whole thing? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Davy Jones 8 years ago Posted 8 years ago. Direct link to Davy Jones's post “For the second part of th...” more For the second part of the new expression of the integral we get at 7:50 , I distributed 5/2 into 1/(x+1) so that I got 5/(2(x+1))=5/(2x+2). Then I took the integral: ∫5/(2x+2)dx = (5/2)∫2/(2x+2)dx. With u-substitution, I get (5/2)ln(2x+2)+C instead of Sal's (5/2)ln(x+1)+C. Where did I go wrong? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Darth Vader 7 years ago Posted 7 years ago. Direct link to Darth Vader's post “Can I use the "dividing e...” more Can I use the "dividing expressions to evaluate integral" method that Sal introduces in the next video here? What kind of expressions can I solve using "dividing expressions" and "partial fraction expansion"? Are there any restrictions on the highest power of x in the num. and the denom.? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer bnadell602 9 years ago Posted 9 years ago. Direct link to bnadell602's post “Where does the 1 infront ...” more Where does the 1 infront of the (x-4)/(x^2 -1) come from? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Arman Gharleghi 10 years ago Posted 10 years ago. Direct link to Arman Gharleghi's post “How would you integrate s...” more How would you integrate something like (x+1)/x^3 ? would you break it into A/x + B/x + C/x ? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Wenxi Chen 9 years ago Posted 9 years ago. Direct link to Wenxi Chen's post “I tried to solve this que...” more I tried to solve this question without the partial fraction expansion. I did ∫(x^2+x-5)/(x^2-1)dx= ∫[(x+2)(x-1)-3]/[(x+1)(x-1)]dx = ∫(x+2)/(x+1)dx - ∫3/(x^2-1)dx Then I used u-substitution to solve ∫(x+2)/(x+1)dx and I used the sec(theta) as x to solve ∫3/(x^2-1)dx. In the end, I got 4x + 1 + ln\|x+1\| + C. Is this a valid solution? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Virgilio Velasco 10 years ago Posted 10 years ago. Direct link to Virgilio Velasco's post “I have a problem understa...” more I have a problem understanding what was done at 6:20 . Why was he able to cancel out the A? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Stefen 10 years ago Posted 10 years ago. Direct link to Stefen's post “Solving a system of equat...” more Solving a system of equations by elimination: Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript [Voiceover] Try to evaluate the following integral. So assuming you've had a go at it, so let's work through this together. And if at any point you get inspired, always feel free to pause the video and continue on with it on your own. So the first thing that might have jumped out at you we have a rational expression. The degree in the numerator is the same as the degree in the nominator, so maybe a little bit of algebraic long division is called for. So let's do that. Let's take X-squared minus one and divide it into X-squared, put that in a different color, divide it into X-squared plus X minus five. So X-squared plus X minus five. And so let's look at the highest degree terms how many times does X-squared go into X-squared? It goes one time, let me write this in a new color. Goes one time. One times X-squared minus one is just going to be X-squared minus one. Now you subtract this green expression from this mauve expression or I could just add the negative on it, so let me just take the negative of it. So it's going to be negative X-squared plus one and we're going to get the X-squared's X-square minus X-square is zero, so those cancel out and we're going to be left with X and the negative five plus one is negative four. So we have X minus four left over. So we can rewrite the expression that we're trying to find the antiderivative of. We can rewrite it as one plus X minus four over X-squared minus one. Maybe I'll do that in that purple color since I already used it as the purple. Over X-squared minus one. So we did one thing, now we have a lower degree in the numerator than we have in the denominator. And obviously this is fairly straight forward, take the antiderivative of, but what do we do now? It's not clear if we look at X-squared minus one it's derivative would be two-X, which is the same degree as this, but it's not X minus four, so it doesn't look like you u-substitution it's going to help us with this. So what can we do now? Now we can take out another tool in our algebraic tool kit, we will do partial fraction expansion. Which is essentially writing this as the sum of two rational expressions that have a lower degree in the denominator. So what do I mean by that? So this term right over here, X minus four over X-squared minus one, we can rewrite that as X minus four over, instead of X-squared minus one, we can factor this. This is X plus one times X minus one. So let's write that, this is X plus one times X minus one. When we think about partial fraction expansion we say okay, can we write this as the sum of something, let's call that A, over X plus one, plus something else, let's call that B, plus B over X minus one. Can we do that? And to attempt to do that, if we had just add these two things, what would we get? Well we would find a common denominator, which would be X plus one times X minus one, and so you would have, if any of this looks unfamiliar I encourage you to review the videos on partial fraction expansion, because that's exactly what we're doing right over here. But this would be equal to if you were to add the two your common denominator would be the product. So it would be X plus one times X minus one. So the first term I would multiply the numerator and the denominator times X minus one. So it would be A times X minus one plus B, the second term I'll multiply the numerator and the denominator times X plus one. And so what do we get? This is going to be equal to AX, maybe I'll do this all in one color. This is going to be equal to AX minus A plus BX plus B and then all of that over this stuff we keep writing over and over again. Actually, let me just copy and paste this. So copy and paste, I can use that over and over again. So we have that over that. So let's see, now we can group the X terms. So we can rewrite this as, so if we take AX plus BX that's going to be A plus B times X. Then we have a negative A and a B. So plus B minus A, and I'll just put parenthesis around that just so I kind of group these constant terms. Then all of that's going to be divided by, good thing I copied and pasted that, X plus one times X minus one. So now this is the crux of partial fraction expansion. We say, okay we kind of went through this whole exercise on the thesis that we could do this, that there is some A and B for which this is true. So if there is some A and B for which this is true, then A plus B must be the coefficient of the X term right over here. So A plus B must be equal to one, must be equal to this coefficient. And B minus A must be equal to the constant, must be equal to negative four. Or if they are then we will found an A and a B, so let's do that. I'll do it up here since I have a little bit of real estate. A plus B is going to be equal to one, and B minus A or I could write that as negative A plus B is equal to negative four. We could add the left hand sides and add the right hand sides and then the A's would disappear. We would get two B is equal to negative three or B is equal to negative three halves. We know that A is equal to one minus B, which would be equal to one plus three halves, since B is negative three halves, which is equal to five halves. A is equal to five halves. B is equal to negative three halves. And just like that we can rewrite this whole integral in a way that is a little bit easier to take the anti or this whole expression so it's easier to integrate. So it's going to be the integral of one plus A over X plus one. A is five halves and so I could just write that as, let me write it this way, five halves times one over X plus one. I wrote it that way because it's very straight forward to take the antiderivative of this. Then plus B over X minus one. Which is going to be negative three halves. So I'll just write it as minus three halves times one over X minus one. That was this right over here, DX. Notice all I did is I took this expression right over here and I did a little bit of partial fraction expansion into these two, I guess you could say, expressions or terms right over there. It's fairly straight forward to integrate this. Antiderivative of one, it's just going to be X. Antiderivative of five halves, one over X plus one, is going to be plus five halves, the natural log of the absolute value of X plus one. We're able to do that because the derivative of X plus one is just one, so the derivative is there so that we can take the antiderivative with respect to X plus one. You could also do u-substitution like we've done in previous examples, U is equal to X plus one. And over here, this is going to be minus three halves times the natural log of the absolute value of X minus one, by the same exact logic with how we were able to take the antiderivative there. And of course we cannot forget our constant. And there we have it. We've been able to integrate, we were able to evaluate this expression. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. 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189188	https://math.stackexchange.com/questions/2408729/is-my-proof-correct-proof-n2-is-odd-then-n-is-odd	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Is my proof correct? Proof: $n^{2}$ is odd then $n$ is odd. I don't think it is but I don't understand why. Conjecture: $n^2 \text{ is odd} \Rightarrow n \text{ is odd}$ () Proof: In general, $$ (A \Rightarrow B) \Leftrightarrow (\neg(B) \Rightarrow \neg(A)) $$ Therefore $(n^2 \text{ is odd } \Rightarrow n \text{ is odd } \Leftrightarrow (\neg(n \text{ is odd}) \Rightarrow \neg(n^2 \text{ is odd})) $ This is the same as $n \text{ is even} \Rightarrow n^2 \text{ is even}$ (✝) Isn't it? If it is, then $$\begin{align} n = 2m &\ &\Leftrightarrow n^2 = 4m^2 \ &\Leftrightarrow n^2 = 2(2m^2)\ \end{align}$$ Therefore $n^2$ is even, by definition of even. And, since we'e proved (✝), surely we have then proved (). I'm a novice to proofs and this is one of my first ones, so I can't notice any obvious problems with it, yet it doesn't seem to me to prove anything - it seems to me to be proving something else entirely. Is it correct? 5 Answers 5 Very well done. Yes, indeed, proving $n\text{ is even}\implies n^2\text{ is even}$ is much easier than the original statement. Alternatively, if $n^2$ is odd, then $n^2-1$ is even. It can be expressed as: $$n^2-1=(n-1)(n+1).$$ If $n$ is even, $n-1$ and $n+1$ are both odd, whose product is also odd and it contradicts the given condition ($n^2-1$ is even). Hence, $n$ is odd. Your idea is correct, but it might be better to write the proof this way: We prove the contrapositive statement, that if $n$ is even then $n^2$ is even. Since $n$ is even, $n=2k$ for some integer $k$. Then $n^2=4k^2=2(2k^2)$ with $2k^2$ an integer, so $n^2$ is even. Notice that in your proof, you did not mention that $m$ or $2m^2$ are integers, which are important details. The last two biconditional arrows also obscure the picture, especially when the nature of $m$ is not specified. (You only need one direction in the proof, and the other direction might not even hold in some cases.) Alt. hint: $\,n^2+n=n(n+1)$ is always even, since one of two consecutive integers must be even. Therefore $n^2$ and $n$ always have the same parity and, in particular, $n^2$ is odd iff $n$ is odd. Your proof is correct as has been pointed by others and perhaps it is the most simple way to go about this, but just for fun, here is somewhat more direct proof (trying to avoid proof by contrapositive or contradiction): If $n^2$ is odd, then $n^2+2n+1$ is even (odd + odd is even). Notice that this is just $(n+1)^2$. This means that $2$ divides $(n+1)^2$ and since $2$ is a prime, it follows $2$ divides $(n+1)$ and so $n$ is odd. Another way to go about this is noticing that if $n^2$ is odd, then it can be written as $n^2=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$ where $p_i$ are odd primes. Now $n$ is divisor of $n^2$ and so it must also consist only of odd primes, hence $n$ is odd. You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Linked Related Hot Network Questions Subscribe to RSS To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Mathematics Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.29.34589
189189	https://www.ama-assn.org/public-health/chronic-diseases/what-doctors-wish-patients-knew-about-rheumatoid-arthritis	Skip to main content My Subscriptions My Topics My Bookmarks Chronic Diseases What doctors wish patients knew about rheumatoid arthritis By Sara Berg, MS , News Editor \| 11 Min Read Subscribe Save Copy Print Share Share via Email Share to Facebook Share to Twitter Share to LinkedIn Share to Doximity What Doctors Wish Patients Knew™ Subscribe AMA News Wire What doctors wish patients knew about rheumatoid arthritis Aug 24, 2023 Among chronic conditions, one stands out for its enigmatic nature and debilitating impact: rheumatoid arthritis (RA). This condition presents an array of challenges and wreaks havoc on the lives of those who live with it. And while there are effective treatments that can prevent or slow the progression of rheumatoid arthritis, many questions remain unanswered, leaving patients and their families searching for better strategies to manage this life-altering condition. What's happening in public health? Get the tools and resources you need to face the latest public health issues. Subscribe Now Rheumatoid arthritis—characterized by painful inflammation and joint deformities—is the most common autoimmune disorder, affecting about 1% of U.S. patients, happening two to three times more frequently in women than men. And while rheumatoid arthritis can strike people of any age, the peak onset is from 50 to 59 years old. The AMA’s What Doctors Wish Patients Knew™ series provides physicians with a platform to share what they want patients to understand about today’s health care headlines. For this installment, two physicians took time to discuss what patients need to know about managing rheumatoid arthritis. They are: Shawn Baca, MD, a rheumatologist at Rheumatology Associates of South Florida and clinical associate professor at Florida Atlantic University Schmidt School of Medicine. He also represents the Florida Medical Association in the AMA House of Delegates. Amish J. Dave, MD, a rheumatologist at Virginia Mason Medical Center in Seattle who also serves in the House of Delegates, representing the Washington State Medical Association. He is also a member of the AMA Ambassador Program. RA is not just bad arthritis “A lot of people think that rheumatoid arthritis is just bad arthritis, and it’s not. There are actually over 100 different kinds of arthritis and rheumatoid is a very specific disease where the immune system attacks the person’s musculoskeletal system,” Dr. Baca said. “But it’s also systemic. It can attack your lungs and kidneys and you can get vasculitis. “So, there are different parts to it that are not just related to the joints, but the joints are predominantly what brings a person to a rheumatologist,” he added. It is a chronic condition Rheumatoid arthritis “doesn’t necessarily go away and a lot of people think that if they do everything right, they can just push it into remission and they don’t have to worry about it,” Dr. Dave said. But “this is a chronic condition. You can get it to a good place, but you won’t cure it.” “The chances of going into remission are very low, probably less than 5%,” Dr. Baca said, noting that “I have only a few patients who have been in remission over my 30 years of practice. But a majority of patients when they start medication, they end up being on some type of therapy for most of their lives.” “There are people who can come off their medication and go into remission for long periods of time, even years, but it flares up again,” Dr. Dave said. “We think of this in the same way we think of Crohn’s or ulcerative colitis. Relapsing rheumatoid arthritis tends to be that way too. “The single best thing you can do is have a good relationship with a thoughtful rheumatologist who's following you over time and recognizing that there will be good months and bad months is important, and that medications can fail,” he added. “It means that being flexible and needing to switch to a new medication might be important for you.” Symptoms must be present for months “Rheumatoid arthritis often tends to be symmetric synovitis, or inflammation of joints. So, often both hands, both wrists, both feet, both ankles have inflammation in the joints, but can affect other joints too within the body depending on that particular individual,” Dr. Dave said. “But the classic person with rheumatoid arthritis has symmetric joint inflammation with more than an hour of morning stiffness.” “Presentation would be somebody who’s young—20s to 40s—and then there’s a second bump somewhere around the 60s. So, we see it in basically young and older age groups,” Dr. Baca said. Rheumatoid arthritis “usually presents with swelling across the hands and wrists and is symmetric. It should almost always be symmetric hands, wrists, knees, feet, shoulders. “About the only places that rheumatoid does not attack is the base of the thumbs or the lower back,” he added, noting “most of the joints are basically prey to it including the temporomandibular joints—TMJs—and cricoarytenoid joint, your voice box, can actually be affected by RA.” “Usually, the symptoms have to be present for at least three months for it to be considered because there are other types of arthritis that can be viral, or following an infection,” Dr. Baca said. “But the difference is that those things will go away whereas RA will never go away.” Related Coverage What doctors wish patients knew about maintaining a healthy weight Cardiovascular risk increases with RA “Patients with rheumatoid arthritis are at an increased risk for cardiovascular disease,” Dr. Baca said, noting that these patients “have inflammation of the pericardium, the myocardium and they can get inflammation of their arteries”. “Some of the risk factors are related to the fact that we used to use more steroids in the past, which raise your cholesterol, blood pressure, weight and all that other stuff,” he added. “But there are strong studies that show that people who are treated aggressively can actually reduce their risk of a heart attack when they’re on RA medications.” “So, the added benefit of taking the medications is that it reduces your risk factors for other diseases,” Dr. Baca said. Join the fight on chronic disease AMA membership offers unique access to savings and resources tailored to enrich the personal and professional lives of physicians, residents and medical students. Join the AMA Environmental triggers play a role “When we think about RA, it’s important to understand that its incidence is increasing over time” as the population ages, Dr. Dave said. “There are almost certainly some environmental triggers—including pollution—that might be contributing to the rise in rheumatoid arthritis.” “Smoking does increase your risk of getting the disease if you're genetically predisposed. And also, if you have RA and you smoke, it makes it more difficult to treat,” Dr. Baca said. “Historically, it was tobacco which was the No. 1 cause of rheumatoid arthritis, but family history and genetics as well as obesity play a role,” Dr. Dave said, noting “there’s also suspected to be a trigger with heavy alcohol intake. So, all those things can combine to increase the risk of rheumatoid arthritis.” There may be long waits for care “How quickly you’re diagnosed and how well you’re taken care of with rheumatoid arthritis really depends on where you live,” Dr. Dave said, noting it is expected that the number of patients per rheumatologist is expected to double by 2025. “If you live in Boston—which is one of the highest concentrations of rheumatologists in the country—versus if you live in Alaska, there’s a huge discrepancy in how long you might have to wait. It’s not uncommon,” he said. “It’s important to recognize for physicians, for family members, for lawmakers, for insurance companies, that the costs of rheumatoid arthritis are really high, and telemedicine has helped with that to some extent, but it doesn’t replace needing to feel or examine the joints, needing to inject or aspirate joints and drain fluid out of it when that’s occurring.” Prior authorization is a roadblock “What type of insurance you have makes such a big impact on the types of medications you can get, your access to medications and whether or not you’ll be stuck following a step therapy method of starting one medicine and then going to another medicine,” Dr. Dave explained. “The medicine that your rheumatologist might want to start you on because he or she or they feel like that’s the medicine for you is not necessarily the medicine that you’re going to get.” “More often than not, the rheumatologist’s staff is arguing over a $40,000 to $60,000 drug per year and that’s really what tests insurance companies,” he said. “The cost of medications for rheumatoid arthritis and other autoimmune diseases is huge. It’s driving up the cost of health care in general.” “Prior authorization and step therapy really affect our patients disproportionately to other illnesses and it’s very difficult sometimes to get medications simply because insurance rules and regulations make things almost impossible,” Dr. Baca said. For example, “one patient who was doing very well on a particular regimen after not doing well with drug X, but when the insurance was changed by their employer, and now drug X is the preferred biologic.” Insurance “wanted a trial within X number of months, meaning that you had to try it again. So, there is nonsense like that, and we try to get around that as much as we can,” he said. There are many medications “If you have good insurance and you have good access to a rheumatologist, there’s never been a better time to have rheumatoid arthritis in the sense that there are so many medications that can be used to treat rheumatoid arthritis,” Dr. Dave said. “For a lot of people, it’s important to recognize that your rheumatologist will pick a medication for you based on what types of antibodies you have, based on how severe your inflammation is.” “We often treat initial attacks of inflammation with medicines like prednisone or steroids, but the goal is not to keep you on prednisone for the rest of your life,” he said. “The goal is to get you on what we call steroid sparing medications or steroid sparing DMARDS—disease-modifying anti-rheumatic drugs.” Additionally, “we often use medicines thinking about the cost to patients and what the insurance providers will approve first,” Dr. Dave said. “We often start with medicines that have been around for a long time like hydroxychloroquine, methotrexate, leflunomide and sulfasalazine. Then we might move to biologic medicines that have been engineered specifically to target or block a specific protein in the immune system.” Related Coverage What doctors wish patients knew about increasing physical activity Lifestyle changes are important too “Medicines are a huge part of caring for patients with rheumatoid arthritis, but lifestyle factors matter so much more too,” Dr. Dave said. That means “stopping smoking if you smoke, cutting back on alcohol intake, especially if medicines like methotrexate are prescribed by your rheumatologist … because it can affect your liver and raise your liver lab tests and cause inflammation in the liver.” “Then, exercise because we know that adipose tissue or fat tissue increases inflammation in the body,” he said, noting that “weight loss has been shown to be super effective at reducing rheumatoid arthritis and hopefully reducing flare ups.” What doctors wish patients knew Subscribe for the answers to the latest questions patients are bringing to the exam room. Subscribe Now Diet alone won’t help “In terms of eating an anti-inflammatory diet, there's not a lot of proof. This was something that was worked out a long time ago before we had treatments,” Dr. Baca said. “My bottom-line answer is if a patient feels better avoiding certain kinds of things, so be it. I'm never going to be against somebody eating a healthier, vegetable-based diet. “But at the same time, I'm not going to tell them that it’s absolutely mandatory, because it doesn’t make as big of a difference as proper pharmaceutical therapy,” he added. “Vitamins and eating a healthy diet are great and super good for your body in so many ways, especially if they promote sleep or help with energy levels, but they don’t stop the inflammation itself,” Dr. Dave said. “Being on medication is what does that. So, really working through a rheumatologist is so important.” Stay up to date on vaccinations “Rheumatologists work very closely with the primary care physician and we—especially in the COVID era—have really had to work closely with them to make sure that we're keeping people up to date on vaccinations,” Dr. Dave said. But “we're facing a lot more vaccine hesitancy than ever before. And that's been challenging. “It used to just be the COVID vaccine and now we're having to remind people with rheumatoid arthritis that they need to keep up to date on their pneumococcal vaccines, tetanus boosters, the annual flu shot and making sure they get their two-shot shingles series,” he added, noting that staying up-to-date on vaccinations is important because “your body’s more focused on attacking your joints and other organs than fighting off infections” such as influenza, COVID-19 and RSV. “Having RA alone increases your risk of infections. And then the flip side of it is having RA alone or being on medicines or RA that suppress your immune system also place you at risk of infection,” Dr. Dave emphasized. The key is preventing progression “If you came to my office 30 years ago when I first started as a young pup, we had probably 30% of the spaces in the waiting room filled with wheelchairs,” Dr. Baca said. “If you come to my office now, you will hardly ever see any wheelchairs in the waiting room. “The difference is that we’re trying to prevent destruction of joints, especially hips, knees and hands with earlier intervention and newer treatments,” he said. There is help out there “Patients need to make sure they understand that there’s always help out there, but they have to look for it in their communities,” Dr. Baca said. “Even if you don’t have health insurance, there are ways that you can get care. Many communities such as ours have free arthritis clinics. Seek out independent local rheumatologists, many are flexible and can affordably see patients without insurance and help one navigate the health system to get the medications they need,” he said. “But you have to look for resources and try to get treated early because that’s how you’re going to prevent crippling, which is the worst outcome that we can have.” Table of Contents RA is not just bad arthritis It is a chronic condition Symptoms must be present for months Cardiovascular risk increases with RA Environmental triggers play a role There may be long waits for care Prior authorization is a roadblock There are many medications Lifestyle changes are important too Diet alone won’t help Stay up to date on vaccinations The key is preventing progression There is help out there Arthritis Nutrition & Diet Prior Authorization Vaccines, Vaccinations & Immunizations Cardiovascular Disease E-cigarettes & Tobacco Alcohol Exercise Patient Resources What Doctors Wish Patients Knew™ Catalog of Topics All Trends FEATURED STORIES ### 109 health systems honored for advancing physician well-being Sep 3, 2025 \| 5 Min Read ### More than 150 scope-creep bills defeated in 2025 Sep 2, 2025 \| 4 Min Read ### Strong care teams are key to boosting physician well-being Sep 9, 2025 \| 8 Min Read ### What doctors wish parents knew about when to seek emergency care Sep 5, 2025 \| 8 Min Read Also of Interest: Health Information Technology Substance Use Disorders Health Care AI
189190	https://sca.district70.org/ourpages/auto/2014/10/29/45969851/8th%20grade%20density_worksheet.pdf	Name:___ Class:___ Date:___ 8th Grade Science Density Worksheet Record all your work on your own paper. Show all work and correct units. 1. State the formula for density in words and mathematical symbols. 2. A rock has a mass of 210 grams and occupies a volume of 70 cm3. What is its density? 3. An unknown liquid occupies a volume of 5 ml and has a mass of 40 grams. Find its density. 4. Define the following vocabulary words: a.) mass b.) volume 5. How does the volume occupied by a cubic centimeter (cm3) compare with the volume occupied by a milliliter (ml)? 6. A rectangular solid of unknown density is 5 meters long, 2 meters high and 4 meters wide. The mass of this solid is 300 grams. Given this information for this homogeneous (alike throughout) material, calculate its density. 7. A rock occupies a volume of 20 cm3 and has a mass of 54 grams. Find the density of this rock. 8. A rock has a density of 4 g/ml and a mass of 16 grams. What is the volume this rock occupies? 9. An unknown substance from planet X has a density of 10 g/ml. It occupies a volume of 80 ml. What is the mass of this unknown substance? 10. Water has a density of 1.0 g/mL and ocean water has a density of 1.025 g/mL. Why are they different? 11. A cube made of an unknown material has a height of 9 cm. The mass of this cube is 3,645 grams. Calculate the density of this cube given this information. Hint: a cube has 6 equal sides 12. Given a large beaker of water at room temperature (20 C), draw the approximate position of the following objects when placed in that beaker of water and explain why you placed them as you did. (In each case draw a beaker of water and the object’s approximate position you are placing in it – you should have 5 drawings). a) styrofoam (D = .05 g/cm3) b) ice (D = .92 g/cm3) c) bone (D = 1.70 g/cm3) d) balsa wood (D = 0.16 g/cm3) e) gold (D = 19.32 g/cm3) 5 m 2 m 4 m Name:___ Class:___ Date:___ 13. A graduated cylinder has 20 ml (cm3) of water placed in it. An irregularly shaped rock is then dropped in the graduated cylinder and the volume of the rock and water in the cylinder now reads 30 ml (cm3). The mass of the rock dropped into the graduated cylinder is 23 grams. a.) Find the volume of the rock dropped into the graduated cylinder. b.) Find the density of the rock dropped into the graduated cylinder. Use the graph below to answer the following questions: 14. What is the density of object A? Does it sink or float in water? 15. What is the density of object B? Does it sink or float in water? 16. What is the density of object C? Does it sink or float in water? 0 5 10 15 20 25 30 25 20 15 10 5 0 Mass (g) Volume (cm3) A B C Name:___ Class:___ Date:_____ Use the table below to answer the following questions: 17. An object with a mass of 24g and a volume of 32mL is most likely what substance? 18. What is the only object in the table that would sink in corn syrup? 19. What is the mass of 100 mL of corn oil? 20. What is the volume of 35g of copper? Substance Density at 20 ◦C Substance Density at 20 ◦C Wood 0.70 g/cm3 Rubber 1.34 g/cm3 Corn oil 0.92 g/cm3 Corn Syrup 1.38 g/cm3 Water 1.00 g/cm3 Copper 8.80 g/cm3
189191	http://lampz.tugraz.at/~hadley/physikm/script/magnetism/loop.en.php	The magnetic field along the axis of a current loop Menu ▼ Outline Formulas Skills Apps Exam questions → Teach Center Exercise NotesPhysik für Geodäsie 511.018 / Physik M 513.805 The magnetic field along the axis of a current loop A current I I [A] flows in a circular wire of radius R R lies in the x−y x−y . To calculate the magnetic field along the z z-axis, the Biot-Savart law can be used. d⃗B(⃗r)=μ 0 4 π I d⃗r w i r e×(⃗r−⃗r w i r e)\|⃗r−⃗r w i r e\|3[T].d B→(r→)=μ 0 4 π I d r→w i r e×(r→−r→w i r e)\|r→−r→w i r e\|3[T]. The contribution to the field of the wire segment d⃗r d r→ along the x x-axis is, d⃗B(⃗r)=μ 0 4 π I d y^y×(z^z−R^x)√R 2+z 2 3[T].d B→(r→)=μ 0 4 π I d y y^×(z z^−R x^)R 2+z 2 3[T]. ^y×^z=^x y^×z^=x^ and ^y×^x=−^z y^×x^=−z^. d⃗B(⃗r)=μ 0 4 π I d y(z^x+R^z)√R 2+z 2 3[T].d B→(r→)=μ 0 4 π I d y(z x^+R z^)R 2+z 2 3[T]. The sum of the horizontal components from all the wire segments around the circle will add to zero so the field at position z z will be, ⃗B(z)=μ 0 4 π I 2 π R 2^z√R 2+z 2 3=μ 0 I R 2^z 2√R 2+z 2 3[T].B→(z)=μ 0 4 π I 2 π R 2 z^R 2+z 2 3=μ 0 I R 2 z^2 R 2+z 2 3[T]. At large distances from the loop, the amplitude of the magnetic field falls like 1 z 3 1 z 3. If such loops are stacked on top of each other so that there are n n wire loops per meter, then the field can be calculated by summing the contributions from all of the loops. ⃗B(z)=∞∫−∞μ 0 n I R 2 d z 2√R 2+z 2 3^z[T].B→(z)=∫−∞∞μ 0 n I R 2 d z 2 R 2+z 2 3 z^[T]. Using the identity, ∞∫−∞d z(R 2+z 2)3/2=2 R 2,∫−∞∞d z(R 2+z 2)3/2=2 R 2, yields the simple result, ⃗B=μ 0 n I^z[T].B→=μ 0 n I z^[T]. This is the result for the magnetic field inside a solenoid with n n windings per meter.
189192	https://bostoncatholic.org/catholic-glossary	\|\|\| Facebook X (Twitter) Instagram Phone Search Catholic Glossary A Absolution - the act by which a priest, acting as an agent of Christ, grants forgiveness of sins in the Sacrament of Reconciliation. Abstinence - the avoidance of a particular type of food, such as meat, as an act of penance or spiritual discipline. Acolyte - person who assists in the celebration of Mass or other liturgical celebration. Adoration - the external acts of reverent admiration or honor given to a thing or person. In the Catholic faith, adoration is reserved to God alone and to Jesus present in the consecrated Eucharist. Adoration of the Blessed Sacrament - Prayer to Christ, who is recognized as being truly present in the Sacrament of Eucharist. Alb - a long, white garment that can be used by all liturgical ministers; it is a reminder of the baptismal garment worn when the new Christian "puts on Christ." Alleluia Acclamation - this acclamation of praise follows the second reading and prepares the assembly for the Gospel. Altar - this is the focal point of the church. It is a table, often made of wood or stone, that is set at the center of the sanctuary and has been consecrated for sacred use. The holy sacrifice of the Mass is offered on the altar, as the gifts of bread and wine become the body and blood of Christ. A relic of a saint is often contained inside the altar. Ambo - looks like a podium and is where the lector proclaims the readings for Mass. The deacon or priest also reads the Gospel from here. Ambry - a recess that holds holy oils that are blessed and consecrated at the Chrism Mass during Holy Week. Amen - a Hebrew word meaning truly, it is true. As a concluding word of prayers, it expressed assent to and acceptance of God's will. Annul - properly called the degree of nullity, this is the declaration by authorities that a marriage is null and void, because it was never valid. Apparition – an appearance to people on earth of a heavenly being – namely, Christ, Mary, or a saint or angel. Apostle - "one sent." This normally refers to the 12 men chosen by Jesus to be the bearers of his teachings to the world. Apostolate - The ministry or work of an apostle. In Catholic usage, this is a term covering all kinds and areas of work and endeavor for the service of God and the Church and the good of people. Apostolic - refers to the 12 apostles. It also characterizes certain documents, appointments or structures initiated by the Pope or the Holy See. Apostolic Nunciature - the offices of the Holy Father's representative to a country or to the Church in that country. Aspergilium - a container or vessel used for sprinkling holy water that is ordinarily made out of metal. Assembly - those gathered to celebrate the liturgy. Associate Pastor - a priest who assists a pastor in the pastoral care of a parish or parishes. Auxiliary Bishop - a bishop assigned to a Catholic diocese or archdiocese, to assist a residential bishop. B Baptismal Font - a receptacle for water that is used in the sacrament of baptism. Basilica - a church to which special privileges are attached. It is a title of honor given to various kinds of Churches. Beatification - final step toward canonization of a saint Benediction Veil - also called the humeral veil; a long, narrow shawl-like vestment used at Benediction. Bishops - the chief priest of a diocese. Bishops are responsible for the pastoral care of their dioceses. In addition, bishops have a responsibility to act in council with other bishops to guide the Church. Blessed Sacrament - the Eucharist, the Body and Blood of Christ, either at Mass or reserved in a special place in the Church. Book of the Gospel - the book which contains the Gospel texts, from which the priests or deacon proclaims the Gospel of the day. Brazier - a metal pan used to hold incense. Bread and Wine - the elements used in the celebration of Eucharist (unleavened bread and natural pure wine). Breaking of the Bread - the celebrant recreates the gestures of Christ at the Last Supper when he broke the bread to give to his disciples. This action signifies that in communion, the many are made one in the one Bread of Life which is Christ. C Canon - Greek for rule, norm, standard, measure. Designates the Canon of Sacred Scripture, the list of books recognized by the Church as inspired by the Holy Spirit. Canon Law - the codified body of general laws governing the Church. Canonization - a declaration by the pope that a person who died a martyr or practiced Christian virtue to a heroic degree is in heaven and is worthy of honor and imitation by the faithful. Verification of miracles is required for canonization (except for martyrs). Cantor – a person who leads the singing during the liturgy. Cardinal - Cardinals are appointed by the pope and constitute a kind of senate of the Church, and aid the pope as his chief counselors. Cassock (KASS-uhk) - a long, black garment worn by altar servers under the surplice; also worn by diocesan priests (black); monsignors (rose); bishops (violet), cardinals (red), and the Pope (white). This is a non-liturgical, full-length robe. Catechesis (cat-UH-key-sis) - religious instruction and formation for persons preparing for baptism and for the faithful in various stages of spiritual development. Catechetical (cat-uh-KIT-uh-kal) - referring to catechesis. Catechetics (cat-uh-KIT-iks) - from the Greek meaning "to sound forth," it is the procedure for teaching religion. Cathedra - the archbishop's chair. It is the symbol of his role of chief teacher and pastor of the local church. The word is Greek and means chair. The word cathedral comes from cathedra, meaning, literally, chair of the bishop. Cathedral - The major church in an archdiocese or diocese. It is the seat of the local Ordinary. Catholic - Greek word for universal. First used in the title Catholic Church in a letter written by St. Ignatius of Antioch to the Christians of Smyrna about 107 A.D. Celebrant - the person who presides over the assembly and consecrates the Eucharist. Celebrant's Chair - the place where the celebrant sits. It expresses his office of presiding over the assembly and of leading the prayer of those present. Chalice (CHAL-is) - the large cup used to hold the wine that becomes the Blood of Christ. It is made of durable material and comes in varies shapes and sizes. Chancellor - the chief archivist of a diocese' official records who is also a notary and secretary of the diocesan curia. Charismatic - Person who believes God endowed them with gifts or graces. Charisms - Gifts or graces given by God to persons for the good of others and the Church. Chasuble (CHAZ-uh-buhl) - the sleeveless, outer garment that when slipped over the head, hangs down from the shoulder covering the alb and stole of the priest. It is worn by the main celebrant and its color varies according to the feast. Chrism - a specially perfumed olive oil that is consecrated for use at baptism, confirmation, and holy orders. Chrism also is used to anoint altars and walls during church or cathedral dedications. This is only time the consecrated oil is not used on a human being. Christ - the title of Jesus, derived from the Greek translation of the Hebrew term Messiah, meaning the Anointed of God. Church - the universal Church that is spread throughout the world; the local Church is that of a particular locality, such as a diocese. The Church embraces all its members--on earth, in heaven, in purgatory. Ciborium (si-BORE-ee-um) - a vessel used to hold the Hosts which will be used for communion; some are cup-like and others are bowl/plate like; they are also used to reserve the Blessed Sacrament in the tabernacle. Cincture (SINGK-sure) - a long cord used for fastening some albs at the waist; it holds the loose-fitting type of alb in place and is used to adjust it to the proper length; it is usually white, although the liturgical color of the day may be used. Cloister - part of a convent or monastery reserved for use by members of the institute. College of Cardinals - the College of Cardinals is made up of the cardinals of the Church, who advise the Pope, assist in the central administration of the Church, head the various curial offices and congregations, administer the Holy See during a vacancy, and elect a new Pope. Collegiality - the shared responsibility and authority that the whole college of bishops, headed by the pope, has for the teaching, sanctification and government of the Church. Communion Cups - when the people receive Christ’s blood at communion, they drink from this chalice-like vessel. These cups are kept on the Credence Table and brought to the Altar at communion time. Communion Song - the music that is used as the consecrated bread and wine is distributed to the faithful. Concelebrants - the priests and bishops who join the celebrant in celebrating the Mass. Concluding Rite - the brief rite at the conclusion of the Mass which consists of the celebrant's greeting to all present, final blessing and dismissal. Confession - part of the Sacrament of Reconciliation, not the term for the sacrament itself. Confirmation - one of the three sacraments of initiation, along with baptism and Eucharist. Conscience - the “interior voice” of a person. It is a God-given and directed sense of what is morally right and wrong. Conscience helps people to be responsible for their actions and to strive to do good and avoid evil. Contemplative Nun - a religious woman who devotes her entire life in the cloister to prayer and reflection. Convent – a house of women religious. Cope (KOPE) - a cape-like garment, open in the front that when placed over the shoulders, hangs to the ankles. It is worn by a priest or deacon in processions at Benediction and other services. Covenant – The relationship between God and human beings which is characterized by mutual commitment and partnership. Creed – An official profession of faith used in the liturgy of the Church. The word means “I believe.” The two most popular Catholic Creeds are the Apostles’ Creed and the Nicene Creed. Crosier (pastoral staff) - The staff which a bishop carries when he presides at the liturgy. Cross bearer - the one who carries the cross in the procession (entrance and recessional). Cross/Crucifix - an object is a crucifix only if it depicts Christ on a cross; otherwise it is a cross. Cult - any act or system of veneration or worship. Cursillo - conducted by priests and laypersons, it consists of a three-day weekend focused on prayer, study, and Christian action, and follow-up program known as the post-cursillo. D Dalmatic (dahl-MAT-ik) - a loose-fitting robe with open sides and wide sleeves worn by a deacon on more solemn feasts; it takes its color from the liturgical feast as listed above. Deacon - an ordained minister who assists the celebrant during the Liturgy of the Word and at the altar for the Liturgy of the Eucharist. Deacons can also provide assistance to the pastor in baptismal and/or marriage ministry. Deacons serve in the ministry of the liturgy, of the word, and of charity. Transitional deacons are men who are preparing for the priesthood. Permanent deacons are men who are not planning to become ordained priests. They can be married and have children. Decanter or Flagon (FLAG-un) - the bottle- or pitcher-like vessel used to hold the wine which will be consecrated at Mass for the communion of the people; it is brought forth with the gifts. Diocese - a particular church; a fully organized ecclesiastical jurisdiction under the pastoral direction of a bishop as local Ordinary. Disciple - one who follows the teachings of Jesus. It is based on a word meaning pupil or student. Dispensation - an exemption from Church law. Doctrine – an official teaching of the church based on the revelation of God by and through Christ. Dogma – Church teachings that are central to the faith, defined by the magisterium, and accorded the fullest weight and authority. Doxology - the response of the people acclaiming the sovereignty of God. E Eastern-rite (Oriental) Church - term used to describe the Catholic Churches which developed in Eastern Europe, Asia, and Africa. They have their own distinctive liturgical and organizational systems. Each is considered equal to the Latin rite within the Church. Ecclesial - having to do with the church in general or the life of the Church Ecclesiastical (ee-CLEE-zee-as-tuh-cal) - refers to official structures or legal and organizational aspects of the Church Ecumenism (eh-KEW-meh-nizm) / Interdenominational / Ecumenical (EK-you-meh-nikal) Movement - a movement for spiritual understanding and unity among Christians and their churches. The term also is extended to apply to efforts toward greater understanding and cooperation between Christians and members of other faiths. Encyclical - a pastoral letter addressed by the Pope to the whole Church. Entrance procession - priest, deacon, altar servers, lectors, enter the church or designated place for celebration of the liturgy. Entrance song/music - the song/music which takes place during the entrance procession. Episcopal - Refers to a bishop or groups of bishops as a form of Church government, in which bishops have authority. Eschatology - doctrine concerning the last things: death, judgment, heaven and hell, and the final state of perfection of the people and the kingdom of God at the end of the world. Eucharistic Prayer - the prayer of thanksgiving and sanctification used during the Mass. It is the center of the celebration. During the Eucharistic Prayer, the Church believes that the bread and wine become the Body and Blood of Jesus Christ. Evangelist - a preacher or revivalist who seeks conversions by preaching to groups. Exarch/Exarchy - A Church jurisdiction, similar to a diocese, established for Eastern-rite Catholics living outside their native land. The head of an exarch, usually a bishop, is an exarch. Excommunication - A penalty of censure by which a baptized person is excluded from the communion of the faithful for committing and remaining obstinate in certain serious offenses specified in canon law. Even though excommunicated, a person still is responsible for fulfillment of the normal obligations of a Catholic. F Final Doxology - a final prayer of praise of God. Free Will - The faculty or capability of making a reasonable choice among several alternatives. Friar - a member of a mendicant community, such as the Dominicans, Franciscans or Carmelites. They live a rule of communal poverty, living primarily from the freewill offerings of the faithful, engage in various forms of pastoral ministry, and belong to a religious order that is a wider community beyond the local house, in contrast to a monastery, which is self-contained, even if in federation with others. G General Intercessions - a prayer of intercession for all of humankind; for the Church, civil authorities, those in various needs, for all peoples, and for the salvation of the world. Great Amen- the acclamation by the people expressing their agreement with all that has been said and done in the Eucharistic prayer. Gloria – an ancient hymn of praise in which the Church glorifies God. It is used on all Sundays, except for those during Advent and Lent, and at solemn celebrations. The text originates from the Christmas narrative in the Gospel of Luke (2:14 - "Glory to God in the highest and on earth peace to those on whom his favor rests.") God - the infinitely perfect Supreme Being, uncaused and absolutely self-sufficient, eternal, the Creator and final end of all things. The one God subsists in three equal Persons, the Father and the Son and the Holy Spirit. Grace - a free gift of God to human beings, grace is a created sharing in the life of God. It is given through the merits of Christ and is communicated by the Holy Spirit. It is necessary for salvation. Greeting - the celebrant greets all present at the liturgy, expressing the presence of the Lord to the assembled community. H Heresy - the conscious and deliberate rejection of a dogma of the Church. Hierarchy - in general, the term refers to the ordered body of clergy, divided into bishops, priests, and deacons. In Catholic practice, the term refers to the bishops of the world or of a particular region. Holy Days of Obligation - feasts in Latin-rite churches on which Catholics are required to attend Mass. Holy See - I) The diocese of the pope, Rome. 2) The pope himself or the various officials and bodies of the Church's central administration--the Roman Curia--which act in the name and by authority of the pope. Holy Communion - after saying a preparatory prayer, the celebrant (or other designated ministers) gives communion (the consecrated bread and wine) to himself and the other ministers at the altar, and then communion is distributed to the congregation. Homily - a reflection by the celebrant or other minister on the Scripture readings and on the application of the texts in the daily lives of the assembled community. Host, The Sacred - the bread under whose appearances Christ is and remains present in a unique manner after the consecration of the Mass. Humanae Vitae - The 1968 encyclical by Pope Paul VI on married love and procreation. Hymnal/Missalette - contains all parts of the Mass for a specific season in the liturgical year, including instructions on when to stand, sit, or kneel. I IHS - in Greek, the first three letters of the name of Jesus. Immaculate Conception - Catholic dogma concerning Mary and the name of a feast in her honor celebrated Dec. 8. It refers to the Catholic belief that Mary was without sin from the moment she was conceived. Incense - material used to produce a fragrant odor when burned and used as a symbol of the Church's offering and prayer going up to God. Used on major feast days and for funerals, it symbolizes communication with God. The image of smoke rising to the heavens in combination with the fragrance it emits, invoke a connection with the divine. Indulgence - the remission before God of the temporal punishment due for sins already forgiven. Infallibility – The gift of the Spirit by which the pope and bishops in union with him are protected from fundamental error when formulating a specific teaching on matters of faith and morals. Intercessions - A series of prayers for the Church, the world, the Pope, clergy and laity, and the dead. Intercommunion - The agreement or practice of two Ecclesial communities by which each admits members of the other communion to its sacraments. J Jesus - The name of Jesus means Savior. K Keys, Power of the - spiritual authority and jurisdiction in the Church, symbolized by the keys of the kingdom of heaven, Christ promised the keys to St. Peter and head-to-be of the Church. - top L Laicization - the process by which a man ordained is relieved of his obligations and is returned to the status of a lay person. Lamb of God - an invocation during the breaking of the bread in which the assembly petitions for mercy and peace. Lay ministries - these are ministries within the church that are carried out by laypersons. Included are altar servers, Eucharistic minister and lectors. Layman, woman, person - any church member who is neither ordained nor a member of a religious order. When the Second Vatican Council spoke of the laity, it used the term in this more common meaning. Lectionary - contains the scripture readings for Mass. Liturgy - The public prayer of the Church. Liturgy of the Word - the occasion during Mass when readings from the Scriptures are proclaimed and reflected upon. On Sundays and major feasts, there are three readings: • First reading - from the Old Testament • Second reading - from the Epistles • Gospel Liturgy of the Hours - this is the preferred term in the Latin rite for the official liturgical prayers sanctifying the parts of each day. Liturgy of the Eucharist - the section of the celebration when the gifts of bread and wine are prepared and the Eucharistic Prayer is proclaimed by the celebrant, and the Blessed Sacrament is distributed to the assembly. Lord's Prayer (Our Father) - the prayer of petition for both daily food (which means the Eucharistic bread for Christians) and the forgiveness of sins. M Magisterium - the official teaching office of the Church. Mass - the common name for the Eucharistic liturgy of the Catholic Church. Also referred to as Eucharist, Celebration of the Liturgy, Eucharistic celebration, Sacrifice of the Mass, Lord's Supper. Master of Ceremonies - one who assists in the preparation of the celebration and is present during it to facilitate the movement of the entire rite. Minister - from the Latin word for "servant," in the ecclesiastical sense a minister is (1) an ordained cleric or (2) one who has the authority to minister to others. Ministers of Communion - Those who assist in the distribution of communion. Miracles - generally miracle is used to refer to physical phenomena that defy natural explanation, such as medically unexplainable cures. An apparition is a supernatural manifestation of God, an angel or a saint to an individual or a group of individuals. Miter (MY-ter) - a headdress worn at solemn liturgical functions by bishops, abbots and, in certain cases, other clerics. Monastery - an autonomous community house of a religious order, which may or may not be a monastic order. The term is used more specifically to refer to a community house of men or women religious in which they lead a contemplative life separate from the world. Monk - a member of a monastic community such as Benedictine, Carthusian, Trappist, etc. Monks tend to live lives more separate from society to pursue, under a formal rule, a life of prayer and work for God's glory, for personal sanctification, and for the good of the Church and world. Monastic communities may have some outside works connected with them, such as a college or retreat house, but their primary ministry is prayer, especially the Liturgy of the Hours. Monsignor - an honorary ecclesiastical title granted by the Pope to some diocesan priests. In the United States, the title is given to the vicar general of a diocese. In Europe, the title is also given to bishops. Mortification - acts of self-discipline, including prayer, hardship, austerities and penances undertaken for the sake of progress in virtue. N National Conference of Catholic Bishops (NCCB) - Episcopal conference of U.S. bishops. The membership is comprised of diocesan bishops and their auxiliary bishops. The conference decides matters of ecclesiastical law and issues policy statements on political and social issues. Newman Apostolate - an apostolate to the Catholic college and university community, now commonly known as "campus ministry." Nun - a member of a religious order of women with solemn vows O Offertory Song - music used during the procession of gifts to the celebrant and as the altar is prepared. Opening prayer - this prayer by the celebrant expresses the general theme of the celebration. Ordain - the proper terms in Catholic usage for references to the conferral of the sacrament of holy orders on a deacon, priest or bishop. Order, Congregation, Society - religious orders is a title loosely applied to all religious groups of men and women. A society is a body of clerics, regular or secular, organized for the purpose of performing an apostolic work. Congregation is any group bound together by common rules. Ordinary - diocesan bishops, religious superiors, and certain other diocesan authorities with jurisdiction over the clergy in a specific geographical area, or the members of a religious order. Ordination - the act that enables a person to act on behalf of the Church through Word, Sacrament, and leadership. A bishop is ordained to represent Christ. Priests share in the bishop's role of representing Christ the Shepherd. Deacons collaborate with the bishop in his role as representative of Christ the Servant. P Pall (PAHL) - the stiff, square, white cover that is placed over the paten when it is on the chalice. Pallium - special stole made of lamb's wool worn over the chasuble by the Pope and archbishops; it signifies communion of archbishops with the Holy See. Papal Infallibility - The end result of divine assistance given the pope, wherefore he is prevented from the possibility and liability of error in teachings on faith or morals. Papal Representatives - the three types of representative of the Roman Pontiff are: 1. Legate - An individual appointed by the Pope to be his personal representative to a nation, international conference, or local church. The legate may be chosen from the local clergy of a country. 2. Apostolic Pro-Nuncio - In the United States, the papal representative is sent by the Pope to both the local church and to the government. His title is Apostolic Pro-Nuncio. Although he holds the title of ambassador, in U.S. law he is not accorded the special privilege of being the dean of the diplomatic corps. In countries where he is dean of the diplomatic corps, his title is Apostolic Nuncio. 3. Permanent Observer to the United Nations - The Apostolic See maintains permanent legates below the ambassadorial level to several world organizations. Since the Papal Legate does not enjoy the right to vote within the organization, his title at the United Nations is that of Observer. Parish - a specific community of the Christian faithful within a diocese which has its own church building and is under the authority of a pastor who is responsible for providing the faithful with ministerial service. Most parishes are formed on a geographic basis, but they may be formed along national or ethnic lines. Parish Administrator - when a parish is without a pastor or a pastor is unable to fulfill his pastoral responsibilities, a priest administrator is appointed by the bishop and is bound by the same obligations and enjoys the same rights as a pastor. Pastor - a priest appointed by a bishop to attend to the pastoral care of one or more parishes. He is responsible for administering the sacraments, instructing the congregation in the doctrine of the Church, and other services to the people of the parish. The pastor fulfills his responsibilities in the areas of teaching, sanctifying and administration with the cooperation of and assistance from other priests as well as deacons and/or lay persons. Pastoral Associate – a member of the laity who is part of the parish ministry team. A certified pastoral associate is a generalist serving in a key parish leadership position. He/she assists the pastor or parish director in the daily operation of the parish. This may include the direct coordination of one or more specific ministries, such as sacramental planning, educational formation, pastoral ministry and parish administration. The pastoral associate is hired directly by the parish and is accountable to the pastor or parish director. Pastoral Council - a group of members of the parish who advise the pastor on parish matters. Pastoral Team - refers to a group of priests assigned to the pastoral care of a parish or parishes with one of them as moderator. All priests who are members of the team have the same responsibilities and rights as a pastor. Paten (PAT-en) - a saucer-like disk that holds the bread that becomes the Body of Christ. Pectoral Cross - a cross worn on a chain about the neck of bishops and abbots as a mark of office. Penitential Rite - a general acknowledgement of sinfulness by the entire assembly, accompanied by requests for God's mercy and forgiveness. Prayer after Communion - The final prayer by the celebrant in which he petitions that the sacrament be beneficial for all. Prayer over the gifts - The prayer by the celebrant asking that the gifts to be offered be made holy and acceptable. Prayer - the raising of the mind and heart to God in adoration, thanksgiving, reparation and petition. The official prayer of the Church as a worshiping community is called liturgy. Preface dialogue - the introductory dialogue between the celebrant and assembly in which all are invited to join in prayer and thanksgiving to God. Preparation of the Gifts - the time in the Mass when the bread and wine to be used in the celebration are brought to the celebrant, usually by representatives of the faithful. Presbyterial Council - Also known as the priests' council, this is the principal consultative body mandated by the Code of Canon Law to advise the diocesan bishop in matters of pastoral governance. It consists of bishops and priests serving the diocese. Primacy - Papal primacy refers to the pope's authority over the whole church. Processional Cross - the cross carried in the processions. Profession of Faith - the assembly joins to recall and proclaim the fundamental teachings of the Roman Catholic faith. The Profession of Faith is also called the Creed. Purgatory - The state or condition in which those who have died in the state of grace, but with some attachment to sin, suffer for a time as they are being purified before they are admitted to the glory and happiness of heaven. Purificator - a white cloth used to cleanse the chalice. R Reader - one who is called upon to proclaim the scriptures during the Liturgy of the Word. A reader may also read the prayers of the faithful at Mass, in the absence of a deacon. Relics - the physical remains and effects of saints, which are considered worthy of veneration inasmuch as they are representative of persons in glory with God. Religion - the adoration and service of God as expressed in divine worship and in daily life. Religious Brother - a man who takes vows and promises to use his talents to serve God wherever the community decides he is needed. Brothers do not get married, live in religious communities, and have many different jobs. They are not ordained. Religious Order - a community of people with a particular charism, as expressed by its founder, and recognized by the Church is a religious order. There are religious orders of priests and brothers, and religious orders of sisters. Religious communities may also have lay associates. Some religious orders are dedicated primarily to prayer (contemplative), while others focus on apostolic (active) ministries. Religious Priest/Diocesan Priest - religious priests are professed members of a religious order or institute. Religious clergy live according to the rule of their respective orders. In pastoral ministry, they are under the jurisdiction of their local bishop, as well as the superiors of their order. Diocesan, or secular, priests are under the direction of their local bishop. They commit to serving their congregations and other institutions. Religious Sister - a woman who belongs to a religious community. Religious sisters make vows and serve God according to the charisms of their communities. Sisters are not married and work in many different jobs, according to the needs of the religious community and/or the needs of the local area. A sister in a cloistered religious community is a nun. Responsorial Psalm - the psalm that is spoken or sung between the first and second readings. The response is repeated after each verse. Retreat - a period of time spent in meditation and religious exercise. Retreats may take various forms, from traditional closed forms, to open retreats which do not disengage the participants from day-to-day life. Both clergy and lay people of all ages participate in retreats. Houses and centers providing facilities for retreats are retreat houses. Rite of Christian Initiation of Adults (RCIA) - the norms and rituals of the Catholic Church for people who wish to join the Church. Part of the book is intended for baptized Christians who wish to become Catholics. The term is used in a general sense to refer to the process of entering the Catholic Church. Roman Curia - the official collective name for the administrative agencies and courts, and their officials, who assist the Pope in governing the Church. Members are appointed and granted authority by the Pope. Rosary - a prayer of meditation primarily on events in the lives of Mary and Jesus, repeating the Our Father and Hail Mary. It is generally said on a physical circlet of beads. S Sacramentary – the book used by the celebrant at Mass, containing all the prayers for the liturgy of the Mass, including the opening prayer, prayer over the gifts, prayer after communion, and solemn blessings, Eucharistic prayers and prefaces for all of the Masses, including special occasions. Sanctuary - the part of the church where the altar is located. Second Vatican Council - a major meeting of the bishops of the world convened by Pope John XXIII to bring about a renewal of the Church for the second half of the 20th century. It ran from 1962 to 1965 and produced important documents involving liturgy, ecumenism, communications and other areas. Secular Institutes - societies of men and women living in the world who dedicate themselves to observe the evangelical counsels and to carry on apostolic works suitable to their talents and opportunities in every day life. See - another name for diocese or archdiocese. Seminary - an educational institute for men preparing for Holy Orders. Shrine - erected to encourage private devotions to a saint, it usually contains a picture, statue or other religious feature capable of inspiring devotions. Sign of Peace - before sharing the body of Christ, the members of the community are invited to express their love and peace with one another. Sign of the Cross - a sign, ceremonial gesture or movement in the form of a cross by which a person confesses faith in the Holy Trinity and Christ, and intercedes for the blessing of himself, other persons, and things. Sodality - a group of laity, established for the promotion of Christian life and worship, or some other religious purpose. Stole - a long, cloth scarf that marks the Office of the priest or deacon according to the manner in which it is worn. A priest wears it around the neck, letting it hang down in front. A deacon wears it over his left shoulder, fastening it at his right side. Superior - the head of a religious order or congregation. He or she may be the head of a province, or an individual house. Surplice (SIR-plis) - worn over the cassock, this is a wide-sleeved garment that when slipped over the head, covers the shoulders and extends down below the hips. Synod - a gathering of designated officials and representatives of a church, with legislative and policymaking powers. T Tabernacle - place in the church where the Eucharist or sacred species is reserved. Theologate - an institution which provides the last four years of study for candidates for priesthood. Theology - the study of God and religion, deriving from and based on the data of Divine Revelation, organized and systematized according to some kind of scientific method. Titular Sees - dioceses where the Church once flourished but which later died out. Bishops without a territorial or residential diocese of their own, e.g., auxiliary bishops, are given titular sees. Tribunal - a tribunal (court) is the name given to the person or persons who exercise the Church's judicial powers. Triduum - the three days of the liturgical year which incorporates the celebrations of Holy Thursday, Good Friday, and the Easter Vigil. V Vatican Councils - councils called by the pope of all bishops of the Church. These councils are usually called to discuss specific matters of interest to the Church. Vatican Congregation – a Vatican body which is responsible for an important area in the life of the Church, such as worship and sacraments, the clergy, and saints causes. Veneration of the altar - the reverencing of the altar with a kiss and the optional use of incense. Vespers - a portion of the Church's divine office recited each day by priests. Also called Evening Prayer. Vestment - the vesture the ministers wear. Vow - a promise made to God with sufficient knowledge and freedom, which has as its object a moral good that is possible and better than its voluntary omission. Vows that religious take include the vows of poverty, chastity, and obedience. W Washing of Hands - an expression of the desire for inward purification. The celebrant washes his hands in symbolic cleansing to prepare himself just as the gifts have been prepared as an offering to the Lord. Witness, Christian - practical testimony or evidence given by Christians of their faith in all circumstances of life--by prayer and general conduct, through good example and good works, etc., being and acting in accordance with Christian belief, actual practice of the Christian faith. Z Zucchetto (zoo-KET-oh) - the skull cap worn by the Pope (red), bishops (purple) and cardinals (red).
189193	https://mathematicallyeducated.com/2017/11/19/why-does-elimination-work/	Why does elimination work? - Mathematically Educated Skip to content Mathematically Educated Thoughts on the teaching and learning of mathematics Widgets Home About Speaking Engagements Follow this Blog Contact Posted on November 19, 2017 November 19, 2017 Why does elimination work? Scroll down to see more content Teachers typically emphasize two techniques (though there are certainly others) for solving systems of equations: substitution and elimination. The other day I asked my class of prospective middle and secondary teachers to explain why elimination works. I asked this because as I was sitting at my kitchen table that morning, My students were not able to answer my question. I realized I too had never thought about why this commonly taught procedure works. Solving systems of linear equations is a pretty important topic in secondary mathematics, so it is definitely a question worth exploring. Consider the system of equations below. 2 x+ 3 y= 14 -2 x− 5 y= -2 As long as the lines are not parallel or the same line, they will intersect at just one point (x,y). Solving this system means that you are searching for that particular coordinate point that satisfies both equations. This is the goal of solving a system of linear equations, to determine whether or not there are points that satisfy both equations. Elimination is one method to do this. With this method, you would add the two equations together in order to eliminate one of the variables so that you are left with a single equation with a single variable. So in the example above, you would get the following: (2 x+ 3 y) + (-2x− 5y)= 14 + -2 -2 y = 12 y = -6 You can then substitute in this y value into either of the original equations to determine the related value of x, like so. 2 x+ 3 y= 14 2 x + 3(-6) = 14 2 x + -18 = 14 2 x = 32 x = 16 Thus, the solution to the system is the point (16, -6). Ordinarily, you cannot add a term from one equation to another because they are not indeed like terms. That is to say, an x in one equation is not necessarily the same x from another equation. For example, consider x + 3 = 5 and x− 2 = 7. The value of x in the first equation is 2 and the value of x in the second equation is 9. Therefore, you can’t simply add the equations together to get x + 3 + x− 2 = 5 + 7 because the x‘s are not like terms. Technically, you have just added 7 to both sides of the x + 3 = 5 equation, however, then you would also have added x + x but, as stated above, the x‘s don’t represent the same unknown value. If you ignore this fact, that would mean, that you would have 2 x + 1 = 12, but that doesn’t make sense because you just added an x with a value of 2 (from the first equation) to an x with a value of 9 (from the second equation). If you add those x’ s, you should have gotten 11 (2 + 9), however, the 2x means it should be double the value of x, so if you take it as the x from the first equation, this would mean 2x should be 4, but if you use the value of x from the second equation, this would mean 2x would be 18. You cannot have 2x simultaneously equal 18 and 4, and this is indeed the wrong answer based on the actual values of each of the x‘s. It would be okay to add these together if you, for example, keep the x from the first equation as x and then rename the x from the second equation to something else like y. This would result in the following:x + 3 + y− 2 = 5 + 7. Back to elimination, we do indeed keep the x‘s the same and add them as if they were like terms because they are indeed like terms. This is because we are looking for the solution for which those x‘s are equal. If this were not the case, we could not add the x‘s together. This is an important point that should be emphasized with students. So in summary, elimination works because we are looking for the values of x and y that are the same in both equations. When we add the two equations together, we are really adding the same value to both sides of one of the equations, which is allowed. If we were to annotate the steps in the process above it would look like this: To solve this system of linear equations 2 x+ 3 y= 14 -2 x− 5 y= -2 Add -2 (in the form of-2 x− 5 y on the left ) to both sides, which is okay because we are looking for the point where x and y are the same for both equations. (2 x+ 3 y) +(-2 x− 5 y)= 14 + -2 Simplify the new equation and solve for y. -2 y = 12 y = -6 Substitute the value of y into either of the original equations to determine the corresponding x-value. 2 x+ 3 y= 14 2 x + 3(-6) = 14 2 x + -18 = 14 2 x = 32 x = 16 Thus, the solution to the system is the point (16, -6). Teachers should encourage their students to discuss these steps and the rationale as they solve to make sure they understand why this works. I hope this was helpful! Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like this: Like Loading... Related Discover more from Mathematically Educated Subscribe to get the latest posts sent to your email. Type your email… Subscribe Published by Zandra de Araujo View all posts by Zandra de Araujo Categories Mathematical ContentTags Algebra3 Comments 3 thoughts on “Why does elimination work?” Pingback: Substitution and Elimination - Mathematically Educated Cassandra Petersonsays: November 4, 2022 at 11:39 am Thank you for this. I’ve been struggling to explain to students why this works and this explanation is far simpler than anything I had come up with. Thank you! Loading... Reply 3. Shaniyal Sohailsays: March 8, 2023 at 10:03 am Extremely helpful. Thankyou soo much. Loading... 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189194	https://www.teacherspayteachers.com/browse?search=solving%20equations%20with%20distributing	Solving Equations With Distributing \| TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn 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It does include parenthesis for the distributive property. Integers / Positive and Negative numbers can be found in this worksheet. 5 th - 9 th Basic Operations, Math, Order of Operations $3.00 Original Price $3.00 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Solving Equations with Variables on Both Sides Coloring Activity Created by Secondary Math Solutions This is a fun coloring activity over Solving Equations with Variables on Both Sides. TEKS Aligned: 8.8c This coloring page contains 12 equations with variables on both sides. The student does not need to use the distributive property or combine like terms. Once the student solves the equation, they find their answer in the tessellation design and color it. Answers are similar to make the activity more difficult. Room to show work is included. All solutions are integers. Coloring is engaging an 7 th - 9 th Algebra, Algebra 2, Other (Math) Also included in:Variables on Both Sides Unit Bundle for 8th Grade Texas TEKS Aligned Curriculum $1.50 Original Price $1.50 Rated 4.85 out of 5, based on 60 reviews 4.9(60) Add to cart Wish List Solving Multi-Step Linear Equations with Distributing Activity Print and Digital Created by Math Beach Solutions This Justify-and-Sequence activity is designed for Google Slides™ and Google Classroom™. The printable scaffolded worksheet version of this activity allows for flexible implementation. Students will solve linear equations, including equations with variables on both sides and those requiring the use of the distributive property to solve. The optional concept check slide for increased accountability is a great option for distance learning. INCLUDES: 2 drag-and-drop slides with 2 multi-step prob 8 th - 9 th Algebra, Math CCSS HSA-REI.A.1 , HSA-REI.B.3 Also included in:Linear Equations and Inequalities - Unit 1 Bundle - Algebra 1 Curriculum $3.00 Original Price $3.00 Rated 4.56 out of 5, based on 9 reviews 4.6(9) Add to cart Wish List Solving Two (2) Step Equations with Distributing Scavenger Hunt Activity Created by Coffee Break Mathematics This differentiated scavenger hunt activity includes 2 separate scavenger hunts that involve solving two (2) step linear equations in one variable with distributing. The first Scavenger Hunt uses coefficients of 1 inside the parentheses while the second Scavenger Hunt uses integer coefficients inside the parentheses. This is a self checking activity. What's Included:Directions for how to use the scavenger hunts10 Scavenger Hunt Cards for each scavenger huntStudent workspacesFull Answer 7 th - 9 th Algebra, Math CCSS 7.EE.B.4a , 8.EE.C.7 , 8.EE.C.7a +1 Also included in:Solving Equations Scavenger Hunt BUNDLE $2.50 Original Price $2.50 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Solving Multi-Step Linear Equations with Distributing Activity Print and Digital Created by Math Beach Solutions This unscramble can be used as a drag-and-drop digital activity designed for Google Slides™ and Google Classroom™ or printed to use as an unscramble worksheet activity. This activity is great for independent practice or at-home learning. Students will get immediate feedback by verifying that their "unscramble" makes a real word or by discovering that their word is nonsense and they need to try again. Students will solve linear equations, including equations with variables on both sides and tho 8 th - 9 th Algebra, Math CCSS HSA-REI.B.3 Also included in:Linear Equations and Inequalities - Unit 1 Bundle - Algebra 1 Curriculum $3.00 Original Price $3.00 Rated 4.67 out of 5, based on 3 reviews 4.7(3) Add to cart Wish List Solving Equations: Variables on BOTH Sides (With Distributing) EDITABLE Notes Created by Fun with Algebra These guided notes notes will help students learn how to solve equations with variables on both sides. Students will need to distribute in this lesson and will have to combine like terms on both sides. There are a total of 2 pages of notes, 2 pages of practice (that can be used as classwork or homework) and a 4 question exit ticket to have students practice and master the order of operations. What's included: PDF download of notes going through solving equations with variables on both sides. D 7 th - 10 th Algebra, Math CCSS HSA-REI.B.3 Also included in:Algebra 1 Curriculum EDITABLE Year of Notes Practice, Exit Tickets & Assessments $3.00 Original Price $3.00 Add to cart Wish List Solving Equations with Distributing Partner Activity Created by Math With Friends In this no-prep Algebra worksheet students will be partnered up to practice Solving Equations with Distributing. Problems are included that require combining like terms! Students will check their work by comparing answers. Both students have a set of 10 problems. Although students have different questions, they should receive the same answers! This activity can be used in a variety of ways: homework, in class practice, or review. Use it however will best help your students master the concept. Th 7 th - 10 th Algebra, Math Also included in:Algebra Fundamentals Partner Activity BUNDLE $3.00 Original Price $3.00 Rated 5 out of 5, based on 3 reviews 5.0(3) Add to cart Wish List Solving Equations with Distributing and Combining Guided Notes 8.EE.7b Created by Middle School Math Camp These guided notes include step-by-step directions for students to solve equations that include distributing and combining like terms. All the examples in this resource have the variables ON ONE SIDE. Two versions of these guided notes are included: a full 8.5 x 11 sheet of paper that can be printed front to back, or a foldable to glue into an interactive notebook. An answer key is also included. Please see the preview to view all the equations included. 8 th Math CCSS 8.EE.C.7b Also included in:Solving Equations BUNDLE 8.EE.7b $3.00 Original Price $3.00 Add to cart Wish List Solving Equations with fractions, decimals, negatives, distributing, like terms Created by Time for Math with Mrs W This practice page includes terms with fractions, decimals and integers. The students need to simplify and solve multi-step equations which involve like terms and distributing. Two word problems are included. I use this as a hw assignment for my 7th graders. 7 th - 8 th Algebra, Math CCSS 7.EE.A.1 , 7.EE.B.3 , 7.EE.B.4a +2 $2.00 Original Price $2.00 Add to cart Wish List Solving Multi-Step Equations Mystery Activity (Scavenger Hunt) Print & Digital Created by Ready-Made Resources Looking for an activity that makes solving multi-step equations FUN? In this CUSTOMIZALBE activity, students solve multistep equations to solve a mystery! You'll love the quick and easy set-up; your students will love solving the case. UPDATED TO INCLUDE TWO VERSIONS: This activity now includes two versions of the mystery. The original version does not require distributing and second version with distributing. (The distributing option is not yet available as a digital activity -- only print. 7 th - 9 th Algebra CCSS 7.EE.B.3 Also included in:Solving Equations Scavenger Hunt BUNDLE! EIGHT Customizable Mystery Activities $3.50 Original Price $3.50 Rated 4.87 out of 5, based on 429 reviews 4.9(429) Add to cart Wish List Solving Equations with Special Cases Activity \| Linear Equations Activity Created by Maneuvering the Middle This Solving Equations with Special Cases Card Sort is great for reinforcing students' knowledge of equations and their solutions. Students will determine whether the equation on each card has one solution, no solutions or infinitely many solutions and record them in the appropriate column. This activity is great for extra practice, as a station or center, and can be used to add variety to homework or independent work. Students can be hands-on, while still applying their understanding of equa 8 th Algebra, Math CCSS 8.EE.C.7a Also included in:Linear Equations Activity Bundle \| Multi-Step Equation Activities \| 8th Grade $2.00 Original Price $2.00 Rated 4.92 out of 5, based on 93 reviews 4.9(93) Add to cart Wish List Solving Multi-Step Equations With Variables on ONE Side Scavenger Hunt Activity Created by Fun with Algebra Students will practice solving multi-step equations with variables on one side. Variables could be on the left or on the right, but there are only variables on one side of the equation. Students will need to combine like terms to solve the equation or distribute then combine like terms. Skills students need to complete the activity: Solve Equations by Combining Like TermsSolve Equations by Distributing then combining like termsSolve equations by distributing twice (like 2(4x-8)+3(x-7)=21) then c 6 th - 9 th Algebra, Basic Operations, Math Also included in:Algebra 1 Year Long Printable Activities Bundle (80+ Activities and Growing!) $3.30 Original Price $3.30 Rated 4.9 out of 5, based on 20 reviews 4.9(20) Add to cart Wish List Multi Step Equations Worksheet Activity (Combine Like Terms &Distribute) Created by Mile High Mathematics A 30-question practice activity/worksheet that covers solving multi-step equations through use of inverse operations and isolating the variable. Students will also have to: use their knowledge of the distributive property and combining like terms to simplify the equation before solving. Perfect resource with a variety of possible applications. It can be used as an assessment, a practice activity to encourage mastery, or an in-class supplement to an engaging lesson. Aligned with the Common Core 5 th - 8 th Algebra, Math CCSS 6.EE.B.7 , 7.EE.B.4 $2.00 Original Price $2.00 Rated 4.8 out of 5, based on 5 reviews 4.8(5) Add to cart Wish List Solving Multi-Step Equations Math Pennant Activity with Distribution Created by Scaffolded Math and Science Students solve multi-step equations with the distributive property and variables on both sides in this activity that doubles as classroom decor. Every problem requires distribution to solve. Included are 20 pennants with distribution on one or both sides, an optional student answer sheet and an answer key. There are 2 pennants on each page. Once a pennant is complete, it can be hung along a string in your classroom to show the world that, "Hey, we know how to distribute!" Each group of students 8 th - 10 th Algebra, Math CCSS 8.EE.C.7 Also included in:All Math Pennants Activity Bundle \| Student-Created Math Classroom Decor $3.00 Original Price $3.00 Rated 4.95 out of 5, based on 34 reviews 5.0(34) Add to cart Wish List Solving Multi Step Equations with the Variable on One Side Maze Created by Hello Algebra This ready to resource includes a printable maze activity for students to practice solving multistep step equations! Each of the equations in this resource requires students to take multiple steps to solve for the variable. NOTE: The variable is only on one side.Students will be required to: expand expressionscombine like termsuse inverse operations to solve for the variableThis resources if great for: small group workindependent workclassworkhomeworkThis resource is ready to go! Once downl 8 th - 10 th Algebra, Math CCSS HSA-REI.B.3 Also included in:Solving Equations Mazes $1.50 Original Price $1.50 Rated 4.5 out of 5, based on 2 reviews 4.5(2) Add to cart Wish List Maze - Solving Multi Step Linear Equations: Distributive Property (One Side) Created by Never Give Up on Math This activity is a good review of understanding how to "Solve Multi-Step Linear Equations: Using the Distributive Property". The solutions of this maze are:☑ Integers The questions of this maze are of this nature:☑ Multi Step with parenthesis on the ONE side ☑ Variable maybe on right or left side Students should feel comfortable with:☑ Using the Distributive Property (distributing integers) ☑ Solving two step equations ☑ Solving one step equations There are 15 equations provided. From start to 6 th - 12 th, Adult Education, Higher Education Algebra, Algebra 2, Math CCSS HSA-CED.A.1 , HSA-REI.A.1 , HSA-REI.B.3 Also included in:Maze - BUNDLE All about Equations $1.50 Original Price $1.50 Rated 4.75 out of 5, based on 8 reviews 4.8(8) Add to cart Wish List Solving Equations with Variables on Both Sides Boom Cards™ Digital Activity Created by Kate's Math Lessons This set of 20 digital task cards is a great way to help students practice solving equations with variables on both sides. Students are shown one question at a time and get instant feedback on their answers! Solutions include positive and negative integers. Some equations require students to distribute first before moving all variables to the same side. Check out the preview to see how they work! Make sure to click on the full-size preview and hit submit to see if your answer is correct. Bo 8 th - 9 th Algebra, Math CCSS 8.EE.C.7 , 8.EE.C.7b , HSA-REI.B.3 Also included in:Solving Equations with Variables on Both Sides BUNDLE Printable & Digital $3.00 Original Price $3.00 Rated 4.74 out of 5, based on 19 reviews 4.7(19) Add to cart Wish List Solving Multi-Step Equations with Fractions Partner Activity Created by The Unique Expressions Students will pair up to solve 7 multi-step equations with fractions. Even though each student will be solving their own problem, all answers will be the same! This activity is differentiated into two parts: A and B involve solving equations with the variable on one or both sides (no distribution required)C and D are more difficult because students will have the variables on both sides and they will need to distribute one or two times in order to solve the equation. Please note: Students shou 7 th - 11 th Algebra, Math CCSS 8.EE.C.7b $3.00 Original Price $3.00 Rated 5 out of 5, based on 7 reviews 5.0(7) Add to cart Wish List Solving Multi-Step Equations with Distributive Property and Combining Like Terms Created by Eddie McCarthy This lesson is meant to be used as guided notes and assessments for an entire class. In the first section, students learn how to solve equations that involve the distributive property on one side or on both. In the second section, they learn how to solve equations that involve distributing and combining. There is helpful 5-step guide at the top of the notes. There are assessments for both sections, which are based off of the guided notes. Answer key is included! Related Activity: Solving Multi 7 th - 10 th Algebra, Algebra 2, Math CCSS 8.EE.C.7a , 8.EE.C.7b , HSA-REI.A.1 +1 $1.50 Original Price $1.50 Rated 5 out of 5, based on 24 reviews 5.0(24) Add to cart Wish List Solving Multi-Step Equations with Distribution & Combining Like Terms Boom Cards Created by Fall In Love With Math This set of 20 BOOM Cards allow students to practice solving Multi-Step Equations involving distribution and combining like terms. All problems require students to distribute once and combine like terms with all variables on one side of the equation. All answers are integer values. Solutions do not include special solutions. BOOM Cards provide instant feedback that students appreciate! You do need a BOOM Card account in order to use. Please be sure you have BOOM set up before purchasing. Studen 6 th - 12 th Algebra, Math $1.00 Original Price $1.00 Rated 4.81 out of 5, based on 26 reviews 4.8(26) Add to cart Wish List Solving Multistep Equations with Fractions Created by A Girl Who Loves Math This set of 28 task cards is for solving multistep linear equations with fractional coefficients. Students will combine fraction operations and algebraic manipulation to solve for the variable. Cards include improper fractions, negative coefficients, and answers that are whole numbers or fractions. Useful for small group instruction, review for assessments, and independent practice. This product is intended for Algebra 1, Algebra 2, and Geometry students and can be used as a review for Pre-Cal 8 th - 12 th Algebra, Algebra 2, PreCalculus $3.00 Original Price $3.00 Rated 5 out of 5, based on 6 reviews 5.0(6) Add to cart Wish List Solving Equations with Variables on Both Sides w/ Distribution Maze Activity Created by The Unique Expressions Students will solve equations with variables on both sides and have to distribute on one side of the equation. Student will solve 11 multi-step equations. If you'd like a digital version of this activity made for Google Drive, please click here! Please Note:Not all boxes will be used in this maze! This is done so that students have to focus on getting the correct answer versus just figuring out the correct route.What's Included: Maze and Answer Key Other Products:If you were looking for simila 8 th - 11 th Algebra, Math CCSS 8.EE.C.7 Also included in:Solving Equations with Variables on Both Sides Maze Activity Bundle $1.50 Original Price $1.50 Rated 4.88 out of 5, based on 8 reviews 4.9(8) Add to cart Wish List Solving Two-Step Equations with Fractions Task Cards Created by A Girl Who Loves Math This set of 28 task cards is for solving linear equations with fractional coefficients. Students will combine fraction operations and algebraic manipulation to solve for the variable. Cards include improper fractions, negative coefficients, and answers that are whole numbers or fractions. Useful for small group instruction, review for assessments, and independent practice. This product is intended for 8th grade math to Algebra 1 and can be used as a review for Geometry and Algebra 2 students. 8 th - 12 th Algebra, Algebra 2, Geometry $3.00 Original Price $3.00 Rated 4.88 out of 5, based on 8 reviews 4.9(8) Add to cart Wish List Combine Like Terms &Solve Multi Step Equations Digital Task Cards Google Slides Created by Middle School in the Mitten A no print, no-prep resource to give your students practice solving multi-step equations that incorporate combining like terms. This independent practice activity builds computer and math skills at the same time! This product includes 20 slides that require students to use combining like terms, inverse operations, and integer strategies to solve multi-step equations. This resource does not include problems that require distributing. This activity includes...With your purchase, you will be able 7 th - 10 th Algebra, Math, Math Test Prep CCSS 7.EE.A.1 , 8.EE.C.7 Also included in:Solving Equations Digital Task Card Bundle \| One Step \| Two Step \| Multi Step $3.49 Original Price $3.49 Rated 4.86 out of 5, based on 7 reviews 4.9(7) Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 1,400+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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189195	https://www.youtube.com/watch?v=wJKEkNdqHNc	find sine or cosine when given sine sin or cosine cos and quadrant of the angle Math Tutorials 1880 subscribers 3 likes Description 546 views Posted: 11 Mar 2021 In this example, we are told the value for sine or cosine and the quadrant where the angle is located. We then use the pythagorean theorem to plug in the value that we know and solve for the other function by creating a power equation. We then check if the function should be positive or negative in the given quadrant by using "all students take calculus". Each part of this process is explained and written out step-by-step. This video contains examples that are from Algebra and Trigonometry, 1st ed, by Abramson, Belloit, Falduto, Gross, Lippman et al. It is an open-source textbook from OpenStax that you may download for free at The text is licensed under the Creative Commons Attribution license. Transcript: okay let's take a look when we're given cosine of theta is negative five over seven and we've been indicated that theta is going to be in the second quadrant and we're asked to find sine of theta to do so we can use the pythagorean identity which just says sine squared plus cosine squared equals one so what i'm going to do is i'm going to plug in negative 5 7 in for cosine of theta and with this formula this really means sine of theta squared plus cosine of theta squared equals 1. so as we plug in i am going to leave it with sine squared of theta but remember it means the same thing as what's written up above plus negative 5 7 squared equals one now we have an equation to solve down so the first thing i'm going to do is go ahead and square negative 5 7 times negative 5 7 is going to be positive 25 49. equals 1. now we can make this into a power equation by moving that fraction to the other side so i'm going to leave sine squared of theta on this side and we're going to subtract so i'm going to get 1 minus 25 49 to move it to the other side now let's combine that 1 minus 25 49's together to do so i need a common denominator so let's rewrite the 1 as 49 over 49 right that means the exact same thing as one but this will give us sine squared of theta equals we already have a common denominator so let's say 49 minus 25 is 24 over 49 all right now it's a power equation what we want to do next is we want to get rid of that square on the left hand side so we'll apply a square root to both sides now we applied the square root to both sides of this equation so we want to keep a positive and negative for the time being but on the left hand side that'll counteract the square those will cancel we're left with sine of theta now we have to make the determination besides simplifying down here we want to determine should it be positive or negative so in my simplifying down i'm going to apply a square root to the numerator and the denominator i'm going to rewrite the 24 as 6 times 4 because 4 is a factor of 24 but it's also the biggest perfect square and in reducing that it'll be nice to be able to clean this up as much as possible now i'm going to clean this up and then we'll make the determination positive or negative so to finish this we have sine of theta that will be 2 square root of 6 over 7 all right just applying these square roots to each one of these factors but now to make that determination should it be positive or negative what we want to do is i always use this phrase all students take calculus so for all students take calculus we go around first second quadrant first second third fourth quadrants in that order in the first quadrant all of our functions are going to be positive these trigonometric functions in the second quadrant it's just sine third is tangent and fourth is cosine alright so we were told that our theta resides in the second quadrant so because it's in the second quadrant sine is positive in the second quadrant so we'll get a positive two square root of 6 divided by 7. all right get the phrasing down plug into the pythagorean identity it's going to be an important one hopefully not that bad to to memorize um and good luck to you on these until next time
189196	https://physics.stackexchange.com/questions/771014/why-must-an-electric-field-be-radial-due-to-symmetry	electrostatics - Why must an electric field be radial due to symmetry? - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why must an electric field be radial due to symmetry? Ask Question Asked 2 years, 2 months ago Modified1 year, 4 months ago Viewed 806 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I'm learning about Gauss's law and was reading Electricity and Magnetism by Purcell and Introduction to Electrodynamics by Griffiths where it said that due to spherical, cylindrical, or planar symmetry, the electric field vector always is in the radial direction with direction in the r vector - because it is the only unique direction. Can anyone explain this in further detail? electrostatics electric-fields symmetry gauss-law Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jul 6, 2023 at 13:02 Qmechanic♦ 222k 52 52 gold badges 636 636 silver badges 2.6k 2.6k bronze badges asked Jul 6, 2023 at 11:22 user364438 user364438 2 What case are you asking about - a point charge?Not_Einstein –Not_Einstein 2023-07-06 12:43:00 +00:00 Commented Jul 6, 2023 at 12:43 If a charge density has, say, spherical symmetric then so too does the potential have that same symmetry. E.g., ρ(→r)=ρ(→r)→ϕ(→r)=ϕ(r). Since the field is the gradient of the potential, the field points in the radial direction (the gradient of a function points in the direction of maximum change).hft –hft 2023-07-06 15:28:10 +00:00 Commented Jul 6, 2023 at 15:28 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Why must an electric field be radial due to symmetry? There is no general requirement for an electric field to be radial. A static electric field is radial when the charge density (the source of the field) has the appropriate symmetry. This follows from the relationship between the charge density and the potential (in Gaussian units): −∇2 ϕ(→r)=4 π ρ(→r), which can be inverted to find the particular solution: ϕ(→r)=∫d 3 r′ρ(→r′)1\|→r−→r′\|. To consider the effect of, say, spherical symmetry, you can expand 1\|→r−→r′\| in the basis of spherical harmonics ("Laplace expansion") to find: ϕ(→r)=∑ℓ,m 4 π(−1)m 2 ℓ+1 Y ℓ,m(ˆ r)∫d 3 r′r ℓY ℓ,−m(ˆ r′)ρ(→r′). If, say, the charge density ρ is spherically symmetric then: ρ(→r)=ρ 0(r)Y 0,0(ˆ r)=ρ 0(r)1√4 π from which Eq. (1) becomes: ϕ(→r)=∑ℓ,m 4 π(−1)m 2 ℓ+1 Y ℓ,m(ˆ r)∫d r′r′2 1 r 1>δ ℓ,0 δ m,0 ρ 0(r′).=√4 π∫d r′r′2 1 r 1>ρ 0(r′). In the further special case where r>r′, we find (surprise, surprise): ϕ(→r)=Q t o t r Or, in SI units: ϕ(→r)=Q t o t 4 π ϵ 0 r Or, in terms of the field: →E(→r)=Q t o t 4 π ϵ 0 r 3→r. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 28, 2024 at 20:21 answered Jul 6, 2023 at 15:22 hfthft 28.8k 2 2 gold badges 35 35 silver badges 83 83 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Consider the work done in bringing unit positive charge from infinity to a distance r from the point charge. Irrespective of the path taken the work done to arrive anywhere at a distance r from the charge is the same. Thus the spherical surface of radius r centred on the charge is an equipotential. Moving a test charge along this equipotential surface requires no work to be done, thus any force on the test charge is at right angles to the equipotential surface and that force is in the direction of an electric field line, ie radial. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 6, 2023 at 14:43 FarcherFarcher 106k 5 5 gold badges 88 88 silver badges 235 235 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Assuming you mean the electric field from a static point charge, you can use a coordinate system with the charge/source at the origin. Since the charge is a single static "point", it has spherical symmetry (no matter how you rotate the system, the system is unchanged). Integrating Gauss' law ∇⋅E=ρ ϵ 0 over the volume of a sphere (and using spherical coordinates, due to the symmetry), we find that the electric field only has a component along ˆ r. (Can be seen from the divergence theorem. I'm certain Griffiths does this.) You can define the charge density ρ using the Dirac delta function. I am not sure what you mean with "unique direction", sorry. This is my first post so I assume someone else can explain better! Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 6, 2023 at 12:02 PhiPhi 11 3 3 bronze badges 2 What about a group of point, static charges user364438 –user364438 2023-07-06 12:05:22 +00:00 Commented Jul 6, 2023 at 12:05 1 @EmilSriram No, a group of two or more (but a finite number) point charges will not produce a spherically symmetric field. The charge distribution of a group of two or more point charges generally does not have spherical symmetry about any point, which leads to the electrostatic potential also not having that symmetry, which means the field doesn't have to be radial.hft –hft 2023-07-06 15:29:59 +00:00 Commented Jul 6, 2023 at 15:29 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Suppose that you are standing in front of the charge and the electric field not only has a radial but also has a tangential field.Now suppose you close your eyes and someone rotates the charge distribution by some certain angle about a diameter which points towards you.Now the radial component will be the same and directed towards you but the tangential component will also have rotates by that certain angle.Now suppose you open your eyes.If the charge distribution has spherical symmetry then the charge distribution would look exactly the same to you as before but now the field will be in a different direction(because the tangential component is rotated by some angle).The same charge distribution cannot produce different electric field,so the tangential component of the field must be zero. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 6, 2023 at 15:50 Niladri SarkarNiladri Sarkar 187 11 11 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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189197	https://www2.lauterbach.com/pdf/trace_tutorial.pdf	MANUAL Release 09.2024 Trace Tutorial Trace Tutorial \| 2 ©1989-2024 Lauterbach Trace Tutorial TRACE32 Online Help TRACE32 Directory TRACE32 Index TRACE32 Debugger Getting Started ..............................................................................................  Trace Tutorial ................................................................................................................................. 1 History ......................................................................................................................................... 3 About the Tutorial ...................................................................................................................... 3 What is Trace? ............................................................................................................................ 3 Trace Use Cases 4 Trace Methods ............................................................................................................................ 5 Simulator Demo .......................................................................................................................... 6 Trace Configuration ................................................................................................................... 7 Trace Recording ......................................................................................................................... 8 Displaying the Trace Results .................................................................................................... 10 Trace List 10 Displaying Function Run-Times 13 Graphical Charts 13 Numerical Statistics and Function Tree 14 Duration Analysis 15 Distance Analysis 16 Variable Display 17 Track Option 18 Searching Trace Results ........................................................................................................... 19 Trace Save and Load ................................................................................................................. 20 Trace Tutorial \| 3 ©1989-2024 Lauterbach Trace Tutorial Version 05-Oct-2024 History 18-Jun-21 New manual. About the Tutorial This tutorial is an introduction to the trace functionality in TRACE32. It shows how to perform a trace recording and how to display the recorded trace information. For simplicity, we use in this tutorial a TRACE32 Instruction Set Simulator, which offers a full trace simulation. The steps and features described in this document are however valid for all TRACE32 products with trace support. The tutorial assumes that the TRACE32 software is already installed. Please refer to “TRACE32 Installation Guide” (installation.pdf) for information about the installation process. Please refer to “Debugger Tutorial” (debugger_tutorial.pdf) for an introduction to debugging in TRACE32 PowerView. What is Trace? Trace is the continuous recording of runtime information for later analysis. In this tutorial, we use the term trace synonymously with core trace. A core trace generates information about program execution on a core, i.e. program flow and data trace. The TRACE32 Instruction Set Simulator used in this tutorial supports a full trace simulation including the full program flow as well as all read and write data accesses to the memory. A real core may not support all types of trace information. Please refer to your Processor Architecture Manual for more information. Trace Tutorial \| 4 ©1989-2024 Lauterbach Trace Use Cases Trace is mainly used in the following cases: 1. Understand the program execution in detail in order to find complex runtime errors more quickly. 2. Analysis of the code performance of the target code 3. Verification of real-time requirements 4. Code-coverage measurements Trace Tutorial \| 5 ©1989-2024 Lauterbach Trace Methods TRACE32 supports various trace methods. The trace method can be selected in the Trace configuration window, which can be opened from the menu Trace > Configuration… If a trace method is not supported by the current hardware/software setup, it is greyed out in the trace configuration window. NONE means that no trace method is selected. We use in this tutorial the trace method Analyzer. Please refer to the description of the command Trace.METHOD for more information about the different trace methods. Trace Tutorial \| 6 ©1989-2024 Lauterbach Simulator Demo We use in this tutorial a TRACE32 Simulator for Arm. The described steps are however valid for the TRACE32 Simulator for other core architectures. To load a demo on the simulator, follow these steps: 1. Start the script search dialog from the menu File > Search for scripts… 2. Enter in the search field “compiler demo” 3. Select a demo from the list with a double click, a PSTEP window will appear. Press the “Continue” button. We will use here the demo “GNU C Example for SRAM”. Trace Tutorial \| 7 ©1989-2024 Lauterbach Trace Configuration In order to set up the trace, follow these steps: 1. Open the menu Trace > Configuration… The trace method Analyzer [A] should be selected per default. If this is not the case, select this trace method 2. Clear the contents of the trace buffer by pressing the Init button [B]. 3. Select the trace operation mode [C]. In mode Fifo, new trace records will overwrite older records. The trace buffer includes thus always the last trace cycles before stopping the recording. In Mode Stack, the recording is stopped if the trace buffer is full. The trace buffer always includes in this case the first cycles after starting the recording. Mode Leash is similar to mode Stack, the program execution is however stopped when the trace buffer is nearly full. TRACE32 supports other trace modes. Some of these modes depend on the core architecture. Please refer to the documentation of the command Trace.Mode for more information. We will keep here the default trace mode selection, which is Fifo. 4. The SIZE field [D] indicates the size of the trace buffer. As we are using a TRACE32 Simulator, the trace buffer is reserved by the TRACE32 PowerView application on the host. It is thus possible to increase the size of this buffer. If a TRACE32 trace hardware is used with a real chip, the size of the trace buffer is limited by the size of the memory available on the trace tool. In order to have a longer trace recording, we will set the trace buffer size to 10000000. B A C D Trace Tutorial \| 8 ©1989-2024 Lauterbach The same configuration steps can be performed using the following PRACTICE script: Trace Recording Press the Go button to start the program execution. The trace recording is automatically started with the program execution. The state in the Trace window changes from OFF to Arm [A]. The used field displays the fill state of the trace buffer [B]. In order to stop the trace recording, stop the program execution with the Break button. The state in the trace window changes to OFF . Trace.METHOD Analyzer Trace.Init Trace.Mode Fifo Trace.SIZE 10000000. B A C Trace Tutorial \| 9 ©1989-2024 Lauterbach The trace recording is automatically started and stopped when starting and stopping the program execution because of the AutoArm [C] setting in the Trace window, which is per default enabled. The trace recording can also be started/stopped manually while the program execution is running using the radio buttons Arm and OFF of the Trace window [A]. Trace Tutorial \| 10 ©1989-2024 Lauterbach Displaying the Trace Results TRACE32 offers different view for displaying the trace results. This document shows some examples. Please note that the trace results can only be displayed if the trace state in the Trace window is OFF . It is not possible to display the trace results while recording. The caption of a TRACE32 window includes the TRACE32 command that can be executed in the TRACE32 command line or in a PRACTICE script to open this window, e.g. here Trace.List Trace List A list view of the trace results can be opened from the menu Trace > List > Default. The same window can be opened from the Trace configuration window by pressing the List button. Trace Tutorial \| 11 ©1989-2024 Lauterbach The Trace.List window displays the recorded trace packets together with the corresponding assembler and source code. In our case, trace packets are program fetches (cycle fetch) or data accesses (e.g. wr-long and rd-long for 32bit write and read accesses). Each trace packet has a record number displayed in the record column. The record number is a negative index for Fifo mode. As we are using a Simulator, each assembly instruction has an own trace packet. This is not the case with a real hardware trace. The displayed information can be reduced using the Less button. By pressing Less three times, only the high-level source code is displayed. This can be reverted using the More button. A double click on a line with an assembly instruction or high-level source code opens a List window showing the corresponding line in the code. Trace Tutorial \| 12 ©1989-2024 Lauterbach Using the TRACE32 menu Trace > List > Tracing with Source, you get a Trace.List and a List /Track window. When doing a simple click on a line in the Trace.List window, the List window will automatically display the corresponding code line. The timing information (see ti.back column) is generated in this case by the TRACE32 Instruction Set Simulator. With a real core trace, timestamps are either generated by the TRACE32 trace hardware or by the onchip trace module. Double click Simple click Trace Tutorial \| 13 ©1989-2024 Lauterbach Displaying Function Run-Times TRACE32 supports nested and flat function run-time analysis based on the trace results. Please refer to the video “Flat vs. Nesting Function Runtime Analysis” for an introduction to function run-time analysis in TRACE32: support.lauterbach.com/kb/articles/trace-based-profiling Graphical Charts By selecting the menu Trace > Chart > Symbols, you can get a graphical chart that shows the distribution of program execution time at different symbols. The displayed results are based on a flat analysis: The corresponding nesting analysis can be displayed using the menu Perf > Function Runtime > Show as Timing. The In and Out buttons can be used to zoom in/out. Alternatively, you can select a position in the window and then use the mouse wheel to zoom in/out. Trace Tutorial \| 14 ©1989-2024 Lauterbach Numerical Statistics and Function Tree The menu entry Perf > Function Runtime >Show Numerical displays numerical statistics for each function with various information as total run-time, minimum, maximum and average run-times, ratio, and number of function calls. Further display options can be selected by doing a right mouse click on a specific function. Parents [A] displays for example a caller tree for the selected function. By doing a right mouse click on func1 and selecting Parents, we see the run-times of the functions func2 and func9, which have called func1 in the trace recording. A B Trace Tutorial \| 15 ©1989-2024 Lauterbach Children [B] displays the run-times of the functions called by the selected function, for example here the function subst called by the function encode. A function call tree view of all function recorded in the trace can be displayed using the menu entries Perf > Function Runtime > Show as Tree or Perf > Function Runtime > Show Detailed Tree. Duration Analysis By doing a right mouse click on a function in the numerical statistics window (Trace.STATistic.Func) then selecting Duration Analysis, you get an analysis of the function run-times between function entry and exit including the time spent in called subroutines, e.g. here for the function subst (P:0x114C corresponds to the start address of the subst function): The time interval can be changed using the Zoom buttons. Trace Tutorial \| 16 ©1989-2024 Lauterbach Distance Analysis By doing a right mouse click on a function in the numerical statistics window (Trace.STATistic.Func) then selecting Distance Analysis, you can get run-times between two consecutive calls of the selected function, e.g. here for the function subst (P:0x114C corresponds to the start address of the subst function): Trace Tutorial \| 17 ©1989-2024 Lauterbach Variable Display The Trace.ListVar command allows to list recorded variables in the trace. If the command is used without parameters all recorded variables are displayed: Y ou can optionally add one or multiple variables as parameters. Example: display all accesses to the variables plot1 and plot2 The Draw button can then be used to plot the displayed variables graphically against time. This corresponds to the following TRACE32 command: Please refer for more information about the Trace.DRAW command to “Application Note for Trace.DRAW” (app_trace_draw.pdf). Trace.ListVar Trace.ListVar %DEFault plot1 plot2 Trace.DRAW.Var %DEFault plot1 plot2 Trace Tutorial \| 18 ©1989-2024 Lauterbach Track Option The /Track options allows to track windows that display the trace results. Y ou just need to add the /Track option after the command that opens a trace window, e.g. The cursor will then follow the movement in other trace windows, e.g. Trace.Chart.Func. Default is time tracking. If no time information is available, tracking to record number is performed. TRACE32 windows that displays the trace results graphically, e.g. Trace.Chart.Func, additionally accept the /ZoomTrack option. If the tracking is performed with another graphical window, the same zoom factor is used in this case. Trace.List /Track Trace.Chart.Func /ZoomTrack Trace Tutorial \| 19 ©1989-2024 Lauterbach Searching Trace Results The Find button allows to search for specific information in the trace results. Example 1: find the first call of function func2 1. Enter “func2” under address / expression 2. Select Program under cycle 3. Press the Find First button. The next entries to func2 in the trace can then be found using the Next button Example 2: Find all write accesses to the variable mstatic1 with the value 0x0 1. Enter “mstatic1” under address / expression 2. Select Write under cycle 3. Enter 0x0 under Data 4. Press the Find All button Please refer to “Application Note for Trace.Find” (app_trace_find.pdf) for more information about Trace.Find. Trace Tutorial \| 20 ©1989-2024 Lauterbach Trace Save and Load The recorded trace can be stored in a file using the command Trace.SAVE, e.g. The saved file can then be loaded in TRACE32 PowerView using the command Trace.LOAD The TRACE32 trace display windows will show in this case a LOAD message in the low left corner Please note that TRACE32 additionally allows to export/import the trace results in different formats. Refer to the documentation of the command groups Trace.EXPORT and Trace.IMPORT for more information. Trace.SAVE file.ad Trace.LOAD file.ad
189198	https://www.thainationalparks.com/species/bar-tailed-godwit	Bar-tailed godwit (Limosa lapponica) Customize your cookie preferences We respect your right to privacy, that's why we don't use any other cookies other than functional cookies. Your cookie preferences will apply across our website. Necessary These cookies are necessary for the website to function properly and cannot be switched off. They help with things like logging in, sending us an enquiry and setting your privacy preferences. [x] Off On Accept functional only Reject non-essential We only use functional cookies on our website. We don't use any other cookies.Cookie Policy. Accept functional only Reject non-essential Preferences Thai National Parks Home Parks National parks Ang Thong Ao Phang-Nga Chae Son Chet Sao Noi Doi Inthanon Doi Phu Kha Doi Suthep-Pui Erawan Huai Yang Kaeng Krachan Khao Chamao - Khao Wong Khao Khitchakut Khao Kho Khao Lak - Lam Ru Khao Luang Khao Phanom Bencha Khao Sam Roi Yot Khao Sok Khao Yai Ko Chang Ko Lanta Kui Buri Mae Wang Mae Wong Nam Nao Pang Sida Phi Phi Islands Phu Hin Rong Kla Phu Kradueng Phu Pha Thoep Phu Suan Sai Sai Yok Similan Islands Sirinat Sri Nakarin Dam Surin Islands Tarutao Tat Mok Thung Salaeng Luang Yong Waterfall Wildlife sanctuaries Huai Kha Khaeng Khao Phra - Bang Khram Khao Phra Thaeo Phu Khiao Non-hunting areas Khao Paeng Ma Tham Pha Tha Phon Forest parks Khao Nang Phanthurat Pran Buri Tours Doi Inthanon National Park Doi Inthanon - day tour Erawan National Park Erawan Falls, cave & historical sites Erawan Falls, Hellfire Pass & historical sites Erawan Falls - day tour from Bangkok Erawan Falls, Sai Yok NP & more - 2D/1N raft house Kaeng Krachan National Park Kaeng Krachan - 1-day wildlife tour Kaeng Krachan - 2-days wildlife tour Kaeng Krachan - 3-days wildlife tour Kaeng Krachan - Birdwatching tours Khao Sok National Park Cheow Lan Lake - 2D/1N private tour Cheow Lan Lake - 3D/2N private tour Shared 2D/1N tour with a small group Khao Yai National Park Khao Yai - 1-day wildlife tour Khao Yai - 1-day wildlife tour with night safari Khao Yai - 2-days wildlife tour Khao Yai - 3-days wildlife tour Khao Yai - Birdwatching tours Khao Yai - Photo tours Kui Buri National Park Kui Buri elephant safari, half day Taxis Taxi services Doi Inthanon National Park Doi Inthanon - DIY taxi tour from Chiang Mai Doi Suthep - Pui National Park Doi Suthep - DIY taxi tour from Chiang Mai Erawan National Park Erawan Falls/Hellfire Pass - DIY taxi tour Khao Sok National Park Khao Sok - taxi service to/from all nearby towns Khao Yai National Park Khao Yai - taxi service from/to Bangkok Khao Yai - taxi day tour from Bangkok Khao Yai - 2D/1N DIY taxi tour from Bangkok Khao Yai - 3D/2N DIY taxi tour from Bangkok Species Birds Mammals Snakes Lizards and crocodiles Amphibians Turtles and tortoises Marine species Species of Thailand Birds Mammals Snakes Lizards & Crocodiles Turtles & tortoises Amphibians Marine species Bar-tailed godwit Limosa lapponica Carolus Linnaeus, 1758 In Thai: นกปากแอ่นหางลาย The bar-tailed godwit (Limosa lapponica) is a large wader in the family Scolopacidae, which feeds on bristle-worms and shellfish on coastal mudflats and estuaries. It has distinctive red breeding plumage, long legs, and a long upturned bill. Bar-tailed godwits breed on Arctic coasts and tundra from Scandinavia to Alaska, and overwinter on coasts in temperate and tropical regions of the Old World, Australia and New Zealand. The migration of the subspecies Limosa lapponica baueri from across the Pacific Ocean from Alaska to New Zealand is the longest known non-stop flight of any bird, and also the longest journey without pausing to feed by any animal. The round-trip migration for this subspecies is over 29, 000 km. Name The genus name Limosa is from Latin and means "muddy", from limus, "mud", referring to its preferred habitat. The specific name lapponica refers to Lapland: it was originally named Scolopax lapponica by Linneaus from a northern Swedish specimen. The English term "godwit" was first recorded in about 1416–17 and may be an imitation of the bird's call, or be derived from the Old English "god whit", meaning "good creature", perhaps referring to its eating qualities. Its English name is taken from the black-and-white barred tail and upper tail coverts in this species. In French it is known as barge rousse, Russian maliy veretennik, Inuit chiuchiuchiak, Yup'ik tevatevaaq, and Māori kūaka. Description The bar-tailed godwit is a relatively short-legged species of godwit. The bill-to-tail length is 37 - 41 cm, with a wingspan of 70 - 80 cm. Males average smaller than females but with much overlap; males weigh 190 - 400 g, while females weigh 260 - 630 g; there is also some regional variation in size (see subspecies, below). The adult has blue-grey legs and a long, tapering, slightly upturned bi-colored bill: pink at the base and black towards the tip. The neck, breast and belly are unbroken brick red in breeding plumage, and dark brown above. Females breeding plumage is much duller than males, with a chestnut to cinnamon belly. Breeding plumage is not fully apparent until the third year, and there are three distinguishable age classes; during their first migration north immature males are noticeably paler in colour than more mature males. Non-breeding birds seen in the Southern Hemisphere are plain grey-brown with darker feather centres, giving them a striped look, and are whitish underneath. Juveniles are similar to non-breeding adults but more buff overall with streaked plumages on flanks and breast. Alaska-breeding bar-tailed godwits show an increase in body size from north to south, but this trend is not apparent in their non-breeding grounds in New Zealand; birds of different sizes mix freely. Limosa lapponica is distinguished from the black-tailed godwit (Limosa limosa) by its black-and-white horizontally-barred (rather than wholly black) tail, and lack of white wing bars. The most similar species is the Asiatic dowitcher (Limnodromus semipalmatus). There are now five generally accepted subspecies of bar-tailed godwit, listed from west to east: L. l. lapponica - (Linnaeus, 1758): Breeds from northern Scandinavia east to the Yamal Peninsula; winters western coasts of Europe and Africa from the British Isles and the Netherlands south to South Africa, and also around the Persian Gulf. Smallest subspecies, males up to 360 g, females to 450 g. L. l. taymyrensis: Engelmoer & Roselaar, 1998: Breeds in central Siberia from the Yamal Peninsula east to the Anabar River delta. L. l. menzbieri - Portenko, 1936: Breeds northeastern Asia from the Anabar River east to the Kolyma River delta; winters in Australia. L. l. anadyrensis: Engelmoer & Roselaar, 1998: from the Anadyr lowlands in far northeastern Asia east of the Kolyma River; the second largest subspecies, smaller than L. l. baueri L. l. baueri - Naumann, 1836: (called ) Breeds in western Alaska; winters in Australia and New Zealand. Largest subspecies. , Diet The birds' main source of food in wetlands is bristle-worms (up to 70%), supplemented by small bivalves and crustaceans. In wet pastures, bar-tailed godwits eat invertebrates. In a major staging site in the northern Yellow Sea, they continue to hunt polychaetes, but most of their food intake is the bivalve mollusc Potamocorbula laevis, which they generally swallow whole. Male bar-tailed godwits are smaller than females and have shorter bills. In a study at the Manawatū Estuary, shorter-billed birds (males) fed mostly on small surface prey like Potamopyrgus snails, half being snail specialists, whereas females consumed more deeply-buried prey such as worms; the birds also displayed some individual food preferences. Breeding The bar-tailed godwit is a non-breeding migrant in Australia and New Zealand. Birds first depart for their northern hemisphere breeding sites at age 2–4. Breeding take place each year in Scandinavia, northern Asia, and Alaska. The nest is a shallow cup in moss sometimes lined with vegetation. Clutch size is from 2 to 5, averaging four. Both sexes share incubation of the eggs for 20 to 21 days, the female during the day and the male at night. Migrations All bar-tailed godwits spend the Northern Hemisphere summer in the Arctic, where they breed, and make a long-distance migration south in winter to more temperate areas. L. l. lapponica make the shortest migration, some only as far as the North Sea, while others travel as far as India. Bar-tailed godwits nesting in Alaska (L. l. baueri) travel all the way to Australia and New Zealand. They undertake the longest non-stop migrations of any bird, and to fuel this carry the greatest fat loads of any migratory bird so far studied, reducing the size of their digestive organs to do so. L. l. bauri breeds in Alaska and spends the non-breeding season in eastern Australia and New Zealand. L. l. menzbieri breeds in Siberia and migrates to northern and western Australia. Birds breeding in Siberia follow the coast of Asia northwards and southwards, but those breeding in Alaska migrate directly across the Pacific to Australasia 11, 000 km away. To track the return journey, seven birds in New Zealand were tagged with surgically-implanted transmitters and tracked by satellite to the Yellow Sea in China, a distance of 9575 km; the actual track flown by one bird was 11026 km, taking nine days. At least three other bar-tailed godwits also appear to have reached the Yellow Sea after non-stop flights from New Zealand. One specific female of the flock, nicknamed "E7", flew onward from China to Alaska and stayed there for the breeding season. Then in August 2007 she departed on an eight-day non-stop flight from western Alaska to the Piako River near Thames, New Zealand, setting a new known flight record of 11680 km. This L. l. bauri female made a 174 day round-trip journey of 29, 280 km with 20 days of flying. To fuel such long journeys, L. l. baueri birds in New Zealand deposit much more fat for their body size than other subspecies, allowing them to fly 6, 000 to 8, 600 km. Both Australasian subspecies head north to their breeding grounds along the coast of Asia to the Yalu Jiang coastal wetland in the north Yellow Sea, the most important staging grounds for godwits and great knots (Calidris tenuirostris) in their northern migration. Baueri birds rested for about 41 days before continuing approximately 7000 km on to Alaska. Menzbieri spent on average 38 days in the Yellow Sea region and flew an additional 4100 km to high Arctic Russia. Birds will often depart early from New Zealand if there are favourable winds; they seem to be able to predict weather patterns that will assist them on the entire migration route. Birds that had nested in southern Alaska were larger and departed New Zealand earliest; this pattern was repeated six months later, with birds departing Alaska in the same order they arrived, and over the same span of days. Birds in southern New Zealand departed on average 9–11 days earlier than birds in more northern sites. Godwits arrive at the Yukon-Kuskokwim Delta in Alaska in two waves; local breeders in early May, and larger flocks in the third week of May en route to breeding grounds further north. Protection The status of the bar-tailed godwit is Near Threatened, and the population is declining. Fewer birds have been using East African estuaries since 1979, and there has been a steady decline in numbers around the Kola Peninsula, Siberia, since 1930. The global population is estimated to number 1, 099, 000–1, 149, 000 individuals. Both L. l. bauri and L. l. menzbieri adult survival rates decreased between 2005 and 2012, probably because of the loss of intertidal staging areas in the Yellow Sea. The construction of seawalls and the reclamation of mudflats have led to a critical reduction in food supplies for migrating birds, particularly subspecies like L. l. menzbieri that rely on the Yalu Jiang estuary on both their northward and southward migrations. Numbers of L. l. baueri have declined in New Zealand from over 100, 000 in the late 1980s to 67, 500 in 2018. The bar-tailed godwit is one of the species to which the Agreement on the Conservation of African-Eurasian Migratory Waterbirds (AEWA) applies. In New Zealand the species is protected under the 1953 Wildlife Act. This article uses material from Wikipedia released under the Creative Commons Attribution-Share-Alike Licence 3.0. Eventual photos shown in this page may or may not be from Wikipedia, please see the license details for photos in photo by-lines. Category / Seasonal Status Wiki listed status (concerning Thai population): Winter visitor BCST Category: Recorded in an apparently wild state within the last 50 years BCST Seasonal status: Non-breeding visitor Scientific classification Kingdom Animalia Phylum Chordata Class Aves Order Charadriiformes Family Scolopacidae Genus Limosa Species Limosa lapponica Common names English:Bar-tailed godwit French:Barge rousse Thai:นกปากแอ่นหางลาย Species in same genus Limosa lapponica, Bar-tailed godwit Limosa limosa, Black-tailed godwit Conservation status Least Concern (IUCN3.1) Near Threatened (BirdLife) Near Threatened (ONEP) Near Threatened (BCST) Photos Please help us review the bird photos if wrong ones are used. We can be reached via our contact us page. Creative Commons Attribution-ShareAlike 4.0 InternationalJJ Harrison Bar-tailed godwit Creative Commons Attribution-ShareAlike 2.5 GenericAndreas Trepte Bar-tailed godwit (breeding plumage) Creative Commons Attribution-ShareAlike 3.0 UnportedDickDaniels ( Bar-tailed Godwit Cairns RWD Creative Commons Attribution-ShareAlike 4.0 InternationalOnioram Bar-tailed godwit Creative Commons Attribution 2.0 GenericIan Kirk, Dorset, UK Bar-tailed godwit Creative Commons Attribution-ShareAlike 3.0 UnportedJJ Harrison Bar-tailed godwit Range Map Amphawa District, Samut Songkhram Ban Laem District, Phetchaburi Bang Pu Recreation Centre Chumphon Coast Hat Chao Mai National Park Khao Lak - Lam Ru National Park Khao Sam Roi Yot National Park Klaeng District, Rayong Ko Libong Ko Samui District, Surat Thani Laem Pak Bia Mueang Chonburi District, Chonburi Mueang Krabi District, Krabi Mueang Pattani District, Pattani Mueang Phetchaburi District, Phetchaburi Mueang Samut Sakhon District, Samut Sakhon Mueang Samut Songkhram District, Samut Songkhram Mueang Satun District, Satun Pak Thale Phi Phi Islands Similan Islands Surat Thani Coast Takua Pa District, Phang Nga Creative Commons Attribution-ShareAlike 4.0 International Thai National Parks Range map of Limosa lapponica in Thailand About our range maps Our range maps are based on limited data we have collected. The data is not necessarily accurate nor complete. Special thanks to Ton Smits, Parinya Pawangkhanant, Ian Dugdale and many others for their contributions. It is free to use this map on various media. See the creative common license terms by clicking "CC" icon below the map. But remember, again; the map may not be accurate or complete. Privacy Policy \| Contact us © Thai National Parks, 2025 T.A.T. License: 14/03405, license issued for GibbonWoot Limited Partnership (managing company) This website is a private initiative, it is not the official website of the national parks in Thailand.
189199	https://quizlet.com/337525071/physics-classroom-velocity-acceleration-flash-cards/	Physics Classroom: Velocity & Acceleration Flashcards \| Quizlet hello quizlet Study tools Subjects Create Log in Physics Classroom: Velocity & Acceleration Save What is a Scalar? Click the card to flip 👆 quantities described by a magnitude (or numerical value) alone Click the card to flip 👆 1 / 33 1 / 33 Flashcards Learn Test Blocks Match SVO1225 Top creator on Quizlet · Students also studied Flashcard sets Study guides Practice tests Physics: Newton's Laws, Force, Motion, and Vectors 24 terms carlos_rodriguez3932 Preview Intro to Audiology: Physics of Sound 29 terms riceisverytasty Preview Radiology Physics images 8 terms stevendthayer Preview Physics: Energy Types, Conservation Law, and Work Principles 7 terms tristan_mclellan1 Preview A Beka: Physics Quiz 12 10 terms shatteredskull Preview Lesson 11: Chapter 11 The Properties of Light 20 terms Alyssa_Oman15 Preview Work and Machines-Chapter 8 21 terms djmelgar Preview Chapter 10 Multiple Choice 29 terms Gabrielle0919 Preview Onsets Restrictions with Universe and Null 6 terms jenniferxu201211 Preview Conceptual Physics Ch 2 10 terms kevin20130619 Preview 4B Physics and Math chapter 4, General chemistry chapter 8 69 terms awesomeMichael5Mucha Preview Waves 11 terms lawrence_addie Preview OAT Review Physics 2 14 terms Lydia_Morgan34 Preview Audio Engineering 37 terms sgs1224 Preview Physics Exam 1 study guide 29 terms Rene_Rodriguez47 Preview APP chapter 20-23 39 terms EM8076 Preview phys ch25 5 terms cherrychailatte Preview LAST PHYSCS 21 terms majesticSwaggyC Preview 2nd Semester Exam Science 12 terms katherine_johnston20 Preview Brainpop: Newton's Laws of Motion Quiz 10 terms kiarajdkdnekem Preview Physics 1.1: Measuring Motion Teacher 9 terms MsP-K Preview Aural Rehabilitation Week 6 Quiz 10 terms pumpkinguts14 Preview Chemistry chapter 4 section 2 & 3 vocab 14 terms kadoodlebug Preview SPLH 463 Practice Questions 36 terms olivia-hansen Preview Mirror Ray 6 terms alietha1 Preview SI Base Units 7 terms Nena_Tiger Preview Physics Segment 15 Teacher 9 terms heathereyles Preview Physics & Math Chapter 1 34 terms lilharr84 Preview Terms in this set (33) What is a Scalar? quantities described by a magnitude (or numerical value) alone What is a Vector? quantities described by both a Magnitude and a Direction. What does 1 dimension mean? when you can go in the direction of forward, backward, up or down What is Distance? Scalar or Vector? refers to "how much ground an object has covered" during its motion Scalar Quantity (No direction only magnitude) What is displacement? refers to "how far out of place an object is"; it is the object's overall change in position Vector quantity ( Has Direction & Magnitude) Consider the motion. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. What is the physics teacher's Displacement & Distance? The physics teacher has walked a total Distance of 12 meters Displacement is 0 meters. During the course of her motion, she has "covered 12 meters of ground" (distance = 12 m). Yet when she is finished walking, she is not "out of place" - i.e Displacement, being a vector quantity, must give attention to direction. The 4 meters east cancels the 4 meters west; and the 2 meters south cancels the 2 meters north. Vector quantities such as displacement are direction aware. Scalar quantities such as distance are ignorant of direction. Image 33: Image: Consider the motion. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. What is the physics teacher's Displacement & Distance? What is Speed? Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. What is Velocity? Velocity is a vector quantity that refers to "the rate at which an object changes its position. velocity is a vector quantity and is direction aware Ex) Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. What is instantaneous speed? the speed at any given instant in time. Ex.) You might think of the instantaneous speed as the speed that the speedometer reads at any given instant in time What is Average Speed? the average of all instantaneous speeds; found simply by a distance/time ratio Ex.) You might think of the average speed as the average of all the speedometer readings during the course of the trip. What is acceleration? The rate at which an object changes its velocity Changing how fast an object is moving Is the Velocity vector always directed in the same direction which an object is moving? Yes A car moving eastward would be described as having an eastward velocity. And a car moving westward would be described as having a westward velocity How can the direction of the acceleration vector be determined? A. When slowing down B. Speeding up Slowing Down: an object which is Slowing down will have an acceleration directed in the direction Opposite of its motion ex.) Applying this rule of thumb would lead us to conclude that an eastward heading car can have a westward directed acceleration vector if the car is slowing down Speeding Up: When a car starts off slow and speeds up its acceleration and velocity are still in the Same direction If an object is changing Velocity is the object accelerating? Yes An object is accelerating if it is changing its velocity If an object has a constant velocity is it accelerating? No it is not accelerating. Acceleration is constant For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel. What's the equation for average acceleration? Velocity(final minus initial) over time UNITS: Acceleration are velocity units divided by time units - thus (m/s)/s or (mi/hr)/s. The (m/s)/s unit can be mathematically simplified to m/s2. The direction of acceleration depends on what 2 things? The direction of the acceleration vector depends on two things: •whether the object is speeding up or slowing down •whether the object is moving in the + or - direction If an object is slowing down, then its acceleration is in the what direction of its motion? Opposite Direction If an object is moving in the positive direction (i.e., has a positive velocity) and is speeding up. The acceleration is in what direction of the velocity When an object is speeding up, the acceleration is in the same direction as the velocity. The position vs. time graphs for the two types of motion: constant velocity and changing velocity (acceleration) can be depicted as what? Draw the graphs The shapes of the graph reveal an important principle. It is often said, "As the slope goes, so goes the velocity." Whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). This very principle can be extended to any motion conceivable Image 42: Image: The position vs. time graphs for the two types of motion: constant velocity and changing velocity (acceleration) can be depicted as what? Draw the graphs On a graph are slope and velocity related? Yes or No Yes-Whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). This very principle can be extended to any motion conceivable The slope of the line on a position versus time graph is equal to the velocity of the object. True or False True The slope equation says that the slope of a line is found by determining the amount of rise of the line between any two points divided by the amount of run of the line between the same two points. •Pick two points on the line and determine their coordinates. •Determine the difference in y-coordinates of these two points (rise). •Determine the difference in x-coordinates for these two points (run). •Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope Draw a position-time graph showing constant Negative Velocity with a slow moving object and fast moving object However, the slope of the graph on the right is larger than that on the left. Once more, this larger slope is indicative of a larger velocity. The object represented by the graph on the right is traveling faster than the object represented by the graph on the left. Draw a position-time graph showing constant Positive velocity with a slow moving object and a fast moving object The graph on the left is representative of an object that is moving with a positive velocity (as denoted by the positive slope), a constant velocity (as denoted by the constant slope) and a small velocity (as denoted by the small slope). However, the slope of the graph on the right is larger than that on the left. This larger slope is indicative of a larger velocity. The object represented by the graph on the right is traveling faster than the object represented by the graph on the left What is the take home message(principle) about the velocity vs. time graph? What information does it reveal? The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). If the acceleration is positive, then the slope is positive (i.e., an upward sloping line). If the acceleration is negative, then the slope is negative (i.e., a downward sloping line). if a line crosses over the x-axis from the positive region to the negative region of the graph (or vice versa), then the object has changed directions. Velocity vs. Time graph showing an object moving with a constant velocity will have what acceleration Moving with a constant velocity is a with zero acceleration Note that a motion described as a constant, positive velocity results in a line of zero slope (a horizontal line has zero slope) How to determine if an object is speeding up or slowing down on a velocity-time graph: If the line is getting further away from the x-axis (the 0-velocity point), then the object is speeding up. And conversely, if the line is approaching the x-axis, then the object is slowing down. How do you determine the area in a Velocity vs. Time graph The shaded regions between the line and the time-axis represent the displacement during the stated time interval The method used to find the area under a line on a velocity-time graph depends upon whether the section bound by the line and the axes is a rectangle, a triangle or a trapezoid. What is another name used for Area on a velocity vs time graph? Displacement A shaded rectangle on the velocity-time graph has a base of 6 s and a height of 30 m/s. What is the Area(displacement) of the shaded rectangle? Since the area of a rectangle is found by using the formula A = b x h, the area is 180 m (6 s x 30 m/s). That is, the object was displaced 180 meters during the first 6 seconds of motion. The shaded trapezoid on the velocity-time graph has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side). What is the area (displacement)? Since the area of trapezoid is found by using the formula A = ½ (b) (h1 + h2), the area is 40 m [½ (2 s) (10 m/s + 30 m/s)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3 seconds What is the alternative method in finding the area/displacement of a trapezoid on the velocity-time graph? An alternative means of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. The areas of the triangle and rectangle can be computed individually; the area of the trapezoid The shaded triangle on the velocity-time graph has a base of 4 seconds and a height of 40 m/s. What is the area (displacement)? Since the area of triangle is found by using the formula A = ½ b h the area is ½ (4 s) (40 m/s) = 80 m. That is, the object was displaced 80 meters during the four seconds of motion. 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