id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
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6700 | https://www.maths.gla.ac.uk/wws/cabripages/misc/misc2.html | hyperbolic page
odds and ends - the isodynamic points
In Kimberling's encyclopedia, the isodynamic points of a triangle ABC are defined as intersections of
three circles. From vertex A, we draw the internal and external bisectors, meeting BC at U,V. We draw
the circle on UV as diameter. Similarly, we draw circles starting from B and C. These three meet in two
points, the isodynamic points. Kimberling points out that the circles are the Apollonian circles of ΔABC.
Here, we define the points in terms of the Apollonian circles. It is then easier to establish that they are
concurrent, and to prove interesting inversive properties. It also avoids the problem with an isosceles
triangle, namely that one external bisector is parallel to the opposite side, so V=∞ and the "circle" on
"diameter" UV must be taken as the perpendicular bisector of the side.
definition
For a triangle ABC, the A-Apollonian is the member of the Apollonian family A(B,C) through A.
Theorem A1
The Apollonians of ΔABC meet at two points inverse with respect to the circumcircle.
They meet at ∞ if and only if ΔABC is equilateral.
proof
Observe that an inversive map takes apollonian families to apollonian families.
We choose an inversive map t taking A to ∞ and B,C to B',C', say.
The B- and C-apollonian circles map to circles through B', centre C' and through C', centre B'
respectively. By symmetry, these meet in two points on the perpendicular bisector of B'C',
and these points are equidistant from B'C'. Thus, they are inverse with respect to B'C'. But
the perpendicular bisector is the member of A(B',C') through ∞, i.e. the ∞-apollonian. Thus,
the three apollonians meet in two points.
Also, the circumcircle of ΔABC maps to the line B'C'. Applying t-1, we have the first part.
In general, the A-apollonian is a line if and only if the perpendicular bisector of BC
passes through A, i.e. the angles at B and C are equal. It follows at once that all the
apollonians are lines - so intersect at ∞ - if and only if ΔABC is equilateral.
definition
The isodynamic points of a triangle are the points of concurrence of its apollonians.
The first isodynamic point is that inside the circumcircle, the second that outside.
main page |
6701 | https://proofwiki.org/wiki/GCD_equals_GCD_with_Product_of_Coprime_Factor | GCD equals GCD with Product of Coprime Factor - ProofWiki
GCD equals GCD with Product of Coprime Factor
From ProofWiki
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Theorem
Let a,b,c∈Z a,b,c∈Z be integers.
Let:
a⊥b a⊥b
where ⊥⊥ denotes coprimality.
Then:
gcd{a c,b}=gcd{c,b}gcd{a c,b}=gcd{c,b}
where gcd gcd denotes greatest common divisor.
Proof
Let a,b,c∈Z a,b,c∈Z such that a a is coprime to b b.
Let d=gcd{c,b}d=gcd{c,b}.
We have:
a a⊥⊥b b
(1):(1):⇝⇝∃x,y∈Z:∃x,y∈Z:1 1==a x+b y a x+b yInteger Combination of Coprime Integers
∃u,v∈Z:∃u,v∈Z:d d==c u+b v c u+b vBézout's Identity
⇝⇝d d==(a x+b y)(c u+b v)(a x+b y)(c u+b v)from (1)(1)
==a c u x+a b v x+b c u y+b 2 v x a c u x+a b v x+b c u y+b 2 v x multiplying out
==a c(u x)+b(a v x+c u y+b v x)a c(u x)+b(a v x+c u y+b v x)rearranging
==gcd{a c,b}gcd{a c,b}Bézout's Identity
■◼
Sources
1980:David M. Burton: Elementary Number Theory(revised ed.)... (previous)... (next): Chapter 2 2: Divisibility Theory in the Integers: 2.2 2.2 The Greatest Common Divisor: Problems 2.2 2.2: 16(c)16(c)
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6702 | https://membean.com/roots/de-off | Word Root: de- (Prefix) | Membean
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de-
off, from
Quick Summary
Prefixes are key morphemes in English vocabulary that begin words. The English prefix de-, which means “off” or “from,” appears in hundreds of English vocabulary words, such as dejected, deduce, and deficient. You can remember that the prefix de- means “from” or “off” via the word descend, or to climb down “from” or “off” a height, such as a mountain.
Deduce Derivatives with De-!
Today we will focus on the prefix de-, which interchangeably means “off” or “from.” Let’s check out the following derivatives that depend upon the Latin preposition de-.
When we study English vocabulary, we find that most English words are derived or come “from” Greek or Latin; these vocabulary words are called derivatives. These Latin and Greek roots help you decide, or cut “off” false meanings of the derivatives to arrive at a decision as to what the word means.
We all depend, or hang “from” the ability of cars to take us from place to place. When you take your foot “off” the gas, your car begins to decelerate, or move down “from” its current speed to a slower velocity. This deceleration may be caused by a traffic jam, making you feel dejected or thrown “off” your customary good mood. There are, however, different degrees of such temporary depression; you could find that when deciduous trees’ leaves begin falling “off” in the autumn, your spirits sink. Or you could be demoted in your job, moved down “from” the position you currently enjoy; perhaps the boss thinks you are deficient in your job performance, doing things “off” of or “from” how they should be done. Worse yet, imagine if you were in Wonderland and the Queen of Hearts gave the order of “Off with his head!” Best to avoid such decapitation at all costs!
Enough doom and gloom about the prefix de- which means “off” or “from.” Some people cannot drink caffeine but still love coffee and tea. Luckily both come in decaffeinated varieties, where the caffeine has been taken “from” the beans or leaves. Say you didn’t know if the coffee you were drinking was decaf or not. You could drink a little, and then deduce, or draw a conclusion “from” your symptoms as to whether or not it contains caffeine. Shaky? Energized?
“Off” with this podcast! Enough of de- to fill up your day—I don’t want to detract or drag you “from” other activities for any longer!
derive: to come “from”
derivative: a word that has come “from” another language
depend: hang “from”
decide: to cut “off” false possibilities or poor options
decision: a cutting “off” of all possibilities but one
decelerate: to move down “from” the current speed to a slower one
dejected: thrown “off” in spirits
depression: pressed “off” or “from” a good mood
deciduous: of leaves falling “from” a tree in autumn
demote: to be moved down “from” a current job status
deficient: of doing tasks “off” from how they should be done
decapitate: to take “off” a head
decaffeinated: state of caffeine having been taken “from” coffee beans or tea leaves
deduce: to arrive at a conclusion by leading evidence “from” a given situation
deduction: a leading “from” evidence to a conclusion
detract: to drag “from”
Related Rootcasts
### The Fascinating Parts of Words Morphology is the study of how words are put together by using morphemes, which include prefixes, roots, and suffixes. Parsing the different morphemes in a word reveals meaning and part of speech. For instance, the word “invention” includes the prefix in- + the root vent + the suffix -ion, from which is formed the noun “invention.”
### Etymology: Word Origins Etymology is that part of linguistics that studies word origins. English vocabulary words are formed from many different sources, especially Latin and Greek. By determining the origins of the morphemes in English words, one is better able to remember and determine the dictionary definitions of words.
Usage
desultory
Something that is desultory is done in a way that is unplanned, disorganized, and without direction.
despondent
If you are despondent, you are extremely unhappy because you are in an unpleasant situation that you do not think will improve.
denouement
A denouement is the end of a book, play, or series of events when everything is explained and comes to a conclusion.
denounce
If you denounce people or actions, you criticize them severely in public because you feel strongly that they are wrong or evil.
delineate
If you delineate something, such as an idea or situation or border, you describe it in great detail.
deference
If you behave with deference towards someone, you show them respect and accept their opinions or decisions, especially because they have an important position.
nondescript
Something is nondescript when its appearance is ordinary, dull, and not at all interesting or attractive.
debonair
A man who is debonair is sophisticated, charming, friendly, and confident.
condescend
When people condescend, they behave in ways that show that they are supposedly more important or intelligent than other people.
indefatigable
If someone is indefatigable, they never show signs of getting tired.
defoliate
Someone defoliates a tree or plant by removing its leaves, usually by applying a chemical agent.
devolve
When something devolves, such as a responsibility or a person's status, it passes along to another person.
desecrate
If you desecrate something that is considered holy or very special, you deliberately spoil or damage it.
decipher
When you decipher a message or piece of writing, you work out what it says, even though it is very difficult to read or understand.
decadent
A decadent person has low moral standards and is more interested in pleasure than serious matters.
despicable
If you say a person's actions are despicable, you think they are extremely unpleasant or nasty.
detritus
Detritus is the small pieces of waste that remain after something has been destroyed or used.
deprecate
If you deprecate something, you disapprove of it strongly.
debunk
When you debunk someone's statement, you show that it is false, thereby exposing the truth of the matter.
deficit
A deficit occurs when a person or government spends more money than has been received.
defunct
Something that is defunct is no longer in existence or does not function.
degenerate
A degenerate person is immoral, wicked, or corrupt.
demented
A demented person is not in their right mind; hence, they are crazy, insane, and highly irrational.
depredation
Depredation can be the act of destroying or robbing a village—or the overall damage that time can do to things held dear.
derivative
A derivative is something borrowed from something else, such as an English word that comes from another language.
derogatory
A derogatory statement about another person is highly offensive, critical, uncomplimentary, and often insulting.
designation
A designation is a name, label, or mark that something is given to identify it.
destitute
Someone who is destitute lives in extreme poverty and thus lacks the basic necessities of life.
deterrent
A deterrent keeps someone from doing something against you.
detrimental
Something detrimental causes damage, harm, or loss to someone or something.
deviate
When someone's behavior is deviating, they do things differently by departing or straying from their usual way of acting.
legerdemain
Legerdemain is the skillful use of one's hands or the employment of another form of cleverness for the purpose of deceiving someone.
predecessor
A predecessor comes before someone else in a job or is an ancestor of someone.
demarcation
Demarcation is the process of setting boundaries or limits; it is also a line that provides a distinct separation between two things.
deplete
When you deplete a supply of something, you use it up or lessen its amount over a given period of time.
deceitful
Someone who is deceitful is tricky, false, or dishonest.
despair
When you are experiencing despair, you feel hopeless; you may believe that you are going through certain and sad defeat that you can do nothing about.
devious
A devious person can act in a dishonest, tricky, or secretive way for their own purposes.
descendant
A descendant is someone's offspring or child; it is also someone who can be traced back to an ancestor.
indeterminate
When a piece of information is indeterminate, it cannot be known with certainty.
deracinate
When someone or something is deracinated, the former is uprooted or removed from their natural environment whereas the latter is eradicated, uprooted, or destroyed.
descry
When you descry something, you detect, discover, discern, or see it.
descend
When you descend something, such as stairs or a mountain, you go down it.
desperate
If you are desperate, you don't know what to do about a very difficult problem you are having; you are very worried about it and willing to do just about anything to make it better.
independent
Someone who is independent takes care of themselves and does not rely on anyone to help them.
desire
A desire is something that you want or wish you had.
decide
When you decide something, you make a choice, usually after thinking about it.
dedicate
When you dedicate yourself to something, you spend a lot of time and effort doing it.
depend
When you depend on another person, you rely on them or need them to do something for you.
decay
When something decays, it breaks down as it is slowly being destroyed.
deceive
When you deceive another person, you trick them in some way, often by making them believe that something is true when it is not.
describe
When you describe something, you say what it is like or explain it.
defend
When you defend someone, you guard or protect them to keep them safe from harm.
depress
Something that depresses someone makes them sad and disappointed with life; they often don’t want to do anything because of it.
detail
A detail of something is a fact about it or one part of it.
design
When you design something, you plan or create it.
degree
The degree of something is the size, amount, level, or strength of it.
detect
When you detect something, you uncover or find it.
destruction
The destruction of something is the causing of so much damage to it that it cannot be fixed.
determine
When you determine something, you figure it out or decide about it.
decapitate
cut the head of
decelerate
lose velocity
deciduous
(of plants and shrubs) shedding foliage at the end of the growing season
decisive
determining or having the power to determine an outcome
deduce
reason by deduction
deficient
inadequate in amount or degree
degradation
changing to a lower state (a less respected state)
demote
assign to a lower position
demystify
make less mysterious or remove the mystery from
deranged
driven insane
detached
showing lack of emotional involvement
detract
take away a part from
desuetude
a state of inactivity or disuse
Related Word Roots
### de god
### de- thoroughly
Back to Word Roots
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Got It! |
6703 | https://www.youtube.com/watch?v=9s8jiskyso0 | Conclusion for a two sample t test using a P value
Khan Academy
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Conclusion for a two sample t test using a P value
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Transcript:
we're told a sociologist studying fertility in argentina and bolivia wanted to test if there was a difference in the average number of babies women in each country have the sociologists obtained a random sample of women from each country here are the results of their test so they take a sample of 75 women in argentina and these women had a mean of 2.4 babies each with a standard deviation of 1.5 and then the standard error of the mean was 0.17 and then they calculated similar statistics for a bolivia and then they give us the t-test for the means being different and we were able to calculate these statistics and they say assume that all conditions for inference have been met at the alpha equals 0.05 level of significance is there sufficient evidence to conclude that there is a difference in the average number of babies women in each country have so pause this video and see if you can answer that all right now let's work through this together so this is classic hypothesis testing right over here where your null hypothesis is actually going to be that your means are the same so that the mean in argentina is equal to the mean in bolivia and then your alternative hypothesis is that your means are different and what you do is you say all right if we assume the null hypothesis what is the probability that we would have gotten means this far apart and that's what our p value tells us that we have a 0.31 probability or 31 percent probability of getting means this far apart now if your probability assuming the null hypothesis is below your level of significance your alpha right over here then you would say all right that seems like such a low probability i'll reject the null hypothesis which suggests the alternative hypothesis but in this situation here if we compare our p to our alpha we see that our p value is for sure greater than our alpha so in this situation i mean you could see it right over here 0.31 is for sure greater than 0.05 so in this situation we cannot reject the null hypothesis cannot reject our null hypothesis and so there is not sufficient evidence to conclude that there is a difference in the average number of babies women in each country have |
6704 | https://openstax.org/books/concepts-biology/pages/3-chapter-summary | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Concepts of Biology
Chapter Summary
Concepts of BiologyChapter Summary
Search for key terms or text.
## 3.1 How Cells Are Studied
A cell is the smallest unit of life. Most cells are so small that they cannot be viewed with the naked eye. Therefore, scientists must use microscopes to study cells. Electron microscopes provide higher magnification, higher resolution, and more detail than light microscopes. The unified cell theory states that all organisms are composed of one or more cells, the cell is the basic unit of life, and new cells arise from existing cells.
## 3.2 Comparing Prokaryotic and Eukaryotic Cells
Prokaryotes are predominantly single-celled organisms of the domains Bacteria and Archaea. All prokaryotes have plasma membranes, cytoplasm, ribosomes, a cell wall, DNA, and lack membrane-bound organelles. Many also have polysaccharide capsules. Prokaryotic cells range in diameter from 0.1–5.0 µm.
Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes, but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its DNA is surrounded by a membrane), and has other membrane-bound organelles that allow for compartmentalization of functions. Eukaryotic cells tend to be 10 to 100 times the size of prokaryotic cells.
## 3.3 Eukaryotic Cells
Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes, but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its DNA is surrounded by a membrane), and has other membrane-bound organelles that allow for compartmentalization of functions. The plasma membrane is a phospholipid bilayer embedded with proteins. The nucleolus within the nucleus is the site for ribosome assembly. Ribosomes are found in the cytoplasm or are attached to the cytoplasmic side of the plasma membrane or endoplasmic reticulum. They perform protein synthesis. Mitochondria perform cellular respiration and produce ATP. Peroxisomes break down fatty acids, amino acids, and some toxins. Vesicles and vacuoles are storage and transport compartments. In plant cells, vacuoles also help break down macromolecules.
Animal cells also have a centrosome and lysosomes. The centrosome has two bodies, the centrioles, with an unknown role in cell division. Lysosomes are the digestive organelles of animal cells.
Plant cells have a cell wall, chloroplasts, and a central vacuole. The plant cell wall, whose primary component is cellulose, protects the cell, provides structural support, and gives shape to the cell. Photosynthesis takes place in chloroplasts. The central vacuole expands, enlarging the cell without the need to produce more cytoplasm.
The endomembrane system includes the nuclear envelope, the endoplasmic reticulum, Golgi apparatus, lysosomes, vesicles, as well as the plasma membrane. These cellular components work together to modify, package, tag, and transport membrane lipids and proteins.
The cytoskeleton has three different types of protein elements. Microfilaments provide rigidity and shape to the cell, and facilitate cellular movements. Intermediate filaments bear tension and anchor the nucleus and other organelles in place. Microtubules help the cell resist compression, serve as tracks for motor proteins that move vesicles through the cell, and pull replicated chromosomes to opposite ends of a dividing cell. They are also the structural elements of centrioles, flagella, and cilia.
Animal cells communicate through their extracellular matrices and are connected to each other by tight junctions, desmosomes, and gap junctions. Plant cells are connected and communicate with each other by plasmodesmata.
## 3.4 The Cell Membrane
The modern understanding of the plasma membrane is referred to as the fluid mosaic model. The plasma membrane is composed of a bilayer of phospholipids, with their hydrophobic, fatty acid tails in contact with each other. The landscape of the membrane is studded with proteins, some of which span the membrane. Some of these proteins serve to transport materials into or out of the cell. Carbohydrates are attached to some of the proteins and lipids on the outward-facing surface of the membrane. These form complexes that function to identify the cell to other cells. The fluid nature of the membrane owes itself to the configuration of the fatty acid tails, the presence of cholesterol embedded in the membrane (in animal cells), and the mosaic nature of the proteins and protein-carbohydrate complexes, which are not firmly fixed in place. Plasma membranes enclose the borders of cells, but rather than being a static bag, they are dynamic and constantly in flux.
## 3.5 Passive Transport
The passive forms of transport, diffusion and osmosis, move material of small molecular weight. Substances diffuse from areas of high concentration to areas of low concentration, and this process continues until the substance is evenly distributed in a system. In solutions of more than one substance, each type of molecule diffuses according to its own concentration gradient. Many factors can affect the rate of diffusion, including concentration gradient, the sizes of the particles that are diffusing, and the temperature of the system.
In living systems, diffusion of substances into and out of cells is mediated by the plasma membrane. Some materials diffuse readily through the membrane, but others are hindered, and their passage is only made possible by protein channels and carriers. The chemistry of living things occurs in aqueous solutions, and balancing the concentrations of those solutions is an ongoing problem. In living systems, diffusion of some substances would be slow or difficult without membrane proteins.
## 3.6 Active Transport
The combined gradient that affects an ion includes its concentration gradient and its electrical gradient. Living cells need certain substances in concentrations greater than they exist in the extracellular space. Moving substances up their electrochemical gradients requires energy from the cell. Active transport uses energy stored in ATP to fuel the transport. Active transport of small molecular-size material uses integral proteins in the cell membrane to move the material—these proteins are analogous to pumps. Some pumps, which carry out primary active transport, couple directly with ATP to drive their action. In secondary transport, energy from primary transport can be used to move another substance into the cell and up its concentration gradient.
Endocytosis methods require the direct use of ATP to fuel the transport of large particles such as macromolecules; parts of cells or whole cells can be engulfed by other cells in a process called phagocytosis. In phagocytosis, a portion of the membrane invaginates and flows around the particle, eventually pinching off and leaving the particle wholly enclosed by an envelope of plasma membrane. Vacuoles are broken down by the cell, with the particles used as food or dispatched in some other way. Pinocytosis is a similar process on a smaller scale. The cell expels waste and other particles through the reverse process, exocytosis. Wastes are moved outside the cell, pushing a membranous vesicle to the plasma membrane, allowing the vesicle to fuse with the membrane and incorporating itself into the membrane structure, releasing its contents to the exterior of the cell.
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6705 | https://blog.csdn.net/lglfa/article/details/80575123 | 样本量对差异性分析（Anova）或者T 检验的影响-CSDN博客
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样本量对差异性分析（Anova）或者T 检验的影响
样本量与显著性差异
最新推荐文章于 2025-07-03 11:06:48 发布
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本文探讨了样本量大小如何影响统计显著性的判断。通过对比大量数据与平均数据的差异性分析，说明了样本量增大时更容易获得显著性差异结论的原因。
样本量大的时候做差异性分析容易得到有显著性差异的结论，原因如下图， 求p值的过程中，n越大，Z0也越大，相对应的p就小了。当然这里默认方差变化不大的情况下，因为一般来说很多数据经过平均后方差不会变化很大，相比平均前后的样本量。
p值计算, from google image
下图中的数据，y1和y2的数据量各为1000，y1m和y2m数据量各为100， 是y1和y2每10个10个数据的平均，所以y1和y1m，y2和y2m的平均值相等，他们的方差也是基本没大变化，我们分别对y1和y2，y1m和y2m做下差异性分析，这里用单因素方差分析（anova1），其实用独立样本t-test会得到相同的结果（计算公式形式虽然不一样，但换汤不换药，结果一样的）.
结果是，y1和y2的p值为0.0189，他们之间有显著性差别；y1m和y2m的p值 0.4603，fail to reject原假设（H0： 他们之间没差别），不能说他们之间有显著性差别的。
标准差
MATLAB 代码
cobol
y1 = sin([0.01:0.01:10])10;y2= y1+0.7;p = anova1([y1;y2]')y1m = mean(reshape(y1,10,100));y2m = mean(reshape(y2,10,100));pm = anova1([y1m;y2m]') subplot(2,1,1)plot(y1,'.')hold on plot(y2, '.')legend('y1','y2')subplot(2,1,2)plot(y1m,'.')hold onplot(y2m,'.')legend('y1m','y2m')
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方差分析 ANOVA、T 检验、卡方 检验 的区别
itslifeng的博客
08-25 4万+
T 检验 和 ANOVA 都是用来看样本之间均值是否相等，但是两者又有什么区别呢？ 卡方 检验 是用来看分类变量之间有没有相关性。 自变量X类型 自变量X组数 因变量Y类型 分析方法 离散（类别） 仅仅2组 连续（均值） t 检验 离散（类别） 2组及以上 连续（均值） ANOVA 离散（类别） 2组及以上 离散（类别） 卡方 检验 进一步细分 方差分析 从理论上讲，方差分析有两个前提条件，一是因变量Y需要满足正态性要求，二是满足方差齐 检验。 方差不齐时可使用‘非参数 检验’，同时还可使用wel
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...SPSS操作全解析_ 样本量 不一致 如何做差异分析
9-27
从理论依据上看,例如Kruskal - Wallis H 检验 ( 用于多组独立样本的非参数 检验 ) 就建立在秩和的基础上。根据文献的研究表明,在多种非正态分布和不同 样本量 的情况下,Kruskal - Wallis H 检验 的I型错误率和 检验 效能都较为稳定可靠。相比于基于正态分布假设的方差分析,在面对异常值或极端偏态数据时更具稳健性。
差异性分析“样本量 不足”导致 检验 效能低的解决办法
9-10
原因:即使 样本量 不足,如果效应量较大,仍然可能检测到显著差异。 解决方法:计算效应量 ( 如Cohen's d ),评估差异的实际意义,而不仅仅依赖于统计显著性。 6.使用更敏感的统计方法 原因:某些统计方法对 样本量 的要求较低,且对差异的敏感性较高。 解决方法:考虑使用更敏感的统计方法,如重复测量方差分析或混合效应模型。
python方差分析 样本量 太大_十五、方差分析--使用Python进行单因素方差分析（ANOVA）...
weixin_39660931的博客
12-13 1798
方差分析方差分析 ( Analysis of Variance，简称 ANOVA )，又称为“变异数分析”， 是由英国统计学家费歇尔 ( Fisher ) 在20世纪20年代提出的，可用于推断两个或两个以上总体均值是否有差异的显著性 检验。由于各种因素的 影响，研究所得的数据呈现波动性。造成波动的原因可分成两类，一是不可控的随机因素，另一是研究中施加的对结果形成 影响 的可控因素。方差分析一般可以分为单因素方差分析和多因...
硬核干货：如果 样本量 不一一样多，或者 不是一一对应关系，如何做差异？相关？...
悟道西方
03-14 8709
写在前面因为我们测定的样本不总是可以匹配上的，但是最少的样本也不能太少，我们测定的三个样本做相关其实还是被质疑，但是测定了5 或者 6个重复，这个时候直接将样本比较多的样本过滤掉比较少的样本不...
SPSS——均值 检验 ( Compare Means )——独立样本T 检验
9-29
指两个样本之间彼此独立没有关联,两个独立样本各自接受相同的测量,主要目的是分析两个独立样本的均值是否有显著差异。 前提 独立性 两个样本相互独立,即从一总体中抽取一批样本对从另一总体中抽取一批样本没有任何 影响,两组样本个案数可以不同。 正态性
SPSS独立样本t 检验 结果分析_独立样本t 检验 结果报告怎么写
9-29
SPSS独立样本t 检验 结果分析 上图为独立样本T 检验。 由下图的基本参数设置生成 结果解读:三步法 第一步:拿到两组核心基本统计量,对于数值变量,核心基本统计量就三个,样本量 N,均值,标准差。然后产生主观意识,发现男生肺活量是3887.16,女生肺活量是2522.57,给人男生肺活量可能比女生高的主观感受 ...
【学习笔记】组间差异比较及相关问题总结
Absolute Zero
03-22 3万+
前 言 最近在对数据进行统计分析时发现自己其实对于统计学并没有一个很系统的了解，尤其是各种假设 检验 组间差异比较的方法过于繁多，实在是让人眼花缭乱，于是便打算把自己的疑惑及思考都总结记录下来，以便彻底搞清楚这些逻辑关系。 目 录 Q1：做组间差异的步骤是什么？方法都有哪些？ Q2：有些资料上提到，当两组服从正态分布的数据进行比较时，小样本（<30）使用t 检验，大样本（>30）使用z 检验，为什么SPSS等一些软件中只有t 检验？ Q3：为什么 Wilcoxon rank-sum tes.
差异性分析 与相关性分析的区别 最新发布
SPSSAU的博客
07-03 330
通过以上对比，可以看出 差异性分析 和相关性分析在定义、应用场景、数据关系、常用方法和结果解读等方面存在显著差异。在实际应用中，应根据研究目的和数据特点选择合适的分析方法。SPSSAU ( 在线SPSS ) 提供了丰富的 差异性分析 和相关性分析工具，帮助用户轻松完成数据分析任务。在数据分析中，差异性分析 和相关性分析是两种常用的统计方法，它们各自具有独特的应用场景和解释方式。差异性分析 与相关性分析的区别。
生存分析方法详解
9-26
寿命表法更适合于 样本量 较大的数据 ( 不过,考虑到现在计算能力的强大,一般程度的数据量并不会对计算速度有太大 影响 ) 3.关注点不同 K-M法关注每一个时点的生存率,重视对生存率规律的细致把握,可以利用K-M的结果去研究 影响 生存率变化 ( 如曲线的突变点 ) 的 影响 因素。 寿命表法则更重视对生存规律的总体把握 ( 如...
深度学习系列笔记之统计基础_cohen's d
9-13
statistic 统计量 发现这些学生都睡6.2个小时 那么这个值叫统计量 parameter 表示总体的数字叫参数 比如3000 在使用一个样本对一个总体进行推理时,样本的平均值可能不会完全等于总体的平均值。样本和总体平均值之间的差异称为: 抽样误差 sampling error 调查的目标要有能力代表整个全体 ...
三种T 检验 的详细区分
super_he_pi的博客
09-12 1万+
之前的文章中SPSSAU已经给大家详细地介绍了方差分析，之后收到的一些反馈以及日常的答疑中，我们发现关于T 检验 三种方法的区分还有很多小伙伴搞不清楚，下面就结合着具体案例详细聊聊T 检验 的那点事。 01. 概念 T 检验 是通过比较不同数据的均值，研究两组数据之间是否存在显著差异。 02. 分类 不同的T 检验 方法适用于不同的分析场景，具体的分类如下： SPSSAU整理 ...
单因素方差分析的条件及应对
zsSSR的博客
06-07 5914
例如，在SPSS中，当 样本量 大于50时，常使用Kolmogorov-Smirnov（K-S）检验。单因素方差分析（One-Way ANOVA）中的方差齐性 检验 是分析前的一个重要步骤，用于 检验 不同水平（或称为组）下观测变量的总体方差是否相等。直方图若呈现“中间高，两头低”的钟形，或者 P-P图和Q-Q图上的点近似呈一条对角直线，则表明数据近似满足正态分布。如果方差齐性 检验 的结果显示各组方差相等，则可以认为满足方差分析的前提假设，进而可以安全地进行单因素方差分析。在实际应用中，如果数据。，并不适用于所有情况。
A/Btest:组间的差异性 检验,统计功效以及反选 样本量,附python底层实现代 ...
9-22
本文深入探讨A/B测试中的统计方法,包括单比率 检验、双比率 检验,统计功效及 样本量 计算。通过Python实现这些功能,帮助理解如何评估两组样本间的差异,设定实验的置信水平和功效。 先概括一下:本文主要阐述了A/Btest中组间差异的比率 检验 ( 单比率 检验,双比率 检验 ),统计功效,以及何通过显著性水平还有统计功效反实验所需选样...
python 方差分析 样本量 太大_单因素 ANOVA 的 样本量 估计
weixin_42644249的博客
02-03 1030
以真实商业案例为数据基础，课程内容围绕scipy.stats和statsmodels包的相关功能展开，从统计分析实战的角度出发详细介绍了如何在Python中完成数据描述、t 检验、单因素方差分析、卡方 检验、相关回归等统计分析操作。通过本课程的学习，学员将深入学习如何正确考察这些方法的适用条件，正确选择所需的方法加以应用，从而既满足了相关统计分析功能的需求，又为进一步学习statsmodels包中的复...
anova.rar_ anova _单因素方差分析_显著性_显著性分析
09-24
在科研和数据分析中，当我们要评估一个独立变量（分类变量）对连续变量的 影响 时，就会用到这种方法。在这个“anova.rar”压缩包中，包含了两个MATLAB文件：“my Anova 1 ( 2 ).m”和“my Anova 1 ( 1 ).m”，它们很可能是...
四台机器生产同一个产品差异性 检验.rar_数理统计分析_显著性差异
09-23
在提供的“四台机器生产同一个产品差异性 检验.doc”文档中，应包含了数据、统计分析的结果以及相关的结论。通过解读这些结果，我们可以得出关于机器日产量差异性的最终判断，并据此为生产效率优化提供依据。
T 检验&ANOVA
打工人 打工魂
06-27 4156
不同分析问题需要使用的分析方法 集合之间平均值校验 集合之间平均值校验 变量关系比较 变量关系比较 变量关系比较 1 ~ 2 个的集合 多个集合 ( 2个以上 ) 范畴型变量独立性校验 数值型变量间线性关系 一种数值型变量与N个变量间的关系 T-test（2个group,比如男女身高差 )
样本不均衡及其解决办法
dfly_zx的博客
02-26 9191
1 什么是类别不均衡 类别不平衡（class-imbalance），也叫数据倾斜，数据不平衡，是指分类任务中不同类别的训练样例数目差别很大的情况。 在现实的分类学习任务中，我们经常会遇到类别不平衡，例如广告点击率预测、故障分析、异常检测等；或者 在通过拆分法解决多分类问题时，即使原始问题中不同类别的训练样例数目相当，在使用OvR（One vs. Rest）、MvM（Many vs. Many）策略后产生的二分类任务仍然可能出现类别不平衡现象。而标准机器学习算法通常假设不同类别的样本数量大致相似，所以类别不
样本不平衡问题分析与部分解决办法
家有代码初写成 的博客
01-20 2万+
最近工作中在处理文本分类问题遇到了分类不均衡的问题，主要还是样本太少还同时非常的不均衡正负样本1:10（类别不平衡比例超过4:1，就会造成偏移），就使用了SMOTE方法。 注意：在进行数据增广的时候一定要将测试集和验证集单独提前分开，扩张只在训练集上进行，否则会造成在增广的验证集和测试集上进行验证和测试，在实际上线后再真实数据中效果可能会非常的差。 目录 什么是样本类别分布不均衡？ 问题描...
解决样本类别分布不均衡的问题
gulie8的博客
06-15 3466
解决样本类别分布不均衡的问题 所谓的不均衡指的是不同类别的 样本量 差异非常大。样本类别分布不均衡主要出现在分类相关的建模问题上。样本类别分布不均衡从数据规模上可以分为大数据分布不均衡和小数据分布不均衡两种。 ·大数据分布不均衡；这种情况下整体数据规模大，只是其中的小样本类的占比较少。但是从每个特征的分布来看，小样本也覆盖了大部分或全部的特征。例如拥有1000万条记录的数据集中，其中占比50万条 的少数分类样本便于属于这种情况。 ·小数据分布不均衡；这种情况下整体数据规模小，并且占据少量 样本比例的分类数量也少
一文解决样本不均衡（全）
Python学习与数据挖掘
06-18 3150
样本（类别）样本不平衡（class-imbalance）指的是分类任务中不同类别的训练样例数目差别很大的情况，一般地，样本类别比例（Imbalance Ratio）（多数类vs少数类）明显大于1:1（如4：1）就可以归为样本不均衡的问题。现实中，样本不平衡是一种常见的现象，如：金融欺诈交易检测，欺诈交易的订单样本通常是占总交易数量的极少部分，而且对于有些任务而言少数样本更为重要。本文将详细介绍如果解决样本不平衡性，喜欢记得点赞、收藏、关注。文末提供技术交流方法。 很多时候我们遇到样本不均衡问题时，很直接的反
python 假设 检验 样本量 太大_ 样本量 大小会 影响 假设 检验 的结果（是否显著）吗？...
weixin_39545805的博客
12-15 1260
今天听课听到这样一个结论：如果假设 检验 的 样本量 很大，那么显著性水平α应该设得小一点。为什么呢？我没想通，于是去网上试图查找答案。结果发现网上很多人还在纠结：如果假设 检验 的 样本量 很大，那么会使假设 检验 的结果非常容易产生显著性。这是不是真的？样本量 太大是不是不好？我：？？？ 很久之前我就知道这种说法没有道理，但是我从来没有仔细去研究过这个问题。这次在知乎和stackexchange上搜罗了一下大家的回...
python流式数据处理_Python - Toolz - 流式分析（Streaming Analytics）工具
weixin_39957265的博客
12-16 2448
Python - Toolz - 流式分析 ( Streaming Analytics ) 工具21 August 2014Toolz 可以用于编写分析大型数据流脚本，它支持通用的分析模式，如通过纯函数来对数据进行筛选 ( Selection )，分组 ( Grouping )，化简 ( Reduction ) 以及连表 ( Joining )。这些函数通常可以模拟类似其他数据分析平台 ( 如 SQL 和 Panda ) 的类似操作行为。...
graphpad prism显著性差异分析_GraphPad Prism统计之单因素 ANOVA 分析 热门推荐
weixin_40009207的博客
11-19 9万+
GraphPad Prism可谓数据分析表现的小能手，能文能武能统计能画图，那它做统计时如何施展拳脚的，这几天让我们由浅入深的来学习下。习题MCF-7细胞过表达MKL-1、STAT3,或MKL-1加上STAT3，检测N-cadherin的相对表达量，pCDNA3.1为对照。请用合适的统计方法对下列模拟数据进行分析：再给个示例图提示本题示例图来源于文献PMID：28499590；Fig 1...
统计学习：ANOVA（方差分析）（1）
去向前方的博客
05-21 5万+
统计学习 最近在处理数据的过程中，越发发觉自己理论知识的薄弱，因此，开始了这一系列的帖子，记录自己的理论补充过程，同时方便后来人理解。 ANOVA（方差分析） 方差分析 ( Analysis of Variance，简称 ANOVA )，又称“变异数分析”，是R.A.Fisher发明的，用于两个及两个以上样本均数差别的显著性 检验。 由于各种因素的 影响，研究所得的数据呈现波动状。造成波动的原因可分...
数据分析中异质性 检验 工具包：地区差异与统计指标
在这份数据包中，通过Stata软件，可以实现对数据异质性的 检验，并对东中西地区的中位数、平均值进行统计分析，以及对不同组之间的系数差异进行 检验。 Stata是一种集成的统计软件，它为数据分析、数据管理以及图形...
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最新评论
pandas之loc深度用法
sarah1993512:讲得很清楚！谢谢啦！
伯努利分布（二项分布）的假设检验
wcy19980920:不会吧。。。这个样本均值最大不超过1，最小不超过0。要让它怎么拟合正态分布呢？
pandas之loc深度用法
花花呼呼 回复 horizon668: 是的
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6706 | https://www.hilotutor.com/archives_bombastic.html | Make Your Point > Archived Issues > BOMBASTIC Send Make Your Point issues straight to your inbox. | | | | | | | | | | | | --- --- --- --- --- | | | | --- | | | | | | | explore the archives: | Bombast is cotton padding, and bombastic people and their words are all puffed-up and overstuffed--full of inflated, overly wordy language. So yes, bombastic is a fancy word for describing fancy words, much like gr_nd_l_q__nt. The latter has a loftier and slightly kinder tone than bombastic, but please use both with discretion. | | | make your point with... | "BOMBASTIC" Someone or something bombastic uses inappropriately fancy language. Pronunciation: bom BASS tick Part of speech: Adjective. (Adjectives are describing words, like "large" or "late." They can be used in two ways: 1. Right before a noun, as in "a bombastic thing" or "a bombastic person." Other common forms: "Bombast" is the noun (meaning "inappropriately fancy language"). The adverb, "bombastically," is rare. How to use it: Use it with discretion: it's a critical, unkind word. So when you must, talk about bombastic people and their bombastic personalities and personas, their bombastic style, their bombastic announcements and pronouncements, their bombastic speeches and remarks, their bombastic exaggeration and rhetoric, etc. Take care to notice that, according to dictionaries, "bombastic" means "pompous, using fancy words," and that it doesn't mean "arrogant," "swaggering," "overly confident," or (brace yourself) "braggadocious." Certainly there's a lot of crossover here: a bombastic speaker can also be all of those things--arrogant, swaggering and so on--but bombastic speakers, by definition, use fancy, lofty, highfalutin language. If you're using simple words and simple phrases, you're not being bombastic. But that's what dictionaries tell us; popular usage says otherwise. Let's keep an eye on this word and how it's being used. Dictionaries might have to acknowledge soon that it does apply to people who brag and boast in very simplistic language. | | | examples: | The worst part of any bombastic blog post is the unnecessarily long introduction. We know where you're going with this; just get there, please. | | | study it now: | Look away from the screen to explain the definitionin your own words. You’ll know you understand what "bombastic" means when you can explain it without saying "puffed-up" or "pompous." | | | try it out: | Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)." Example: "Dr. Terminus is the bombastic villain in Pete's Dragon." | | | before you review: | |
| | | | | | | | | | | |
--- --- --- --- ---
| | | | --- | | | | | | | explore the archives: | Bombast is cotton padding, and bombastic people and their words are all puffed-up and overstuffed--full of inflated, overly wordy language. So yes, bombastic is a fancy word for describing fancy words, much like gr_nd_l_q__nt. The latter has a loftier and slightly kinder tone than bombastic, but please use both with discretion. | | | make your point with... | "BOMBASTIC" Someone or something bombastic uses inappropriately fancy language. Pronunciation: bom BASS tick Part of speech: Adjective. (Adjectives are describing words, like "large" or "late." They can be used in two ways: 1. Right before a noun, as in "a bombastic thing" or "a bombastic person." Other common forms: "Bombast" is the noun (meaning "inappropriately fancy language"). The adverb, "bombastically," is rare. How to use it: Use it with discretion: it's a critical, unkind word. So when you must, talk about bombastic people and their bombastic personalities and personas, their bombastic style, their bombastic announcements and pronouncements, their bombastic speeches and remarks, their bombastic exaggeration and rhetoric, etc. Take care to notice that, according to dictionaries, "bombastic" means "pompous, using fancy words," and that it doesn't mean "arrogant," "swaggering," "overly confident," or (brace yourself) "braggadocious." Certainly there's a lot of crossover here: a bombastic speaker can also be all of those things--arrogant, swaggering and so on--but bombastic speakers, by definition, use fancy, lofty, highfalutin language. If you're using simple words and simple phrases, you're not being bombastic. But that's what dictionaries tell us; popular usage says otherwise. Let's keep an eye on this word and how it's being used. Dictionaries might have to acknowledge soon that it does apply to people who brag and boast in very simplistic language. | | | examples: | The worst part of any bombastic blog post is the unnecessarily long introduction. We know where you're going with this; just get there, please. | | | study it now: | Look away from the screen to explain the definitionin your own words. You’ll know you understand what "bombastic" means when you can explain it without saying "puffed-up" or "pompous." | | | try it out: | Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)." Example: "Dr. Terminus is the bombastic villain in Pete's Dragon." | | | before you review: | |
| | |
--- |
| | |
| |
| explore the archives: |
Bombast is cotton padding, and bombastic people and their words are all puffed-up and overstuffed--full of inflated, overly wordy language. So yes, bombastic is a fancy word for describing fancy words, much like gr_nd_l_q__nt. The latter has a loftier and slightly kinder tone than bombastic, but please use both with discretion.
| |
| make your point with... |
"BOMBASTIC"
Someone or something bombastic uses inappropriately fancy language.
Pronunciation:
bom BASS tick
Part of speech:
Adjective.
(Adjectives are describing words, like "large" or "late."
They can be used in two ways:
1. Right before a noun, as in "a bombastic thing" or "a bombastic person."
Other common forms:
"Bombast" is the noun (meaning "inappropriately fancy language").
The adverb, "bombastically," is rare.
How to use it:
Use it with discretion: it's a critical, unkind word. So when you must, talk about bombastic people and their bombastic personalities and personas, their bombastic style, their bombastic announcements and pronouncements, their bombastic speeches and remarks, their bombastic exaggeration and rhetoric, etc.
Take care to notice that, according to dictionaries, "bombastic" means "pompous, using fancy words," and that it doesn't mean "arrogant," "swaggering," "overly confident," or (brace yourself) "braggadocious." Certainly there's a lot of crossover here: a bombastic speaker can also be all of those things--arrogant, swaggering and so on--but bombastic speakers, by definition, use fancy, lofty, highfalutin language. If you're using simple words and simple phrases, you're not being bombastic. But that's what dictionaries tell us; popular usage says otherwise. Let's keep an eye on this word and how it's being used. Dictionaries might have to acknowledge soon that it does apply to people who brag and boast in very simplistic language.
| |
| examples: |
The worst part of any bombastic blog post is the unnecessarily long introduction. We know where you're going with this; just get there, please.
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "bombastic" means when you can explain it without saying "puffed-up" or "pompous."
| |
| try it out: |
Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)."
Example: "Dr. Terminus is the bombastic villain in Pete's Dragon."
| |
| before you review: |
| |
| make your point with... |
Someone or something bombastic uses inappropriately fancy language.
Pronunciation:
bom BASS tick
Part of speech:
Adjective.
(Adjectives are describing words, like "large" or "late."
They can be used in two ways:
1. Right before a noun, as in "a bombastic thing" or "a bombastic person."
Other common forms:
"Bombast" is the noun (meaning "inappropriately fancy language").
The adverb, "bombastically," is rare.
How to use it:
Use it with discretion: it's a critical, unkind word. So when you must, talk about bombastic people and their bombastic personalities and personas, their bombastic style, their bombastic announcements and pronouncements, their bombastic speeches and remarks, their bombastic exaggeration and rhetoric, etc.
Take care to notice that, according to dictionaries, "bombastic" means "pompous, using fancy words," and that it doesn't mean "arrogant," "swaggering," "overly confident," or (brace yourself) "braggadocious." Certainly there's a lot of crossover here: a bombastic speaker can also be all of those things--arrogant, swaggering and so on--but bombastic speakers, by definition, use fancy, lofty, highfalutin language. If you're using simple words and simple phrases, you're not being bombastic. But that's what dictionaries tell us; popular usage says otherwise. Let's keep an eye on this word and how it's being used. Dictionaries might have to acknowledge soon that it does apply to people who brag and boast in very simplistic language.
| |
| examples: |
The worst part of any bombastic blog post is the unnecessarily long introduction. We know where you're going with this; just get there, please.
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "bombastic" means when you can explain it without saying "puffed-up" or "pompous."
| |
| try it out: |
Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)."
Example: "Dr. Terminus is the bombastic villain in Pete's Dragon."
| |
| before you review: |
| |
| study it now: |
Look away from the screen to explain the definitionin your own words. You’ll know you understand what "bombastic" means when you can explain it without saying "puffed-up" or "pompous."
| |
| try it out: |
Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)."
Example: "Dr. Terminus is the bombastic villain in Pete's Dragon."
| |
| before you review: |
| |
| try it out: |
Think of a real or fictional person who talks in an overly fancy way. Fill in the blanks: "(Someone) is (a/the) bombastic (type of person)."
Example: "Dr. Terminus is the bombastic villain in Pete's Dragon."
| |
| before you review: |
| |
| before you review: |
Spend at least 20 seconds occupying your mind with the game below. Then try the review questions. Don’t go straight to the review now—let your working memory empty out first.
| |
| review today's word: |
1. The opposite of BOMBASTIC is
| |
| a final word: |
To be a sponsor and include your ad in an issue, please contact Liesl at Liesl@HiloTutor.com
Subscribe to "Make Your Point" for a daily vocabulary boost.
© Copyright 2016 | All rights reserved. |
6707 | https://www.prontuarionline.it/da-kg-a-newton/ | Da Kg a Newton: come convertire Massa e Peso
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21/01/2025
conversioni
Da Kg a Newton: come convertire Massa e Peso
/ conversioni / Da Kg a Newton: come convertire Massa e Peso
Riassunto:
Come passare da kg a Newton? Se hai bisogno di convertire i chilogrammi (kg) in Newton (N) e stai cercando una guida chiara, sei nel posto giusto. A differenza delle conversioni tra [continua]
Tabella dei Contenuti
Come passare da kg a Newton?
La conversione da kg a Newton
Come passare da Newton a kg?
Facciamo un esempio pratico
Cos’è il Newton: il concetto di Peso
Cos’è il kg: concetto di Massa
Relazione fisica tra Newton e Kg
Regola generale
Tabella di conversione rapida da kg a Newton
Esempi di calcolo: due esercizi per convertire Kg in Newton
Esempio 1: la spesa di Marco
Esempio 2: la gru al lavoro
Riassumendo questo articolo
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Come passare da kg a Newton?
Se hai bisogno di convertire i chilogrammi (kg) in Newton (N) e stai cercando una guida chiara, sei nel posto giusto. A differenza delle conversioni tra unità di peso o volume, passare da kg a Newton richiede di considerare un fattore fondamentale: l’accelerazione gravitazionale.
In questo articolo, ti spiegheremo come svolgere questa conversione e ti forniremo esempi pratici per facilitare il calcolo.
La conversione da kg a Newton
Per passare dai chilogrammi (kg) ai Newton (N), dobbiamo tenere conto della relazione tra massa, accelerazione gravitazionale e forza. La formula è:
N = kg × g
Dove:
N è la forza in Newton;
kg è la massa in chilogrammi;
g è l’accelerazione gravitazionale, che sulla Terra vale circa 9,81 m/s² (può variare leggermente a seconda del luogo).
Ad esempio, 1 kg di massa corrisponde a circa 9,81 N sulla superficie terrestre.
Come passare da Newton a kg?
Il procedimento inverso è altrettanto semplice. Per calcolare la massa in chilogrammi partendo dai Newton, devi dividere la forza per l’accelerazione gravitazionale:
kg = N ÷ g
Questa conversione è particolarmente utile in contesti di ingegneria, fisica o in applicazioni pratiche che richiedono di conoscere il peso effettivo in chilogrammi basato sulla forza gravitazionale.
Facciamo un esempio pratico
Esempio 1: Convertire 10 kg in NewtonSe hai un oggetto con una massa di 10 kg e vuoi sapere qual è la forza gravitazionale esercitata su di esso, il calcolo è:
10 kg × 9,81 m/s² = 98,1 N
Risultato: 10 kg equivalgono a 98,1 N sulla Terra.
Esempio 2: Convertire 196,2 N in kgSe invece hai una forza di 196,2 Newton e vuoi conoscere la massa corrispondente:
196,2 N ÷ 9,81 m/s² = 20 kg
Risultato: Una forza di 196,2 N corrisponde a una massa di 20 kg.
Cos’è il Newton: il concetto di Peso
Il Newton (N) è l’unità di misura della forza nel Sistema Internazionale (SI). Prende il nome da Isaac Newton e si definisce come la forza necessaria per accelerare una massa di 1 kg con un’accelerazione di 1 m/s². La relazione è descritta dalla formula:
1 N = 1 kg × 1 m/s²
Passiamo quindi al concetto di forza peso, che si definisce come la forza esercitata dalla gravità su un corpo con massa. Si calcola moltiplicando la massa per l’accelerazione gravitazionale (g), la cui formula fisica è:
Peso (N) = Massa (kg) × g (9,81 m/s²)
Nanonewton nN 10−9 1 nN = 10−9 N
Micronewton μN 10−6 1 μN = 10−6 N
Millinewton mN 10−3 1 mN = 10−3 N
Centinewton cN 10−2 1 cN = 10−2 N
Decinewton dN 10−1 1 dN = 10−1 N
NewtonN10 0 1 N = 1 N
Decanewton daN 10 1 1 daN = 10 N
Hectonewton hN 10 2 1 hN = 100 N
Kilonewton kN 10 3 1 kN = 1000 N
Meganewton MN 10 6 1 MN = 10 6 N
Giganewton GN 10 9 1 GN = 10 9 N
Cos’è il kg: concetto di Massa
Il chilogrammo (kg) è l’unità base della massa nel Sistema Internazionale (SI). Si definisce come la massa del prototipo internazionale del chilogrammo, un cilindro di platino-iridio conservato presso il Bureau International des Poids et Mesures (BIPM).
La massa, invece, rappresenta la quantità di materia di un corpo ed è una proprietà intrinseca, indipendente dalla gravità. Non va confusa con il peso, che dipende dalla forza gravitazionale.
| Unità | Abbreviazione | Fattore | Equivalente in kg |
--- --- |
| Milligrammo | mg | 10−6 | 1 mg = 10−6 kg |
| Centigrammo | cg | 10−5 | 1 cg = 10−5 kg |
| Decigrammo | dg | 10−4 | 10−4 kg |
| Grammo | g | 10−3 | 1 g = 10−3 kg |
| Decagrammo | dag | 10−2 | 1 dag = 10−2 kg |
| Ettogrammo | hg | 10−1 | 1 hg = 10−1 kg |
| Chilogrammo | kg | 10 0 | 1 kg = 1 kg |
| Quintale | q | 10 2 | 1 q = 100 kg |
| Tonnellata | t | 10 3 | 1 t = 1000 kg |
Relazione fisica tra Newton e Kg
Quando passiamo dai chilogrammi (unità di massa) ai Newton (unità di forza), stiamo applicando la seconda legge di Newton:
F = m × a
In questo caso, l’accelerazione è la gravità terrestre, che ha un valore medio di 9,81 m/s². Questo spiega perché il peso in Newton varia in base al valore di g, che non è sempre uguale in tutto il mondo (ad esempio, ai poli è leggermente maggiore rispetto all’equatore).
Per ottenere il peso in Newton, basta moltiplicare la massa in kg per il valore di g. Viceversa, dividendo i Newton per g, ottieni la massa in chilogrammi.
Regola generale
Ora che conosci la formula di base, ecco un riepilogo per eseguire rapidamente le conversioni:
Da kg a N: Moltiplica i chilogrammi (kg) per 9,81 m/s².
Formula: N = kg × 9,81
Da N a kg: Dividi i Newton (N) per 9,81 m/s².
Formula: kg = N ÷ 9,81
Se lavori in un contesto in cui l’accelerazione gravitazionale è diversa (ad esempio, su un altro pianeta), sostituisci il valore di 9,81 con il valore locale di g.
Tabella di conversione rapida da kg a Newton
Massa (kg)Forza (N)
1 9,81
5 49,05
10 98,1
20 196,2
50 490,5
100 981
500 4.905
Questa tabella può essere utilizzata come riferimento rapido per calcoli di routine.
Esempi di calcolo: due esercizi per convertire Kg in Newton
Esempio 1: la spesa di Marco
Marco ha deciso di fare una grande spesa per una festa di compleanno e si ritrova a trasportare un sacco di cibo e bevande. Quando pesa i sacchetti sul bilanciere, nota che il totale è di 50 kg. Curioso di sapere con quanta forza sta “sfidando” la gravità per trasportare tutto, si chiede: quanti Newton equivalgono a questi 50 kg?
Calcolo: Per calcolare la forza peso, Marco utilizza la formula: Peso (N) = Massa (kg) × g (9,81 m/s²)
Moltiplica quindi: 50 kg x 9,81 m/s²
Risultato:La spesa di Marco “pesa” ben 490,5 Newton.
Esempio 2: la gru al lavoro
In un cantiere edile, una gru sta sollevando un blocco di cemento. Il misuratore di forza della gru segna un carico di 2.452,5 N. L’operatore vuole verificare la massa del blocco per assicurarsi che non superi il limite consentito della gru. Si chiede quindi: quanti chilogrammi corrispondono a 2.452,5 Newton?
Calcolo: Per trovare la massa, l’operatore utilizza la formula inversa: Massa (kg) = Peso (N) / g (9,81 m/s²)
Divide quindi: 2.452,5 N : 9,81 9,81 m/s²
Risultato:Il blocco di cemento ha una massa di 250 kg, ed è sicuro per la gru sollevarlo.
Riassumendo questo articolo
Con questa guida, hai a disposizione tutti gli strumenti per passare rapidamente da kg a Newton (e viceversa). Ricorda:
Moltiplica i kgper 9,81 per ottenere i Newton
Dividi i Newton per 9,81 per ottenere i kg
Se vuoi effettuare calcoli più complessi o hai bisogno di un aiuto rapido, salva il prossimo linke scopri come semplificare il tutto con il nostro Convertitore Unità di Misura.
Convertitore Unità di Misura
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2: Gas Laws
Thermodynamics and Chemical Equilibrium (Ellgen)
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Fri, 28 Apr 2023 06:59:20 GMT
2.2: Charles' Law
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Contents
1. Home
2. Bookshelves
3. Physical & Theoretical Chemistry
4. Thermodynamics and Chemical Equilibrium (Ellgen)
5. 2: Gas Laws
6. 2.2: Charles' Law
Expand/collapse global location
Thermodynamics and Chemical Equilibrium (Ellgen)
Front Matter
1: Introduction - Background and a Look Ahead
2: Gas Laws
3: Distributions, Probability, and Expected Values
4: The Distribution of Gas Velocities
5: Chemical Kinetics, Reaction Mechanisms, and Chemical Equilibrium
6: Equilibrium States and Reversible Processes
7: State Functions and The First Law
8: Enthalpy and Thermochemical Cycles
9: The Second Law - Entropy and Spontaneous Change
10: Some Mathematical Consequences of the Fundamental Equation
11: The Third Law, Absolute Entropy, and the Gibbs Free Energy of Formation
12: Applications of the Thermodynamic Criteria for Change
13: Equilibria in Reactions of Ideal Gases
14: Chemical Potential - Extending the Scope of the Fundamental Equation
15: Chemical Potential, Fugacity, Activity, and Equilibrium
16: The Chemical Activity of the Components of a Solution
17: Electrochemistry
18: Quantum Mechanics and Molecular Energy Levels
19: The Distribution of Outcomes for Multiple Trials
20: Boltzmann Statistics
21: The Boltzmann Distribution Function
22: Some Basic Applications of Statistical Thermodynamics
23: The Ensemble Treatment
24: Indistinguishable Molecules - Statistical Thermodynamics of Ideal Gases
25: Bose-Einstein and Fermi-Dirac Statistics
26: Appendices
Back Matter
2.2: Charles' Law
Last updated Apr 28, 2023
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2.1: Boyle's Law
2.3: Avogadro's Hypothesis
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Quantitative experiments establishing the law were first published in 1802 by Gay-Lussac, who credited Jacques Charles with having discovered the law earlier. Charles’ law relates the volume and temperature of a gas when measurements are made at constant pressure. We can imagine rediscovering Charles’ law by trapping a sample of gas in a tube and measuring its volume as we change the temperature, while keeping the pressure constant. This presumes that we have a way to measure temperature, perhaps by defining it in terms of the volume of a fixed quantity of some other fluid—like liquid mercury. At a fixed pressure, P 1, we observe a linear relationship between the volume of a sample of gas and its temperature, like that in Figure 2. If we repeat this experiment with the same gas sample at a higher pressure, P 2, we observe a second linear relationship between the volume and the temperature of the gas. If we extend these lines to their intersection with the temperature axis at zero volume, we make a further important discovery: Both lines intersect the temperature axis at the same point.
Figure 2: Gas volume versus temperature.
We can represent this behavior mathematically as
V=β∗(n,P)T∗+γ∗(n,P)
where we recognize that both the slope and the V-axis intercept of the graph depend on the pressure of the gas and on the number of moles of gas in the sample. A little reflection shows that here too the slope and intercept must be directly proportional to the number of moles of gas, so that we can rewrite our equation as
V=nβ(P)T∗+nγ(P)
When we repeat these experiments with different gaseous substances, we discover an additional important fact: β(P) and γ(P) are the same for any gas. This means that the temperature at which the volume extrapolates to zero is the same for any gas and is independent of the constant pressure we maintain as we vary the temperature (Figure 2).
This page titled 2.2: Charles' Law is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform.
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6710 | https://pubmed.ncbi.nlm.nih.gov/24525983/ | Pulmonary dysfunction and its management in post-polio patients - PubMed
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. 1997;8(2):139-53.
doi: 10.3233/NRE-1997-8207.
Pulmonary dysfunction and its management in post-polio patients
J R Bach1,M Tilton2
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Affiliations
1 Department of Physical Medicine and Rehabilitation, UMDNJ-New Jersey Medical School, 150 Bergen Street, Newark, NJ 07103, USA Center for Ventilator Management Alternatives, University Hospital, Newark, NJ, USA.
2 Department of Physical Medicine and Rehabilitation, UMDNJ-New Jersey Medical School, 150 Bergen Street, Newark, NJ 07103, USA Kessler Institute for Rehabilitation, West Orange, NJ, USA.
PMID: 24525983
DOI: 10.3233/NRE-1997-8207
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Pulmonary dysfunction and its management in post-polio patients
J R Bach et al. NeuroRehabilitation.1997.
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. 1997;8(2):139-53.
doi: 10.3233/NRE-1997-8207.
Authors
J R Bach1,M Tilton2
Affiliations
1 Department of Physical Medicine and Rehabilitation, UMDNJ-New Jersey Medical School, 150 Bergen Street, Newark, NJ 07103, USA Center for Ventilator Management Alternatives, University Hospital, Newark, NJ, USA.
2 Department of Physical Medicine and Rehabilitation, UMDNJ-New Jersey Medical School, 150 Bergen Street, Newark, NJ 07103, USA Kessler Institute for Rehabilitation, West Orange, NJ, USA.
PMID: 24525983
DOI: 10.3233/NRE-1997-8207
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Abstract
Respiratory dysfunction is extremely common and entails considerable risk of morbidity and mortality for individuals with past poliomyelitis. Although it is usually primarily due to respiratory muscle weakness, post- poliomyelitis individuals also have a high incidence of scoliosis, obesity, sleep disordered breathing, and bulbar muscle dysfunction. Although these factors can result in chronic alveolar hypoventilation (CAH) and frequent pulmonary complications and hospitalizations, CAH is usually not recognized until acute respiratory failure complicates an otherwise benign upper respiratory tract infection. The use of non-invasive inspiratory and expiratory muscle aids, however, can decrease the risk of acute respiratory failure, hospitalizations for respiratory complications, and need to resort to tracheal intubation. Timely introduction of non-invasive intermittent positive pressure ventilation (IPPV), manually assisted coughing, and mechanical insufflation-exsufflation (MI-E) and non-invasive blood gas monitoring which can most often be performed in the home setting, are the principle interventions for avoiding complications and maintaining optimal quality of life.
Keywords: Artificial ventilation; Obstructive sleep apnea syndrome; Poliomyelitis; Rehabilitation.
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Management of post-polio respiratory sequelae.Bach JR.Bach JR.Ann N Y Acad Sci. 1995 May 25;753:96-102. doi: 10.1111/j.1749-6632.1995.tb27535.x.Ann N Y Acad Sci. 1995.PMID: 7611664
Airway secretion clearance by mechanical exsufflation for post-poliomyelitis ventilator-assisted individuals.Bach JR, Smith WH, Michaels J, Saporito L, Alba AS, Dayal R, Pan J.Bach JR, et al.Arch Phys Med Rehabil. 1993 Feb;74(2):170-7.Arch Phys Med Rehabil. 1993.PMID: 8431102
Mechanical exsufflation, noninvasive ventilation, and new strategies for pulmonary rehabilitation and sleep disordered breathing.Bach JR.Bach JR.Bull N Y Acad Med. 1992 Mar-Apr;68(2):321-40.Bull N Y Acad Med. 1992.PMID: 1586868 Free PMC article.Review.
Pulmonary rehabilitation in patients with neuromuscular disease.Kang SW.Kang SW.Yonsei Med J. 2006 Jun 30;47(3):307-14. doi: 10.3349/ymj.2006.47.3.307.Yonsei Med J. 2006.PMID: 16807978 Free PMC article.Review.
Non-invasive assessment of respiratory muscle strength in patients with previous poliomyelitis.Soliman MG, Higgins SE, El-Kabir DR, Davidson AC, Williams AJ, Howard RS.Soliman MG, et al.Respir Med. 2005 Oct;99(10):1217-22. doi: 10.1016/j.rmed.2005.02.035. Epub 2005 Apr 12.Respir Med. 2005.PMID: 16140221
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Comparison of activity and fatigue of the respiratory muscles and pulmonary characteristics between post-polio patients and controls: A pilot study.Shoseyov D, Cohen-Kaufman T, Schwartz I, Portnoy S.Shoseyov D, et al.PLoS One. 2017 Jul 27;12(7):e0182036. doi: 10.1371/journal.pone.0182036. eCollection 2017.PLoS One. 2017.PMID: 28750019 Free PMC article.
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6711 | https://stackoverflow.com/questions/8804975/time-complexity-of-modular-arithmetic | Stack Overflow
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Time complexity of modular arithmetic
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Say I wanted to calculate a (mod n). What is the time complexity of this? I'm using Matlab, and am not sure how Matlab calculates it. Does it divide a by n, subtract the integer part, and then multiply by n?
Does it make sense to ask 'what is the time complexity of this'?
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edited Jan 10, 2012 at 14:47
Andrey Rubshtein
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It makes sense to ask about the time complexity for long numbers a and/or n. The related field is called Computational Number Theory. For example, see this book.
The usual integer arithmetic, which most likely is used by Matlab, is an ALU operation (or multiple operations) performed in a constant time. In this case, one has to remember that the size of the integer is restricted.
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answered Jan 10, 2012 at 14:47
amit kumaramit kumar
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According to the built in help Matlab calculates MOD(x,y) as:
MOD(x,y) = x - floor(x./y).y
where the floor function rounds towards minus infinity (that is strips the decimal part).
Runtime will be constant as long as you don't calculate mod(X, y) where X is a vector, in that case it will scale linearly with the number of elements in the vector
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edited Sep 11, 2020 at 17:50
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answered Jan 10, 2012 at 15:10
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6712 | https://www.nccn.org/patients/guidelines/content/PDF/ovarian-patient.pdf | 2024 Ovarian Cancer NCCN GUIDELINES FOR PATIENTS ® Presented with support from FOUNDATION Guiding Treatment. Changing Lives.
NATIONAL COMPREHENSIVE CANCER NETWORK ® Available online at NCCN.org/patientguidelines Ü 1 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Ovarian Cancer Find an NCCN Cancer Center near you NCCN.org/cancercenters About the NCCN Guidelines for Patients® Did you know that top cancer centers across the United States work together to improve cancer care? This alliance of leading cancer centers is called the National Comprehensive Cancer Network® (NCCN®). Cancer care is always changing. NCCN develops evidence-based cancer care recommendations used by health care providers worldwide. These frequently updated recommendations are the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®). The NCCN Guidelines for Patients plainly explain these expert recommendations for people with cancer and caregivers.
These NCCN Guidelines for Patients are based on the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Ovarian Cancer/Fallopian Tube Cancer/Primary Peritoneal Cancer, Version 3.2024 — July 15, 2024.
View the NCCN Guidelines for Patients free online NCCN.org/patientguidelines Connect with us 2 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Ovarian Cancer Supporters Additional support is provided by To make a gift or learn more, visit online or email The National Ovarian Cancer Coalition (NOCC) is an influential advocate for those experiencing ovarian cancer. NOCC is committed to providing tools and resources for patients and caregivers by offering virtual, evidence-based educational programing, peer-to-peer support groups, and direct support services using our regional model throughout the United States. NOCC’s community-focused approach is at the heart of everything we do, from funding innovative research that will lead to improved quality of life outcomes to promoting advocacy in action through disease awareness and outreach events in communities like yours. For more information, please visit ovarian.org or call 888-OVARIAN.
Ovarian Cancer Research Alliance is committed to curing ovarian cancer, advocating for patients, and supporting survivors. OCRA is the largest ovarian cancer charity, with over $122M invested in research. Our conference, website, clinical trial finder, and doctor finder connect patients to information on diagnosis, treatment and living with gynecologic cancers. Our support programs include our patient support line, peer mentor program, and Staying Connected workshops. OCRA builds community through advocacy, research, and support.ocrahope.org or call 212-268-1002. NCCNFoundation.org/donate PatientGuidelines@ NCCN.org NCCN Foundation gratefully acknowledges the following corporate supporters for helping to make available these NCCN Guidelines for Patients: AstraZeneca and GSK.
NCCN independently adapts, updates, and hosts the NCCN Guidelines for Patients. Our corporate supporters do not participate in the development of the NCCN Guidelines for Patients and are not responsible for the content and recommendations contained therein.
NCCN Guidelines for Patients are supported by funding from the NCCN Foundation® FOUNDATION Guiding Treatment. Changing Lives.
NATIONAL COMPREHENSIVE CANCER NETWORK ® 3 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Ovarian Cancer Contents © 2024 National Comprehensive Cancer Network, Inc. All rights reserved. NCCN Guidelines for Patients and illustrations herein may not be reproduced in any form for any purpose without the express written permission of NCCN. No one, including doctors or patients, may use the NCCN Guidelines for Patients for any commercial purpose and may not claim, represent, or imply that the NCCN Guidelines for Patients that have been modified in any manner are derived from, based on, related to, or arise out of the NCCN Guidelines for Patients. The NCCN Guidelines are a work in progress that may be redefined as often as new significant data become available. NCCN makes no warranties of any kind whatsoever regarding its content, use, or application and disclaims any responsibility for its application or use in any way.
NCCN Foundation seeks to support the millions of patients and their families affected by a cancer diagnosis by funding and distributing NCCN Guidelines for Patients. NCCN Foundation is also committed to advancing cancer treatment by funding the nation’s promising doctors at the center of innovation in cancer research. For more details and the full library of patient and caregiver resources, visit NCCN.org/patients. National Comprehensive Cancer Network (NCCN) and NCCN Foundation 3025 Chemical Road, Suite 100, Plymouth Meeting, PA 19462 USA 4 Ovarian cancer basics 8 Testing for ovarian cancer 17 Treatment for common ovarian cancers 41 Treatment for less common ovarian cancers 52 Survivorship 56 Making treatment decisions 64 Words to know 68 NCCN Contributors 69 NCCN Cancer Centers 72 Index 4 NCCN Guidelines for Patients® Ovarian Cancer, 2024 1 Ovarian cancer basics 5 The ovaries 6 Types of ovarian cancer 7 Cancer care plan 7 Key points 5 NCCN Guidelines for Patients® Ovarian Cancer, 2024 1 Ovarian cancer basics » The ovaries Most ovarian cancers are found in the surface layer of tissue surrounding the ovaries, called the epithelium. These cancers can also start in the fallopian tube, close to where the tube meets the ovary. This guide provides treatment recommendations for common and rare ovarian cancers.
The ovaries The ovaries are part of the female reproductive system. In addition to the 2 ovaries, this system includes the fallopian tubes, uterus, cervix, and vagina. The ovaries make eggs needed for sexual reproduction. They also release hormones that affect breast growth, body shape, and the menstrual cycle (periods). Each ovary is about the size and shape of a grape. One is on the left side of the uterus and the other is on the right. Each is surrounded by a long, thin tube called a fallopian tube. After an egg is pushed out by the ovary, it is caught by the fallopian tube and travels into the uterus. Here the fetus grows and develops during pregnancy. If the egg is not fertilized, this leads to a period. The uterus and at least 1 ovary and fallopian tube are needed for menstruation and pregnancy. The female reproductive system The female reproductive system includes the ovaries, fallopian tubes, uterus, cervix, and vagina.
6 NCCN Guidelines for Patients® Ovarian Cancer, 2024 1 Ovarian cancer basics » Types of ovarian cancer Types of ovarian cancer Most ovarian cancers are found in the outer surface of the ovaries, called the epithelium. There are more than 5 types of epithelial ovarian cancer. The most common forms are: h High-grade serous carcinoma (HGSC) h High-grade endometrioid carcinoma Less common ovarian cancers Rare types of ovarian cancer are called less common ovarian cancers (LCOCs) or less common ovarian histologies (LCOHs). They can start in the epithelium, in tissues that support the ovaries, or in the reproductive (egg) cells of the ovary.
Less common epithelial ovarian cancers include: h Low-grade serous carcinoma h Low-grade endometrioid carcinoma h Carcinosarcoma (also called malignant mixed Mullerian tumors) h Clear cell carcinoma h Mucinous carcinoma h Borderline epithelial tumor (also called low malignant potential tumors) Less common non-epithelial ovarian cancers include: h Malignant sex-cord stromal tumors h Malignant germ cell tumors How is the type determined?
To diagnose ovarian cancer and determine the cancer type, tumor tissue needs to be removed from your body and tested. If surgery is planned first, the tumor and other tissues removed during surgery will be tested. If chemotherapy is planned first, a biopsy will be performed to remove a sample of the tumor. A physician expert called a pathologist determines the type of ovarian cancer by examining the cancerous tissue.
The pathologist also determines the cancer grade. The grade is a rating of how abnormal the cancer cells look under a microscope. High-grade cancers grow and spread more quickly than low-grade cancers. The cancer grade is different than the stage.
Stages are categories that describe where the cancer has or hasn't spread from the ovary. Surgery is needed in order to know exactly how much cancer is in the body. Testing can provide a best guess of how far the cancer has spread before surgery. Primary peritoneal cancer Primary peritoneal cancer forms in the the tissue that lines the abdominal wall and covers the abdominal organs. The treatment information in this guide also applies to primary peritoneal cancer and fallopian tube cancer.
7 NCCN Guidelines for Patients® Ovarian Cancer, 2024 1 Ovarian cancer basics » Cancer care plan Cancer care plan Your treatment team Treatment for ovarian cancer takes a team of experts. When possible, a gynecologic oncologist should perform the initial surgery. This type of doctor is an expert in surgery and chemotherapy for gynecologic cancers. Your care team may also include a medical oncologist. This doctor is an expert in treating cancer with chemotherapy and other medicines.
You may also receive care from registered nurses, nurse practitioners, physician assistants, social workers, genetic counselors, sexual health experts, and others. Ask for the names and contact information of your care providers to be included in the treatment plan.
Cancer treatment may be improved if your primary care provider is involved. They can help manage other health conditions that may be affected by your cancer treatment.
Your treatment plan There isn't one treatment plan that is best for everyone. There is often more than one option, including clinical trials. Clinical trials study the safety and effectiveness of investigational treatments. The treatment that you and your care team agree on should be noted in the treatment plan, along with possible side effects. Keep in mind that your plan may change. Testing may provide new information. Your feelings about treatment may change. Side effects or other health conditions may prompt a change of plan. Key points h The ovaries are a pair of grape-sized organs in the pelvis. They make hormones and eggs for sexual reproduction. h Most ovarian cancers affect the layer of tissue surrounding the ovaries, called the epithelium. h High-grade serous carcinoma (HGSC) and high-grade endometrioid carcinoma are the most common types of ovarian tumors. h Less common ovarian cancers can start in the epithelium, in tissues that support the ovaries, or in the reproductive (egg) cells of the ovary. h Treating ovarian cancer takes a team of experts. Gynecologic oncologists and medical oncologists often work together to plan your treatment.
8 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer 9 Abdominal and pelvic exam 10 Imaging tests 12 Biopsy 13 Family history and genetic testing 14 Nutritional and digestive tract health 14 Blood tests 15 Biomarker testing 16 Key points 9 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Abdominal and pelvic exam This chapter describes the testing used to learn more about suspected ovarian cancer, including whether surgery is possible.
Ovarian cancer can cause changes in the body that you can feel or notice. But you might not have symptoms until the tumor has grown large or the cancer has spread. Common symptoms include: h Feeling bloated h Heartburn and indigestion h Pain or pressure in the pelvis or belly h Trouble eating or feeling full fast h Having to urinate often or urgently h Pain during sex These symptoms can also be caused by hormonal changes or other common health problems. Ovarian cancer is more likely to be the cause if the symptoms: h Began less than 1 year ago, h Occur more than 12 days per month, and h Are becoming more severe over time.
If your provider suspects ovarian cancer based on your symptoms, you will have testing as described in this chapter. Testing helps determine the clinical (pre-treatment) stage. The clinical stage provides a best guess of how far the cancer has spread. It is a best guess because surgery is needed in order to know exactly how much cancer is in the body.
Testing also helps determine whether surgery first is the best treatment. Having surgery first may not be possible based on the size and location of the tumor, or because of other health factors. Abdominal and pelvic exam Your provider will feel different areas of your belly. This is called an abdominal exam. They are checking to see if organs are of normal size, are soft or hard, or cause pain when touched. Your doctor will also feel for signs of fluid buildup (ascites) in the belly area or around the ovaries.
Your provider will also feel for abnormal changes in the size, shape, or position of your ovaries, cervix, and uterus. This is called a pelvic exam. A widening instrument, called a speculum, is used to view your vagina and cervix. A sample of cells may be removed for testing. This is known as a Pap test. It is used to detect cervical cancer or pre-cancer, not ovarian cancer.
Sometimes, a biopsy of the uterine lining (an endometrial biopsy) may be part of the initial evaluation during the pelvic exam. This test can rule out a uterine cancer.
An exam of the rectum and vagina together may also be done to check for cancer in the space between the rectum and vagina. This is called a rectovaginal exam.
10 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Imaging tests Imaging tests Imaging tests can show the location, size, and shape of an ovarian tumor. They can also show if the cancer has spread beyond the ovaries. Your care team will tell you how to prepare for your imaging tests.
Ultrasound Ultrasound is often the first imaging test used to look for ovarian cancer. It uses sound waves to make pictures of areas inside of the body. Ultrasound is good at showing the size, shape, and location of the ovaries, fallopian tubes, uterus, and nearby tissues. It can also show if there is a mass in the ovary and whether the mass is solid or filled with fluid.
The 2 types of ultrasounds that may be used to look for ovarian cancer are described next. Both are done using a hand-held device called an ultrasound probe. Ultrasounds are generally painless, but you may feel some discomfort when the probe is inserted.
For a transabdominal ultrasound, a gel will be spread on your abdomen and pelvis. The gel helps to make the pictures clearer. Your doctor will place the probe on your skin and guide it back and forth in the gel.
For a transvaginal ultrasound, your doctor will insert the probe into your vagina. This may help the doctor see your ovaries more clearly.
Transvaginal ultrasound Ultrasound uses sound waves to make pictures of the inside of the body. For a transvaginal ultrasound, a probe is inserted into the vagina. Ultrasounds are generally painless, but you may feel some discomfort when the probe is inserted.
11 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Imaging tests CT Imaging to look for ovarian cancer may include computed tomography (CT) scans of your abdomen, pelvis, and chest. CT scans are good at showing if the cancer has spread outside of the ovaries. They may also show if nearby lymph nodes are bigger than normal. This can be a sign that the cancer has spread.
If you can have it, a substance called contrast is used to make the pictures clearer. Before the scan you will be asked to drink a large glass of oral contrast. A contrast agent will also be injected into your vein. It may cause you to feel flushed or get hives. Rarely, allergic reactions can occur. Tell your team if you've had an allergic reaction to contrast in the past.
A CT scanner is a large machine with a tunnel in the middle. While you lie on a table that moves through the tunnel, the scanner will rotate an x-ray beam around you to take pictures from many angles. MRI If the ultrasound images are unclear, you may have magnetic resonance imaging (MRI) of your abdomen and pelvis. An MRI of your chest or liver may be used to look for signs of cancer spread. MRI doesn’t use radiation. It uses radio waves and powerful magnets to take pictures of areas inside the body.
Getting an MRI scan is similar to getting a CT scan but takes longer. Like a CT scan, a contrast agent may be used to make the pictures clearer. You will lie on a table that moves through a large tunnel in the scanning machine. The machine is more enclosed than a CT scan. Tell your care team if you get nervous in enclosed spaces. You may be given a type of medicine called a sedative to help you relax. CT scan A CT scan is a more detailed kind of x-ray. It takes a lot of pictures, or images, from different angles. A computer then combines the images to make 3-D pictures.
12 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Biopsy PET CT or MRI are sometimes combined with positron emission tomography (PET). A PET scan shows how your cells are using a simple form of sugar, which can be helpful for identifying cancer. A sugar radiotracer is put into your body through a vein. The radiotracer puts out a small amount of energy that is detected by the machine that takes pictures. Cancer cells appear brighter because they use sugar faster than normal cells.
PET is very good at showing small groups of cancer cells. This test may also be useful for showing if ovarian cancer has spread.
Chest x-ray A chest x-ray can show if cancer has spread to your lungs. It may be ordered with other tests when ovarian cancer is first suspected or found. It may also be used to check treatment results. A chest x-ray is fast and painless and uses small amounts of radiation.
Diagnostic laparoscopy If the cancer is advanced, you may have a diagnostic laparoscopy before treatment. The goal is to learn how much cancer is in the abdomen. It helps your doctors to decide whether surgery can be the first treatment, or if chemotherapy is needed first. This minimally invasive procedure involves making a tiny cut in the abdomen. A thin tube with a light and a camera (laparoscope) is used to view the lining of the abdomen and the surface of organs in the abdomen. Tissue samples are taken and tested for cancer cells in a lab.
Biopsy To diagnose ovarian cancer, a sample of tissue must be removed from your body for testing. This is called a biopsy. The biopsy is usually done during initial surgery to remove the cancer.
But sometimes a biopsy is done to diagnose ovarian cancer before surgery or other planned treatment. This may be the case if the cancer has spread too much to be removed by surgery initially and chemotherapy is needed first. In such cases, a fine-needle aspiration (FNA) biopsy, core biopsy, or paracentesis may be used. FNA uses a very thin needle to remove a small sample of tissue from the tumor. A core biopsy removes tissue samples with a hollow needle. For paracentesis, a long, thin needle is inserted through the skin of the belly to remove a sample of fluid. The biopsy samples are sent to a pathologist for testing. A pathologist is a doctor who is an expert in testing cells to find disease. The pathologist views the samples with a microscope to look for cancer cells. If the cells are cancerous, the pathologist notes their appearance and other features.
Prior surgery or biopsy The cancer may have been found during a surgery or biopsy performed by another doctor. In this case, your treatment team will need to review the prior surgery and testing results. A pathologist will examine the tumor tissue with a microscope to make sure it is ovarian cancer. Your doctors will also want to know if any cancer was left in your body after surgery. 13 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Family history and genetic testing Family history and genetic testing Ovarian cancer usually occurs for unknown reasons. However, about 1 in 6 ovarian cancers is caused by mutations (changes) in genes that are passed down from parent to child. This is called hereditary ovarian cancer. It is most often caused by mutations in either breast cancer gene 1 (BRCA1) or breast cancer gene 2 (BRCA2). Everyone has BRCA1 and BRCA2 genes. When working properly, they are helpful and repair damaged cells. But mutations in these genes increase the risk of developing ovarian, breast, and some other cancers.
Another cause of hereditary ovarian cancer is Lynch syndrome. Lynch syndrome is the most common cause of hereditary colon and uterine cancers but can also cause ovarian and other cancers. Ovarian cancer associated with a BRCA mutation or Lynch syndrome usually starts at a younger age than non-hereditary ovarian cancer. Using your age, health history, and family history, your doctor will assess how likely you are to have hereditary ovarian cancer.
Genetic testing can tell if you have a mutation in the BRCA genes, or in other genes that play a role in hereditary cancer. It is recommended for everyone diagnosed with ovarian cancer. If initial treatment works well, BRCA status (whether you have a BRCA mutation) plays an important role in guiding decisions about maintenance therapy. Genetic testing may be done through your gynecology or oncology care team, or by a genetic counselor. The testing is done on normal tissue–either blood, saliva, or a cheek swab. Those with a positive genetic test or who have a strong cancer family history should see a health expert. This is typically a genetic counselor. A genetic counselor has special training to help patients understand changes in genes that are related to disease.
There are many other hereditary syndromes besides BRCA and Lynch. Genetic testing typically tests for all of them. Commercial genetic tests currently available over the counter look for the most common gene mutations, but aren't comprehensive. In addition to testing for germline (inherited) BRCA mutations, the tumor itself should be tested for mutations in the BRCA and related genes. These are known as somatic or tumor mutations. Genetic testing is recommended for everyone diagnosed with ovarian cancer. It can determine if you have a mutation in the BRCA genes, or in other genes that play a role in hereditary cancer.
14 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Nutritional and digestive tract health 14 2 Testing for ovarian cancer » Nutritional and digestive tract health » Blood tests Nutritional and digestive tract health Your provider may ask about your diet and eating habits. Symptoms of ovarian cancer include bloating, pain in the pelvis or abdomen, difficulty eating, and feeling full quickly. These symptoms may cause you to eat less in general, or to eat foods lacking in nutrients. Your overall health and nutrition level can have an impact on the success of surgery and other treatment outcomes. If you need help planning healthy meals or have questions about your diet, ask your provider for a referral to a registered dietitian or nutritionist.
Your doctor may want to check your gastrointestinal (GI) tract using an imaging test. The GI tract is made of the organs that food passes through when you eat. This includes your stomach, small bowel, and large bowel (rectum and colon). An imaging tool called a scope is used to examine these organs. A scope is a long, thin tube with a light and a camera that can be guided into your body. A colonoscopy is used to examine the large bowel. This involves inserting a scope into your anus and guiding it through the rectum and colon. To examine the upper GI tract, a scope is guided down the throat into the esophagus, stomach, and small bowel. This is called an upper endoscopy.
Blood tests The following tests are not used alone to diagnose ovarian cancer, but abnormal results may signal health problems.
General health A complete blood count (CBC) measures the number of red blood cells, white blood cells, and platelets in a sample of blood. Red blood cells carry oxygen throughout the body. White blood cells fight infection. Platelets help to control bleeding. Your blood counts may be too low or too high because of cancer or other health problems.
A blood chemistry profile measures the levels of different chemicals that are affected by your kidneys, bones, and other organs and tissues. Levels that are too high or too low may be a sign that an organ isn’t working well. Abnormal levels may also be caused by the spread of cancer or by other diseases. This test can also provide information about nutrient intake, such as protein levels. This can help guide treatment decisions.
The liver is an organ that does many important jobs, such as removing toxins from your blood. Liver function tests measure chemicals that are made or processed by the liver. Levels that are too high or low may be a sign of liver damage or cancer spread.
Tumor markers A tumor marker is a substance found in body tissue or fluid that may be a sign of cancer. Along with other information, tumor markers can help diagnose ovarian cancer and monitor response to treatment.
15 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Biomarker testing A cancer antigen-125 (CA-125) test is the most common tumor marker test for ovarian cancer. High levels of this protein in the blood may be a sign of ovarian or other cancers. A CA-125 test alone cannot diagnose ovarian cancer.
Health problems that are not cancer, such as endometriosis and diverticulitis, can raise your CA-125 level. Some ovarian cancers don’t cause CA-125 to rise.
Your blood may also be tested for the following tumor markers. These may be found in higher-than-normal amounts in people with less common ovarian cancers (LCOCs). h Inhibin (typically inhibin A and inhibin B) h Beta-human chorionic gonadotropin (β-hCG) h Alpha-fetoprotein (AFP) h Lactate dehydrogenase (LDH) h Carcinoembryonic antigen (CEA) h CA 19-9 h HE4 Biomarker testing Biomarkers are features of the tumor that can help guide your treatment. They are often mutations (changes) in the DNA of the cancer cells. Testing for biomarkers involves analyzing a piece of tumor tissue in a lab or testing a sample of blood. The results can be used to determine whether you can join certain clinical trials, and whether you may benefit from specific maintenance therapies.
Other names for biomarker testing include molecular testing, tumor profiling, genomic testing, tumor gene testing, next-generation sequencing (NGS), mutation testing, liquid biopsy, and precision oncology.
BRCA and HRD A BRCA mutation is the most important biomarker used to plan ovarian cancer treatment. Everyone diagnosed with ovarian cancer should have the tumor tested for mutations in the BRCA genes, and in other similar genes important in DNA repair.
This is different than genetic testing of the blood for inherited (germline) BRCA mutations. Mutations in the tumor itself are known as somatic or simply “tumor” mutations.
BRCA mutations are a form of homologous recombination deficiency (HRD). This means that if you have a BRCA mutation, the cancer is also homologous recombination deficient or HRD positive. However, you can also have an HRD-positive tumor without a BRCA mutation.
Other changes in the tumor’s DNA can make it homologous recombination deficient. The tumor's BRCA and HRD status are used to 16 NCCN Guidelines for Patients® Ovarian Cancer, 2024 2 Testing for ovarian cancer » Key points guide decisions about maintenance therapy after initial treatment.
Other biomarkers The timing of testing for the biomarkers described next can vary. Some providers test for these (in addition to BRCA) early in the treatment process. Others may only test for BRCA and wait to see if therapies that require other biomarkers are needed. However, testing for these biomarkers is generally recommended for ovarian cancer that returns after treatment (recurrent). Testing is performed on removed tumor tissue. h Microsatellite instability (MSI) h Mismatch repair (MMR) h HER2 expression h Tumor mutational burden (TMB) h BRAF V600E mutation h Folate receptor alpha (FRα) expression h RET mutations h NTRK gene fusion Key points h The biopsy to diagnose ovarian cancer is usually done during initial surgery. If your team recommends chemotherapy before surgery, a biopsy will be done before starting chemotherapy h Ultrasound is often the first imaging test performed for suspected ovarian cancer. h Blood tests for suspected ovarian cancer include a CBC, chemistry profile, liver function tests, and tumor marker tests. h Hereditary ovarian cancer is most often caused by a mutation in the BRCA genes. h Families with a history of Lynch syndrome may also be at risk for ovarian and other cancers. h Everyone diagnosed with ovarian cancer should have genetic testing of the blood for inherited (germline) BRCA mutations. h Biomarker testing looks for features of the cancer, such as mutations, that can help guide your treatment. Everyone diagnosed with ovarian cancer should have their tumor tested for mutations in the BRCA genes and others important in DNA repair.
17 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers 18 Surgery 21 If surgery first isn’t an option 22 Staging 29 Chemotherapy 31 Maintenance therapy 33 Surveillance 34 Recurrence 38 Clinical trials 40 Key points 18 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Surgery The most common types of ovarian cancer are high-grade serous carcinoma and high-grade endometrioid carcinoma. These cancers are treated with surgery and chemotherapy. If treatment works well, maintenance therapy may be an option for more advanced cancers.
Surgery Surgery is often the first treatment if you are willing and able to have it. Sometimes chemotherapy is given first. Surgery should be performed by a gynecologic oncologist. This is a surgeon who is an expert in cancers that start in the female reproductive organs. If your team recommends chemotherapy before surgery, see page 21.
The main goals of surgery are to: h Remove all or as much of the cancer as possible, and h Learn how far the cancer has spread.
Hysterectomy with BSO The most common surgery for ovarian cancer is hysterectomy and bilateral salpingo-oophorectomy (BSO). A hysterectomy is surgery to remove the uterus. When the cervix is removed in addition to the uterus, it is called a total or complete hysterectomy. A BSO removes both ovaries and both fallopian tubes.
Pregnancy isn’t possible after a hysterectomy. Fertility-sparing surgery (described below) may be an option for some very early ovarian cancers that haven't spread beyond the ovaries.
If cancer has spread outside the ovaries, your surgeon will attempt to remove as much of it as possible. This is called debulking or cytoreductive surgery. The extent of the surgery depends on how far the cancer has spread. It may involve removing all or part of nearby organs. Lymph nodes that look abnormal or that are larger than normal will also be removed when possible.
Fertility-sparing surgery Pregnancy isn’t possible after the uterus is removed. This is difficult for those wishing to get pregnant in the future. Fertility-sparing surgery may be an option. This involves removing one or both ovaries and fallopian tubes but leaving the uterus in place. Surgery to remove one ovary and its fallopian tube is called a unilateral salpingo-oophorectomy (USO). USO is only an option if the cancer is in 1 ovary and the cancer is appropriate for this procedure. 19 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Surgery After a USO, you may still be able to become pregnant naturally if you haven’t entered menopause.
If the cancer is in both ovaries, a BSO (without hysterectomy) may be an option. While you can't become pregnant naturally after a BSO, pregnancy may be possible using assisted reproductive approaches. One such approach is in vitro fertilization (IVF). In IVF, eggs are fertilized with sperm in a lab to create embryos. The embryos are implanted into the uterus or frozen for future use. The eggs used for IVF may be yours (removed from your ovary before surgery) or donor eggs. Donor eggs are removed from women who have volunteered to go through hormone treatment to stimulate egg production in the ovaries.
Surgical methods A laparotomy is the most common method for ovarian cancer surgery. A laparotomy is a long surgical cut in the abdomen. It is often an up-and-down (vertical) cut from the top of the belly button down to the pelvic bone.
This lets your doctor see the tumor and other organs and tissues in your abdomen and pelvis. This method is recommended most often when surgical staging (described next) or cytoreductive surgery is planned.
Less often, a minimally invasive type of surgery called laparoscopy may be used. The surgery is performed through a few small cuts in the abdomen. Laparoscopy may be used in select cases, such as when cancer is only in the ovaries. This surgery should only be done by a gynecologic oncologist experienced in this method.
Hysterectomy and BSO The most commonly used surgery for ovarian cancer removes the uterus, both ovaries, and both fallopian tubes. 20 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Surgery Surgical staging If it does not look like the cancer has spread, surgical staging should be performed. Surgical staging is the most accurate way to stage ovarian cancer. This involves taking samples during surgery from organs and tissues where ovarian cancer often spreads. The samples are tested for cancer cells. Your surgeon will also take samples from nearby tissues where it looks like cancer hasn’t spread. This is done to check for cancer cells that have spread outside the ovaries or pelvis and can only be seen with a microscope. These are called microscopic metastases.
The omentum and sometimes nearby lymph nodes will be removed. The omentum is the fatty layer of tissue that covers organs in the belly (abdomen). Lymph nodes are groups of disease-fighting cells where cancer can also spread. If there is fluid buildup in the abdomen, the fluid will also be sampled. If there isn’t fluid buildup, your doctor may “rinse” the space inside your belly with a special liquid. This is called a peritoneal washing. Samples of the liquid will then be tested for cancer cells.
Preparing for surgery Your treatment team will give you instructions on how to prepare for surgery. You may be asked to stop taking some medicines for a short time. You also should not eat or drink after midnight the night before the surgery.
On the day of your surgery, you will be given medicine to put you into a deep sleep so you won’t feel pain. This is called general anesthesia. Surgery may take 3 or more hours to complete. More or less time may be needed depending on how much tissue is removed.
After the surgery, expect to stay in the hospital for a few days or weeks to recover. You may feel some pain and tenderness in your belly and pelvis. It may last for a few days or weeks. You may be able to return to normal activities in a few weeks. The time it takes to fully recover varies from person to person. It also varies depending on the extent of the surgery.
Premature menopause If you have not entered menopause, surgery that removes both ovaries will cause it. This is known as surgical menopause. It is caused by the sudden drop in estrogen in the body. This drop can cause symptoms of menopause, including: h Hot flashes h Sleeping problems h Night sweats h Weight gain h Changes in mood h Thinning, drying, and irritation of the vaginal lining (vaginal atrophy) When caused by surgery, the symptoms of menopause may be sudden and more severe. There are also long-term risks of not having enough estrogen. They include heart or blood vessel problems (cardiovascular disease) and bone loss (osteoporosis).
If you have symptoms of surgical menopause, your doctor may suggest non-hormonal medicine or hormone replacement therapy (HRT). Discussion with a menopausal symptom team is recommended to determine whether HRT is right for you.
21 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » If surgery first isn’t an option Other risks and side effects With any type of surgery, there are health risks and side effects. Common side effects include pain, swelling, and scars. Common side effects of ovarian cancer surgery include leg swelling, trouble urinating, and constipation. Cancer and recent abdominal surgery are risk factors for developing blood clots, also known as deep vein thrombosis (DVT). Many patients are placed on blood thinners (either oral medications or injections) for up to 4 weeks after surgery to help prevent blood clots.
If surgery first isn’t an option Having surgery first may not be an option. This could be due to the size or location of the tumor, other health conditions, or your overall health. In this case, chemotherapy is given first to shrink the cancer. You will need a biopsy to confirm that the tumor is ovarian cancer before starting chemotherapy. At this time, preferred regimens include: h Paclitaxel and carboplatin h Paclitaxel, carboplatin, bevacizumab, and maintenance bevacizumab While the above regimens are preferred, there are other recommended options for chemotherapy. Your team will take into account any medical conditions and your overall health. These regimens may change as new information becomes available. After a few cycles of chemotherapy (2 to 3 months), your doctor will check to see how well chemotherapy worked and if surgery is an option. The goal of surgery is to remove as much of the cancer as possible, as well as the ovaries, fallopian tubes, and uterus. Surgery performed after chemotherapy is called interval cytoreductive surgery (ICS).
For stage 3 disease, hyperthermic intraperitoneal chemotherapy (HIPEC) may be used during ICS. HIPEC is a technique in which chemotherapy is warmed and then circulated in the spaces between the organs of the abdomen during surgery.
If cancer improves after several cycles of chemotherapy, surgery is usually recommended. If cancer stays the same, your doctor may recommend proceeding with surgery or continuing chemotherapy to see if there is improvement.
After surgery, more chemotherapy is usually given. Maintenance therapy may follow once your cancer is in remission.
22 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Staging The information gained during surgery and surgical staging is used to determine the pathologic (post-surgery) stage. The pathologic stage provides the most accurate picture of how far the cancer has spread. It is used to guide treatment after surgery.
A staging system is a standard way of describing the extent of cancer in the body. There are 2 staging systems for ovarian cancer. One was developed by the American Joint Committee on Cancer (AJCC), the other by the International Federation of Gynecology and Obstetrics (FIGO). They are similar but the FIGO system is used most often.
In the FIGO system, the cancer stage is defined by 3 main areas of cancer growth: h The extent of the first (primary) tumor h The spread of cancer to nearby lymph nodes h The spread of cancer to distant sites Ovarian cancer stages are numbered from 1 to 4. Doctors write cancer stages as I, II, III, and IV. The stages are also divided into smaller groups, called substages. This helps to describe the extent of cancer in more detail.
The FIGO stages of ovarian cancer are described on the following pages. Cancers of the same stage tend to have similar outcomes. Early-stage cancers tend to have better outcomes than more advanced cancers. Other factors not used for staging, such as your general health, are also important.
Stage 1A Cancer is in one ovary. The outer sac (capsule) of the ovary is intact. There is no cancer on the outside surface of the ovary. No cancer cells are found in ascites or washings.
23 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 1B Cancer is in both ovaries. The capsules are intact and there is no cancer on the outside surface of the ovaries. No cancer cells are found in ascites or washings.
Stage 1C Cancer is in one or both ovaries and one or more of the following has also happened: • Stage 1C1 – The capsule of the ovary broke open during surgery. This is called surgical spill.
• Stage 1C2 – The capsule of the ovary broke open before surgery, or there is cancer on the outer surface of the ovary or fallopian tube.
• Stage 1C3 – Cancer cells are found in ascites or washings 24 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 2A There is cancer in one or both ovaries. Cancer has grown into and/or spread implants on the uterus and/or fallopian tubes.
Stage 2B Cancer is in one or both ovaries.
Cancer has grown into and/or spread implants on other organs or tissues in the pelvis, such as the bladder, colon, or rectum.
25 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 3A1 There is cancer in one or both ovaries. Cancer has spread to lymph nodes in the back of the abdomen (retroperitoneal lymph nodes).
• Stage 3A1 (i) – Cancer in the lymph nodes is 10 mm (millimeters) or smaller.
• Stage 3A1 (ii) – Cancer in the lymph nodes is larger than 10 mm.
Stage 3A2 Cancer has spread to the tissue lining the abdomen. The cancer is so small it can only be seen with a microscope. There may also be cancer in lymph nodes in the back of the abdomen.
26 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 3B There is visible cancer on the tissue lining the abdomen. The area of cancer is smaller than a peanut (about 2 centimeters or smaller). There may also be cancer in lymph nodes in the back of the abdomen.
27 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 3C There is visible cancer on the tissue lining the abdomen. The area of cancer is larger than 2 cm. There might be cancer in lymph nodes in the back of the abdomen. The cancer may have also spread to the outer surface of the liver or spleen.
28 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Staging Stage 4 Cancer has spread to other parts of the body.
• Stage 4A – There are cancer cells in the fluid around the lungs. This is called a malignant pleural effusion.
• Stage 4B – Cancer has spread to the inside of the liver or spleen, to distant lymph nodes, or to other organs outside the abdomen.
29 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Chemotherapy Chemotherapy Chemotherapy is the use of medicine(s) to kill cancer cells. It is a type of systemic therapy. When given before surgery, chemotherapy is called neoadjuvant therapy. When given after surgery, it is called primary or adjuvant chemotherapy. Platinum-based chemotherapy is recommended for ovarian cancer. These medicines contain the metal platinum. Carboplatin, cisplatin, and oxaliplatin are examples. One of these is often given with a different type of chemotherapy called a taxane. Paclitaxel and docetaxel are taxanes.
Your options for chemotherapy will depend on your age and overall health. Your provider will also consider your risk for nerve damage, called peripheral neuropathy. This common side effect of paclitaxel causes pain, tingling, and numbness, often in the hands and feet. Chemotherapy is given in cycles of treatment followed by days of rest. This allows the body to recover before the next treatment. The cycles vary in length depending on which drugs are used.
Stage 1 Chemotherapy is recommended after surgery for most newly diagnosed stage 1 cancers. Observation may be an option for a stage 1A or 1B, low-grade tumor. Ask your doctor if this applies to your cancer.
At this time, the preferred chemotherapy regimen is paclitaxel with carboplatin, given every 3 weeks. If you can't have this regimen, there are other recommended options.
Six cycles of chemotherapy are recommended for high-grade serous tumors. Between 3 and 6 cycles are recommended for all other stage 1 tumors. The specific number of cycles needed depends on the tumor type and other factors.
Stages 2, 3, and 4 For common tumor types, chemotherapy is recommended after surgery for all newly diagnosed stage 2, 3, and 4 ovarian cancers.
At this time, the preferred chemotherapy regimen is paclitaxel with carboplatin, given every 3 weeks. Six cycles are given for stage 2, 3, and 4 cancers. If you can't have this regimen, there are other recommended options. A drug called bevacizumab (Avastin) may be added to your chemotherapy. It stops the growth of new blood vessels that feed the tumor.
If chemotherapy works well, the next step may include maintenance therapy. See page 31 for more information. How chemotherapy is given Most chemotherapy for ovarian cancer is given intravenously. This means the medicine is put directly into your bloodstream through a vein. You may get a port to receive chemotherapy. This is a small, round disc that is usually placed under your skin in the upper chest. It is inserted during a minor surgery and stays in the body until treatment is complete. After treatment the port can be easily removed. Once the port is removed, the skin will heal.
30 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Chemotherapy Chemotherapy medicine can also be slowly injected into the abdomen. This is called intraperitoneal (IP) chemotherapy. When given this way, higher doses of the drugs are delivered directly to the cancer cells in the belly area. IP chemotherapy is given through a thin tube called a catheter. The catheter is usually connected to a port placed inside the abdomen during surgery.
Monitoring during chemotherapy Your doctor will monitor how well the chemotherapy is working and assess for side effects. Expect to have a physical exam every 1 to 3 cycles. A pelvic and rectovaginal exam may be done at the same time. Imaging and blood tests are ordered as needed. Testing for CA-125 or other tumor markers may be performed before each cycle of chemotherapy.
Side effects of chemotherapy Common side effects of chemotherapy include: h Loss of appetite h Nausea and vomiting h Mouth sores h Hair loss h Fatigue h Increased risk of infection h Bleeding or bruising easily h Nerve damage (neuropathy) The side effects of chemotherapy depend on the specific medicines being used, the dose, and other factors. In general, side effects are caused by the death of fast-growing cells, which are found in the bowel, mouth, and blood.
Intraperitoneal chemotherapy tends to cause more severe side effects than intravenous chemotherapy. This includes infections, kidney damage, pain in the belly, and nerve damage. Rare but serious side effects include myelodysplastic syndrome (MDS) and acute myeloid leukemia (AML). Ask your care team for a full list of common and rare side effects of the drugs you receive. Tell your care team about any medications or supplements you are taking. Some may interact with chemotherapy.
Side effects Managing side effects is a shared effort between you and your care team. It is important to speak up about bothersome side effects, such as nausea and vomiting. Ask about your options for managing or relieving the effects of treatment.
More information on nausea and vomiting is available at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app. 31 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Maintenance therapy Maintenance therapy Maintenance therapy is the use of systemic therapy after successful initial treatment for ovarian cancer. It can reduce the risk of cancer returning or extend the time until it returns or gets worse. Maintenance therapy is an option for stage 2, 3, and 4 cancers that respond well to surgery and platinum-based chemotherapy.
PARP inhibitors (PARPi) are a newer option for maintenance therapy after initial treatment. These oral targeted therapies work best in homologous recombination deficiency (HRD)- positive cancers, including those caused by a BRCA mutation.
PARP inhibitors currently used for maintenance therapy after initial treatment of ovarian cancer include: h Olaparib (Lynparza) h Niraparib (Zejula) h Rucaparib (Rubraca) The most common side effects of PARP inhibitors are similar to those caused by chemotherapy. They include fatigue, nausea, vomiting, and low blood cell counts. Rare but serious side effects include myelodysplastic syndrome (MDS) and acute myeloid leukemia (AML).
Myelodysplastic syndrome is a cancer in which the bone marrow does not make enough healthy blood cells. There are abnormal cells in the blood and/or bone marrow. Acute myeloid leukemia is a fast-growing disease in which too many immature white blood cells are found in the bone marrow and blood. In some cases MDS can become AML.
If chemotherapy included bevacizumab For HR deficient cancers and BRCA-mutated cancers, maintenance therapy with both bevacizumab and olaparib (a PARP inhibitor) is a recommended option. If you can't have olaparib, niraparib is given instead. For BRCA-mutated cancers, maintenance therapy with a PARP inhibitor alone is also an option.
Maintenance therapy with bevacizumab alone is also an option for cancers not caused by a BRCA mutation, or whose BRCA status is unknown. If chemotherapy didn’t include bevacizumab If you have a BRCA mutation and chemotherapy didn't include bevacizumab, maintenance therapy with a PARP inhibitor alone is recommended. For some stage 2 cancers with a BRCA mutation, observation may be an option.
If you don't have a BRCA mutation (or have not had a BRCA test), maintenance therapy with niraparib or rucaparib may be an option, especially if the cancer is HRD-positive. Observation is also an option if there was a complete response to chemotherapy. This means there are no signs of cancer in the body.
32 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Maintenance therapy How long does maintenance therapy last?
The length of maintenance therapy after initial treatment depends on the specific drug(s). Olaparib alone can be given for up to 2 years. Olaparib + bevacizumab together can be given for up to 15 months, with the olaparib continued for up to 2 years total. Niraparib alone can be given for up to 3 years. Rucaparib alone can be given for up to 2 years. These recommendations can change with ongoing research.
Keep in mind that any maintenance therapy will be stopped if the cancer grows or spreads. It will also be stopped if the side effects become too harsh or make it unsafe to continue.
There isn’t much research on maintenance therapy with a PARP inhibitor after initial treatment for stage 2 ovarian cancer. If your cancer is stage 2 and you are eligible for maintenance therapy, talk to your provider about your options.
33 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Surveillance Surveillance When there are no signs of cancer after treatment, expect to see your oncologist on a regular basis for physical and pelvic exams.
First 2 years: Every 2 to 4 months Next 3 years: Every 3 to 6 months After 5 years: Once a year Your provider may order blood and imaging tests if you develop symptoms or if there are other reasons to suspect relapse. If your CA-125 level (or other tumor marker) was high originally, it may be checked on a regular basis after treatment.
In addition to surveillance testing, a range of other care is important for cancer survivors. See Chapter 5: Survivorship for more information.
"I truly believe that you have to go through something life changing, to gain something life affirming." – Ovarian cancer survivor 34 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Recurrence Recurrence The return of cancer after treatment is called a recurrence, or a relapse. Symptoms can be a sign of recurrence. Tell your care team if you have any of these symptoms: h Pain or bloating in your pelvis or belly h Unexplained weight loss h Upset stomach h Constipation h Trouble eating or feeling full fast h Fatigue h Needing to urinate often or urgently The presence of specific biomarkers helps guide treatment for recurrent ovarian cancer. If testing for the following biomarkers hasn't already been done, it is recommended now: h BRCA1 and BRCA2 mutations h Homologous recombination deficiency (HRD) status h HER2 expression h Microsatellite instability (MSI) h Mismatch repair (MMR) h Tumor mutational burden (TMB) h BRAF V600E mutation h Folate receptor alpha (FRα) h RET mutations h NTRK gene fusions Your doctor may choose to test for even more biomarkers than those listed.
Everyone with persistent or recurrent ovarian cancer is encouraged to consider a clinical trial for treatment.
Platinum-resistant cancer Ovarian cancer is called platinum-resistant if: h It doesn’t improve or worsens during platinum-based chemotherapy, or h It returns less than 6 months after successful treatment with platinum-based chemotherapy.
Because platinum-based chemotherapy didn’t improve your cancer, a different type of recurrence treatment is recommended. Non-platinum chemotherapy is usually given first. Another preferred option is bevacizumab (Avastin). This may also be added to your chemotherapy.
For tumors with the folate receptor alpha (FRα) biomarker, the targeted therapy mirvetuximab soravtansine-gynx (Elahere) is preferred for recurrence treatment. It is a type of antibody drug conjugate (ADC).
Other options may include endocrine therapy, targeted therapy, or immunotherapy. These options are described more next. Enrolling in a clinical trial is encouraged if you are eligible.
Platinum-sensitive cancer If you enter complete remission after platinum-based chemotherapy and cancer returns more than 6 months later, the cancer is considered platinum-sensitive. This means that platinum-based chemotherapy drugs work well against the cancer.
35 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Recurrence Because it worked well before, platinum chemotherapy is typically recommended for recurrent platinum-sensitive disease. This is especially true for the first recurrence. The targeted therapy bevacizumab may be added to chemotherapy. In certain circumstances, before starting recurrence treatment, your doctor may suggest surgery to remove all visible cancer. This is called secondary cytoreductive surgery.
If recurrence treatment with platinum-based chemotherapy works well or very well, maintenance therapy is an option. If bevacizumab was included in your recurrence chemotherapy regimen, it can be continued alone as maintenance therapy. A PARP inhibitor may also be an option for maintenance therapy, if you haven’t already been treated with one and there is a BRCA mutation. After successful chemotherapy for recurrent cancer, maintenance therapy with a PARP inhibitor can be continued until the cancer grows or spreads, or until the side effects make it intolerable or unsafe to continue. When used after recurrence treatment, the safety of maintenance therapy with a PARP inhibitor for longer than 2 years is unknown.
Hypersensitivity reactions With repeat use of carboplatin and/or cisplatin, you are at increased risk of a hypersensitivity (allergic) reaction. This can be life-threatening. If your treatment team hasn’t brought it up, below are some questions you can ask to learn about this risk. 3 How likely is it that I will have an allergic reaction to chemotherapy? 3 How will I know if I’m having an allergic reaction? What are the symptoms? 3 Does the staff on hand know how to manage hypersensitivity reactions? 3 Will the right medical equipment be available in case I have an allergic reaction?
36 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Recurrence Biomarker-based treatment If the cancer has any of the biomarkers listed in Guide 3 below, targeted therapy or immunotherapy may be an option. For information on the side effects of immunotherapy, see the NCCN Guidelines for Patients Immunotherapy Side Effects: Immune Checkpoint Inhibitors at NCCN.org/ patientguidelines and on the NCCN Patient Guides for Cancer app.
Guide 3 Biomarker-based treatment Biomarker Available targeted therapies dMMR/MSI-H • Dostarlimab-gxly (Jemperli) • Pembrolizumab (Keytruda) BRAF V600E mutation Dabrafenib (Tafinlar) + trametinib (Mekinist) FRα (FOLR1)-expression Mirvetuximab soravtansine-gynx (Elahere) + bevacizumab TMB-H Pembrolizumab (Keytruda) RET gene fusion Selpercatinib (Retevmo) HER2 expression Fam-trastuzumab deruxtecan-nxki (Enhertu) NTRK gene fusion • Entrectinib (Rozlytrek) • Larotrectinib (Vitrakvi) • Repotrectinib (Augtyro) 37 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Recurrence Endocrine therapy Estrogen and progesterone are hormones made by the ovaries until menopause. Hormones are sometimes offered to help with symptoms of menopause, such as hot flashes. This is known as menopausal hormone therapy. It used to be called hormonal replacement therapy (HRT). This may help some ovarian cancers grow.
In some cases, treatment can be used to block these hormones from working, or to lower hormone levels. The goal is to help slow ovarian cancer growth. This is called endocrine therapy or anti-estrogen therapy. It may be used for persistent or recurrent ovarian cancer, most often for low-grade tumors.
Endocrine therapy often causes symptoms of menopause, including: h Hot flashes h Changes in mood h Vaginal dryness h Trouble sleeping h Night sweats h Vaginal discharge h Weight gain Other side effects include swelling in the hands and feet, fatigue, and less interest in sex. Blood clots are a rare but serious side effect of tamoxifen. Aromatase inhibitors can weaken your bones and may also cause joint and muscle pain.
Radiation therapy to help with symptoms Depending on the specific recurrence treatment planned, radiation therapy may also be given to help with symptoms. It can be used treat vaginal bleeding, areas of cancer in bone, and isolated areas causing pain. Radiation treatment to the pelvis can cause the vagina to become shorter and narrower (vaginal stenosis). This can make it uncomfortable or even painful to have sex, or to have vaginal exams by a doctor. Using a vaginal dilator can prevent or treat vaginal stenosis. This is a device used to gradually stretch or widen the vagina. You can start using one as soon as 2 to 4 weeks after radiation therapy has ended, and continue to use it for as long as you want.
Advance care planning Talking with your doctor about your prognosis can help with treatment planning. If the cancer cannot be controlled or cured, a care plan for the end of life can be made. Benefits of advance care planning include: • Knowing what to expect • Making the most of your time • Lowering the stress of caregivers • Having your wishes followed • Having a better quality of life Advance care planning starts with an honest talk between you and your doctors. Just having a general idea of your prognosis will help you decide at what point you may want to stop treatment, if at all.
38 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Clinical trials Clinical trials A clinical trial is a type of medical research study. After being developed and tested in a laboratory, potential new ways of fighting cancer need to be studied in people. If found to be safe and effective in a clinical trial, a drug, device, or treatment approach may be approved by the U.S. Food and Drug Administration (FDA). Everyone with cancer should carefully consider all of the treatment options available for their cancer type, including standard treatments and clinical trials. Talk to your doctor about whether a clinical trial may make sense for you.
Phases Most cancer clinical trials focus on treatment. Treatment trials are done in phases. h Phase 1 trials study the dose, safety, and side effects of an investigational drug or treatment approach. They also look for early signs that the drug or approach is helpful. h Phase 2 trials study how well the drug or approach works against a specific type of cancer. h Phase 3 trials test the drug or approach against a standard treatment. If the results are good, it may be approved by the FDA. h Phase 4 trials study the long-term safety and benefit of an FDA-approved treatment.
Who can enroll?
Every clinical trial has rules for joining, called eligibility criteria. The rules may be about age, cancer type and stage, treatment history, or general health. These requirements ensure that participants are alike in specific ways and that the trial is as safe as possible for the participants. Informed consent Clinical trials are managed by a group of experts called a research team. The research team will review the study with you in detail, including its purpose and the risks and benefits of joining. All of this information is also provided in an informed consent form. Read the form carefully and ask questions before signing it. Take time to discuss it with family, friends, or others you trust. Keep in mind that you can leave and seek treatment outside of the clinical trial at any time.
Start the conversation Don’t wait for your doctor to bring up clinical trials. Start the conversation and learn about all of your treatment options. If you find a study that you may be eligible for, ask your treatment team if you meet the requirements. If you have already started standard treatment, you may not be eligible for certain clinical trials. Try not to be discouraged if you cannot join. New clinical trials are always becoming available. Frequently asked questions There are many myths and misconceptions surrounding clinical trials. The possible benefits and risks are not well understood by many with cancer. 39 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Clinical trials Will I get a placebo?
Placebos (inactive versions of real medicines) are almost never used alone in cancer clinical trials. It is common to receive either a placebo with a standard treatment or a new drug with a standard treatment. You will be informed, verbally and in writing, if a placebo is part of a clinical trial before you enroll.
Are clinical trials free?
There is no fee to enroll in a clinical trial. The study sponsor pays for research-related costs, including the study drug. You may, however, have costs indirectly related to the trial, such as the cost of transportation or child care due to extra appointments. During the trial, you will continue to receive standard cancer care. This care is billed to—and often covered by— insurance. You are responsible for copays and any costs for this care that are not covered by your insurance.
Finding a clinical trial In the United States NCCN Cancer Centers NCCN.org/cancercenters The National Cancer Institute (NCI) cancer.gov/about-cancer/treatment/ clinical-trials/search Worldwide Ovarian Cancer Research Alliance (OCRA) ocra.careboxhealth.com The U.S. National Library of Medicine (NLM) clinicaltrials.gov Need help finding a clinical trial? NCI’s Cancer Information Service (CIS) 1.800.4.CANCER (1.800.422.6237) cancer.gov/contact 40 NCCN Guidelines for Patients® Ovarian Cancer, 2024 3 Treatment for common ovarian cancers » Key points Key points h Hysterectomy with BSO is the recommended first treatment for ovarian cancer whenever possible. Fertility-sparing surgery may be an option if the cancer hasn't spread beyond the ovary. h Ovarian cancer is staged during surgery to remove the cancer. This is called surgical staging. h For common ovarian tumor types, platinum-based chemotherapy is recommended after surgery for most stage 1 cancers and for all stage 2, 3, and 4 cancers. A targeted therapy called bevacizumab may be added. h Maintenance therapy is recommended for many stage 2, 3, and 4 cancers that respond well to initial treatment. PARP inhibitors are often an option. They work best in cancers with a BRCA mutation and/or HRD-positive cancers. h If it was included in chemotherapy, bevacizumab may be given alone or with a PARP inhibitor for maintenance therapy. h If not already performed, tumor biomarker testing is recommended for everyone with recurrent ovarian cancer. h Clinical trials give people access to investigational treatments that may, in time, be approved by the FDA.
I may not be grateful for my cancer, but I am certainly grateful for the lessons it has taught me and the wonderful people that I have met along the way” - Ovarian cancer survivor “ 41 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers 42 Carcinosarcoma 42 Clear cell carcinoma 43 Mucinous carcinoma 44 Grade 1 endometrioid carcinoma 44 Low-grade serous carcinoma 45 Ovarian serous borderline epithelial tumors 47 Malignant germ cell tumors 50 Malignant sex cord-stromal tumors 51 Key points 42 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Carcinosarcoma 42 4 Treatment for less common ovarian cancers » Carcinosarcoma » Clear cell carcinoma Less common ovarian cancers are often diagnosed during a surgery or other procedure. Treatment for these rare cancers is often individualized. When possible, receiving treatment as part of a clinical trial is strongly recommended. Less common ovarian cancers can start in the epithelium, in tissues that support the ovaries, or in the reproductive (egg) cells of the ovary. If not already done, you may have surgery first to remove any remaining cancer or to stage the cancer.
If receiving treatment as part of a clinical trial isn’t an option, treatment for less common ovarian cancers should be individualized. Whether the cancer has any biomarkers helps guide treatment decisions. Biomarkers are features of a cancer, such as gene mutations (changes).
Carcinosarcoma Carcinosarcomas are the most aggressive type of ovarian tumor. Fertility-sparing surgery isn’t a treatment option, regardless of your age or the cancer stage. These cancers are also known as malignant mixed Müllerian tumors. Treatment with platinum-based chemotherapy is recommended. A preferred regimen for all stages is paclitaxel and carboplatin, given every 3 weeks. For stage 2, 3, or 4 tumors, the targeted therapy bevacizumab may be given with chemotherapy. If so, it is often continued as maintenance therapy.
When chemotherapy is over, follow-up care will begin for stage 1 tumors. For stage 2, 3 or 4 tumors with a BRCA mutation, maintenance therapy may follow chemotherapy. For information on follow-up care and recurrence, see pages 33 and 34. Clear cell carcinoma Clear cell carcinomas are the most common of the less common ovarian cancers. They are considered high-grade (fast-growing) tumors. Most don't have estrogen receptors.
Platinum-based chemotherapy is recommended for most clear cell carcinomas. A preferred regimen for all stages is paclitaxel and carboplatin, given every 3 weeks. For stage 2, 3, or 4 tumors, the targeted therapy bevacizumab may be given with chemotherapy. If so, it is often continued as maintenance therapy.
When chemotherapy is over, follow-up care will begin for stage 1 tumors. For stage 2, 3, or 4 tumors with a BRCA mutation, maintenance therapy may follow chemotherapy. For information on follow-up care and recurrence, see pages 33 and 34. 43 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Mucinous carcinoma Mucinous carcinoma Mucinous tumors are often found at an earlier age than more common ovarian cancers. These tumors can grow so large that they fill the abdomen and pelvis. Most people have early-stage disease at the time of diagnosis. Mucinous tumors usually respond well to treatment.
Testing often involves a check of the gastrointestinal (GI) tract. This can help tell the difference between a true mucinous cancer of the ovary versus a cancer that may have spread to the ovary from the GI tract. Blood tests measuring carcinoembryonic antigen (CEA) and CA 19-9 are recommended.
If not already done, you may have surgery to remove any remaining cancer and to surgically stage the cancer. For stage 1A and 1B tumors, observation is recommended. Stage 1C tumors may be observed or treated with systemic therapy. Chemotherapy is recommended for all stage 2, 3, and 4 mucinous neoplasms. The preferred chemotherapy regimens are listed in Guide 4.
For information on follow-up care and recurrence, see pages 33 and 34.
Guide 4 Mucinous tumors – preferred options for chemotherapy Note: These regimens may change as new information becomes available.
Stage 1C, grades 1 to 3 • 5-FU, leucovorin, oxaliplatin • Capecitabine and oxaliplatin • Paclitaxel and carboplatin (every 3 weeks) Stages 2, 3, and 4 • 5-FU, leucovorin, and oxaliplatin with or without bevacizumab • Capecitabine and oxaliplatin with or without bevacizumab • Paclitaxel and carboplatin (every 3 weeks) • Paclitaxel, carboplatin, bevacizumab, and maintenance bevacizumab 44 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Grade 1 endometrioid carcinoma Grade 1 endometrioid carcinoma Testing for a biomarker (feature) called mismatch repair deficiency (dMMR)/ high microsatellite instability (MSI-H) is recommended for all grade 1 endometrioid carcinomas.
Stage 1A and 1B tumors are observed without treatment. While observation is also an option for stage 1C tumors, these cancers are often treated with chemotherapy or endocrine therapy. For all stage 2, 3, and 4 tumors, treatment with either chemotherapy or hormonal therapy is recommended.
For chemotherapy, paclitaxel and carboplatin (given every 3 weeks) is recommended. A targeted therapy called bevacizumab may be added to chemotherapy for stage 2, 3, or 4 tumors. If so, it may be continued as maintenance therapy.
If treatment with chemotherapy is planned, your doctor may suggest maintenance endocrine therapy afterward.
For information on follow-up care and recurrence, see pages 33 and 34.
Low-grade serous carcinoma Low-grade serous carcinoma isn’t the same as the more commonly diagnosed high-grade serous carcinoma. Low-grade serous carcinomas tend to be diagnosed at an earlier age. A little over half are linked with borderline serous tumors, also called low malignant potential tumors.
Observation is recommended for all stage 1A and 1B tumors. While observation is also an option for stage 1C tumors, these cancers are often treated with chemotherapy or hormonal therapy. Treatment with either chemotherapy or hormonal therapy is recommended for all stage 2, 3, and 4 tumors.
For chemotherapy, paclitaxel and carboplatin (given every 3 weeks) is recommended. A targeted therapy called bevacizumab may be added to chemotherapy for stage 2, 3, Supportive care is available for everyone with cancer. It isn't meant to treat the cancer, but rather to help with symptoms and make you more comfortable.
45 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Ovarian serous borderline epithelial tumors or 4 tumors. If so, it may be continued as maintenance therapy.
If treatment with chemotherapy is planned, your doctor may suggest maintenance endocrine therapy afterward.
After treatment When treatment is over, follow-up visits are recommended every 2 to 4 months for the first 2 years, then every 3 to 6 months for 3 years. After year 5, they are scheduled once a year.
Physical exams, including pelvic exam, are performed as needed. If biomarker testing hasn't been done yet, it is recommended now. Physical exams, imaging, and blood tests are performed as needed. If CA-125 or other tumor markers were high before treatment, they will be monitored after treatment. If the cancer returns, treatment options include: h Targeted therapy with a kinase inhibitor h Endocrine therapy h Chemotherapy (if not already received) h Biomarker-based targeted therapy (if the cancer has biomarkers) h Observation Ovarian serous borderline epithelial tumors Ovarian serous borderline epithelial tumors are also called low malignant potential (LMP) tumors. These rare tumors have cancer-like features, but don't invade other tissues like most cancers do.
Borderline epithelial tumors are slow-growing. Compared to more invasive types of ovarian cancer, those diagnosed with a borderline epithelial tumor tend to be younger and often have stage 1 disease. They are also often candidates for fertility-sparing surgery.
Surgery is the main treatment for borderline epithelial tumors. Both standard surgery and fertility-sparing surgery may be options. You should be evaluated by a gynecologic oncologist for this decision.
If cancer remains after prior surgery Prior surgery for a borderline epithelial tumor may be considered incomplete. This is the case if the cancer wasn’t fully removed, or fully staged. If your doctor suspects that cancer remains in the body, another surgery is recommended if possible. This may not be possible if you are otherwise not healthy enough, or if the cancer can’t be surgically removed.
If you want to keep your fertility, fertility-sparing surgery may be an option. This involves removing only the ovary with cancer and its ovarian tube, along with any remaining visible cancer. This is called a unilateral salpingo-oophorectomy (USO). For some, removing 46 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Ovarian serous borderline epithelial tumors both ovaries and fallopian tubes but keeping the uterus intact may be an option.
If you don’t desire fertility-sparing surgery, completion surgery (hysterectomy and removal of the opposite ovary and fallopian tube) is performed. Any remaining cancer will also be resected. Removal and testing of lymph nodes during surgery is considered on a case-by-case basis.
After surgery (fertility-sparing or completion), the pathology team will test the removed tissues. Sometimes the tumor type changes as a result of this testing. If the final results confirm that it is borderline, follow-up care is described next.
Prior complete surgery If the cancer was completely resected and no low-grade serous carcinoma was found, observation is recommended. If low-grade serous carcinoma was found, more treatment is recommended.
Follow-up Physical exams are given every 3 to 12 months for the first 5 years after treatment. After that, they are given as needed. Your doctor may also do a pelvic exam at these visits.
If CA-125 or other tumor marker levels were high at diagnosis, they should be checked at each follow-up visit. Other blood tests and imaging are performed as needed. Imaging may be ordered if your doctor suspects that cancer has returned.
If you had fertility-sparing surgery, you may have ultrasounds after treatment. This can help catch recurrence early. Talk to your doctor about completion surgery after you’ve had the baby.
Relapse The return of cancer after treatment is called a relapse or a recurrence. In the case of a relapse, debulking surgery is often recommended when possible. This surgery aims to remove all of the cancer that the surgeon can see.
The results of surgery may show that the tumor is a different type than previously thought. Treatment based on the updated tumor type is recommended. 47 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Malignant germ cell tumors Malignant germ cell tumors Although very rare, germ cell tumors are the most common type of ovarian cancer diagnosed in people ages 16 to 20 years. These non-epithelial tumors are usually diagnosed at the earliest stage and have excellent treatment outcomes. Types of germ cell tumors include: h Dysgerminomas h Immature teratomas h Embryonal tumors h Endodermal sinus (yolk sac) tumors.
The following tumor markers tend to be found in higher-than-normal amounts in those with a malignant germ cell tumor: h Alpha-fetoprotein (AFP) h Beta-human chorionic gonadotropin (β-hCG) h Lactate dehydrogenase (LDH) Treatment If you want the option of having children after treatment, fertility-sparing surgery is recommended. The cancer can be any stage. Full surgical staging is performed at the same time. Surveillance after fertility-sparing surgery involves having ultrasounds on a regular basis. Talk to your doctor about completion surgery after you’ve had the baby.
For those who don’t desire fertility preservation, completion surgery with full surgical staging is recommended. Malignant germ cell tumors are staged with the same system used for common ovarian cancers. In children or adolescents with early-stage germ cell tumors, full surgical staging may be skipped.
After surgery, chemotherapy is recommended for most malignant germ cell tumors. This includes: h Any stage embryonal tumor h Any stage endodermal sinus tumor (yolk sac tumor) h Stage 2, 3, or 4 dysgerminoma h Stage 1, grade 2 or 3 immature teratoma h Stage 2, 3, or 4 immature teratoma h Any stage nongestational choriocarcinoma Some germ cell tumors don't need chemotherapy after surgery. Observation with surveillance is recommended for: h Stage 1 dysgerminomas, and h Stage 1, grade 1 immature teratomas. For tumors that need chemotherapy, 3 to 4 cycles of the BEP regimen (bleomycin, etoposide, and cisplatin) is preferred. Bleomycin can damage the lungs. Expect to have tests to check how well your lungs work before chemotherapy starts. In some cases bleomycin can't be used. Your team will speak with you about other options.
After chemotherapy, imaging will be ordered to see how the cancer responded. If you have a complete response to chemotherapy, expect to 48 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Malignant germ cell tumors have follow-up checks every 2 to 4 months for 2 years. If the levels of AFP and beta-HCG were high originally, these tumor markers will also be checked with blood tests. See Guide 5.
Residual or recurrent disease Sometimes the tumor doesn’t go away completely with treatment. This is called residual disease. Or the tumor may return after treatment. This is called recurrent disease. If the tumor can still be seen on imaging tests after surgery and chemotherapy and tumor Guide 5 Surveillance for malignant germ cell tumors Year 1 Year 2 Year 3 Years 4 to 5 After 5 years Dysgerminoma Physical exam and tumor marker blood tests Every 2 to 3 months Every 3 to 4 months Every 6 months Every 6 months Every year CT of abdomen and pelvis Every 3 to 4 months Every 6 months Every year Every year As needed Non-dysgerminoma Physical exam and tumor marker blood tests Every 2 months Every 2 months Every 4 to 6 months Every 6 months Every year Imaging CT of chest, abdomen, and pelvis every 3 to 4 months CT of chest, abdomen, and pelvis every 4 to 6 months CT of abdomen and pelvis every 4 to 6 months CT of abdomen and pelvis every 6 to 12 months As needed 49 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Malignant germ cell tumors marker levels are normal, your doctor may suggest surgery to remove the remaining tumor tissue. Observation with imaging is also an option.
If surgery is planned, next steps depend on the results of surgery. If all of the cancer could not be removed, your doctor may recommend 2 more cycles of platinum-based chemotherapy.
For those with confirmed cancer (either residual or recurrent) after first-line chemotherapy and abnormal tumor markers (AFP and/or β-hCG), options to try to cure the cancer include: h TIP chemotherapy (paclitaxel + ifosfamide + cisplatin) h High-dose chemotherapy with hematopoietic cell transplant (HCT) For some people, a hematopoietic cell transplant will cure the cancer. If your doctor thinks a cure is possible, you should be referred to a specialized care center for a consultation about high-dose chemotherapy and stem cell rescue. The specific high-dose chemotherapy regimens used vary between cancer centers.
If treatment with TIP or high-dose chemotherapy isn’t possible or desired, palliative chemotherapy is an option. The goal of care is to make you more comfortable and improve your quality of life. There are many options for palliative chemotherapy. Talk to your doctor about which may be right for you.
For cancers with the following biomarkers, immunotherapy with a checkpoint inhibitor may also be an option. h Microsatellite instability-high (MSI-H) h Mismatch repair deficient (dMMR) h Tumor mutational burden-high (TMB-H) Radiation therapy targeting the tumor area can help relieve symptoms caused by the cancer. Also keep in mind that receiving only supportive care without other treatment is always an option.
50 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Malignant sex cord-stromal tumors Malignant sex cord-stromal tumors Malignant sex cord-stromal tumors are nonepithelial. These rare tumors include granulosa cell tumors (most common) and Sertoli-Leydig cell tumors. The prognosis (outlook) for both types is good. Most people diagnosed with a malignant granulosa cell tumor have early-stage disease, and the cancer is typically slow-growing. Surgery to stage the cancer is recommended for malignant sex cord-stromal tumors. Lymph node dissection (removal) is generally not included in surgical staging for these tumors.
If fertility is desired and the cancer hasn't spread beyond the ovary, fertility-sparing surgery with full staging is an option instead. If this is planned, talk to your doctor about having completion surgery after childbearing is finished. Completion surgery removes the uterus and the remaining ovary and fallopian tube. Next steps of care depend on the cancer stage, as determined by surgery. Malignant sex cord stromal tumors are staged with the same system used for common ovarian cancers.
Stage 1 Observation is recommended after surgery for low-risk stage 1 tumors. Medium- or high-risk stage 1 tumors may be observed or treated with platinum-based chemotherapy. The preferred chemotherapy regimen at this time is paclitaxel and carboplatin.
Stage 2, 3, or 4 For those with a stage 2, 3, or 4 tumor, treatment options include platinum-based chemotherapy and radiation therapy. Radiation is an option only if there is a limited amount of cancer in the body. Otherwise, chemotherapy is usually given. Surveillance Granulosa cell tumors can return decades after treatment. Long-term surveillance is recommended after treatment for these tumors. Physical exams are given as needed based on the cancer stage. Exams are often given once or twice a year for early-stage and low-risk cancers. For high-risk disease, exams are given more often (about every 4 to 6 months).
Sex cord-stromal tumors, especially granulosa cell tumors, can make a protein called inhibin. If the level of inhibin in your blood was high at the time of diagnosis, your doctor may continue to check it after treatment. If the level goes up, it could be a sign of relapse. Keep in mind that a blood test alone cannot confirm that the cancer has returned.
Testing for CA-125 and other tumor markers is individualized. If your doctor recommends it, how often the testing is needed is also based on stage. Blood tests may be ordered once or twice a year if the cancer is early-stage and low-risk. For high-risk disease, testing may be ordered every 4 to 6 months.
Imaging isn’t needed on a regular basis after treatment. It may be ordered if you develop cancer symptoms, if tumor marker levels are 51 NCCN Guidelines for Patients® Ovarian Cancer, 2024 4 Treatment for less common ovarian cancers » Key points high, or if there are concerning physical exam findings.
Relapse A relapse (also called recurrence) is the return of cancer after being cancer-free. For those with a stage 2, 3, or 4 tumor who have a relapse, options include: h Enrolling in a clinical trial h Chemotherapy h Hormone therapy Chemotherapy is most often used. The regimen preferred at this time is paclitaxel and carboplatin. There are other recommended regimens that your provider may suggest.
If endocrine therapy is planned instead of chemotherapy, options include aromatase inhibitors (anastrazole, exemestane, letrozole), leuprolide or goserelin acetate (for granulosa cell tumors), and tamoxifen.
Your doctor may suggest another surgery to remove as much of the cancer as possible. Radiation therapy targeting the tumor area can help relieve symptoms caused by the cancer. Also keep in mind that receiving only supportive care without other treatment is always an option.
Key points h Less common ovarian cancers (LCOCs) are often diagnosed during a surgery or other procedure. h Receiving treatment as part of a clinical trial is strongly recommended, if there is an open trial you are eligible for. h If a clinical trial isn’t an option, treatment for these rare cancers is individualized and often involves chemotherapy. h If not already done, you may have surgery first to remove any remaining cancer and to stage the cancer.
52 NCCN Guidelines for Patients® Ovarian Cancer, 2024 5 Survivorship 53 Your primary care provider 53 Paying for care 54 Healthy habits 55 More information 55 Key points 53 NCCN Guidelines for Patients® Ovarian Cancer, 2024 5 Survivorship » Your primary care provider 53 5 Survivorship » Your primary care provider » Paying for care Survivorship begins on the day you learn you have ovarian cancer. It focuses on the physical, emotional, and financial issues unique to cancer survivors.
Your primary care provider After finishing cancer treatment, your primary care provider (PCP) and oncologist will work together to make sure you get recommended follow-up care. Ask your oncologist for a written survivorship care plan that includes: h A summary of your cancer treatment history h A description of possible short-term, late, and long-term side effects h A schedule of follow-up cancer tests h Information on when your care will be transferred to your PCP h Recommendations for general health and well-being For many survivors, the end of treatment signals a time of celebration but also of anxiety. This is normal. You may need support to address issues that arise from not having regular visits with your cancer care team. Paying for care Cancer survivors face a unique financial burden. Paying for doctor visits, tests, and treatments can become unmanageable, especially for those with little or no health insurance. You may also have costs not directly related to treatment, such as travel expenses and the cost of childcare or missed work. The term financial toxicity is used to describe the problems patients face related to the cost of medical care.
Financial toxicity can affect your quality of life and access to needed health care. If you need help paying for your cancer care, financial assistance may be available. Talk with a patient navigator, your treatment team’s social worker, and your hospital’s financial services department. 54 NCCN Guidelines for Patients® Ovarian Cancer, 2024 5 Survivorship » Healthy habits Healthy habits Steps you can take to help prevent other health issues and to improve your quality of life are described next.
Cancer screening Get screened for other types of cancer, such as breast, colorectal, lung, and skin cancer. Your primary care doctor should tell you what cancer screening tests you should have based on your age and risk level.
Other health care Get other recommended health care for your age, such as blood pressure screening, hepatitis C screening, and immunizations (such as the flu shot).
Diet and exercise Leading a healthy lifestyle includes maintaining a healthy body weight. Try to exercise at a moderate intensity for at least 150 minutes per week. All patients should have a discussion with their doctor before starting a new exercise regimen. Eat a healthy diet with lots of plant-based foods, including vegetables, fruits, and whole grains.
Alcohol may increase the risk of certain cancers. Drink little to no alcohol. Quit smoking If you use tobacco products, quit! Counseling and other resources are available. Your treatment team can help.
Complementary and alternative therapies Complementary and alternative therapies may help with side effects and improve comfort and well-being during and after cancer treatment. Some of these practices and products include: 3 Acupuncture 3 Dietary supplements 3 Eastern medicine 3 Medical marijuana 3 Herbal teas and preparations 3 Homeopathy 3 Hypnosis 3 Meditation 3 Reiki 3 Yoga 3 Massage therapy If you have questions or are curious about complementary therapies, talk to your treatment team. Many cancer centers have integrative oncology programs. This approach to cancer care combines conventional (standard) cancer treatment with complementary and alternative therapies. Be sure to tell your care team if you are taking any herbal supplements. Some can interact with chemotherapy.
55 NCCN Guidelines for Patients® Ovarian Cancer, 2024 5 Survivorship » More information 55 5 Survivorship » More information » Key points More information For more information on cancer survivorship, the following are available at NCCN.org/ patientguidelines: h Survivorship Care for Healthy Living h Survivorship Care for Cancer-Related Late and Long-Term Effects These resources address topics relevant to ovarian cancer survivors, including: h Anxiety, depression, and distress h Fatigue h Pain h Sexual health h Sleep problems h Healthy lifestyles h Immunizations h Working, insurance, and disability concerns Key points h Survivorship focuses on the physical, emotional, and financial issues unique to cancer survivors. h Survivorship care is improved if your oncologist and primary care provider (PCP) work together to get the long-term care you need. h A survivorship care plan is helpful in transitioning your care to your primary care doctor. h Healthy habits play a key role in helping to prevent other diseases and second cancers. h If you have concerns about paying for your cancer care, financial help may be available.
56 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions 57 It’s your choice 57 Questions to ask 62 Resources 57 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » It’s your choice 57 6 Making treatment decisions » It’s your choice » Questions to ask It’s important to be comfortable with the cancer treatment you choose. This choice starts with having an open and honest conversation with your care team.
It’s your choice In shared decision-making, you and your care team share information, discuss the options, and agree on a treatment plan. It starts with an open and honest conversation between you and your care team.
Treatment decisions are very personal. What is important to you may not be important to someone else. Some things that may play a role in your decision-making: h What you want and how that might differ from what others want h Your religious and spiritual beliefs h Your feelings about certain treatments h Your feelings about pain or side effects h Cost of treatment, travel to treatment centers, and time away from school or work h Quality of life and length of life h How active you are and the activities that are important to you Think about what you want from treatment. Discuss openly the risks and benefits of specific treatments and procedures. Weigh options and share concerns with your care team. If you take the time to build a relationship with your care team, it will help you feel supported when considering options and making treatment decisions.
Second opinion It is normal to want to start treatment as soon as possible. While cancer can’t be ignored, there is time to have another doctor review your test results and suggest a treatment plan. This is called getting a second opinion, and it’s a normal part of cancer care. Even doctors get second opinions! Things you can do to prepare: h Check with your insurance company about its rules on second opinions. There may be out-of-pocket costs to see doctors who are not part of your insurance plan. h Make plans to have copies of all your records sent to the doctor you will see for your second opinion.
Support groups Many people diagnosed with cancer find support groups to be helpful. Support groups often include people at different stages of treatment. Some people may be newly diagnosed, while others may be finished with treatment. If your hospital or community doesn’t have support groups for people with cancer, check out the websites listed in this book.
Questions to ask Possible questions to ask your care team are listed on the following pages. Feel free to use these questions or come up with your own. 58 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » Questions to ask Questions about treatment options 1. Am I able to have surgery first? Why or why not?
2. Is a clinical trial an option for me? (also see next page) 3. How do my age, general health, and other factors affect my treatment options?
4. Which type of surgery do you recommend for me? How soon do I need it?
5. What if I am pregnant, or planning to get pregnant in the future?
6. Will I need chemotherapy after surgery? For how long?
7. Does my cancer have any biomarkers? How does this affect my treatment?
8. Are you suggesting options other than what NCCN recommends? If so, why?
9. Can you refer to me specialists in nutrition, sexual health, and/or mental health?
59 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » Questions to ask Questions about clinical trials 1. Do you recommend that I consider a clinical trial for treatment?
2. How do I find clinical trials that I can participate in?
3. What is the treatment used in the clinical trial? Has it been used for other types of cancer?
4. What are the risks and benefits of this treatment?
5. What side effects should I expect and how will they be managed?
6. How long will I be in the clinical trial?
7. Will I be able to get other treatment if this doesn’t work?
8. How will you know if the treatment is working?
9. Will the clinical trial cost me anything?
10. What additional procedures or tests are required?
60 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » Questions to ask Questions about survivorship and late effects 1. How likely is it that I will be cancer free after treatment?
2. What late effects are caused by this treatment? How will these be screened?
3. What are the chances the cancer will return, or that I will get another type of cancer?
4. What follow-up tests will I need, and who is responsible for scheduling them? 5. Who do I see for follow-up care? How often? For how many years?
6. What should I do if I have trouble paying for follow-up care?
7. I am looking for a survivor support group. What supportive services or other resources can you recommend?
8. If I move after treatment, will you help me find a new doctor?
61 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » Questions to ask Questions about resources and support 1. Who can I talk to about help with housing, food, and other basic needs?
2. What help is available for transportation, childcare, and home care?
3. How much will treatment cost? How much will my insurance company cover?
4. What help is available to pay for medicines and treatment?
5. What other services are available to me and my caregivers?
6. How can I connect with others and build a support system?
7. Who can help me with my concerns about missing work or school?
8. Who can I talk to if I don’t feel safe at home, at work, or in my neighborhood?
9. How can I get help to stop smoking or vaping?
62 NCCN Guidelines for Patients® Ovarian Cancer, 2024 6 Making treatment decisions » Resources Resources Bone Marrow & Cancer Foundation bonemarrow.org CancerCare cancercare.org/ Cancer Hope Network Cancerhopenetwork.org FORCE: Facing Our Risk of Cancer Empowered facingourrisk.org Imerman Angels Imermanangels.org National Coalition for Cancer Survivorship www.canceradvocacy.org National Ovarian Cancer Coalition (NOCC) Ovarian.org Ovarcome ovarcome.org Ovarian Cancer Research Alliance (OCRA) ocrahope.org Sharsheret sharsheret.org Triage Cancer Triagecancer.org Unite for HER uniteforher.org U.S. National Library of Medicine Clinical Trials Database clinicaltrials.gov/ 63 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Ü Ü 64 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Words to know Words to know abdomen The belly area between the chest and pelvis.
adjuvant chemotherapy Chemotherapy given after surgery.
ascites Abnormal fluid buildup in the belly (abdomen) or pelvis.
bilateral salpingo-oophorectomy (BSO) Surgery to remove both ovaries and both fallopian tubes.
biomarker Features of a cancer or tumor that can help guide treatment. Biomarkers in some ovarian cancers include somatic BRCA mutations, homologous recombination deficiency, MSI, MMR, HER2 expression, TMB, BRAF V600E mutation, FRα expression, RET mutations, and NTRK gene fusion.
biopsy Removal of small amounts of tissue from the body to be tested for disease.
BRCA1 or BRCA2 genes Genes involved in DNA repair. Abnormal changes (mutations) in either of these genes increases the risk of developing breast and ovarian cancer.
cancer antigen-125 (CA-125) A substance that may be found in high amounts in the blood of patients with ovarian cancer. cancer grade A rating of how abnormal the cancer cells look under a microscope. High-grade cancers grow and spread more quickly than low-grade cancers. cancer stage A rating of the growth and spread of cancer in the body, as determined by surgery.
capsule The thin layer of tissue that surrounds the ovaries.
cervix The lower part of the uterus that connects to the vagina.
chemotherapy Drugs that kill fast-growing cells throughout the body, including normal cells and cancer cells.
clear cell carcinoma of the ovary A rare type of epithelial ovarian cancer, in which the insides of the cells look clear when viewed under a microscope. A less common ovarian cancer (LCOC).
clinical stage The pre-treatment stage of a cancer. The clinical stage provides a best guess of how far the cancer has spread.
clinical trial Research on an investigational test or treatment to assess its safety or how well it works.
cytoreductive surgery Surgery to remove as much cancer as possible. Also called debulking surgery.
debulking surgery Surgery to remove as much cancer as possible. Also called cytoreductive surgery.
65 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Words to know endocrine therapy Treatment that adds, blocks, or removes hormones. The goal is to help slow ovarian cancer growth. It may be used for persistent or recurrent ovarian cancer, most often for low-grade tumors. Also called anti-estrogen therapy.
endometrioid carcinoma of the ovary A type of epithelial ovarian cancer. Grade 2 and 3 endometrioid tumors are common. Grade 1 endometrioid tumors are less common ovarian cancers (LCOCs).
epithelial ovarian cancer Cancer that starts in the cells that form the outer layer of tissue around the ovaries.
fallopian tube A thin tube through which an egg travels from the ovary to the uterus.
fertility-sparing surgery Surgery that removes one ovary and the attached fallopian tube.
genetic counseling A discussion with a health expert about the risk for a disease caused by changes in genes.
genetic testing Testing of the blood or saliva for germline (inherited) mutations that cause ovarian cancer. Recommended for everyone diagnosed with ovarian cancer.
germline mutation A gene change that is passed from a parent to their biological child(ren).
gynecologic oncologist A surgeon who is an expert in cancers that start in the female reproductive organs. They can also give chemotherapy. A gynecologic oncologist should perform ovarian cancer surgery. hereditary ovarian cancer Ovarian cancer caused by gene mutations passed down from parent to child.
homologous recombination deficiency (HRD) A feature of some ovarian cancers that may help guide treatment. BRCA mutations are 1 form of HRD. You can also be HRD positive without a BRCA mutation.
hot flashes A health condition of intense body heat and sweat for short periods.
hyperthermic intraperitoneal chemotherapy (HIPEC) A cancer treatment that involves filling the abdominal cavity with warmed chemotherapy drugs.
hysterectomy Surgery to remove the uterus.
implant Cancer cells that broke away from the first tumor and formed new tumors on the surface of nearby organs and tissues.
intraperitoneal (IP) chemotherapy Chemotherapy drugs given directly into the belly (abdomen) through a small tube.
laparotomy Surgery with a long, up-and-down cut through the wall of the belly (abdomen).
less common ovarian cancers (LCOC) Rare types of ovarian cancer, some of which are epithelial cancers. Includes carcinosarcoma, clear cell carcinoma, mucinous neoplasms, grade 1 endometrioid, low-grade serous, borderline epithelial, malignant sex-cord stromal, and malignant germ cell tumors. Also called less common ovarian histologies (LCOHs).
66 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Words to know low malignant potential (LMP) tumor A tumor formed by abnormal cells that start in the epithelial cells of the ovary. This tumor type is slow growing and does not invade other tissue. A less common ovarian cancer (LCOC). Also called a borderline epithelial tumor.
lymph A clear fluid containing white blood cells that fight infection and disease.
lymph nodes Small groups of special disease-fighting cells located throughout the body.
Lynch syndrome Abnormal changes within genes that increase the chances of developing colon, rectal, endometrial, ovarian, and other cancers. Also called hereditary non-polyposis colorectal cancer (HNPCC).
maintenance therapy Treatment given to continue (maintain) good results of prior treatment.
medical oncologist A doctor who is an expert in treating cancer with drugs such as chemotherapy.
menopause The point in time when menstrual periods end.
metastasis The spread of cancer cells from the first tumor to another body part.
microscopic metastases Cancer cells that have spread from the first tumor to another body part and are too small to be seen with the naked eye.
mucinous carcinoma of the ovary One of 4 types of epithelial cancer. A less common ovarian cancer (LCOC).
neoadjuvant chemotherapy Chemotherapy given before surgery.
neuropathy A nerve problem that causes pain, tingling, and numbness in the hands and feet.
omentum The layer of fatty tissue that covers organs in the belly (abdomen).
ovary One of a pair of organs that make hormones and eggs for reproduction.
pathologic stage Pathologic stage or surgical stage is determined by examining tissue removed during surgery.
pathologist A doctor who is an expert in testing cells and tissue to find disease.
peritoneum The layer of tissue that lines the inside of the belly (abdomen) and pelvis and covers most organs in this space.
platinum-based chemotherapy Treatment with two or more chemotherapy drugs and the main drug is made with platinum. Such drugs include cisplatin and carboplatin.
platinum-resistant When cancer drugs made with platinum, such as cisplatin and carboplatin, do not work well against the cancer.
platinum-sensitive When cancer drugs made with platinum, such as cisplatin and carboplatin, work well against the cancer.
67 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Words to know PARP inhibitor A type of oral targeted therapy used for maintenance therapy in some ovarian cancers. relapse The return of cancer after treatment. Also called a recurrence.
serous A type of epithelial ovarian cancer. Grade 2 and 3 (high-grade) serous tumors are the most common ovarian cancers. Grade 1 (low-grade) serous tumors are less common ovarian cancers (LCOCs).
somatic mutation A non-hereditary change in DNA. Also called tumor mutation.
supportive care Treatment given to relieve the symptoms of a disease. Also called palliative care.
surgical menopause The onset of menopause caused by surgery. Results from a sudden drop in estrogen in the body.
surgical stage The extent of the cancer, as determined by surgery.
surgical stage The extent of the cancer, as determined by examining the tissues removed during surgery. targeted therapy Treatment with drugs that target a specific or unique feature of cancer cells.
taxane A type of chemotherapy drug. Often given with a platinum chemotherapy drug to treat ovarian cancer.
tumor marker A substance found in body tissue or fluid that may be a sign of cancer.
unilateral salpingo-oophorectomy (USO) Surgery that removes one ovary and the attached fallopian tube. Take our survey and help make the NCCN Guidelines for Patients better for everyone!
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68 NCCN Guidelines for Patients® Ovarian Cancer, 2024 NCCN Contributors The NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Ovarian Cancer/Fallopian Tube Cancer/ Primary Peritoneal Cancer, Version 3.2024 were developed by the following NCCN Panel Members: Deborah K. Armstrong, MD/Chair The Sidney Kimmel Comprehensive Cancer Center at Johns Hopkins Ronald D. Alvarez, MD, MBA/Vice Chair Vanderbilt-Ingram Cancer Center Floor J. Backes, MD The Ohio State University Comprehensive Cancer Center - James Cancer Hospital and Solove Research Institute Lisa Barroilhet, MD University of Wisconsin Carbone Cancer Center Kian Behbakht, MD University of Colorado Cancer Center Andrew Berchuck, MD Duke Cancer Institute Lee-may Chen, MD UCSF Helen Diller Family Comprehensive Cancer Center Joshua Cohen, MD City of Hope National Medical Center Marie DeRosa, RN Patient Advocate Eric L. Eisenhauer, MD Mass General Cancer Center David M. Gershenson, MD The University of Texas MD Anderson Cancer Center Heidi J. Gray, MD Fred Hutchinson Cancer Center Rachel Grisham, MD Memorial Sloan Kettering Cancer Center Ardeshir Hakam, MD Moffitt Cancer Center Angela Jain, MD Fox Chase Cancer Center Gottfried E. Konecny, MD UCLA Jonsson Comprehensive Cancer Center Charles A. Leath III, MD, MSPH O'Neal Comprehensive Cancer Center at UAB Gary Leiserowitz, MD UC Davis Comprehensive Cancer Center Babak Litkouhi, MD Stanford Cancer Institute Joyce Liu, MD, MPH Dana-Farber/Brigham and Women’s Cancer Center Lainie Martin, MD Abramson Cancer Center at the University of Pennsylvania Daniela Matei, MD Robert H. Lurie Comprehensive Cancer Center of Northwestern University Michael McHale, MD UC San Diego Moores Cancer Center David S. Miller, MD UT Southwestern Simmons Comprehensive Cancer Center John Moroney, MD The UChicago Medicine Comprehensive Cancer Center Sanja Percac-Lima, MD, PhD Mass General Cancer Center Elena Ratner, MD, MDA Yale Cancer Center/Smilow Cancer Hospital Sharon Robertson, MD, MPH Indiana University Melvin and Bren Simon Comprehensive Cancer Center Kerry Rodabaugh, MD Fred & Pamela Buffett Cancer Center John Schorge, MD St. Jude Children's Research Hospital/ The University of Tennessee Health Science Center Premal H. Thaker, MD Siteman Cancer Center at Barnes- Jewish Hospital and Washington University School of Medicine Shitanshu Uppal, MD University of Michigan Rogel Cancer Center Roberto Vargas, MD Case Comprehensive Cancer Center/ University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute Andrea Wahner Hendrickson, MD Mayo Clinic Comprehensive Cancer Center Theresa L. Werner, MD Huntsman Cancer Institute at the University of Utah Emese Zsiros, MD, PhD Roswell Park Comprehensive Cancer Center NCCN Frankie Jones Guidelines Layout Specialist Emily Kovach Guidelines Layout Specialist Swathi Ramakrishnan, PhD Oncology Scientist/Medical Writer NCCN Contributors This patient guide is based on the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Ovarian Cancer/Fallopian Tube Cancer/Primary Peritoneal Cancer, Version 3.2024. It was adapted, reviewed, and published with help from the following people: Dorothy A. Shead, MS Senior Director Patient Information Operations Erin Vidic, MA Senior Medical Writer, Patient Information Laura Phillips Graphic Artist Reviewed this patient guide. For disclosures, visit NCCN.org/disclosures.
69 NCCN Guidelines for Patients® Ovarian Cancer, 2024 NCCN Cancer Centers NCCN Cancer Centers Abramson Cancer Center at the University of Pennsylvania Philadelphia, Pennsylvania 800.789.7366 • pennmedicine.org/cancer Case Comprehensive Cancer Center/ University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute Cleveland, Ohio UH Seidman Cancer Center 800.641.2422 • uhhospitals.org/services/cancer-services CC Taussig Cancer Institute 866.223.8100 • my.clevelandclinic.org/departments/cancer Case CCC 216.844.8797 • case.edu/cancer City of Hope National Medical Center Duarte, California 800.826.4673 • cityofhope.org Dana-Farber/Brigham and Women’s Cancer Center | Mass General Cancer Center Boston, Massachusetts 877.442.3324 • youhaveus.org 617.726.5130 • massgeneral.org/cancer-center Duke Cancer Institute Durham, North Carolina 888.275.3853 • dukecancerinstitute.org Fox Chase Cancer Center Philadelphia, Pennsylvania 888.369.2427 • foxchase.org Fred & Pamela Buffett Cancer Center Omaha, Nebraska 402.559.5600 • unmc.edu/cancercenter Fred Hutchinson Cancer Center Seattle, Washington 206.667.5000 • fredhutch.org Huntsman Cancer Institute at the University of Utah Salt Lake City, Utah 800.824.2073 • healthcare.utah.edu/huntsmancancerinstitute Indiana University Melvin and Bren Simon Comprehensive Cancer Center Indianapolis, Indiana 888.600.4822 • www.cancer.iu.edu Mayo Clinic Comprehensive Cancer Center Phoenix/Scottsdale, Arizona Jacksonville, Florida Rochester, Minnesota 480.301.8000 • Arizona 904.953.0853 • Florida 507.538.3270 • Minnesota mayoclinic.org/cancercenter Memorial Sloan Kettering Cancer Center New York, New York 800.525.2225 • mskcc.org Moffitt Cancer Center Tampa, Florida 888.663.3488 • moffitt.org O’Neal Comprehensive Cancer Center at UAB Birmingham, Alabama 800.822.0933 • uab.edu/onealcancercenter Robert H. Lurie Comprehensive Cancer Center of Northwestern University Chicago, Illinois 866.587.4322 • cancer.northwestern.edu Roswell Park Comprehensive Cancer Center Buffalo, New York 877.275.7724 • roswellpark.org Siteman Cancer Center at Barnes-Jewish Hospital and Washington University School of Medicine St. Louis, Missouri 800.600.3606 • siteman.wustl.edu St. Jude Children’s Research Hospital/ The University of Tennessee Health Science Center Memphis, Tennessee 866.278.5833 • stjude.org 901.448.5500 • uthsc.edu Stanford Cancer Institute Stanford, California 877.668.7535 • cancer.stanford.edu The Ohio State University Comprehensive Cancer Center - James Cancer Hospital and Solove Research Institute Columbus, Ohio 800.293.5066 • cancer.osu.edu The Sidney Kimmel Comprehensive Cancer Center at Johns Hopkins Baltimore, Maryland 410.955.8964 www.hopkinskimmelcancercenter.org The UChicago Medicine Comprehensive Cancer Center Chicago, Illinois 773.702.1000 • uchicagomedicine.org/cancer The University of Texas MD Anderson Cancer Center Houston, Texas 844.269.5922 • mdanderson.org UC Davis Comprehensive Cancer Center Sacramento, California 916.734.5959 • 800.770.9261 health.ucdavis.edu/cancer UC San Diego Moores Cancer Center La Jolla, California 858.822.6100 • cancer.ucsd.edu UCLA Jonsson Comprehensive Cancer Center Los Angeles, California 310.825.5268 • uclahealth.org/cancer 70 NCCN Guidelines for Patients® Ovarian Cancer, 2024 NCCN Cancer Centers UCSF Helen Diller Family Comprehensive Cancer Center San Francisco, California 800.689.8273 • cancer.ucsf.edu University of Colorado Cancer Center Aurora, Colorado 720.848.0300 • coloradocancercenter.org University of Michigan Rogel Cancer Center Ann Arbor, Michigan 800.865.1125 • rogelcancercenter.org University of Wisconsin Carbone Cancer Center Madison, Wisconsin 608.265.1700 • uwhealth.org/cancer UT Southwestern Simmons Comprehensive Cancer Center Dallas, Texas 214.648.3111 • utsouthwestern.edu/simmons Vanderbilt-Ingram Cancer Center Nashville, Tennessee 877.936.8422 • vicc.org Yale Cancer Center/Smilow Cancer Hospital New Haven, Connecticut 855.4.SMILOW • yalecancercenter.org 71 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Notes Notes 72 NCCN Guidelines for Patients® Ovarian Cancer, 2024 Index Index biomarker testing 15–16, 34, 36, 42, 45, 49 BRAF V600E 16, 34, 36 BRCA genes 13, 15, 31, 34, 42 CA-125 15, 30, 33, 45–46, 50 carcinoembryonic antigen (CEA) 15, 43 clinical trial 7, 38–39, 42, 51 endocrine therapy 34, 36–37 fertility-sparing surgery 18, 45–47, 50 folate receptor alpha (FRα) 16, 34 genetic counseling 13 HER2 16, 34 homologous recombination deficiency (HRD) 15, 31, 34 HIPEC 21 immunotherapy 36, 49 maintenance therapy 31–32, 35, 44 microsatellite instability (MSI) 16, 34, 36, 44, 49 mismatch repair (MMR) 16, 34, 36, 44, 49 NTRK gene fusion 16, 34, 36 PARP inhibitor 31–32, 35 RET mutations 16, 34 supportive care 44, 49, 51 surgical menopause 20–21 tumor mutational burden (TMB) 16, 34, 36, 49 Ü PAT-N-1793-0924 3025 Chemical Road, Suite 100 Plymouth Meeting, PA 19462 215.690.0300 NCCN.org/patients – For Patients | NCCN.org – For Clinicians Ovarian Cancer 2024 NCCN GUIDELINES FOR PATIENTS ® To support the NCCN Guidelines for Patients, visit NCCNFoundation.org/Donate |
6713 | https://math.stackexchange.com/questions/1729209/changing-a-finite-number-of-terms-in-a-sequence-does-not-affect-the-limit | real analysis - Changing a finite number of terms in a sequence does not affect the limit - Mathematics Stack Exchange
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Changing a finite number of terms in a sequence does not affect the limit
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In Introduction to Infinite Series by Bonar and Khoury, the following are given as "facts" but left up to the reader to prove. Can you please let me know if I made any errors in my attempts and if the proof is completely wrong give the correct proof? Also you are more than welcomed to give a shorter proof.
Theorem:
Changing a finite number of terms in a sequence has no effect on the convergence, divergence or the limit if it exists.
For example, the sequences
1,1 2,1 3,1 4,1 5,1 6,1 7⋯,1 n,⋯
1,1 2,1 3,1 4,1 5,1 6,1 7⋯,1 n,⋯
and
2,7,5,1 10,1 5,1 6,1 7,⋯,1 n,⋯
2,7,5,1 10,1 5,1 6,1 7,⋯,1 n,⋯
both converge 0 0.
Proof. By definition, a sequence converges to a real number A A if , for each ϵ>0 ϵ>0, there exists an integer N N such that for all n>N n>N, |a n−A|<ϵ|a n−A|<ϵ. As we can see from the definition changing any terms |a N||a N| for all N<n N<n does not affect the limit and therefore convergence because |a N| is independent on the choice of ϵ.
And by definition, a sequence diverges to ∞ if, for any M>0, there exists an integer N such that all n>N , a n>M. Similarly, we can change any a N<M, and it does not affect the choice of ϵ for a n.A similar argument can be used for the case of divergence to −∞.◼
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edited Mar 19, 2023 at 17:08
Arctic Char
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Why do you keep changing the tags?user169852 –user169852 2016-04-06 02:05:45 +00:00 Commented Apr 6, 2016 at 2:05
....yes but. What if the finite terms we change are bigger than N?fleablood –fleablood 2016-04-06 02:40:26 +00:00 Commented Apr 6, 2016 at 2:40
@fleablood N is a variable that represents an a N beyond which there are no finite terms that change as I understand it. If there are we can then choose N accordingly so that all different finite terms are in the sequence as less than a n.Bunny –Bunny 2016-04-06 23:39:10 +00:00 Commented Apr 6, 2016 at 23:39
1 Precise definitions for proofs can be frustrating. The "idea" is that if there are only a finite changes of difference then there are infinite that are the same and the finite ones "don't really matter". Because it's the "tail" that determines the limit And the tail can "always" "start" beyond any specific finite number of terms. But try putting that into correct terms...fleablood –fleablood 2016-04-07 18:18:01 +00:00 Commented Apr 7, 2016 at 18:18
1 I definitely knew what you were going for but I don't think you specifically addressed it. The n < N do not affect the limit. True. But you need to point out that there are only a finite number n that are changed (we should probably use the variable i rather than n). The i < N do not affect the limit but we need to point out that as there are only a finite number of i that are changed, that we can find an N > any of the i. That's all. That seemed to be lacking in your proof. Or so is my opinion.fleablood –fleablood 2016-04-07 18:23:10 +00:00 Commented Apr 7, 2016 at 18:23
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Your idea is right. You can make it a bit more formal as follows.
Suppose we start with a sequence a n. If we change finitely many terms, then this results in a new sequence b n. Since we only changed finitely many terms, there is some M such that a n=b n for all n>M.
Now suppose that a n converges to A. Let ϵ>0. There is some N such that |a n−A|<ϵ for all n>N.
Then, for all n>max{N,M}, we have b n=a n, so |b n−A|=|a n−A|<ϵ
which shows that b n also converges to A.
We have shown that if two sequences differ in only finitely many terms, and one sequence converges, then the other also converges (to the same limit). The contrapositive: if one diverges, then the other must also diverge.
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edited Apr 5, 2016 at 17:33
answered Apr 5, 2016 at 17:14
user169852 user169852
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Can you necessarily say that the sequences must agree, i.e. a n=b n, when n>M, for some M? What if the very last terms are different? Or maybe there's no "last term", since the sequences are infinite?mavavilj –mavavilj 2016-10-01 17:24:58 +00:00 Commented Oct 1, 2016 at 17:24
@mavavilj There is no last term. If you tell me you found the „last“ term, say a n, I can always find a later one a n+1, or I should rather say infinitely many later ones. That is the whole weirdness of the concept of infinity.mdcq –mdcq 2018-01-07 19:03:08 +00:00 Commented Jan 7, 2018 at 19:03
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6714 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9777900/ | Pulmonary Fibrosis Related to Amiodarone—Is It a Standard Pathophysiological Pattern? A Case-Based Literature Review - PMC
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Pulmonary Fibrosis Related to Amiodarone—Is It a Standard Pathophysiological Pattern? A Case-Based Literature Review
Corina Eugenia Budin
Corina Eugenia Budin
1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
2 Pneumology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
Find articles by Corina Eugenia Budin
1,2, Iuliu Gabriel Cocuz
Iuliu Gabriel Cocuz
1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
3 Pathology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
Find articles by Iuliu Gabriel Cocuz
1,3,, Adrian Horațiu Sabău
Adrian Horațiu Sabău
1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
3 Pathology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
Find articles by Adrian Horațiu Sabău
1,3, Raluca Niculescu
Raluca Niculescu
1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
3 Pathology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
Find articles by Raluca Niculescu
1,3, Ingrid Renata Ianosi
Ingrid Renata Ianosi
4 Faculty of Medicine, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
Find articles by Ingrid Renata Ianosi
4, Vladimir Ioan
Vladimir Ioan
4 Faculty of Medicine, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
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4, Ovidiu Simion Cotoi
Ovidiu Simion Cotoi
1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
3 Pathology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
Find articles by Ovidiu Simion Cotoi
1,3
Editor: Chiara Martini
Author information
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1 Pathophysiology Department, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
2 Pneumology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
3 Pathology Department, Mures Clinical County Hospital, 540142 Targu Mures, Romania
4 Faculty of Medicine, George Emil Palade University of Medicine, Pharmacy, Science and Technology of Târgu Mures, 540139 Targu Mures, Romania
Correspondence: iuliu.cocuz@umfst.ro
Roles
Chiara Martini: Academic Editor
Received 2022 Nov 12; Accepted 2022 Dec 16; Collection date 2022 Dec.
© 2022 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC9777900 PMID: 36553223
Abstract
Amiodarone hydrochloride is an antiarrhythmic drug, with proven efficacy in prevention and treatment of numerous arrhythmias, atrial fibrillation especially, or ventricular arrhythmias, with a long half-life (55–60 days). The increased risk of developing amiodarone-induced pulmonary fibrosis is directly related to the dose and the duration of the intake. Amiodarone-induced pulmonary toxicity is conditioned by dose, patient’s age, and pre-existent pulmonary pathologies. The pattern for drug-induced lung injury may vary in many forms, but the amiodarone can cause polymorphous injuries such as diffuse alveolar damage, chronical interstitial pneumonia, organizing pneumonia, pulmonary hemorrhage, lung nodules or pleural disease. The pathological mechanism of pulmonary injury induced by amiodarone consists of the accumulation of phospholipid complexes in histocytes and type II pneumocytes. Differential diagnosis of pulmonary fibrosis induced by amiodarone is made mainly with idiopathic pulmonary fibrosis, left ventricular failure or infectious disease. Before starting treatment with amiodarone, patients should be informed of potential adverse effects and any new respiratory symptoms should promptly be reported to their family physician or attending physician. The assessment carried out at the initiation of amiodarone treatment should include at least chest X-ray and respiratory function tests and extrapulmonary evaluation.
Keywords: amiodarone, pulmonary toxicity, interstitial lung disease, diffuse parenchymal lung disease, radiology
1. Introduction
1.1. Clinical Pharmacology
Amiodarone hydrochloride is an antiarrhythmic drug with proven efficacy in the prevention and treatment of numerous arrhythmias, atrial fibrillation especially, or ventricular arrhythmias , with a long half-life (55–60 days) [1,2]. Although it is classified as a class III antiarrhythmic, it has been shown that it can affect all phases of the action potential . Amiodarone is metabolized to desethylamiodarone (DEA) by cytochrome P450 in the liver . After the hepatic metabolism and biliary excretion, a small amount of amiodarone and DEA is found in the urine [1,4].
After the first dose, amiodarone reaches its plasmatic peak levels in 3 to 7 h. The onset of action can take from a few days to a few weeks. Biodisponibility can be influenced by age, liver pathology and interactions with other drugs or substances that can inhibit or stimulate cytochrome P 450 [5,6].
Because of the lipophilic structure, both amiodarone and its metabolites accumulate in high quantities in tissues and also interact with the phospholipid’s metabolism [1,7]. These tissues are represented by adipose tissue and well-perfused organs: liver, lung or skin tissue. The most frequent affected organs by the accumulation of amiodarone and, respectively, the high risk of developing injuries induced by amiodarone are the eyes (cornea deposits, photophobia), the thyroid gland (hypo/hyperthyroidism), the liver (drug-induced hepatitis, dyspeptic syndromes), the skin (photosensitivity) and the nervous system (peripheral neuropathy) [1,8].
Although the lungs are rarely affected (approximately 4–6% of all complications) , pulmonary injury has the most clinically significant impact, which can lead to the patient’s demise [8,10]. Amiodarone-induced pulmonary toxicity is conditioned by dose, patient’s age and pre-existent pulmonary pathologies . Those effects reach a plateau at a cumulative dose bigger than 150 g. Patient’s comorbidities, oxigenotherapy, invasive procedures or surgical interventions can trigger the pulmonary symptoms induced by amiodarone toxicity [1,4].
Amiodarone therapy also interferes with other drug classes, such as warfarin, simvastatin, atorvastatin as well as antiretroviral medication used in patients with HIV [3,11]. Taking into account these considerations and the frequent use of amiodarone in medical practice, physicians must know the indications, contraindications, dosage, adverse effects and drug interactions of amiodarone treatment [3,11,12].
Usual doses between 200–600 mg/day have minimal hemodynamic adverse effects. They cause a negative inotropic effect related to the administered dose by reducing systemic vascular resistance . It has no effect on the ejection fraction of the left ventricle, and arterial hypotension rarely occurs during oral treatment with amiodarone .
1.2. Pathology
Pulmonary fibrosis is characterized by pulmonary destruction and remodeling due to the accumulation of collagen and extracellular matrix at the tissue level . They cause an irreversible decrease in lung capacity, impairment of gas exchange and hypoxemia [4,13]. Systemic or inhaled toxic agents such as bleomycin, nitrofurantoin, amiodarone or ionizing radiations can be involved as etiological factors [13,14].
Evidence from the scientific literature has highlighted the fact that cells of the innate immune system as well as the adaptive one and the mediators that these cells release cause the appearance of interstitial changes . Macrophages are phagocytic cells belonging to the innate immune system . Present in all tissues of the body, most commonly in the lung and liver, macrophages function as immune sentinels, with the aim of defending the body against pathogens and injuries . Resident macrophages are distinct from bone marrow-derived inflammatory macrophages that accumulate in tissues in response to injury or infection [13,14]. Inflammatory macrophages are mainly involved in the development of pulmonary fibrosis . Macrophages have been classified into M1-pro-inflammatory/cytotoxic and M2-anti-inflammatory/reparative , which develop in response to signals present in the tissue microenvironment .
The pattern for drug-induced lung injury may vary in many forms, but amiodarone can cause polymorphous injuries such as diffuse alveolar damage (DAD), chronical interstitial pneumonia (CIP), organizing pneumonia, pulmonary hemorrhage, lung nodules or pleural disease [20,21].
The pathological mechanism of pulmonary injury induced by amiodarone consists of the accumulation of phospholipid complexes, which contain amiodarone, in histocytes and type II pneumocytes . The accumulation of phospholipids is determined by the amiodarone’s suppression of phospholipases [1,22]. The occurrence of amiodarone-induced pulmonary lesions is correlated with cumulative doses and pre-existent respiratory diseases. The majority of lesions appear at doses greater than 400 mg/day, but there are documented cases in which these can occur at doses of 200 mg/day or even less . Pulmonary lesions appear mostly after 2 months of administration . However, a correlation between the administered dose of amiodarone and the severity of the lesions could not be found . The presence of type II pneumocytes and macrophages with vacuolated cytoplasm indicates a history of amiodarone administration, but it does not support the amiodarone-induced acute pulmonary lesion diagnosis . In the bronchoalveolar lavage fluid of patients with amiodarone-induced pulmonary injury we can observe cytotoxic T cells. Amiodarone can determine the production of oxygen free radicals, which lead to cellular lysis. .
Characteristically, the histopathological diagnostic of amiodarone-induced lung injury is established on tissue samples, in the routine staining of hematoxylin–eosin. Microscopically, upon lung tissue analysis of the pulmonary parenchyma we can highlight diffuse interstitial pneumonitis . It is described by a type II pneumocytes hyperplasia, inflammatory infiltrate within the alveolar septa and variable degrees of pulmonary fibrosis . Vacuolated cytoplasm can be hilighted in alveolar pneumocytes, bronchial epithelium and endothelial cells [4,24]. The accumulation of foamy alveolar macrophages is specific to APT (amiodarone-induced pulmonary toxicity). Electron microscopy examination displays membrane-bound lamellar bodies and lipid particles such as surfactant due to accumulation of drug accumulation into the lungs [1,4,23]. Less frequent pathological manifestations are patchy bronchiolitis obliterans, organizing pneumonia or, in severe cases, diffuse alveolar injury with formation of hyaline membranes or alveolar hemorrhage induced by amiodarone . For establishing a histopathological diagnostic of amiodarone-induced lung toxicity, the correlation between the microscopic aspect and the clinical data must be corelated.
1.3. Clinical Presentation
From a clinical point of view, the patients with amiodarone-induced pulmonary fibrosis can present different degrees of progressive dyspnea, unproductive cough, fever or, rarely, pleural pain. On auscultation, bilaterally basal “Velcro-like” inspiratory crackles are found . Acute or over-acute forms are presented as acute respiratory failure, with possible onset of acute respiratory distress syndrome [24,25]. Among cases with diffuse alveolar damage, the phenomena of respiratory failure dominate the clinical picture. In cases of bronchiolitis obliterans-organizing pneumonia (BOOP), symptoms may mimic bacterial pneumonia [26,27].
1.4. Diagnostic
From a functional point of view, a moderately restrictive type of pattern is frequently highlighted, with a decrease in forced vital capacity (FVC) and a moderate decrease in the diffusing capacity for carbon monoxide (DLCO) in approximately 45% of patients [5,10]. A nonspecific inflammatory syndrome can also be highlighted, characterized by a mild leukocytosis, increased erythrocyte sedimentation rate and increased C-reactive protein (CRP) value, but these are nonspecific and are associated with interstitial inflammation [7,13].
Bronchoalveolar lavage (BAL) is essential to confirming the diagnosis of interstitial damage induced by amiodarone. The decrease in the number of macrophages in BAL is characteristic of amiodarone-induced alveolitis. It is a sensitive method, but not very specific. Moreover, the presence of foamy macrophages suggests amiodarone-related alveolitis but without complications.
1.5. Imagistic Features
Interstitial lung damage has been described as a severe adverse effect associated with some drug therapies, such as bleomycin, amiodarone or methotrexate. The mechanism by which amiodarone produces this type of lung damage is through the accumulation of phospholipids at the level of alveolar macrophages . Pulmonary involvement is diverse . It can be like the ground glass opacities type, with peripheral localization. Interstitial, alveolar or mixed infiltrates located bilaterally or high attenuation areas in infiltrates, lung nodules or “masses”, may be single or multiple, often peripheral in location. Dense bilateral basal reticular opacities and traction bronchiectasis suggests pulmonary fibrosis [1,4]. Upon CT examination, a severity score can be calculated, depending on the number of affected regions (right and left; upper, middle and lower; and central and peripheral). This severity score correlates with symptomatology and pulmonary functional impairment [30,31].
If amiodarone-induced lung damage is BOOP type, bilateral consolidation areas are present, located subpleural and peribronchovascularly, with patchy areas of ground glass [8,32]. The location and intensity of these lesions may vary over time, either with treatment or spontaneously . In non-specific interstitial pneumonia (NSIP) pulmonary involvement, HRCT examination reveals areas of ground glass, reticular opacities and linear fibrosis. Associated, subpleural areas of honeycombing and traction bronchiectasis are present . A chest X-ray reveals an interstitial lesion. On HRCT (high resolution CT) interstitial damage, reticular or reticulonodular opacities and traction bronchiectasis are present [1,25]. Honeycombing is less common than in idiopathic pulmonary fibrosis . The most dramatic manifestation of amiodarone-induced alveolitis is rapidly progressive diffuse pneumonitis with acute respiratory failure and ARDS-like changes [24,25].
1.6. Differential Diagnosis
The main pathologies with which the differential diagnosis (Table 1) of amiodarone-induced interstitial lung damage is made are idiopathic pulmonary fibrosis, hypersensitivity pneumonitis, infectious pathologies or heart failure [33,34]. Comorbidities such as diabetes or stroke are common in these patients, which increases the risk of respiratory infections [8,35].
Table 1.
Main differential diagnostics of amiodarone-induced pulmonary fibrosis.
Idiopathic pulmonary fibrosis Usual interstitial pneumonia (UIP). High-resolution CT often shows honeycomb changes, traction bronchiectasis and a reticular pattern that is predominantly in the periphery of the lower lobes.
Hypersensitivity pneumonitis Exposure history. Specific IgG antibodies. Pulmonary infiltrates and suspected nonspecific interstitial pneumonia or idiopathic pulmonary fibrosis.
Infectious pathologies Purulent sputum, unilateral localization. Procalcitonin value elevated.
Heart failure. Acute cardiogenic pulmonary oedema Kerley lines and peri bronchial oedema. Subpleural patchy areas in pulmonary oedema.
BOOP Procalcitonin value unchanged.
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The differential diagnosis between bacterial pneumonia and BOOP is difficult to perform because the clinical and radiological manifestations are superimposed [8,11]. The sudden onset, purulent sputum, unilateral localization advocates the diagnosis of bacterial pneumonia. Procalcitonin is a useful marker for differential diagnosis, being elevated in bacterial pneumonia and unchanged in BOOP .
Early stages of interstitial damage are difficult to differentiate from cardiac stasis because patients indicated for amiodarone treatment have chronic cardiac pathologies [8,36].
Heart failure and pulmonary thromboembolism are pathologies that must be considered in patients with interstitial involvement . Pulmonary thromboembolism superimposed on amiodarone-induced interstitial lung damage is difficult to diagnose and contributes to the poor prognosis of these patients . Repetitive pulmonary thromboembolism in the small branches of the pulmonary artery may not have a characteristic clinical expression, but may be manifested only by mild hemoptysis, progressive dyspnea in the context of the establishment of pulmonary hypertension . At the same time, subpleural consolidations due to pulmonary infarcts are difficult to differentiate from the imaging changes of BOOP .
Acute cardiogenic pulmonary oedema is a possible complication in patients with decompensated heart failure or pulmonary stenosis. The imaging appearance is confirmed by the appearance of Kerley lines and peribronchial oedema . Subpleural patchy areas also occur in pulmonary oedema. The clinical expression is what differentiates the diagnosis at patient presentation .
1.7. Treatment
The first therapeutic measure that is required is to stop the administration of amiodarone. The next steps are dependent on the patient’s clinical condition. Some of the studies in the scientific literature recommend observing the patient and revaluating after one month. If the condition is good and the patient is stable, the next evaluation is performed at 3 months, and subsequently at 6 months [8,37,38].
In patients with respiratory failure or significant lung damage, the administration of systemic corticosteroids is indicated [25,37,38]. Prednisolone is used in a dose of 40–60 mg/day. Due to the increased half-life of amiodarone, 2 months of oral corticosteroid treatment is recommended, followed by a dose reduction period, summing up total period of treatment for at least 6 months. Even after administration of this type of treatment, pulmonary recovery, both imaging and functional, is not completely reversible [8,33].
The recommendation to use antifibrotic medication is very recent. Nintedanib is currently indicated for use in pulmonary fibrosis with a progressive phenotype, including in this category pulmonary fibrosis induced by amiodarone .
2. Case Presentation
The presented cases are from patients admitted to the Pneumology Department of the Mures Clinical County Hospital. Informed consent was obtained from all patients involved in the study. This study was conducted in accordance with the Declaration of Helsinki and approved by the Ethics Committee of Mures Clinical County Hospital, Romania (protocol code (IRB number) 16051, on 7 November 2022).
2.1. Case Presentation 1
We present the case of a 67-year-old male patient, with a history of atrial fibrillation, aortic insufficiency III/IV and tricuspid insufficiency I/IV from 2015. The patient was treated with dabigatran etexilate 75 mg, perindopril and amiodarone for one month.
In September 2019, the atrial fibrillation relapsed and was treated with amiodarone, but in October 2019 he came back accusing dyspnea present on rest and clinically presenting hypoxemia. It should be noted that the lung X-ray (Figure 1) from July 2019 is without changes.
Figure 1.
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Chest X-ray—2019—Case 1. July 2019: no signs of interstitial damage; October 2019: reticular opacities associated with ground glass opacities with heterogeneous distribution, located subpleural especially. (a) July 2019; (b) October 2019.
The HRCT (Figure 2) showed bilaterally, subpleural honeycombing in the superior half, nodular lesions, linear fibrosis, traction bronchiectasis, reticulation, ground glass opacities, with associated minimal emphysema and bilaterally, subpleural cryptogenic organizing pneumonia foci and diffuse pulmonary infiltrates in the inferior half. The treatment regimen was consequently initiated with corticosteroids (prednisone 45 mg/day and amiodarone discontinuation. As the patient was known to have diabetes, the dose of prednisone was reduced to 20 mg/day for 5 months, and subsequently the treatment with prednisone was gradually stopped. Since the symptoms reappeared, prednisone 15 mg/day continued for another 3 months. Afterwards, the patient had a favorable evolution with 90% oxygen saturation and dyspnea during ordinary activities. In 2020, another high-resolution computed tomography was performed and it revealed a better pulmonary aspect with persistent and reduced ground glass opacities, minimum bronchiectasis associated with minimum fibrosis and bilateral emphysema (Figure 3). The evolution remained favorable until the next evaluation in 2022, without treatment with oral corticosteroids (Figure 4).
Figure 2.
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HRCT—July 2019—Case 1. Subpleural honeycombing, nodular lesions, linear fibrosis, traction bronchiectasis, reticulation and ground glass opacities with associated minimal emphysema. Diffuse pulmonary infiltrates in the inferior half.
Figure 3.
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HRCT—2020—Case 1. In comparison to 2019 HRCT, reduced ground glass opacities, minimum bronchiectasis associated with minimum fibrosis and bilateral emphysema.
Figure 4.
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Chest X-ray—2022—Case 1. Diffuse reticular opacities, especially in the basal regions.
2.2. Case Presentation 2
We present the case of a 71-year-old male patient, with a history of atrial fibrillation, aortic valve prosthesis and mitral insufficiency from 2016. The patient was treated with amiodarone for 5 years.
The imaging appearance on the thoracic CT was performed 2 years after the initiation of amiodarone treatment, and although the patient was not symptomatic from a respiratory point of view, it describes a more pronounced honeycombing appearance in the right lung field. Functional exploration reveals a restrictive syndrome with FVC 45%, FEV1 71% and an FEV1/FVC ratio of 89. The patient was not referred to a Pulmonology service.
After 5 years of treatment with amiodarone 200 mg/day (2021), the patient was sent for a pulmonary control before performing a coronary angiography procedure. The HRCT examination (Figure 5) reveals an appearance of pulmonary fibrosis with a probable UIP (usual interstitial pneumonia) pattern, with reticulation plaques, traction bronchiectasis, subpleural air microcysts, without the typical appearance of honeycombing, in places with crazy paving.
Figure 5.
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HRCT—2021—Case 2. Reticulation plaques, traction bronchiectasis, subpleural air microcysts. No honeycombing. Probable UIP pattern.
After this imaging interpretation, at the recommendation of the pulmonologist, the treatment with amiodarone was stopped, and although the patient had no respiratory symptoms, and treatment with metoprolol, acenocumarolum and candesartan was initiated.
Spirometry describes a restrictive syndrome with FVC 64.6%, FEV1 72% and FEV1/FVC of 86.34.
At pulmonary revaluation after 1 year, the patient reported that the 2-month amiodarone treatment had been resumed because the cardiologist intended to perform an ablation procedure. HRCT further describes pulmonary changes with the appearance of probable UIP. Body plethysmography: total resistance 93%, residual volume (RV) 60%, total lung capacity (TLC) 62%, FEV1 78%, DLCO 58. Although the patient presented a progressive fibrotic phenotype, he did not accept to initiate antifibrotic treatment.
2.3. Case Presentation 3
We present the case of a 75-year-old female patient, with a history of atrial fibrillation and arterial hypertension. The patient was treated with amiodarone (200 mg/day) for 6 years. The presentation in the Pulmonology Service was made because the patient complained of dyspnea at low/medium efforts.
After one year since initiation of amiodarone treatment, a chest X-ray (Figure 6) was initially performed, followed by a native thoracic CT (2017), which revealed in the lower lobes extensive ground-glass areas with bronchiectasis, fibrosis changes in honeycombs, numerous micronodules and centrilobular pulmonary nodules (maximum 5 mm), some with a tree-in bud disposition.
Figure 6.
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Chest X-ray—2017—Case 3. Nodular and micronodular opacities localized mainly in the lower lobes. Veiling of the pulmonary bases. Reticular opacities on both pulmonary areas. Enlarged cardiac silhouette.
The patient did not present then to the Pulmonology Service, and the CT examination (Figure 7) was repeated 5 years later, in 2022. Extensive areas of pulmonary condensation with ground glass appearance predominantly in the basal segments were described bilaterally, associating fibrotic lesions and bronchiectatic dilations.
Figure 7.
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HRCT—2022—Case 3. Extensive areas of pulmonary condensation with ground glass. Bronchiectatic dilations.
The pulmonary functional examination revealed normal lung volumes and flows (simple spirometry and body plethysmography), DLCO 80, although the imaging aspect revealed important pulmonary changes of fibrotic type, for which reason it was decided only to monitor the patient and she was directed to cardiology for evaluation the possibility of excluding amiodarone from chronic treatment.
3. Discussion and Conclusions
Amiodarone-induced interstitial lung damage includes varying forms of presentation, from mild to moderate/severe . These include organizing pneumonia, interstitial pneumonitis or respiratory failure. The typical presentation of amiodarone-induced lung damage is subacute, with dry cough, progressive dyspnea, low-grade fever and weight loss . Case 2 fits this description, while case 1 presented severe symptomatology with acute respiratory failure, and case 3 presented poor and nonspecific symptomatology. Associated chronic medication or occupational exposure could not be associated with the imaging lesions for any of the patients. The increased risk of developing amiodarone-induced pulmonary fibrosis is directly related to the dose and the duration of the intake. Case 1 followed treatment with 400 mg/day, considered to be a high dose, but for a short period (only one month), and the other two cases followed treatment with a low dose of amiodarone (200 mg/day). Although pulmonary-adverse effects occur at increased cumulative dose associated with long duration of administration, imaging lesions were significant in these patients. Although amiodarone is a potent antiarrhythmic, studies in the specialized literature have demonstrated the occurrence of pulmonary toxicity associated with this treatment [4,21]. The incidence of these complications has decreased considerably with the use of reduced doses, but nevertheless, in some cases, the clinical presentation is acute and may be life-threatening [8,37].
Although the adverse effects produced using amiodarone are known, the adherence of medical staff and patients to the guidelines for monitoring therapy is poor [31,41]. Worldwide, there are medical centers where it is possible to determine the serum level of amiodarone. Values higher than 2.5 mg/L are indicators of a high level of toxicity [31,42]. In our hospital, there is no possibility of this serum’s determination. This determination is useful both for the determination of amiodarone-induced toxicity, in case the patient is symptomatic, or in case of recurrence of an arrhythmia and the possibility of discontinuing the treatment is discussed [31,43]. Differential diagnosis of pulmonary fibrosis induced by amiodarone is made mainly with idiopathic pulmonary fibrosis, left ventricular failure or infectious disease [33,44]. The differential diagnosis in these patients with left ventricular failure is difficult to achieve because cardiac pathology is predominant, and many of the patients have multiple pathological personal antecedents from a cardiological point of view [33,45].
Therefore, preventive and monitoring measures have an important place in the management of these patients. Before starting treatment with amiodarone, patients should be informed of potential adverse effects and prompt reporting of any new respiratory symptoms to their family physician or attending physician [4,9,38].
The assessment carried out at the initiation of amiodarone treatment should include at least a chest X-ray and respiratory function tests, including DLCO [21,45].
None of the patients included in our case series had this assessment performed, which made it difficult to fully assess the extent of lung damage following the period of amiodarone treatment.
Moreover, the assessment at the initiation of amiodarone treatment should also include extrapulmonary evaluation : assessment of lung function, monitoring of liver enzymes or warning patients about the photosensitivity induced by the treatment [29,37]. Because these three patients presented to our Pulmonology Service, we do not have data on extrapulmonary toxicity assessment. The clinical presentation of the three presented cases was different. The first case was presented to pulmonology in an acute manner, with significant lung damage from an imaging point of view and signs of acute respiratory failure. Fortunately, the response to the exclusion of amiodarone from the treatment regimen and the administration of systemic corticosteroids was a very good one. Pulmonary toxicity occurs both in clinically acute forms and in chronic forms , such as in cases 2 and 3.
Radiological changes unaccompanied by significant clinical impact can often go unnoticed [1,47]. In cases where imaging damage is accompanied by severe clinical impact, therapeutic treatment with oral corticosteroids is also required. The evolution of case 1 is consistent with the results of the study conducted by Mankikian in 2014, who claimed that, from an imaging point of view, an improvement in alveolar opacities is observed upon HRCT examination in the medium term. This trend is also maintained in the long-term, with an improvement in the severity score on HRCT of 40%, but without restitutio ad integrum . Pulmonary functional exploration highlights a restrictive syndrome. DLCO decreased by 15% advocates pulmonary toxicity in a patient under amiodarone treatment [1,47]. In the first two cases presented, we do not have the DLCO value available, and case 3 presents with a DLCO of 80%, which practically excludes the phenomena of diffusion alteration. Case 2 would have been eligible according to the latest recommendations for antifibrotic therapy, but the patient refused, being scheduled for an ablation procedure.
In cases with no significant clinical impact, the simple exclusion of amiodarone from the treatment scheme is sufficient and no other therapeutic measures are necessary [1,47,48].
Bronchoalveolar lavage was not performed in any of the three cases presented. The diagnosis of pulmonary toxicity induced by amiodarone is difficult, it is a diagnosis of exclusion, based on clinical phenomena of respiratory insufficiency, imaging interstitial-type affection, sometimes even usual interstitial pneumonia, which makes the differential diagnosis with idiopathic pulmonary fibrosis and/or biological [25,46,48]. Early recognition of respiratory complications induced by amiodarone treatment and intensive treatment can cause a favorable evolution of the patient. Any delay in discontinuing amiodarone treatment when there is clinical suspicion may lead to an unfavorable prognosis for the patient . Further imaging and functional monitoring (volumes and respiratory flows) is mandatory.
The prevention of adverse effects is the responsibility of the entire team that interacts with the patient: the attending physician, the one who prescribes the treatment, primary care physician, specialist physician and pharmacist . The multidisciplinary approach remains essential to improving the quality of life and the patient’s outcome. Effective follow-up of the patient after initiation of amiodarone therapy involves responsibility on the part of the entire medical team as well as the patient . Current information and effective communication between patient and doctor are essential for further development .
Author Contributions
Conceptualization, C.E.B. and I.G.C.; methodology, C.E.B. and R.N.; validation, C.E.B., I.G.C. and O.S.C.; formal analysis, C.E.B., A.H.S. and I.R.I.; investigation, C.E.B., I.G.C. and V.I.; resources, C.E.B. and A.H.S.; writing—original draft preparation, C.E.B. and I.G.C.; writing—review and editing, C.E.B., I.G.C. and O.S.C. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
The study was conducted in accordance with the Declaration of Helsinki and approved by the Ethics Committee of Mures Clinical County Hospital, Romania (protocol code 16051, on 7 November 2022).
Informed Consent Statement
Informed consent was obtained from all subjects involved in the study.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research received no external funding.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Back to Top |
6715 | https://www.youtube.com/watch?v=Mgp2vrQeLEw | Complex Numbers as Rotation Matrices
Math Easy Solutions
56900 subscribers
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Description
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Posted: 18 Nov 2022
In this video I go over a very interesting concept and that is in interpreting complex or imaginary numbers as and equivalent rotational matrix form. This concept was brought to my attention by Lori Gardi from the @FractalWoman YouTube channel. Rewriting complex numbers as rotation matrices allows for looking at complex numbers from a different perspective, and this case that perspective is in simply rotating vectors. The imaginary number "i" is thus viewed as simply a 90 degree rotation when viewed as a rotation matrix. I also rewrite the famous Euler's formula into a rotation matrix form and that turns out to just be a 180 degree rotation. Since complex numbers appear often in mathematics and physics, it may be insightful to reinterpret such complex equations as simply vector rotations!
Note that while I only covered 2D rotation matrices, a similar concept can be applied to 3D rotation matrices, and which I may cover in future videos so stay tuned.
The topics covered as well as their timestamps are listed below.
Introduction: 0:00
Topics to Cover: 0:41
Fractal Woman and Other References: 1:28
Rotating a Vector: 3:01
Matrix Multiplication: 12:47
Matrix Addition: 25:29
Rotation Matrix: 27:13
Complex Numbers: 36:58
Relating the Complex Plane with the Rotation Matrix: 43:30
Complex Numbers as Vector Rotations: 57:12
Reinterpreting Euler's Formula: 1:10:02
Applications to Other Fields: 1:20:51
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View video notes on the Hive blockchain:
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View the full Vectors and Geometry of Space video series:
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3D Coordinate Systems: .
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Transcript:
Introduction uh it's time for another math easy solution uh today I'm gonna go over a very interesting video on complex numbers as rotational matrices and basically show we could write complex numbers in an in a matrix format and then show how it could basically be used to rotate vectors which is quite fascinating uh as we dive deep into this all right let's just uh scroll down here and also just a note if you don't have time to watch this whole video you could play this video at a faster speed and uh and you can download the notes and yeah view them on the hive blockchain and article format and so on and all the time stamps of all the parts are in the description all right let's uh Dive Right In so let's look at the topics to cover and uh Topics to Cover note here the topics covered are listed below and their timestamps will be included in the video description so we have a fractal woman and other references uh then we'll go over rotating a vector and then look at matrix multiplication Matrix addition then look at the rotation Matrix and then look at complex numbers and and then basically relating the complex plane with the rotation Matrix in other words we could write complex numbers as Vector rotations and and then we get a reinterpret or reinterpreting uh Euler's formula and then applications to other fields just uh yes very interesting um you have to think about all right let's 1. Fractal Woman and Other References Dive Right In so fractal woman and other references so I got the inspiration to make this video from Lori gardy's YouTube channel called fractal woman which I've linked below and here's a link and I just went to her channel here in just another searched a complex number so you could see the the ones that yeah the videos that she has up regarding them just quite interesting so yes um and also another note is that I know that Laurie uh herself states that she got this idea from Robert descente from his YouTube channel and right now the name is www.thecenty.com and here's a link below here and also a bit of search uh so you can go to this channel search complex numbers and then see those that he has all right and now it's another note listed below are some some sources uh which I've used in the making of this video and this is a simphiz video derivation and this is uh derives the yeah a complex number uh to uh rotation Matrix um yeah relationship and also Euler's formulas over here Wikipedia rotation Matrix from Wiki and this is my vectors video on uh yeah and on the hive blockchain is put the article uh format there with all the notes and the links to the videos are in there as well and uh here's another uh another reference uh matrix multiplication the Wikipedia pages and yeah these Stars Wikipedia once Matrix Edition complex number Etc imaginary unit I and Euler's formula here and most of these are Wikipedia Pages because I haven't covered matrices in in depth in my channel yet but 2. Rotating a Vector anyways let's go uh further let's look at rotating a vector so let's consider what happens when we rotate a vector by an angle Theta all right so if you recall from earlier video on vectors um basically we have a vector like this let's say we have the X Y axis here is the X this is the Y and here I just uh stretched out the y a bit more let's say we have a vector here it goes from the origin zero or o so let's make it like this okay so this is the vector a and a with an arrow there so basically it has a magnitude and Direction the length is the magnitude and we'll have an arrow on tops indicating it's a vector and then we'll put this kind of bracket this triangle bracket X Y so those are the coordinates of it oh yeah it reaches all the way over to here and we'll put that indicating as a vector and it has a y and this is a let's draw a triangle it's a right angle triangle it has a y component here which equals 2 and then we'll say this angle right here is a so let's say here this angle's a all right yeah so that's angles a and then this y would be R and let's say the the it has a radius of or the length r this is basically polar coordinates uh but here what this is R times sine of a and then this bottom part is going to be well x equals two the x equals two let's move it over here actually because we're gonna need some room once we rotate this x equals to R the cosine now of a all right so that's what we have this now let's say we rotate it because you're rotating this vector by an angle Theta so it's going to be over across here this is exactly the same length r we'll say it's over here and this is the length R and now we're rotating it by Theta okay so that's Theta there and this is right here b equals to uh yeah and we'll call that Vector B so B is the rotation of the of the A and we'll call this x prime or Y prime or Y dash like that to indicate that it's different from this one so it's rotating across over to here all right so now it's components if I draw this dashed line I'll look all the way across here in red like that and then draw a right angle sign there so this one here so its height is going to be its y component uh Y dash is equal to or the sine of a plus now we combine these two angles a plus Theta a plus Theta and the next one here is yes over here is going to be our X component x dash this equals 2 R cosine let's write this better cosine Theta all right so we have R cosine I mean not that a plus Theta because we the angle is is bigger than that so we've moved over so a plus Theta now just using a basic trigonomic yeah uh the concept zero definition is basically we're using a basic trigonomic definitions and just uh put a note here recall just so how we did the triangle setup now so recall if you had uh just a triangle like this if this uh this is the adjacent a this is B and this angle is Theta and the hypotenuse is C and then basically sine is defined as the opposite over uh is defined as the opposite of a hypotenuse so B over C and then we can move the C over there we will get this b equals 2 c times sine Theta or in this case r sine a and likewise cosine Theta is equal to adjacent over hypotenuse and then we move the C over to the other side and so on and then yeah in that case we'll just get this right here R cosine a in this case where R is C and A is Theta there all right so now that we've recapped on that now what we have is let's put these all together let's put the Y that all the components together so thus we have X is equal to just R cosine a so that's just X is here R cosine a y is R sine a y equals R sine a and then the X X Prime and then y Prime this has the a plus Theta so we have X Prime equals R sine a plus Theta and then y Prime is equal to r i mean this is cosine R sine a plus Theta all right so now that we have this I'm just going to box this out just to make it uh nice neat package all right so we have boxed that out so now uh we could uh simplify this by using the identities for these addition cosine and sine addition formulas so recall the following trigger trigonometric identities from my earlier videos so this is cosine a uh plus minus B is this link to it and then sine a plus or minus B and then basically yes we have here so for cosine plus or minus B we'll get cosine a cosine B then minus plus sine a sine B this means basically if we have if you're adding these angles then we're going to subtract this term and then sine is going to be a plus or minus B and then sine a COS B and then plus or minus cosine a sine B in other words if it's plus here it's plus here it's minus it's minus but for cosine switches if it's minus it's plus and if it's uh and if it's plus it goes to minus all right so thus utilizing the addition the addition identities we obtain and we'll look at this right here so we're going to look at X Prime and Y Prime so we have X Prime is equal to R times cosine a plus Theta all right now we just plug this into that formula here and then this one here I derive both in these two videos so yes make sure to watch that if you want to see the proof so now this becomes r we're going to get R times it by cosine now this is uh a is this a and then B is this Theta there so we'll get cosine a cosine of B I mean no yeah cosine Theta then we're going to go minus because it's plus turns into a minus for the cosine identity all right and now just minus then we're going to do sine a cosine or sine again sine Theta like that and now you actually are here uh we also need the the r in both because you could expand that out yeah so the inside will be there and we said we also have to have that R there because we just foil it out it's gonna be an R there all right so now what we have here is well if you notice this part here R cosine a that's just this part here that's this R cosine is just X so this equals to X now this one here are sine a this equals to y that's y let's erase these so yeah those are just y r sine a so in other words this equals two equals to x cosine Theta minus y sine Theta all right so now let's bring this down and write X Prime and then box this out so that is a very interesting uh formula Yeah so basically in this case we don't even need to know the angle a we just need to know the coordinates here and then uh basically then we'll have this Theta there but you could also always solve for the the angle a if you know these two so they're using basic trig so then we'll get yeah just a formula in terms of x y and then this Theta there to get the rotation uh rotation Vector there so we rotate over here we can get the components accordingly so that's for the X now let's take a look at the Y one so y Prime is the sine one so y Prime equals to R sine a plus Theta all right so that's the r uh sine a plus Theta and make this data nicer all right and now it will get is for the identity so sine a COS B and then plus cos a sine B this equals 2 R times it by sine a COS B which is Theta and plus we're going to get a COS or a multiplied by the R so cosine uh a and then sine b or Theta and that is what we get there and now likewise this part right here this is our sine sine a that's just our y is equals to Y now this equals to X all right so now let's put this together okay so R cosine is X and we'll put the X component first so we'll put X notice the difference this one's cosine on top this one's going to be sine so X sine Theta X sine Theta and then we're gonna have a plus and then we'll write this one second so y cosine Theta cosine Theta and notice the r is gone there's no uh yeah so we replace the r if we know X and Y we don't need to know r or Theta I mean or or a at the angle a so this is going to be bring this down and we've got a nice neat little equation here y Prime all right so now we have this all right so we have this and remember yeah these are just the components of the same Vector that's rotated by an angle Theta so that's rotating a vector all right so now we have that under our bag and now 3. Matrix Multiplication let's look at matrix multiplication so a matrix plural is matrices is just a rectangular array or table of number symbols or Expressions such as the following two by two Matrix so here's just a matrix with that is two rows and two columns I'll call this a inside here b c d so this is two by two Matrix like that all right and uh yeah that's pretty much all a matrix is just a uh a table or array of data points in this case there's four for this two by two Matrix and you could uh make them whatever you want and so on so now let's take a look at multiplying two matrices such as a and b this is defined such that it produces a second Matrix C and it is denoted by a b and follows the rules listed below so the number of columns of the first Matrix a a must equal to the number of rows in the second Matrix B and the resulting Matrix has the same number of rows of the first a and the number of columns of the second Matrix B so the same as the second Matrix B columns and another one a rows and the mes node typically the notation for Matrix is to bold the symbol such as a but I'll use the following notation as well where it is more clear with this rectangle rectangular bracket and bold we'll have the latest up bracket a and an N bracket and this notation uh yeah this one Lori guardi uses it and she also got it from descente Robert Cindy yes which is quite interesting and uh let's go now and illustrate these rules or this multiplication mainly these rules that we are defining for multiplication so let's say we have a matrix that's this this much rows we'll call this so we have five rows we'll call this I those eyes five rows like that eyes five and this is we'll go here one two this is three this is gonna be four it's gonna be fine we have one two three four five five rows like that just erase this erase erase erase erase so that's five and let's say we have uh three columns in this first one all right so that's our first Matrix and this is the three one two one two three and so on and this is gonna be it'll call this m equals to three and this is our Matrix a I got and we're going to multiply this Matrix a by B so no let's look at the first rule the number of columns so this is this is the columns yeah this right here is the columns m is three of the first one so the number of columns of the first Matrix a must equal to the number of rows of the second Matrix B all right so then these ones have to match with B so in other words we need an m is three so we're going to have to have uh three right here so so one two three rows for the second Matrix B but otherwise we can't multiply it according to our definition this is going to be we'll call this also going to be ms3 so that's the second Matrix and there's no other rules for the length of this one so we'll make this four so it's different so we'll call this one two and then there's there's four now yeah that means there's four whoops all right so this is one two three four like that so there's four and we'll call this n for the number of rows of the second this equals to four and this is our B so a matrix times B this has to equal now let's take a look at the rules the resulting Matrix has the number of rows of the first a so in other words this has to have eyes five and the number of columns of the second Matrix B no number of columns four so in other words we're going to get uh yeah we're going to get i5 and we're gonna get N4 so we're going to go here this is going to be one two make us longer one two three four and this is going to be like that like this and like that and uh here's the fix it I've moved it around so that's four so we go one two three four and it's four and then we have eyes five one two three four five and this is going to be I equals to five so that is this is equals to our C or a b or a B like that so yeah that's our C Matrix there all right so now that we've gone over these rules for the size of these matrices when you multiply in the rules that you need these ones to be the same so this needs to be the same here and then the the following one has the lengths accordingly so the rows of the first and then Columns of the second and now we'll look at the uh the rules for the actual values inside this uh Matrix C so the values in C or a B involve multiplying each corresponding row of a so we'll look at the row of Vari we get with each corresponding column of B and summing them up as followers follows all right yeah so to illustrate the rules uh for uh the values inside of this I will look at a simple um yeah Matrix we'll look at a three by Two Times by two by two so in other words uh the first one let's take let's use three rows so I is three and what I'll do is well it's gonna be three rows so the First Column is going to be uh we'll label it such as this where the first number one indicates the row and the second number indicates the column so a11 and then this is going to be two now is the second row a21 still this First Column and this next one's going to be a three one and then we'll use a just a two by two so I mean a three by two so then this is the next one is going to be well first row second column this next one is going to be first second row second column next one is going to be uh second I mean third row second column and this is going to be m is equal to two so if m is 2 in this example if we're going to multiply by another Matrix it needs to be the same so M needs to be 2 as well for the rows yeah for the rows times the by and then we're going to have to have the uh m is 2 as well it has to be m is two and let's say the first one is going to be let's use B now for this Matrix we'll call this B one one first row and First Column second row First Column and then let's just use a simple two by two so then we'll just stick over here so B1 first row second column uh second row second column so B two two e one one B one two and B two two and this is just our n is equal to two so now if we multiply yeah when we multiply we're gonna get well number the same we're going to get I if this is I equals five this is I equals five so in other words it's going to be three it's going to be 3 I equals to 3. so 3 rows and it's going to be then it's going to be n is 4 n is 4. so n is going to be 2 uh two columns and the values inside yeah they're going to be well according to this uh these rules so multiplying each row of a so we'll take a row of a and then by the column of B so this is a column of B like that so and they're going to be like so a11 multiplies with this one here so this multiplies with this and we're going to get a 1 1 times by b 1 1 and then we sum them up so then we sum up and we're just going to be added two and then the next one a one two Times by B21 so these are the rules for matrix multiplication it's going to be b or a12 times it by B uh two one yeah so two one so the rows and columns and that is basically the pattern we're going to follow for every single setup here so that's the yeah so that's the corresponding row with the column so that's the First Column so this is just one data point there all right and here's move things over to the left just to get more room for the second uh second column so that's the first one there and then the second one is going to be well we're going to multiply now this row by this column so in other words we're going to get a 1 1 again I'll just do this one last time for this just to illustrate this a11 multiplies with this a 1 1 and then B one two and then we're gonna get um this multiplied by this and then plus yeah this is going to be plus a12 1 2 and then b 2 2. I mean yeah B22 that's right and I'm going to move this a bit closer we're just going to box that in now draw the box here all right here let's move this over and yeah so that's the first one and yes we got the first two call uh Columns of the data on this one here yeah I mean uh the first uh two columns of the first row of this uh Matrix here and again I'll put this here n is equal to two so the same as this one here and this 2 I is 3. this is the number of columns all right let's go back here so now that the second row is basically going to use this row and again the columns that's why it that's why the number of rows here match this one and number of columns match the second Matrix and then we're going to multiply that again so it's going to be a two one times uh b11 a21 times B one one and notice the pattern here so this a11 is a b one one and the next one is going to be B21 as well and this one's going to be just a 2 2. plus a 2 2 and then this one here B21 like that and then the next right here is going to be uh now we remove this and now we're going to multiply by this side this one stays the same as the same row and this is going to be a 2 1 b 1 2 and these ones always the same on these ones Plus a 2 2 then b 2 2. all right yeah so Matrix uh multiplication or just matrices in in general aren't that complex they're they're just basically uh doing a repeating a whole bunch of tedious calculations into a compact form so that's pretty much a matrix algebra and when you go to linear algebra and so on it's pretty much the idea of it is to streamline so it'll be easier for computers to do all right so now we have that and then the next one is here a31 a32 this is third row and then we'll look at the First Column of the third row or multiply this so a31 multiplies b11 then a32 B21 and so on so notice these ones all match so we could just simplify this whole thing b11 B21 if that's going to happen and then this one here all the same ones that are exact so B12 and this one's B22 those ones aren't going to change and then multiply this inside here a31 and the next one is a32 Plus a32 and then this one here notice this one this one's the same the rows are the same and then this one here the columns are the same so notice the the Symmetry there or the pattern so the next one is going to be just a three one move this here multiply by that in here a31 plus a uh three two like that all right all right now that we've covered matrix multiplication now let's quickly look at Matrix addition this one is easier or 4. Matrix Addition just a simpler than this so Matrix addition requires that two matrices have equal number of rows and columns and the result is that each corresponding entry simply adds up together as a following example illustrates so basically two matrices have to have the equal number of rows and columns so they need to be the exact same size so in other words if we have this right here so this is how it's defined a11 we'll do it simple two by two and we just add up each corresponding one so this is a one one a one two and a21 a22 so the second uh yes the number the first number is the row second one is the column so a one one a one two or first row single and multiple or added by we're going to add this to second Matrix b 1 1 b 1 2 b 2 1 b 2 2 like that all right so now this equals two and then uh yeah basically we're gonna add up each corresponding entry in other words this adds up with the corresponding one instead of the exact same size so this is going to be a one and this would be the same size both two by two so a one one plus uh B one one like that and the next one is going to be a12 adds up with B12 so yeah this one's much simpler than multiplication a12 plus B12 and then likewise here is going to be a 2 1 plus b 2 1 and then a two two plus b 2 2. like that all right now now that we've covered addition because we're 5. Rotation Matrix going to use this later on the video uh now let's take a look at rotation Matrix or rotation matrices so now let's get back to our earlier vector rotation and summarize what we have so if you scroll back up and take a look at here so this is what we have and then we wrote it down like this and then we solved for these values these x x Prime and Y prime those are the rotations so let's just summarize what we have and this the fundamental stuff that we covered and so what we basically had was a rotation so we had a this is X Y axis X Y this is the origin and let's say you have a vector here a and this has components right it's in a triangle bracket indicating it's a vector x y and then we rotated this and put it like this a bit longer yeah so we rotated it by an angle Theta rotated uh counterclockwise so we'll have the angle on the vector B with components X Prime y Prime and uh yeah let's write this here actually let's put it all all as complex as we can where X Prime is uh is just going to be X cosine Theta minus y sine Theta well actually it's yeah it's not gonna it's gonna run out of room there I'm just going to move it over here so I wrote it down actually here so here I just move it moved it down over here so this is the X Prime equal to x cosine Theta minus y sine thetas again as double check that so we have uh X Prime is equal to x cosine Theta minus y sine Theta and then the Y Prime is X sine theta plus y cosine Theta all right so we have that then we're going to get um y Prime is equal to X sine Theta and then becomes a plus y cosine the cosine switch cosine sine switch sides on in this case and then we're gonna have a Plus instead of minus all right so now that we have this we summarized we covered and now let's apply the Matrix stuff that we learned the multiplication and addition and so on but first let's uh try to convert our vector rotation into an equivalent Matrix form so here's our vectors and let's say what we have let's say we have our first Vector a a like that is equal to we'll put a um triangle bracket indicating it's a vector so you could write in this notation but let's write this as a matrix and this will write this as a basically a two by one the way this is an x y and put it like that so that's a two by one Matrix like that or one by two depending on which one you're counting first so it has two rows one column and for I just added added the clarity I equals to 2 m equals to one here and yeah well that's pretty obvious I'm just going to raise it so that's a matrix there for it and now the B vector B Vector is over here x Prime y Prime that is darker so this equals two and component form X Prime y Prime and then Matrix form this equals 2 x Prime y Prime all right so now what we have is uh well I'm going to write now we have this I'm going to take this and write this out or expand it and by itself I'm just going to put X Prime y Prime and plug in the actual values for them or the formulas that we have this equals 2 in a two bar in a uh yeah two by one Matrix this right here x cosine Theta minus y sine Theta so this is going to be x cosine Theta minus y sine Theta yes sine Theta like that and the next one is uh this is going to be X sine Theta for the Y Prime plus y cosine Theta like this all right and now what I'll do is well I'm going to put the X on the other side so we're gonna have a matrix um yes we'll have the cosines in the front I'm just going to switch them over so we'll write this as uh cosine Theta times x and then minus sine Theta times y and then this one's going to be uh sine Theta times x Plus cosine Theta times y we did that just because we're gonna have to have a matrix a bigger Matrix in the front the rotation Matrix a little straight that in a bit cosine all right so we have this and now what we can do is just break this down into a uh yeah a matrix multiplication so a notice right here today uh in this case right here we have the rows is going to be M uh we'll call this I here I is two and then it has n is equal to one because these are just a subtraction so but it's actually just one there and then this what happens is this equals two and we're going to need uh so if the number of row a number of of rows is two so we're going to multiply it let's go back to our setup here so we have a setup like this so we have this one has to match the first one so we're gonna have a two row and then the second one uh the number of columns must match the second one and then then this is going to have a row of the same here like that so in other words we have to have so we have to have I is 2. like that yeah let's put this like this I equals 2 and then this is n equals to one and we're gonna have to when we multiply it by you're gonna have to multiply this by something with um the same this is M we have to have m then we're gonna have to have n equals to one there so matches that so let's just start off with what we have here well it should be pretty straightforward we just look at this right here this is let's just take out the cosine Theta and this Y and a sign minus sine Theta we'll put that as we'll just look at two by two Matrix X and the next one sine Theta so yeah this is pretty straightforward like that this is going to be our m is equal to 2. yeah here are some of the things around to move this uh here actually I equals to 2 here so it matches up with this or just the same method we were using this is going to be m equals to two then you multiply this by well it has to have m equals to two and this is just our X Y vector and this is n equals to one so this matches this matches and this match yeah so that's pretty interesting they'll match up all right so yeah so that that is what has to happen and when we multiply this out well we obviously get this so we multiply the rows by this in other words put the X in here and then the Y and we just get this now this is erase that so yeah it's pretty straightforward yeah the next one is multiply this by this just basically matrix multiplication as we defined above and we're going to get the sine Theta cosine Theta yeah so this uh this X and this can be Y and that's this part uh in this part right there yeah so in other words what we have is erase everything whoops no actually not that yeah that's right so this is what we have and uh let's sum this up and write this out so thus we have we have the rotation vector rotation as a matrix multiplication so X Prime y Prime equals two cosine Theta let's try this better cosine Theta minus sine Theta and then we'll have a sine Theta then we have a cosine Theta like that and multiplied by x y now this is absolutely fascinating all right and now this is box this whole thing in so we have all right so now we have this and uh now let's go further so the above two by two Matrix this one right here the setup right there uh is called the rotation Matrix and which we denote by r or r as a function of theta since it rotates the vector a by Theta so in other words R the rotation Matrix is equal to well R Theta which equals two I'll just put oligon because yeah but but sometimes I'll use the r Theta when we put in a value for Theta it's going to be cosine Theta minus sine Theta then sine Theta here cosine Theta here all right now that we have this and we 6. Complex Numbers have basically rotation Matrix under our belt let's jump right into complex numbers and then we'll go over a relationship between the two so note that I had done a video on complex or imaginary numbers 11 years ago but it is a bit outdated and unpolished as it was one of my first videos on the channel you can see the difference 11 years makes so there it is like that and this is all in one one page I got yes very interesting stuff here so December 21st 2011 very very up I guess a lot upvotes there Paul bees for that anyways let's Jump Right In so let's start yeah so let's now start fresh here because this is just not as clear so a complex number is an extension of real numbers by using the imaginary unit or imaginary number unit I the imaginary unit I is a solution to the equation so if you have the equation x squared plus one equals to zero so real numbers can't solve this because uh Square when you squared you always get a positive number so in other words uh we move this over let's just move it over and then what we get is and then we're going to get I'll just move that over then we get uh x squared equals to negative one and then because you can't use a real number so we'll just say the numbers I so I squared is equal to negative one like that or you could Square both sides and we're going to get I is equal to well plus or minus square root of negative 1. maybe plus or minus because we're going to square root um there and square root of it just means you're multiplying yeah two numbers to get this Square there and then when you multiply two negatives those cancel up and becomes positive so similar to just the positive becomes positive and negative negative all right so that's plus or minus like that let's move this over here and uh yeah so uh because no real number satisfies the above equation x squared plus one equals zero I was called an end quote imaginary number and still let's call that in many places so uh complex numbers are written in the form yeah so that's that's imaginary number or imaginary unit uh num or number unit I so yes Wikipedia just says imagine your unit now for some reason so anyways I've also called the imaginary number so complex numbers are written in the form Zed equals a plus b i where A and B are real numbers and I is an imaginary unit the a term is considered the real part and the B uh and the B term this is considered the uh the imaginary part yeah so the real one is this a and the this is even though A and B are both real numbers it's just indicating that the A is for the real Part B is the Imaging a complex number can be visually represented as a pair of numbers A and B forming a vector on a diagram called an argand or a argand diagram where the imaginary axis I'll call that IM is the vertical axis and the real axis re is the horizontal here so in other words basically a complex number is just a two a two-part number instead of just one so let's just draw this out so if you have make this a bit bigger all right so we have this is the real part we'll call this re axis there's the imaginary i m and this is the origin right you have a vector here and this is a vector is a complex number so it's going to be called Z is equal to a plus b i where the components are a is the real part and then the vertical part is the imaginary RB and so it's not imaginary it's just uh saying that there's a second part to it so each number will have a second part of this number and an example of adding uh such a like a second part or property is for example let's say you had all the real numbers and then you assign a color to all of them so you could be like all the real numbers that are blue red Etc so instead of saying they have an imaginary uh property which they know they have color for example yeah so in this case you're just adding an another property and you're graphing it on this axis is called imaginary axis and so on all right so let's go further so now let's break down the complex number Z into its component parts via the angle Theta and only consider the unit length of one since we only want to consider rotation for now so we don't we don't want to deal with the length and so on we just care about the rotation Etc from this axis there so let's write this out all right so let's draw the imaginary and the reals the real is the imaginary I am and now we have a origin right there so let's say we have the unit length here's unit length is one just one to be easier to deal with and this is our Z equals 2 A plus b i all right and then this is the angle Theta so from the real axis and this is going to go down DOT like that dot dot and now this these components are this is the B this is the vertical so it's a vertical component and that just makes it with the well the hypotenuse there is one uh so we just do one times sine Theta R to sine Theta and then a is equal to well cosine Theta as B4 all right so that's we have that all right so that's for this unit uh length the complex number Z is equal to a plus b i uh or this equals to well cosine a is cosine Theta plus uh B is sine Theta I so in other words we have Z like that and box that in all right so now that we have the complex number here the unit complex number written such as uh using these components cosine Theta and sine Theta now we've gotten here we've gone over the Matrix and rotation Matrix so 7. Relating the Complex Plane with the Rotation Matrix now let's rotate or relate all of them all together so relating the complex plane which is basically the set of all the complex numbers over here uh but with the rotation Matrix so let's try to manipulate our previous rotation Matrix equation to try to get it in a form similar to the above complex equation so this is the complex number uh and now let's take a look at the r Matrix so here R equals to cosine Theta this is the Matrix cosines on the uh on the on these column or oh yeah on the forward diagonal and the signs on the on the opposite uh diagonal over there and the backward diagram we'll call that backward diagonal all right and then there's these negative one for this sign there so let's just write this down and try to break it down using our Matrix addition setup all right so we're going to have R equals 2 cosine theta's on the top and cosine Theta is on the far column and then we have a negative sine Theta just notice the pattern and sine Theta like that all right so what we could do now is we can manipulate this setup using our addition setup remember when we do Matrix addition we just scroll back above here symmetric condition basically uh these are just added terms so we can separate this big one into separate ones here and to do that what we'll do is uh notice the pattern here so we have the cosine cosine we want those in in its own Matrix and the sine and signs in their own Matrix so what we could do is write this out as cosine Theta and then this one's zero this one's zero this one cosine Theta in other words we have that and then what we're going to do it now is plus and then we're going to take only the signs now zero and this can be negative sine Theta and it's going to be sine Theta is going to be zero like that and we could see basically by a matrix addition well this would just get the above set up here because we would add all the corresponding terms so cosine theta plus zero is just cosine Theta and then zero plus negative signs that I just get this Etc so this is the same as this up top and now we'll do is we're just going to factor out the cosine because then yeah when you have a constant there multiplied inside a matrix you just put that in every term so we'll multiply or factor out this cosine Theta out of there and now we're going to be left with uh one zero zero one I'm going to multiply cosine inside we just get the cosine on the diagonal there and the next one is going to be sine Theta so take out the sine and we're going to be left with a 0 all right put this like this 0 negative 1 1 0. like that all right so now let's call this Matrix right here I'm going to call this I this is a this is called an identity Matrix whenever you have a diagonal forward one like this with all ones and everything else zero and I'll put the bracket on top just just so uh yeah just so it's more clear that it's a matrix and this one here I'm going to call this X because we don't know what this is yet let's put that uh as x with the bracket indicating it's a matrix and what we get is let's put R well R equals toast put this here in this notation uh Matrix like that because we have this cosine it's not a matrix and then I and then plus sine Theta X like this and yeah so this is what we have here I'm going to put this all the way here and I'm going to box this out all right so we have uh the rotation Matrix equals to cosine times this I Matrix called the identity Matrix and then plus sine Theta times this x Matrix right here we'll call that X like that because it's uh this one is identity just uh yeah because it's one across there this one has a negative one there that is interesting properties so let's go further all right and uh here yeah I just added node note that the I Matrix here or I bracket uh inside the bracket a matrix above is called the identity Matrix let's put this in bold so let's compare our above expression so this one here with that of the complex number or unit complex number it's a unit length one uh yeah complex number equation we derived earlier compare that with this setup here all right so that's Z equals cosine theta plus sine Theta I and this one's here so let's write this out I'm going to write this in this notation r uh Matrix equals to cosine Theta Times by the identity Matrix I plus sine Theta and then X as a big one let's put a big X like that the versus I'm going to put this as right here vs I'll put this here versus we put it here vs versus uh the complex number the unit complex number Z equals to cosine Theta is by itself so this is cosine Theta but I'm going to put a 1 I'm going to put Times by one this one times by I Times by one plus sine Theta Times by I like that all right so these look very very similar uh now let's compare the properties of well one and I so let's compare the properties of I and x with that of 1 and I the imaginary number or imaginary unit yeah so what I'll do is I'll look at this I and let's Square it so if we have I the I or bracket i i squared this equals 2 uh well this is number of the I the identity Matrix this is one uh one one on the diagonal zero everywhere else and let's Square this in other words we're going to do matrix multiplication so if we Square this all right here fix that up squared this just equals to well just matrix multiplication 0 1. times it by 1 0 0 1. all right so again matrix multiplication so we just multiply this the row with the um the column we go one times one plus zero times zero so in other words we get a 1. and the next one is and also these are the same size so the the resulting Matrix will be the same as well and the next thing we're going to do is well this is going to be a one next one is the the first row in the second column so one times zero plus zero times one in other words we get a zero and then we shift it over to here and this row so zero times one is zero plus one times zero zero and then the next one we get zero times one and one times one we get one all right in other words we get back to our identity Matrix so this equals to I so in other words we have I squared is equal to one I mean this I I squared is equal to I like that it's quite amazing versus now we'll compare it with vs um yeah compared this with the imaginary number I squared I mean not ice if the imaginary numbers is one I mean not the imaginary this one here compared with one so this is the identity Matrix let's compare the identity number or just number one one squared is equal to well it's one times one equals to one so these are these check out so I'll do a check so these work the same so now that's the first one and let's check x squared so let's do this uh x squared So This One X is a zero negative one one zero so let's do that x squared is equal to zero negative one one zero this equals 2 and then we'll be squaring it so this equals to zero negative one one zero multiplied by zero negative one one zero this equals two and here's move things around so if it fits now do the matrix multiplication so this row and this column so zero times zero is zero plus uh negative one times by one that's going to be negative one now the next one is um the first row second column like this so we go zero times negative one is zero negative one times zero is zero so those add up to zero and then the next one is over here over here and here and um no that's the column this row column one times zero is zero and then plus zero times one is zero and then the last one is here one times negative one is negative one plus zero times zero is zero so we get a negative one here yes in other words this this is just well negative I remember this is just one one so we put a negative in there just becomes negative 1 negative one yeah so this uh yes equals to negative I so in in other words x squared is equal to negative well a negative I or the negative identity Matrix like that all right yeah so notice the similarities and first I'm going to box this out and yeah so box that and now compare this with the well vs uh the I squared equals to well negative one yeah which is yeah let's put over here which is that imaginary number setup and this is I'm just gonna box this out like that and this checks out compare it out with this in other words this checks out so this uh and notice they're exactly the same this is the same property uh when you square this it's still one when you uh Square this one here you get negative negative uh yeah negative of the identity Matrix and this one is uh but yeah similar to this one when you square the eyes it becomes negative one so this is some interesting stuff here and this deserves a hashtag wow so I was actually blown away when I found some all right yeah so thus uh we could say that our rotation Matrix is equivalent to an unit complex number hashtag amazing so we can do is write this out as r yeah R equals two okay R equals to uh cosine Theta cosine Theta negative sine Theta sine Theta now here's fix that up and now this equals two we could write it in this format cosine of theta Times by the identity Matrix I'll put this in the bracket like that I plus sine and then the random X put an X Matrix and now switch over that X and put this as cosine Theta identity Matrix I Plus sine Theta and now recall this I'm going to put this as lowercase i in other words it's an imaginary number uh but as a matrix so which is just a this right here just this right here that's just I as a as a matrix form as a rotation Matrix all right so we have that in other words and now we know that in the complex plane so when you go back to the complex plane yeah this complex plane right here so number a is cosine B is sine so in other words uh this just equals to A uh identity Matrix plus b and then the I Matrix all right and I just want to make a note so this is typically used for the identity Matrix the capital I but another way it possibly is look at a cap just a one with a bracket inside but then that probably uh adds other confusion but just to match up with this here but nonetheless uh capital i as a good symmetry or pattern on its own because so Capital indicates the real number part and then the lowercase i is the imaginary one so anyways again this is for the unit complex number uh length and now you could write this again this is the uh this is the identity Matrix and there is the imaginary Matrix and then when you plug in those inside so then the cosine sine Etc this is cosine is just a sine is just B so we could write this as a matrix a negative B b a yes absolutely fasting stuff there's a whole bunch of ways of writing this Matrix rotation Matrix and relating it to equivalent form of the complex number as a rotation Matrix which is quite amazing all right so now that we have this relationship here let's continue further and now look at complex Complex Numbers as Vector Rotations numbers as rotations so the identity Matrix I um yeah I and the imaginary or the unit output unit here and the unit Matrix I can be a unit imaginary Matrix I can be viewed as rotations of zero degrees for the I the capital I and this is lowercase i and the 90 degrees Yeah and 90 degrees 4 lowercase I so remember we're dealing with a rotation Matrix so note that I'll recall that at Theta equal to zero we have cosine zero is this equal to one and sine zero is equal to zero and I'll illustrate this in a bit so thus what we have is R of theta equals to zero of a rotation Matrix this equals 2 well cosine zero negative sine zero sine zero I mean yeah sine zero this is cosine zero so in other words we get a uh yeah we get a one I'll put this actually lower let's put this right here this equals to well cosine zero is one and I assign 0 negative sine zero is zero zero one now this is just the identity Matrix like that and now just recall just uh quick recap recall trig functions or how they are so if you have if this is the angle Theta this is y and this is our sign so sine looks like that where this height is one this is negative one here all right and then this is at zero the origin zero there so that's for sine uh sine Theta so at sine at Theta is equal to zero we get zero so that's how it looks like and then if we draw in red it's going to look something like this red looks like this this is a cosine one all right so this is cosine here let's put the arrow there this is cosine cosine Theta and notice how the value is one at the origin so we have one there so it looks like that and there's the one one yeah so that's we have all right so let's continue further so this is here for the identity Matrix now let's look at the imaginary Matrix and uh this one is at uh at 90 degrees turn so at Theta is equal to zero uh and then this one now if we look further at theta equals to 90 degrees which is just equal to pi over 2 radians so the different units there radians uh what we get is cosine yeah so we get cosine pi over 2 yeah cosine pi over two this equals to zero that's this value right here this is at pi over 2. so that's cosine pi over 2 and then the sine becomes 1. and then sine pi over 2 equals to 1. all right so thus if we plug this into our rotation Matrix formula so we get thus yeah so thus R of for this here R of theta equals to pi over two and this equals two yeah so this is theta equals zero is pi over two we're doing it dealing with radians and this is going to be cosine pi over 2 negative sine pi over two sine pi over 2 and then cosine pi over 2 this equals two plug this in we get 0 negative one zero I mean not zero sine is one zero yeah and this is again that's just our imaginary Matrix I'll put it like that you put it like that or just leave it as I because we're dealing with Matrix sees it so it's pretty straightforward here yeah so we're dealing with that just similar here I should put the without the uh just put the I capital I without the brackets because it's pretty straightforward uh but yeah might as well just box these in let's box this in because it's uh we just sold that out and this equals to and solve this um all right going further so now we have these as angles or just rotations there because we plug in the rotation angle inside this rotation Matrix and we get our Identity or imaginary Matrix so equating unit again now top of the unit there because it's length one unit complex numbers to rotation matrices effectively means that we can interpret complex numbers or unit complex numbers because uh because if it's not the same length I think there's gonna be some stretching involved and so on so unit complex numbers as applying a rotation to another Vector so yeah so we're going to be if you look at a vector here say we have X as the origin the Y like that and let's say we have this Vector a like that's a vector and and then we're going to have and then you rotate it by they don't start to make it exactly the same rotated by uh let's make this a bit longer one more try all right that's perfect all right it's rotated by Theta so this Vector B is equal to the rotation Matrix R uh we'll put this Theta there just because we're using that exact angle and then put that in a bracket so it indicates the Matrix Times by a so you rotate it across so that's you could view that as complex numbers all right so this is a generic angle Theta but if we want to rotate by zero that's just multiplying it by one or uh I so this means that put it like this make it bigger here move it down here and we're gonna put these two together so we get this is the origin this is the Y this is the X like that and now this is the this is a vector and this is vectors A and this angle here uh Theta is equal to zero so let's say we didn't wrote it or rotate it by zero so the angle is zero degrees or just a zero so zero degrees of radians so it's all zero so this is going to be equal to this is going to be R of zero uh which equals two there are zero times it by or and we know R of 0 that's just I this just becomes the I capital I so the identity Matrix times it by a like that and this equals to B like that so it's the same thing so nothing changed this equals to itself and versus now let's move this uh here yeah versus shift that over or shifted that over let's move this here and this is going to be our new axis this is our X this is our Y and now we're dealing with well if you have a like this I'll make it a let's make it bigger so that's a and now we're going to be rotating this by 90 degrees so this is our new a like that let's say let's make this a bit more steep all right so this would be very very similar over here just fix that up it's uh somewhat like it and this angle this is 90 degrees I put a right angle there I'll make this bit better right angle all right so that's a right angle and this is theta equals to 90 degrees and this one is zero degrees let's put the yes put that there it's for uh completeness all right so now that we have B this equals 2 the unit identity Matrix times it by a so yeah so this Vector times about that we get that so that is it that's all it is then yeah so the complex number as a rotation Matrix is just shifting this by 90 degrees counterclockwise and then the identity Matrix just doesn't do anything just leaves it as is all right so now that we have this and let's go further so note that if we rotate by a full 360 degrees the degrees or two Pi radians we get back to where we started so let's uh illustrate that so we'll have let's say we have it here and if we have this um make this more horizontal x y and we have a vector a this a right here uh and now we start off here so we go by a full 360. so we go 360 degrees like that this equals two R and then this is going to be 2 pi radians put it like that 2 pi times it by a uh yeah the Matrix comes by a vector this equals this has to equal itself this has to be just the identity Matrix Times by a which again this is our B so they all they all equal each other and we can actually see this further by Computing this or calculating it out so a note uh note that R of 2 pi is equal to I'll put in our identity Matrix I mean our rotation Matrix we get cosine 2 pi negative sine 2 pi and then sine 2 pi then cosine 2 pi this equals two this equals two let's put it over here let's let's go back up to our graph here so at 2 pi should be somewhere across uh this is a bit uh outstretched this should be right here actually two pies there and and here just quickly shift this over let's fix this up this is a 2 pi so 2 pi cosine I'm assign 0 and this one is going to be at one Z1 there all right yeah so then this is yeah cosine 2 pi equals one and this is a sine 2 pi equals zero like that all right let's continue back here and plug that in we get a zero optional it's a one one uh sine of zero cosine is one one zero zero one and this just equals to the identity Matrix and now here is boxed that and and uh so note that and also let's just do the we haven't done this calculation actually but it's pretty straightforward that's why it's called identity Matrix so if you have this right here I times a so yeah so if you have let's say we have I times a like that and let's say this equals 2 we have our 1 0 0 1 and let's say a has uh components x y like that so we're multiplying by the identity so then this row Times by this column so x times it uh this equals 2 well x times one I'm going to x dot yeah 1 times x is just X because then zero plus zero times one and this is going to be um yeah this is going to have the same rows as here this is I is 2 is I is 2 Etc and then it has n is one here and there's one n is one that's so I'm just going to erase these so that's straightforward and then the next one is only one column so zero times x is zero plus one one times Y is y in other words nothing changes and this just equals to A so multiplying by the identity Matrix just gets a equals all this and this equals a so nothing changed Reinterpreting Euler's Formula all right so now we've covered all this and now let's do some interesting stuff with what we've learned so reinterpreting Euler's formula so now let's consider the famous Euler's formula by Leonard Euler and reinterpret it as a two by two rotation Matrix so this is the famous Formula E to the i x is equal to cosine X plus I times sine X and you could prove this using power series for uh for e to the X plug in i x and then separate them you get cosine X sine X and so on I may do that I'm not sure if I done it or not in my videos but it's pretty straightforward anyway so this is the famous formula let's box this in or and now write it over vs write it in the different format e to the i x but we don't do a circle e to the i x equals 2 in notation format I'm in a two by two Matrix format as a two by two this is a number this is all it is cosine X and then there's I sine X in other words that's imaginary component this is the complex number that's the real part and there's a is exactly that's exactly the complex number and here I scroll all the way back up so Z equals a plus b i yeah uh but uh yeah the Wikipedia just has the i in front but they use that notation but anyways this is just a rotation Matrix cosine x negative sine X and this is going to be sine X and this is going to be cosine X like that so that's fascinating stop all right and now let's take a look at Euler's identity I'll just write this formula bolt this one so that's a bolded so when when X is equals to Pi we get Euler's identity and Euler's identity well let's write this here so at x equals to Pi this is radians equals to the same thing as saying 180 degrees and yet degrees so what we get is uh for cosine Pi this equals to negative one then sine of pi equals to zero and we could see this from our trusty uh setup over here so there's sine and this is pi right there so sine is zero sine pi equals zero and then this one here is at the bottom there that's cosine cosine pi equals negative one all right so we have those values and now we can throw them inside our rotation Matrix and we get uh e to the I pi equals 2 cosine well actually before the rotation meter this is just the uh mainstream or just the standard derivation we're just using this right here for now [Music] all right so e to the i e to the i i Pi where X is pi equals to cosine pi plus I sine Pi so in other words this vanishes at zero so this just goes to we have zero I'll just put this here this equals to uh one I mean negative one plus zero and this is interpreted as well we could move this over yeah so what we get is basically e to the I pi equals to negative one or you can move this over to the other side and this is how it's typically written e to the I Pi or actually here I just erase that move this over here so move it here and just write e to I Pi so move the plot of the one over here so it would be plus one e to the I pi plus 1 equals to zero and it's it's uh famous because it has all of the interesting Parts there it has an i it has pi it has e it has one and a zero all right so a lot of interesting stuff here all right in here let's move the arrow on top so it's easier and we could box this in all right so that's what we have either I pi plus y equals zero so it looks elegant and and so on but I think the two by two Matrix is uh more Talent all right so that's what we have and also put this vs or actually instead of saying versus run you're going against anyone else with or write it like this so or and two by two Matrix format or uh yeah in the um in the rotation Matrix format we get or we could write this as e to the I pi equals to cosine pi times one uh first let's write it down like that plus is write the notation we're used to sine pi times I so just rearrange this put it here put the 1 there and now this equals to negative one yeah negative 1 times um yeah this is cosine Pi is negative one sine Pi is negative one so negative one times one plus the next one is zero times I so you could write it like this so that you're talking about complex numbers and then equals 2 and I switch it over into Matrix format identity Matrix plus zero times it by uh or just race the Times by same thing oh actually I just think that zero Times by I Matrix like that so this is equal or equivalent to let's put it just equal to so we're switching it over and now this equals two and uh what this equals to this vanishes so all we're left with is a negative of the identity Matrix so in other words we're going to have a negative one zero zero negative one like that and again plug this into the identity Matrix format or I mean uh into the rotation Matrix remember this was just the cosine Pi which equals to negative one negative sine pi to zero where we have sine pi and then cosine pi that's what we have and so in other words we could write e to the I pi equals to negative identity Matrix there's a few if you erase all this just uh just across it all out if you want but I think when you cross out stuff like this you're removing information so in this case removing that it's uh you're dealing with complex numbers in other words a two-dimensional number you're at you have multiple properties of it so just a basic number so this is like that and if you shift this over to the other side I'll put it on the top Arrow move it over here not we're not putting it into the exponents putting it here all right let's fix that let's move it over to this side we get uh the famous equation and so e to the i i pi equals negative uh this is I Matrix or uh put it here e to the I pi plus identity Matrix equals to zero like that absolutely fascinating stuff like that yeah and just for completeness let's put the negative side as well because it's interesting stuff here so equals to uh negative I um yeah negative the identity Matrix like that and uh similar to here or e to the I pi equals to negative one like that yeah so in other words uh e to the I Pi is just a rotation Matrix It rotates a vector by 180 degrees so let's take a basic Vector like here so if you have uh set up here and here's fixed up this is the x is the Y and uh you have let's say you have to draw this a thick line like this let's say you have an arrow that's on the axis this is let's say you have the vector a like that and uh here's the origin and now it starts off from the origin goes to here and now you're going to rotate this by 180 degrees 180 degrees is just equals to Pi radians foreign so you're going to get a vector across the here I'm going to cross it like that oops same thing here so basically you're going to go negative on it and this is going to be negative a like that's negative a and this just equals to negative identity Matrix of a I'll put this higher it's going to be negative a right like that which equals to negative identity Matrix of a yes fascinating stuff so that's on the axis to make it a bit more visible I'll put this like this all right so let's say we have an angle like our a vector like that x y and we're going to rotate this it's in other words we'll just go backwards hence that negative one it just means you're going backwards on this vector so you're going backwards uh this is 180. this is yeah going going backwards by pi and uh I'll just put 180. 180 degrees and then this is if that's a this is going to be well a negative a which equals 2 negative uh the bracket I a vector I got so yes that is and there's the origin zero so this is actually absolutely fascinating stuff and uh here I just move things around so that it all fits nicely together horizontally move this up over here and then finally yeah this one just equals to again this is uh it's like you're yes when you have this negative uh a uh this a vector and this is a equals to negative the identity Matrix Times by this Times by a vector over here and you're rotating 180 degrees it's uh again this is since they're equal you're just multiplying it by either the e to the I pi times the by the vector so we multiply that Vector out it's like shifting it over to the Sun and basically you're multiplying it by one yeah it's a negative I so if you have a components like this is uh let's say you had a one or let's say you had components a and B now you're going to go backwards now you're gonna go negative B or I'll actually this one back a B you're going to have negative B negative a you're just going backwards it's fascinating faster and stuff all Applications to Other Fields right uh now we're pretty much uh done the video just the last section right here so let's just keep going further so uh application to other fields uh so since complex numbers are applied uh to many different fields in mathematics and physics including electromagnetism and quantum mechanics uh reinterpreting such equations as rotational matrices May provide unique Insight so yes try investigating yourself and let me know what you discover so the same way that we changed up Euler's formula into a rotation Matrix is basically a 180 degree rotation uh so yeah so let me know what you discover for especially electromagnetism and so on so uh finally uh we cover 2D rotational matrices by the concept was similar when dealing with 3D which is something I may cover in future videos so stay tuned anyways that is all for today hopefully you enjoyed this uh this is pretty epic videos a lot longer than I expected but uh pretty in-depth um just to illustrate the the basically I also have to cover matrices and so on and yeah it's very interesting how you get view complex numbers as rotational matrices anyways that's all for today hopefully you learned and uh yeah you can download these notes these exact notes and Link the description below as well as viewing them as an article on the hive blockchain and the notes will be also in the uh the article link will also be in the description below anyways thanks for watching stay tuned for another math easy solution |
6716 | https://www.freecodecamp.org/news/euclidean-algorithm-in-golang/ | / #algorithms
How the Euclidean Algorithm Works – with Code Examples in Go
By Otavio Ehrenberger
The Euclidean Algorithm is a well-known and efficient method for finding the greatest common divisor (GCD) of two integers. The GCD is the largest number that can divide both integers without leaving a remainder.
The algorithm is named after the ancient Greek mathematician Euclid, who presented it in his book "Elements" around 300 BCE.
You can use this algorithm to solve Diophantine equations, to tackle the shortest-vector problem which is the foundation of lattice-based cryptography, and also to detect common patterns of pixels in images. This is, among other things, applied to optimize rendering processes and detect different objects in images.
How Does the Euclidean Algorithm Work?
Here's a step-by-step explanation of how the Euclidean Algorithm works:
Start with two positive integers, a and b, where a >= b. If a < b, simply swap their values. Note that this is meant for a convenient mathematical demonstration, as the implementation also works for a < b.
Divide a by b and find the remainder, r (use the modulo operation, represented as a % b). If r is 0, the GCD is b, and the algorithm terminates.
If r is not 0, set a to b and b to r. Then, repeat step 2.
The algorithm continues to iterate until the remainder is 0. At that point, the last non-zero remainder is the GCD of the original two numbers.
The Euclidean Algorithm works because the GCD of two numbers remains unchanged when the larger number is replaced by its remainder when divided by the smaller number.
Example of Euclidean Algorithm
Here's an example to illustrate the algorithm:
Let's find the GCD of 30 and 9:
a = 30, b = 9
Calculate the remainder: r = a % b = 30 % 9 = 3 (since 3 is not 0, continue to step 3)
Update the values: a = 9, b = 3
Calculate the new remainder: r = a % b = 9 % 3 = 0 (r is now 0)
The GCD of 30 and 9 is 3.
Why Does the Euclidean Algorithm Work?
The greatest common divisor of two integers is the largest positive integer that divides both of them without leaving a remainder. So the algorithm is based on the following key property:
If a and b are two integers, then the GCD of a and b is the same as the GCD of b and a % b, where % represents the modulo operator (the remainder after division).
Mathematically, the key property of the algorithm can be justified using the division algorithm:
Let a and b be two positive integers, such that a >= b. We can write the division algorithm as:
a = bq + r, where q is the quotient and r is the remainder.
Now, let d be a common divisor of a and b. Then, a = d m1 and b = d m2 for some integers m1 and m2. We can rewrite the division algorithm as:
d m1 = (d m2) q + r.
Rearranging the equation, we get:
r = d (m1 - m2 q).
Since d is a factor of both a and b, and r can also be written as a multiple of d, we can conclude that d is also a divisor of r. This means that the GCD of a and b is also a divisor of r. So, we can replace b with r and keep finding the GCD using this algorithm until b becomes 0.
The Euclidean Algorithm is particularly useful due to its efficiency and simplicity, making it easy to implement in computer algorithms and programming languages.
Let's see some different ways to implement it in Go:
Recursive Implementation of the Euclidean Algorithm in Go
This implementation of the Euclidean Algorithm in Golang is a recursive version that finds the GCD of two integers. Let's go through it step by step:
The function is defined as GCD(a, b int) int. It takes two integer inputs, a and b, and returns an integer output.
The base case of the recursion is checked with if b == 0. If b is 0, the function returns the value of a as the GCD.
If b is not 0, a temporary variable tmp is created and assigned the value of a. This temporary variable is used to store the value of a before updating its value in the next step.
The values of a and b are updated as follows:
a is assigned the current value of b.
b is assigned the value of the remainder when tmp (the previous value of a) is divided by the new value of a (which was b before the update).
The function calls itself recursively with the updated values of a and b as input, return GCD(a, b).
The algorithm continues to call itself recursively until the base case is reached, that is b becomes 0. At this point, the function returns the GCD, which is the value of a.
// Recursive approach:func GCD(a, b int) int func GCD(a, b int) int int if 0 return return
For example, let's say we want to find the GCD of 56 and 48:
First call: GCD(56, 48)
Since b (48) is not 0, update a and b:
a becomes 48
b becomes 56 % 48 = 8
The function calls itself with the new values: GCD(48, 8)
Second call: GCD(48, 8)
Since b (8) is not 0, update a and b:
a becomes 8
b becomes 48 % 8 = 0
The function calls itself with the new values: GCD(8, 0)
Third call: GCD(8, 0)
Now, b (0) is 0, so the function returns a (8) as the GCD.
Iterative Implementation of the Euclidean Algorithm in Go
This implementation of the Euclidean Algorithm in Golang is an iterative version using a loop to find the GCD of two integers. Let's go through the code step by step:
The function is defined as GCD(a, b int) int. It takes two integer inputs, a and b, and returns an integer output.
A loop is used to iterate as long as b is not equal to 0. The loop condition is b != 0. Note that this for loop construction in Go is essentially a while loop in many other languages.
Inside the loop, the values of a and b are updated simultaneously using a tuple assignment: a, b = b, a%b. This line does the following:
a is assigned the current value of b.
b is assigned the value of the remainder when a is divided by b.
When the loop exits (that is, b becomes 0), the value of a is returned as the GCD.
The algorithm iterates until the remainder (b) is 0, at which point the GCD is the last non-zero remainder, which is the value of a.
func GCD(a, b int) int func GCD(a, b int) int int for 0 return
For example, let's say we want to find the GCD of 100 and 64:
Initialize a as 100 and b as 64. Check the loop condition: b (64) is not 0.
Inside the loop, update a and b:
a becomes 64
b becomes 100 % 64 = 36Check the loop condition again: b (36) is not 0.
Inside the loop, update a and b:
a becomes 36
b becomes 64 % 36 = 28Check the loop condition again: b (28) is not 0.
Inside the loop, update a and b:
a becomes 28
b becomes 36 % 28 = 8Check the loop condition again: b (8) is not 0.
Inside the loop, update a and b:
a becomes 8
b becomes 28 % 8 = 4Check the loop condition again: b (4) is not 0.
Inside the loop, update a and b:
a becomes 4
b becomes 8 % 4 = 0Check the loop condition again: Now, b (0) is 0, so the loop exits.
The function returns the value of a (4) as the GCD.
Testing the Solutions
These tests were created by Jon Calhoun in his free Go Algorithms course. Assuming you defined your GCD function in a file named gcd.go, place these tests in a file named gcd_test.go. Read below to see how the tests will work:
The function TestGCD is defined, taking a single parameter t testing.T. The testing.T is a pointer to a testing.T object that provides methods for reporting test failures and logging additional information.
An array of anonymous structs is defined, called tests. Each struct has three fields: a, b, and want. These structs represent test cases, where a and b are input values for the GCD function, and want is the expected result (correct GCD).
Several test cases are defined in the tests array, covering different scenarios.
A for loop iterates through the tests array. In each iteration, a single test case (struct) is assigned to the variable tc.
The t.Run function is called to run a subtest for the current test case. The first argument is a formatted string that describes the test case (using the input values tc.a and tc.b). The second argument is an anonymous function that takes a testing.T parameter, similar to the main test function.
Inside the subtest function, the GCD function is called with the input values tc.a and tc.b, and the result is assigned to the got variable.
The got result is compared to the expected result tc.want. If they are not equal, the t.Fatalf function is called to report the test failure and provide an error message with the incorrect result and the expected result.
This test function helps ensure that the GCD function works correctly for various input values and edge cases. Running this test function with the go test command will execute all the test cases and report any failures, which can help identify potential issues with the GCD function implementation.
import "fmt" "testing"func TestGCD(t testing.T) func TestGCD(t testing.T) struct int int 10 5 5 25 5 5 30 15 15 30 9 3 100 9 1 2 2 3 3 5 2 3 5 7 13 2 3 5 2 2 3 3 13 2 3 5 7 13 2 3 13 2 3 5 7 11 13 17 19 3 3 7 7 11 11 17 17 3 7 11 17 for range"(%v,%v)"func(t testing.T) func(t testing.T) if"GCD() = %v; want %v"
Run the tests, creating different cases as you wish:
test # verbose flag
Conclusion
Hope you had a good time learning about the Euclidean algorithm and its implementation in Go.
If you are interested in how these clever tricks work, take a look at the field of Number Theory in mathematics, which is particularly focused on the properties of integers, especially involving prime numbers.
If you read this far, thank the author to show them you care.
Learn to code for free. freeCodeCamp's open source curriculum has helped more than 40,000 people get jobs as developers. Get started |
6717 | https://books.google.com/books/about/Stell_Maran_s_Textbook_of_Head_and_Neck.html?id=O2zNBQAAQBAJ | Stell & Maran's Textbook of Head and Neck Surgery and Oncology - John Watkinson, Ralph Gilbert - Google Books
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Stell & Maran's Textbook of Head and Neck Surgery and Oncology ============================================================== John Watkinson, Ralph Gilbert CRC Press, Dec 30, 2011 - Medical - 1184 pages First written by Philip Stell and Arnold Maran in 1972, Stell & Maran's Textbook of Head and Neck Surgery and Oncology has been revised in both content and approach over the years to reflect the enormous progress made in the area. Now in its fifth edition, the book remains a key textbook for trainees in otolaryngology and head and neck surgery. More » Preview this book »
Selected pages
Title Page
Table of Contents
Index
References
Contents
1 The history of head and neck surgery 1
INTRODUCTION TO HEAD AND NECK SURGERY 7
BENIGN DISEASE 215
ENDOCRINE DISEASE 307
MALIGNANT DISEASE 545
RECONSTRUCTION 905
DIRECTIONS FOR FUTURE RESEARCH AND DEVELOPMENTS IN HEAD AND NECK CANCER SURGERY 1097
Index1113
Copyright
Other editions - View all
Stell and Maran's Textbook of Head and Neck Surgery and Oncology
John C. Watkinson,Ralph W. Gilbert
No preview available - 2012
Stell and Maran's Textbook of Head and Neck Surgery and Oncology
Philip M. Stell
No preview available - 2012
Common terms and phrases
adenomaairwayarteryassessmentassociatedbenignbiopsybonecalcitonincarotidcartilagecent of patientscervicalClinical EndocrinologycystsdiagnosisdiseaseendocrineEndocrinology and MetabolismendoscopicevaluationexcisionfactorsFigureflapFNACfollicularfunctiongenegeneticgoitrehead and neckhistologicalHNSCChormonehyperparathyroidismhyperthyroidismhypocalcaemiaimagingincreasedinjuryinvasioninvolvementiodineJournal of ClinicalJournal of SurgerylaryngectomyLaryngoscopelesionslevelslymph nodesmalignantmandiblemedullary thyroid cancermedullary thyroid carcinomametastasesmucosalmusclemutationsneck cancerneck dissectionNeck SurgeryneoplasmsnormaloncocyticOncologyoral cavityOtolaryngologyoutcomeparagangliomasparathyroid glandspartial laryngectomyPathologypituitarypostoperativepreoperativeprognosticradiationradioiodineradiologicalradiotherapyreceptorreconstructionrecurrent laryngeal nerveresectionrisksalivary glandscanserumsoft tissuesquamous cell carcinomastagesurgeonsurgicalsymptomssyndrometechniquetherapythyroid glandthyroid nodulesthyroidectomytracheatracheostomytraumatreatmenttumourultrasoundusuallyvascular
About the author(2011)
John C Watkinson MSc MS FRCS DLO
Consultant Head and Neck and Thyroid Surgeon, Queen Elizabeth Hospital, University of Birmingham NHS Trust. One-time Hunterian Professor, Royal College of Surgeons of England, UK
Ralph W Gilbert MD FRCSC
Deputy Chief, Otolaryngology-Head and Neck Surgery, University Health Network; and Professor, Department of Otolaryngology-Head and Neck Surgery, University of Toronto, Ontario, Canada
Bibliographic information
Title Stell & Maran's Textbook of Head and Neck Surgery and Oncology
AuthorsJohn Watkinson, Ralph Gilbert
Edition 5, illustrated
Publisher CRC Press, 2011
ISBN 1444128752, 9781444128758
Length 1184 pages
SubjectsMedical
›
Oncology
›
General
Medical / General
Medical / Oncology / General
Medical / Otolaryngology
Medical / Surgery / General
Export CitationBiBTeXEndNoteRefMan
About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home |
6718 | https://artofproblemsolving.com/wiki/index.php/Titu%27s_Lemma?srsltid=AfmBOopMfujmE2ex-qANdRGJlDc-owOWsoqjvOt4rePHxpo3KphPN0vs | Art of Problem Solving
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Titu's Lemma
Titu's lemma states that:
It is a direct consequence of Cauchy-Schwarz inequality.
Equality holds when for .
Titu's lemma is named after Titu Andreescu and is also known as T2 lemma, Engel's form, or Sedrakyan's inequality.
Contents
[hide]
1 Examples
1.1 Example 1
1.1.1 Solution
1.2 Example 2
1.2.1 Solution
1.3 Example 3
1.3.1 Solution
2 Problems
2.1 Introductory
2.2 Intermediate
2.3 Olympiad
Examples
Example 1
Given that positive reals , , and are subject to , find the minimum value of . (Source: cxsmi)
Solution
This is a somewhat standard application of Titu's lemma. Notice that When solving problems with Titu's lemma, the goal is to get perfect squares in the numerator. Now, we can apply the lemma.
Example 2
Prove Nesbitt's Inequality.
Solution
For reference, Nesbitt's Inequality states that for positive reals , , and , We rewrite as follows. This is the application of Titu's lemma. This step follows from .
Example 3
Let , , , , , , , be positive real numbers such that . Show that (Source)
Solution
By Titu's Lemma,
Problems
Introductory
There exists a smallest possible integer such that for all real sequences . Find the sum of the digits of . (Source)
Intermediate
Prove that, for all positive real numbers (Source)
Olympiad
Let be positive real numbers such that . Prove that (Source)
Let be positive real numbers such that . Prove that
(Source)
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6719 | https://www.youtube.com/watch?v=R21YsHUr4Iw | When to Use Absolute Value to Simplify Radicals
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In this math video we simplify radicals with variables in the radicand showing our answers in absolute value form. Simplifying is defined as writing entire radicals as mixed radicals. Absolute value brackets are used only for even roots. Square roots and cube roots are shown. Math tutoring on radicals using examples converting between mixed and entire radicals. Vocabulary includes radicals, radicands, index numbers, simplifying radicals, ordering radicals in ascending and descending order.
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Intro welcome back to my channel today we're going to take a look at radicals and absolute value in the answer if you're taking an enriched math course or more challenging one your teacher may require you to know this or sometimes teachers just sort of skip over this because it can be a little bit confusing to get into but let's go through sort of the motivation for absolute value making sure that you know absolute value is a set of brackets they're vertical bars and we'll see what that means in a second Radicals okay so radicals here is a radical it looks ugly it's got a power on the inside m and it's got a nth root on the outside and what we have to be be careful with here is when you take for instance if i take the square root of a number say 9 then the idea here is is that the square root is actually the principal square root or the positive square root in fact if you put that into the calculator it's not going to tell you negative 3. it's going to tell you positive 3. so the answer is positive okay and that has to do with the definition of that square root symbol on its own it means what's the positive square root what's the number that when multiplied by itself will give the number under the radical the nine now we know that negative three times negative three is also 9 but it doesn't want the negative case it wants the positive case okay so this is true whenever the index here is even now in this case it's square root but this what i'm about to tell you is any case where the index is even okay where n is even um where n is even if n is not even then what we're about to talk about doesn't hold okay it has to be even um Examples and so the idea here is that when n is even the answer the answer that we place has to be positive because it wants the principle and through it or the positive nth root and again this is only true when n is even 2 4 6 8 and so on so let's take a look at some examples that make this point now here's the square root of 9 x squared so when we do this most of you are going to know that you go through and you say what's the square root of 9 and you say the answer is 3 not negative 3 because we have to ensure that the answer is positive okay and then we say the square root of x squared well look the index is 2. so what i could do here is say the answer is x this here gets divided by that so 2 over 2 or the square root of x squared is x right and you would then say that the answer is 3 x x to the 1. now you have to be careful here because remember this has to balance for any possible input on x if i was to put a square box around the x call it the xbox if i was to type in a number here that's negative it will take the negative value because look what's happening you're squaring the number before you square root it so this actually accepts negative values now the so in other words say x is equal to negative 2. well negative 2 is negative 2 squared which is 4. 4 times 9 is 36 the square root of 36 is 6. so this original question not simplified this original blue question um works for negative 2. now if i put negative 2 into here i get an answer of negative 6. well that's not balancing right so this has to be equal to this not only that but we want the positive answer so in order to ensure that we have a positive answer for x for cases where x is negative we have to put an absolute value set of brackets around the x this only is true when the exponent on the answer is odd okay so 1 3 5 and so on let's take a look at this next one here if i was to now this is square root land again so to build my answer which i want to be positive right because i'm asking for the principal square root that's going to be x 4 over 2. so 4 divided by 2 and then the y times y 6 divided by 2. okay now break this down so that's going to be equal to well x squared x squared and then the y to the 6 over 2 is going to be y to the 3. now just double check make sure that the answer is going to be positive the this here is okay x is okay because look x squared the square is going to ensure that the answer stays positive but look here y is being cubed and so the cube doesn't guarantee that the answer stays positive so in this case you have to put absolute value around the y you can put it around the x but it's redundant it's not needed let's take a look at the next one here so we get into a little bit of a roll here check here notice now the outside index is three this rule is only applicable when n is even so in this case here we don't use absolute value brackets you don't need them for odd indices so here the answer would be x 6 divided by 3 which is 2 and then y 3 divided by 3 which is 1. now even though y is being raised to an odd exponent you don't want to put the absolute value here because when you take the cube root of a number you can get a negative value so we don't try to ensure positiveness in our answer for uh for indices that are odd okay so you leave it just like that and now the last one here this one looks like it could be important to consider because look the index is four so if the index is four again n is even four is even so we we tread carefully here so x four over four is one and y to the 8 8 divided by 4 is 2. so in this case we check our exponents in the answer and if either of them are odd and in this case it is see that's a 1 then you have to install your absolute value brackets okay i don't need to worry about the y because the square on the y is going to ensure that the answer stays positive remember when the index is even you want a positive answer when the index is odd the answer can be negative right don't forget the cube root of negative eight is equal to negative two so with with um indices that are are not even you can get a negative answer all right i hope that helps leave a like on this video if it did help you and don't forget to subscribe and i'll see you right back here in the next video |
6720 | https://faculty.fiu.edu/~meziani/Lecture11.pdf | Real Analysis MAA 6616 Lecture 11 Egorov and Luzin Theorems Egorov’s Theorem Egorov’s Theorem states that if a sequence of measurable functions converges pointwise a.e. on a set of finite measure to a function that is a.e. finite, then it converges uniformly except on a subset with arbitrarily small measure.
Start with an example. Consider the sequence of piecewise linear functions {fn} defined on [0, 2] as follows: ▶For n even: fn(0) = fn( 1 n) = fn(1 −1 2n ) = fn(1 + 1 2n) = fn(2 −1 n ) = fn(2) = 0; fn( 1 2n ) = fn(2 −1 2n) = 1 and f(1) = 1 −1 n and fn is the linear function connecting any two consecutive points so that f(x) = 2nx for x ∈[0, 1/2n]; ▶For n odd: fn(0) = fn( 1 n) = fn(1 −1 2n ) = fn(1 + 1 2n) = fn(2 −1 n ) = fn(2) = 0; fn( 1 2n) = fn(2 −1 2n ) = −1 and f(1) = −1 + 1 n and fn is the linear function connecting any two consecutive points so that f(x) = −2nx for x ∈[0, 1/2n].
The sequence {fn} converges pointwise on [0, 1) ∪(1, 2] to the function f = 0 and the sequence diverges for x = 1.
Indeed f2j(1) = 1 −(1/(2j)) converges to 1 while f2j+1(1) = −1 + (1/(2j + 1)) converges to −1. If x = 0 or x = 2, then fn(x) = 0 and the sequence converges to 0. If x ̸= 0, 1, 2, let δ = min(|x| , |1 −x| , |2 −x|). Let N ∈N such that N ≥1 δ . Then for every n ≥N we have fn(x) = 0 and fn(x) converges to 0.
Now for any ϵ > 0 (no matter how small), let N ≥1 4ϵ , then {fn} converges uniformly to f = 0 on the set [ ϵ 4 , 1 −ϵ 4] ∪[1 + ϵ 4, 2 −ϵ 4 ] with measure 2 −ϵ.
Lemma (1) Let {fn} be a sequence of measurable functions on a measurable set E ⊂Rq with finite measure. Assume that {fn} converges pointwise a.e. on E to a function f such that f is finite a.e.
on E. Then for every ϵ > 0 and η > 0, there exists a measurable set A ⊂E and an integer N ∈N such that m(E\A) < η and |fn(x) −f(x)| < ϵ for all x ∈A and for all n > N.
Proof.
Let Z1 be the subset of E where f is not finite and let Z2 be the subset of E where {fn} fails to converge to f. Let Z = Z1 ∪Z2.
Then m(Z) = 0 by hypothesis. For every j ∈N, let Aj = {x ∈E\Z : |f(x) −fk(x)| < ϵ for all k ≥j}. The set Aj is measurable since it can be expressed as a countable intersection of measurable sets: Aj = \ k≥j {|f −fk| < ϵ} ∩(E\Z).
Note that Aj ⊂Aj+1 and S j≥1 Aj = E\Z (since fn − →f on E\Z). It follows from the continuity of the Lebesgue measure that lim n→∞m(An) = m(E\Z) = m(E). Therefore for the given η > 0 there exists N ∈N such that m(E\AN) < η and for every x ∈A = AN and for every n ≥N we have |fn(x) −f(x)| < ϵ.
Egorov’s Theorem Theorem (1) Let {fn} be a sequence of measurable functions on a measurable set E ⊂Rq with finite measure. Assume that {fn} converge pointwise a.e. on E to a function f such that f is finite a.e.
on E. Then for every η > 0 there exists a closed set A ⊂E such that m(E\A) < η and {fn} converges uniformly to f on A.
Proof.
Let m ∈N. It follows from Lemma 1 that there exists a measurable set Am ⊂E and N(m) ∈N such that m(E\Am) < η 2m+1 and |fn −f| < 1 m for all n ≥N(m).
Let ˜ A = ∞ \ m=1 Am. The set ˜ A is measurable and m(E\˜ A) = m ∞ [ m=1 (E\Am) ≤ ∞ X m=1 η 2m+1 = η 2 .
Now we prove that {fn} converges uniformly to f on ˜ A. Let ϵ > 0 and let m ∈N such that 1 m < ϵ. It follows from the definition of ˜ A and of Am that for every n ≥N(m) and for every x ∈˜ A ⊂Am we have |fn −f| < 1 m < ϵ. This proves the uniform convergence on ˜ A Next, since ˜ A is measurable we can find a closed set A ⊂˜ A such that m(˜ A\A) < η/2. The sequence is uniformly convergent on A and m(E\A) = m (E\˜ A) ∪(˜ A\A) ≤m(E\˜ A) + m(˜ A\A) ≤ η 2 + η 2 = η Luzin’s Theorem states that a measurable function on E is "nearly continuous" in the following sense: For any ϵ > 0, there is a subset whose measure is within ϵ to that of E and on which the function is continuous. We start with the case of simple functions.
Proposition (1) Let ϕ be a simple and measurable function defined on a set E. Then for every ϵ > 0 there exists a closed F ⊂E such that m(E\F) < ϵ and ϕ continuous on F.
Proof.
We use the canonical representation of simple function to express ϕ as ϕ = n X j=1 ajχEj for some collection of n disjoint measurable sets E1, · · · , En in E. For ϵ > 0, there exist closed sets F1, · · · , Fn with Fj ⊂Ej and m(Ej\Fj) < ϵ/n.
Then F = Sn j=1 Fj is a closed subset in E and m(E\F) < ϵ. Note that the restriction of ϕ to each Fj is continuous since it is constant on each Fj. It remains to verify that ϕ is continuous on their disjoint union F.
For N ∈N, let BN = BN(0) be the ball in Rq with center 0 and radius N. Let FN j = Fj ∩BN and FN = Sn j=1 FN j . To prove the continuity of ϕ on F it is enough to prove continuity on FN for an arbitrary N (since F = S N FN). To prove continuity on FN, it is enough to verify that the closed sets FN j ’s are separated. That is, there exists δ0 > 0 such that for j ̸= k, and for every x ∈FN j and y ∈FN k we have |x −y| ≥δ0.
For j ̸= k, let δj,k = inf{|x −y| : x ∈FN j , y ∈FN k }. We claim that δj,k > 0. Indeed, if δj,k = 0, then there would be sequences {xp}p ⊂FN j and {yp}p ⊂FN k such that lim p→∞ xp −yp = 0. Since FN j and FN k are compact, then we can extract convergent subsequences {xpm }m ⊂FN j and {ypm }m ⊂FN k that converge to the same limit z ∈FN j ∩FN k (since δj,k = 0) and this is a contradiction since Fj and Fk are disjoint. Hence δj,k > 0 and δ0 = min{δj,k : j ̸= k} is strictly positive. Therefore the FN j ’s are separated and ϕ is continuous on FN.
Luzin’s Theorem Theorem (2) Let f : E ⊂Rq − →R be measurable and finite a.e. on E. Then for every ϵ > 0, there exists a closed F ⊂E such that m(E\F) < ϵ and f continuous on F.
Proof.
Since f is measurable, by the Simple Approximation Theorem, there exists a sequence {ϕn} of simple functions on E which converges pointwise on E to f. According to Egorov’s Theorem, given any ϵ > 0, there exists a closed set C ⊂E with m(E\C) < ϵ/2 and {ϕn} converges uniformly on C. For each n ∈N, there exists a closed set Fn ⊂E with m(E\Fn) < (ϵ/2n+1) such that the simple function ϕn is continuous on Fn. Consider the closed set F = C ∩ T∞ n=1 Fn .
We have E\F = (E\C) ∪ S∞ n=1(E\Fn) so that m(E\F) ≤m(E\C) + ∞ X n=1 m (E\Fn) ≤ ϵ 2 + ∞ X n=1 ϵ 2n+1 = ϵ .
Since each function ϕn is continuous on F ⊂Fn and {ϕn} converges uniformly on F, then the limit f is also continuous on F. |
6721 | https://ca.ixl.com/math/grade-5/mean-find-the-missing-number | IXL | Mean: find the missing number | Grade 5 math
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Bruce has the following data:
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7
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If the mean is 8, which number could u be?
6
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17
key idea
The mean is the average of the numbers. You find the mean by adding the numbers together and then dividing by the number of numbers in the group.
solution
There are 5 numbers in the group.
Try x = 2.
Find the sum of the numbers.
7 +2 +16 +13 +7 =45
Divide by 5.
45 ÷ 5 = 9
When x = 2, the mean is 9.
Try x = 17.
Find the sum of the numbers.
7 +17 +16 +13 +7 =60
Divide by 5.
60 ÷ 5 = 12
When x = 17, the mean is 12.
The mean is 12, so x = 17.
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6722 | https://mathematica.stackexchange.com/questions/230225/maximize-an-expression-with-respect-to-a-variable-after-its-minimization-with-re | mathematical optimization - Maximize an expression with respect to a variable after its minimization with respect to other variables - Mathematica Stack Exchange
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Maximize an expression with respect to a variable after its minimization with respect to other variables
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I started to use Mathematica a few time ago. I want to minimize the following expression (function of l,p,q,r,c l,p,q,r,c) with respect to variables l,p,q,r l,p,q,r and then maximize the result obtained with respect to variable c c. However, when I try to obtain an expression function of c c to maximize later using Minimize, I do not get any result because it takes too long. How can I solve this issue?
mathematica
Minimize[{(((l^2/2)(1-(1/4-c))+(lp)(1-1/4)+(lq)(1-(1/4+c))+(lr)(1-1/2)+(p^2/2)(1-c)+(pq)(1-2c)+(pr)(1-(1/4+c))+(q^2/2)(1-c)+(qr)(1-1/4)+(r^2/2)(1-(1/4-c)))/((l^2/2)(1-(1/4-c))+(lp)(1-1/4)+(lq)(1-(1/4-c))+(lr)(1-0)+(p^2/2)(1-c)+(pq)(1-(1/4-c))+(pr)(1-0)+(q^2/2)(1-(1/4-c))+(qr)(1-1/4)+(r^2/2)(1-0))), l >= 1, p >= 0, q >= 0, r >= 0, l + p + q + r == 1000000, 1/7<c<1/5}, {l, p, q, r}]
Why does the above command never end?
mathematical-optimization
maximum
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edited Sep 17, 2020 at 5:29
user64494
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asked Sep 16, 2020 at 16:51
Penelope BenenatiPenelope Benenati
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This can be done numerically (only numerically in my opinion) in a standard way which takes a lot of time:
mathematica
f[c_?NumericQ] := NMinimize[{(((l^2/2)(1 - (1/4 - c)) + (lp)(1 - 1/4) + (l
q)(1 - (1/4 + c)) + (lr)(1 - 1/2) + (p^2/2)(1 -
c) + (pq)(1 - 2c) + (pr)(1 - (1/4 + c)) + (q^2/2)(1 -
c) + (qr)(1 - 1/4) + (r^2/2)(1 - (1/4 - c)))/((l^2/
2)(1 - (1/4 - c)) + (lp)(1 - 1/4) + (l
q)(1 - (1/4 - c)) + (lr)(1 - 0) + (p^2/2)(1 - c) + (p
q)(1 - (1/4 - c)) + (pr)(1 - 0) + (q^2/
2)(1 - (1/4 - c)) + (qr)(1 - 1/4) + (r^2/2)(1 - 0))),
l >= 1, p >= 0, q >= 0, r >= 0, l + p + q + r == 1000000
}, {l, p, q, r}, Method -> "DifferentialEvolution"]
f[0.196]
(0.732468)
Plot[f[c], {c, 0.19, 0.20}]
mathematica
NMaximize[{f[c], 1/7 <= c && c <= 1/5}, c, Method -> "DifferentialEvolution"]
({0.733816, {c -> 0.2}})
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answered Sep 17, 2020 at 5:26
user64494user64494
1
4
Thank you a lot! This is really helpful! I also wanted to find the maximum where c∈[1/8,1/4]c∈[1/8,1/4], but it seems that it never ends. How long should it take in your opinion (working online)?Penelope Benenati –Penelope Benenati 2020-09-17 12:27:23 +00:00 Commented Sep 17, 2020 at 12:27
1 @PenelopeBenenati: A fast approach n = 20; Table[f[1/8 + i/8/n], {i, 0, n}] // AbsoluteTiming results in {35.0773, {0.707272, 0.709595, 0.711898, 0.714179, 0.716439, 0.71868, 0.7209, 0.7231, 0.725283, 0.727443, 0.729585, 0.73171, 0.733816, 0.73141, 0.721986, 0.712454, 0.702283, 0.691479, 0.680106, 0.668219, 0.655869}}.user64494 –user64494 2020-09-17 15:31:28 +00:00 Commented Sep 17, 2020 at 15:31
1 @PenelopeBenenati: The long approach NMaximize[{f[c], 1/8 <= c && c <= 1/4}, c, Method -> "DifferentialEvolution"] // AbsoluteTiming results in {5589.06, {0.738147, {c -> 0.201805}}}.user64494 –user64494 2020-09-17 16:36:40 +00:00 Commented Sep 17, 2020 at 16:36
That's very useful! Thank you a lot!Penelope Benenati –Penelope Benenati 2020-09-17 16:41:03 +00:00 Commented Sep 17, 2020 at 16:41
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6723 | https://brainly.com/question/43078486 | [FREE] Suppose that \theta is an acute angle of a right triangle and \tan(\theta) = \sqrt{3} . Find - brainly.com
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Suppose that θ is an acute angle of a right triangle and tan(θ)=3. Find sin(θ) and cos(θ).
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To find sin(θ) and cos(θ) when tan(θ) = √3, we assume a right triangle with sides of 1 and √3. Using the Pythagorean theorem, we determine the hypotenuse is 2. Therefore, sin(θ) = √3 / 2 and cos(θ) = 1 / 2.
Explanation
You have been given that the tangent of an acute angle θ in a right triangle equals √3, which means tan(θ) = √3. The goal is to find sin(θ) and cos(θ) based on this information.
In trigonometry, the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. Since θ is an acute angle and we know that tan(θ) = Яω, we can assume that the opposite side (y) is √3 and the adjacent side (x) is 1, which are the simplest numbers that keep the ratio of √3 to 1. We do this to simplify our calculations.
Using the Pythagorean theorem, which states that in a right triangle the square of the hypotenuse (h) is equal to the sum of the squares of the other two sides, we can find the length of the hypotenuse:
x² + y² = h²
1² + (√3)² = h²
1 + 3 = h²
h² = 4
h = √4
h = 2
Now that we have the length of the hypotenuse, we can find sin(θ) and cos(θ):
sin(θ) = opposite / hypotenuse = √3 / 2
cos(θ) = adjacent / hypotenuse = 1 / 2
Therefore, sin(θ) = √3 / 2 and cos(θ) = 1 / 2.
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Given that tan(θ)=3, we found that sin(θ)=2 3 and cos(θ)=2 1. These values were obtained using a right triangle where the opposite side is 3 and the adjacent side is 1. The hypotenuse was calculated using the Pythagorean theorem, resulting in a hypotenuse of 2.
Explanation
To find sin(θ) and cos(θ) when tan(θ)=3, we can visualize this using a right triangle. tan(θ), or the tangent of the angle θ, is defined as the ratio of the opposite side to the adjacent side. Since tan(θ)=3, we can express this relationship as follows:
Let the length of the opposite side (which we'll call y) be 3.
Let the length of the adjacent side (which we'll call x) be 1.
Now we have:
tan(θ)=x y=1 3
Next, we need to find the length of the hypotenuse (h) using the Pythagorean theorem, which states that:
h 2=x 2+y 2
Substituting our values for x and y: h 2=1 2+(3)2
h 2=1+3
h 2=4
h=4=2
Now that we have the lengths of all three sides, we can find sin(θ) and cos(θ):
sin(θ) is given by the ratio of the opposite side to the hypotenuse:
sin(θ)=h y=2 3
cos(θ) is given by the ratio of the adjacent side to the hypotenuse:
cos(θ)=h x=2 1
In summary, we have:
sin(θ)=2 3
cos(θ)=2 1
Examples & Evidence
For instance, if you consider a 60-degree angle in a right triangle, you can visualize that the opposite side is 3 and the adjacent side is 1, leading to the same values for sine and cosine as calculated above. This is a common 30-60-90 triangle where the ratios follow consistently for those angle measures.
The relationships used in this solution stem from fundamental trigonometric definitions and the Pythagorean theorem, which applies to all right triangles. Furthermore, it corresponds with known trigonometric values for standard angles in right triangles.
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### Free Mathematics solutions and answers
Community Answer What is the value of θ for the acute angle in a right triangle? sin(θ)=cos(53°) Enter your answer in the box. θ= °
Community Answer Sketch a triangle that has acute angle θ. tan(θ)= 5 4 Find the other five trigonometric ratios of θ. sin(θ)= cos(θ)= csc(θ)= sec(θ)= cot(θ)= × × × × ×
Community Answer Let θ be an acute angle such that cos(θ) = 0.8. Find the value of tan(θ).
Community Answer Select all the correct statements. A. If sin(θ) = 23, then sec(θ) = 32 B. If cot(θ) = 13, then tan(θ) = 3. C. If cos(θ) = sin(18°), then θ = 32°. D. If sin(θ) = cos(θ), then θ = 45°. E. If θ = 68°, then tan(θ) = cot(22°).
Community Answer Sketch a right triangle corresponding to the trigonometric function of the acute angle θ. Then find the exact values of the other five trigonometric functions of θ. Cos(θ) = 15 sin(θ) = cos(θ) tan(θ) = sec(θ) = cot(θ) =
Community Answer Consider a right triangle and let θ be one of the acute angles in the triangle. Suppose the adjacent leg has length 12 and the hypotenuse has length 15. Compute the values of sin(θ),cos(θ),tan(θ),csc(θ),sec(θ), and cot(θ).
Community Answer Sketch a right triangle corresponding to the trigonometric function of the acute angle θ. Use the Pythagorean Theorem to determine the third side of the triangle and then find the other five trigonometric functions of θ. cot(θ) = 3 sin(θ) = cos(θ) = tan(θ) = csc(θ) = sec(θ) =
Community Answer cos(θ) sin(θ) =cot(θ) sin(θ) cos(θ) =tan(θ) sec(θ) csc(θ) =tan(θ) sin(θ) cos(θ) =cot(θ) cos(θ) sin(θ) =tan(θ) sin(θ) cos(θ) =cot(θ)
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)?
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6724 | https://www.khanacademy.org/math/ap-statistics/random-variables-ap/binomial-mean-standard-deviation/v/expected-value-of-binomial-variable | Expected value of a binomial variable (video) | Khan Academy
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AP®︎/College Statistics
Course: AP®︎/College Statistics>Unit 8
Lesson 6: Parameters for a binomial distribution
Expected value of a binomial variable
Variance of a binomial variable
Finding the mean and standard deviation of a binomial random variable
Mean and standard deviation of a binomial random variable
Math>
AP®︎/College Statistics>
Random variables and probability distributions>
Parameters for a binomial distribution
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Expected value of a binomial variable
AP.STATS: UNC‑3 (EU), UNC‑3.C (LO), UNC‑3.C.1 (EK)
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Deriving and using the expected value (mean) formula for binomial random variables.
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Andrea Menozzi 4 years ago Posted 4 years ago. Direct link to Andrea Menozzi's post “where is explained the su...” more where is explained the sum of independent variables? E(X+Y) does not makes sense to me
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Jerry Nilsson 4 years ago Posted 4 years ago. Direct link to Jerry Nilsson's post “Let 𝑋 = {𝑥₁, 𝑥₂, 𝑥₃} ...” more Let
𝑋 = {𝑥₁, 𝑥₂, 𝑥₃}
𝑌 = {𝑦₁, 𝑦₂}
Thereby,
𝐸(𝑋) = (𝑥₁ + 𝑥₂ + 𝑥₃)∕3
𝐸(𝑌) = (𝑦₁ + 𝑦₂)∕2
Also,
𝑋 + 𝑌 = {𝑥₁ + 𝑦₁, 𝑥₂ + 𝑦₁, 𝑥₃ + 𝑦₁, 𝑥₁ + 𝑦₂, 𝑥₂ + 𝑦₂, 𝑥₃ + 𝑦₂}
This gives us,
𝐸(𝑋 + 𝑌) = (𝑥₁ + 𝑦₁ + 𝑥₂ + 𝑦₁ + 𝑥₃ + 𝑦₁ + 𝑥₁ + 𝑦₂ + 𝑥₂ + 𝑦₂ + 𝑥₃ + 𝑦₂)∕6
= (2𝑥₁ + 2𝑥₂ + 2𝑥₃ + 3𝑦₁ + 3𝑦₂)∕6
= (𝑥₁ + 𝑥₂ + 𝑥₃)∕3 + (𝑦₁ + 𝑦₂)∕2
= 𝐸(𝑋) + 𝐸(𝑌)
This example can quite easily be generalized to where 𝑋 has 𝑚 elements and 𝑌 has 𝑛 elements.
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alphadirect99 5 years ago Posted 5 years ago. Direct link to alphadirect99's post “Notice that X is a binom...” more Notice that X is a binomial variable, whereas Y is a bernoulli variable, the simplest case of a binomial variable.
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daniella a year ago Posted a year ago. Direct link to daniella's post “You're correct that X is ...” more You're correct that X is a binomial variable representing the number of successes in a series of n independent Bernoulli trials. Each Y represents a single Bernoulli trial, where success occurs with probability p and failure occurs with probability 1 − p. X is indeed calculated as the sum of n independent Bernoulli random variables (Ys).
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bsoni1660 4 years ago Posted 4 years ago. Direct link to bsoni1660's post “Why is X taken as a sum o...” more Why is X taken as a sum of Y's?
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daniella a year ago Posted a year ago. Direct link to daniella's post “X is taken as the sum of ...” more X is taken as the sum of Ys because each Y represents the outcome of a single trial, where success or failure is determined by the Bernoulli distribution with probability p. Since X represents the total number of successes in n trials, it makes sense to sum up the outcomes of all individual trials (Ys) to get the total number of successes.
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Tobey 7 years ago Posted 7 years ago. Direct link to Tobey's post “As the mean/expected valu...” more As the mean/expected value of a Bernoulli distribution is p and the mean/expected value of a binomial variable is np, is a binomial variable a multiple of a Bernoulli distribution?
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𝜏 Is Better Than 𝝅 5 years ago Posted 5 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “Yes, the Bernoulli distri...” more Yes, the Bernoulli distribution is the simplest case of the binomial distribution where n = 1, meaning one trial.
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Rowain Hardby 5 years ago Posted 5 years ago. Direct link to Rowain Hardby's post “What is the expected valu...” more What is the expected value of a variable like:
"Flip a fair coin until you get tails. X = the number of heads you flipped."
I realize this wouldn't be a binomial variable, but it seemed pretty similar.
Note: P(H) = P(T) = 0.5
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Rowain Hardby 5 years ago Posted 5 years ago. Direct link to Rowain Hardby's post “Never mind. I found it: h...” more Never mind. I found it:
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glorioussekao 5 years ago Posted 5 years ago. Direct link to glorioussekao's post “Let X~Bin(n,p),find E(e^(...” more Let X~Bin(n,p),find E(e^(tx) where t is a constant
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Ian Pulizzotto 5 years ago Posted 5 years ago. Direct link to Ian Pulizzotto's post “Nice question! The plan ...” more Nice question! The plan is to use the definition of expected value, use the formula for the binomial distribution, and set up to use the binomial theorem in algebra in the final step.
We have
E(e^(tx))
= sum over all possible k of P(X=k)e^(tk)
= sum k from 0 to n of p^k (1-p)^(n-k) (n choose k) e^(tk)
= sum k from 0 to n of (pe^t)^k (1-p)^(n-k) (n choose k)
= (pe^t + 1 - p)^n, from the binomial theorem in algebra.
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Larissa Ford 6 years ago Posted 6 years ago. Direct link to Larissa Ford's post “The thing I get caught up...” more The thing I get caught up on is the Expected value of Y at 5:40 . Could someone give me a link to the logic behind E(Y)=p? Specifically, when he talked about the probability of weighted outcomes?
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𝜏 Is Better Than 𝝅 5 years ago Posted 5 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “By assuming that E(nY) ca...” more By assuming that E(nY) can be dissected into E(Y)+E(Y)+E(Y)+...+E(Y) n times, we have assumed that Y is independent with respect to itself, but clearly knowing what Y is would constrain what Y is, so are all random variables somehow independent with respect to themselves, and if so why?
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Ian Pulizzotto 5 years ago Posted 5 years ago. Direct link to Ian Pulizzotto's post “The explanation given cou...” more The explanation given could be better. X should really be written as the sum Y_1+Y_2+Y_3+...+Y_n, where Y_k is 1 if the kth trial is a success, and is 0 if the kth trial is a failure. The variables Y_1,Y_2,Y_3,...Y_n are independent, and E(Y_k)=p for all k with 1<=k<=n.
So E(X) = E(Y_1+Y_2+Y_3+...+Y_n) = E(Y_1)+E(Y_2)+E(Y_3)+....+E(Y_n) = sum of n copies of p = np.
(By the way, the assumption of independence is actually not needed in this type of situation. The expected value of the sum of any random variables is always the sum of their expected values. However, the assumption of independence would be needed to make a similar statement about variance.)
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N N 3 years ago Posted 3 years ago. Direct link to N N's post “Why is Y=0 or Y=1? If Y ...” more Why is Y=0 or Y=1?
If Y is not either 0 or 1, what kind of formula should we use?
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Jerry Nilsson 3 years ago Posted 3 years ago. Direct link to Jerry Nilsson's post “In this case we want 𝑌 t...” more In this case we want 𝑌 to represent whether a trial is a success or not.
So it needs to have two outcomes – one for "Success" and one for "Not Success".
The reason we choose the outcomes to be either 0 or 1 is because it allows us to easily count the number of successes after 𝑛 trials:
𝑌₁ + 𝑌₂ + 𝑌₃ + ... + 𝑌ₙ
– – –
The way we define a variable depends on what we want it to represent.
For example, if you and friend were competing in a game you might want to keep track of who has won more often after 𝑛 rounds.
Then we might want to define 𝑌 = −1 if you lose a round, 𝑌 = 0 if a round ends in a draw, and 𝑌 = 1 if you win a round.
If the sum 𝑌₁ + 𝑌₂ + 𝑌₃ + ... + 𝑌ₙ is negative you lost more rounds than you won,
If the sum is 0, then both of you won equally often.
And if the sum is positive you won more rounds than you lost.
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Bryan 5 years ago Posted 5 years ago. Direct link to Bryan's post “Does the expected value o...” more Does the expected value of a binomial variable represent the most likely amount of successes in n trials? Would it always be the tallest point on a graph of the probability distribution of the variable then?
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daniella a year ago Posted a year ago. Direct link to daniella's post “The expected value of a b...” more The expected value of a binomial variable represents the average number of successes you would expect to see in a given number of trials. While it's often a useful measure, especially for making predictions, it doesn't necessarily represent the most likely amount of successes. The most likely amount of successes would correspond to the mode of the binomial distribution, which may or may not coincide with the expected value. In some cases, the mode may be different from the expected value, especially if the distribution is skewed.
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Video transcript
[Tutor] So I've got a binomial variable X and I'm gonna describe it in very general terms, it is the number of successes after n trials, after n trials, where the probability of success, success for each trial is P and this is a reasonable way to describe really any random, any binomial variable, we're assuming that each of these trials are independent, the probability stays constant, we have a finite number of trials right over here, each trial results in either a very clear success or a failure. So what we're gonna focus on in this video is well, what would be the expected value of this binomial variable? What would the expected value, expected value of X be equal to? And I will just cut to the chase and tell you the answer and then later in this video, we'll prove it to ourselves a little bit more mathematically. The expected value of X, it turns out, is just going to be equal to the number of trials times the probability of success for each of those trials and so if you wanted to make that a little bit more concrete, imagine if a trial is a Free Throw, taking a shot from the Free Throw line, success, success is made shot, so you actually make the shot, the ball went in the basket, your probability is, use this yellow color, your probability, this would be your Free Throw percentage, so let's say it's 30% or 0.3 and let's say for the sake of argument, that we're taking 10 Free Throws, so n is equal to 10, so this is making it all a lot more concrete, so in this particular scenario, your expected value, your expected value, if X is the number of made Free Throws, after taking 10 Free Throws with a Free Throw percentage of 30%, well, based on what I just told you, it would be n times p, it would be the number of trials times the probability of success in any one of those trials, times 0.3, which is just going to be, of course equal to three. Now does that make intuitive sense? Well, if you're taking 10 shots with a 30% Free Throw percentage, it actually does feel natural that I would expect to make three shots. Now with that out of the way, let's make ourselves feel good about this mathematically and we're gonna leverage some of our expected value properties, in particular, we're gonna leverage the fact that if I have the expected value of the sum of two independent random variables, let's say X plus Y, it's going to be equal to the expected value of X plus the expected value of Y, that we talk about in other videos and so assuming this right over here, let's construct a new random variable, let's call our random variable Y and we know the following things about Y, the probability that Y is equal to one is equal to p and the probability that Y is equal to zero is equal to one minus p and these are the only two outcomes for this random variable and so you might be seeing where this is going, you could view this random variable, it's really representing one trial, it becomes one in its success, zero when you don't have a success and so you could view our original random variable, X right over here as being equal to Y plus Y and we're gonna have 10 of these, so we're gonna have 10 Ys, in the concrete sense, you could view the random variable Y as equaling one, if you make a Free Throw and equaling zero, if you don't make a Free Throw, it's really just representing one of those trials and you can view X as the sum of n of those trials, well now actually, let me be very clear here, I immediately went to the concrete, but I really should be saying n Ys, 'cause I wanna stay general right over here, so there are n, n Ys right over here, this was just a particular example, but I am going to try to stay general for the rest of the video, because now we are really trying to prove this result right over here, so let's just take the expected value of both sides, so what is it going to be? So we get the expected value of X, of X is equal to well, it's the expected value of all of this thing, but by that property right over here is going to be the expected value of Y plus the expected value of Y plus, and we're gonna do this n times, plus the expected value of Y and we're gonna have n of these, so we have n and so you could rewrite this as being equal to, this is our n right over here, this is n times the expected value of Y. Now what is the expected value of Y? Well, this is pretty straightforward, we can actually just do it directly, the expected value of Y, let me just write it over here, the expected value of Y is just the probability-weighted outcome and since there's only two discrete outcomes here, it's pretty easy to calculate, we have a probability of p of getting a one, so it's p times one plus we have a probability of one minus p of getting a zero, well, what does this simplify to? Well, zero times anything, that's zero and then you have one times p, this is just equal to p, so expected value of Y is just equal to p and so there you have it, we get the expected value of X is 10 times the expected value, or the expected value of X is n times the expected value of Y and the expected value of Y is p, so the expected value of X is equal to np, hope you feel good about that.
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Ch.1 - Chemical Tools: Experimentation & Measurement
All textbooksMcMurry 8th EditionCh.1 - Chemical Tools: Experimentation & MeasurementProblem 148
Chapter 1, Problem 148
The Rankine temperature scale used in engineering is to the Fahrenheit scale as the Kelvin scale is to the Celsius scale. That is, 1 Rankine degree is the same size as 1 Fahrenheit degree, and 0 °R = absolute zero. (a) What temperature corresponds to the freezing point of water on the Rankine scale?
Verified step by step guidance
1
Identify the freezing point of water in Fahrenheit, which is 32°F.
Understand that the Rankine scale is to Fahrenheit as Kelvin is to Celsius, meaning 0°R is absolute zero.
Use the relationship between Fahrenheit and Rankine: °R = °F + 459.67.
Substitute the freezing point of water in Fahrenheit into the equation: °R = 32 + 459.67.
Simplify the equation to find the temperature in Rankine at the freezing point of water.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Rankine Temperature Scale
The Rankine scale is an absolute temperature scale used primarily in engineering. It is similar to the Kelvin scale but uses Fahrenheit degrees instead of Celsius. On this scale, absolute zero is defined as 0 °R, and the size of one Rankine degree is equivalent to one Fahrenheit degree.
Recommended video:
Guided course
02:33
The pH Scale
Freezing Point of Water
The freezing point of water is the temperature at which water transitions from liquid to solid, typically defined as 0 °C or 32 °F. In the Rankine scale, this temperature must be converted from Celsius or Fahrenheit to find its equivalent, which is essential for solving the problem presented.
Recommended video:
Guided course
02:18
Freezing Point Example
Temperature Conversion
Temperature conversion involves changing a temperature value from one scale to another, such as Celsius to Rankine. The formula provided in the image indicates a specific conversion factor, which is crucial for accurately determining the freezing point of water in the Rankine scale. Understanding these conversions is fundamental in thermodynamics and engineering applications.
Recommended video:
Guided course
02:10
Temperature Conversion Example
Related Practice
Textbook Question
The element gallium (Ga) has the second-largest liquid range of any element, melting at 29.78 °C and boiling at 2204 °C at atmospheric pressure. (a) What is the density of gallium in g/cm3 at 25 °C if a 1 in. cube has a mass of 0.2133 lb?
639
viewsTextbook Question
The element gallium (Ga) has the second-largest liquid range of any element, melting at 29.78 °C and boiling at 2204 °C at atmospheric pressure. (b) Assume that you construct a thermometer using gallium as the fluid instead of mercury and that you define the melting point of gallium as 0 °G and the boiling point of gallium as 1000 °G. What is the melting point of sodium chloride (801 °C) on the gallium scale?
1058
viewsTextbook Question
An 8.894 g block of aluminum was pressed into a thin square of foil with 36.5 cm edge lengths. (a) If the density of Al is 2.699 g>cm3, how thick is the foil in centimeters?
406
views |
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16.4: Acid Strength and the Acid Dissociation Constant (Ka)
Last updated
: Aug 14, 2020
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16.3: Definitions of Acids and Bases
16.5: Autoionization of Water and pH
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: 47028
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
To know the relationship between acid or
base
strength and the magnitude of , , , and .
To understand the
leveling effect
.
The magnitude of the
equilibrium
constant for an ionization reaction can be used to determine the relative strengths of acids and
bases
. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate
base
, is as follows:
The
equilibrium
constant for this dissociation is as follows:
As we noted earlier, because water is the solvent, it has an activity equal to 1, so the term in Equation is actually the , which is equal to 1.
Again, for simplicity, can be written as in Equation .
Keep in mind, though, that free does not exist in aqueous
solutions
and that a proton is transferred to in all acid ionization reactions to form hydronium
ions
, . The larger the , the stronger the acid and the higher the
concentration
at
equilibrium
. Like all
equilibrium
constants, acid–
base
ionization constants are actually measured in terms of the activities of or , thus making them unitless. The values of for a number of common acids are given in Table .
Table : Values of , , , and for Selected Acids ( and Their Conjugate
Bases
()
| Acid | | | | | | |
| The number in parentheses indicates the ionization step referred to for a polyprotic acid. | | | | | | |
| hydroiodic acid | | | −9.3 | | | 23.26 |
| sulfuric acid (1) | | | −2.0 | | | 16.0 |
| nitric acid | | | −1.37 | | | 15.37 |
| hydronium ion | | | 0.00 | | | 14.00 |
| sulfuric acid (2) | | | 1.99 | | | 12.01 |
| hydrofluoric acid | | | 3.20 | | | 10.80 |
| nitrous acid | | | 3.25 | | | 10.75 |
| formic acid | | | 3.750 | | | 10.25 |
| benzoic acid | | | 4.20 | | | 9.80 |
| acetic acid | | | 4.76 | | | 9.24 |
| pyridinium ion | | | 5.23 | | | 8.77 |
| hypochlorous acid | | | 7.40 | | | 6.60 |
| hydrocyanic acid | | | 9.21 | | | 4.79 |
| ammonium ion | | | 9.25 | | | 4.75 |
| water | | | 14.00 | | | 0.00 |
| acetylene | | | 26.0 | | | −12.0 |
| ammonia | | | 35.0 | | | −21.0 |
Weak bases
react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent
base
and BH+ is its conjugate acid:
The
equilibrium
constant for this reaction is the
base ionization constant
(Kb), also called the
base
dissociation constant:
Once again, the activity of water has a value of 1, so water does not appear in the
equilibrium constant expression
. The larger the , the stronger the
base
and the higher the
concentration
at
equilibrium
. The values of for a number of common
weak bases
are given in Table .
Table : Values of , , , and for Selected
Weak Bases
(B) and Their Conjugate Acids (BH+)
| Base | | | | | | |
| As in Table . | | | | | | |
| hydroxide ion | | | 0.00 | | | 14.00 |
| phosphate ion | | | 1.68 | | | 12.32 |
| dimethylamine | | | 3.27 | | | 10.73 |
| methylamine | | | 3.34 | | | 10.66 |
| trimethylamine | | | 4.20 | | | 9.80 |
| ammonia | | | 4.75 | | | 9.25 |
| pyridine | | | 8.77 | | | 5.23 |
| aniline | | | 9.13 | | | 4.87 |
| water | | | 14.00 | | | 0.00 |
There is a simple relationship between the magnitude of for an acid and for its conjugate
base
. Consider, for example, the ionization of hydrocyanic acid () in water to produce an acidic solution, and the reaction of with water to produce a basic solution:
The
equilibrium constant expression
for the ionization of HCN is as follows:
The corresponding expression for the reaction of cyanide with water is as follows:
If we add Equations and , we obtain the following:
add Equations and , we obtain
| Reaction | Equilibrium Constants |
| | |
| | |
| | |
In this case, the sum of the reactions described by and is the equation for the autoionization of water, and the product of the two
equilibrium
constants is :
Thus if we know either for an acid or for its conjugate
base
, we can calculate the other
equilibrium
constant for any conjugate acid–
base
pair.
Just as with , , and pKw, we can use negative logarithms to avoid exponential notation in writing acid and
base
ionization constants, by defining as follows:
and as
Similarly, Equation , which expresses the relationship between and , can be written in logarithmic form as follows:
At 25 °C, this becomes
The values of and are given for several common acids and
bases
in Tables and , respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (), with a of 3.25, is about a million times stronger acid than hydrocyanic acid (HCN), with a of 9.21. Conversely, smaller values of correspond to larger
base
ionization constants and hence stronger
bases
.
The relative strengths of some common acids and their conjugate
bases
are shown graphically in Figure . The conjugate acid–
base
pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of . This order corresponds to decreasing strength of the conjugate
base
or increasing values of . At the bottom left of Figure are the common strong acids; at the top right are the most common strong
bases
. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate
base
. Thus the conjugate
base
of a strong acid is a very weak
base
, and the conjugate
base
of a very weak acid is a strong
base
.
The conjugate
base
of a strong acid is a weak
base
and vice versa.
We can use the relative strengths of acids and
bases
to predict the direction of an acid–
base
reaction by following a single rule: an acid–
base
equilibrium
always favors the side with the weaker acid and
base
, as indicated by these arrows:
In an acid–
base
reaction, the proton always reacts with the stronger
base
.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute
aqueous solution
to produce and ; only negligible amounts of molecules remain undissociated. Hence the ionization
equilibrium
lies virtually all the way to the right, as represented by a single arrow:
In contrast, acetic acid is a weak acid, and water is a weak
base
. Consequently, aqueous
solutions
of acetic acid contain mostly acetic acid molecules in
equilibrium
with a small
concentration
of and acetate
ions
, and the ionization
equilibrium
lies far to the left, as represented by these arrows:
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong
base
, and ammonia is a weak
base
, whereas the ammonium ion is a stronger acid than water. Hence this
equilibrium
also lies to the left:
All acid–
base
equilibria favor the side with the weaker acid and
base
. Thus the proton is bound to the stronger
base
.
Example : Butyrate and Dimethylammonium Ions
Calculate and of the butyrate ion (). The of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
Calculate and of the dimethylammonium ion (). The
base ionization constant
of dimethylamine () is at 25°C.
Given: and
Asked for: corresponding and , and
Strategy:
The constants and are related as shown in Equation . The and for an acid and its conjugate
base
are related as shown in Equations and . Use the relationships pK = −log K and K = 10−pK (Equations and ) to convert between and or and .
Solution:
We are given the for butyric acid and asked to calculate the and the for its conjugate
base
, the butyrate ion. Because the value cited is for a temperature of 25°C, we can use Equation : + = pKw = 14.00. Substituting the and solving for the ,
Because , is .
In this case, we are given for a
base
(dimethylamine) and asked to calculate and for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is rather than , we can use Equation : . Substituting the values of and at 25°C and solving for ,
Because = −log , we have . We could also have converted to to obtain the same answer:
If we are given any one of these four quantities for an acid or a
base
(, , , or ), we can calculate the other three.
Exercise : Lactic Acid
Lactic acid () is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its is 3.86 at 25°C. Calculate for lactic acid and and for the lactate ion.
Answer
: for lactic acid;
= 10.14 and
for the lactate ion
A Video Calculating
pH
in Strong Acid or Strong
Base
Solutions
: Calculating
pH
in Strong Acid or Strong
Base
Solutions [youtu.be]
Solutions of Strong Acids and Bases : The Leveling Effect
You will notice in Table that acids like and lie above the
hydronium ion
, meaning that they have values less than zero and are stronger acids than the ion. Recall from Chapter 4 that the acidic proton in virtually all
oxoacids
is bonded to one of the oxygen
atoms
of the oxoanion. Thus nitric acid should properly be written as . Unfortunately, however, the formulas of
oxoacids
are almost always written with hydrogen on the left and oxygen on the right, giving instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 have values less than zero, which means that they have a greater tendency to lose a proton than does the ion. Conversely, the conjugate
bases
of these strong acids are weaker
bases
than water. Consequently, the proton-transfer equilibria for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the ion and the conjugate
base
of the acid.
Although for is about 108 greater than for , the reaction of either or with water gives an essentially stoichiometric solution of and I− or . In fact, a 0.1 M
aqueous solution
of any strong acid actually contains 0.1 M , regardless of the identity of the strong acid. This phenomenon is called the
leveling effect
: any species that is a stronger acid than the conjugate acid of water () is leveled to the strength of in
aqueous solution
because is the strongest acid that can exist in
equilibrium
with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in
aqueous solution
, and an alternative approach must be used to determine their relative acid strengths.
One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker
base
, with a lesser tendency to accept a proton than . Measurements of the conductivity of 0.1 M
solutions
of both HI and in acetic acid show that HI is completely dissociated, but is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than . The relative order of acid strengths and approximate and values for the strong acids at the top of Table were determined using measurements like this and different nonaqueous solvents.
In aqueous
solutions, is the strongest acid and is the strongest
base
that can exist in
equilibrium
with .
The
leveling effect
applies to
solutions
of strong
bases
as well: In
aqueous solution
, any
base
stronger than OH− is leveled to the strength of OH− because OH− is the strongest
base
that can exist in
equilibrium
with water. Salts such as , (sodium methoxide), and (sodamide, or sodium amide), whose
anions
are the conjugate
bases
of species that would lie below water in Table , are all strong
bases
that react essentially completely (and often violently) with water, accepting a proton to give a solution of and the corresponding cation:
Other examples that you may encounter are potassium hydride () and organometallic compounds such as methyl lithium ().
Polyprotic Acids and Bases
As you learned, polyprotic acids such as , , and contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the increases. Consider , for example:
The
equilibrium
in the first reaction lies far to the right, consistent with being a strong acid. In contrast, in the second reaction, appreciable quantities of both and are present at
equilibrium
.
For a polyprotic acid, acid strength decreases and the increases with the sequential loss of each proton.
The hydrogen sulfate ion () is both the conjugate
base
of and the conjugate acid of . Just like water, HSO4− can therefore act as either an acid or a
base
, depending on whether the other reactant is a stronger acid or a stronger
base
. Conversely, the sulfate ion () is a polyprotic
base
that is capable of accepting two protons in a stepwise manner:
Like any other conjugate acid–
base
pair, the strengths of the conjugate acids and
bases
are related by + = pKw. Consider, for example, the conjugate acid–
base
pair. From Table , we see that the of is 1.99. Hence the of is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak
base
, whereas is a strong
base
, so the
equilibrium
shown in Equation lies to the left. The ion is also a very weak
base
( of = 2.0, of ), which is consistent with what we expect for the conjugate
base
of a strong acid.
Example
Predict whether the
equilibrium
for each reaction lies to the left or the right as written.
Given: balanced chemical equation
Asked for:
equilibrium
position
Strategy:
Identify the conjugate acid–
base
pairs in each reaction. Then refer to Tables and and Figure to determine which is the stronger acid and
base
.
Equilibrium
always favors the formation of the weaker acid–
base
pair.
Solution:
The conjugate acid–
base
pairs are and . According to Tables and , is a stronger acid ( ) than (pKa = 12.32), and is a stronger
base
( ) than ( ). The
equilibrium
will therefore lie to the right, favoring the formation of the weaker acid–
base
pair:
The conjugate acid–
base
pairs are and . According to Table , HCN is a weak acid (pKa = 9.21) and is a moderately weak
base
(pKb = 4.79). Propionic acid () is not listed in Table , however. In a situation like this, the best approach is to look for a similar compound whose acid–
base
properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid ( versus ), so we might expect the two compounds to have similar acid–
base
properties. In particular, we would expect the of propionic acid to be similar in magnitude to the of acetic acid. (In fact, the of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than . Because the stronger acid forms the weaker conjugate
base
, we predict that cyanide will be a stronger
base
than propionate. The
equilibrium
will therefore lie to the right, favoring the formation of the weaker acid–
base
pair:
Exercise
Predict whether the
equilibrium
for each reaction lies to the left or the right as written.
Answer a
Answer b
: left
A Video Discussing Polyprotic Acids: Polyprotic Acids [youtu.be]
Summary
base
reactions always contain two conjugate acid–
base
pairs. Each acid and each
base
has an associated ionization constant that corresponds to its acid or
base
strength. Two species that differ by only a proton constitute a conjugate acid–
base
pair. The magnitude of the
equilibrium
constant for an ionization reaction can be used to determine the relative strengths of acids and
bases
. For an
aqueous solution
of a weak acid, the dissociation constant is called the acid ionization constant (). Similarly, the
equilibrium
constant for the reaction of a weak
base
with water is the
base ionization constant
(). For any conjugate acid–
base
pair, . Smaller values of correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of correspond to larger
base
ionization constants and hence stronger
bases
. At 25°C, . Acid–
base
reactions always proceed in the direction that produces the weaker acid–
base
pair. No acid stronger than and no
base
stronger than can exist in
aqueous solution
, leading to the phenomenon known as the
leveling effect
. Polyprotic acids (and
bases
) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest
base
.
Key Equations
Acid ionization constant:
Base ionization constant
:
Relationship between and of a conjugate acid–
base
pair:
Definition of :
Definition of :
Relationship between and of a conjugate acid–
base
pair:
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
16.3: Definitions of Acids and Bases
16.5: Autoionization of Water and pH |
6727 | https://tomas.rokicki.com/rubik20.pdf | SIAM J. DISCRETE MATH.
c ⃝2013 Society for Industrial and Applied Mathematics Vol. 27, No. 2, pp. 1082–1105 THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY∗ TOMAS ROKICKI†, HERBERT KOCIEMBA‡, MORLEY DAVIDSON§, AND JOHN DETHRIDGE¶ Abstract.
We give an expository account of our computational proof that every position of Rubik’s Cube can be solved in 20 moves or less, where a move is defined as any twist of any face.
The roughly 4.3 × 1019 positions are partitioned into about two billion cosets of a specially chosen subgroup, and the count of cosets required to be treated is reduced by considering symmetry. The reduced space is searched with a program capable of solving one billion positions per second, using about one billion seconds of CPU time donated by Google. As a byproduct of determining that the diameter is 20, we also find the exact count of cube positions at distance 15.
Key words. group theory, algorithm performance, Rubik’s Cube AMS subject classifications. 20-04, 05C12, 20B40 DOI. 10.1137/120867366 1. Introduction. In this paper we give an expository account of our proof that every position of Rubik’s Cube1 can be solved in 20 moves or less, where a move is defined as any twist of any face. This manner of counting moves, known as the half-turn metric (HTM), is by far the most popular move-count metric for the cube and is understood to be the underlying metric when discussions around this problem, also known as finding “God’s number” for the puzzle, arise in various online forums devoted to the cube.
The problem has been of keen interest ever since the cube appeared on shelves three decades ago. In group theory language, the problem we solve is to determine the diameter, i.e., maximum edge-distance between vertices, of the HTM-associated Cayley graph of the Rubik’s Cube group. As summarized in the next section, many researchers have found increasingly tight upper and lower bounds for the HTM diameter of the cube. The present work explains the computational aspects of our proof that it equals 20.
While the technological improvements of increased memory capacity and greater CPU power were necessary for this result, it was primarily a series of mathematical and algorithmic improvements that allowed us to settle the question in the current time frame. Our decomposition of the problem, along with certain insights, let us solve it with a program capable of doing about 65 billion group operations per second on a single common desktop CPU, which works out to be about 23 Rubik’s Cube group operations per CPU cycle.
The decomposition is based on a subgroup H of the Cube group.
Following the section on the problem’s history, our paper recommences with a description of this subgroup and its properties, including the notion of an H-wise relabeling of the cube for illustration purposes. We then introduce the notion of canonical sequences, ∗Received by the editors February 24, 2012; accepted for publication (in revised form) February 19, 2013; published electronically June 19, 2013.
†725 B Loma Verde, Palo Alto, CA 94303 (rokicki@gmail.com).
‡Bismarckstr. 31, 64293 Darmstadt, Germany (kociemba@t-online.de).
§Department of Mathematical Sciences, Kent State University, Kent, OH 44242 (davidson@ math.kent.edu).
¶Google Inc., Mountain View, CA 94043 (johndethridge@gmail.com).
1Rubik’s Cube is a registered trademark of Seven Towns Limited.
1082 THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1083 relate them to the count of positions at a given distance from the solved position, and describe how we can use these ideas to simultaneously and quickly solve many positions at a time. We discuss the ideas and implementation of our solving program itself, first giving an overview of the two main phases of the coset solving program: search and prepass. We then present some empirically derived heuristic improvements that were key to the success of the final runs. Next we describe how we set up the actual final runs and give some statistics and details from those runs, followed by some ideas for future work.
In particular, the diameter question can be posed relative to move-count metrics other than the HTM. Both the quarter-turn metric (where a half-twist of a face constitutes two moves) and the slice-turn metric (where a middle slice may be rotated in a single move) are of significant popular interest and can be approached with techniques like the ones we use.
Our new and essential code optimizations were rather intricate and so are merely summarized, using short code excerpts, in the search and prepass sections. Our critical inner loop as compiled by the GNU compiler collection is given as an appendix. The source code includes multiple implementations of its key algorithms at increasing levels of sophistication; this was critical to verify the correct operation of the code.
In addition, a completely independent (sharing no code) high-performance version of the solver was created by one of the authors, and many comparisons were made between the results of the two for verification.
Annotated source code and many further optimization details are available at
It is worth mentioning one improvement not detailed here since it would take us too far afield, namely, our treatment of a set cover problem which allowed us to further reduce, beyond what standard combinatorial arguments give, the number of cosets of H that had to be processed.
This gave an additional speedup factor of about 2.5, incremental in relative terms and of interest here only in being our last major improvement prior to seeking a computing sponsor. The mathematical and computational derivation of our set cover solution will be treated in a separate publication, but the actual solution is posted with our source code and may be easily verified, as three of us did independently from scratch. Despite all of our efforts to keep the total runtime low, the need for substantial computing resources was inevitable using the current approach on current machinery.
2. Previous work. Despite widespread interest in the mathematics of Rubik’s Cube, not only among mathematicians and computer scientists but also in the popular press, the basic question of finding the diameter of the group, in the half-turn or any other nontrivial metric, has remained unsolved for more than 30 years.
By 1980 it was understood from a very simple counting argument (see ) that at least 18 moves were required for some positions, but no positions were yet proved to require more moves. Thistlethwaite proved in July 1981 that 52 moves suffice.
For most of the next decade, these two results were all that were published.
In 1990, Kloosterman lowered the upper bound to 42 . In 1992, this bound was lowered to 39 by Reid , and then to 37 by Winter . Also in 1992, Kociemba introduced his two-phase algorithm, surprising the cube world with an algorithm that found near-optimal solutions rapidly, on consumer hardware (the Atari ST) with only a megabyte of memory.
In 1995, Reid lowered the upper bound to 29 and also increased the lower bound to 20 by proving that the “superflip” position (all corners solved, all edges flipped in their home positions) requires 20 moves to solve. Two years later, Korf 1084 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE Table 2.1 The solution rate, in positions per second, for four different algorithms for solving the cube.
The first row shows existing position-at-a-time techniques, and the second row shows the rate when solving an entire coset at a time using our coset solver.
Optimal Near-optimal Individual position 2.0 3900 Coset of H 2 × 106 109 introduced the first practical computer program for optimally solving arbitrary positions in the Rubik’s Cube group with a typical run time of about a day.
The next seven years were relatively quiet, but the 2004 reincarnation of the “cube-lovers” mailing list as an internet forum by Mark Longridge unleashed a flurry of activity. In 2005 Radu lowered the upper bound to 28 and then in 2006 lowered it again to 27 , both times using only a small amount of computer time. In 2007, Kunkle and Cooperman lowered the bound to 26 using a computer cluster; see and .
It was during this period that we began the work presented in this paper . In 2008, as our technique matured, we were able to lower the bound to 25 (using only a few home machines) and then 23 and 22 (with John Welborn using idle time at Sony Pictures Imageworks). Further improvements in efficiency led to the present work.
To give an idea of the size of the problem, there are about 4.3×1019 (43 quintillion) positions in the cube group. We can immediately reduce this number by about a factor of 48 by considering the spatial symmetries of the cube: 24 rotational symmetries about various axes (e.g., the line through the centers of the up and down faces), 12 reflectional symmetries about various planes (e.g., the up-down “equatorial” plane), and 12 additional composite symmetries (e.g., up-down reflection followed by a 90-degree rotation about the up-down axis).
Further reduction by about a factor of 2 is realized by group inversion, noting that when a solution sequence is inverted (i.e., the moves applied in reverse order with each move individually reversed), one generally obtains a solution sequence to a different scrambling. The fastest optimal solver we know of, based on Korf’s ideas and requiring 33GB of RAM for tables, can solve positions on our Nehalem baseline processor at a rate of 2.0 positions per second.
Applying this optimal solver to each position after the reductions above would require seven billion CPU years. The fastest existing algorithm for finding near-optimal solutions, Kociemba’s two-phase algorithm, when optimized carefully for batch speed, can find length-20 solutions for random positions at a rate of 3900 per second. The time required to solve all positions near-optimally this way would be about 3.7 million CPU years. These two solvers are listed in the first row of Table 2.1.
The second row gives our contribution. When solving an entire coset of positions at once, we can solve about two million per second optimally and about a billion per second near-optimally. One revealing aspect of our proof that the HTM diameter equals 20 is that it used fewer core-cycles (about 1.2 × 1019) than there are cube positions.
3. Overview. Our determination of just how scrambled a cube can get is based on a single key idea: partition the space of cube positions into subsets such that each subset can be solved quickly and independently. In order to set the stage for the details of the actual program given in later sections, in this overview section we consider how our overall approach would apply to a much simpler and more familiar decomposition of the space.
THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1085 Given a cube to play with, most people will attempt to solve the top layer first and proceed with the rest of the cube only after they have solved that layer. This subgoal is reasonably achievable by most people with a little bit of effort. It is also quite simple for a computer. There are only about 26 billion different “top-layer positions,” i.e., configurations of the top eight cubies (or movable pieces) when distributed throughout the entire cube, so a table can be constructed in memory giving how many moves each of these positions is from having the top layer solved. To find a solution to the top layer, we look up this top-layer distance for the current position, and then try each move one by one on the current position to see which of them bring us closer to having the top layer solved. Such a table, called a pattern database , allows us to generate solution sequences, often including several “optimal” (i.e., shortest-length) solution sequences for any top-layer position very rapidly.
The vast majority of those who solve the top layer for the first time will still have much work to do, with the remaining cubies in a state of disarray. But every so often one lucky person may find that their solution to the top layer also fortuitously solved all the remaining cubies as well. Each solution to the top layer is also a solution to some full-cube position, and different solution sequences to the top layer are typically full-cube solutions to different initial positions. Indeed, it is easy to compute which full-cube position is solved by a given solution to the top layer: simply start with a solved cube and “undo” the given solution, reversing each individual move and the order of the move sequence.
Since we can use a pattern database to generate many solutions to a given top-layer position, and since each such solution solves some full-cube position, it follows that we can keep track of which full-cube positions we have seen and therefore quickly find solutions to many different positions. This is the key idea behind our effort.
Given a set of move sequences that solve some particular top-layer position, the full-cube positions that they solve are related in that they all have the same initial position of the top cubies. Thus, we can use the position of the top-layer cubies to partition the cube space into subsets, and then separately solve each subset, i.e., solve all positions in the subset. The number of subsets in the partition is the count of top-layer positions, or about 26 billion, and the number of positions in each subset is the count of reachable positions of the remaining cubies, or about 1.7 billion; the product of these two counts is the total number of reachable cube positions.
To generate nonoptimal solutions to the top layer using the pattern database, we need something slightly more complicated than we described above, since not all moves in such sequences will necessarily take us closer to the solution. But we do know that at any point with, say, only n moves left, if our current position is more than n moves from start according to the pattern database, we cannot find a solution from that position and should backtrack. This leads to a straightforward depth-first search with pruning: function search(position p, sequence sofar, int togo) { if (dist[p] > togo) return ; if (togo == 0) record(sofar) ; else for (move m in moves) search(p.move(m), sofar.append(m), togo-1) ; } 1086 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE Because the average distance returned in our pattern database is about ten, this depth-first search can handle very deep searches quickly. Indeed, it takes only O(d) time per solution, where d is the number of moves in the solution.
In order to solve an entire subset in our partition, we need to keep track of which positions in the subset have been solved and those that remain unsolved. For all solutions generated, the above algorithm performs a record() operation, which is responsible for keeping track of the positions solved by the sequences generated by the search. We assign an index to each position in the subset and then use a Boolean array (stored densely as a bitmap) to track which positions we have seen.
If we generate solution sequences ordered by length, then the first time we see a solution for a particular position, we know it is an optimal solution to that position. This is the principle behind the Iterated Deepening A algorithm (IDA) , except, instead of searching for the solution to a single position, we search for solutions to all the sequences in a large set. Once all the bits in the Boolean array are set, we have found a solution to all positions in the subset, and we know precisely how many were of each length. In one run of this algorithm, for a given partition, searching out to whatever depth is needed, we solve 1.7 billion distinct cube positions.
The problem with this approach is that one might see millions of solutions to an easy position before seeing the first solution to one of the more difficult (deeper) positions. One fix is to just run the algorithm for a certain amount of time, or until a certain relatively small number of positions are left, and then solve all remaining positions using a different technique. For instance, one might enumerate all solutions in the subset through length 19 and then solve the very few remaining positions with another program.
But there is another fix that is even more effective. Once the top layer is solved, no twist of the bottom layer will change the top layer. Therefore, any solution to the top layer can be extended by a move of the bottom layer to create a new, different solution to the top layer. With some careful indexing and layout of the bitmap, we can extend an entire set of already-solved positions by one of these bottom layer moves using fast table lookups and logical bit operations. (How this can be done will be described in a later section.) For example, if we have explored all solutions through length 13, we can then extend this to all solutions of length 14 that end in a bottom-layer move in a single fast pass over the bitmap, with no need for additional search. For each original position in the bitmap, there are three possible bottom-layer moves to extend them, essentially multiplying the effectiveness of our search. We call this operation a prepass because we run it before our search, which then generates all sequences at each level: for (int depth=0; ; depth++) { prepass() ; // extend position[a] set at depth d-1 to depth d search(inputpos, empty-sequence, depth) ; // search for new positions at distance d } With this structure we can also change our search routine to never generate solu-tion sequences that end in a bottom layer move; this is a somewhat subtle optimization but improves the search speed. In practice, the prepass eliminates a level of search; a full depth-18 search followed by the fast prepass operation solves a very large frac-tion of distance-19 positions. Without the prepass operation we would need a good fraction of a depth-19 search to solve an equivalent number of positions.
4. The subgroup H and its cosets. The partitioning described above is a partitioning of a group into the cosets of a subgroup. The group is the full space THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1087 of cube positions; the subgroup is the subset of positions that solve the subgoal (for instance, with the top layer solved). If the move sequence m solves the subgoal for an initial position p, then it also solves the subgoal for all positions that are in the same coset as p.
The primary difficulty with exploring any state space this large is keeping track of which states have been seen so far and which have not. By partitioning the space into subsets small enough to fit in memory and solving each subset independently, we can explore much larger state spaces.
Let us consider some subgroup H. To examine a specific right coset C of this subgroup we take an arbitrary representative p of C, so C = H · p. With a search at depth d we find all move sequences m of length d that take the initial position p into the subgroup, that is, p · m ∈H. For each of these move sequences m there is exactly one cube position of C which is solved by this move sequence, namely, m′, where m′ is the inverse of m. The sequence m′ is in C because p · m is some h ∈H with p = h · m′, and therefore h′ · p = m′ so m′ ∈H · p.
We keep track of solved cube positions with a bitmap. When the search at depth d is complete, we know which cube positions of C are solvable within d moves. When we use the term search in the following text we mean this procedure. A carefully chosen subgroup H has the property that there are some single moves that, when applied to a position in H, result in a position still in H. We use these moves to extend the bitmap obtained from the depth-d search to find all cube positions in C which are solved by sequences of length d + 1 which end in such a move. We call this a prepass because we apply it before the search at depth d + 1.
When we talk about the subgroup H from now on, we refer to that well-known subgroup that is the second goal state of Thistlethwaite’s four-stage algorithm , and the first goal state of Kociemba’s two-phase algorithm . As we shall see in this section, this subgroup choice provides unique advantages to our prepass technique, so much so that the bulk of the computational work of our proof was done with the prepass, and not with search. Furthermore, this subgroup is small enough that the bitmap used in the search fits into memory.
The top-layer-solved subgroup we used in the prior section is easy to visualize; we just ignore cubies not originally from the top layer. An informal way to understand the H subgroup is to change a few of the stickers on the solved cube and remove some others, as shown in Figure 4.1(b). First, we give opposite faces of the cube the same color stickers, and second, we remove the stickers with a vertical orientation from the first and third layers of the cube. In this restickering, all the corner cubies are identical, and there are only two distinguishable types of edges: middle edges and top/bottom edges. This relabeled puzzle is much simpler to solve than the full cube puzzle, much like solving only the top layer is much easier than solving the whole cube.
We are relabeling cubies, and not positions; another description of the same relabeling that applies to any scrambled position goes as follows. Let us assume that the colors of opposite faces of the solved cube are yellow and white, blue and green, and red and orange and that yellow and white are oriented so they are the top and bottom faces, respectively. We relabel all yellow stickers to white, all blue stickers to green, and all orange stickers to red. Next, any sticker that is adjacent to a white sticker on the same cubie, we remove.
The unique solved position of this relabeled cube is just the relabeled full cube: top and bottom faces are solid white, and the four other faces have a single horizontal stripe of a single color. The reason the prepass is so effective when this subgroup is used is that so few moves affect this solved position; no twist of the top or bottom 1088 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE (a) (b) (c) (d) Fig. 4.1. (a) The normal coloring of a solved cube, (b) H-wise relabeling of the solved position, (c) the position generated by R1L3U2L1R3, and (d) the coset H · R1.
face, nor any half-twist of any of the other four faces, changes the appearance of the relabeled cube. This means our prepass can use ten of the 18 moves to extend the set of found positions, rather than just three when using the top-layer subgroup. If our search phase examined n sequences to determine the found positions, then a single prepass operation on those positions is essentially generating 10n additional sequences, and possibly many additional positions, all in a single pass over the bitmap.
It is convenient to provide short names to the faces and moves.
Viewing the puzzle’s surface from its exterior as usual, a clockwise quarter-twist of the “up face” (U) is denoted here by U1, a half-twist by U2, and a counterclockwise quarter-twist by U3. We do likewise for the remaining faces: down (D), right (R), left (L), front (F), and back (B). This defines 18 moves; we refer to this set of moves as S. The cube group G can then be viewed as the action of the free group ⟨S⟩on the set of cubie (individual block) stickers, also known as facelets, and the state of the puzzle is determined by the placement and orientation of the noncenter cubies (8 corners and 12 edges). The subgroup H is defined as being generated by all moves of the U and D faces and half-turns of the remaining faces. We define this subset of ten moves as (4.1) A = {U1,U2,U3,D1,D2,D3,F2,B2,L2,R2} and the subgroup so generated as H = ⟨A⟩. We divide the 43,252,003,274,489,856,000 distinct cube positions of the cube group G into 2,217,093,120 cosets of H in G.
We will simply call them cosets of H since, as with all later coset spaces introduced in this paper, the parent subgroup will always be G. Each coset of H is composed of 19,508,428,800 cube positions, which is small enough that a check-offtable can comfortably fit in the RAM on a good modern desktop system. We chose this approach because we are able to solve every element of a coset of H with at most 20 moves and, crucially, at an extremely rapid speed, by a method closely related to Kociemba’s two-phase algorithm.
If we relabel a solved cube as described earlier, the moves from A—which generate H—do not change the relabeled cube. So all cube positions in H show this color THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1089 Fig. 4.2. Three positions in the coset H · R1.
pattern after relabeling. But can we also be sure that all cube positions that show this color pattern after relabeling are elements of H?
For example, the position generated by R1L3U2L1R3, shown in Figure 4.1(c), shows the color pattern of the solved cube after relabeling, but since its generator uses moves not in A it is not clear that it is an element of H.
In fact, this position can also be generated by F2U2R2F2R2U2R2F2R2, so it is an element of H. The fact that any position with H-wise relabeling identical to that of the solved cube can be generated using only moves defining H is not obvious but can be verified either by a brute force calculation or by using stabilizer subgroup algorithms (e.g., as built-in to the computer algebra system GAP). It can also be proved with pen and paper alone by adapting well known group-theoretic arguments, e.g., as presented in sections 11.2 and 12.1 of .
There are 8! ways to permute the corner (cubies) without changing the color pat-tern of the relabeled cube, while the 4 red/green edges of the middle slice can be permuted in 4! ways, and the remaining 8 edges with the white facelets can be per-muted in 8! ways. So H would have 8!8!4! elements if we were allowed to scramble the puzzle by disassembling and then reassembling the cubies. Since we are constrained to only turning the faces, and since each face turn is an even permutation on the cubies, we find that |H| = 8!8!4!/2 = 19,508,428,800. As it is impossible to twist a corner or flip an edge on the relabeled cube without destroying the color pattern, flips and twists do not contribute to enumerating the elements of H.
The cosets of H can be visualized in a similar way, by a specific color pattern of the H-wise relabeled cube, such as in Figure 4.1(d). The three positions depicted in Figure 4.2 belong to the same coset because their H-wise relabelings are identical. In fact, they are identical to the same pattern shown in Figure 4.1(d), which depicts the right coset H · R1 = {h · R1 : h ∈H}.
We now calculate the number of possible color patterns (i.e., the number of cosets of H) for an H-wise relabeled cube. First note that there are 12 4 ways to assign the location of the 4 red-green edges. Flipping an edge or twisting a corner always gives a different pattern, as we noted above, so if disassembling the cube were permitted, we would get 21238 possibilities for flips and twists. But as before, the move constraints allow only 21137 ways, since the flip/twist states of the final edge and corner are forced. As a result, there are 12 4 21137 = 2,217,093,120 different cosets of H.
The subgroup H exhibits 16-way symmetry.
When we define an equivalence relation on the cosets of H by this symmetry, we find there are only 138,639,780 such equivalence classes, so we only need to solve a single representative of each. Careful consideration of a decomposition of the cosets of H into cosets of a smaller subgroup permits us to take advantage of more of the natural 48-way symmetry of the cube and reduce the number of cosets to be solved down to 55,882,296. Details of this reduction will be published separately.
1090 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE 5. Sequences of moves. This section explains the relationship between move sequences and positions, defines the notion of canonical sequences from which an effi-cient search method is devised, and extends this notion to sequences ending in moves from A. Our coset solver exploits the fact that the set A of generators of the subgroup H is a relatively large subset of the set S of all moves. For a set of positions in H, right-multiplication by moves in A results in positions in H. We can perform this operation (called a prepass) in bulk extremely efficiently, and this is the key to our overall result.
The performance of the coset solver is driven by practical considerations, so we must specify a hardware baseline. All performance numbers in this paper are measured on a Nehalem X3460 CPU running at 2.8GHz with four cores and hyperthreading enabled.
We distinguish core seconds (execution on one core) and CPU seconds (execution on the entire CPU). Our primary metric throughout is CPU seconds.
Our coset solver maintains a bitmap indicating which of the 19,508,428,800 el-ements of the coset have been solved at every point. Two primary techniques are used in the coset solver to check offpositions in the bitmap. The first technique is an extremely fast search that enumerates canonical sequences of a particular length and marks the resulting positions in the bitmap. This search runs at about 25 million positions per second. At small lengths, when there are few canonical sequences (as described below), this completes quickly, while at larger lengths, with exponentially more positions, it can take much longer. The second technique is the prepass; it takes the set of positions already found and extends that set by all moves in A using a quick scan over memory and some logical operations. It is called a prepass because we use it for a particular depth before a search at that depth. The prepass is much faster than search on a per-position basis because it processes 24 positions and ten moves at a time in a short inner loop composed of just 61 machine instructions; this executes group operations at a rate of 65 billion per CPU second. It forms the heart of our program, where most of the time is spent and where the most positions are found.
Combined judiciously, these two techniques process a coset in, on average, about 20 seconds, or about a billion positions per second.
When searching for solutions to positions, it is important to minimize redundancy in the search tree.
Separately searching sequences such as U1D2 and D2U1 that reach the same position in the same number of moves can slow down the search dramatically. In addition, sequences such as U1U2 are clearly suboptimal and should also be avoided. We define a canonical sequence as a sequence that obeys a specific order for adjacent individual moves that commute and is free of consecutive moves of the same face. These two restrictions alone are surprisingly effective.
We begin with a standard technique from linear algebra to determine the number, denoted here qn(S), of canonical sequences of length n from the set S of all 18 moves.
(We could use a simpler approach here, but explicit matrix arithmetic makes it easier to extend these ideas to other sets of sequences, as we do later.) Ordering this set as S = {U1, U2, U3, F1, F2, F3, R1, R2, R3, D1, D2, D3, B1, B2, B3, L1, L2, L3}, we decree that a two-move sequence sisj of elements from S is canonical if and only if the (i, j)th entry of the following 18 × 18 incidence matrix C is 1: C = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ Z W W W W W W Z W W W W W W Z W W W Z W W Z W W W Z W W Z W W W Z W W Z ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1091 where W = ⎛ ⎝ 1 1 1 1 1 1 1 1 1 ⎞ ⎠and Z = ⎛ ⎝ 0 0 0 0 0 0 0 0 0 ⎞ ⎠.
Any move is permitted as the first move in a canonical sequence and serves as the base case for the following inductive procedure to specify and enumerate the canonical sequences of a given length. To initialize the process, we let a be the 18-entry column vector composed solely of 1’s, and denote its transpose by a⊤= (1, . . . , 1 18 1’s ).
The product a⊤C is then a row vector which encodes, for each of the 18 possible moves, the number of canonical two-move sequences ending with that move: a⊤C = (12, . . . , 12 nine 12’s , 18, . . ., 18 nine 18’s ).
For example, the first entry in this row vector indicates that there are 12 canonical two-move sequences that have U1 as the second move. To be canonical, their first move must be one of the 12 possible moves of F, R, B, or L, because consecutive moves of the same (U) face must be avoided and we require twists of the U face before twists of the D face whenever these faces occur consecutively. To get the total number q2(S) of two-move canonical sequences, we multiply a⊤C by the column vector a, giving the 1 × 1 matrix a⊤Ca = (243). Extending this and dropping the matrix parentheses as is usual for 1 × 1 matrices, we have qn(S) = a⊤Cn−1a.
Since all of our restrictions can be categorized according to faces, we can collapse rows and columns together and instead redefine the matrices a⊤= (1, . . . , 1 six 1’s ), C = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 3 3 3 3 3 3 0 3 3 3 3 3 3 0 3 3 3 0 3 3 0 3 3 3 0 3 3 0 3 3 3 0 3 3 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , in which case we get qn(S) = 3a⊤Cn−1a.
Using this formula one can readily derive the recurrence relation qn(S) = 12qn−1(S) + 18qn−2(S) (n ≥3), which can also be seen combinatorially (see p. 34 of ). The matrix approach allows easy adaptation to further restrictions on allowed moves, which we use when analyzing the prepass.
When we compare the count of canonical sequences of length d to the known count of Rubik’s Cube distance-d positions in Table 5.1, we find the values are close 1092 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE Table 5.1 The count of canonical sequences of length d and positions at distance d. Note how close the values are through d = 16. The count of cube positions at distance 15 is a new result from this work.
The approximate values are from and .
d Canonical sequences Positions 0 1 1 1 18 18 2 243 243 3 3,240 3,240 4 43,254 43,239 5 577,368 574,908 6 7,706,988 7,618,438 7 102,876,480 100,803,036 8 1,373,243,544 1,332,343,288 9 18,330,699,168 17,596,479,795 10 244,686,773,808 232,248,063,316 11 3,266,193,870,720 3,063,288,809,012 12 43,598,688,377,184 40,374,425,656,248 13 581,975,750,199,168 531,653,418,284,628 14 7,768,485,393,179,328 6,989,320,578,825,358 15 103,697,388,221,736,960 91,365,146,187,124,313 16 1,384,201,395,738,071,424 ≈1,100,000,000,000,000,000 17 18,476,969,736,848,122,368 ≈12,000,000,000,000,000,000 18 246,639,261,965,462,754,048 ≈29,000,000,000,000,000,000 19 3,292,256,598,848,819,251,200 ≈1,500,000,000,000,000,000 20 43,946,585,901,564,160,587,264 ≈300,000,000 Table 5.2 Search time for one coset of H up to a given depth, count of sequences with n moves from S followed by m moves from A, time for m prepass operations, and total time (prepass plus search).
Note how the total time increases much more rapidly than the sequence count as n increases past 15.
n Search time m qn,m(S, A) Prepass time Total time 14 0.14 seconds 6 8.10 × 1020 18.6 seconds 18.7 seconds 15 1.9 seconds 5 1.59 × 1021 15.5 seconds 17.4 seconds 16 25 seconds 4 3.13 × 1021 12.4 seconds 37.4 seconds 17 5.6 minutes 3 6.18 × 1021 9.3 seconds 5.8 minutes 18 1.2 hours 2 1.21 × 1022 6.2 seconds 1.2 hours 19 16 hours 1 2.44 × 1022 3.1 seconds 16 hours 20 9 days 0 4.39 × 1022 0 9 days through distance 16; this indicates that basing a search on canonical sequences is sufficient to nearly eliminate any inefficient repeated search of the same position.
The count of canonical sequences allows us to estimate the runtime of a search through a particular depth. Empirically, our search implementation can visit about 25 million sequences per second. There are 2113712 4 = 2,217,093,120 cosets of H, and the canonical sequences are split among these cosets. Our time required to process a single coset by canonical sequences at depth n, in seconds, is therefore on average about qn(S)/(2,217,093,120 × 25,000,000).
Below depth 14, the time is dominated by the setup time. For depths d ≥14, we get the search times per coset given in the first two columns of Table 5.2 by summing the above for n from 1 to d.
One technique to prove a diameter of 20 for a given coset, then, would be to do a full search to depth 19 within the coset, thus finding all distance-19 positions in the coset. Any remaining positions must be of distance 20 or more. We then execute a THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1093 single prepass, which will almost certainly prove that all remaining positions in the coset are of distance 20. But if it does not, and there are remaining positions, we can solve them separately using Kociemba’s two-phase solver. This technique for a single coset would take about 16 CPU hours. All 138,639,780 distinct cosets of H would require about 280,000 CPU years, a substantial improvement over the 3.6 million CPU years calculated earlier, but clearly still not feasible.
There is a better approach than this for finding optimal solutions for every position in a coset. A full depth-18 search followed by a prepass finds all distance-18 positions and the majority of distance-19 positions in that coset. Any leftover positions are distance-19 or deeper, and they are thought to be very few . On these leftover positions, we first use Kociemba’s two-phase algorithm, limited to a few seconds per position. This succeeds a majority of the time. For the remaining positions, we apply an optimal solver, which runs at about 2×106 solutions per second. This combination of techniques reduces the search time for finding the overall diameter of the group to 21,000 CPU years.
For the purposes of proving that the diameter is 20, positions need only be solved in at most 20 moves each, that is, we do not need to find optimal solutions for every position. We do this by searching to a lesser depth (say, n) and then performing 20−n prepasses to find distance-20 solutions. Choosing the right value for n is critical; a smaller n will run more quickly but increases the chance of having many remaining positions without solutions of 20 moves or less.
We know that a depth-19 search eliminates almost all positions, and fewer than one in a billion positions have distance 20. We also know that the number of length-19 canonical sequences is about 3.29 × 1021. If we do not search past depth n, but only use prepasses, this restricts the canonical sequences to those where all moves past the nth move must come from A. We can use our matrix representation to compute how many such canonical sequences there are and compare this with q19(S) to find out how deep we need to search to likely cover the space adequately. To this end we create a modified transition matrix CA from C by removing those moves not in A, i.e., the quarter-twists of F, R, B, and L. The resulting matrix, corresponding to the element ordering in equation 4.1, is CA = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 1 1 3 1 1 3 0 1 3 1 1 3 1 0 3 1 1 0 1 1 0 1 1 3 0 1 3 0 1 3 1 0 3 1 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .
As before, we see that the number of canonical sequences with the first n moves from S and the next m moves from A, denoted qn,m(S, A), is given by qn,m(S, A) = 3a⊤Cn−1Cm A a.
The results of this calculation are shown in Table 5.2. Since we are proving a diameter of 20, we have n + m = 20. A complete search at depth 16, extended by four moves from A, may generate about as many positions as a complete search to depth 19.
So we only need do a search to depth 16, requiring about 27 seconds total, followed by four prepasses, requiring about 3 seconds each, to solve the vast majority of the positions in each coset.
1094 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE Empirically, we have found that doing a full search to depth 15 followed by five prepasses usually eliminates all but a few dozen positions. The canonical sequences are not distributed evenly over the cosets, so for some cosets there are significantly fewer length-15 canonical sequences; for these cosets, we do a partial depth-16 search.
In sum, we average about 3s per coset on the search, and 15s for the five prepasses, plus about 1.5s of overhead, for a total of about 19.5s per coset. On average we find 345 positions left per coset, which we can then quickly solve in 20 or fewer moves using Kociemba’s two-phase algorithm. Using this technique, our CPU time requirements are reduced to about 87 CPU years. We were able to further reduce this to 35 CPU years by reducing the count of cosets needed. We plan to describe this improvement in a separate publication.
6. The coset solver.
6.1. Search phase. Our coset solver relies on a fast IDA search routine to seed the prepass. This search routine must enumerate canonical sequences of a given length that bring a particular coset representative p into the group H, calculate which element of H the representative p brings it to, and check and set bits in the subgroup bitmap corresponding to that element. We use a large exact pattern database to guide the search. The pattern database is indexed by the coset of H represented by the current position, and it gives the number of moves required to reach the subgroup H.
6.2. Search optimizations. The set of canonical sequences can be thought of as a tree, with the empty sequence at the root, the sequences of length 1 at the next level down, and so on. Every node represents a particular sequence. Iterated depth-first search walks this tree at increasingly deeper levels. For our coset search with representative position p, we associate with each node the position reached by executing the sequence for that node from the position p. Each of these positions has a distance from the group H. Our pattern database is used to look up that distance.
For a given search depth, we consider all the nodes in the search tree at that depth as leaves, because we do not visit deeper nodes in that iteration of the search.
The pattern database is indexed by cosets of H. There are 2,217,093,120 such cosets, but we exploit the 16-way symmetry of the H group and its cosets to reduce the size of the table to about 170 million entries. For fast move execution and fast indexing, we use a coordinate representation for the coset position, much as Kociemba does with his Cube Explorer program .
For a search to a particular depth, there are only certain canonical sequences of that length that take a coset representative p to the group H. If we enumerate all of these sequences, we must visit all of the corresponding nodes and their ancestors in the search tree. Any time we visit a node that is not one of those solving sequences or a prefix of a solving sequence, we waste execution time. We want to minimize the number of false nodes, whose sequences are neither solving sequences nor prefixes of solving sequences.
Normal IDA search, using a pattern database to provide the heuristic function, will usually explore many more false nodes than useful ones. Every useful node must be visited, and normal IDA search visits all immediate successors of every useful node, most of which are not useful.
The dominant cost of visiting a node in our type of IDA search is querying the pattern database for the value of the heuristic function; the table is relatively large, and thus accesses will likely miss the processor cache and perhaps the translation lookaside buffer (TLB) as well. Every main memory access fetches 64 bytes on our THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1095 Nehalem processor, so we might as well make use of a few more bytes. We do this by including information on the distance of all neighboring nodes, as well as the node it-self; we use this distance information to decide which neighbors are useful even before we calculate their position.
Unlike a normal IDA search where the first successful sequence terminates the search, we want to enumerate all successful sequences. Using the information about the neighbors allows us to only visit useful nodes, walking down the tree to precisely those leaves that correspond to sequences that bring positions p into H. For the leaves, we do not perform a heuristic function lookup as we know all useful leaves correspond to positions in H.
One advantage of using H as the subgroup from which to compute cosets is the useful subtree of the search tree has a much higher ratio of leaves to nonleaf nodes than other subgroups. This means the ratio of successful sequences (one per leaf node, each of which is a new candidate solution) to useful nodes (a basic unit of work in our tree search) is much higher. Consider, by comparison, the subgroup E that fixes edges, so the cosets are defined by the edge positions. The shortest canonical sequence between any two different positions in the subgroup E is quite large, because every move from a position in E messes up the edges, which then need to be put back into place.
In the useful subtree of the search tree for any coset of H, any ancestor node that has useful leaves has at least two useful leaves. The sequence for a given leaf must either end in a move in A or a move not in A. If the sequence ends in a move in A, that means the position associated with the ancestor node is also in H, and thus, it has several children in H, one for each move in A that is permitted after the move that precedes that move. If the predecessor move is a move of the U face, it will have seven useful leaves from moves in A of F2, R2, D2, B2, L1, L2, and L3. If it is a move of the D face, it will have four useful leaves from the moves in A of F2, R2, B2, and L2. If it is a move of F or R, it will have eight useful leaves from any move of U or of D, or a half twist of the opposite face. If it is a move of B or L, it will have six useful leaves from any move of U or D.
If the sequence for a given leaf does not end in a move in A, there are only eight possible moves (those not in A) it could end with: quarter twists of the faces F, R, B, and L. Let us assume without loss of generality (by symmetry) that the last move from the ancestor node to this leaf is F1. Then the move F3 from the ancestor node is also a position in H and hence is a leaf. This is because F1 and F3 differ by F2 which leaves an H position in H. Thus, if the sequence does not end in a move in A, but there are useful leaves (that is, child nodes in H), then there must be exactly two of them.
For depth-16 searches or deeper, we use the prepass to handle all those canonical sequences that end in moves in A, rather than explicitly enumerating them.
At the beginning of a search at depth d, the coset bitmap represents all those elements of H that can be reached by move sequences of length d −1 applied to the coset representative p, so the prepass calculates exactly the same set of positions as a normal search phase at depth d restricted to sequences that end in moves from H. As a consequence, at depth d, we only need to explore sequences that end in a move not in H. This typically saves about 10/18 or 56% of the sequences (since there are ten moves in A but 18 in S). It does not save quite this much execution time because the search nodes it eliminates are exactly those that have more than two useful leaves per ancestor node. Despite this, it is a useful optimization for the deeper searches.
In order to confine our search to sequences not ending in a move from H, we observe that the shortest canonical sequences that take a position that is in H out of H and then back into H have length 5 (such a sequence is R1L3U2L1R3). Any 1096 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE sequence that ends with a move not in A, but ends with a position in H, must enter it on the last move. Thus, any time we enter H early in the search, we must have at least five moves remaining for us to exit H and reenter it; further, we must exit H with at least four moves remaining or we will not be able to get back into H in time.
There are two technical details worthy of mention. First, the use of 2MB TLB pages rather than the default 4K TLB pages can significantly reduce the cost of TLB misses; the libhugetblfs library can enable this easily on modern Linux kernels.
Second, search required to find the next position can be overlapped with the memory access for the previous bit. Normally our search-driven bitmap updating might look like something like void checkpos(position p) { unsigned char addr = calcaddr(p) ; unsigned char b = calcbit(p) ; if (0 == (addr & b)) { // new position found!
count++ ; addr |= b ; } } Modern processors include prefetching instructions that tell the memory system that the processor intends to use a particular memory location in the near future, so it should be loaded into the cache, but without stalling the execution pipeline to do so. We can take advantage of this by issuing such an instruction when we have found a new address to check, but not actually performing the check until we have found another position: unsigned char addr = 0 ; unsigned char b = 0 ; void checkpos(position p) { if (addr != 0 && 0 == (addr & b)) { // new position found!
count++ ; addr |= b ; } addr = calcaddr(p) ; b = calcbit(p) ; prefetch(addr) ; } This idea can be extended to enable good scalability when multithreading the search. The pattern database itself does not need any sort of locking because it is read-only during the search, but updating the coset bitmap must be carefully managed so threads do not obliterate bits set by other threads. To do this, each thread builds a local queue of addresses and bits to check in the table. As it calculates each address, it prefetches the memory address to get the data into the cache. After a thread has accumulated, say, 64 such addresses, it then obtains a global lock, checks and sets all those 64 addresses, and releases the global lock. This eliminates any possibility of concurrent access to the coset bitmap, which guarantees correctness, yet permits the actual search for those addresses to occur in parallel, yielding high scalability.
By making the queue reasonably short, no thread is held up too long waiting for the global lock, but by making it as long as we do, we amortize the cost of getting the global lock. In our testing, we see almost perfect scalability in the search phase up to eight threads.
THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1097 6.3. Prepass phase. Our coset solver maintains a large bitmap that represents the set of positions in that coset for which a solution has already been found. Every position in the coset is assigned an ordinal number, and we use an indexing algorithm to convert the position to that assigned ordinal number, and another algorithm to convert an ordinal number to that position.
The coset positions that are being solved are elements of the coset H · p, where p is a coset representative. Our search constructs sequences m such that p · m is in H (i.e., the sequence m brings the position p back into the coset H). It is much easier to index the elements of the group H than it is to index an arbitrary coset of H, because all positions in H share some common properties that are not shared by the positions of a coset of H. So instead of indexing the position solved by the sequence m, we index the position reached by p · m, which is in H.
Positions in H all have the solved orientation on both edges and corners, so we do not need to consider orientation in our indexing operation. Further, all positions in H have the middle edges always in the middle slots. Thus, we can separate the state of a position in H into three portions: the permutation of the corners, the permutation of the up/down edges, and the permutation of the middle edges. These have sizes 8!, 8!, and 4!, respectively. The overall parities of the three of these must have an even sum, so the total size of the group is 8!8!4!/2; a bitmap of this size requires 2.3GB of memory. We index each of these components separately and combine the results into a single index.
Overall, our prepass simply iterates over the range of the ordinals for the positions in H, and for each bit set in the bitmap, computes all neighbors to that position (by moves in H) and marks them in a new copy of the bitmap. The code essentially does the following: newbitmap = oldbitmap ; for (i : ordinals of H) { if (oldbitmap[i]) for (move m : A) { cubepos p = unindex(i) ; p.move(m) ; newbitmap[index(p)] = 1 ; } } With careful organization of the bitmaps and a handful of lookup tables, we can optimize this routine to execute at a rate of about 65 billion potentially new positions per second on a single CPU. The next section describes how this is done.
6.4. Prepass optimizations. Making the prepass fast primarily involves orga-nizing the bitmap in such a way that group operations can be performed in parallel with bitwise logical operations and relatively small lookup tables. To enable this, we split the overall ordinal indexing of the bitmap into three coordinates: the permuta-tion of the corners, the permutation of the up and down edges, and the permutation of the middle edges (from most significant to least significant).
Because 8! = 40, 320 is so small, for the corners we simply construct an array nextcorner[i][m] that gives the ordinal value for the position of the corners after taking the position for ordinal i and making move m. We do the same thing for the eight up/down edges if we restrict moves to H, and the same thing for the four middle edges via arrays nextudedge[i][m] and nextmedge[i][m]. To reflect parity 1098 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE considerations, where the corner permutation parity must match the edge permutation parity, we do not distinguish the permutation of two of the up/down edges (chosen arbitrarily), since these are forced by the parity of the permutation.
With these three arrays in place, our prepass now looks like for (c : ordinals of corners) { for (eud: ordinals of up/down edges) { for (em: ordinals of middle edges) { int i = (c 8!/2 + eud) 4! + em ; if (oldbitmap[i]) for (move m : A) newbitmap[ (nextcorner[c][m] 8!/2 + nextudedge[eud][m]) 4! + nextmedge[em][m]] = 1 ; } } } If we premultiply the nextcorner array by 8!/2 ∗4! and the nextudedge array by 4!, the inner expression simplifies to newbitmap[nextcorner[c][m] + nextudedge[eud][m] + nextmedge[em][m]] = 1 ; We can hoist the indexing of newbitmap and nextcorner using a temporary array that holds the following sums: for (c : ordinals of corners) { char cptrs[] ; for (move m : A) cptrs[m] = newbitmap + nextcorner[c][m] ; for (eud: ordinals of up/down edges) { for (em: ordinals of middle edges) { int i = (c 8!/2 + eud) 4! + em ; if (oldbitmap[i]) for (move m : A) cptrs[m][nextudedge[eud][m]+nextmedge[em][m]] = 1 ; } } } We now focus on moves of the U and D faces, which are six of the ten moves in A. These moves do not affect the middle layer, so nextmedge[em][m] == em. There are 24 values of em, and we want one bit per value, so we recast our bitmap as an array of 24-bit integers, one integer per (c, eud) combination, with one bit of each integer corresponding to a value of em. We use the bitwise logical “or” operation to perform the test of the source and conditional setting of the destination. We modify the constant multiplications we have done in nextudedge and nextcorner to take this into account, and for the up/down moves only, we end up with for (c : ordinals of corners) { int24 cptrs[] ; for (move m : A) cptrs[m] = newbitmap + nextcorner[c][m] ; for (eud: ordinals of up/down edges) { int emset = oldbitmap[c8!/2 + eud] ; THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1099 for (move m : {U1, U2, U3, D1, D2, D3}) cptrs[m][nextudedge[eud][m]] |= emset ; } } For the remaining moves in A, which are F2, R2, B2, and L2, the middle edge position is modified, so we need to compute this modification. This modification to the middle edges, when a subset of 24 permutations is represented as a bitmask, is just a permutation of those bits. In practice, one must consider that modern commodity CPUs do not include instructions for arbitrary bit permutations. We could use a large lookup table (224 entries for each of these four moves with each entry three bytes, requiring a total of 200MB) to perform the permutation of bits, but the resulting cache misses would destroy the performance we hope to attain. So instead we split the source value into two, 12-bit subwords and use several lookup tables, each much smaller, to perform the bit permutation.
Our mapping of permutations to bits is arbitrary, so we put all the even permu-tations in the low 12 bits of the 24-bit word, and the odd permutations in the high 12 bits. Since all of F2, R2, B2, and L2 are odd permutations on the middle edges, this means low 12 bits from the source will all map to high 12 bits from the destination, and vice versa. We use eight arrays to perform this operation—an odd array and an even array for each of the four moves we handle.
We need additional code to extract the high and low word from the source, do the array lookups, and then combine the result. The code then for just these four moves looks like int evenempermute[1<<12], oddempermute[1<<12] ; for (c : ordinals of corners) { int24 cptrs[] ; for (move m : A) cptrs[m] = newbitmap + nextcorner[c][m] ; for (eud: ordinals of up/down edges) { int emset = oldbitmap[c8!/2 + eud] ; int emodd = emset >> 12 ; int emeven = (emset & 0xfff) ; for (move m : {F2, R2, B2, L2}) cptrs[m][nextudedge[eud][m]] |= (evenpermute[m][emeven] << 12) | oddpermute[m][emodd] ; } } With tiny innermost loop bodies like this, it is advantageous to combine the two loops and unroll the inner bodies.
This inner loop is reading the source bitmap, calculating from it 10 different parts of the destination bitmap, and modifying them in the process. This generates a lot of memory bus traffic to write the updated values back to memory, so we instead turn the loop around and compute the new value for the destination bitmap based on ten different portions of the source bitmap it reaches.
We can do this transformation because A is closed under inversion. This significantly reduces the write traffic. We integrate this operation with the bitmap copy, yielding the following code: int evenempermute[1<<12], oddempermute[1<<12] ; for (c : ordinals of corners) { 1100 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE int24 cptrs[] ; for (move m : A) cptrs[m] = oldbitmap + nextcorner[c][m] ; for (eud: ordinals of up/down edges) { int F2set = cptrs[F2][nextudedge[eud][F2]] ; int R2set = cptrs[R2][nextudedge[eud][R2]] ; int B2set = cptrs[B2][nextudedge[eud][B2]] ; int L2set = cptrs[L2][nextudedge[eud][L2]] ; newbitmap[c 8! / 2 + eud] = oldbitmap[c 8! / 2 + eud] | cptrs[U1][nextedge[eud][U1]] | cptrs[U2][nextedge[eud][U2]] | cptrs[U3][nextedge[eud][U3]] | cptrs[D1][nextedge[eud][D1]] | cptrs[D2][nextedge[eud][D2]] | cptrs[D3][nextedge[eud][D3]] | ((evenpermute[F2][F2set & 0xfff] | evenpermute[R2][R2set & 0xfff] | evenpermute[B2][B2set & 0xfff] | evenpermute[L2][L2set & 0xfff]) << 12) | oddpermute[F2][F2set >> 12] | oddpermute[R2][R2set >> 12] | oddpermute[B2][B2set >> 12] | oddpermute[L2][L2set >> 12] ; } } The straight-line code in the inner loop above seems lengthy, but it performs ten moves on 24 different positions, and it does so with very simple instructions that execute quickly, using small lookup arrays that fit either in the cache or are accessed sequentially.
There are two more improvements we have made to the above code. First, since our arrangement of the mapping of middle edge permutation to bit position is arbi-trary, except that the even positions must be in the lower 12 bits, we arrange it so that the evenpermute and oddpermute arrays for a single one of the moves represent the identity operation. This allows us to remove two of the eight arrays and their lookups.
Further, we do not need to include the oldbitmap[] lookup (the first element of the large disjunction), because at any depth greater than 0, whenever a bit is set in the array, so is a different bit that is related to this bit by a move in H. This is due to the shape of the search trees (as discussed in the previous section), where all leaf nodes always have immediate siblings related by moves in A.
Modern CPU architectures do not have a native 24-bit integer value, but a 32-bit int works fairly well as its replacement. This yields a bitmap that is too large, as eight of the 32 bits are wasted. In our testing on modern Intel and AMD processors, unaligned 32-bit accesses on an array, with the 24-bit integers packed densely, actually performs somewhat better than using the sparser layout.
Our final prepass consists of an inner loop containing 61 instructions including the test and branch at the end. For each three-second prepass operation, this inner loop is executed 8!∗8!/2 or 812,851,200 times. On our Nehalem processor, this is executed across eight logical threads in four physical cores, giving an approximate execution THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1101 rate of 1.5 instructions per cycle per core, which is fairly good, especially considering how much memory traffic is generated. This inner loop handles 240 group operations each execution, so our operation rate is 65 billion group operations per second on this single CPU. The GNU compiled code for this crucially fast loop is provided as an appendix.
There are a few additional but minor optimizations we have performed.
We allocate portions of the new bitmap only as they are needed, and we free portions of the old bitmap upon their last use, so the overall memory impact is significantly lower than twice the size of a single bitmap. We evaluate the corner positions not in ordinal order, but in a different order that enhances the cache hit rate and also minimizes the maximum amount of memory allocated at once.
7. Heuristics. Much of the speed of our approach comes from using only the prepass at greater depths, but the prepass only considers moves in A.
Thus, not finding solutions of length 20 or less for some positions does not mean those positions require more than 20 moves to solve. We can call these positions that remain after a coset search leftover positions; we must consider these separately.
Luckily, the Kociemba two-phase search algorithm, especially when extended to six-axis search , can find length-20 or shorter solutions quickly, at an average rate of about 3900 positions per second. Thus, we want to balance the amount of time spent doing search in the coset solver against the expected count of leftover positions.
In this section we discuss empirically derived heuristics that gave us a reasonable balance.
So far we have done our analysis based on the count of canonical sequences of a particular length divided by the count of cosets, that is, based on averages. In reality, some cosets have many more sequences of a particular length, while some have many fewer. For most cosets, a full search to depth 15, followed by five prepasses, finds solutions of length 20 or less to all positions except a few.
For some cosets this strategy leaves an impractical number of positions remaining to solve with the two-phase algorithm. So for about a third of the cosets we do a partial depth-16 search. We performed a full search to at least depth 15 to compute, as a secondary and corroborating result, the total number of positions at a distance of exactly 15.
On average, a full depth-16 search would require about 27 seconds. Since without the depth-16 search the time per coset is less than 20 seconds, doing a full depth-16 search would more than double the time for those cosets.
For depth 16, we always first do a prepass. In our initial testing, we tried to find a particular count of total positions found at which to terminate the depth-16 search.
However, we found that for the same total position count, the number of leftover positions differed significantly across cosets. Let us explain why.
After the depth-16 search, we only do prepasses. These prepasses only execute moves from A. Let us determine how many next moves there are for a canonical sequence for moves in A and moves not in A. For U1, U2, and U3, we have 7 possible successor moves. For F2 and R2, we have 10 possible successors. For D1, D2, and D3, we have only 4 possible successors, and for B2 and L2, we have 8 possible successors.
The average over these 10 moves is 6.9 successors.
For moves not in A, we have F1, F3, R1, and R3, each with 9 successors; for B1, B3, L1, and L3, we have 8 successors. The average here is 8.5 successors. So moves ending in A have, on average, fewer successors, and are therefore less valuable towards our goal of solving all positions in at most 20 moves.
1102 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE As a solution, we derate positions already found by the depth-16 prepass when calculating how many positions at which to terminate the search. As a result, almost all positions found at depth 16 were found by the prepass. We performed a sequence of experiments to determine how much to derate these positions and at what point to terminate the search. If x denotes the number of positions found in total after the depth-16 prepass, we perform a depth-16 search until the total number of positions found is 167,000,000 + x/3.
8. The big run. Managing nearly 56 million independent coset runs on a large cluster can be difficult. To amortize the overhead of calculating or loading the required pattern database, we program the coset solver to iteratively execute a sequence of cosets in a single execution.
By executing enough cosets in sequence in a single execution, the time to calculate the pattern database was made negligible, and all considerations of loading, storing, or distributing the large table were eliminated. To improve the credibility of our results, we calculate a hash of each pattern database a priori and validate that hash in memory at the beginning and end of each run.
To ease management of which cosets need to be solved, which are pending, and which are completed, we integrate the list of which cosets were needed directly into the source of the coset solver, assign each an ordinal, and make each execution be responsible for a contiguous range in that ordinal.
We also use the integrated coset list to compute the exact count of positions at each distance from 0 to 15. As Korf’s article indicates, at depth 10 a breadth-first search approach is already challenging. We include logic that, for the set of positions at every level, computes which of those positions this particular coset was the canonical coset (the lowest-indexed coset) for, and which of the positions are duplicated in an earlier coset.
Since the count of positions through distance 14 is already known , verifying the sums of these counts across all 56 million runs provides a certain amount of validation of our results. We always perform a depth-15 search to completion, so we can also compute precisely the number of positions at that distance.
Almost all positions solved in one coset have a corresponding inverse position of the same distance in a different coset. This additional redundancy also helps validate our result.
All cosets and leftover positions were run on Google servers during July of 2010.
There were a total of 19,260,301,834 leftover positions (an average of 345 leftover posi-tions per coset) that were individually solved using our six-axis two-phase algorithm.
Using a simple Perl script to iterate through the logs and sum up the positions at each depth, we were relieved to find that the counts confirmed the known results for distances 0 through 14. We announced our new result for the count of positions at distance 15 (91,365,146,187,124,313) in July, 2010 . This result was independently confirmed by Scheunemann in August, 2010 .
Because performance and specifications of the Google computers are confidential, and also to partially validate the results, we ran a subset of the run (80,100 of the cosets, or about one in every 700) on an individual Nehalem machine and compared the results against those from the Google runs. We used performance numbers from these local runs to extrapolate the overall runtime had the full run been done on Nehalem machines. This is the basis for the performance numbers throughout.
Our sample cosets averaged 19.622 seconds each to run, including the time re-quired to solve the leftover positions, so we extrapolate the total run as 55,882,296 times this or 34.75 CPU years. Of this time, 98.99% was in the coset solver, and THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1103 1.01% was in the two-phase solver for the leftover positions. About 16.0% of the total time was used for search, with 9.45% of the total time used for just depth-15 search.
Prepasses used 81.14% of the total time, with the average single prepass taking 3.119s.
We note that leftover positions generally took a bit longer than random positions for the two-phase solver to solve, a rate of only 1660 (vs. 3900) per second.
9. Future work and conclusions. We have computed the diameter of the Rubik’s Cube group, but there is much still unknown about this group. In particular, the exact counts of positions at distances 16 through 20 are still unknown. Without an exact distance distribution, many may consider this puzzle still unsolved.
The diameter of the group in both the quarter turn and the slice turn metric is still unknown. The best known results on the quarter turn metric are a lower bound of 26 and an upper bound of 29 ; for the slice turn metric we know a lower bound of 18 and an upper bound of 20 (this paper). We anticipate God’s number for the quarter turn metric to be 26, and for the slice turn metric to be 18.
Our approach of partitioning a large search space by cosets of a subgroup provides significant breathing room over standard in-memory breadth-first search, without introducing the latency or bandwidth limitations of keeping the bitmap in secondary storage. We plan to use this technique on additional problems.
10. Appendix. The inner loop of the prepass phase as compiled by the Gnu compiler collection is shown here. By handling 65 billion group operations per second per CPU, its efficiency was key to our main result.
.L13: movzwl 6(%r10), %eax movq -48(%rsp), %rdx movzwl 2(%r10), %ecx movl (%rdx,%rax), %r9d movzwl 8(%r10), %eax movq -40(%rsp), %rdx movl (%rdx,%rax), %esi movzwl 16(%r10), %eax movq -32(%rsp), %rdx movl (%rdx,%rax), %edi movzwl 18(%r10), %eax movq -24(%rsp), %rdx movl (%rdx,%rax), %r8d movzwl 4(%r10), %eax movl (%r15,%rax), %edx movq -16(%rsp), %rax orl (%rax,%rcx), %edx movzwl (%r10), %eax orl (%r14,%rax), %edx movzwl 10(%r10), %eax orl (%r13,%rax), %edx movzwl 12(%r10), %eax orl (%r12,%rax), %edx movzwl 14(%r10), %eax addq $20, %r10 orl (%rbp,%rax), %edx movl %r9d, %eax andl $4095, %r9d 1104 ROKICKI, KOCIEMBA, DAVIDSON, AND DETHRIDGE sarl $12, %eax andl $4095, %eax orl %eax, %edx movl %esi, %eax andl $4095, %esi sarl $12, %eax andl $4095, %eax movswl rearrange+24576(%rax,%rax),%eax orl %eax, %edx movl %edi, %eax andl $4095, %edi sarl $12, %eax movswl rearrange+8192(%rdi,%rdi),%ecx andl $4095, %eax movswl rearrange+8192(%rax,%rax),%eax orl %eax, %edx movl %r8d, %eax andl $4095, %r8d sarl $12, %eax andl $4095, %eax movswl rearrange+40960(%rax,%rax),%eax orl %eax, %edx movswl rearrange(%rsi,%rsi),%eax orl %ecx, %eax movswl rearrange+16384(%r8,%r8),%ecx orl %r9d, %eax orl %ecx, %eax sall $12, %eax orl %eax, %edx movl %edx, (%r11) addq $3, %r11 cmpq %rbx, %r11 jne .L13 Acknowledgments. We extend our sincere thanks to Google for providing the CPU power required to finally put this question to rest.
We also owe a debt of gratitude to the two anonymous referees for their advice, corrections, and suggested improvements to this paper.
REFERENCES J. Culberson and J. Schaeffer, Pattern databases, Comput. Intelligence, 14 (1998), pp. 318– 334.
D. Joyner, Adventures in Group Theory: Rubik’s Cube, Merlin’s Magic & Other Mathematical Toys, The John Hopkins University Press, Baltimore, MD, 2008.
H. Kociemba, Close to God’s algorithm, Cubism For Fun, 28 (April 1992), pp. 10–13.
H. Kociemba, Cube Explorer (Windows program),
H. Kloosterman, Rubik’s cube in 42 moves, Cubism For Fun, 25 (1990), pp. 19–22.
R.E. Korf, Depth-first iterative-deepening: An optimal admissible tree search, Artificial Intel-ligence, 27 (1985), pp. 97–109.
R.E. Korf, Finding optimal solutions to Rubik’s cube using pattern databases, in Proceedings of the Fourteenth National Conference on Artificial Intelligence (AAAI-97), Providence, RI, 1997, pp. 700–705.
R.E. Korf, Linear-time disk-based implicit graph search, J. ACM, 55 (2008), pp. 1–40.
D. Kunkle and G. Cooperman, Twenty-six moves suffice for Rubik’s cube, in Proceedings of the International Symposium on Symbolic and Algebraic Computation (ISSAC ’07), 2007, ACM Press.
THE DIAMETER OF THE RUBIK’S CUBE GROUP IS TWENTY 1105 D. Kunkle and G. Cooperman, Harnessing parallel disks to solve Rubik’s cube, J. Symbolic Comput., 44 (2009), pp. 872–890.
S. Radu, Solving Rubik’s Cube in 28 Face Turns, view/37 (2005).
S. Radu, New Upper Bounds on Rubik’s cube, download/risc 3122/uppernew3.ps (2005).
M. Reid, New Upper Bound, (1992).
M. Reid, New Upper Bounds, (1995).
M. Reid, Superflip Requires 20 Face Turns, (1995).
M. Reid, Superflip Composed with Four Spot, (1998).
T. Rokicki, In Search of: 21fs and 20fs; A Four Month Odyssey,
org/drupal/?q=node/view/56 (2006).
T. Rokicki, Twenty-five moves suffice for Rubik’s Cube, arxiv.org/abs/0803.3435, 2008.
T. Rokicki, Twenty-Three Moves Suffice, 117 (2008).
T. Rokicki, New estimate for 20f: 300,000,000, view/167 (2009).
T. Rokicki, 1,000,000 Cubes Optimally Solved, view/172 (2010).
T. Rokicki, Twenty-two moves suffice for Rubik’s cube, Math. Intelligencer, 32 (2010), pp. 33– 40.
T. Rokicki, 15f = 91365146187124313, 197 (2010).
T. Scheunemann, God’s Algorithm out to 14f, view/191 (2010).
T. Scheunemann, God’s Algorithm out to 15f, view/201 (2010).
D. Singmaster, Notes on Rubik’s Magic Cube, 5th ed., Enslow Publishers, Hillside, NJ, 1981.
M. Thistlethwaite, 52-move algorithm for Rubik’s Cube, thistle.htm (1981).
D.T. Winter, New upper bound on God’s algorithm for Rubik’s cube,
ucf.edu/˜reid/Rubik/Cubelovers/ (1992). |
6728 | https://www.fldoe.org/core/fileparse.php/7576/urlt/Grade3_MultandDivUnit.pdf | Grade 3 Multiplication and Division Unit of Study This is a progressive unit of instruction beginning with students exploring multiplication as a set through the literature Amanda Bean’s Amazing Dream. Students then explore multiplication as an array which takes them to understanding the distributive property. Although these lessons are multiplication and division specific as they were written, it’s very easy to teach your students the concepts of multiplication and division side by side as they manipulate the objects and write the equations. In doing so you are walking your students from concrete to representational and finally abstract. Teaching your students the distributive property and encouraging them to compose and decompose numbers will support their multiplication fact fluency. Operations and Algebraic Thinking Students develop an understanding of the meanings of multiplication and division of whole numbers through activities and problems involving equal-sized groups, arrays, and area models; multiplication is finding an unknown product, and division is finding an unknown factor in these situations. For equal-sized group situations, division can require finding the unknown number of groups or the unknown group size. Students use properties of operations to calculate products of whole numbers, using increasingly sophisticated strategies based on these properties to solve multiplication and division problems involving single-digit factors. By comparing a variety of solution strategies, students learn the relationship between multiplication and division. A bibliography of children's literature with a focus on multiplication is provided, which can be integrated so that students can connect through literature. 1. Amanda Bean’s Amazing Dream, Cindy Neuschwander 2. The Grapes of Math, Greg Tang 3. Each Orange Had 8 Slices, Paul Giganti 4. The Doorbell Rang, Pat Hutchins 5. One Grain of Rice, Demi 6. Sea Squares, Joy Hulme 7. The Hershey’s Multiplication Book, Jerry Pallotta 8. The Lion’s Share, Matthew McElliot 9. The Best of Times, Greg Tang 10. 7 x 9 = Trouble, Claudia Mills 11. 2 x 2 = Boo!, Loreen Leedy 12. Math Attack!, Joan Horton & Krysten Brooker 13. The King’s Chessboard, David Birch & Devis Grebu 14. Ten Times Better, Richard Michelson Cluster 1: Represent and solve problems involving multiplication division Cluster 2: Understand properties of multiplication and the relationship between multiplication and division Cluster 3: Multiply and divide within 100 Cluster 4: Solve problems involving the four operations, and identify and explain patterns in arithmetic Parent Resources How to Teach the Multiplication Tables to Your Child Using Arrays to Multiply 354 The Multiplication Game 15. Divide and Ride, Stuart Murphy 16. One Hundred Hungry Ants, Elinor Pinczes 17. One Hungry Cat, Joanne Rocklin & Rowane Murphy This literature is referenced in a lesson below. Introduction to Multiplication using Literature This lesson plan can be used to introduce the concept of multiplication to students through the use of literature. The story Amanda Bean's Amazing Dream is used to demonstrate the different ways to count items and how multiplication can make that process much faster. MAFS.3.OA.3.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. All About Multiplication This four-lesson unit from NCTM's Illuminations has students explore several meaning and representations of multiplication (number line, equal sets, arrays, and balanced equations). Other multiplication topics covered include: the commutative (order) property, the results of multiplying by 1 and 0, and the inverse property. Students will write and solve multiplication story problems, and convert word problems into equations. MAFS.3.OA.1.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. MAFS.3.OA.3.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. How Many Circles? How Many Stars? 250 How many Circles? How many Stars? is an activity that will give students a visual representation of multiplication and repeated addition. It will also help students see multiplication as the combining of equal-size groups that can be represented with a multiplication equation. MAFS.3.OA.1.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. Each Orange Had 8 Slices: Multiplying Equal Groups 278 Students will learn how to represent and count equal groups through the use of literature and situational story problems. Using the story Each Orange Had 8 Slices, students will use manipulatives to create arrays to assist calculation of equal groups. Students will learn to write corresponding addition and multiplication sentences for the arrays. MAFS.3.OA.1.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. Skip Counting to Multiply (2’s, 3’s, 5’s and 10’s) Students will build a conceptual understanding of multiplication by creating a hundreds chart, using different colors to assist them with skip counting by 2, 3, 5 and 10. Students will discuss; "How many groups of (2, 3, 5 and/or 10)?" are in each number MAFS.3.OA.3.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Introduction to Division "The lesson will help students develop an initial understanding of division and clarify how the four operations of addition, subtraction, multiplication, and division relate to and are separate from each other. The lesson begins with a brainstorming discussion which builds background and fosters comprehension. A big book, The Doorbell Rang, by Pat Hutchins, is used along with manipulatives to provide instruction at concrete and pictorial levels. Students will demonstrate what they have learned by writing a short story incorporating simple division." (ALEX - Alabama Learning Exchange) MAFS.3.OA.1.2: Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Pet Store Partitive Division 840 In this lesson students will model partitive division through the real-world activity of a pet store owner. MAFS.3.OA.1.2: Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Grandma Wants to Know! 284 Help Mom and Dad tell Grandma about Cindy's trip to the carnival using bar models and arrays to relate division to multiplication with an unknown factor. MAFS.3.OA.2.6: Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8 Cheezy Arrays 412 This lesson is a hands on activity that includes multiplication using arrays. The lesson also serves as a great transition from repeated addition to multiplication. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.1.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Amazing Arrays This is a hands-on lesson for introducing and practicing building arrays to create models that represent the distributive property of multiplication, and then using those arrays to draw models of the equations they represent. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.3.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. MAFS.3.OA.2.5: Apply properties of operations as strategies to multiply and divide Discovering the Mystery Factor Through Arrays 450 Students will begin with the use of manipulatives to solve for unknown factors by building arrays. They will progress to drawn models as mastery is shown with manipulatives. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.1.4: Determine the unknown whole number in a multiplication or division equation relating three whole numbers. Three is NOT a Crowd! This lesson will provide students with practical application activities to help them understand how division is simply solving a problem with an unknown factor. Students will be given snacks in which they must share with their group members equally. This activity includes opportunities for students to use fact families when identifying unknown factors and a tic-tac-toe game to provide whole group practice as well as to be used at a center, or for independent reinforcement of the skills. MAFS.3.OA.2.6: Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8 The Array Frame, your best friend 733 In this lesson, students will learn to use the structure of array frames to build familiarity and fluency with the array as a tool. Students will solve several multiplication word problems using the array as a representation. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Giddy Up, Round Up: Relating Division to Multiplication 479 In this lesson, students will learn to solve division problems by relating them to multiplication facts. Practice materials focus on the 6's and 8's multiplication facts. MAFS.3.OA.1.4: Determine the unknown whole number in a multiplication or division equation relating three whole numbers. Making Sense of Multiplication to Build Fluency of 6’s, 7’s, 8’s, and 9’s This lesson will help students multiply numbers with factors of 6, 7, 8, or 9 through decomposing numbers in an array and applying the distributive property. Many times, these factors are difficult for students to recall from memory. Teaching students how to use an array can give them a visual representation of the final product. This visual can also help students to make the connection that multiplying whole numbers is a sum of equal groups. Decomposing the numbers and using the distributive property is a strategy for students to use who are having trouble solving these higher factor multiplication facts. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.3.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. Apples, Oranges, and Bananas of Math? In this lesson, the students will work in independently or in small groups to write their own math riddles around the concepts of multiplication. The teacher will use the book, The Grapes of Math by Greg Tang. MAFS.3.OA.1.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Tang, to support this lesson. Cookies for All 560 This lesson allows students to use everyday objects in order to understand equal shares. The lesson uses "The Doorbell Rang" by Pat Hutchins to engage students during the lesson and to make a connection by using literacy in mathematics. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Make Your Way With Arrays 450 This lesson walks the students from representational to the abstract very nicely! Students will solve multiplication and division word problems by drawing arrays and writing the related equation. Use Graph paper to show the arrays – concrete to representational. MAFS.3.OA.1.3: Use multiplication and division within 100 to solve work problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Tasty Algebra: Using toasted O cereal to find the missing factor in a multiplication equation 376 In this lesson students will use Cheerios to solve multiplication equations relating 3 whole numbers from world problems that include missing factors ranging from one through ten. Students will also argue the validity of multiplication equations that include missing factors and products with corresponding word problems. MAFS.3.OA.1.4: Determine the unknown whole number in a multiplication or division equation relating three whole numbers. Getting the hang of two-step word problems 238 Students will solve two-step word problems involving subtraction and division and represent these problems using equations with a letter standing for the unknown quantity. MAFS.3.OA.4.8: Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Water Park Fun Day 701 This third grade MEA asks students to work as a team to figure out which activities they would like to do at the water park with a given amount of tickets and time. Students will make informed decisions about which activities and food and beverage items on which to spend their allotted tickets. MAFS.3.OA.4.8: Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Mix It Up- Algebraic Thinking Math Project Using fruit salad problems "students will be introduced to proportional reasoning through modeling, sharing, and questioning techniques." MAFS.3.OA.4.8: Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. MAFS.4.OA.3.5: Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule “Add 3” and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way. Tricky Rice Math Patterns MEA 603 This is a 3rd grade MEA that requires students to use mathematical patterns to solve the problem, along with the analysis of data. After reading One Grain of Rice by Demi, students will look for ways to help Rani's relative find a new pattern so she can secure a large supply of rice to feed the people of her province in India. The twist is likely to cause controversy, so prepare for some strong debates. MAFS.3.OA.4.9: Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends. Patterns in the multiplication table 3321 The goal is to look for structure and identify patterns and then try to find the mathematical explanation for this. This problem examines the ''checkerboard'' pattern of even and odd numbers in a single digit multiplication table. The even numbers in the table are examined in depth using a grade appropriate notion of even, namely the possibility of reaching the number counting by 2's or expressing the number as a whole number of pairs. MAFS.3.OA.4.9: Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends. The Power of Patterns Students will work a real world problem to discover similarities and differences between the patterns of adding 2 to a number and doubling a number. The problem is set in the real world context of twin brothers who choose different patterning strategies given by their grandma to save for buying a car. MAFS.3.OA.4.9: Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends. Supply List Suggested class supplies for each student Math Notebook/Journal Math Folder Scissors Construction paper Colored pencils Pencils Index cards Place value charts Blank hundreds charts Number lines (can use string and index cards) Number strips (1 per 2 or 3 students) Egg cartons and small objects to show “groups of things” Empty containers to show liquid volume (milk jugs, cans, jars, etc.) Rulers Operations and Algebraic Thinking (class set) two colored circles (class set) square tiles graph paper (class set) place value blocks box of toasted O cereal 2 boxes of Cheeze Its hundreds chart for each student |
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Noise Exposure Limits for Emergency Responders
Understanding OSHA Standards and Other Key Guidelines
Noise exposure is an often-overlooked occupational hazard faced by emergency responders, including police officers, firefighters, and EMS professionals. Whether it's the wail of sirens, the roar of firefighting equipment, or the sudden blast of a firearm, these professionals are routinely subjected to high noise levels that can lead to significant auditory damage over time. Establishing and adhering to noise exposure limits is critical to protecting their hearing and overall well-being.
OSHA Noise Exposure Standards
The Occupational Safety and Health Administration (OSHA) has established guidelines to protect workers from the harmful effects of excessive noise. For most industries, OSHA’s Occupational Noise Exposure Standard (29 CFR 1910.95) sets the permissible exposure limit (PEL) for noise at 90 decibels (dB) over an 8-hour time-weighted average (TWA). If noise levels exceed this threshold, employers are required to implement hearing conservation programs. In addition to the PEL, OSHA specifies an action level of 85 dB TWA. When noise exposure reaches or exceeds this level, employees must be enrolled in a program that includes regular noise monitoring, hearing tests, and the provision of hearing protection devices (HPDs). Emergency responders often encounter noise levels well above these thresholds, particularly during high-stress and equipment-intensive situations.
NIOSH Recommended Exposure Limits (RELs)
The National Institute for Occupational Safety and Health (NIOSH) provides more stringent recommendations in its Noise Criteria Document. The recommended exposure limit (REL) is set at 85 dB over an 8-hour TWA. For every 3-dB increase in noise level, the allowable exposure time is halved. For example, exposure to 100 dB noise should not exceed 15 minutes per NIOSH guidelines. This stricter standard aims to reduce the risk of noise-induced hearing loss (NIHL), which remains prevalent among emergency personnel.
International Standards for Noise Exposure
Noise exposure limits and guidelines vary globally, with many countries adopting standards similar to OSHA and NIOSH. Here are a few examples:
United Kingdom: The HSE's Control of Noise at Work Regulations 2005 sets the exposure limit at 87 dB (A-weighted) for an 8-hour TWA, considering the use of hearing protection. The lower exposure action value is 80 dB, and the upper exposure action value is 85 dB.
Australia: According to Safe Work Australia's Managing Noise and Preventing Hearing Loss at Work Code of Practice, the exposure standard for noise is 85 dB (A-weighted) over an 8-hour TWA. Peak noise should not exceed 140 dB(C).
China: The Occupational Exposure Limits for Hazardous Agents in the Workplace (GBZ 2.2-2007) specifies a limit of 85 dB over an 8-hour workday, aligning closely with NIOSH recommendations.
Unique Challenges for Emergency Responders
Unlike workers in controlled environments, emergency responders face unpredictable and dynamic noise hazards. Some of the common noise sources include:
Sirens: Reaching levels as high as 120 dB, sirens are a constant presence for police officers and EMS professionals. Extended exposure during response times can be particularly harmful.
Firefighting Equipment: Chainsaws, hydraulic tools, and portable pumps generate noise levels ranging from 100 dB to 120 dB, often in enclosed spaces that amplify sound.
Law enforcement officers routinely train with firearms, which can produce impulse noise exceeding 150 dB, a level that can cause immediate hearing damage without adequate protection. For example, live-fire training exercises or use-of-force scenarios expose police officers to repeated high-decibel bursts that pose significant hearing risks.
State and Local Regulations
In addition to federal guidelines, some states and municipalities enforce their own noise exposure standards for public safety workers. These may include stricter monitoring requirements or specialized provisions for high-risk groups like firefighters and police. Agencies are encouraged to stay informed of these regulations to ensure compliance and improve workforce safety.
Noise Monitoring for Compliance and Safety
Implementing noise monitoring programs is an essential step for agencies aiming to protect their personnel. Modern noise dosimeters and sound level meters provide real-time data on exposure levels, enabling proactive measures to reduce risks. By tracking cumulative exposure, these tools also help agencies ensure compliance with OSHA, NIOSH or other standards. Noise monitoring is particularly beneficial in:
Identifying High-Risk Scenarios: By pinpointing environments and activities with the greatest noise hazards, agencies can focus mitigation efforts more effectively.
Educating Personnel: Providing responders with data about their exposure empowers them to take protective measures, such as using HPDs or rotating tasks to limit exposure time.
Addressing noise exposure is not only a regulatory requirement but a critical component of safeguarding the health and effectiveness of emergency responders. By adhering to established noise limits and leveraging modern monitoring technologies, agencies can create safer work environments for those who protect and serve the public.
Discover Sound and Noise Monitoring Solutions
Related Resources
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Respirator fit testing is crucial for first responders to verify safety, comply with regulations, and maintain operational effectiveness. Regular testing and training are essential for proper protection.
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Hearing Loss and Noise Exposure in Law Enforcement and Fire Services
Noise-induced hearing loss is a significant risk for law enforcement and firefighters. Preventive measures and TSI's noise monitoring tools can help protect their hearing while helping to ensure job performance and safety.
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6730 | https://courses.lumenlearning.com/suny-openstax-calculus1/chapter/defining-the-derivative/ | 3. Derivatives
3.1 Defining the Derivative
Learning Objectives
Recognize the meaning of the tangent to a curve at a point.
Calculate the slope of a tangent line.
Identify the derivative as the limit of a difference quotient.
Calculate the derivative of a given function at a point.
Describe the velocity as a rate of change.
Explain the difference between average velocity and instantaneous velocity.
Estimate the derivative from a table of values.
Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727) and the German Gottfried Leibniz (1646–1716), whose images appear in (Figure). When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today.
Figure 1. Newton and Leibniz are credited with developing calculus independently.
Tangent Lines
We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point latex[/latex] to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of [latex]x[/latex] near [latex]a[/latex] and drawing a line through the points latex[/latex] and latex[/latex], as shown in (Figure). The slope of this line is given by an equation in the form of a difference quotient:
We can also calculate the slope of a secant line to a function at a value [latex]a[/latex] by using this equation and replacing [latex]x[/latex] with [latex]a+h[/latex], where [latex]h[/latex] is a value close to [latex]0[/latex]. We can then calculate the slope of the line through the points latex[/latex] and latex[/latex]. In this case, we find the secant line has a slope given by the following difference quotient with increment [latex]h[/latex]:
Let [latex]f[/latex] be a function defined on an interval [latex]I[/latex] containing [latex]a[/latex]. If [latex]x\ne a[/latex] is in [latex]I[/latex], then
is a difference quotient.
Also, if [latex]h\ne 0[/latex] is chosen so that [latex]a+h[/latex] is in [latex]I[/latex], then
is a difference quotient with increment [latex]h[/latex].
View several Java applets on the development of the derivative.
These two expressions for calculating the slope of a secant line are illustrated in (Figure). We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation.
Figure 2. We can calculate the slope of a secant line in either of two ways.
In (Figure)(a) we see that, as the values of [latex]x[/latex] approach [latex]a[/latex], the slopes of the secant lines provide better estimates of the rate of change of the function at [latex]a[/latex]. Furthermore, the secant lines themselves approach the tangent line to the function at [latex]a[/latex], which represents the limit of the secant lines. Similarly, (Figure)(b) shows that as the values of [latex]h[/latex] get closer to 0, the secant lines also approach the tangent line. The slope of the tangent line at [latex]a[/latex] is the rate of change of the function at [latex]a[/latex], as shown in (Figure)(c).
Figure 3. The secant lines approach the tangent line (shown in green) as the second point approaches the first.
You can use this site to explore graphs to see if they have a tangent line at a point.
In (Figure) we show the graph of [latex]f(x)=\sqrt{x}[/latex] and its tangent line at latex[/latex] in a series of tighter intervals about [latex]x=1[/latex]. As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of [latex]x[/latex] close to 1. In fact, the graph of [latex]f(x)[/latex] itself appears to be locally linear in the immediate vicinity of [latex]x=1[/latex].
Figure 4. For values of [latex]x[/latex] close to 1, the graph of [latex]f(x)=\sqrt{x}[/latex] and its tangent line appear to coincide.
Formally we may define the tangent line to the graph of a function as follows.
Definition
Let [latex]f(x)[/latex] be a function defined in an open interval containing [latex]a[/latex]. The tangent line to [latex]f(x)[/latex] at [latex]a[/latex] is the line passing through the point latex[/latex] having slope
[latex]m_{\tan}=\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}[/latex]
provided this limit exists.
Equivalently, we may define the tangent line to [latex]f(x)[/latex] at [latex]a[/latex] to be the line passing through the point latex[/latex] having slope
[latex]m_{\tan}=\underset{h\to 0}{\lim}\frac{f(a+h)-f(a)}{h}[/latex]
provided this limit exists.
Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines.
Finding a Tangent Line
Find the equation of the line tangent to the graph of [latex]f(x)=x^2[/latex] at [latex]x=3[/latex].
Show Solution
First find the slope of the tangent line. In this example, use (Figure).
[latex]\begin{array}{lllll}m_{\tan} & =\underset{x\to 3}{\lim}\frac{f(x)-f(3)}{x-3} & & & \text{Apply the definition.} \ & =\underset{x\to 3}{\lim}\frac{x^2-9}{x-3} & & & \text{Substitute} \, f(x)=x^2 \, \text{and} \, f(3)=9. \ & =\underset{x\to 3}{\lim}\frac{(x-3)(x+3)}{x-3}=\underset{x\to 3}{\lim}(x+3)=6 & & & \text{Factor the numerator to evaluate the limit.} \end{array}[/latex]
Next, find a point on the tangent line. Since the line is tangent to the graph of [latex]f(x)[/latex] at [latex]x=3[/latex], it passes through the point latex[/latex]. We have [latex]f(3)=9[/latex], so the tangent line passes through the point latex[/latex].
Using the point-slope equation of the line with the slope [latex]m=6[/latex] and the point latex[/latex], we obtain the line [latex]y-9=6(x-3)[/latex]. Simplifying, we have [latex]y=6x-9[/latex]. The graph of [latex]f(x)=x^2[/latex] and its tangent line at [latex]x=3[/latex] are shown in (Figure).
Figure 5. The tangent line to [latex]f(x)[/latex] at [latex]x=3[/latex].
The Slope of a Tangent Line Revisited
Use (Figure) to find the slope of the line tangent to the graph of [latex]f(x)=x^2[/latex] at [latex]x=3[/latex].
Show Solution
The steps are very similar to (Figure). See (Figure) for the definition.
[latex]\begin{array}{lllll}m_{\tan} & =\underset{h\to 0}{\lim}\frac{f(3+h)-f(3)}{h} & & & \text{Apply the definition.} \ & =\underset{h\to 0}{\lim}\frac{(3+h)^2-9}{h} & & & \text{Substitute} \, f(3+h)=(3+h)^2 \, \text{and} \, f(3)=9. \ & =\underset{h\to 0}{\lim}\frac{9+6h+h^2-9}{h} & & & \text{Expand and simplify to evaluate the limit.} \ & =\underset{h\to 0}{\lim}\frac{h(6+h)}{h}=\underset{h\to 0}{\lim}(6+h)=6 \end{array}[/latex]
We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.
Finding the Equation of a Tangent Line
Find the equation of the line tangent to the graph of [latex]f(x)=1/x[/latex] at [latex]x=2[/latex].
Show Solution
We can use (Figure), but as we have seen, the results are the same if we use (Figure).
[latex]\begin{array}{lllll}m_{\tan} & =\underset{x\to 2}{\lim}\frac{f(x)-f(2)}{x-2} & & & \text{Apply the definition.} \ & =\underset{x\to 2}{\lim}\frac{\frac{1}{x}-\frac{1}{2}}{x-2} & & & \text{Substitute} \, f(x)=\frac{1}{x} \, \text{and} \, f(2)=\frac{1}{2}. \ & =\underset{x\to 2}{\lim}\frac{\frac{1}{x}-\frac{1}{2}}{x-2} \cdot \frac{2x}{2x} & & & \begin{array}{l}\text{Multiply numerator and denominator by} \, 2x \, \text{to} \ \text{simplify fractions.} \end{array} \ & =\underset{x\to 2}{\lim}\frac{(2-x)}{(x-2)(2x)} & & & \text{Simplify.} \ & =\underset{x\to 2}{\lim}\frac{-1}{2x} & & & \text{Simplify using} \, \frac{2-x}{x-2}=-1, \, \text{for} \, x\ne 2. \ & =-\frac{1}{4} & & & \text{Evaluate the limit.} \end{array}[/latex]
We now know that the slope of the tangent line is [latex]-\frac{1}{4}[/latex]. To find the equation of the tangent line, we also need a point on the line. We know that [latex]f(2)=\frac{1}{2}[/latex]. Since the tangent line passes through the point latex[/latex] we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation [latex]y=-\frac{1}{4}x+1[/latex]. The graphs of [latex]f(x)=\frac{1}{x}[/latex] and [latex]y=-\frac{1}{4}x+1[/latex] are shown in (Figure).
Figure 6. The line is tangent to [latex]f(x)[/latex] at [latex]x=2[/latex]
Find the slope of the line tangent to the graph of [latex]f(x)=\sqrt{x}[/latex] at [latex]x=4[/latex].
Show Solution
[latex]\frac{1}{4}[/latex]
Hint
Use either (Figure) or (Figure). Multiply the numerator and the denominator by a conjugate.
The Derivative of a Function at a Point
The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation.
Definition
Let [latex]f(x)[/latex] be a function defined in an open interval containing [latex]a[/latex]. The derivative of the function [latex]f(x)[/latex] at [latex]a[/latex], denoted by [latex]f^{\prime}(a)[/latex], is defined by
[latex]f^{\prime}(a)=\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}[/latex]
provided this limit exists.
Alternatively, we may also define the derivative of [latex]f(x)[/latex] at [latex]a[/latex] as
[latex]f^{\prime}(a)=\underset{h\to 0}{\lim}\frac{f(a+h)-f(a)}{h}[/latex]provided this limit exists.
Estimating a Derivative
For [latex]f(x)=x^2[/latex], use a table to estimate [latex]f^{\prime}(3)[/latex] using (Figure).
Show Solution
Create a table using values of [latex]x[/latex] just below 3 and just above 3.
| [latex]x[/latex] | [latex]\frac{x^2-9}{x-3}[/latex] |
--- |
| 2.9 | 5.9 |
| 2.99 | 5.99 |
| 2.999 | 5.999 |
| 3.001 | 6.001 |
| 3.01 | 6.01 |
| 3.1 | 6.1 |
After examining the table, we see that a good estimate is [latex]f^{\prime}(3)=6[/latex].
For [latex]f(x)=x^2[/latex], use a table to estimate [latex]f^{\prime}(3)[/latex] using (Figure).
Show Solution
6
Hint
Evaluate [latex]\frac{(x+h)^2-x^2}{h}[/latex] at [latex]h=-0.1,-0.01,-0.001,0.001,0.01,0.1[/latex]
Finding a Derivative
For [latex]f(x)=3x^2-4x+1[/latex], find [latex]f^{\prime}(2)[/latex] by using (Figure).
Show Solution
Substitute the given function and value directly into the equation.
[latex]\begin{array}{lllll}f^{\prime}(x)& =\underset{x\to 2}{\lim}\frac{f(x)-f(2)}{x-2} & & & \text{Apply the definition.} \ & =\underset{x\to 2}{\lim}\frac{(3x^2-4x+1)-5}{x-2} & & & \text{Substitute} \, f(x)=3x^2-4x+1 \, \text{and} \, f(2)=5. \ & =\underset{x\to 2}{\lim}\frac{(x-2)(3x+2)}{x-2} & & & \text{Simplify and factor the numerator.} \ & =\underset{x\to 2}{\lim}(3x+2) & & & \text{Cancel the common factor.} \ & =8 & & & \text{Evaluate the limit.} \end{array}[/latex]
Revisiting the Derivative
For [latex]f(x)=3x^2-4x+1[/latex], find [latex]f^{\prime}(2)[/latex] by using (Figure).
Show Solution
Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in (Figure).
[latex]\begin{array}{lllll}f^{\prime}(2) & =\underset{h\to 0}{\lim}\frac{f(2+h)-f(2)}{h} & & & \text{Apply the definition.} \ & =\underset{h\to 0}{\lim}\frac{(3(2+h)^2-4(2+h)+1)-5}{h} & & & \begin{array}{l}\text{Substitute} \, f(2)=5 \, \text{and} \ f(2+h)=3(2+h)^2-4(2+h)+1. \end{array} \ & =\underset{h\to 0}{\lim}\frac{3h^2+8h}{h} & & & \text{Simplify the numerator.} \ & =\underset{h\to 0}{\lim}\frac{h(3h+8)}{h} & & & \text{Factor the numerator.} \ & =\underset{h\to 0}{\lim}(3h+8) & & & \text{Cancel the common factor.} \ & =8 & & & \text{Evaluate the limit.} \end{array}[/latex]
The results are the same whether we use (Figure) or (Figure).
For [latex]f(x)=x^2+3x+2[/latex], find [latex]f^{\prime}(1)[/latex].
Show Solution
[latex]f^{\prime}(1)=5[/latex]
Hint
Use either (Figure), (Figure), or try both. Use either (Figure) or (Figure) as a guide.
Velocities and Rates of Change
Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if [latex]s(t)[/latex] is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [latex][a,t][/latex] if [latex]t>a[/latex] or [latex][t,a][/latex] if [latex]t[latex]v_{\text{avg}}=\frac{s(t)-s(a)}{t-a}[/latex].
As the values of [latex]t[/latex] approach [latex]a[/latex], the values of [latex]v_{\text{avg}}[/latex] approach the value we call the instantaneous velocity at [latex]a[/latex]. That is, instantaneous velocity at [latex]a[/latex], denoted [latex]v(a)[/latex], is given by
[latex]v(a)=s^{\prime}(a)=\underset{t\to a}{\lim}\frac{s(t)-s(a)}{t-a}[/latex].
To better understand the relationship between average velocity and instantaneous velocity, see (Figure). In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time [latex]t=a[/latex] whose position at time [latex]t[/latex] is given by the function [latex]s(t)[/latex]. The slope of the secant line (shown in green) is the average velocity of the object over the time interval [latex][a,t][/latex].
Figure 7. The slope of the secant line is the average velocity over the interval [latex][a,t][/latex]. The slope of the tangent line is the instantaneous velocity.
We can use (Figure) to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using (Figure).
Estimating Velocity
A lead weight on a spring is oscillating up and down. Its position at time [latex]t[/latex] with respect to a fixed horizontal line is given by [latex]s(t)= \sin t[/latex] ((Figure)). Use a table of values to estimate [latex]v(0)[/latex]. Check the estimate by using (Figure).
Figure 8. A lead weight suspended from a spring in vertical oscillatory motion.
Show Solution
We can estimate the instantaneous velocity at [latex]t=0[/latex] by computing a table of average velocities using values of [latex]t[/latex] approaching 0, as shown in (Figure).
Average velocities using values of [latex]t[/latex] approaching 0
| [latex]t[/latex] | [latex]\frac{\sin t - \sin 0}{t-0}=\frac{\sin t}{t}[/latex] |
| -0.1 | 0.998334166 |
| -0.01 | 0.9999833333 |
| -0.001 | 0.999999833 |
| 0.001 | 0.999999833 |
| 0.01 | 0.9999833333 |
| 0.1 | 0.998334166 |
From the table we see that the average velocity over the time interval [latex][-0.1,0][/latex] is 0.998334166, the average velocity over the time interval [latex][-0.01,0][/latex] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is [latex]v(0)=1[/latex].
By using (Figure), we can see that
[latex]v(0)=s^{\prime}(0)=\underset{t\to 0}{\lim}\frac{\sin t- \sin 0}{t-0}=\underset{t\to 0}{\lim}\frac{\sin t}{t}=1[/latex].
Thus, in fact, [latex]v(0)=1[/latex].
A rock is dropped from a height of 64 feet. Its height above ground at time [latex]t[/latex] seconds later is given by [latex]s(t)=-16t^2+64, \, 0\le t\le 2[/latex]. Find its instantaneous velocity 1 second after it is dropped, using (Figure).
Show Solution
-32 ft/sec
Hint
[latex]v(t)=s^{\prime}(t)[/latex]. Follow the earlier examples of the derivative using (Figure).
As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.
Definition
The instantaneous rate of change of a function [latex]f(x)[/latex] at a value [latex]a[/latex] is its derivative [latex]f^{\prime}(a)[/latex].
Chapter Opener: Estimating Rate of Change of Velocity
Figure 9. (credit: modification of work by Codex41, Flickr)
Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?
Show Solution
First observe that 60 mph = 88 ft/s, 100 mph [latex]\approx 146.67[/latex] ft/s, 200 mph [latex]\approx 293.33[/latex] ft/s, and 229.9 mph [latex]\approx 337.19[/latex] ft/s. We can summarize the information in a table.
[latex]v(t)[/latex] at different values of [latex]t[/latex]
| [latex]t[/latex] | [latex]v(t)[/latex] |
| 0 | 0 |
| 3.05 | 88 |
| 5.88 | 147.67 |
| 14.51 | 293.33 |
| 19.96 | 337.19 |
Now compute the average acceleration of the car in feet per second on intervals of the form [latex][t,19.96][/latex] as [latex]t[/latex] approaches 19.96, as shown in the following table.
Average acceleration
| [latex]t[/latex] | [latex]\frac{v(t)-v(19.96)}{t-19.96}=\frac{v(t)-337.19}{t-19.96}[/latex] |
| 0.0 | 16.89 |
| 3.05 | 14.74 |
| 5.88 | 13.46 |
| 14.51 | 8.05 |
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).
Rate of Change of Temperature
A homeowner sets the thermostat so that the temperature in the house begins to drop from [latex]70^{\circ}\text{F}[/latex] at 9 p.m., reaches a low of [latex]60^{\circ}[/latex] during the night, and rises back to [latex]70^{\circ}[/latex] by 7 a.m. the next morning. Suppose that the temperature in the house is given by [latex]T(t)=0.4t^2-4t+70[/latex] for [latex]0\le t\le 10[/latex], where [latex]t[/latex] is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.
Show Solution
Since midnight is 3 hours past 9 p.m., we want to compute [latex]T^{\prime }(3)[/latex]. Refer to (Figure).
[latex]\begin{array}{lllll}T^{\prime}(3) & =\underset{t\to 3}{\lim}\frac{T(t)-T(3)}{t-3} & & & \text{Apply the definition.} \ & =\underset{t\to 3}{\lim}\frac{0.4t^2-4t+70-61.6}{t-3} & & & \begin{array}{l}\text{Substitute}T(t)=0.4t^2-4t+70 \, \text{and} \ T(3)=61.6. \end{array} \ & =\underset{t\to 3}{\lim}\frac{0.4t^2-4t+8.4}{t-3} & & & \text{Simplify.} \ & =\underset{t\to 3}{\lim}\frac{0.4(t-3)(t-7)}{t-3} & & & =\underset{t\to 3}{\lim}\frac{0.4(t-3)(t-7)}{t-3} \ & =\underset{t\to 3}{\lim}0.4(t-7) & & & \text{Cancel.} \ & =-1.6 & & & \text{Evaluate the limit.} \end{array}[/latex]
The instantaneous rate of change of the temperature at midnight is [latex]-1.6^{\circ}\text{F}[/latex] per hour.
Rate of Change of Profit
A toy company can sell [latex]x[/latex] electronic gaming systems at a price of [latex]p=-0.01x+400[/latex] dollars per gaming system. The cost of manufacturing [latex]x[/latex] systems is given by [latex]C(x)=100x+10,000[/latex] dollars. Find the rate of change of profit when 10,000 games are produced. Should the toy company increase or decrease production?
Show Solution
The profit [latex]P(x)[/latex] earned by producing [latex]x[/latex] gaming systems is [latex]R(x)-C(x)[/latex], where [latex]R(x)[/latex] is the revenue obtained from the sale of [latex]x[/latex] games. Since the company can sell [latex]x[/latex] games at [latex]p=-0.01x+400[/latex] per game,
[latex]R(x)=xp=x(-0.01x+400)=-0.01x^2+400x[/latex].
Consequently,
[latex]P(x)=-0.01x^2+300x-10,000[/latex].
Therefore, evaluating the rate of change of profit gives
[latex]\begin{array}{ll}P^{\prime}(10000)& =\underset{x\to 10000}{\lim}\frac{P(x)-P(10000)}{x-10000} \ & =\underset{x\to 10000}{\lim}\frac{-0.01x^2+300x-10000-1990000}{x-10000} \ & =\underset{x\to 10000}{\lim}\frac{-0.01x^2+300x-2000000}{x-10000} \ & =100 \end{array}[/latex]
Since the rate of change of profit [latex]P^{\prime}(10,000)>0[/latex] and [latex]P(10,000)>0[/latex], the company should increase production.
A coffee shop determines that the daily profit on scones obtained by charging [latex]s[/latex] dollars per scone is [latex]P(s)=-20s^2+150s-10[/latex]. The coffee shop currently charges [latex]\$3.25[/latex] per scone. Find [latex]P^{\prime}(3.25)[/latex], the rate of change of profit when the price is [latex]\$3.25[/latex] and decide whether or not the coffee shop should consider raising or lowering its prices on scones.
Show Solution
[latex]P^{\prime}(3.25)=20>0[/latex]; raise prices
Hint
Use (Figure) for a guide.
Key Concepts
The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment [latex]h[/latex].
The derivative of a function [latex]f(x)[/latex] at a value [latex]a[/latex] is found using either of the definitions for the slope of the tangent line.
Velocity is the rate of change of position. As such, the velocity [latex]v(t)[/latex] at time [latex]t[/latex] is the derivative of the position [latex]s(t)[/latex] at time [latex]t[/latex]. Average velocity is given by
[latex]v_{\text{avg}}=\frac{s(t)-s(a)}{t-a}[/latex].
Instantaneous velocity is given by
[latex]v(a)=s^{\prime}(a)=\underset{t\to a}{\lim}\frac{s(t)-s(a)}{t-a}[/latex].
We may estimate a derivative by using a table of values.
Key Equations
Difference quotient [latex]Q=\frac{f(x)-f(a)}{x-a}[/latex]
Difference quotient with increment [latex]h[/latex] [latex]Q=\frac{f(a+h)-f(a)}{a+h-a}=\frac{f(a+h)-f(a)}{h}[/latex]
Slope of tangent line [latex]m_{\tan}=\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}[/latex] [latex]m_{\tan}=\underset{h\to 0}{\lim}\frac{f(a+h)-f(a)}{h}[/latex]
Derivative of [latex]f(x)[/latex] at [latex]a[/latex] [latex]f^{\prime}(a)=\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}[/latex] [latex]f^{\prime}(a)=\underset{h\to 0}{\lim}\frac{f(a+h)-f(a)}{h}[/latex]
Average velocity [latex]v_{\text{avg}}=\frac{s(t)-s(a)}{t-a}[/latex]
Instantaneous velocity [latex]v(a)=s^{\prime}(a)=\underset{t\to a}{\lim}\frac{s(t)-s(a)}{t-a}[/latex]
For the following exercises, use (Figure) to find the slope of the secant line between the values [latex]x_1[/latex] and [latex]x_2[/latex] for each function [latex]y=f(x)[/latex].
1.[latex]f(x)=4x+7; \, x_1=2, \, x_2=5[/latex]
Show Solution
4
2.[latex]f(x)=8x-3; \, x_1=-1, \, x_2=3[/latex]
3.[latex]f(x)=x^2+2x+1; \, x_1=3, \, x_2=3.5[/latex]
Show Solution
8.5
4.[latex]f(x)=\text{−}{x}^{2}+x+2;{x}_{1}=0.5,{x}_{2}=1.5[/latex]
5.[latex]f(x)=\frac{4}{3x-1}; \, x_1=1, \, x_2=3[/latex]
Show Solution
[latex]-\frac{3}{4}[/latex]
6.[latex]f(x)=\frac{x-7}{2x+1}; \, x_1=-2, \, x_2=0[/latex]
7.[latex]f(x)=\sqrt{x}; \, x_1=1, \, x_2=16[/latex]
Show Solution
0.2
8.[latex]f(x)=\sqrt{x-9}; \, x_1=10, \, x_2=13[/latex]
9.[latex]f(x)=x^{1/3}+1; \, x_1=0, \, x_2=8[/latex]
Show Solution
0.25
10.[latex]f(x)=6x^{2/3}+2x^{1/3}; \, x_1=1, \, x_2=27[/latex]
For the following functions,
use (Figure) to find the slope of the tangent line [latex]m_{\tan}=f^{\prime}(a)[/latex], and
find the equation of the tangent line to [latex]f[/latex] at [latex]x=a[/latex].
11.[latex]f(x)=3-4x, \, a=2[/latex]
Show Solution
a. -4 b. [latex]y=3-4x[/latex]
12.[latex]f(x)=\frac{x}{5}+6, \, a=-1[/latex]
13.[latex]f(x)=x^2+x, \, a=1[/latex]
Show Solution
a. 3 b. [latex]y=3x-1[/latex]
14.[latex]f(x)=1-x-x^2, \, a=0[/latex]
15.[latex]f(x)=\frac{7}{x}, \, a=3[/latex]
Show Solution
a. [latex]\frac{-7}{9}[/latex] b. [latex]y=\frac{-7}{9}x+\frac{14}{3}[/latex]
16.[latex]f(x)=\sqrt{x+8}, \, a=1[/latex]
17.[latex]f(x)=2-3x^2, \, a=-2[/latex]
Show Solution
a. 12 b. [latex]y=12x+14[/latex]
18.[latex]f(x)=\frac{-3}{x-1}, \, a=4[/latex]
19.[latex]f(x)=\frac{2}{x+3}, \, a=-4[/latex]
Show Solution
a. -2 b. [latex]y=-2x-10[/latex]
20.[latex]f(x)=\frac{3}{x^2}, \, a=3[/latex]
For the following functions [latex]y=f(x)[/latex], find [latex]f^{\prime}(a)[/latex] using (Figure).
21.[latex]f(x)=5x+4, \, a=-1[/latex]
Show Solution
5
22.[latex]f(x)=-7x+1, \, a=3[/latex]
23.[latex]f(x)=x^2+9x, \, a=2[/latex]
Show Solution
13
24.[latex]f(x)=3x^2-x+2, \, a=1[/latex]
25.[latex]f(x)=\sqrt{x}, \, a=4[/latex]
Show Solution
[latex]\frac{1}{4}[/latex]
26.[latex]f(x)=\sqrt{x-2}, \, a=6[/latex]
27.[latex]f(x)=\frac{1}{x}, \, a=2[/latex]
Show Solution
[latex]-\frac{1}{4}[/latex]
28.[latex]f(x)=\frac{1}{x-3}, \, a=-1[/latex]
29.[latex]f(x)=\frac{1}{x^3}, \, a=1[/latex]
Show Solution
-3
30.[latex]f(x)=\frac{1}{\sqrt{x}}, \, a=4[/latex]
For the following exercises, given the function [latex]y=f(x)[/latex],
find the slope of the secant line [latex]PQ[/latex] for each point [latex]Q(x,f(x))[/latex] with [latex]x[/latex] value given in the table.
Use the answers from a. to estimate the value of the slope of the tangent line at [latex]P[/latex].
Use the answer from b. to find the equation of the tangent line to [latex]f[/latex] at point [latex]P[/latex].
31. [T][latex]f(x)=x^2+3x+4, \, P(1,8)[/latex] (Round to 6 decimal places.)
| [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] | [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] |
--- --- |
| 1.1 | (i) | 0.9 | (vii) |
| 1.01 | (ii) | 0.99 | (viii) |
| 1.001 | (iii) | 0.999 | (ix) |
| 1.0001 | (iv) | 0.9999 | (x) |
| 1.00001 | (v) | 0.99999 | (xi) |
| 1.000001 | (vi) | 0.999999 | (xii) |
Show Solution
a. (i) [latex]5.100000[/latex], (ii) [latex]5.010000[/latex], (iii) [latex]5.001000[/latex], (iv) [latex]5.000100[/latex], (v) [latex]5.000010[/latex], (vi) [latex]5.000001[/latex],(vii) [latex]4.900000[/latex], (viii) [latex]4.990000[/latex], (ix) [latex]4.999000[/latex], (x) [latex]4.999900[/latex], (xi) [latex]4.999990[/latex], (xii) [latex]4.999999[/latex]
b. [latex]m_{\tan}=5[/latex]c. [latex]y=5x+3[/latex]
32. [T][latex]f(x)=\frac{x+1}{x^2-1}, \, P(0,-1)[/latex]
| [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] | [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] |
--- --- |
| 0.1 | (i) | -0.1 | (vii) |
| 0.01 | (ii) | -0.01 | (viii) |
| 0.001 | (iii) | -0.001 | (ix) |
| 0.0001 | (iv) | -0.0001 | (x) |
| 0.00001 | (v) | -0.00001 | (xi) |
| 0.000001 | (vi) | -0.000001 | (xii) |
33. [T][latex]f(x)=10e^{0.5x}, \, P(0,10)[/latex] (Round to 4 decimal places.)
| [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] |
--- |
| -0.1 | (i) |
| -0.01 | (ii) |
| -0.001 | (iii) |
| -0.0001 | (iv) |
| -0.00001 | (v) |
| −0.000001 | (vi) |
Show Solution
a. (i) [latex]4.8771[/latex], (ii) [latex]4.9875[/latex], (iii) [latex]4.9988[/latex], (iv) [latex]4.9999[/latex], (v) [latex]4.9999[/latex], (vi) [latex]4.9999[/latex]b. [latex]m_{\tan}=5[/latex]c. [latex]y=5x+10[/latex]
34. [T][latex]f(x)= \tan (x), \, P(\pi,0)[/latex]
| [latex]x[/latex] | Slope [latex]m_{PQ}[/latex] |
--- |
| 3.1 | (i) |
| 3.14 | (ii) |
| 3.141 | (iii) |
| 3.1415 | (iv) |
| 3.14159 | (v) |
| 3.141592 | (vi) |
For the following position functions [latex]y=s(t)[/latex], an object is moving along a straight line, where [latex]t[/latex] is in seconds and [latex]s[/latex] is in meters. Find
the simplified expression for the average velocity from [latex]t=2[/latex] to [latex]t=2+h[/latex];
the average velocity between [latex]t=2[/latex] and [latex]t=2+h[/latex], where (i) [latex]h=0.1[/latex], (ii) [latex]h=0.01[/latex], (iii) [latex]h=0.001[/latex], and (iv) [latex]h=0.0001[/latex]; and
use the answer from a. to estimate the instantaneous velocity at [latex]t=2[/latex] seconds.
35. [T][latex]s(t)=\frac{1}{3}t+5[/latex]
Show Solution
a. [latex]\frac{1}{3}[/latex];b. (i) [latex]0.\bar{3}[/latex] m/s, (ii) [latex]0.\bar{3}[/latex] m/s, (iii) [latex]0.\bar{3}[/latex] m/s, (iv) [latex]0.\bar{3}[/latex] m/s;c. [latex]0.\bar{3}=\frac{1}{3}[/latex] m/s
36. [T][latex]s(t)=t^2-2t[/latex]
37. [T][latex]s(t)=2t^3+3[/latex]
Show Solution
a. [latex]2(h^2+6h+12)[/latex];b. (i) 25.22 m/s, (ii) 24.12 m/s, (iii) 24.01 m/s, (iv) 24 m/s;c. 24 m/s
38. [T][latex]s(t)=\frac{16}{t^2}-\frac{4}{t}[/latex]
39.Use the following graph to evaluate a. [latex]f^{\prime}(1)[/latex] and b. [latex]f^{\prime}(6)[/latex].
Show Solution
a. [latex]1.25[/latex]; b. 0.5
40.Use the following graph to evaluate a. [latex]f^{\prime}(-3)[/latex] and b. [latex]f^{\prime}(1.5)[/latex].
For the following exercises, use the limit definition of derivative to show that the derivative does not exist at [latex]x=a[/latex] for each of the given functions.
41.[latex]f(x)=x^{1/3}, \, x=0[/latex]
Show Solution
[latex]\underset{x\to 0^-}{\lim}\frac{x^{1/3}-0}{x-0}=\underset{x\to 0^-}{\lim}\frac{1}{x^{2/3}}=\infty[/latex]
42.[latex]f(x)=x^{2/3}, \, x=0[/latex]
43.[latex]f(x)=\begin{cases} 1 & \text{if} \, x<1 \ x & \text{if} \, x \ge 1 \end{cases}, \, x=1[/latex]
Show Solution
[latex]\underset{x\to 1^-}{\lim}\frac{1-1}{x-1}=0\ne 1=\underset{x\to 1^+}{\lim}\frac{x-1}{x-1}[/latex]
44.[latex]f(x)=\frac{|x|}{x}, \, x=0[/latex]
45. [T] The position in feet of a race car along a straight track after [latex]t[/latex] seconds is modeled by the function [latex]s(t)=8t^2-\frac{1}{16}t^3[/latex].
Find the average velocity of the vehicle over the following time intervals to four decimal places:
[4, 4.1]
[4, 4.01]
[4, 4.001]
[4, 4.0001]
Use a. to draw a conclusion about the instantaneous velocity of the vehicle at [latex]t=4[/latex] seconds.
Show Solution
a. (i) 61.7244 ft/s, (ii) 61.0725 ft/s, (iii) 61.0072 ft/s, (iv) 61.0007 ft/sb. At 4 seconds the race car is traveling at a rate/velocity of 61 ft/s.
46. [T] The distance in feet that a ball rolls down an incline is modeled by the function [latex]s(t)=14t^2[/latex], where [latex]t[/latex] is seconds after the ball begins rolling.
Find the average velocity of the ball over the following time intervals:
[5, 5.1]
[5, 5.01]
[5, 5.001]
[5, 5.0001]
Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at [latex]t=5[/latex] seconds.
47.Two vehicles start out traveling side by side along a straight road. Their position functions, shown in the following graph, are given by [latex]s=f(t)[/latex] and [latex]s=g(t)[/latex], where [latex]s[/latex] is measured in feet and [latex]t[/latex] is measured in seconds.
Which vehicle has traveled farther at [latex]t=2[/latex] seconds?
What is the approximate velocity of each vehicle at [latex]t=3[/latex] seconds?
Which vehicle is traveling faster at [latex]t=4[/latex] seconds?
What is true about the positions of the vehicles at [latex]t=4[/latex] seconds?
Show Solution
a. The vehicle represented by [latex]f(t)[/latex], because it has traveled 2 feet, whereas [latex]g(t)[/latex] has traveled 1 foot.b. The velocity of [latex]f(t)[/latex] is constant at 1 ft/s, while the velocity of [latex]g(t)[/latex] is approximately 2 ft/s.c. The vehicle represented by [latex]g(t)[/latex], with a velocity of approximately 4 ft/s.d. Both have traveled 4 feet in 4 seconds.
48. [T] The total cost [latex]C(x)[/latex], in hundreds of dollars, to produce [latex]x[/latex] jars of mayonnaise is given by [latex]C(x)=0.000003x^3+4x+300[/latex].
Calculate the average cost per jar over the following intervals:
[100, 100.1]
[100, 100.01]
[100, 100.001]
[100, 100.0001]
Use the answers from a. to estimate the average cost to produce 100 jars of mayonnaise.
49. [T] For the function [latex]f(x)=x^3-2x^2-11x+12[/latex], do the following.
Use a graphing calculator to graph [latex]f[/latex] in an appropriate viewing window.
Use the ZOOM feature on the calculator to approximate the two values of [latex]x=a[/latex] for which [latex]m_{\tan}=f^{\prime}(a)=0[/latex].
Show Solution
b. [latex]a\approx -1.361, \, 2.694[/latex]
50. [T] For the function [latex]f(x)=\frac{x}{1+x^2}[/latex], do the following.
Use a graphing calculator to graph [latex]f[/latex] in an appropriate viewing window.
Use the ZOOM feature on the calculator to approximate the values of [latex]x=a[/latex] for which [latex]m_{\tan}=f^{\prime}(a)=0[/latex].
51.Suppose that [latex]N(x)[/latex] computes the number of gallons of gas used by a vehicle traveling [latex]x[/latex] miles. Suppose the vehicle gets 30 mpg.
Find a mathematical expression for [latex]N(x)[/latex].
What is [latex]N(100)[/latex]? Explain the physical meaning.
What is [latex]N^{\prime}(100)[/latex]? Explain the physical meaning.
Show Solution
a. [latex]N(x)=\frac{x}{30}[/latex]b. [latex]\sim 3.3[/latex] gallons. When the vehicle travels 100 miles, it has used 3.3 gallons of gas.c. [latex]\frac{1}{30}[/latex]. The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled 100 miles.
52. [T] For the function [latex]f(x)=x^4-5x^2+4[/latex], do the following.
Use a graphing calculator to graph [latex]f[/latex] in an appropriate viewing window.
Use the [latex]\text{nDeriv}[/latex] function, which numerically finds the derivative, on a graphing calculator to estimate [latex]f^{\prime}(-2), \, f^{\prime}(-0.5), \, f^{\prime}(1.7)[/latex], and [latex]f^{\prime}(2.718)[/latex].
53. [T] For the function [latex]f(x)=\frac{x^2}{x^2+1}[/latex], do the following.
Use a graphing calculator to graph [latex]f[/latex] in an appropriate viewing window.
Use the [latex]\text{nDeriv}[/latex] function on a graphing calculator to find [latex]f^{\prime}(-4), \, f^{\prime}(-2), \, f^{\prime}(2)[/latex], and [latex]f^{\prime}(4)[/latex].
Show Solution
a.
b. [latex]-0.028, \, -0.16, \, 0.16, \, 0.028[/latex]
Glossary
derivative
: the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivative
difference quotient
: of a function [latex]f(x)[/latex] at [latex]a[/latex] is given by
[latex]\frac{f(a+h)-f(a)}{h}[/latex] or [latex]\frac{f(x)-f(a)}{x-a}[/latex]
differentiation
: the process of taking a derivative
instantaneous rate of change
: the rate of change of a function at any point along the function [latex]a[/latex], also called [latex]f^{\prime}(a)[/latex], or the derivative of the function at [latex]a[/latex]
Candela Citations
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Find an equation of the tangent line to the hyperbolax^2/a^2 - y^2/b^2 = 1 at the point (x0, y0).
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Question: Find an equation of the tangent line to the hyperbolax^2/a^2 - y^2/b^2 = 1 at the point (x0, y0).
Find an equation of the tangent line to the hyperbola
x^2/a^2 - y^2/b^2 = 1 at the point (x0, y0).
There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 Consider the following equation of a hyperbola
x 2 a 2−y 2 b 2=1 View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question
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6732 | https://askfilo.com/user-question-answers-smart-solutions/let-and-be-positive-integers-such-that-banana-and-corona-are-3339333335393231 | Question asked by Filo student
Let n and k be positive integers such that k≤2n. Banana and Corona are playing the following variant of the guessing game. First, Banana secretly picks an integer x such that 1≤x≤n. Corona will attempt to determine x by asking some questions, which are described as follows. In each turn, Corona chooses k distinct subsets of {1,2,…,n} and, for each chosen set S, asks the question "Is x in the set S?''. Banana picks one of these k questions and tells both the question and its answer to Corona, who can then start another turn. Find all pairs (n,k) such that, regardless of Banana's actions, Corona could determine x in finitely many turns with absolute certainty.
Views: 5,394 students
Updated on: Sep 25, 2025
Text SolutionText solutionverified iconVerified
Concepts
Combinatorics, Information Theory, Adversarial Games, Set Theory, Binary Representation
Explanation
Banana picks x∈{1,2,…,n}. Each turn, Corona picks k distinct subsets of {1,2,…,n}, and for each subset S, asks: "Is x in S?" Banana then chooses one of these k questions and reveals both the question and its answer. Corona then continues to the next turn.
Corona's goal is to determine x with certainty, regardless of Banana's choices. We are to find all pairs (n,k) with k≤2n such that this is always possible.
Key Observations
Reformulation
Extreme Cases
The Critical Case
Formal Argument
Suppose k<2n. At each turn, Corona presents k subsets. There are 2n possible subsets, so at least one subset T is never asked about. Banana can always avoid giving information about T. Thus, for any x, there is always at least one other y such that the answers so far are consistent with both x and y. Thus, Corona cannot determine x with certainty.
Conclusion
Step-By-Step Solution
Step 1
Consider the case k=2n. Corona can ask about all possible subsets. No matter which subset Banana picks, the answer uniquely determines x.
Step 2
Consider the case k<2n. At each turn, there is at least one subset T not among the k subsets. Banana can always avoid giving information about T, so Corona can never learn the answer to "Is x in T?". Thus, Corona cannot guarantee to determine x.
Final Answer
All pairs (n,k) such that k=2n (with n≥1) are the only solutions.
That is, Corona can determine x in finitely many turns with absolute certainty if and only if k=2n. For all k<2n, it is not possible.
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| Question Text | Let n and k be positive integers such that k≤2n. Banana and Corona are playing the following variant of the guessing game. First, Banana secretly picks an integer x such that 1≤x≤n. Corona will attempt to determine x by asking some questions, which are described as follows. In each turn, Corona chooses k distinct subsets of {1,2,…,n} and, for each chosen set S, asks the question "Is x in the set S?''. Banana picks one of these k questions and tells both the question and its answer to Corona, who can then start another turn. Find all pairs (n,k) such that, regardless of Banana's actions, Corona could determine x in finitely many turns with absolute certainty. |
| Updated On | Sep 25, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Answer Type | Text solution:1 |
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6733 | https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces | 5.6 Common Forces - University Physics Volume 1 | OpenStax
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University Physics Volume 1
5.6 Common Forces
University Physics Volume 15.6 Common Forces
Contents Contents
Highlights
Table of contents
Preface
Mechanics
1 Units and Measurement
2 Vectors
3 Motion Along a Straight Line
4 Motion in Two and Three Dimensions
5 Newton's Laws of Motion
Introduction
5.1 Forces
5.2 Newton's First Law
5.3 Newton's Second Law
5.4 Mass and Weight
5.5 Newton’s Third Law
5.6 Common Forces
5.7 Drawing Free-Body Diagrams
Chapter Review
6 Applications of Newton's Laws
7 Work and Kinetic Energy
8 Potential Energy and Conservation of Energy
9 Linear Momentum and Collisions
10 Fixed-Axis Rotation
11 Angular Momentum
12 Static Equilibrium and Elasticity
13 Gravitation
14 Fluid Mechanics
Waves and Acoustics
A | Units
B | Conversion Factors
C | Fundamental Constants
D | Astronomical Data
E | Mathematical Formulas
F | Chemistry
G | The Greek Alphabet
Answer Key
Index
Search for key terms or text.
Close
Learning Objectives
By the end of this section, you will be able to:
Define normal and tension forces
Distinguish between real and fictitious forces
Apply Newton’s laws of motion to solve problems involving a variety of forces
Forces are given many names, such as push, pull, thrust, and weight. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. Several of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text.
A Catalog of Forces: Normal, Tension, and Other Examples of Forces
A catalog of forces will be useful for reference as we solve various problems involving force and motion. These forces include normal force, tension, friction, and spring force.
Normal force
Weight (also called the force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 5.21(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 5.21(b)? When the bag of dog food is placed on the table, the table sags slightly under the load. This would be noticeable if the load were placed on a card table, but even a sturdy oak table deforms when a force is applied to it. Unless an object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or a trampoline or diving board). The greater the deformation, the greater the restoring force. Thus, when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point, the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly and the sag is slight, so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it.
Figure 5.21(a) The person holding the bag of dog food must supply an upward force F⃗hand F→hand F→hand equal in magnitude and opposite in direction to the weight of the food w⃗w→w→ so that it doesn’t drop to the ground. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N⃗N→N→ equal in magnitude and opposite in direction to the weight of the load.
We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting the weight of an object, or a load, is perpendicular to the surface of contact between the load and its support, this force is defined as a normal force and here is given by the symbol N⃗.N→.N→. (This is not the newton unit for force, or N.) The word normal means perpendicular to a surface. This means that the normal force experienced by an object resting on a horizontal surface can be expressed in vector form as follows:
N⃗=−m g⃗.N→=−m g→.N→=−m g→.
In scalar form, this becomes
N=m g.N=m g.N=m g.
The normal force can be less than the object’s weight if the object is on an incline.
Example 5.12
Weight on an Incline
Consider the skier on the slope in Figure 5.22. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is 45.0 N?
Figure 5.22 Since the acceleration is parallel to the slope and acting down the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular to it (axes shown to the left of the skier). N⃗N→N→ is perpendicular to the slope and f⃗f→f→ is parallel to the slope, but w⃗w→w→ has components along both axes, namely, w y w y w y and w x w x w x. Here, w⃗w→w→ has a squiggly line to show that it has been replaced by these components. The force N⃗N→N→ is equal in magnitude to w y w y w y, so there is no acceleration perpendicular to the slope, but f is less than w x w x w x, so there is a downslope acceleration (along the axis parallel to the slope).
Strategy
This is a two-dimensional problem, since not all forces on the skier (the system of interest) are parallel. The approach we have used in two-dimensional kinematics also works well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two one-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Motions along mutually perpendicular axes are independent.) We use x and y for the parallel and perpendicular directions, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and the acceleration is downslope. Regarding the forces, friction is drawn in opposition to motion (friction always opposes forward motion) and is always parallel to the slope, w x w x w x is drawn parallel to the slope and downslope (it causes the motion of the skier down the slope), and w y w y w y is drawn as the component of weight perpendicular to the slope. Then, we can consider the separate problems of forces parallel to the slope and forces perpendicular to the slope.
Solution
The magnitude of the component of weight parallel to the slope is
w x=w sin 25°=m g sin 25°,w x=w sin 25°=m g sin 25°,w x=w sin 25°=m g sin 25°,
and the magnitude of the component of the weight perpendicular to the slope is
w y=w cos 25°=m g cos 25°.w y=w cos 25°=m g cos 25°.w y=w cos 25°=m g cos 25°.
a. Neglect friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the component of the skier’s weight parallel to slope w x w x w x and friction f. Using Newton’s second law, with subscripts to denote quantities parallel to the slope,
a x=F net x m a x=F net x m a x=F net x m
where F net x=w x=−m g sin 25°,F net x=w x=−m g sin 25°,F net x=w x=−m g sin 25°, assuming no friction for this part. Therefore,
a x=F net x m=m g sin 25°m=g sin 25°(9.80 m/s 2)(0.4226)=4.14 m/s 2 a x=F net x m=m g sin 25°m=g sin 25°(9.80 m/s 2)(0.4226)=4.14 m/s 2 a x=F net x m=m g sin 25°m=g sin 25°(9.80 m/s 2)(0.4226)=4.14 m/s 2
is the acceleration.
b. Include friction. We have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is
F net x=w x−f.F net x=w x−f.F net x=w x−f.
Substituting this into Newton’s second law, a x=F net x/m,a x=F net x/m,a x=F net x/m, gives
a x=F net x m=w x−f m=m g sin 25°−f m.a x=F net x m=w x−f m=m g sin 25°−f m.a x=F net x m=w x−f m=m g sin 25°−f m.
We substitute known values to obtain
a x=(60.0 kg)(9.80 m/s 2)(0.4226)−45.0 N 60.0 kg.a x=(60.0 kg)(9.80 m/s 2)(0.4226)−45.0 N 60.0 kg.a x=(60.0 kg)(9.80 m/s 2)(0.4226)−45.0 N 60.0 kg.
This gives us
a x=3.39 m/s 2,a x=3.39 m/s 2,a x=3.39 m/s 2,
which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.
Significance
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. It is a general result that if friction on an incline is negligible, then the acceleration down the incline is a=g sin θ a=g sin θ a=g sin θ, regardless of mass. As discussed previously, all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).
When an object rests on an incline that makes an angle θ θ θ with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, w y w y w y, and a force acting parallel to the plane, w x w x w x (Figure 5.23). The normal force N⃗N→N→ is typically equal in magnitude and opposite in direction to the perpendicular component of the weight w y w y w y. The force acting parallel to the plane, w x w x w x, causes the object to accelerate down the incline.
Figure 5.23 An object rests on an incline that makes an angle θ θ θ with the horizontal.
Be careful when resolving the weight of the object into components. If the incline is at an angle θ θ θ to the horizontal, then the magnitudes of the weight components are
w x=w sin θ=m g sin θ w x=w sin θ=m g sin θ w x=w sin θ=m g sin θ
and
w y=w cos θ=m g cos θ.w y=w cos θ=m g cos θ.w y=w cos θ=m g cos θ.
We use the second equation to write the normal force experienced by an object resting on an inclined plane:
N=m g cos θ,N=m g cos θ,N=m g cos θ,
object at rest or sliding on an inclined plane. Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, we draw the right angle formed by the three weight vectors. The angle θ θ θ of the incline is the same as the angle formed between w and w y w y w y. Knowing this property, we can use trigonometry to determine the magnitude of the weight components:
cos θ=w y w,w y=w cos θ=m g cos θ sin θ=w x w,w x=w sin θ=m g sin θ.cos θ=w y w,w y=w cos θ=m g cos θ sin θ=w x w,w x=w sin θ=m g sin θ.cos θ=w y w,w y=w cos θ=m g cos θ sin θ=w x w,w x=w sin θ=m g sin θ.
Check Your Understanding 5.8
A force of 1150 N acts parallel to a ramp to push a 250-kg gun safe into a moving van. The ramp is frictionless and inclined at 17°.17°.17°. (a) What is the acceleration of the safe up the ramp? (b) If we consider friction in this problem, with a friction force of 120 N, what is the acceleration of the safe?
Tension
A tension is a force along the length of a medium; in particular, it is a pulling force that acts along a stretched flexible connector, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons.
Any flexible connector, such as a string, rope, chain, wire, or cable, can only exert a pull parallel to its length; thus, a force carried by a flexible connector is a tension with a direction parallel to the connector. Tension is a pull in a connector. Consider the phrase: “You can’t push a rope.” Instead, tension force pulls outward along the two ends of a rope.
Consider a person holding a mass on a rope, as shown in Figure 5.24. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero and the net force is zero. The only external forces acting on the mass are its weight and the tension supplied by the rope. Thus,
F net=T−w=0,F net=T−w=0,F net=T−w=0,
where T and w are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. As we proved using Newton’s second law, the tension equals the weight of the supported mass:
T=w=m g,T=w=m g,T=w=m g,
object hanging at rest. Thus, for a 5.00-kg mass (neglecting the mass of the rope), we see that
T=m g=(5.00 kg)(9.80 m/s 2)=49.0 N.T=m g=(5.00 kg)(9.80 m/s 2)=49.0 N.T=m g=(5.00 kg)(9.80 m/s 2)=49.0 N.
If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.
Figure 5.24 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T⃗,T→,T→, that force must be parallel to the length of the rope, as shown. By Newton’s third law, the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.
Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a tendon, or a bicycle brake cable. If there is no friction, the tension transmission is undiminished; only its direction changes, and it is always parallel to the flexible connector, as shown in Figure 5.25.
Figure 5.25(a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force’s direction but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the brake lever on the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed.
Example 5.13
What Is the Tension in a Tightrope?
Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 5.26.
Figure 5.26 The weight of a tightrope walker causes a wire to sag by 5.0°5.0°5.0°. The system of interest is the point in the wire at which the tightrope walker is standing.
Strategy
As you can see in Figure 5.26, the wire is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows that have the same direction as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w⃗w→w→ and the two tensions T⃗L T→L T→L (left tension) and T⃗R T→R T→R (right tension). It is reasonable to neglect the weight of the wire. The net external force is zero, because the system is static. We can use trigonometry to find the tensions. One conclusion is possible at the outset—we can see from Figure 5.26(b) that the magnitudes of the tensions T L T L T L and T R T R T R must be equal. We know this because there is no horizontal acceleration in the rope and the only forces acting to the left and right are T L T L T L and T R T R T R. Thus, the magnitude of those horizontal components of the forces must be equal so that they cancel each other out.
Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case, the best coordinate system has one horizontal axis (x) and one vertical axis (y).
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to look at a new free-body diagram showing all horizontal and vertical components of each force acting on the system (Figure 5.27).
Figure 5.27 When the vectors are projected onto vertical and horizontal axes, their components along these axes must add to zero, since the tightrope walker is stationary. The small angle results in T being much greater than w.
Consider the horizontal components of the forces (denoted with a subscript x):
F net x=T R x−T L x.F net x=T R x−T L x.F net x=T R x−T L x.
The net external horizontal force F net x=0,F net x=0,F net x=0, since the person is stationary. Thus,
F net x T L x==0=T R x−T L x T R x.F net x=0=T R x−T L x T L x=T R x.F net x=0=T R x−T L x T L x=T R x.
Now observe Figure 5.27. You can use trigonometry to determine the magnitude of T L T L T L and T R T R T R:
cos 5.0°cos 5.0°==T L x T L,T L x=T L cos 5.0°T R x T R,T R x=T R cos 5.0°.cos 5.0°=T L x T L,T L x=T L cos 5.0°cos 5.0°=T R x T R,T R x=T R cos 5.0°.cos 5.0°=T L x T L,T L x=T L cos 5.0°cos 5.0°=T R x T R,T R x=T R cos 5.0°.
Equating T L x and T R x:
T L cos 5.0°=T R cos 5.0°.T L cos 5.0°=T R cos 5.0°.T L cos 5.0°=T R cos 5.0°.
Thus,
T L=T R=T,T L=T R=T,T L=T R=T,
as predicted. Now, considering the vertical components (denoted by a subscript y), we can solve for T. Again, since the person is stationary, Newton’s second law implies that F net y=0 F net y=0 F net y=0. Thus, as illustrated in the free-body diagram,
F net y=T L y+T R y−w=0.F net y=T L y+T R y−w=0.F net y=T L y+T R y−w=0.
We can use trigonometry to determine the relationships among T Ly,T Ry,T Ly,T Ry,T Ly,T Ry, and T. As we determined from the analysis in the horizontal direction, T L=T R=T T L=T R=T T L=T R=T:
sin 5.0°sin 5.0°==T L y T L,T L y=T L sin 5.0°=T sin 5.0°T R y T R,T R y=T R sin 5.0°=T sin 5.0°.sin 5.0°=T L y T L,T L y=T L sin 5.0°=T sin 5.0°sin 5.0°=T R y T R,T R y=T R sin 5.0°=T sin 5.0°.sin 5.0°=T L y T L,T L y=T L sin 5.0°=T sin 5.0°sin 5.0°=T R y T R,T R y=T R sin 5.0°=T sin 5.0°.
Now we can substitute the vales for T Ly T Ly T Ly and T Ry T Ry T Ry, into the net force equation in the vertical direction:
F net y F net y 2 T sin 5.0°−w 2 T sin 5.0°====T L y+T R y−w=0 T sin 5.0°+T sin 5.0°−w=0 0 w F net y=T L y+T R y−w=0 F net y=T sin 5.0°+T sin 5.0°−w=0 2 T sin 5.0°−w=0 2 T sin 5.0°=w F net y=T L y+T R y−w=0 F net y=T sin 5.0°+T sin 5.0°−w=0 2 T sin 5.0°−w=0 2 T sin 5.0°=w
and
T=w 2 sin 5.0°=m g 2 sin 5.0°,T=w 2 sin 5.0°=m g 2 sin 5.0°,T=w 2 sin 5.0°=m g 2 sin 5.0°,
so
T=(70.0 kg)(9.80 m/s 2)2(0.0872),T=(70.0 kg)(9.80 m/s 2)2(0.0872),T=(70.0 kg)(9.80 m/s 2)2(0.0872),
and the tension is
T=3930 N.T=3930 N.T=3930 N.
Significance
The vertical tension in the wire acts as a force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, so most of the tension in the wire is not used to support the weight of the tightrope walker.
If we wish to create a large tension, all we have to do is exert a force perpendicular to a taut flexible connector, as illustrated in Figure 5.26. As we saw in Example 5.13, the weight of the tightrope walker acts as a force perpendicular to the rope. We saw that the tension in the rope is related to the weight of the tightrope walker in the following way:
T=w 2 sin θ.T=w 2 sin θ.T=w 2 sin θ.
We can extend this expression to describe the tension T created when a perpendicular force (F⊥)(F⊥)(F⊥) is exerted at the middle of a flexible connector:
T=F⊥2 sin θ.T=F⊥2 sin θ.T=F⊥2 sin θ.
The angle between the horizontal and the bent connector is represented by θ θ θ. In this case, T becomes large as θ θ θ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., θ=0 θ=0 θ=0 and sin θ=0 θ=0 θ=0). For example, Figure 5.28 shows a situation where we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as straight as possible. The tension in the chain is given by T=F⊥2 sin θ,T=F⊥2 sin θ,T=F⊥2 sin θ, and since θ θ θ is small, T is large. This situation is analogous to the tightrope walker, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F⊥F⊥F⊥ is applied.
Figure 5.28 We can create a large tension in the chain—and potentially a big mess—by pushing on it perpendicular to its length, as shown.
Check Your Understanding 5.9
One end of a 3.0-m rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls sideways on the midpoint of the rope, displacing it a distance of 0.25 m. If he exerts a force of 200.0 N under these conditions, determine the force exerted on the car.
In Applications of Newton’s Laws, we extend the discussion on tension in a cable to include cases in which the angles shown are not equal.
Friction
Friction is a resistive force opposing motion or its tendency. Imagine an object at rest on a horizontal surface. The net force acting on the object must be zero, leading to equality of the weight and the normal force, which act in opposite directions. If the surface is tilted, the normal force balances the component of the weight perpendicular to the surface. If the object does not slide downward, the component of the weight parallel to the inclined plane is balanced by friction. Friction is discussed in greater detail in the next chapter.
Spring force
A spring is a special medium with a specific atomic structure that has the ability to restore its shape, if deformed. To restore its shape, a spring exerts a restoring force that is proportional to and in the opposite direction in which it is stretched or compressed. This is the statement of a law known as Hooke’s law, which has the mathematical form
F⃗=−k x⃗.F→=−k x→.F→=−k x→.
5.11
The constant of proportionality k is a measure of the spring’s stiffness. The line of action of this force is parallel to the spring axis, and the sense of the force is in the opposite direction of the displacement vector (Figure 5.29). The displacement must be measured from the relaxed position; x=0 x=0 x=0 when the spring is relaxed.
Figure 5.29 A spring exerts its force proportional to a displacement, whether it is compressed or stretched. (a) The spring is in a relaxed position and exerts no force on the block. (b) The spring is compressed by displacement Δ x⃗1 Δ x→1 Δ x→1 of the object and exerts restoring force −k Δ x⃗1.−k Δ x→1.−k Δ x→1. (c) The spring is stretched by displacement Δ x⃗2 Δ x→2 Δ x→2 of the object and exerts restoring force −k Δ x⃗2.−k Δ x→2.−k Δ x→2.
Real Forces and Inertial Frames
There is another distinction among forces: Some forces are real, whereas others are not. Real forces have some physical origin, such as a gravitational pull. In contrast, fictitious forces arise simply because an observer is in an accelerating or noninertial frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s Northern Hemisphere, then to an observer on Earth, it will appear to experience a force to the west that has no physical origin. Instead, Earth is rotating toward the east and moves east under the satellite. In Earth’s frame, this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). We can identify a fictitious force by asking the question, “What is the reaction force?” If we cannot name the reaction force, then the force we are considering is fictitious. In the example of the satellite, the reaction force would have to be an eastward force on Earth. Recall that an inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter.
Earth’s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On a large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed (Figure 5.30).
Figure 5.30 Hurricane Fran is shown heading toward the southeastern coast of the United States in September 1996. Notice the characteristic “eye” shape of the hurricane. This is a result of the Coriolis effect, which is the deflection of objects (in this case, air) when considered in a rotating frame of reference, like the spin of Earth. This hurricane shows a counter-clockwise rotation because it is a low pressure storm. (credit "runner": modification of work by "Greenwich Photography"/Flickr)
The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are in inertial frames.
The forces discussed in this section are real forces, but they are not the only real forces. Lift and thrust, for example, are more specialized real forces. In the long list of forces, are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as you will see in the treatment of modern physics later in the text.
Interactive
Explore forces and motion in this interactive simulation as you push household objects up and down a ramp. Lower and raise the ramp to see how the angle of inclination affects the parallel forces. Graphs show forces, energy, and work.
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Stretch and compress springs in the simulation below to explore the relationships among force, spring constant, and displacement. Investigate what happens when two springs are connected in series and in parallel.
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6734 | https://www.anderson.ucla.edu/documents/areas/fac/dotm/bio/pdf_KM11.pdf | Bundling Retail Products: Models and Analysis Kevin F. McCardle, Kumar Rajaram, Christopher S. Tang∗ UCLA Anderson Graduate School of Management Los Angeles, CA 90095-1481 November 3, 2005 Abstract We consider the impact of bundling products on retail merchandising. We consider two broad classes of retail products: basic and fashion. For these product classes, we develop models to calculate the optimal bundle prices, order quantities, and profits under bundling.
We use this analysis to establish conditions and insights under which bundling is profitable.
Our analysis confirms that bundling profitability depends on individual product demands, bundling costs, and the nature of the relationship between the demands of the products to be bundled. We also provide detailed numerical examples.
Key Words: Retailing, Product Bundling, Pricing, Inventory Management, Basic Prod-ucts, Fashion Products.
∗The authors thank Professor Steven Lippman and 3 anonymous referees for their comments and suggestions.
Direct correspondence to any of the authors. Email: kmccardl@anderson.ucla.edu, krajaram@anderson.ucla.edu, ctang@anderson.ucla.edu.
0 1 Introduction An increasing number of separate retail products are being bundled together and sold. Examples can be found across a range of products from several industries including food (cans of chicken broth), apparel (tee shirts), cosmetics (shampoo and conditioner), entertainment (concert tickets and CDs), and electronics (computers and printers). Sometimes different manufacturers of similar products employ different bundling strategies: for example, inexpensive dinner ware by Pfaltzgraff, Mikasa, or Corelle, is often sold in bundles (complete place settings for 4 or 8) while high-end dinner ware items by Christofe, Versace, Villeroy and Boch, or Wedgwood are sold individually. There are several reasons why retailers bundle products. These include reducing logistics, packaging, and transaction costs, increasing market share and sales, and improving customer service; all of which could eventually contribute to increased profitability. However, to realize this potential, it is critical for retailers to determine optimal bundle prices and quantities, and to then decide whether bundling a given set of products would be profitable or not.
In this paper, we address the issues of determining the optimal bundle prices, order quantities, and profits, and of assessing whether or not it is profitable to bundle. We perform this analysis on two broad categories of retail products: basic and fashion. Basic products like men’s underwear and dress shirts typically have more predictable demands and longer life cycles than fashion products like women’s dresses, jewelry, sportswear etc. We also determine how product demand, costs, and the relationship of demand between products affect optimal prices, profits, and the bundling decision. Finally, we compare the results associated with the basic and fashion products to provide insights into how product category affects bundling decisions.
There have been several streams of research on bundling in the economics and marketing litera-ture. Starting with Stigler (1960), economists have studied bundling from the perspective of the customer. In Stigler’s paper, a vector of reservation prices captures customer demand informa-tion, and it is assumed that customers choose products that maximize the difference between these reservation prices and product prices. Stigler uses this framework to show that motion picture dis-tributors would prefer to pure bundle or lease movies in multiple movie packages rather than lease individual movies. Adams and Yellen (1976) utilize this framework and consider three different sales strategies: unbundled sales in which the two products are priced and sold separately, pure bundling in which only a bundle consisting of one unit of each product is sold, and mixed bundling 1 in which both the bundle and the two products are offered. They discuss the possible implications of switching from one strategy to another, but do not determine the optimal bundle prices and conditions under which bundling increases profits. Schmalensee (1984) extends their model by assuming that the distribution of reservation prices follows a bivariate normal distribution. This assumption precludes closed form analytical solutions, and the results are based on a numerical analysis. Also extending the Stigler/Adams and Yellen model, McAfee, McMillan, and Whinston (1989) provide a general sufficient condition for when mixed bundling is optimal in a two-product case. Hanson and Martin (1990) construct and solve a mixed integer program to determine the optimal bundle and prices when the number of customers in each segment and their reservation prices are known. Eppen, Hanson and Martin (1991) consider the strategic use of bundles to reduce cost or increase demand. They advocate that for bundling to be effective, producers should treat bundles as new products, but not as marketing gimmicks. They also propose a managerial frame-work to implement product bundling successfully. Salinger (1995) conducts a graphical analysis of bundling and illustrates the benefits of bundling depends on the costs of the individual products, its relative magnitude to the reservation prices, the cost savings from bundling and the correlation of demands between products. In addition, this analysis is used to show that bundling is profitable only if the optimal bundle price results in the bundle demand being greater than the aggregate demand across products and bundling can increase customer surplus. Armstrong (1996) develops a model to analyze the multi-product nonlinear pricing problem and shows that pure bundles are optimal. This paper also develops conditions on the demand function under which this problem is solvable. However, the analysis does not include ways to calculate the optimal order quantities.
Recent work on bundling in the management science literature includes Bakos and Brynjolfsson (1999, 2000), Ernst and Kouvelis (1999) and Jedidi, Jagpal and Manchanda (2003). Bakos and Brynjolfsson (1999, 2000) focus on the bundling of information products. However, due to the unique characteristics of information products, they noted these products tend to have zero or very low marginal costs of production. As such, their model would not be applicable to retail products. Ernst and Kouvelis (1999) consider the impact of mixed bundling on the inventory decisions of a firm. They provide insights on the degree of sub optimality in profits when inventory decisions are made without explicit consideration of demand substitution between the bundle and the individual products. They develop a stylized two-product model and present the necessary and sufficient optimal inventory conditions. Their model is solved using an efficient numerical search 2 to obtain the optimal stocking levels for the individual products and the bundle. Jedidi, Jagpal and Manchanda (2003) develop and test a model to capture heterogeneity in the joint distribution of reservation process for individual products and bundles. They also develop a procedure that uses the inputs of the model to develop an optimal product-line pricing policy. Stermersch and Tellis (2002) provide a comprehensive summary of the literature in bundling, present a framework to classify the various types of bundling, and also discuss the legality of bundling in each case.
Our paper differs from these papers in several aspects. First, we consider bundling from a retail operations-management perspective. We consider two broad classes of retail products: basic and fashion. For basic products, we capture the notion that aggregate demand is stable by assum-ing that the size of the market is known. We capture the variation in disaggregate demand by modeling consumers’ reservation prices as a uniform random variable. For fashion products, we capture the notion that aggregate demand is unknown by modeling the size of the market as a uniform random variable. Second, in both cases, we provide closed-form analytical solutions for the optimal prices, order quantities, and profits under bundling. Third, we use these results to establish conditions under which bundling is optimal. These conditions provide precise guidance on when a retailer should consider forming a product bundle. Fourth, we show how individual product costs, individual product demands, bundle costs, and the relationship between the de-mands of the individual products affect optimal prices, profits, and the decision of whether or not to bundle. Finally, we compare prices and profits for basic and fashion products under bundling.
This comparison provides insights on the impact of product type on the efficacy of bundling.
This paper is organized as follows. In the next section, we present our model for pure bundling of basic products.
We use this model to determine the optimal prices, order quantities, and profits under bundling. (Henceforth, we shall refer the term ”bundling” as ”pure bundling” in which the retailer offers only the bundle or the individual products, but not a combination of both.) We establish conditions under which bundling is profitable and also determine how product costs, demand and the demand correlation between bundled products affects optimal prices, order quantities and the bundling decision. In Section 3, we develop our model for bundling of fashion products and repeat this analysis. In Section 4, we compare our results between basic and fashion products and provide insight on how product category affects bundling decisions.
A detailed numerical example is presented in Section 5. In the concluding section, we summarize our key results and provide future research directions.
3 2 Basic Products In this section we introduce the preliminary model and notation. The firm produces two products, A and B, at constant marginal costs cA and cB, respectively. We assume 0 fixed costs. Demand for each of the products is specified by consumer reservation prices rA and rB. There is a potential market of size M for each of the products. In this section we consider basic products, such as men’s white dress shirts, black socks, undergarments, etc., and assume that M is known. (In the next section, on fashion products, we model the market size M as a random variable.) The assumption that M is known for basic products seems reasonable because basic products tend to have long life cycles, more stable aggregate demand (i.e., market size), and more historical data for generating more accurate forecasts. Uncertainty occurs at the disaggregate level due to variablity in customer reservation prices. That is, for basic products we know how many potential consumers there are, but there is uncertainty regarding how many of the potential consumers will convert to purchasers. For tractability, the distribution of the consumers’ reservation prices for product i is assumed to be uniform between rl i and ru i . That is, the distribution of reservation prices for product A is uniform between rl A and ru A, and the distribution of reservation prices for product B is uniform between rl B and ru B. We assume without loss of generality that rl A = 0 and ru A = 1. Of course, rl B ≥0, cA ≤1, and cB ≤ru B. Finally, we assume ru B ≤1.
2.1 No Bundle For a baseline comparison, we first determine the optimal prices for the case with no bundling.
If the firm sets prices pA and pB, then all customers with reservation prices above pA and pB, respectively, purchase the products. Thus, the demands DA and DB when the market for the products has size M are given by: DA = M Z 1 pA 1dx = M(1 −pA), and (1) DB = M Z ru B pB 1 ru B −rl B dx = M ru B −pB ru B −rl B !
.
(2) The firm’s objective is to choose prices pA and pB in order to maximize the profit function 4 π(A,B)(pA, pB): π∗ (A,B) = max pA,pB M " (pA −cA)(1 −pA) + (pB −cB) ru B −pB ru B −rl B !# .
(3) Differentiating (3) and solving for the optimal prices, quantities, and profit yields: p∗ A = 1 + cA 2 ≥ cA, (4) p∗ B = ru B + cB 2 ≥ cB, (5) Q∗ A = M 1 −cA 2 , (6) Q∗ B = M ru B −cB 2(ru B −rl B) !
, and (7) π∗ (A,B) = M "1 −cA 2 2 + 1 ru B −rl B ! ru B −cB 2 2# .
(8) The optimal price of an unbundled good is increasing in the associated cost, and the profit function decreases in the costs.
2.2 Pure Bundle – Independent Demands If the firm offers only the bundled product AB at price p, a consumer will purchase the bundle only if rA + rB ≥p; that is, if the sum of his reservation prices exceeds the bundle price. Let DAB be the demand of consumers in the market of size M whose sum of reservation prices exceeds the bundle price p. Then, by considering the fact that rA ∼U[0, 1] and rB ∼U[rl B, ru B], we have: DAB = M Z ru B max(p−1,rl B) Z 1 max(p−xB,0) 1dxA !
1 ru B −rl B dxB = M 1 − 1 ru B−rl B (p−rl B)2 2 if p < ru B 1 + ru B+rl B 2 −p if ru B ≤p ≤1 + rl B 1 ru B−rl B (1−p+ru B)2 2 if p > 1 + rl B.
(9) As can be seen from the above equation, the changes in the formula representing the demand for the bundle are generated by the changes in the limits of integration. Let cAB be the unit cost of the bundled product. The firm chooses the bundle price p to maximize profit πAB(p), where: π∗ AB = max p (p −cAB)DAB.
5 For simplicity in what follows, we will restrict the bundle cost cAB so that the optimal bundle price falls in the middle of the three regions listed in (9). That is, we want ru B ≤p ≤1 + rl B, though similar results hold when p falls in the other regions. To achieve the condition on p, we assume 1.5ru B −.5rl B −1 ≤cAB ≤1 + 1.5rl B −.5ru B. Let RB = 1 + [(ru B + rl B)/2]. Solving for the optimal bundle price p∗ AB, the optimal quantity Q∗ AB, and substituting in the profit function yields p∗ AB = [RB + cAB] /2, (10) Q∗ AB = M [RB −cAB]/2, and (11) π∗ AB = M {[RB −cAB] /2}2 .
(12) Note that our assumption on cAB results in Q∗ AB ≥0 and πAB ≥0. As in the case with no bundling, an increase in the cost leads to an increase in the optimal price and a decrease in the optimal profit. Our interest, though, is in a comparison of the prices and profits with and without bundling.
Ordering the quantities is problematic. To elaborate, recall that Q∗ A is units of A produced and Q∗ B is units of B produced in the unbundled state. On the other hand, Q∗ AB is the bundled units, each including one unit of A and one unit of B. Therefore, unlike with the prices, it makes no sense to compare the bundle quantity with the sum of the unbundled quantities. For this reason, we shall analyze the ordering of the profits rather than the (intermediate) quantities.
Proposition 1 Assume demands for the two products are independent and there are no dis-economies of bundling; i.e., cAB ≤cA + cB. Then I : p∗ AB < p∗ A + p∗ B, and II : π∗ AB > π∗ (A,B) ⇐ ⇒(1 −cA)2 + (ru B −cB)2 ru B −rl B < (RB −cAB)2 .
Furthermore, if the two products have identically distributed reservation prices, rA ∼U[0, 1] and rB ∼U[0, 1], and cAB = cA + cB, then II′ : π∗ AB < π∗ (A,B).
Proof: All proofs are provided in the Appendix.
The claim in Part (I) of Proposition 1 is to be expected when there are economies to bundling as a reduction in the aggregate cost allows for a concomitant reduction in the bundle price while 6 maintaining profit margins. Part (I) shows that even with no economies of bundling, the optimal bundle price is lower than the sum of the optimal prices of the individual goods without bundling.
As noted below, this is partly explained by an increase in quantity sold. Because the optimal bundle price p∗ AB is continuous in the cost cAB, there must also be cases with small dis-economies of bundling, i.e., when cAB > cA + cB, where the optimal bundle price is less than the sum of the prices of the unbundled products.
The condition in Part (II) of Proposition 1 warrants further discussion. Assume that there are no economies to bundling, i.e., cAB = cA + cB, fix cAB = c, and consider the condition as a function of x = cB. Then bundling is optimal if and only if (1 −(c −x))2 + (ru B −x)2 ru B −rl B < (RB −c)2 .
The left-hand side of this condition is quadratic in x, while the right-hand side is a constant. If the minimum of the quadratic function on the left (which occurs when x = [ru B + (c −1)(ru B − rl B)]/[1 + (ru B −rl B)]) is less than the value on the right, then there is a range of values for cA and cB where bundling is optimal. If the minimum of the quadratic function on the left is greater than the value on the right, then bundling is not optimal for any costs. Roughly speaking, if the costs of the two products are dis-similar (i.e., x near 0 or c), no bundling dominates bundling; but if the products have similar costs, bundling is optimal. We provide a numerical example in a later section.
For the special case considered in Part (II’), with no economies to bundling, rl B = 0, ru B = 1, and cAB = 0.5, the minimum of the quadratic function is attained at x = 0.25. It is then easy to check that the left-hand side is always greater than the right; hence, bundling is not optimal. That is, if basic products have identical costs and identical distributions of reservation prices, then no bundling is more profitable than bundling.
2.3 Pure Bundle – Perfectly Correlated Demands In the previous section we analyzed the case where the distributions of reservation prices for the two products were independent. In this section we analyze the case where the distributions of reservation prices are perfectly correlated. The analysis of the unbundled case is exactly as in section 2.1 because the analysis does not rely on the independence of the demands. We begin with 7 the positively correlated case: rA ∼U[0, 1], and rB = [rl B + (ru B −rl B)rA] ∼U[rl B, ru B].
(13) The reservation prices can be perfectly positively correlated when the products are complements and the preference for the second item is determined by the preference for the first (shampoo and conditioner); when the products are identical (bars of soap or sets of sport socks); or when the second unit has the same reservation price as the first unit. For the case where the firm offers only the bundled product at price p, consumers whose sum of reservation prices for the individual products exceeds p will purchase the product. That is, if rA + rB ≥p the consumer will choose to purchase the bundle. Substituting from (13) above, it is easy to show that the consumer will purchase the bundle if and only if rA ≥max 0, p −rl B 1 + ru B −rl B !
.
Let ˆ DAB be the demand of consumers in the market who will choose to purchase. When p ≤rl B, DAB = M and the optimal p∗= rl B. Hence, it suffices to consider the case with p ≥rl B, whence ˆ DAB = M " 1 + rl B 1 + ru B −rl B !
− 1 1 + ru B −rl B !
p # .
The first order condition yields the optimal price ˆ p∗, quantity ˆ Q∗ AB, and profit ˆ π∗ AB as follows: ˆ p∗ AB = (1 + ru B + cAB) /2, (14) ˆ Q∗ AB = M 1 + ru B −cAB 2(1 + ru B −rl B) !
and (15) ˆ π∗ AB = M 1 + ru B −rl B 1 + ru B −cAB 2 2 .
(16) Note that the optimal bundle price for the positively dependent case (14) is strictly greater than the optimal bundle price for the independent case (10).
Proposition 2 Assume demands for the two products are perfectly positively correlated. If there are no dis-economies to bundling, cAB ≤cA + cB, then I : ˆ p∗ AB ≤p∗ A + p∗ B and II : ˆ π∗ AB > π∗ (A,B) ⇐ ⇒(1 −cA)2 + (ru B −cB)2 ru B −rl B < [1 + ru B −cAB]2 1 + ru B −rl B .
8 If there are neither economies nor dis-economies to bundling, i.e., cAB = cA + cB, then I′ : ˆ p∗ AB = p∗ A + p∗ B and II′ : ˆ π∗ AB ≤π∗ (A,B).
As parts (I) and (II) with possible economies of bundling follow the same logic for the similar parts of Proposition 1, we shall limit our discussion to parts (I’) and (II’). Unlike the independent de-mands case, when reservation prices are perfectly positively correlated and there are no economies to bundling, Proposition 2 part (I’) shows that the bundled good is neither discounted nor marked up relative to the prices of the unbundled goods. It follows then that ˆ p∗ AB > p∗ AB, that is, the price of the bundled good when demands are perfectly positively correlated is greater than the price of the bundled good when demands are independent. When the demands are perfectly positively correlated, if a consumer purchases one product, then he will certainly purchase the other; and if he does not purchase one product, then he will not purchase the other. Hence, discounting the bundle price yields no benefit. Furthermore, part (II’) of Proposition 2 proves that the bundle profit in the positively correlated case (16) are strictly less than unbundled profit (8). To finish the comparisons, ˆ π∗ AB < π∗ AB if and only if cAB < 1 − √ 0.5; that is, the bundle profit in the positively correlated case are less than the bundle profit in the independent case as long as the unit cost of the bundled goods cAB is sufficiently small. The reservation price rA + rB for the bundled good in the independent case has a lower variance than in the positively correlated case and the means are the same. When costs are low, prices are low, but the independent demand case makes up for it in quantity sold.
Now consider the case of perfectly negatively correlated reservation prices: rA ∼U[0, 1], and rB = [ru B −(ru B −rl B)rA] ∼U[rl B, ru B].
For instance, suppose that the goods are close substitutes, for example, a DVD and a VHS tape of the same movie, and once a consumer has one of the products, the demand for the other goes down. Again, the analysis of the unbundled case remains unchanged. But for the bundle, the sum of the reservation prices exceeds the bundle price if rA ≥max ( 0, p −ru B [1 −(ru B −rl B)] ) .
9 When p ≤ru B, DAB = M, and the optimal price is p∗= ru B. As previously, we are interested in the case when p is not at the boundary, i.e., p ≥ru B. Solving for the optimal price, quantity, and profits as before, we have: ˜ p∗ AB = 1 + rl B + cAB 2 , (17) ˜ Q∗ AB = M 1 + rl B −cAB 2[1 −(ru B −rl B)] !
and (18) ˜ π∗ AB = M 1 −(ru B −rl B) "1 + rl B −cAB 2 #2 .
(19) We state Proposition 3 only for the case of no economies or diseconomies of bundling. The existence of economies to bundling would only act to enlarge the parameter range over which bundling is optimal.
Proposition 3 Assume demands for the two products are perfectly negatively correlated. If there are no economies or dis-economies to bundling, i.e., cAB = cA + cB, then I : ˜ p∗ AB < p∗ A + p∗ B.
If, in addition, rB ∼U[d, 1] for d > 0, then IIa : ˜ π∗ AB > π∗ (A,B) ⇐ ⇒ 1 d 1 + d −cA −cB 2 !2 > 1 −cA 2 2 + 1 1 −c 1 −cB 2 2 ; Alternatively, if rB ∼U[0, d] for d < 1, then IIb : ˜ π∗ AB > π∗ (A,B) ⇐ ⇒ 1 1 −d 1 −cA −cB 2 2 > 1 −cA 2 2 + 1 d d −cB 2 !2 .
Because the bundle profit in the perfectly positive correlated case with no economies to bundling are strictly less than the unbundled profit, a basic intuition might lead one to believe that the opposite would be true for perfect negative correlation; that is, ˜ π∗ AB ≥π∗ (A,B). Parts (IIa) and (IIb) of Proposition 3 shows that this intuition is incorrect: unlike the positively correlated case, there is not an unambiguous ordering of profits. Because the bundle price is lower with negative correlation, the firm must have a more-than-offsetting increase in volume to increase profit.
To summarize, with no economies to bundling and perfectly positive correlated reservation prices, the optimal bundle price is equal to the sum of the unbundled prices, and the bundle profit is 10 lower than the unbundled profit. As such, the firm should not bundle. With no economies to bundling and perfect negative correlation, the optimal bundle price is lower than the sum of the unbundled prices, and there are cases where the bundle profit exceeds the unbundled profit and alternative cases where the opposite holds. Therefore, the firm may, but need not, find it optimal to offer the bundle. With no economies to bundling and independent reservation prices, again the bundle price is less than the sum of the unbundled prices, and again, there are cases where the bundle profit exceeds the unbundled profit and alternative cases where the opposite holds. Similar results for bi-variate normally distributed reservation prices have been derived by Schmalensee (1984). But the intractability of the resulting model does not permit analytical comparisons with economies of scale.
The economy to bundling is given by cA+cB−cAB. Greater economies to bundling leads to a higher bundle profit, increasing the range wherein bundling is optimal. Similarly, greater diseconomies to bundling leads to a lower bundle profit, shrinking the range wherein bundling is optimal. This is true irrespective of the correlation between the demands.
The above results highlight the importance of economies to bundling to the bundling decision, that is, the bundling decision is as much an operational decision as it is a marketing one. Next, we consider the case of fashion products.
3 Fashion Products In this section we generalize from the previous assumption that the market size M is known, substituting instead an assumption about the distribution of the possible (random) market size M. This can be expected for fashion products for which the aggregate demand is often unknown and highly variable. If the market size (i.e., demand) is random but decreasing in price, and the firm makes its bundling (i.e., production) decision before demand is known, we have, in essence, a newsboy model with pricing, but not amenable to analytic, closed-form solutions (c.f., Petruzzi and Dada (1999)). To overcome some technical challenges, we make several simplifying assumptions.
Rather than introduce a new set of notation for this section, we reuse earlier notation: the meaning should be clear from the context. For example, π∗ AB will represent the optimal profit from bundling with independent demands, in the current section with the additional assumption that the market size is unknown, while in the previous section with the assumption that market size is known.
11 As per a newsboy model, let Qi be the order quantity and Di the demand distribution of product i = A, B, and AB. (While it is straightforward to also include a salvage value si, for ease of notation, we assume the salvage value is 0 for all products.) If the firm produces Q units of a good at a unit cost of c for sale at price p when the realized demand is x, then the firm’s profit π(Q, x) is given by: π(Q, x) = (p −c) min{Q, x} −c[Q −x]+.
When demand X has distribution F, it is easy to show that the expected profit π(Q) = EXπ(Q, X) is maximized at Q∗, where: Q∗ = F −1 p −c p !
, (20) π∗ = p Z Q∗ 0 xdF(x).
(21) If the demand is uniformly distributed in the interval [0, 2d(p)] where d(p) is some downward sloping function of price, equations (20) and (21) become Q∗(p) = 2d(p) p −c p !
, and (22) π∗(p) = d(p) (p −c)2 p !
.
(23) That is, if price is given, the optimal order quantity is Q∗(p) and the optimal profits are π∗(p) as given in (22) and (23).
Suppose that the demand is linear in price, i.e., d(p) = δ −αp, (24) for some scalars δ and α, where δ represents the size of the potential market and α the price sensitivity. Then we can establish the following Proposition.
Proposition 4 When the demand is linear in price, the profit function (23) is unimodal in price.
In light of Proposition 4, there is a unique price p∗that maximizes profits. It can be found by substituting (24) in (23) and solving for the first order condition to get: p∗= δ + √ δ2 + 8αδc 4α .
(25) 12 Note that the optimal price is increasing in the market size δ, decreasing in the price sensitivity α, and increasing in the cost c. Substituting the optimal price (25) into the quantity (22) and profit (23), we have: Q∗ = 3δ − √ δ2 + 8αδc 2 ! 1 − 4αc δ + √ δ2 + 8αδc !
, and (26) π∗ = 1 4 3δ − √ δ2 + 8αδc δ + √ δ2 + 8αδc −4αc 2 4α(δ + √ δ2 + 8αδc) .
(27) Notice that both the optimal order quantity Q∗and the optimal profit π∗are decreasing in the cost c. A numerical example in an upcoming section will confirm this.
In order to draw a parallel with the analysis of the previous section on basic products, we assume that the market size M is uniformly distributed in the range [0, 2M] where M is defined as before.
Thus, E(M) = M. When market size is known, demand for product A is characterized in (1), for product B in (2), and for the bundle AB in (9). With uncertain market size, we extend the demand characterizations in (1), (2), and (9), and define new demands as follows: DA = M(1 −pA) (28) DB = M(ru B −pB)/(ru B −rl B) and (29) DAB = M (RB −pAB) .
(30) As in (9), the demand DAB for the bundled good given by (30) assumes the demands for the individual goods are independent of each other, and the cost of the bundled good is in a middle range. Note that (28), (29), and (30), all have the linear form of (24). That is, for product A, δA = M and αA = M; for product B, δB = Mru B/(ru B −rl B) and αB = M/(ru B −rl B); and for the bundled product AB, δAB = M[1 + ((rl B + ru B)/2)] and αAB = M.
3.1 No Bundle If the firm only offers the individual products, A and B, with random demand for the goods, DA and DB, given by (28) and (29), respectively, substitution in the optimal price, quantity, and profit equations (25) – (27) yields the following values for product A: p∗ A = 1 + √1 + 8cA 4 (31) 13 Q∗ A = M 2 3 − √ 1 + 8cA 1 − 4cA 1 + √1 + 8cA !
(32) π∗ A = M 4 3 −√1 + 8cA 1 + √1 + 8cA −4cA 2 4(1 + √1 + 8cA) .
(33) Similarly for product B: p∗ B = ru B + q (ru B)2 + 8ru BcB 4 (34) Q∗ B = M 2(ru B −rl B) 3ru B − q (ru B)2 + 8ru BcB 1 − 4cB ru B + q (ru B)2 + 8ru BcB (35) π∗ B = M 4(ru B −rl B) 3ru B − q (ru B)2 + 8ru BcB ru B + q (ru B)2 + 8ru BcB −4cB 2 4 ru B + q (ru B)2 + 8ru BcB .
(36) At this point all that can be said is that, as would be expected, the optimal prices p∗ A and p∗ B are increasing, and the optimal quantities Q∗ A and Q∗ B and optimal profits π∗ A and π∗ B are decreasing, in the costs cA and cB, respectively. The unbundled prices, quantities, and profits are important, however, for a comparison with the bundled values.
3.2 Pure Bundle – Independent Demands Recall that RB = 1 + [(ru B + rl B)/2]. If the firm only offers the bundled good AB produced at cost cAB, with 1.5ru B −.5rl B −1 ≤cAB ≤1 + 1.5rl B −.5ru B, and with random demand DAB given in (30), substitution in (25) – (27) yields the following for the optimal bundle price, p∗ AB = 1 4 RB + q R2 B + 8RBcAB , (37) the optimal order quantity, Q∗ AB = M 2 3RB − q R2 B + 8RBcAB 1 − 4cAB RB + q R2 B + 8RBcAB , (38) and the optimal profits, π∗ AB = M 4 3RB − q R2 B + 8RBcAB RB + q R2 B + 8RBcAB −4cAB 2 4 RB + q R2 B + 8RBcAB .
(39) 14 As in the unbundled case, the optimal price is increasing and the optimal profit is decreasing in the cost of production. In addition, note from (37) that price is increasing in RB, or ru B and rl B. This is because it is optimal for the retailer to increase prices to extract the surplus from customers who have higher reservation prices.
3.3 Comparing No Bundling with Pure Bundling So far, we have derived the optimal prices, order quantities, and profits for the separate cases of no bundling and pure bundling. We now compare the optimal prices and profits between no bundling and pure bundling. As in the case of known demand, comparing quantities without additional assumptions is problematic, and so we focus on profits.
In Section 2, we proved that, when the reservation prices for the individual products are indepen-dent, the optimal bundle price is less than the sum of the unbundled prices. A similar result holds for fashion products when the market size is unknown.
Proposition 5 Assume market size is random and demands for the two products are independent.
If there are no economies to bundling, i.e., cAB = cA + cB, and the range of reservation prices for product B is U[0, 1]. Then I : p∗ AB < p∗ A + p∗ B.
If, in addition, the costs of the individual products are the same, i.e., cA = cB, then II : π∗ AB < π∗ (A,B).
As the optimal bundle price is lower than the sum of the unbundled prices when there are no economies to bundling, in order for bundling to be a profitable undertaking, it must be that the firm “increases” the quantity sold. Proposition 5 shows that for the special case when cA = cB, the firm cannot increase the quantity sold enough to compensate for the lower price. Moreover, Proposition 5 mirrors the results of Proposition 1.
15 4 Comparison of Known with Uncertain Market Size Our specification of the demand for the products with random market size is such that the expected market size E(M) equals the market size M in the non-random case. It is reasonable to ask, then, what effect uncertainty has on optimal prices and profits. First consider the case of the individual products.
Proposition 6 I: The prices of the individual basic products with known market size are lower than that of the individual fashion products with unknown market size.
II: The profits of the individual basic products with known market size are greater than the indi-vidual fashion products with unknown market size.
Part I is consistent with actual practice where basic products are usually priced lower than com-parable fashion products. This lower price is offset by an increase in demand, in turn, increasing overall profits. Furthermore, as can be seen in the proof of Proposition 6 in the appendix, the difference in the prices between individual fashion and basic products is increasing in the difference between the reservation price and the production cost. As the production cost of an individual product increases, the difference between the prices between the fashion product and the basic product also increases.
We now examine the optimal prices and profits for the case of bundled goods.
Proposition 7 I: The price of the bundled basic product with known market size is lower than the price of the bundled fashion product with unknown market size.
II: The profit of the bundled basic product with known market size is greater than the profit of the bundled fashion product with unknown market size.
Proposition 7 shows that the results of Proposition 6 carry over to the bundled product. We omit the details.
16 5 Numerical Examples In this section, we construct numerical examples to illustrate the ideas developed in this paper.
The first example relates to basic products. Here, we let M = 100, rA ∼U[0, 1] and rB ∼U[0, 1], cA = cB = 0.25, and cAB = cA + cB = 0.5 = c. Substituting these values in (4) to (8), (10) to (12) and (14) to (16), we get the optimal prices, order quantities and profits under no bundling and bundling with independent demand and perfectly positive correlated demand. These results are summarized in Table 1. As expected, by Proposition 1, these results confirm that p∗ AB = 1.05 < (p∗ A + p∗ B) = (0.625 + 0.625). In addition, since c = 0.5, π∗ AB = 25 < π∗ (A,B) = 28.125. Finally, as implied by Proposition 2, we have ˆ p∗ AB = 1.25 = p∗ A + p∗ B and ˆ π∗ AB = 28.125 ≤π∗ (A,B) = 28.125.
INSERT TABLE 1 HERE To consider the impact of perfectly negatively correlated demand, we set rA ∼U[0, 1] and rB ∼ U[0.2, 1] and recompute the optimal price, order quantities and profits under no bundling and the various cases of bundling (i.e., independent, positive and negative dependent demand). These results are summarized in Table 2. Consistent with Proposition 3, these results show that ˜ p∗ AB = 0.85 < p∗ A + p∗ B = 1.25 and ˜ π∗ AB = 61.25 > π∗ (A,B) = 31.64. Finally, in this example, observe that ˜ π∗ AB = 61.25 > ˆ π∗ AB = 31.25 > π∗ AB = 30.25.
INSERT TABLE 2 HERE To illustrate the impact of costs on optimal prices and profits, we set rA ∼U[0, 1] and rB ∼ U[0.2, 1], x = cA = cB, cAB = cA + cB, and vary x from 0.2 to 0.4 in increments of 0.02.
We consider 4 scenarios, namely, no bundling, bundling with independent demands, positively correlated demands and negatively correlated demands. The impact of these changes on optimal prices is summarized in Figure 1, while the impacts on optimal profits are summarized in Figure 2. Figure 1 shows that, as expected, prices are increasing in costs. In addition, this figure shows that prices under bundling with negative correlated and independent demands are lower than prices of the sum of the individual products, results that are consistent with Propositions 1 and 3, respectively. In addition, as expected from Proposition 2, prices under positively correlated demand are always equal to the prices of the unbundled quantities. Figure 2 shows the impact of costs on profits for the various scenarios: no bundling, bundling under independent, perfectly positive and negatively correlated demands. This figure illustrates that when costs are relatively 17 low (i.e., 0.2 < c < 0.4) the optimal strategy is to bundle perfectly negatively correlated products.
From this analysis, we also confirmed that profit from bundling under perfectly positive correlated demands is dominated by not bundling.
INSERT FIGURES 1 AND 2 HERE To understand the impact of asymmetric product costs on profits for basic products (i.e., when cA ̸= cB ), we set cA +cB = 0.5 and varied the ratio cA/cB from 0.25 to 4. These results are shown in Figure 3 for the case of no bundling and bundling with independent and positively correlated demands. This figure shows that with increased asymmetry in product costs, the greater is the difference in profits between the unbundled and bundled case.
INSERT FIGURE 3 HERE The second example deals with fashion products. Here again, we let M = 100, rA ∼U[0, 1] and rB ∼U[0, 1], cA = cB = 0.25 and cAB = cA + cB = 0.5. Substituting these values in (31) to (39), we get the optimal prices, order quantities and profits under no bundling and bundling with independent demand. These results are consistent with Proposition 5: p∗ AB = 1.09 < (p∗ A + p∗ B) = (0.68 + 0.68) and π∗ AB = 13.09 < π∗ (A,B) = 17.4.
To understand the impact of asymmetric product costs on profits for fashion products, here again we set cA + cB = 0.5 and varied the ratio cA/cB from 0.25 to 4. These results are shown in Figure 4 and show that with increased asymmetry in product costs, the greater is the difference in profits between the unbundled and bundled case.
INSERT FIGURE 4 HERE To analyze the impact of reservation price variability on optimal prices and profits, we set rA ∼ U[0, 1] and rB ∼U[0, ru B], c = cA = cB = 0.15 and cAB = cA + cB and vary ru B from 0 to 0.8 in increments of 0.1. These results for optimal prices and profits are summarized in Figures 5 and 6, respectively, showing that p∗ AB < (p∗ A + p∗ B) and π∗ AB < π∗ (A,B) across the entire range of ru B. This is expected from Proposition 5. In addition, as expected from (37) prices are increasing in ru B.
However, Figure 5 also shows that the rate of increase of optimal price varies with and without bundling. It is higher for no bundling as the retailer can increase the price on one product without affecting the chances that the customer would buy the other product. The rate of increase is smaller for the bundled product, as the increase in price for one product, affects the chance the 18 customer would buy the other product due to the higher bundled price. Similarly, observe from Figure 6 that profits are increasing in ru B. This is because the gain in prices offsets the loss in sales due to higher prices, which in turn increases overall profits. Since the rate of increase in prices differ with and without bundling, so does the rate of increase in profits as shown in Figure 6.
INSERT FIGURES 5 AND 6 HERE Finally, by considering the case in which the demands are independent, we compared prices and profits between basic and fashion products when M = 100, rA ∼U[0, 1] and rB ∼U[0, 1], c = cA = cB = 0.25 and cAB = cA + cB = 0.5. These results are summarized in Table 3 and confirm Proposition 6. For instance, prices of individual basic products are lower than prices of the corresponding fashion products, while profits from the individual basic products are higher.
In addition, as expected from Proposition 7, prices of the bundled basic good are lower than the bundled fashion good, while profits of the basic bundle are higher.
INSERT TABLE 3 HERE 6 Conclusion In this paper, we have analyzed the impact of product bundling on retail merchandising. To perform this analysis, we considered two broad classes of retail products: basic and fashion. For basic products, demand is specified by uniformly distributed reservation prices and the market size is assumed to be known. For fashion products, we generalize this assumption by treating the market size as a random variable. For both basic and fashion products, we consider pairs of products and develop models to calculate the optimal bundle prices, order quantities and profits under bundling. Our analysis confirms that bundling profitability depends on individual product demands, bundling costs and the nature of the relationship between demands of the products to be bundled. We also provide detailed numerical examples.
For basic products, we find that with no economies to bundling and when product demands are independent or perfectly negatively correlated, the optimal bundle price is lower than the sum of the unbundled prices. We also show that there are cases where the bundled profit exceeds the unbundled profit and alternative cases where the opposite holds. Therefore, the firm might or 19 might not bundle. For products with perfectly positive correlated demand, the optimal bundle price is equal to the sum of the optimal unbundled prices, but the optimal profit is always lower.
Hence, the firm should not bundle. Thus, the firm needs to carefully understand costs, demand and the relationship between demands between the products before making the bundling decision.
Our paper provides a framework to determine if bundling is profitable, and also to determine optimal prices, order quantities and profits under bundling. Our numerical example shows how optimal prices and profits are affected by costs.
For fashion products, when there are no economies to bundling and when product demands are independent, we find that the optimal bundle price is lower than the sum of the optimal unbundled prices. Our numerical example illustrates the impact of reservation price variability on the optimal prices and profits. Finally, we compare basic and fashion products and find that the optimal unbundled prices and profit of individual basic products are higher. These results also hold when the products are bundled.
This paper provides several new avenues for future research. First, it would be constructive to consider the case of mixed bundling, where the retailer offers the individual products and the bundle. While this is a realistic setting in many situations, the exact analysis of this problem is complex due to intractability of the joint profit function. Second, one could consider different distributions of reservation prices, such as the multi-nomial logit distribution that could also capture the impact of difference in market share among the bundled products. Third, it could be instructive to examine product bundling under retail competition. Fourth, it could be important to identify and incorporate the additional financial, operational and marketing constraints that may be needed to implement bundling across various product lines. Finally, future work could include the problem of forming bundles with basic and fashion products and dynamically updating and changing bundle prices with realized sales information. We hope the ideas developed in this paper will be used as building blocks to address these extensions.
In conclusion, we believe that the methods developed in this paper provide a simple and useful framework to understand and analyze bundling in the context of retail merchandising.
20 7 Appendix: Proofs of Propositions Proof of Proposition 1. Part (I) follows from (4), (5), and (10). Compare the profits without bundling (8) and the profits with bundling (12) to establish parts (II) and (II’)of the proposition.
Proof of Proposition 2, parts (I) and (I’) follow from (4), (5), and (14). Proof of Proposition 2, part (II) follows from comparing (8) with (16). Proposition 2, part (II’) requires proof: bundle profits (16) in the positively correlated case, known market size, no economies or dis-economies, are less than the unbundled profits (8).
π∗ (A,B) : ˆ π∗ AB (1 −cA)2 + 1 ru B −rl B (ru B −cB)2 : [(1 −cA) + (ru B −cB)]2 1 + ru B −rl B (1 + ru B −rl B)[(1 −cA)2 + 1 ru B −rl B (ru B −cB)2] : (1 −cA)2 + (ru B −cB)2 +2(1 −cA)(ru B −CB) (ru B −rl B)(1 −cA)2 + 1 ru B −rl B (ru B −cB)2 : 2(1 −cA)(ru B −cB) (ru B −rl B)(1 −cA)2 −2(1 −cA)(ru B −cB) + 1 ru B −rl B (ru B −cB)2 : 0 q ru B −rl B(1 −cA) − 1 q ru B −rl B (ru B −cB) 2 > 0.
Thus, the profits in the known market size unbundled case (the left column) dominate the known market size bundled case (the right column) when there are no economies to bundling (cAB = cA + cB) and reservation prices are perfectly positively correlated.
Proof of Proposition 3, part (I) follows from (4), (5), and (17). Proof of Proposition 3 parts (IIa) and (IIb) are proved similarly; we provide the proof for (IIb). Consider when the range of the reservation price for product B is [0, d] where d < 1. Thus, rB = d(1 −rA). The optimal bundle price given in (17) shows that the bundle price offers a discount over the unbundled prices. The bundle profits for the known market size, perfectly negatively correlated case, ˜ π∗ AB given by (19), 21 are larger than the known market size unbundled profits π∗ (A,B) given by (8) if and only if 1 1 −d 1 −cA −cB 2 2 > 1 −cA 2 2 + 1 d d −cB 2 !2 ⇐ ⇒ d(1 −cA −cB)2 > d(1 −d)(1 −cA)2 + (1 −d)(d −cB)2.
Rearranging terms, bundle profits exceed unbundle profits if and only if d2(2 −2(cA + cB) + c2 A) + d(2c2 B + 2cAcB −1) −c2 B > 0.
For example, if cA = cB = c, calculations show that this inequality is satisfied if d = 0.8 and c ≤0.3, but that it is not satisfied for any value of c when d = 0.4.
Proof of Proposition 4: If demand is linear, then profits are unimodal in p. The profit function is given by: π(p) = (δ −αp)(p −c)2/p.
Note that at p = c, profits are 0. Differentiating the profit function and simplifying yields π′(p) = (1/p)2(p −c)[−2αp2 + δp + δc].
When p = c, π′(p) = 0, but the third term on the right-hand side is positive. Thus, when p is slightly above c, π′(p) is positive. There are two other zeros of π′, one negative and one positive.
The positive zero is p = δ + √ δ2 + 8αδc 4α .
Thus, π(p) is unimodal for p ≥c.
Proof of Proposition 5: To validate part (I) of Proposition 5, compare the bundle price (37) with the sum of the unbundled prices, (31) plus (34). The bundle price is p∗ AB = 1.5 + q (1.5)2 + 8(1.5)(cA + cB) 4 (40) while the sum of the unbundled prices is p∗ A + p∗ B = 2 + √1 + 8cA + √1 + 8cB 4 .
(41) 22 Because cA < 1 and cB < 1, it follows that the bundle price (40) is less than (41). This is easiest to see if cA = cB = c, in which case cAB = 2c.
Assume the range of reservation prices for product B is [0, 1], there are no economies to bundling, and the costs of the unbundled goods are the same, i.e. cA = cB = c, whence cAB = 2c. The profit for the bundled good (39) is π∗ AB = M 4 3(1.5) − q 1.52 + 8(1.5)2c 1.5 + q 1.52 + 8(1.5)2c −4(2c) 2 4 1.5 + q 1.52 + 8(1.5)2c , (42) while the sum of the unbundled profits is π∗ (A,B) + 2M 4 3 − √ 1 + 8c 1 + √1 + 8c −4c 2 4 1 + √1 + 8c .
(43) Some algebraic manipulations show that the bundle profits (42) exceed the sum of the unbundle profits (43), i.e., bundling is optimal, if and only if 9 3 − s 1 + 8 4 3 c 1 + r 1 + 8 4 3 c −4 4 3 c 2 1 + r 1 + 8 4 3 c > 8 3 − √ 1 + 8c 1 + √1 + 8c −4c 2 1 + √ 1 + 8c .
Calculations show that this last condition is satisfied only when the cost c < 0.09, which violates the assumption regarding the value of cAB, hence bundling is not optimal.
Proof of Proposition 6 (I): The prices, p∗ A and p∗ B, of the individual products are lower when the market size is known.
We begin by showing that the price for product B with known market size is less than the price for product B when the market size is random. The price of product B with known demand given in (5) is less than the price of product B with random demand given in (34) if and only if (ru B −cB) 2 < ru B + q (ru B)2 + 8ru BcB 4 ⇐ ⇒ ru B + 2cB < q (ru B)2 + 8ru BcB ⇐ ⇒ (ru B)2 + 4ru BcB + 4c2 B < (ru B)2 + 8ru BcB ⇐ ⇒ cB < ru B.
23 This last condition holds by assumption: the production cost of the product must be less than the highest reservation price lest there be no demand whatsoever. Following the same line of logic (with ru A = 1) leads to an equivalent result for product A: the price is lower when the market size is known.
Proof of Proposition 6 (II): The profit to the individual products π∗ A and π∗ B are greater when demand is known than when it is random.
We establish this proposition for product A; the proof for B is similar. The profit to product A in the known demand case is greater than the profit to product A in the random demand case if and only if: M 1 −cA 2 2 > M 4 3 − √ 1 + 8cA 1 + √1 + 8cA −4cA 2 4 1 + √1 + 8cA .
Several tedious algebraic steps and some simple calculations show that for all costs such that 0 < cA < 1, this inequality is satisfied. Thus, the firm earns greater profits from product A when demand is known. A similar formulation shows the same is true for product B. That is, the firm earns greater (expected) profits for the unbundled goods when demand is known.
Proof or Proposition 7 (I): The price, p∗ AB, of the bundled good is lower with known market size.
Let x = [1 + (ru B + rl B)/2]. Follow the same path outlined following Proposition 6: the bundle price with known market size given in (10) is less than the bundle price with random market size given in (37) if and only if x + cAB 2 < 1 4 x + q x2 + 8xcAB ⇐ ⇒ x + 2cAB < q x2 + 8xcAB ⇐ ⇒ x2 + 4xcAB + 4c2 AB < x2 + 8xcAB ⇐ ⇒ cAB < x = [1 + (ru B + rl B)/2].
The last inequality holds by assumption.
24 Proof of Proposition 7 (II): The profit to the bundled good π∗ AB when demand is known is greater than when demand is random.
The bundle profit, known demand, given in (12) is greater than the bundle profit, random demand, given in (39) if and only if (x −cAB)2 > 3x − q x2 + 8xcAB x + √x2 + 8xcAB −4cAB 2 4 x + √x2 + 8xcAB , where x = [1+(ru B +rl B)/2]. As long as x > cAB, this inequality holds. This is the same inequality which establishes when the price of the bundled good in the known demand case is less than the bundle price in the random demand case.
25 References Adams, W.J., and Yellen, J., 1976. “Commodity Bundling and the Burden of Monopoly.” Quar-terly Journal Of Economics 40, 475-498.
Armstrong, M., 1996. ”Multiproduct Nonlinear Pricing.” Econometrica, 64 (1), 51-75.
Bakos, J.Y., and Brynjolfsson, E., 1999.
“Bundling Information Goods: Pricing, Profits and Efficiency.” Management Science 45 (12), 1613-1630.
Bakos, J.Y., and Brynjolfsson, E., 2000. “Bundling and Competition on the Internet.” Marketing Science 19 (1), 63-82.
Eppen, G.D., Hanson, W. A., and Martin R.K., 1991. ”Bundling - New Products, New Markets, Low Risk.” Sloan Management Review, 32 (4), 7-14.
Ernst, R. and Kouvelis, K., 1999. ”The Effects of Selling Packaged Goods on Inventory Decisions.” Management Science, 45 (8), 1142-1155.
Hanson, W.A., and Martin, R.K., 1990. “Optimal Bundle Pricing.” Management Science 36 (2), 155-74.
Jedidi, K., Jagpal, S., and Manchanda, P., 2003. ”Measuring Heterogeneous Reservation Prices for Product Bundles.” Marketing Science, 22 (1), 107-130.
McAfee, R.P., McMillan, J., and Whinston, M.D., 1989. “Multiproduct Monopoly, Commodity Bundling, and Correlation of Values.” Quarterly Journal of Economics 103, 371-383.
Petruzzi N.C. and Dada M., 1999. ”Pricing and the Newsvendor Problem: A Review with Exten-sions.” Operations Research, Volume: 47 (2), 183-194 Salinger, M.A., 1995. ”A Graphical Analysis of Bundling.” Journal of Business, 68 (1), 85-98.
Schmalensee, R., 1984. “Gaussian Demand and Commodity Bundling.” Journal Of Business 57, S211-231.
Stremersch, S., and Tellis, G.J., 2002.
“Strategic Bundling of Products and Prices: A New Synthesis for Marketing.” Journal Of Marketing 66, 55-72.
26 Stigler, G., 1963. “United States v. Loew’s, Inc: A Note on Block Booking.” Supreme Court Rev., 152.
27 Table 1: Results For Basic Products With Symmetric Demand No Bundling Product A Product B Bundling with Independent Demand Bundling with Perfectly Positively Correlated Demand Optimal Prices 0.625 0.625 1.05 1.25 Optimal Order Quantities 37.5 37.5 50 37.5 Optimal Profits 14.0625 14.0625 25 28.125 Table 2: Results For Basic Products With Asymmetric Demand No Bundling Product A Product B Bundling with Independent Demand Bundling with Perfectly Positively Correlated Demand Bundling with Perfectly Negatively Correlated Demand Optimal Prices 0.625 0.625 1.05 1.25 0.85 Optimal Order Quantities 37.5 46.875 55 41.66 175 Optimal Profits 14.06 17.58 30.25 31.25 61.25 Table 3: Comparing Basic And Fashion Products Basic: No Bundling Fashion: No Bundling Product A Product B Product A Product B Bundled Basic Products with Independent Demand Bundled Fashion Products with Independent Demand Optimal Prices 0.625 0.625 0.68 0.68 1 1.09 Optimal Order Quantities 37.5 37.5 20.1 20.1 50 22.08 Optimal Profits 14.0625 14.0625 8.7 8.7 25 13.09 Figure 1: Profits for Basic Products with Asymmetric Costs 24 25 26 27 28 29 30 0 1 2 3 4 Product Cost Ratio Profits No Bundling Independent Positive Figure 2: Prices Versus Costs For Basic Products 0 0.5 1 1.5 2 2.5 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 Costs Prices No Bundling Independent Positive Negative Figure 3: Profits Versus Costs For Basic Products 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 80.00 90.00 100.00 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 Costs Profits No Bundling Independent Positive Negative Figure 4: Profits for Fashion Products with Asymmetric Costs 0 5 10 15 20 25 0 1 2 3 4 Ratio Profit No Bundling Independent Figure 5: Price Vs. Reservation Price Variability for Fashion Products 0.7 0.8 0.9 1 1.1 1.2 1.3 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Reservation Price Varability Price Price A+B Price AB Figure 6: Expected Profit Vs. Reservation Price Variability for Fashion Products 5 10 15 20 25 30 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Reservation Price Varability Profit Profit A+B Profit AB |
6735 | https://personal.math.ubc.ca/~CLP/CLP2/clp_2_ic_problems.pdf | CLP–2 INTEGRAL CALCULUS EXERCISES Elyse YEAGER Joel FELDMAN Andrew RECHNITZER THIS DOCUMENT WAS TYPESET ON FRIDAY 16TH AUGUST, 2024.
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2 HOW TO USE THIS BOOK IJ Introduction First of all, welcome to Calculus!
This book is written as a companion to the CLP-2 Integral Calculus textbook.
§ § How to Work Questions This book is organized into four sections: Questions, Hints, Answers, and Solutions. As you are working problems, resist the temptation to prematurely peek at the back! It’s important to allow yourself to struggle for a time with the material. Even professional mathematicians don’t always know right away how to solve a problem. The art is in gathering your thoughts and figuring out a strategy to use what you know to find out what you don’t.
If you find yourself at a real impasse, go ahead and look for a hint in the Hints section.
Think about it for a while, and don’t be afraid to read back in the notes to look for a key idea that will help you proceed. If you still can’t solve the problem, well, we included the Solutions section for a reason! As you’re reading the solutions, try hard to understand why we took the steps we did, instead of memorizing step-by-step how to solve that one particular problem.
If you struggled with a question quite a lot, it’s probably a good idea to return to it in a few days. That might have been enough time for you to internalize the necessary ideas, and you might find it easily conquerable. Pat yourself on the back–sometimes math makes you feel good! If you’re still having troubles, read over the solution again, with an emphasis on understanding why each step makes sense.
One of the reasons so many students are required to study calculus is the hope that it will improve their problem-solving skills. In this class, you will learn lots of concepts, and be asked to apply them in a variety of situations. Often, this will involve answering one really big problem by breaking it up into manageable chunks, solving those chunks, then i HOW TO USE THIS BOOK putting the pieces back together. When you see a particularly long question, remain calm and look for a way to break it into pieces you can handle.
§ § Working with Friends Study buddies are fantastic! If you don’t already have friends in your class, you can ask your neighbours in lecture to form a group. Often, a question that you might bang your head against for an hour can be easily cleared up by a friend who sees what you’ve missed.
Regular study times make sure you don’t procrastinate too much, and friends help you maintain a positive attitude when you might otherwise succumb to frustration. Struggle in mathematics is desirable, but suffering is not.
When working in a group, make sure you try out problems on your own before coming together to discuss with others. Learning is a process, and getting answers to questions that you haven’t considered on your own can rob you of the practice you need to master skills and concepts, and the tenacity you need to develop to become a competent problem-solver.
§ § Types of Questions Q: Questions outlined in blue make up the representative question set. This set of questions is intended to cover the most essential ideas in each section. These questions are usually highly typical of what you’d see on an exam, although some of them are atypical but carry an important moral. If you find yourself unconfident with the idea behind one of these, it’s probably a good idea to practice similar questions.
This representative question set is our suggestion for a minimal selection of questions to work on. You are highly encouraged to work on more.
Q2: In addition to original problems, this book contains problems pulled from quizzes and exams given at UBC for Math 101 and 105 (second-semester calculus) and Math 121 (honours second-semester calculus). These problems are marked with a star. The authors would like to acknowledge the contributions of the many people who collaborated to produce these exams over the years.
Instructions and other comments that are attached to more than one question are written in this font. The questions are organized into Stage 1, Stage 2, and Stage 3.
§ § Stage 1 The first category is meant to test and improve your understanding of basic underlying concepts. These often do not involve much calculation. They range in difficulty from very basic reviews of definitions to questions that require you to be thoughtful about the concepts covered in the section.
ii HOW TO USE THIS BOOK § § Stage 2 Questions in this category are for practicing skills. It’s not enough to understand the philo-sophical grounding of an idea: you have to be able to apply it in appropriate situations.
This takes practice!
§ § Stage 3 The last questions in each section go a little farther than Stage 2. Often they will combine more than one idea, incorporate review material, or ask you to apply your understanding of a concept to a new situation.
In exams, as in life, you will encounter questions of varying difficulty. A good skill to practice is recognizing the level of difficulty a problem poses. Exams will have some easy questions, some standard questions, and some harder questions.
iii CONTENTS How to use this book i I The questions 1 1 Integration 2 1.1 Definition of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 Basic properties of the definite integral . . . . . . . . . . . . . . . . . . . . . .
10 1.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . .
13 1.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 1.5 Area between curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 1.6 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 1.7 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 1.8 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 1.9 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 1.10 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 1.11 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 1.12 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 1.13 More Integration Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 2 Applications of Integration 60 2.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60 2.2 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67 2.3 Centre of Mass and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72 2.4 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . .
81 3 Sequences and Series 91 3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 3.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97 3.3 Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 3.4 Absolute and Conditional Convergence . . . . . . . . . . . . . . . . . . . . . 112 iv CONTENTS CONTENTS 3.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 II Hints to problems 127 1.1 Definition of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 1.2 Basic properties of the definite integral . . . . . . . . . . . . . . . . . . . . . 131 1.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 132 1.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 1.5 Area between curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 1.6 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 1.7 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 1.8 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 1.9 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 1.10 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 1.11 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 1.12 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 1.13 More Integration Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 2.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 2.2 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 2.3 Centre of Mass and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 2.4 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 156 3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 3.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 3.3 Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 3.4 Absolute and Conditional Convergence . . . . . . . . . . . . . . . . . . . . . 165 3.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 3.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 III Answers to problems 170 1.1 Definition of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 1.2 Basic properties of the definite integral . . . . . . . . . . . . . . . . . . . . . 175 1.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 176 1.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 1.5 Area between curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 1.6 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 1.7 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 1.8 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 1.9 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 1.10 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 1.11 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 1.12 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 1.13 More Integration Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 2.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 2.2 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 2.3 Centre of Mass and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 v CONTENTS CONTENTS 2.4 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 209 3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 3.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 3.3 Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 3.4 Absolute and Conditional Convergence . . . . . . . . . . . . . . . . . . . . . 221 3.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 3.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 IV Solutions to problems 227 1.1 Definition of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 1.2 Basic properties of the definite integral . . . . . . . . . . . . . . . . . . . . . 263 1.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 273 1.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 1.5 Area between curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 1.6 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 1.7 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 1.8 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 1.9 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 1.10 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 1.11 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 1.12 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 1.13 More Integration Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 2.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 2.2 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 2.3 Centre of Mass and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527 2.4 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 562 3.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 3.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 3.3 Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637 3.4 Absolute and Conditional Convergence . . . . . . . . . . . . . . . . . . . . . 662 3.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 670 3.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689 vi THE QUESTIONS Part I 1 INTEGRATION Chapter 1 1.1IJ Definition of the Integral Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 For Questions 1 through 5, we want you to develop an understanding of the model we are using to define an integral: we approximate the area under a curve by bounding it between rectangles. Later, we will learn more sophisticated methods of integration, but they are all based on this simple concept.
Q: Give a range of possible values for the shaded area in the picture below.
x y 1 3 0.75 1.25 2 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q: Give a range of possible values for the shaded area in the picture below.
x y 1 2 3 4 0.75 1.25 0.25 2.25 1.75 Q: Using rectangles, find a lower and upper bound for ż 3 1 1 2x dx that differ by at most 0.2 square units.
x y y = 1 2x 1 3 Q: Let f (x) be a function that is decreasing from x = 0 to x = 5. Which Riemann sum approximation of ż 5 0 f (x)dx is the largest–left, right, or midpoint?
Q: Give an example of a function f (x), an interval [a, b], and a number n such that the midpoint Riemann sum of f (x) over [a, b] using n intervals is larger than both the left and right Riemann sums of f (x) over [a, b] using n intervals.
In Questions 6 through 10, we practice using sigma notation. There are many ways to write a given sum in sigma notation. You can practice finding several, and deciding which looks the clearest.
Q: Express the following sums in sigma notation: (a) 3 + 4 + 5 + 6 + 7 (b) 6 + 8 + 10 + 12 + 14 (c) 7 + 9 + 11 + 13 + 15 (d) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 3 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q: Express the following sums in sigma notation: (a) 1 3 + 1 9 + 1 27 + 1 81 (b) 2 3 + 2 9 + 2 27 + 2 81 (c) ´2 3 + 2 9 ´ 2 27 + 2 81 (d) 2 3 ´ 2 9 + 2 27 ´ 2 81 Q: Express the following sums in sigma notation: (a) 1 3 + 1 3 + 5 27 + 7 81 + 9 243 (b) 1 5 + 1 11 + 1 29 + 1 83 + 1 245 (c) 1000 + 200 + 30 + 4 + 1 2 + 3 50 + 7 1000 Q: Evaluate the following sums. You might want to use the formulas from Theorems 1.1.5 and 1.1.6 in the CLP-2 text.
(a) 100 ÿ i=0 3 5 i (b) 100 ÿ i=50 3 5 i (c) 10 ÿ i=1 i2 ´ 3i + 5 (d) b ÿ n=1 1 e n + en3 , where b is some integer greater than 1.
Q: Evaluate the following sums. You might want to use the formulas from Theorem 1.1.6 in the CLP-2 text.
(a) 100 ÿ i=50 (i ´ 50) + 50 ÿ i=0 i (b) 100 ÿ i=10 (i ´ 5)3 (c) 11 ÿ n=1 (´1)n (d) 11 ÿ n=2 (´1)2n+1 Questions 11 through 15 are meant to give you practice interpreting the formulas in Definition 1.1.11 of the CLP-2 text. The formulas might look complicated at first, but if you understand what each piece means, they are easy to learn.
4 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q: In the picture below, draw in the rectangles whose (signed) area is being computed by the midpoint Riemann sum 4 ÿ i=1 b ´ a 4 ¨ f a + i ´ 1 2 b ´ a 4 .
x y b a y = f (x) Q12: 4 ÿ k=1 f (1 + k) ¨ 1 is a left Riemann sum for a function f (x) on the interval [a, b] with n subintervals. Find the values of a, b and n.
Q: Draw a picture illustrating the area given by the following Riemann sum.
3 ÿ i=1 2 ¨ (5 + 2i)2 Q: Draw a picture illustrating the area given by the following Riemann sum.
5 ÿ i=1 π 20 ¨ tan π(i ´ 1) 20 Q15: Fill in the blanks with right, left, or midpoint; an interval; and a value of n.
3 ř k=0 f (1.5 + k) ¨ 1 is a Riemann sum for f on the interval [ , ] with n = .
Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż 5 0 x dx 5 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż 5 ´2 x dx § § Stage 2 Q18: Use sigma notation to write the midpoint Riemann sum for f (x) = x8 on [5, 15] with n = 50. Do not evaluate the Riemann sum.
Q19: Estimate ż 5 ´1 x3 dx using three approximating rectangles and left hand end points.
Q20: Let f be a function on the whole real line. Express ż 7 ´1 f (x) dx as a limit of Riemann sums, using the right endpoints.
Q21: The value of the following limit is equal to the area below a graph of y = f (x), integrated over the interval [0, b]: lim nÑ8 n ÿ i=1 4 n sin 2 + 4i n 2 Find f (x) and b.
Q22: For a certain function f (x), the following equation holds: lim nÑ8 n ÿ k=1 k n2 c 1 ´ k2 n2 = ż 1 0 f (x) dx Find f (x).
Q23: Express lim nÑ8 n ÿ i=1 3 ne´i/n cos 3i n as a definite integral.
Q24: Let Rn = n ÿ i=1 iei/n n2 . Express lim nÑ8 Rn as a definite integral. Do not evaluate this integral.
Q25: Express lim nÑ8 n ÿ i=1 e´1´2i/n ¨ 2 n as an integral in three different ways.
Questions 26 and 27 use the formula for a geometric sum, Equation 1.1.3 in the CLP-2 text.
6 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q: Evaluate the sum 1 + r3 + r6 + r9 + ¨ ¨ ¨ + r3n.
Q: Evaluate the sum r5 + r6 + r7 + ¨ ¨ ¨ + r100.
Remember that a definite integral is a signed area between a curve and the x-axis. We’ll spend a lot of time learning strategies for evaluating definite integrals, but we already know lots of ways to find area of geometric shapes. In Questions 28 through 33, use your knowledge of geometry to find the signed areas described by the integrals given.
Q28: Evaluate ż 2 ´1 |2x| dx.
Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż 5 ´3 |t ´ 1| dt Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż b a x dx where 0 ď a ď b.
Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż b a x dx where a ď b ď 0.
Q: Evaluate the following integral by interpreting it as a signed area, and using geometry: ż 4 0 a 16 ´ x2 dx Q33: Use elementary geometry to calculate ż 3 0 f (x) dx, where f (x) = # x, if x ď 1, 1, if x ą 1.
7 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q34: A car’s gas pedal is applied at t = 0 seconds and the car accelerates continuously until t = 2 seconds. The car’s speed at half-second intervals is given in the table below. Find the best possible upper estimate for the distance that the car traveled during these two seconds.
t (s) 0 0.5 1.0 1.5 2 v (m/s) 0 14 22 30 40 Q: True or false: the answer you gave for Question 34 is definitely greater than or equal to the distance the car travelled during the two seconds in question.
Q: An airplane’s speed at one-hour intervals is given in the table below. Approximate the distance travelled by the airplane from noon to 4pm using a midpoint Riemann sum.
time 12:00 pm 1:00 pm 2:00 pm 3:00 pm 4:00 pm speed (km/hr) 800 700 850 900 750 § § Stage 3 Q37: (a) Express lim nÑ8 n ÿ i=1 2 n d 4 ´ ´2 + 2i n 2 as a definite integal.
(b) Evaluate the integral of part (a).
Q38: Consider the integral: ż 3 0 (7 + x3) dx.
(˚) (a) Approximate this integral using the left Riemann sum with n = 3 intervals.
(b) Write down the expression for the right Riemann sum with n intervals and calculate the sum. Now take the limit n Ñ 8 in your expression for the Riemann sum, to evaluate the integral (˚) exactly.
You may use the identity n ÿ i=1 i3 = n4 + 2n3 + n2 4 Q39: Using a limit of right–endpoint Riemann sums, evaluate ż 4 2 x2 dx.
You may use the formulas n ř i=1 i = n(n+1) 2 and n ř i=1 i2 = n(n+1)(2n+1) 6 .
8 INTEGRATION 1.1 DEFINITION OF THE INTEGRAL Q40: Find ż 2 0 (x3 + x) dx using the definition of the definite integral. You may use the summation formulas n ř i=1 i3 = n4+2n3+n2 4 and n ř i=1 i = n2+n 2 .
Q41: Using a limit of right–endpoint Riemann sums, evaluate ż 4 1 (2x ´ 1) dx. Do not use anti-differentiation, except to check your answer. You may use the formula n ř i=1 i = n(n+1) 2 .
You’ll learn about this method starting in Section 1.3 of the CLP-2 text. You can also check this answer using geometry.
Q: Give a function f (x) that has the following expression as a right Riemann sum when n = 10, ∆(x) = 10 and a = ´5: 10 ÿ i=1 3(7 + 2i)2 sin(4i) .
Q: Using the method of Example 1.1.2 in the CLP-2 text, evaluate ż 1 0 2x dx Q: (a) Using the method of Example 1.1.2 in the CLP-2 text, evaluate ż b a 10x dx (b) Using your answer from above, make a guess for ż b a cx dx where c is a positive constant. Does this agree with Question 43?
Q: Evaluate ż a 0 a 1 ´ x2 dx using geometry, if 0 ď a ď 1.
Q: Suppose f (x) is a positive, decreasing function from x = a to x = b. You give an upper and lower bound on the area under the curve y = f (x) using n rectangles and a left and right Riemann sum, respectively, as in the picture below.
9 INTEGRATION 1.2 BASIC PROPERTIES OF THE DEFINITE INTEGRAL x y y = f (x) a b x y y = f (x) a b (a) What is the difference between the lower bound and the upper bound? (That is, if we subtract the smaller estimate from the larger estimate, what do we get?) Give your answer in terms of f, a, b, and n.
(b) If you want to approximate the area under the curve to within 0.01 square units using this method, how many rectangles should you use? That is, what should n be?
Q: Let f (x) be a linear function, let a ă b be integers, and let n be a whole number. True or false: if we average the left and right Riemann sums for ż b a f (x) dx using n rectangles, we get the same value as the midpoint Riemann sum using n rectangles.
1.2IJ Basic properties of the definite integral Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: For each of the following properties of definite integrals, draw a picture illustrating the concept, interpreting definite integrals as areas under a curve.
For simplicity, you may assume that a ď c ď b, and that f (x), g(x) give positive values.
(a) ż a a f (x) dx = 0 (Theorem 1.2.3.a in the CLP-2 text) (b) ż b a f (x) dx = ż c a f (x) dx + ż b c f (x)dx (Theorem 1.2.3.c in the CLP-2 text) (c) ż b a ( f (x) + g(x)) dx = ż b a f (x) dx + ż b a g(x) dx (Theorem 1.2.1.a in the CLP-2 text) Q: If ż b 0 cos x dx = sin b, then what is ż b a cos x dx?
10 INTEGRATION 1.2 BASIC PROPERTIES OF THE DEFINITE INTEGRAL Q3: Decide whether each of the following statements is true or false. If false, provide a counterexample. If true, provide a brief justification.
(a) ż ´2 ´3 f (x)dx = ´ ż 2 3 f (x)dx.
(b) If f (x) is an odd function, then ż ´2 ´3 f (x) dx = ż 3 2 f (x) dx.
(c) ż 1 0 f (x) ¨ g(x) dx = ż 1 0 f (x) dx ¨ ż 1 0 g(x) dx.
Q: Suppose we want to make a right Riemann sum with 100 intervals to approximate 0 ş 5 f (x) dx, where f (x) is a function that gives only positive values.
(a) What is ∆x?
(b) Are the heights of our rectangles positive or negative?
(c) Is our Riemann sum positive or negative?
(d) Is the signed area under the curve y = f (x) from x = 0 to x = 5 positive or negative?
§ § Stage 2 Q5: Suppose ż 3 2 f (x) dx = ´1 and ż 3 2 g(x) dx = 5. Evaluate ż 3 2 6f (x) ´ 3g(x) dx.
Q6: If ż 2 0 f (x) dx = 3 and ż 2 0 g(x) dx = ´4, calculate ż 2 0 2f (x) + 3g(x) dx.
Q7: The functions f (x) and g(x) obey ż ´1 0 f (x) dx = 1 ż 2 0 f (x) dx = 2 ż 0 ´1 g(x) dx = 3 ż 2 0 g(x) dx = 4 Find ş2 ´1 3g(x) ´ f (x) dx.
Q: In Question 45, Section 1.1, we found that ż a 0 a 1 ´ x2 dx = π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2 when 0 ď a ď 1.
Using this fact, evaluate the following: (a) ż 0 a a 1 ´ x2 dx, where ´1 ď a ď 0 (b) ż 1 a a 1 ´ x2 dx, where 0 ď a ď 1 11 INTEGRATION 1.2 BASIC PROPERTIES OF THE DEFINITE INTEGRAL Q9: Evaluate ż 2 ´1 |2x| dx.
You may use the result from Example 1.2.6 in the CLP-2 text that b ş a x dx = b2´a2 2 .
Q: Evaluate ż 5 ´5 x|x| dx .
Q: Suppose f (x) is an even function and ż 2 ´2 f (x)dx = 10. What is ż 0 ´2 f (x)dx?
§ § Stage 3 Q12: Evaluate ż 2 ´2 5 + a 4 ´ x2 dx.
Q13: Evaluate ż +2012 ´2012 sin x log(3 + x2)dx.
Q14: Evaluate ż +2012 ´2012 x1/3 cos x dx.
Q: Evaluate ż 6 0 (x ´ 3)3 dx .
Q: We want to compute the area of an ellipse, (ax)2 + (by)2 = 1 for some (let’s say positive) constants a and b.
(a) Solve the equation for the upper half of the ellipse. It should have the form “y = ¨ ¨ ¨ ” (b) Write an integral for the area of the upper half of the ellipse. Using properties of integrals, make the integrand look like the upper half of a circle.
(c) Using geometry and your answer to part (b), find the area of the ellipse.
Q: Fill in the following table: the product of an (even/odd) function with an (even/odd) function is an (even/odd) function. You may assume that both functions are defined for all real numbers.
ˆ even odd even odd Q: Suppose f (x) is an odd function and g(x) is an even function, both defined at x = 0. What are the possible values of f (0) and g(0)?
12 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS Q: Suppose f (x) is a function defined on all real numbers that is both even and odd.
What could f (x) be?
Q: Is the derivative of an even function even or odd? Is the derivative of an odd function even or odd?
1.3IJ The Fundamental Theorem of Calculus Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q1: Suppose that f (x) is a function and F(x) = e(x2´3) + 1 is an antiderivative of f (x).
Evaluate the definite integral ż ?
5 1 f (x) dx.
Q2: For the function f (x) = x3 ´ sin 2x, find its antiderivative F(x) that satisfies F(0) = 1.
Q3: Decide whether each of the following statements is true or false. Provide a brief justification.
(a) If f (x) is continuous on [1, π] and differentiable on (1, π), then ż π 1 f 1(x) dx = f (π) ´ f (1).
(b) ż 1 ´1 1 x2 dx = 0.
(c) If f is continuous on [a, b] then ż b a x f (x) dx = x ż b a f (x) dx.
Q: True or false: an antiderivative of 1 x2 is log(x2) (where by log x we mean logarithm base e).
Q: True or false: an antiderivative of cos(ex) is sin(ex) ex .
Q: Suppose F(x) = ż x 7 sin(t2) dt. What is the instantaneous rate of change of F(x) with respect to x?
13 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS Q: Suppose F(x) = ż x 2 e1/t dt. What is the slope of the tangent line to y = F(x) when x = 3?
Q: Suppose F1(x) = f (x). Give two different antiderivatives of f (x).
Q: In Question 45, Section 1.1, we found that ż a 0 a 1 ´ x2 dx = π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2.
(a) Verify that d da "π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2 = a 1 ´ a2.
(b) Find a function F(x) that satisfies F1(x) = ?
1 ´ x2 and F(0) = π.
Q: Evaluate the following integrals using the Fundamental Theorem of Calculus Part 2, or explain why it does not apply.
(a) ż π ´π cos x dx.
(b) ż π ´π sec2 x dx.
(c) ż 0 ´2 1 x + 1 dx.
Questions 11 through 14 are meant to help reinforce key ideas in the Fundamental Theorem of Calculus and its proof.
Q: As in the proof of the Fundamental Theorem of Calculus, let F(x) = şx a f (t) dt. In the diagram below, shade the area corresponding to F(x + h) ´ F(x).
t y a x x + h y = f (t) 14 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS Q: Let F(x) = ż x 0 f (t)dt, where f (t) is shown in the graph below, and 0 ď x ď 4.
(a) Is F(0) positive, negative, or zero?
(b) Where is F(x) increasing and where is it decreasing?
t y 1 2 3 4 y = f (t) Q: Let G(x) = ż 0 x f (t)dt, where f (t) is shown in the graph below, and 0 ď x ď 4.
(a) Is G(0) positive, negative, or zero?
(b) Where is G(x) increasing and where is it decreasing?
t y 1 2 3 4 y = f (t) Q: Let F(x) = ż x a t dt. Using the definition of the derivative, find F1(x).
Q: Give a continuous function f (x) so that F(x) = ż x 0 f (t)dt is a constant.
So far, we have been able to guess many antiderivatives. Often, however, antiderivatives are very difficult to guess. In Questions 16 through 19, we will find some antiderivatives that might appear in a table of integrals. Coming up with the antiderivative might be quite difficult (strategies to do just that will form a 15 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS large part of this semester), but verifying that your antiderivative is correct is as simple as differentiating.
Q: Evaluate and simplify d dxtx log(ax) ´ xu, where a is some constant and log(x) is the logarithm base e. What antiderivative does this tell you?
Q: Evaluate and simplify d dxtex x3 ´ 3x2 + 6x ´ 6 u. What antiderivative does this tell you?
Q: Evaluate and simplify d dx !
log ˇ ˇ ˇx + ?
x2 + a2 ˇ ˇ ˇ ) , where a is some constant. What antiderivative does this tell you?
Q: Evaluate and simplify d dx "b x(a + x) ´ a log ?x + ?
a + x , where a is some constant. What antiderivative does this tell you?
§ § Stage 2 Q20: Evaluate ż 2 0 x3 + sin x) dx.
Q21: Evaluate ż 2 1 x2 + 2 x2 dx.
Q: Evaluate ż 1 1 + 25x2dx.
Q: Evaluate ż 1 ?
2 ´ x2dx.
Q: Evaluate ż tan2 x dx.
Q: Evaluate ż 3 sin x cos x dx.
Q: Evaluate ż cos2 x dx.
Q27: If F(x) = ż x 0 log(2 + sin t) dt and G(y) = ż 0 y log(2 + sin t) dt find F1 π 2 and G1 π 2 .
Q28: Let f (x) = ż x 1 100(t2 ´ 3t + 2)e´t2 dt. Find the interval(s) on which f is increasing.
16 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS Q29: If F(x) = ż cos x 0 1 t3 + 6 dt, find F1(x).
Q30: Compute f 1(x) where f (x) = ż 1+x4 0 et2dt.
Q31: Evaluate d dx #ż sin x 0 (t6 + 8)dt + .
Q32: Let F(x) = ż x3 0 e´t sin πt 2 dt. Calculate F1(1).
Q33: Find d du #ż 0 cos u dt 1 + t3 + .
Q34: Find f (x) if x2 = 1 + ż x 1 f (t) dt.
Q35: If x sin(πx) = ż x 0 f (t) dt where f is a continuous function, find f (4).
Q36: Consider the function F(x) = ż x2 0 e´t dt + ż 0 ´x e´t2 dt.
(a) Find F1(x).
(b) Find the value of x for which F(x) takes its minimum value.
Q37: If F(x) is defined by F(x) = ż x x4´x3 esin t dt, find F1(x).
Q38: Evaluate d dx " ż ´x2 x5 cos et dt .
Q39: Differentiate ż ex x ?
sin t dt for 0 ă x ă log π.
Q40: Evaluate ż 5 1 f (x) dx, where f (x) = # 3 if x ď 3 x if x ě 3.
§ § Stage 3 Q41: If f 1(1) = 2 and f 1(2) = 3, find ż 2 1 f 1(x) f 2(x) dx.
17 INTEGRATION 1.3 THE FUNDAMENTAL THEOREM OF CALCULUS Q42: A car traveling at 30 m/s applies its brakes at time t = 0, its velocity (in m/s) decreasing according to the formula v(t) = 30 ´ 10t. How far does the car go before it stops?
Q43: Compute f 1(x) where f (x) = ż 2x´x2 0 log 1 + et dt. Does f (x) have an absolute maximum? Explain.
Q44: Find the minimum value of ż x2´2x 0 dt 1 + t4. Express your answer as an integral.
Q45: Define the function F(x) = ż x2 0 sin( ?
t) dt on the interval 0 ă x ă 4. On this interval, where does F(x) have a maximum?
Q46: Evaluate lim nÑ8 π n n ÿ j=1 sin jπ n by interpreting it as a limit of Riemann sums.
Q47: Use Riemann sums to evaluate the limit lim nÑ8 1 n n ÿ j=1 1 1 + j n .
Q: Below is the graph of y = f (t), ´5 ď t ď 5. Define F(x) = ż x 0 f (t) dt for any x in [´5, 5]. Sketch F(x).
x y y = f (x) ´5 ´3 ´1 1 3 5 Q49: Define f (x) = x3 ż x3+1 0 et3dt.
(a) Find a formula for the derivative f 1(x). (Your formula may include an integral sign.) (b) Find the equation of the tangent line to the graph of y = f (x) at x = ´1.
18 INTEGRATION 1.4 SUBSTITUTION Q: Two students calculate ş f (x) dx for some function f (x).
• Student A calculates ş f (x) dx = tan2 x + x + C • Student B calculates ş f (x) dx = sec2 x + x + C • It is a fact that d dxttan2 xu = f (x) ´ 1 Who ended up with the correct answer?
Q: Let F(x) = ż x 0 x3 sin(t) dt.
(a) Evaluate F(3).
(b) What is F1(x)?
Q: Let f (x) be an even function, defined everywhere, and let F(x) be an antiderivative of f (x). Is F(x) even, odd, or not necessarily either one? (You may use your answer from Section 1.2, Question 20. ) 1.4IJ Substitution Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: (a) True or False: ż sin(ex) ¨ ex dx = ż sin(u) du ˇ ˇ ˇ ˇ u=ex = ´ cos(ex) + C (b) True or False: ż 1 0 sin(ex) ¨ ex dx = ż 1 0 sin(u) du = 1 ´ cos(1) Q: Is the following reasoning sound? If not, fix it.
Problem: Evaluate ż (2x + 1)2dx.
Work: We use the substitution u = 2x + 1. Then: ż (2x + 1)2dx = ż u2 du = 1 3u3 + C = 1 3 (2x + 1)3 + C 19 INTEGRATION 1.4 SUBSTITUTION Q: Is the following reasoning sound? If not, fix it.
Problem: Evaluate ż π 1 cos(log t) t dt.
Work: We use the substitution u = log t, so du = 1 t dt. Then: ż π 1 cos(log t) t dt = ż π 1 cos(u)du = sin(π) ´ sin(1) = ´ sin(1) .
Q: Is the following reasoning sound? If not, fix it.
Problem: Evaluate ż π/4 0 x tan(x2) dx.
Work: We begin with the substitution u = x2, du = 2xdx: ż π/4 0 x tan(x2) dx = ż π/4 0 1 2 tan(x2) ¨ 2xdx = ż π2/16 0 1 2 tan u du = 1 2 ż π2/16 0 sin u cos udu Now we use the substitution v = cos u, dv = ´ sin u du: = 1 2 ż cos(π2/16) cos 0 ´1 vdv = ´1 2 ż cos(π2/16) 1 1 vdv = ´1 2 [log |v|]cos(π2/16) 1 = ´1 2 log cos(π2/16) ´ log(1) = ´1 2 log cos(π2/16) Q5: What is the integral that results when the substitution u = sin x is applied to the integral ż π/2 0 f (sin x) dx?
Q: Let f and g be functions that are continuous and differentiable everywhere.
Simplify ż f 1(g(x))g1(x) dx ´ f (g(x)).
20 INTEGRATION 1.4 SUBSTITUTION § § Stage 2 Q7: Use substitution to evaluate ż 1 0 xex2 cos(ex2) dx.
Q8: Let f (t) be any function for which ż 8 1 f (t) dt = 1. Calculate the integral ż 2 1 x2 f (x3) dx.
Q9: Evaluate ż x2 (x3 + 1)101dx.
Q10: Evaluate ż e4 e dx x log x.
Q11: Evaluate ż π/2 0 cos x 1 + sin x dx.
Q12: Evaluate ż π/2 0 cos x ¨ (1 + sin2 x) dx.
Q13: Evaluate ż 3 1 (2x ´ 1)ex2´x dx.
Q14: Evaluate ż (x2 ´ 4)x ?
4 ´ x2 dx.
Q: Evaluate ż e ?
log x 2x a log x dx .
§ § Stage 3 Q16: Calculate ż 2 ´2 xex2 dx.
Q17: Calculate lim nÑ8 n ÿ j=1 j n2 sin 1 + j2 n2 .
Questions 18 through 22 can be solved by substitution, but it may not be obvious which substitution will work. In general, when evaluating integrals, it is not always immediately clear which methods are appropri-ate. If this happens to you, don’t despair, and definitely don’t give up! Just guess a method and try it. Even if it fails, you’ll probably learn something that you can use to make a better guess.1 1 This is also pretty decent life advice.
21 INTEGRATION 1.5 AREA BETWEEN CURVES Q: Evaluate ż 1 0 u3 u2 + 1 du.
Q: Evaluate ż tan3 θ dθ .
Q: Evaluate ż 1 ex + e´x dx Q: Evaluate ż 1 0 (1 ´ 2x) a 1 ´ x2 dx Q: Evaluate ż tan x ¨ log (cos x) dx Q23: Evaluate lim nÑ8 n ÿ j=1 j n2 cos j2 n2 .
Q24: Calculate lim nÑ8 n ÿ j=1 j n2 c 1 + j2 n2.
Q: Using Riemann sums, prove that ż b a 2f (2x)dx = ż 2b 2a f (x)dx 1.5IJ Area between curves Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: We want to approximate the area between the graphs of y = cos x and y = sin x from x = 0 to x = π using a left Riemann sum with n = 4 rectangles.
(a) On the graph below, sketch the four rectangles.
(b) Calculate the Riemann approximation.
22 INTEGRATION 1.5 AREA BETWEEN CURVES x y π π 2 π 4 3π 4 y = cos x y = sin x Q: We want to approximate the bounded area between the curves y = arcsin 2x π and y = cπx 2 using n = 5 rectangles.
(a) Draw the five (vertical) rectangles on the picture below corresponding to a right Riemann sum.
(b) Draw five rectangles on the picture below we might use if we were using horizontal rectangles.
x y π 2 y = arcsin 2x π y = b xπ 2 Q3: Write down a definite integral that represents the finite area bounded by the curves y = x3 ´ x and y = x for x ě 0. Do not evaluate the integral explicitly.
Q4: Write down a definite integral that represents the area of the finite region bounded by the line y = ´x 2 and the parabola y2 = 6 ´ 5x 4 . Do not evaluate the integral explicitly.
Q5: Write down a definite integral that represents the area of the finite plane region bounded by y2 = 4ax and x2 = 4ay, where a ą 0 is a constant. Do not evaluate the integral explicitly.
23 INTEGRATION 1.5 AREA BETWEEN CURVES Q6: Write down a definite integral that represents the area of the finite region bounded between the line x + 12y + 5 = 0 and the curve x = 4y2. Do not evaluate the integral explicitly.
§ § Stage 2 Q7: Find the area of the region bounded by the graph of f (x) = 1 (2x ´ 4)2 and the x–axis between x = 0 and x = 1.
Q8: Find the area of the finite region between the curves y = x and y = 3x ´ x2, by first identifying the points of intersection and then integrating.
Q9: Calculate the area of the finite region enclosed by y = 2x and y = ?x + 1.
Q10: Find the area of the finite region bounded between the two curves y = ?
2 cos(πx/4) and y = |x|.
Q11: Find the area of the finite region that is bounded by the graphs of f (x) = x2?
x3 + 1 and g(x) = 3x2.
Q12: Find the area to the left of the y–axis and to the right of the curve x = y2 + y.
Q: Find the area of the finite region below y = ?
9 ´ x2 and above both y = |x| and y = ?
1 ´ x2.
§ § Stage 3 Q14: The graph below shows the finite region between y = 4 + π sin x and y = 4 + 2π ´ 2x.
y x π 2 π 3π 2 2π 2 4 6 8 y = 4 + π sin(x) y = 4 + 2π −2x Find the area of this region.
24 INTEGRATION 1.6 VOLUMES Q15: Compute the area of the finite region bounded by the curves x = 0, x = 3, y = x + 2 and y = x2.
Q16: Find the total area between the curves y = x ?
25 ´ x2 and y = 3x, on the interval 0 ď x ď 4.
Q: Find the area of the finite region below y = ?
9 ´ x2 and y = x, and above y = a 1 ´ (x ´ 1)2.
Q: Find the area of the finite region bounded by the curve y = x(x2 ´ 4) and the line y = x ´ 2.
1.6IJ Volumes Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Consider a right circular cone.
What shape are horizontal cross-sections? Are the vertical cross-sections the same?
25 INTEGRATION 1.6 VOLUMES Q: Two potters start with a block of clay h units tall, and identical square cookie cutters. They form columns by pushing the square cookie cutter straight down over the clay, so that its cross-section is the same square as the cookie cutter. Potter A pushes their cookie cutter down while their clay block is sitting motionless on a table; Potter B pushes their cookie cutter down while their clay block is rotating on a potter’s wheel, so their column looks twisted. Which column has greater volume?
Column A Column B Q: Let R be the region bounded above by the graph of y = f (x) shown below and bounded below by the x-axis, from x = 0 to x = 6. Sketch the washers that are formed by rotating R about the y-axis. In your sketch, label all the radii in terms of y, and label the thickness.
x y y = f (x) 2 1 4 6 1 3 Q4: Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.
(a) The volume of the solid obtained by rotating around the x–axis the region between the x–axis and y = ?x ex2 for 0 ď x ď 3.
(b) The volume of the solid obtained by revolving the region bounded by the curves 26 INTEGRATION 1.6 VOLUMES y = x2 and y = x + 2 about the line x = 3.
Q5: Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.
(a) The volume of the solid obtained by rotating the finite plane region bounded by the curves y = 1 ´ x2 and y = 4 ´ 4x2 about the line y = ´1.
(b) The volume of the solid obtained by rotating the finite plane region bounded by the curve y = x2 ´ 1 and the line y = 0 about the line x = 5.
Q6: Write down a definite integral that represents the volume of the solid obtained by rotating around the line y = ´1 the region between the curves y = x2 and y = 8 ´ x2. Do not evaluate the integrals explicitly.
Q: A tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. (You might have seen this shape as a 4-sided die; think of a pyramid with a triangular base.) Using the methods from this section, calculate the volume of a tetrahedron with side-length ℓ. You may assume without proof that the height of a tetrahedron with side-length ℓis b 2 3ℓ.
ℓ b 2 3ℓ § § Stage 2 Q8: Let a ą 0 be a constant. Let R be the finite region bounded by the graph of y = 1 + ?xex2, the line y = 1, and the line x = a. Using vertical slices, find the volume generated when R is rotated about the line y = 1.
Q9: Find the volume of the solid generated by rotating the finite region bounded by y = 1/x and 3x + 3y = 10 about the x–axis.
Q10: Let R be the region inside the circle x2 + (y ´ 2)2 = 1. Let S be the solid obtained by rotating R about the x-axis.
(a) Write down an integral representing the volume of S.
(b) Evaluate the integral you wrote down in part (a).
Q11: The region R is the portion of the first quadrant which is below the parabola y2 = 8x and above the hyperbola y2 ´ x2 = 15.
(a) Sketch the region R.
(b) Find the volume of the solid obtained by revolving R about the x axis.
27 INTEGRATION 1.6 VOLUMES Q12: The region R is bounded by y = log x, y = 0, x = 1 and x = 2. (Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.) (a) Sketch the region R.
(b) Find the volume of the solid obtained by revolving this region about the y axis.
Q13: The finite region between the curves y = cos( x 2) and y = x2 ´ π2 is rotated about the line y = ´π2. Using vertical slices (disks and/or washers), find the volume of the resulting solid.
Q14: The solid V is 2 meters high and has square horizontal cross sections. The length of the side of the square cross section at height x meters above the base is 2 1+x m.
Find the volume of this solid.
Q15: Consider a solid whose base is the finite portion of the xy–plane bounded by the curves y = x2 and y = 8 ´ x2. The cross–sections perpendicular to the x–axis are squares with one side in the xy–plane. Compute the volume of this solid.
Q16: A frustum of a right circular cone (as shown below) has height h. Its base is a circular disc with radius 4 and its top is a circular disc with radius 2. Calculate the volume of the frustum.
h 2 4 § § Stage 3 Q: The shape of the earth is often approximated by an oblate spheroid, rather than a sphere. An oblate spheroid is formed by rotating an ellipse about its minor axis (its shortest diameter).
(a) Find the volume of the oblate spheroid obtained by rotating the upper (positive) half of the ellipse (ax)2 + (by)2 = 1 about the x-axis, where a and b are positive constants with a ě b.
(b) Suppose2 the earth has radius at the equator of 6378.137 km, and radius at the poles of 6356.752 km. If we model the earth as an oblate spheroid formed by rotating the upper half of the ellipse (ax)2 + (by)2 = 1 about the x-axis, what are a and b?
(c) What is the volume of this model of the earth? (Use a calculator.) 2 Earth Fact Sheet, NASA, earthfact.html, accessed 2 July 2017 28 INTEGRATION 1.6 VOLUMES (d) Suppose we had calculated the volume of the earth by modelling it as a sphere with radius 6378.137 km. What would our absolute and relative errors be, compared to our oblate spheroid calculation?
Q18: Let R be the bounded region that lies between the curve y = 4 ´ (x ´ 1)2 and the line y = x + 1.
(a) Sketch R and find its area.
(b) Write down a definite integral giving the volume of the region obtained by rotating R about the line y = 5. Do not evaluate this integral.
Q19: Let R = ␣(x, y) : (x ´ 1)2 + y2 ď 1 and x2 + (y ´ 1)2 ď 1 ( .
(a) Sketch R and find its area.
(b) If R rotates around the y–axis, what volume is generated?
Q20: Let R be the plane region bounded by x = 0, x = 1, y = 0 and y = c ?
1 + x2, where c ě 0 is a constant.
(a) Find the volume V1 of the solid obtained by revolving R about the x–axis.
(b) Find the volume V2 of the solid obtained by revolving R about the y–axis.
(c) If V1 = V2, what is the value of c?
Q21: The graph below shows the region between y = 4 + π sin x and y = 4 + 2π ´ 2x.
y x π 2 π 3π 2 2π 2 4 6 8 y = 4 + π sin(x) y = 4 + 2π −2x The region is rotated about the line y = ´1. Express in terms of definite integrals the volume of the resulting solid. Do not evaluate the integrals.
29 INTEGRATION 1.7 INTEGRATION BY PARTS Q: On a particular, highly homogeneous planet, we observe that the density of the atmosphere h kilometres above the surface is given by the equation ρ(h) = c2´h/6 kg m3, where c is the density on the planet’s surface.
(a) What is the mass of the atmosphere contained in a vertical column with radius one metre, sixty kilometres high?
(b) What height should a column be to contain 3000cπ log 2 kilograms of air?
This is clearly a simplified model: air density changes all the time, and depends on lots of complicated factors aside from altitude.
However, the equation we’re using is not so far off from an idealized model of the earth’s atmosphere, taken from Pressure and the Gas Laws by H.P. Schmid, GasPressWeb/PressGasLaws.html, accessed 3 July 2017.
1.7IJ Integration by parts Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: The method of integration by substitution comes from the rule for differentiation.
The method of integration by parts comes from the rule for differentiation.
Q: Suppose you want to evaluate an integral using integration by parts. You choose part of your integrand to be u, and part to be dv. The part chosen as u will be: (differenti-ated, antidifferentiated). The part chosen as dv will be: (differentiated, antidifferentiated).
Q: Let f (x) and g(x) be differentiable functions. Using the quotient rule for differenti-ation, give an equivalent expression to ż f 1(x) g(x) dx.
Q: Suppose we want to use integration by parts to evaluate ż u(x) ¨ v1(x)dx for some differentiable functions u and v. We need to find an antiderivative of v1(x), but there are infinitely many choices. Show that every antiderivative of v1(x) gives an equivalent final answer.
Q: Suppose you want to evaluate ż f (x)dx using integration by parts. Explain why dv = f (x)dx, u = 1 is generally a bad choice.
30 INTEGRATION 1.7 INTEGRATION BY PARTS Note: compare this to Example 1.7.8 of the CLP-2 text, where we chose u = f (x), dv = 1dx.
§ § Stage 2 Q6: Evaluate ż x log x dx.
Q7: Evaluate ż log x x7 dx.
Q8: Evaluate ż π 0 x sin x dx.
Q9: Evaluate ż π 2 0 x cos x dx.
Q: Evaluate ż x3exdx.
Q: Evaluate ż x log3 x dx.
Q: Evaluate ż x2 sin x dx.
Q: Evaluate ż (3t2 ´ 5t + 6) log t dt.
Q: Evaluate ż ?se ?sds.
Q: Evaluate ż log2 xdx.
Q: Evaluate ż 2xex2+1dx.
Q17: Evaluate ż arccos y dy.
§ § Stage 3 Q18: Evaluate ż 4y arctan(2y) dy.
31 INTEGRATION 1.8 TRIGONOMETRIC INTEGRALS Q: Evaluate ż x2 arctan x dx.
Q: Evaluate ż ex/2 cos(2x)dx.
Q: Evaluate ż sin(log x)dx.
Q: Evaluate ż 2x+log2 xdx.
Q: Evaluate ż ecos x sin(2x)dx.
Q: Evaluate ż xe´x (1 ´ x)2dx.
Q25: A reduction formula.
(a) Derive the reduction formula ż sinn(x) dx = ´sinn´1(x) cos(x) n + n ´ 1 n ż sinn´2(x) dx.
(b) Calculate ż π/2 0 sin8(x) dx.
Q26: Let R be the part of the first quadrant that lies below the curve y = arctan x and between the lines x = 0 and x = 1.
(a) Sketch the region R and determine its area.
(b) Find the volume of the solid obtained by rotating R about the y–axis.
Q27: Let R be the region between the curves T(x) = ?xe3x and B(x) = ?x(1 + 2x) on the interval 0 ď x ď 3. (It is true that T(x) ě B(x) for all 0 ď x ď 3.) Compute the volume of the solid formed by rotating R about the x-axis.
Q28: Let f (0) = 1, f (2) = 3 and f 1(2) = 4. Calculate ż 4 0 f 2 ?x dx.
Q: Evaluate lim nÑ8 n ÿ i=1 2 n 2 ni ´ 1 e 2 ni´1 .
1.8IJ Trigonometric Integrals Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
32 INTEGRATION 1.8 TRIGONOMETRIC INTEGRALS Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: Suppose you want to evaluate ż π/4 0 sin x cosn x dx using the substitution u = cos x.
Which of the following need to be true for your substitution to work?
(a) n must be even (b) n must be odd (c) n must be an integer (d) n must be positive (e) n can be any real number Q: Evaluate ż secn x tan xdx, where n is a strictly positive integer.
Q: Derive the identity tan2 x + 1 = sec2 x from the easier-to-remember identity sin2 x + cos2 x = 1.
§ § Stage 2 Questions 4 through 10 deal with powers of sines and cosines. Review Section 1.8.1 in the CLP-2 text for integration strategies.
Q4: Evaluate ż cos3 x dx.
Q5: Evaluate ż π 0 cos2 x dx.
Q6: Evaluate ż sin36 t cos3 t dt.
Q: Evaluate ż sin3 x cos4 x dx.
Q: Evaluate ż π/3 0 sin4 x dx.
Q: Evaluate ż sin5 x dx.
Q: Evaluate ż sin1.2 x cos x dx.
33 INTEGRATION 1.8 TRIGONOMETRIC INTEGRALS Questions 12 through 21 deal with powers of tangents and secants. Review Section 1.8.2 in the CLP-2 text for strategies.
Q: Evaluate ż tan x sec2 xdx.
Q12: Evaluate ż tan3 x sec5 x dx.
Q13: Evaluate ż sec4 x tan46 x dx.
Q: Evaluate ż tan3 x sec1.5 x dx.
Q: Evaluate ż tan3 x sec2 x dx.
Q: Evaluate ż tan4 x sec2 x dx.
Q: Evaluate ż tan3 x sec´0.7 x dx.
Q: Evaluate ż tan5 x dx.
Q: Evaluate ż π/6 0 tan6 x dx.
Q: Evaluate ż π/4 0 tan8 x sec4 x dx.
Q: Evaluate ż tan x?sec x dx.
Q: Evaluate ż sec8 θ tane θ dθ.
34 INTEGRATION 1.9 TRIGONOMETRIC SUBSTITUTION § § Stage 3 Q23: A reduction formula.
(a) Let n be a positive integer with n ě 2. Derive the reduction formula ż tann(x) dx = tann´1(x) n ´ 1 ´ ż tann´2(x) dx.
(b) Calculate ż π/4 0 tan6(x) dx.
Q: Evaluate ż tan5 x cos2 x dx.
Q: Evaluate ż 1 cos2 θdθ.
Q: Evaluate ż cot x dx.
Q: Evaluate ż ex sin(ex) cos(ex) dx.
Q: Evaluate ż sin(cos x) sin3 x dx.
Q: Evaluate ż x sin x cos x dx.
1.9IJ Trigonometric Substitution Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
35 INTEGRATION 1.9 TRIGONOMETRIC SUBSTITUTION § § Stage 1 Q1: For each of the following integrals, choose the substitution that is most beneficial for evaluating the integral.
(a) ż 2x2 ?
9x2 ´ 16 dx (b) ż x4 ´ 3 ?
1 ´ 4x2 dx (c) ż (25 + x2) ´5/2 dx Q: For each of the following integrals, choose a trigonometric substitution that will eliminate the roots.
(a) ż 1 ?
x2 ´ 4x + 1 dx (b) ż (x ´ 1)6 (´x2 + 2x + 4)3/2 dx (c) ż 1 ?
4x2 + 6x + 10 dx (d) ż a x2 ´ x dx Q: In each part of this question, assume θ is an angle in the interval [0, π/2].
(a) If sin θ = 1 20, what is cos θ ?
(b) If tan θ = 7, what is csc θ ?
(c) If sec θ = ?
x ´ 1 2 , what is tan θ ?
Q: Simplify the following expressions.
(a) sin arccos x 2 (b) sin arctan 1 ?
3 (c) sec (arcsin (?x)) § § Stage 2 Q5: Evaluate ż 1 (x2 + 4)3/2 dx.
Q6: Evaluate ż 4 0 1 (4 + x2)3/2 dx. Your answer may not contain inverse trigonometric functions.
36 INTEGRATION 1.9 TRIGONOMETRIC SUBSTITUTION Q7: Evaluate ż 5/2 0 dx ?
25 ´ x2.
Q8: Evaluate ż dx ?
x2 + 25 . You may use that ż sec x dx = log ˇ ˇ sec x + tan x ˇ ˇ + C.
Q: Evaluate ż x + 1 ?
2x2 + 4x dx.
Q10: Evaluate ż dx x2?
x2 + 16 .
Q11: Evaluate ż dx x2?
x2 ´ 9 for x ě 3. Do not include any inverse trigonometric func-tions in your answer.
Q12: (a) Show that ż π/4 0 cos4 θ dθ = (8 + 3π)/32.
(b) Evaluate ż 1 ´1 dx (x2 + 1)3.
Q: Evaluate ż π/12 ´π/12 15x3 (x2 + 1)(9 ´ x2)5/2 dx.
Q14: Evaluate ż ?
4 ´ x2 dx.
Q15: Evaluate ż ?
25x2 ´ 4 x dx for x ą 2 5.
Q: Evaluate ż ?
17 ?
10 x3 ?
x2 ´ 1 dx.
Q17: Evaluate ż dx ?
3 ´ 2x ´ x2.
Q: Evaluate ż 1 (2x ´ 3)3?
4x2 ´ 12x + 8 dx for x ą 2.
Q: Evaluate ż 1 0 x2 (x2 + 1)3/2dx.
You may use that ş sec xdx = log | sec x + tan x| + C.
37 INTEGRATION 1.9 TRIGONOMETRIC SUBSTITUTION Q: Evaluate ż 1 (x2 + 1)2 dx.
§ § Stage 3 Q: Evaluate ż x2 ?
x2 ´ 2x + 2 dx.
You may assume without proof that ż sec3 θ dθ = 1 2 sec θ tan θ + 1 2 log | sec θ + tan θ| + C.
Q: Evaluate ż 1 ?
3x2 + 5x dx.
You may use that ş sec xdx = log | sec x + tan x| + C.
Q: Evaluate ż (1 + x2)3/2 x dx. You may use the fact that ż csc θ dθ = log | cot θ ´ csc θ| + C.
Q: Below is the graph of the ellipse x 4 2 + y 2 2 = 1. Find the area of the shaded region using the ideas from this section.
x y ´1 1 Q: Let f (x) = |x| 4 ?
1 ´ x2, and let R be the region between f (x) and the x-axis over the interval [´1 2, 1 2].
(a) Find the area of R.
(b) Find the volume of the solid formed by rotating R about the x-axis.
Q: Evaluate ż ?
1 + ex dx. You may use the antiderivative ż csc θdθ = log | cot θ ´ csc θ| + C.
38 INTEGRATION 1.9 TRIGONOMETRIC SUBSTITUTION Q: Consider the following work.
ż 1 1 ´ x2 dx = ż 1 1 ´ sin2 θ cos θ dθ using x = sin θ, dx = cos θ dθ = ż cos θ cos2 θ dθ = ż sec θ dθ = log | sec θ + tan θ| + C Example 1.8.19 in the CLP-2 text = log ˇ ˇ ˇ ˇ 1 ?
1 ´ x2 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ + C θ ?
1 ´ x2 x 1 = log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ + C (a) Differentiate log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ.
(b) True or false: ż 3 2 1 1 ´ x2 dx = log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ x=3 x=2 (c) Was the work in the question correct? Explain.
Q: (a) Suppose we are evaluating an integral that contains the term ?
a2 ´ x2, where a is a positive constant, and we use the substitution x = a sin u (with inverse u = arcsin(x/a)), so that a a2 ´ x2 = a a2 cos2 u = |a cos u| Under what circumstances is |a cos u| ‰ a cos u?
(b) Suppose we are evaluating an integral that contains the term ?
a2 + x2, where a is a positive constant, and we use the substitution x = a tan u (with inverse u = arctan(x/a)), so that a a2 + x2 = a a2 sec2 u = |a sec u| Under what circumstances is |a sec u| ‰ a sec u?
(c) Suppose we are evaluating an integral that contains the term ?
x2 ´ a2, where a is a positive constant, and we use the substitution x = a sec u (with inverse u = arcsec(x/a) = arccos(a/x)), so that a x2 ´ a2 = a a2 tan2 u = |a tan u| Under what circumstances is |a tan u| ‰ a tan u?
39 INTEGRATION 1.10 PARTIAL FRACTIONS 1.10IJ Partial Fractions Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: Below are the graphs of four different quadratic functions. For each quadratic function, decide whether it is: (i) irreducible, (ii) the product of two distinct linear factors, or (iii) the product of a repeated linear factor (and possibly a constant).
x y (a) x y (b) x y (c) x y (d) Q2: Write out the general form of the partial-fractions decomposition of x3 + 3 (x2 ´ 1)2(x2 + 1). You need not determine the values of any of the coefficients.
Q3: Find the coefficient of 1 x ´ 1 in the partial fraction decomposition of 3x3 ´ 2x2 + 11 x2(x ´ 1)(x2 + 3).
Q: Re-write the following rational functions as the sum of a polynomial and a rational function whose numerator has a strictly smaller degree than its denominator. (Remember our method of partial fraction decomposition of a rational function only works when the degree of the numerator is strictly smaller than the degree of the denominator.) (a) x3 + 2x + 2 x2 + 1 (b) 15x4 + 6x3 + 34x2 + 4x + 20 5x2 + 2x + 8 (c) 2x5 + 9x3 + 12x2 + 10x + 30 2x2 + 5 Q: Factor the following polynomials into linear and irreducible factors.
(a) 5x3 ´ 3x2 ´ 10x + 6 40 INTEGRATION 1.10 PARTIAL FRACTIONS (b) x4 ´ 3x2 ´ 5 (c) x4 ´ 4x3 ´ 10x2 ´ 11x ´ 6 (d) 2x4 + 12x3 ´ x2 ´ 52x + 15 Q: Here is a fact: Suppose we have a rational function with a repeated linear factor (ax + b)n in the denominator, and the degree of the numerator is strictly less than the degree of the denominator. In the partial fraction decomposition, we can replace the terms A1 ax + b + A2 (ax + b)2 + A3 (ax + b)3 + ¨ ¨ ¨ + An (ax + b)n (1) with the single term B1 + B2x + B3x2 + ¨ ¨ ¨ + Bnxn´1 (ax + b)n (2) and still be guaranteed to find a solution.
Why do we use the sum in (1), rather than the single term in (2), in partial fraction decom-position?
§ § Stage 2 Q7: Evaluate ż 2 1 dx x + x2.
Q8: Calculate ż 1 x4 + x2 dx.
Q9: Calculate ż 12x + 4 (x ´ 3)(x2 + 1) dx.
Q10: Evaluate the following indefinite integral using partial fraction: F(x) = ż 3x2 ´ 4 (x ´ 2)(x2 + 4) dx.
Q11: Evaluate ż x ´ 13 x2 ´ x ´ 6dx.
Q12: Evaluate ż 5x + 1 x2 + 5x + 6dx.
Q: Evaluate ż 5x2 ´ 3x ´ 1 x2 ´ 1 dx.
41 INTEGRATION 1.10 PARTIAL FRACTIONS Q: Evaluate ż 4x4 + 14x2 + 2 4x4 + x2 dx.
Q: Evaluate ż x2 + 2x ´ 1 x4 ´ 2x3 + x2 dx.
Q: Evaluate ż 3x2 ´ 4x ´ 10 2x3 ´ x2 ´ 8x + 4 dx.
Q: Evaluate ż 1 0 10x2 + 24x + 8 2x3 + 11x2 + 6x + 5 dx.
§ § Stage 3 In Questions 18 and 19, we use partial fraction to find the antiderivatives of two important functions: cosecant, and cosecant cubed.
Q: Using the method of Example 1.10.5 in the CLP-2 text, integrate ż csc x dx.
Q: Using the method of Example 1.10.6 in the CLP-2 text, integrate ż csc3 x dx.
The purpose of performing a partial fraction decomposition is to manipulate an integrand into a form that is easily integrable. These “easily integrable” forms are rational functions whose denominator is a power of a linear function, or of an irreducible quadratic function. In Questions 20 through 23, we explore the integration of rational functions whose denominators involve irreducible quadratics.
Q: Evaluate ż 2 1 3x3 + 15x2 + 35x + 10 x4 + 5x3 + 10x2 dx.
Q: Evaluate ż 3 x2 + 2 + x ´ 3 (x2 + 2)2 dx.
Q: Evaluate ż 1 (1 + x2)3 dx.
Q: Evaluate ż 3x + 3x + 1 x2 + 5 + 3x (x2 + 5)2 dx.
In Questions 24 through 26, we use substitution to turn a non-rational integrand into a rational integrand, then evaluate the resulting integral using partial fraction. Till now, the partial fraction problems you’ve seen have all looked largely the same, but keep in mind that a partial fraction decomposition can be a small step in a larger problem.
Q: Evaluate ż cos θ 3 sin θ + cos2 θ ´ 3 dθ.
42 INTEGRATION 1.11 NUMERICAL INTEGRATION Q: Evaluate ż 1 e2t + et + 1 dt.
Q: Evaluate ż ?
1 + ex dx using partial fraction.
Q27: The region R is the portion of the first quadrant where 3 ď x ď 4 and 0 ď y ď 10 ?
25 ´ x2.
(a) Sketch the region R.
(b) Determine the volume of the solid obtained by revolving R around the x–axis.
(c) Determine the volume of the solid obtained by revolving R around the y–axis.
Q: Find the area of the finite region bounded by the curves y = 4 3 + x2, y = 2 x(x + 1), x = 1 4, and x = 3.
Q: Let F(x) = ż x 1 1 t2 ´ 9dt.
(a) Give a formula for F(x) that does not involve an integral.
(b) Find F1(x).
1.11IJ Numerical Integration Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: Suppose we approximate an object to have volume 1.5m3, when its exact volume is 1.387m3. Give the relative error, absolute error, and percent error of our approximation.
Q: Consider approximating ż 10 2 f (x) dx, where f (x) is the function in the graph below.
43 INTEGRATION 1.11 NUMERICAL INTEGRATION x y 2 10 (a) Draw the rectangles associated with the midpoint rule approximation and n = 4.
(b) Draw the trapezoids associated with the trapezoidal rule approximation and n = 4.
You don’t have to give an approximation.
Q: Let f (x) = ´ 1 12x4 + 7 6x3 ´ 3x2.
(a) Find a reasonable value M such that |f 2(x)| ď M for all 1 ď x ď 6.
(b) Find a reasonable value L such that |f (4)(x)| ď L for all 1 ď x ď 6.
Q: Let f (x) = x sin x + 2 cos x. Find a reasonable value M such that |f 2(x)| ď M for all ´3 ď x ď 2.
Q: Consider the quantity A = ż π ´π cos x dx.
(a) Find the upper bound on the error using Simpson’s rule with n = 4 to approximate A using Theorem 1.11.12 in the CLP-2 text.
(b) Find the Simpson’s rule approximation of A using n = 4.
(c) What is the (actual) absolute error in the Simpson’s rule approximation of A with n = 4?
Q: Give a function f (x) such that: • f 2(x) ď 3 for every x in [0, 1], and • the error using the trapezoidal rule approximating ż 1 0 f (x) dx with n = 2 intervals is exactly 1 16.
Q: Suppose my mother is under 100 years old, and I am under 200 years old.3 Who is older?
Q: 3 We’re going somewhere with this.
44 INTEGRATION 1.11 NUMERICAL INTEGRATION (a) True or False: for fixed positive constants M, n, a, and b, with b ą a, M 24 (b ´ a)3 n2 ď M 12 (b ´ a)3 n2 (b) True or False: for a function f (x) and fixed constants n, a, and b, with b ą a, the n-interval midpoint approximation of ż b a f (x) dx is more accurate than the n-interval trapezoidal approximation.
Q9: Decide whether the following statement is true or false. If false, provide a counterexample. If true, provide a brief justification.
When f (x) is positive and concave up, any trapezoidal rule approximation for ż b a f (x) dx will be an upper estimate for ż b a f (x) dx.
Q: Give a polynomial f (x) with the property that the Simpson’s rule approximation of ż b a f (x) dx is exact for all a, b, and n.
§ § Stage 2 Questions 11 and 12 ask you to approximate a given integral using the formulas in Equations 1.11.2, 1.11.6, and 1.11.9 in the CLP-2 text.
Q: Write out all three approximations of ż 30 0 1 x3 + 1 dx with n = 6. (That is: midpoint, trapezoidal, and Simpson’s.) You do not need to simplify your answers.
Q12: Find the midpoint rule approximation to ż π 0 sin x dx with n = 3.
Questions 13 though 17 ask you to approximate a quantity based on observed data.
Q13: The solid V is 40 cm high and the horizontal cross sections are circular disks.
The table below gives the diameters of the cross sections in centimeters at 10 cm intervals. Use the trapezoidal rule to estimate the volume of V.
height 0 10 20 30 40 diameter 24 16 10 6 4 Q14: A 6 metre long cedar log has cross sections that are approximately circular. The diameters of the log, measured at one metre intervals, are given below: metres from left end of log 0 1 2 3 4 5 6 diameter in metres 1.2 1 0.8 0.8 1 1 1.2 Use Simpson’s Rule to estimate the volume of the log.
45 INTEGRATION 1.11 NUMERICAL INTEGRATION Q15: The circumference of an 8 metre high tree at different heights above the ground is given in the table below. Assume that all horizontal cross-sections of the tree are circular disks.
height (metres) 0 2 4 6 8 circumference (metres) 1.2 1.1 1.3 0.9 0.2 Use Simpson’s rule to approximate the volume of the tree.
Q16: By measuring the areas enclosed by contours on a topographic map, a geologist determines the cross sectional areas A in m2 of a 60 m high hill. The table below gives the cross sectional area A(h) at various heights h. The volume of the hill is V = ş60 0 A(h) dh.
h 0 10 20 30 40 50 60 A 10,200 9,200 8,000 7,100 4,500 2,400 100 (a) If the geologist uses the Trapezoidal Rule to estimate the volume of the hill, what will be their estimate, to the nearest 1,000m3?
(b) What will be the geologist’s estimate of the volume of the hill if they use Simpson’s Rule instead of the Trapezoidal Rule?
Q17: The graph below applies to both parts (a) and (b).
x y 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 (a) Use the Trapezoidal Rule, with n = 4, to estimate the area under the graph between x = 2 and x = 6. Simplify your answer completely.
(b) Use Simpson’s Rule, with n = 4, to estimate the area under the graph between x = 2 and x = 6.
In Questions 18 through 24, we practice finding error bounds for our approximations.
Q18: The integral ż 1 ´1 sin(x2) dx is estimated using the Midpoint Rule with 1000 intervals. Show that the absolute error in this approximation is at most 2 ¨ 10´6.
You may use the fact that when approximating şb a f (x) dx with the Midpoint Rule using n 46 INTEGRATION 1.11 NUMERICAL INTEGRATION points, the absolute value of the error is at most M(b ´ a)3/24n2 when |f 2(x)| ď M for all x P [a, b].
Q19: The total error using the midpoint rule with n subintervals to approximate the integral of f (x) over [a, b] is bounded by M(b ´ a)3 (24n2) , if |f 2(x)| ď M for all a ď x ď b.
Using this bound, if the integral ż 1 ´2 2x4 dx is approximated using the midpoint rule with 60 subintervals, what is the largest possible error between the approximation M60 and the true value of the integral?
Q20: Both parts of this question concern the integral I = ż 2 0 (x ´ 3)5 dx.
(a) Write down the Simpson’s Rule approximation to I with n = 6. Leave your answer in calculator-ready form.
(b) Which method of approximating I results in a smaller error bound: the Midpoint Rule with n = 100 intervals, or Simpson’s Rule with n = 10 intervals? You may use the formulas |EM| ď M(b ´ a)3 24n2 and |ES| ď L(b ´ a)5 180n4 , where M is an upper bound for |f 2(x)| and L is an upper bound for |f (4)(x)|, and EM and ES are the absolute errors arising from the midpoint rule and Simpson’s rule, respectively.
Q21: Find a bound for the error in approximating ż 5 1 1 x dx using Simpson’s rule with n = 4. Do not write down the Simpson’s rule approximation S4.
In general the error in approximating şb a f (x) dx using Simpson’s rule with n steps is bounded by L(b ´ a) 180 (∆x)4 where ∆x = b ´ a n and L ě |f (4)(x)| for all a ď x ď b.
Q22: Find a bound for the error in approximating ż 1 0 e´2x + 3x3 dx using Simpson’s rule with n = 6. Do not write down the Simpson’s rule approximation Sn.
In general, the error in approximating şb a f (x) dx using Simpson’s rule with n steps is bounded by L(b ´ a) 180 (∆x)4 where ∆x = b ´ a n and L ě |f (4)(x)| for all a ď x ď b.
Q23: Let I = ż 2 1 (1/x) dx.
47 INTEGRATION 1.11 NUMERICAL INTEGRATION (a) Write down the trapezoidal approximation T4 for I. You do not need to simplify your answer.
(b) Write down the Simpson’s approximation S4 for I. You do not need to simplify your answer.
(c) Without computing I, find an upper bound for |I ´ S4|. You may use the fact that if ˇ ˇf (4)(x) ˇ ˇ ď L on the interval [a, b], then the error in using Sn to approximate şb a f (x) dx has absolute value less than or equal to L(b ´ a)5/180n4.
Since 1 x5 is a decreasing function when x ą 0, look for its maximum value when x is as small as possible.
Q24: A function s(x) satisfies s(0) = 1.00664, s(2) = 1.00543, s(4) = 1.00435, s(6) = 1.00331, s(8) = 1.00233. Also, it is known to satisfy ˇ ˇs(k)(x) ˇ ˇ ď k 1000 for 0 ď x ď 8 and all positive integers k.
(a) Find the best Trapezoidal Rule and Simpson’s Rule approximations that you can for I = ż 8 0 s(x) dx.
(b) Determine the maximum possible sizes of errors in the approximations you gave in part (a). Recall that if a function f (x) satisfies ˇ ˇf (k)(x) ˇ ˇ ď Kk on [a, b], then ˇ ˇ ˇ ˇ ż b a f (x) dx ´ Tn ˇ ˇ ˇ ˇ ď K2(b ´ a)3 12n2 and ˇ ˇ ˇ ˇ ż b a f (x) dx ´ Sn ˇ ˇ ˇ ˇ ď K4(b ´ a)5 180n4 Q25: Consider the trapezoidal rule for making numerical approximations to ż b a f (x) dx. The error for the trapezoidal rule satisfies |ET| ď M(b ´ a)3 12n2 , where |f 2(x)| ď M for a ď x ď b. If ´2 ă f 2(x) ă 0 for 1 ď x ď 4, find a value of n to guarantee the trapezoidal rule will give an approximation for ż 4 1 f (x) dx with absolute error, |ET|, less than 0.001.
§ § Stage 3 Q26: A swimming pool has the shape shown in the figure below. The vertical cross-sections of the pool are semi-circular disks. The distances in feet across the pool are given in the figure at 2–foot intervals along the sixteen–foot length of the pool. Use Simpson’s Rule to estimate the volume of the pool.
48 INTEGRATION 1.11 NUMERICAL INTEGRATION 10’ 12’ 10’ 8’ 6’ 8’ 10’ 2’ Q27: A piece of wire 1m long with radius 1mm is made in such a way that the density varies in its cross-section, but is radially symmetric (that is, the local density g(r) in kg/m3 depends only on the distance r in mm from the centre of the wire). Take as given that the total mass W of the wire in kg is given by W = 2π10´6 ż 1 0 rg(r) dr Data from the manufacturer is given below: r 0 1/4 1/2 3/4 1 g(r) 8051 8100 8144 8170 8190 (a) Find the best Trapezoidal Rule approximation that you can for W based on the data in the table.
(b) Suppose that it is known that |g1(r)| ă 200 and |g2(r)| ă 150 for all values of r.
Determine the maximum possible size of the error in the approximation you gave in part (a). Recall that if a function f (x) satisfies |f 2(x)| ď M on [a, b], then |I ´ Tn| ď M(b ´ a)3 12n2 where I = şb a f (x) dx and Tn is the Trapezoidal Rule approximation to I using n subintervals.
Q28: Simpson’s rule can be used to approximate log 2, since log 2 = ż 2 1 1 x dx.
(a) Use Simpson’s rule with 6 subintervals to approximate log 2.
(b) How many subintervals are required in order to guarantee that the absolute error is less than 0.00001?
Note that if En is the error using n subintervals, then |En| ď L(b ´ a)5 180n4 where L is the maximum absolute value of the fourth derivative of the function being integrated and a and b are the end points of the interval.
49 INTEGRATION 1.11 NUMERICAL INTEGRATION Q29: Let I = ż 2 0 cos(x2) dx and let Sn be the Simpson’s rule approximation to I using n subintervals.
(a) Estimate the maximum absolute error in using S8 to approximate I.
(b) How large should n be in order to ensure that |I ´ Sn| ď 0.0001?
Note: The graph of f 4(x), where f (x) = cos(x2), is shown below. The absolute error in the Simpson’s rule approximation is bounded by L(b ´ a)5 180n4 when |f 4(x)| ď L on the interval [a, b].
0.5 1.0 1.5 2.0 −300 −200 −100 0 100 x Q30: Define a function f (x) and an integral I by f (x) = ż x2 0 sin( ?
t) dt, I = ż 1 0 f (t) dt Estimate how many subdivisions are needed to calculate I to five decimal places of accuracy using the trapezoidal rule.
Note that if En is the error using n subintervals, then |En| ď M(b ´ a)3 12n2 , where M is the maximum absolute value of the second derivative of the function being integrated and a and b are the limits of integration.
Q: Let f (x) be a function4 with f 2(x) = x2 x + 1.
(a) Show that |f 2(x)| ď 1 whenever x is in the interval [0, 1].
(b) Find the maximum value of |f 2(x)| over the interval [0, 1].
(c) Assuming M = 1, how many intervals should you use to approximate ż 1 0 f (x) dx to within 10´5?
4 For example, f (x) = 1 6x3 ´ 1 2x2 + (1 + x) log |x + 1| will do, but you don’t need to know what f (x) is for this problem.
50 INTEGRATION 1.12 IMPROPER INTEGRALS (d) Using the value of M you found in (b), how many intervals should you use to approximate ż 1 0 f (x) dx to within 10´5?
Q: Approximate the function log x with a rational function by approximating the integral ż x 1 1 t dt using Simpson’s rule. Your rational function f (x) should approximate log x with an error of not more than 0.1 for any x in the interval [1, 3].
Q: Using an approximation of the area under the curve 1 x2 + 1, show that the constant arctan 2 is in the interval hπ 4 + 0.321, π 4 + 0.323 i .
You may assume use without proof that d4 dx4 " 1 1 + x2 = 24(5x4 ´ 10x2 + 1) (x2 + 1)5 . You may use a calculator, but only to add, subtract, multiply, and divide.
1.12IJ Improper Integrals Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: For which values of b is the integral ż b 0 1 x2 ´ 1 dx improper?
Q: For which values of b is the integral ż b 0 1 x2 + 1 dx improper?
51 INTEGRATION 1.12 IMPROPER INTEGRALS Q: Below are the graphs y = f (x) and y = g(x). Suppose ż 8 0 f (x) dx converges, and ż 8 0 g(x) dx diverges. Assuming the graphs continue on as shown as x Ñ 8, which graph is f (x), and which is g(x)?
x y Q4: Decide whether the following statement is true or false. If false, provide a counterexample. If true, provide a brief justification. (Assume that f (x) and g(x) are continuous functions.) If ż 8 1 f (x) dx converges and g(x) ě f (x) ě 0 for all x, then ż 8 1 g(x) dx converges.
Q: Let f (x) = e´x and g(x) = 1 x + 1. Note ş8 0 f (x) dx converges while ş8 0 g(x) dx diverges.
For each of the functions h(x) described below, decide whether ş8 0 h(x) dx converges or diverges, or whether there isn’t enough information to decide. Justify your decision.
(a) h(x), continuous and defined for all x ě 0, h(x) ď f (x).
(b) h(x), continuous and defined for all x ě 0, f (x) ď h(x) ď g(x).
(c) h(x), continuous and defined for all x ě 0, ´2f (x) ď h(x) ď f (x).
§ § Stage 2 Q6: Evaluate the integral ż 1 0 x4 x5 ´ 1 dx or state that it diverges.
Q7: Determine whether the integral ż 2 ´2 1 (x + 1)4/3 dx is convergent or divergent. If it is convergent, find its value.
Q8: Does the improper integral ż 8 1 1 ?
4x2 ´ x dx converge? Justify your answer.
52 INTEGRATION 1.12 IMPROPER INTEGRALS Q9: Does the integral ż 8 0 dx x2 + ?x converge or diverge? Justify your claim.
Q: Does the integral ż 8 ´8 cos x dx converge or diverge? If it converges, evaluate it.
Q: Does the integral ż 8 ´8 sin x dx converge or diverge? If it converges, evaluate it.
Q: Evaluate ż 8 10 x4 ´ 5x3 + 2x ´ 7 x5 + 3x + 8 dx, or state that it diverges.
Q: Evaluate ż 10 0 x ´ 1 x2 ´ 11x + 10 dx, or state that it diverges.
Q14: Determine (with justification!) which of the following applies to the integral ż +8 ´8 x x2 + 1dx: (i) ż +8 ´8 x x2 + 1dx diverges (ii) ż +8 ´8 x x2 + 1dx converges but ż +8 ´8 ˇ ˇ ˇ ˇ x x2 + 1 ˇ ˇ ˇ ˇ dx diverges (iii) ż +8 ´8 x x2 + 1dx converges, as does ż +8 ´8 ˇ ˇ ˇ ˇ x x2 + 1 ˇ ˇ ˇ ˇ dx Remark: these options, respectively, are that the integral diverges, converges condition-ally, and converges absolutely. You’ll see this terminology used for series in Section 3.4.1 of the CLP-2 text.
Q15: Decide whether I = ż 8 0 | sin x| x3/2 + x1/2 dx converges or diverges. Justify.
Q16: Does the integral ż 8 0 x + 1 x1/3(x2 + x + 1) dx converge or diverge?
53 INTEGRATION 1.12 IMPROPER INTEGRALS § § Stage 3 Q: We craft a tall, vuvuzela-shaped solid by rotating the line y = 1 x from x = a to x = 1 about the y-axis, where a is some constant between 0 and 1.
x y y = 1 x 1 a ´1 1 a True or false: No matter how large a constant M is, there is some value of a that makes a solid with volume larger than M.
Q18: What is the largest value of q for which the integral ż 8 1 1 x5q dx diverges?
Q: For which values of p does the integral ż 8 0 x (x2 + 1)p dx converge?
Q: Evaluate ż 8 2 1 t4 ´ 1dt, or state that it diverges.
Q: Does the integral ż 5 ´5 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx converge or diverge?
Q: Evaluate ż 8 0 e´x sin x dx, or state that it diverges.
Q23: Is the integral ż 8 0 sin4 x x2 dx convergent or divergent? Explain why.
Q: Does the integral ż 8 0 x ex + ?x dx converge or diverge?
54 INTEGRATION 1.13 MORE INTEGRATION EXAMPLES Q25: Let Mn,t be the Midpoint Rule approximation for ż t 0 e´x 1 + x dx with n equal subintervals. Find a value of t and a value of n such that Mn,t differs from ş8 0 e´x 1+x dx by at most 10´4. Recall that the error En introduced when the Midpoint Rule is used with n subintervals obeys |En| ď M(b ´ a)3 24n2 where M is the maximum absolute value of the second derivative of the integrand and a and b are the end points of the interval of integration.
Q: Suppose f (x) is continuous for all real numbers, and ż 8 1 f (x) dx converges.
(a) If f (x) is odd, does ż ´1 ´8 f (x) dx converge or diverge, or is there not enough information to decide?
(b) If f (x) is even, does ż 8 ´8 f (x) dx converge or diverge, or is there not enough information to decide?
Q: True or false: There is some real number x, with x ě 1, such that ż x 0 1 et dt = 1.
1.13IJ More Integration Examples Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: Match the integration method to a common kind of integrand it’s used to antidifferentiate.
(A) u = f (x) substitution (I) a function multiplied by its derivative (B) trigonometric substitution (II) a polynomial function times an exponential function (C) integration by parts (III) a rational function (D) partial fractions (IV) the square root of a quadratic function § § Stage 2 Q: Evaluate ż π/2 0 sin4 x cos5 x dx.
55 INTEGRATION 1.13 MORE INTEGRATION EXAMPLES Q: Evaluate ż a 3 ´ 5x2 dx.
Q: Evaluate ż 8 0 x ´ 1 ex dx.
Q: Evaluate ż ´2 3x2 + 4x + 1dx.
Q: Evaluate ż 2 1 x2 log x dx.
Q7: Evaluate ż x x2 ´ 3 dx.
Q8: Evaluate the following integrals.
(a) ż 4 0 x ?
9 + x2 dx (b) ż π/2 0 cos3 x sin2 x dx (c) ż e 1 x3 log x dx Q9: Evaluate the following integrals.
(a) ż π/2 0 x sin x dx (b) ż π/2 0 cos5 x dx Q10: Evaluate the following integrals.
(a) ż 2 0 xex dx (b) ż 1 0 1 ?
1 + x2 dx (c) ż 5 3 4x (x2 ´ 1)(x2 + 1) dx Q11: Calculate the following integrals.
(a) ż 3 0 a 9 ´ x2 dx (b) ż 1 0 log(1 + x2) dx (c) ż 8 3 x (x ´ 1)2(x ´ 2) dx 56 INTEGRATION 1.13 MORE INTEGRATION EXAMPLES Q: Evaluate ż sin4 θ ´ 5 sin3 θ + 4 sin2 θ + 10 sin θ sin2 θ ´ 5 sin θ + 6 cos θ dθ.
Q13: Evaluate the following integrals. Show your work.
(a) ż π 4 0 sin2(2x) cos3(2x) dx (b) ż 9 + x2´ 3 2 dx (c) ż dx (x ´ 1)(x2 + 1) (d) ż x arctan x dx Q14: Evaluate the following integrals.
(a) ż π/4 0 sin5(2x) cos(2x) dx (b) ż a 4 ´ x2 dx (c) ż x + 1 x2(x ´ 1) dx Q15: Calculate the following integrals.
(a) ż 8 0 e´x sin(2x) dx (b) ż ?
2 0 1 (2 + x2)3/2 dx (c) ż 1 0 x log(1 + x2) dx (d) ż 8 3 1 (x ´ 1)2(x ´ 2) dx Q16: Evaluate the following integrals.
(a) ż x log x dx (b) ż (x ´ 1) dx x2 + 4x + 5 (c) ż dx x2 ´ 4x + 3 (d) ż x2 dx 1 + x6 Q17: Evaluate the following integrals.
57 INTEGRATION 1.13 MORE INTEGRATION EXAMPLES (a) ż 1 0 arctan x dx.
(b) ż 2x ´ 1 x2 ´ 2x + 5 dx.
Q18: (a) Evaluate ż x2 (x3 + 1)101 dx.
(b) Evaluate ż cos3x sin4x dx.
Q: Evaluate ż π π/2 cos x ?
sin x dx.
Q20: Evaluate the following integrals.
(a) ż ex (ex + 1)(ex ´ 3) dx (b) ż 4 2 x2 ´ 4x + 4 ?
12 + 4x ´ x2 dx Q21: Evaluate these integrals.
(a) ż sin3 x cos3 x dx (b) ż 2 ´2 x4 x10 + 16 dx Q: Evaluate ż x ?
x ´ 1 dx.
Q: Evaluate ż ?
x2 ´ 2 x2 dx for x ě ?
2.
You may use that ş sec x dx = log | sec x + tan x| + C.
Q: Evaluate ż π/4 0 sec4 x tan5 x dx.
Q: Evaluate ż 3x2 + 4x + 6 (x + 1)3 dx.
Q: Evaluate ż 1 x2 + x + 1 dx.
Q: Evaluate ż sin x cos x tan x dx.
Q: Evaluate ż 1 x3 + 1 dx.
58 INTEGRATION 1.13 MORE INTEGRATION EXAMPLES Q: Evaluate ż (3x)2 arcsin x dx.
§ § Stage 3 Q: Evaluate ż π/2 0 ?
cos t + 1 dt.
Q: Evaluate ż e 1 log ?x x dx.
Q: Evaluate ż 0.2 0.1 tan x log(cos x) dx.
Q33: Evaluate these integrals.
(a) ż sin(log x) dx (b) ż 1 0 1 x2 ´ 5x + 6 dx Q34: Evaluate (with justification).
(a) ż 3 0 (x + 1) a 9 ´ x2 dx (b) ż 4x + 8 (x ´ 2)(x2 + 4) dx (c) ż +8 ´8 1 ex + e´x dx Q: Evaluate ż c x 1 ´ x dx.
Q: Evaluate ż 1 0 e2xeex dx.
Q: Evaluate ż xex (x + 1)2 dx.
Q: Evaluate ż x sin x cos2 x dx.
You may use that ş sec xdx = log | sec x + tan x| + C.
Q: Evaluate ż x(x + a)n dx, where a and n are constants.
Q: Evaluate ż arctan(x2) dx.
59 APPLICATIONS OF INTEGRATION Chapter 2 2.1IJ Work Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Find the work (in joules) required to lift a 3-gram block of matter a height of 10 centimetres against the force of gravity (with g = 9.8 m/sec2).
Q: A rock exerts a force of 1 N on the ground where it sits due to gravity. Use g = 9.8 m/sec2.
What is the mass of the rock?
How much work (in joules) does it take to lift that rock one metre in the air?
Q: Consider the equation W = ż b a F(x) dx where x is measured in metres and F(x) is measured in kilogram-metres per second squared (newtons).
For some large n, we might approximate W « n ÿ i=1 F(xi)∆x where ∆x = b´a n and xi is some number in the interval [a + (i ´ 1)∆x, a + i∆x]. (This is just the general form of a Riemann sum).
(a) What are the units of ∆x?
60 APPLICATIONS OF INTEGRATION 2.1 WORK (b) What are the units of F(xi)?
(c) Using your answers above, what are the units of W?
Remark: we already know the units of W from the text, but the Riemann sum illustrates why they make sense arising from this particular integral.
Q: Suppose f (x) has units smoot megaFonzie, and x is measured in barns. What are the units of the quantity ş1 0 f (x) dx?
For this problem, it doesn’t matter what the units measure, but a smoot is a silly measure of length; a megaFonzie is an apocryphal measure of coolness; and a barn is a humorous (but actu-ally used) measure of area.
For explanations (and entertainment) see
org/wiki/List_of_humorous_units_of_measurement and wiki/List_of_unusual_units_of_measurement (accessed 27 July 2017).
Q: You want to weigh your luggage before a flight. You don’t have a scale or balance, but you do have a heavy-duty spring from your local engineering-supply store. You nail it to your wall, marking where the bottom hangs. You hang a one-litre bag of water (with mass one kilogram) from the spring, and observe that the spring stretches 1 cm. Where on the wall should you mark the bottom of the spring corresponding to a hanging mass of 10kg?
bottom of unloaded spring bottom of spring with 10kg mass You may assume that the spring obeys Hooke’s law.
Q: The work done by a force in moving an object from position x = 1 to x = b is W(b) = ´b3 + 6b2 ´ 9b + 4 for any b in [1, 3]. At what position x in [1, 3] is the force the strongest?
§ § Stage 2 Q7: A variable force F(x) = a ?x Newtons moves an object along a straight line when 61 APPLICATIONS OF INTEGRATION 2.1 WORK it is a distance of x meters from the origin. If the work done in moving the object from x = 1 meters to x = 16 meters is 18 joules, what is the value of a? Don’t worry about the units of a.
Q: A tube of air is fitted with a plunger that compresses the air as it is pushed in. If the natural length of the tube of air is ℓ, when the plunger has been pushed x metres past its natural position, the force exerted by the air is c ℓ´x N, where c is a positive constant (depending on the particulars of the tube of air) and x ă ℓ.
x ℓ (a) What are the units of c?
(b) How much work does it take to push the plunger from 1 metre past its natural position to 1.5 metres past its natural position? (You may assume ℓą 1.5.) Questions 9 through 16 offer practice on two broad types of calculations covered in the text: lifting things against gravity, and stretching springs. You may make the same physical assumptions as in the text: that is, springs follow Hooke’s law, and the acceleration due to gravity is a constant ´9.8 metres per second squared.
Q9: Find the work (in joules) required to stretch a spring 10 cm beyond equilibrium, if its spring constant is k = 50 N/m.
Q10: A force of 10 N (newtons) is required to hold a spring stretched 5 cm beyond its natural length. How much work, in joules (J), is done in stretching the spring from its natural length to 50 cm beyond its natural length?
Q11: A 5-metre-long cable of mass 8 kg is used to lift a bucket off the ground. How much work is needed to raise the entire cable to height 5 m? Ignore the mass of the bucket and its contents.
Q: A tank 1 metre high has pentagonal cross sections of area 3 m2 and is filled with water. How much work does it take to pump out all the water?
You may assume the density of water is 1 kg per 1000 cm3.
62 APPLICATIONS OF INTEGRATION 2.1 WORK Q13: A sculpture, shaped like a pyramid 3m high sitting on the ground, has been made by stacking smaller and smaller (very thin) iron plates on top of one another. The iron plate at height z m above ground level is a square whose side length is (3 ´ z) m. All of the iron plates started on the floor of a basement 2 m below ground level.
Write down an integral that represents the work, in joules, it took to move all of the iron from its starting position to its present position. Do not evaluate the integral. (You can use 9.8 m/s2 for the acceleration due to gravity and 8000 kg/m3 for the density of iron.) Q: Suppose a spring extends 5 cm past its natural length when one kilogram is hung from its end. How much work is done to extend the spring from 5 cm past its natural length to 7 cm past its natural length?
Q: Ten kilograms of firewood are hoisted on a rope up a height of 4 metres to a second-floor deck. If the total work done is 400 joules, what is the mass of the 4 metres of rope?
You may assume that the rope has the same density all the way along.
Q: A 5 kg weight is attached to the middle of a 10-metre long rope, which dangles out a window. The rope alone has mass 1 kg. How much work does it take to pull the entire rope in through the window, together with the weight?
Q: A box is dragged along the floor. Friction exerts a force in the opposite direction of motion from the box, and that force is equal to µ ˆ m ˆ g, where µ is a constant, m is the mass of the box and g is the acceleration due to gravity. You may assume g = 9.8 m/sec2.
(a) How much work is done dragging a box of mass 10 kg along the floor for three metres if µ = 0.4?
(b) Suppose the box contains a volatile substance that rapidly evaporates. You pull the box at a constant rate of 1 m/sec for three seconds, and the mass of the box at t seconds (0 ď t ď 3) is (10 ´ ?
t) kilograms. If µ = 0.4, how much work is done pulling the box for three seconds?
For Questions 18 and 19, use the principle (introduced after Definition 2.1.1 in the CLP-2 text and utilized in Example 2.1.5) that the work done on a particle by a force over a distance is equal to the change in kinetic energy of that particle.
Q: A ball of mass 1 kg is attached to a spring, and the spring is attached to a table.
The ball moves with some initial velocity, and the spring slows it down. At its farthest, 63 APPLICATIONS OF INTEGRATION 2.1 WORK the spring stretches 10 cm past its natural length. If the spring constant is 5 N/m, what was the initial velocity of the ball?
You may assume that the ball starts moving with initial velocity v0, and that the only force slowing it down is the spring. You may also assume that the spring started out at its natural length, it follows Hooke’s law, and when it is stretched its farthest, the velocity of the ball is 0 m/sec.
Q: A mild-mannered university professor who is definitely not a spy notices that when their car is on the ground, it is 2 cm shorter than when it is on a jack. (That is: when the car is on a jack, its struts are at their natural length; when on the ground, the weight of the car causes the struts to compress 2 cm.) The university professor calculates that if they were to jump a local neighborhood drawbridge, their car would fall to the ground with a speed of 4 m/sec. If the car can sag 20 cm before important parts scrape the ground, and the car has mass 2000 kg unoccupied (2100 kg with the professor inside), can the professor, who is certainly not involved in international intrigue, safely jump the bridge?
Assume the car falls vertically, the struts obey Hooke’s law, and the work done by the struts is equal to the change in kinetic energy of the car + professor. Use 9.8 m/sec2 for the acceleration due to gravity.
§ § Stage 3 Q: A disposable paper cup has the shape of a right circular cone with radius 5 cm and height 15 cm, and is completely filled with water. How much work is done sucking all the water out of the cone with a straw?
You may assume that 1 m3 of water has mass 1000 kilograms, the acceleration due to gravity is ´9.8 m/sec2, and that the water moves as high up as the very top of the cup and no higher.
64 APPLICATIONS OF INTEGRATION 2.1 WORK Q21: A spherical tank of radius 3 metres is half–full of water. It has a spout of length 1 metre sticking up from the top of the tank. Find the work required to pump all of the water in the tank out the spout. The density of water is 1000 kilograms per cubic metre.
The acceleration due to gravity is 9.8 metres per second squared.
1m 3m Q: A 5-metre cable is pulled out of a deep hole, where it was dangling straight down.
The cable has density ρ(x) = (10 ´ x) kg/m, where x is the distance from the bottom end of the rope. (So, the bottom of the cable is denser than the top.) How much work is done pulling the cable out of the hole?
Q: A rectangular tank is fitted with a plunger that can raise and lower the water level by decreasing and increasing the length of its base, as in the diagrams below. The tank has base width 1 m (which does not change) and contains 3 m3 of water.
3 m 1 m 1 m The force of the water acting on any tiny piece of the plunger is PA, where P is the pressure of the water, and dA is the area of the tiny piece. The pressure varies with the depth of the piece (below the surface of the water). Specifically, P = cD, where D is the depth of the tiny piece and c is a constant, in this case c = 9800 N/m3.
(a) If the length of the base is 3 m, give the force of the water on the entire plunger. (You can do this with an integral: it’s the sum of the force on all the tiny pieces of the plunger.) (b) If the length of the base is x m, give the force of the water on the entire plunger.
(c) Give the work required to move the plunger in so that the base length changes from 3 m to 1 m.
65 APPLICATIONS OF INTEGRATION 2.1 WORK Q: A leaky bucket picks up 5 L of water from a well, but drips out 1 L every ten seconds. If the bucket was hauled up 5 metres at a constant speed of 1 metre every two seconds, how much work was done?
Assume the rope and bucket have negligible mass and one litre of water has 1 kg mass, and use 9.8 m/sec2 for the acceleration due to gravity.
Q: The force of gravity between two objects, one of mass m1 and another of mass m2, is F = Gm1m2 r2 , where r is the distance between them and G is the gravitational constant.
How much work is required to separate the earth and the moon far enough apart that the gravitational attraction between them is negligible?
Assume the mass of the earth is 6 ˆ 1024 kg and the mass of the moon is 7 ˆ 1022 kg, and that they are currently 400 000 km away from each other. Also, assume G = 6.7 ˆ 10´11 m3 kg¨sec2, and the only force acting on the earth and moon is the gravity between them.
Q: True or false: the work done pulling up a dangling cable of length ℓand mass m (with uniform density) is the same as the work done lifting up a ball of mass m a height of ℓ/2.
ℓ ℓ/2 Q: A tank one metre high is filled with watery mud that has settled to be denser at the bottom than at the top.
At height h metres above the bottom of the tank, the cross-section of the tank has the shape of the finite region bounded by the two curves y = x2 and y = 2 ´ h ´ 3x2. At height h metres above the bottom of the tank, the density of the liquid is 1000 ?
2 ´ h kilograms per cubic metre.
How much work is done to pump all the liquid out of the tank?
You may assume the acceleration due to gravity is 9.8 m/sec2.
Q: An hourglass is 0.2 m tall and shaped such that that y metres above or below its vertical centre it has a radius of y2 + 0.01 m. It is exactly half-full of sand, which has mass M = 1 7 kilograms.
How much work is done on the sand by quickly flipping the hourglass over?
66 APPLICATIONS OF INTEGRATION 2.2 AVERAGES Assume that the work done is only moving against gravity, with g = 9.8 m/sec2, and the sand has uniform density. Also assume that at the instant the hourglass is flipped over, the sand has not yet begun to fall, as in the picture above.
Q: Suppose at position x a particle experiences a force of F(x) = ?
1 ´ x4 N.
Approximate the work done moving the particle from x = 0 to x = 1/2, accurate to within 0.01 J.
2.2IJ Averages Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it is often denoted ln x.
§ § Stage 1 Q: Below is the graph of a function y = f (x). Its average value on the interval [0, 5] is A. Draw a rectangle on the graph with area ş5 0 f (x) dx.
x y 5 A 67 APPLICATIONS OF INTEGRATION 2.2 AVERAGES Q: Suppose a car travels for 5 hours in a straight line, with an average velocity of 100 kph. How far did the car travel?
Q: A force F(x) acts on an object from position x = a metres to position x = b metres, for a total of W joules of work. What was the average force on the object?
Q: Suppose we want to approximate the average value of the function f (x) on the interval [a, b]. To do this, we cut the interval [a, b] into n pieces, then take n samples by finding the function’s output at the left endpoint of each piece, starting with a. Then, we average those n samples. (In the example below, n = 4.) x y a b average these y-values (a) Using n samples, what is the distance between two consecutive sample points xi and xi+1?
(b) Assuming n ě 4, what is the x-coordinate of the fourth sample?
(c) Assuming n ě 4, what is the y-value of the fourth sample?
(d) Write the approximation of the average value of f (x) over the interval [a, b] using sigma notation.
Q: Suppose f (x) and g(x) are functions that are defined for all numbers in the interval [0, 10].
(a) If f (x) ď g(x) for all x in [0, 10], then is the average value of f (x) is less than or equal to the average value of g(x) on the interval [0, 10], or is there not enough information to tell?
(b) Suppose f (x) ď g(x) for all x in [0.01, 10]. Is the average value of f (x) less than or equal to the average value of g(x) over the interval [0, 10], or is there not enough information to tell?
Q: Suppose f is an odd function, defined for all real numbers. What is the average of f on the interval [´10, 10]?
§ § Stage 2 Q7: Find the average value of f (x) = sin(5x) + 1 over the interval ´π/2 ď x ď π/2.
Q8: Find the average value of the function y = x2 log x on the interval 1 ď x ď e.
68 APPLICATIONS OF INTEGRATION 2.2 AVERAGES Q9: Find the average value of the function f (x) = 3 cos3 x + 2 cos2 x on the interval 0 ď x ď π 2 .
Q10: Let k be a positive constant. Find the average value of the function f (x) = sin(kx) on the interval 0 ď x ď π/k.
Q11: The temperature in Celsius in a 3 m long rod at a point x metres from the left end of the rod is given by the function T(x) = 80 16´x2. Determine the average temperature in the rod.
Q12: What is the average value of the function f (x) = log x x on the interval [1, e]?
Q13: Find the average value of f (x) = cos2(x) over 0 ď x ď 2π.
Q: The carbon dioxide concentration in the air at a particular location over one year is approximated by C(t) = 400 + 50 cos t 12π + 200 cos t 4380π parts per million, where t is measured in hours.
(a) What is the average carbon dioxide concentration for that location for that year?
(b) What is the average over the first day?
(c) Suppose measurements were only made at noon every day: that is, when t = 12 + 24n, where n is any whole number between 0 and 364. Then the daily variation would cease: 50 cos (12+24n) 12 π = 50 cos (π + 2πn) = 50 cos π = ´50. So, the approximation for the concentration of carbon dioxide in the atmosphere might be given as N(t) = 350 + 200 cos t 4380π ppm What is the relative error in the yearly average concentration of carbon dioxide involved in using N(t), instead of C(t)?
You may assume a day has exactly 24 hours, and a year has exactly 8760 hours.
Q: Let S be the solid formed by rotating the parabola y = x2 from x = 0 to x = 2 about the x-axis.
(a) What is the average area of the circular cross-sections of S? Call this value A.
(b) What is the volume of S?
(c) What is the volume of a cylinder with circular cross-sectional area A and length 2?
For Questions 16 through 18, let the root mean square of f (x) on [a, b] be d 1 b ´ a ż b a f 2(x) dx. This is the formula used in Example 2.2.6 in the CLP-2 text.
Q: Let f (x) = x.
(a) Calculate the average of f (x) over [´3, 3].
(b) Calculate the root mean square of f (x) over [´3, 3].
69 APPLICATIONS OF INTEGRATION 2.2 AVERAGES Q: Calculate the root mean square of f (x) = tan x over ´π 4 , π 4 .
Q: A force acts on a spring, and the spring stretches and contracts. The distance beyond its natural length at time t is f (t) = sin (tπ) cm, where t is measured in seconds.
The spring constant is 3 N/cm.
(a) What is the force exerted by the spring at time t, if it obeys Hooke’s law?
(b) Find the average of the force exerted by the spring from t = 0 to t = 6.
(c) Find the root mean square of the force exerted by the spring from t = 0 to t = 6.
§ § Stage 3 Q19: A car travels two hours without stopping. The driver records the car’s speed every 20 minutes, as indicated in the table below: time in hours 0 1/3 2/3 1 4/3 5/3 2 speed in km/hr 50 70 80 55 60 80 40 (a) Use the trapezoidal rule to estimate the total distance traveled in the two hours.
(b) Use the answer to part (a) to estimate the average speed of the car during this period.
Q: Let s(t) = et.
(a) Find the average of s(t) on the interval [0, 1]. Call this quantity A.
(b) For any point t, the difference between s(t) and A is s(t) ´ A. Find the average value of s(t) ´ A on the interval [0, 1].
(c) For any point t, the absolute difference between s(t) and A is |s(t) ´ A|. Find the average value of |s(t) ´ A| on the interval [0, 1].
Q: Consider the two functions f (x) and g(x) below, both of which have average A on [0, 4].
x y 4 A y = f (x) y = g(x) (a) Which function has a larger average on [0, 4]: f (x) ´ A or g(x) ´ A?
(b) Which function has a larger average on [0, 4]: |f (x) ´ A| or |g(x) ´ A|?
70 APPLICATIONS OF INTEGRATION 2.2 AVERAGES Q: Suppose the root mean square of a function f (x) on the interval [a, b] is R. What is the volume of the solid formed by rotating the portion of f (x) from a to b about the x-axis?
x y a b y = f (x) As in Example 2.2.6 of the CLP-2 text, let the root mean square of f (x) on [a, b] be d 1 b ´ a ż b a f 2(x) dx.
Q: Suppose f (x) = ax2 + bx + c, and the average value of f (x) on the interval [0, 1] is the same as the average of f (0) and f (1). What is a?
Q: Suppose f (x) = ax2 + bx + c, and the average value of f (x) on the interval [s, t] is the same as the average of f (s) and f (t). Is it possible that a ‰ 0?
That is– does the result of Question 23 generalize?
Q: Let f (x) be a function defined for all numbers in the interval [a, b], with average value A over that interval. What is the average of f (a + b ´ x) over the interval [a, b]?
Q: Suppose f (t) is a continuous function, and A(x) is the average of f (t) on the interval from 0 to x.
(a) What is the average of f (t) on [a, b], where a ă b? Give your answer in terms of A.
(b) What is f (t)? Again, give your answer in terms of A.
Q: (a) Find a function f (x) with average 0 over [´1, 1] but f (x) ‰ 0 for all x in [´1, 1], or show that no such function exists.
(b) Find a continuous function f (x) with average 0 over [´1, 1] but f (x) ‰ 0 for all x in [´1, 1], or show that no such function exists.
Q: Suppose f (x) is a positive, continuous function with lim xÑ8 f (x) = 0, and let A(x) be the average of f (x) on [0, x].
True or false: lim xÑ8 A(x) = 0.
Q: Let A(x) be the average of the function f (t) = e´t2 on the interval [0, x]. What is lim xÑ8 A(x)?
71 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE 2.3IJ Centre of Mass and Torque Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Using symmetry, find the centroid of the finite region between the curves y = (x ´ 1)2 and y = ´x2 + 2x + 1.
x y y = (x ´ 1)2 y = ´x2 + 2x + 1 Q: Using symmetry, find the centroid of the region inside the unit circle, centred on the origin, and outside a rectangle, also centred on the origin, with width 1 and height 0.5.
x y Q: A long, straight, thin rod has a number of weights attached along it. True or false: if it balances at position x, then the mass to the right of x is the same as the mass to the left of x.
Q: A straight rod with negligible mass has the following weights attached to it: • A weight of mass 1 kg, 1m from the left end, 72 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE • a weight of mass 2 kg, 3m from the left end, • a weight of mass 2 kg, 4m from the left end, and • a weight of mass 1 kg, 6m from the left end.
Where is the centre of mass of the weighted rod?
Q: For each picture below, determine whether the centre of mass is to the left of, to the right of, or along the line x = a, or whether there is not enough information to tell. The shading of a region indicates density: darker shading corresponds to a denser area. In part (d), the right hand side of the right hand B ˆ A rectangle has x = 2a.
x y a (a) x y a (b) a x y (c) A A B B a x y (d) (e) A A/2 B 2B a x y Q: Tank A is spherical, of radius 1 metre, and filled completely with water. The bottom of tank A is three metres above the ground, where Tank B sits. Tank B is tall and rectangular, with base dimensions 2 metres by 1 metre, and empty. Calculate the work done by gravity to drain all the water from Tank A to Tank B by modelling the situation as a point mass, of the same mass as the water, being moved from the height of the centre of mass of A to the height of the centre of mass of the water after it has been moved to B.
73 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE 2 m 1 m B 1 m A 3 m You may use 1000 kg/m3 for the density of water, and g = 9.8 m/sec2 for the acceleration due to gravity.
Q: Let S be the region bounded above by y = 1 x and and below by the x-axis, 1 ď x ď 3. Let R be a rod with density ρ(x) = 1 x at position x, 1 ď x ď 3.
(a) What is the area of a thin slice of S at position x with width dx?
(b) What is the mass of a small piece of R at position x with length dx?
(c) What is the total area of S?
(d) What is the total mass of R?
(e) What is the x-coordinate of the centroid of S?
(f) What is the centre of mass of R?
In Questions 8 through 10, you will derive the formulas for the centre of mass of a rod of variable density, and the centroid of a two-dimensional region using vertical slices (Equations 2.3.2 and 2.3.3 in the CLP-2 text).
Knowing the equations by heart will allow you to answer many questions in this section; understanding where they came from will you allow to generalize their ideas to answer even more questions.
74 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE Q: Suppose R is a straight, thin rod with density ρ(x) at a position x. Let the left endpoint of R lie at x = a, and the right endpoint lie at x = b.
(a) To approximate the centre of mass of R, imagine chopping it into n pieces of equal length, and approximating the mass of each piece using the density at its midpoint.
Give your approximation for the centre of mass in sigma notation.
m1 m2 mn a b (b) Take the limit as n goes to infinity of your approximation in part (a), and express the result using a definite integral.
Q: Suppose S is a two-dimensional object and at (horizontal) position x its height is T(x) ´ B(x). Its leftmost point is at position x = a, and its rightmost point is at position x = b.
To approximate the x-coordinate of the centroid of S, we imagine it as a straight, thin rod R, where the mass of R from a ď x ď b is equal to the area of S from a ď x ď b.
(a) If S is the sheet shown below, sketch R as a rod with the same horizontal length, shaded darker when R is denser, and lighter when R is less dense.
x y T(x) B(x) a b (b) If we cut S into strips of very small width dx, what is the area of the strip at position x?
(c) Using your answer from (b), what is the density ρ(x) of R at position x?
(d) Using your result from Question 8(b), give the x-coordinate of the centroid of S. Your answer will be in terms of a, b, T(X), and B(x).
Q: Suppose S is flat sheet with uniform density, and at (horizontal) position x its height is T(x) ´ B(x). Its leftmost point is at position x = a, and its rightmost point is at position x = b.
To approximate the y-coordinate of the centroid of S, we imagine it as a straight, thin, vertical rod R. We slice S into thin, vertical strips, and model these as weights on R with: • position y on R, where y is the centre of mass of the strip, and • mass in R equal to the area of the strip in S.
(a) If S is the sheet shown below, slice it into a number of vertical pieces of equal length, 75 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE approximated by rectangles. For each rectangle, mark its centre of mass. Sketch R as a rod with the same vertical height, with weights corresponding to the slices you made of S.
x y T(x) B(x) a b a1 b1 (b) Imagine a thin strip of S at position x, with thickness dx. What is the area of the strip? What is the y-value of its centre of mass?
(c) Recall the centre of mass of a rod with n weights of mass Mi at position yi is given by n ř i=1 (Mi ˆ yi) n ř i=1 Mi Considering the limit of this formula as n goes to infinity, give the y-coordinate of the centre of mass of S.
Q11: Express the x–coordinate of the centroid of the triangle with vertices (´1, ´3), (´1, 3), and (0, 0) in terms of a definite integral. Do not evaluate the integral.
§ § Stage 2 Use Equations 2.3.2 and 2.3.3 in the CLP-2 text to find centroids and centres of mass in Questions 12 through 23.
Q: A long, thin rod extends from x = 0 to x = 7 metres, and its density at position x is given by ρ(x) = x kg/m. Where is the centre of mass of the rod?
Q: A long, thin rod extends from x = ´3 to x = 10 metres, and its density at position x is given by ρ(x) = 1 1+x2 kg/m. Where is the centre of mass of the rod?
Q14: Find the y-coordinate of the centroid of the region bounded by the curves y = 1, y = ´ex, x = 0 and x = 1. You may use the fact that the area of this region equals e.
Q15: Consider the region bounded by y = 1 ?
16´x2, y = 0, x = 0 and x = 2.
(a) Sketch this region.
(b) Find the y–coordinate of the centroid of this region.
76 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE Q16: Find the centroid of the finite region bounded by y = sin(x), y = cos(x), x = 0, and x = π/4.
Q17: Let A denote the area of the plane region bounded by x = 0, x = 1, y = 0 and y = k ?
1 + x2, where k is a positive constant.
(a) Find the coordinates of the centroid of this region in terms of k and A.
(b) For what value of k is the centroid on the line y = x?
Q18: The region R is the portion of the plane which is above the curve y = x2 ´ 3x and below the curve y = x ´ x2.
(a) Sketch the region R (b) Find the area of R.
(c) Find the x coordinate of the centroid of R.
Q19: Let R be the region where 0 ď x ď 1 and 0 ď y ď 1 1+x2. Find the x–coordinate of the centroid of R.
Q20: Find the centroid of the region below, which consists of a semicircle of radius 3 on top of a rectangle of width 6 and height 2.
y x −2 −1 1 2 3 3 2 1 −1 −2 −3 Q21: Let D be the region below the graph of the curve y = ?
9 ´ 4x2 and above the x-axis.
(a) Using an appropriate integral, find the area of the region D; simplify your answer completely.
(b) Find the centre of mass of the region D; simplify your answer completely. (Assume it has constant density ρ.) Q: The finite region S is bounded by the lines y = arcsin x, y = arcsin(2 ´ x), and y = ´π 2 . Find the centroid of S.
Q: Calculate the centroid of the figure bounded by the curves y = ex, y = 3(x ´ 1), y = 0, x = 0, and x = 2.
77 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE § § Stage 3 Q24: Find the y-coordinate of the centre of mass of the (infinite) region lying to the right of the line x = 1, above the x–axis, and below the graph of y = 8/x3.
Q25: Let A be the region to the right of the y-axis that is bounded by the graphs of y = x2 and y = 6 ´ x.
(a) Find the centroid of A, assuming it has constant density ρ = 1. The area of A is 22 3 (you don’t have to show this).
(b) Write down an expression, using horizontal slices (disks), for the volume obtained when the region A is rotated around the y-axis. Do not evaluate any integrals; simply write down an expression for the volume.
Q26: (a) Find the y–coordinate of the centroid of the region bounded by y = ex, x = 0, x = 1, and y = ´1.
(b) Calculate the volume of the solid generated by rotating the region from part (a) about the line y = ´1.
Q: Suppose a rectangle has width 4 m, height 3 m, and its density x metres from its left edge is x2 kg/m2. Find the centre of mass of the rectangle.
x y 4 3 Q: Suppose a circle of radius 3 m has density (2 + y) kg/m2 at any point y metres above its bottom. Find the centre of mass of the circle.
3 ´3 x y 78 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE Q: A right circular cone of uniform density has base radius r m and height h m. We want to find its centre of mass. By symmetry, we know that the centre of mass will occur somewhere along the straight vertical line through the tip of the cone and the centre of its base. The only question is the height of the centre of mass.
x y h r We will model the cone as a rod R with height h, such that the mass of the section of the rod from position a to position b is the same as the volume of the cone from height a to height b. (You can imagine that the cone is an umbrella, and we’ve closed it up to look like a cane.) x y h r h 0 (a) Using this model, calculate how high above the base of the cone its centre of mass is.
(b) If we cut off the top h ´ k metres of the cone (leaving an object of height k), how high above the base is the new centre of mass?
This analogy isn’t exact: if the cone were an umbrella, closing it would move the outside fabric verti-cally. A more accurate, but less familiar, image might be vacuum-wrapping an umbrella, watching it shrivel towards the middle but not move vertically.
Q: An hourglass is shaped like two identical truncated cones attached together. Their base radius is 5 cm, the height of the entire hourglass is 18 cm, and the radius at the thinnest point is .5 cm. The hourglass contains sand that fills up the bottom 6 cm when 79 APPLICATIONS OF INTEGRATION 2.3 CENTRE OF MASS AND TORQUE it’s settled, with mass 600 grams and uniform density. We want to know the work done flipping the hourglass smoothly, so the sand settles into a truncated, inverted-cone shape before it starts to fall down.
18 cm 5 cm 0.5 6 cm 8.8 cm Using the methods of Section 2.1 to calculate the work done would be quite tedious.
Instead, we will model the sand as a point of mass 0.6 kg, being lifted from the centre of mass of its original position to the centre of mass of its upturned position. Using the results of Question 29, how much work was done on the sand?
To simplify your calculation, you may assume that the height of the upturned sand (that is, the distance from the skinniest part of the hourglass to the top of the sand) is 8.8 cm.
(Actually, it’s 3 ?
937 ´ 1 « 8.7854 cm.) So, the top 0.2 cm of the hourglass is empty.
80 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS Q: Tank A is in the shape of half a sphere of radius 1 metre, with its flat face resting on the ground, and is completely filled with water. Tank B is empty and rectangular, with a square base of side length 1 m and a height of 3 m.
1 m A 1 m B 3 m (a) To pump the water from Tank A to Tank B, we need to pump all the water from Tank A to a height of 3 m. How much work is done to pump all the water from Tank A to a height of 3 m? You may model the water as a point mass, originally situated at the centre of mass of the full Tank A.
(b) Suppose we could move the water from Tank A directly to its final position in Tank B without going over the top of Tank B. (For example, maybe tank A is elastic, and Tank B is just Tank A after being smooshed into a different form.) How much work is done pumping the water? (That is, how much work is done moving a point mass from the centre of mass of Tank A to the centre of mass of Tank B?) (c) What percentage of work from part (a) was “wasted” by pumping the water over the top of Tank B, instead of moving it directly to its final position?
You may assume that the only work done is against the acceleration due to gravity, g = 9.8 m/sec2, and that the density of water is 1000 kg/m3.
Remark: the answer from (b) is what you might think of as the net work involved in pumping the water from Tank A to Tank B. When work gets “wasted,” the pump does some work pumping water up, then gravity does equal and opposite work bringing the water back down.
Q: Let R be the region bounded above by y = 2x sin(x2) and below by the x-axis, 0 ď x ď b π 2 . Give an approximation of the x-value of the centroid of R with error no more than 1 100.
You may assume without proof that ˇ ˇ ˇ d4 dx4 ␣ 2x2 sin(x2) (ˇ ˇ ˇ ď 415 over the interval h 0, b π 2 i .
2.4IJ Separable Differential Equations Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
Recall that we are using log x to denote the logarithm of x with base e. In other courses it 81 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS is often denoted ln x.
§ § Stage 1 Q: Below are pairs of functions y = f (x) and differential equations. For each pair, decide whether the function is a solution of the differential equation.
function differential equation (a) y = 5(ex ´ 3x2 ´ 6x ´ 6) dy dx = y + 15x2 (b) y = ´2 x2 + 1 y1(x) = xy2 (c) y = x3/2 + x dy dx 2 + dy dx = y Q: Following Definition 2.4.1 in the CLP-2 text, a separable differential equation has the form dy dx(x) = f (x) g y(x) .
Show that each of the following equations can be written in this form, identifying f (x) and g(y).
(a) 3y dy dx = x sin y (b) dy dx = ex+y (c) dy dx + 1 = x (d) dy dx 2 ´ 2x dy dx + x2 = 0 Q: Suppose we have the following functions: • y is a differentiable function of x • f is a function of x, with ş f (x) dx = F(x) • g is a nonzero function of y, with ş 1 g(y) dy = G(y) = G(y(x)).
In the work below, we set up a solution to the separable differential equation dy dx = f (x) g(y) = f (x) g(y(x)) without using the mnemonic of Equation 2.4.1 in the CLP-2 text.
By deleting some portion of our work, we can create the solution as it would look using the mnemonic. What portion can be deleted?
Remark: the purpose of this exercise is to illuminate what, exactly, the mnemonic is a shortcut for. Despite its peculiar look, it agrees with what we already know about integration.
82 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS dy dx = f (x)g(y(x)) Since g(y(x)) is a nonzero function, we can divide both sides by it.
1 g(y(x)) ¨ dy dx = f (x) If these functions of x are the same, then they have the same antiderivative with respect to x.
ż 1 g(y(x)) ¨ dy dx dx = ż f (x) dx The left integral is in the correct form for a change of variables to y. To make this easier to see, we’ll use a u-substitution, since it’s a little more familiar than a y-substitution. If u = y, then du dx = dy dx, so du = dy dxdx.
ż 1 g(u) du = ż f (x) dx Since u was just the same as y, again for cosmetic reasons, we can swap it back. (Formally, you could have skipped the step above–we just included it to be extra clear that we’re not using any integration techniques we haven’t seen before.) ż 1 g(y) dy = ż f (x) dx We’re given the antiderivatives in question.
G(y) + C1 = F(x) + C2 G(y) = F(x) + (C2 ´ C1) where C1 and C2 are arbitrary constants. Then also C2 ´ C1 is an arbitrary constant, so we might as well call it C.
G(y) = F(x) + C Q: Suppose y = f (x) is a solution to the differential equation dy dx = xy.
True or false: f (x) + C is also a solution, for any constant C.
Q: Suppose a function y = f (x) satisfies |y| = Cx, for some constant C ą 0.
(a) What is the largest possible domain of f (x), given the information at hand?
(b) Give an example of function y = f (x) with the following properties, or show that none exists: 83 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS • |y| = Cx, • dy dx exists for all x ą 0, and • y ą 0 for some values of x, and y ă 0 for others.
Q: Express the following sentence1 as a differential equation. You don’t have to solve the equation.
About 0.3 percent of the total quantity of morphine in the bloodstream is eliminated every minute.
Q: Suppose a particular change is occurring in a language, from an old form to a new form. Let p(t) be the proportion (measured as a number between 0, meaning none, and 1, meaning all) of the time that speakers use the new form. Piotrowski’s law† predicts the following.
Use of the new form over time spreads at a rate that is proportional to the product of the proportion of the new form and the proportion of the old form.
Express this as a differential equation. You do not need to solve the differential equation.
An example is the change in German from “wollt” to “wollst” for the second-person conjugation of the verb “wollen.” This example is provided by the site Laws in Quantitative Linguistics, “Change in Language” accessed 18 August 2017.
† Piotrowski’s law is paraphrased from the page Piotrowski-Gesetz on Glottopedia,
glottopedia.org/index.php/Piotrowski-Gesetz, accessed 18 August 2017. According to this source, the law was based on work by the married couple R. G. Piotrowski and A. A. Piotrowskaja, later generalized by G. Altmann.
1 The sentence is paraphrased from the Pharmakokinetics website of Universit´ e de Lausanne, “Elimina-tion Kinetics,” at . The half-life of morphine is given on the same website at
id=85 . Accessed 12 August 2017.
84 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS Q: Consider the differential equation y1 = y 2 ´ 1.
(a) When y = 0, what is y1?
(b) When y = 2, what is y1?
(c) When y = 3, what is y1?
(d) On the axes below, interpret the marks we have made, and use them to sketch a possible solution to the differential equation.
x y 1 1 Q: Consider the differential equation y1 = y ´ x 2.
(a) If y(1) = 0, what is y1(1)?
(b) If y(1) = 2, what is y1(1)?
(c) If y(1) = ´2, what is y1(1)?
(d) Draw a sketch similar to that of Question 8(d) showing the derivatives of y at the points with integer values for x in [0, 6] and y in [´3, 3].
(e) Sketch a possible graph of y.
§ § Stage 2 Q10: Find the solution to the separable initial value problem: dy dx = 2x ey , y(0) = log 2 Express your solution explicitly as y = y(x).
Q11: Find the solution y(x) of dy dx = xy x2 + 1, y(0) = 3.
Q12: Solve the differential equation y1(t) = e y 3 cos t. You should express the solution 85 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS y(t) in terms of t explicitly.
Q13: Solve the differential equation dy dx = xex2´log(y2) Q14: Let y = y(x). Find the general solution of the differential equation y1 = xey.
Q15: Find the solution to the differential equation yy1 ex ´ 2x = 1 y that satisfies y(0) = 3.
Solve completely for y as a function of x.
Q16: Find the function y = f (x) that satisfies dy dx = ´xy3 and f (0) = ´1 4 Q17: Find the function y = y(x) that satisfies y(1) = 4 and dy dx = 15x2 + 4x + 3 y Q18: Find the solution y(x) of y1 = x3y with y(0) = 1.
Q19: Solve the initial value problem xdy dx + y = y2 y(1) = ´1 Q20: A function f (x) is always positive, has f (0) = e and satisfies f 1(x) = x f (x) for all x. Find this function.
Q21: Solve the following initial value problem: dy dx = 1 (x2 + x)y y(1) = 2 Q22: Find the solution of the differential equation 1 + a y2 ´ 4 tan x y1 = sec x y that satisfies y(0) = 2. You don’t have to solve for y in terms of x.
86 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS Q23: The fish population in a lake is attacked by a disease at time t = 0, with the result that the size P(t) of the population at time t ě 0 satisfies dP dt = ´k ?
P where k is a positive constant. If there were initially 90,000 fish in the lake and 40,000 were left after 6 weeks, when will the fish population be reduced to 10,000?
Q24: An object of mass m is projected straight upward at time t = 0 with initial speed v0. While it is going up, the only forces acting on it are gravity (assumed constant) and a drag force proportional to the square of the object’s speed v(t). It follows that the differential equation of motion is mdv dt = ´(mg + kv2) where g and k are positive constants. At what time does the object reach its highest point?
Q25: A motor boat is traveling with a velocity of 40 ft/sec when its motor shuts off at time t = 0. Thereafter, its deceleration due to water resistance is given by dv dt = ´k v2 where k is a positive constant. After 10 seconds, the boat’s velocity is 20 ft/sec.
(a) What is the value of k?
(b) When will the boat’s velocity be 5 ft/sec?
Q26: Consider the initial value problem dx dt = k(3 ´ x)(2 ´ x), x(0) = 1, where k is a positive constant. (This kind of problem occurs in the analysis of certain chemical reactions.) (a) Solve the initial value problem. That is, find x as a function of t.
(b) What value will x(t) approach as t approaches +8.
Q27: The quantity P = P(t), which is a function of time t, satisfies the differential equation dP dt = 4P ´ P2 and the initial condition P(0) = 2.
(a) Solve this equation for P(t).
(b) What is P when t = 0.5? What is the limiting value of P as t becomes large?
Q28: An object moving in a fluid has an initial velocity v of 400 m/min. The velocity is decreasing at a rate proportional to the square of the velocity. After 1 minute the velocity is 200 m/min.
87 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS (a) Give a differential equation for the velocity v = v(t) where t is time.
(b) Solve this differential equation.
(c) When will the object be moving at 50 m/min?
§ § Stage 3 Q29: An investor places some money in a mutual fund where the interest is compounded continuously and where the interest rate fluctuates between 4% and 8%.
Assume that the amount of money B = B(t) in the account in dollars after t years satisfies the differential equation dB dt = 0.06 + 0.02 sin t B (a) Solve this differential equation for B as a function of t.
(b) If the initial investment is $1000, what will the balance be at the end of two years?
Q30: An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem B1(t) = aB ´ m for t ě 0, with B(0) = B0 . The constant a reflects the annual interest rate, m is the annual rate of withdrawal, and B0 is the initial balance in the account.
(a) Solve the initial value problem with a = 0.02 and B(0) = B0 = $30, 000. Note that your answer depends on the constant m.
(b) If a = 0.02 and B(0) = B0 = $30, 000, what is the annual withdrawal rate m that ensures a constant balance in the account?
Q31: A certain continuous function y = y(x) satisfies the integral equation y(x) = 3 + ż x 0 y(t)2 ´ 3y(t) + 2 sin t dt (˚) for all x in some open interval containing 0. Find y(x) and the largest interval for which (˚) holds.
Q32: A cylindrical water tank, of radius 3 meters and height 6 meters, is full of water when its bottom is punctured. Water drains out through a hole of radius 1 centimeter. If • h(t) is the height of the water in the tank at time t (in meters) and • v(t) is the velocity of the escaping water at time t (in meters per second) then • Torricelli’s law states that v(t) = a 2gh(t) where g = 9.8 m/sec2. Determine how long it takes for the tank to empty.
Q33: A spherical tank of radius 6 feet is full of mercury when a circular hole of radius 1 inch is opened in the bottom. How long will it take for all of the mercury to drain from the tank?
88 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS Use the value g = 32 feet/sec2. Also use Torricelli’s law, which states when the height of mercury in the tank is h, the speed of the mercury escaping from the tank is v = a 2gh.
Q34: Consider the equation f (x) = 3 + ż x 0 f (t) ´ 1 f (t) ´ 2 dt (a) What is f (0)?
(b) Find the differential equation satisfied by f (x).
(c) Solve the initial value problem determined in (a) and (b).
Q35: A tank 2 m tall is to be made with circular cross–sections with radius r = yp. Here y measures the vertical distance from the bottom of the tank and p is a positive constant to be determined. You may assume that when the tank drains, it obeys Torricelli’s law, that is A(y)dy dt = ´c?y for some constant c where A(y) is the cross–sectional area of the tank at height y. It is desired that the tank be constructed so that the top half (y = 2 to y = 1) takes exactly the same amount of time to drain as the bottom half (y = 1 to y = 0). Determine the value of p so that the tank has this property. Note: it is not possible or necessary to find c for this question.
Q: Suppose f (t) is a continuous, differentiable function and the root mean square of f (t) on [a, x] is equal to the average of f (t) on [a, x] for all x. That is, 1 x ´ a ż x a f (t) d(t) = d 1 x ´ a ż x a f 2(t) dt (˚) You may assume x ą a.
(a) Guess a function f (t) for which the average of f (t) is the same as the root mean square of f (t) on any interval.
(b) Differentiate both sides of the given equation.
(c) Simplify your answer from (b) by using Equation (˚) to replace all terms containing şx a f 2(t) dt with terms containing şx a f (t) dt.
(d) Let Y(x) = şx a f (t) dt, so the equation from (c) becomes a differential equation. Find all functions that satisfy it.
(e) What is f (t)?
89 APPLICATIONS OF INTEGRATION 2.4 SEPARABLE DIFFERENTIAL EQUATIONS Q: Find the function y(x) such that d2y dx2 = 2 y3 ¨ dy dx and if x = ´ 1 16 log 3, then y = 1 and dy dx = 3.
You do not need to solve for y explicitly.
90 SEQUENCES AND SERIES Chapter 3 3.1IJ Sequences Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
91 SEQUENCES AND SERIES 3.1 SEQUENCES § § Stage 1 Q: Assuming the sequences continue as shown, estimate the limit of each sequence from its graph.
x y 1 (a) x y 1 (b) x y 1 (c) Q: Suppose an and bn are sequences, and an = bn for all n ě 100, but an ‰ bn for n ă 100.
True or false: lim nÑ8 an = lim nÑ8 bn.
Q: Let tanu8 n=1, tbnu8 n=1, and tcnu8 n=1, be sequences with lim nÑ8 an = A, lim nÑ8 bn = B, and lim nÑ8 cn = C. Assume A, B, and C are nonzero real numbers.
Evaluate the limits of the following sequences.
(a) an ´ bn cn (b) cn n (c) a2n+5 bn Q: Give an example of a sequence tanu8 n=1 with the following properties: 92 SEQUENCES AND SERIES 3.1 SEQUENCES • an ą 1000 for all n ď 1000, • an+1 ă an for all n, and • lim nÑ8 an = ´2 Q: Give an example of a sequence tanu8 n=1 with the following properties: • an ą 0 for all even n, • an ă 0 for all odd n, • lim nÑ8 an does not exist.
Q: Give an example of a sequence tanu8 n=1 with the following properties: • an ą 0 for all even n, • an ă 0 for all odd n, • lim nÑ8 an exists.
Q: The limits of the sequences below can be evaluated using the squeeze theorem. For each sequence, choose an upper bounding sequence and lower bounding sequence that will work with the squeeze theorem.
(a) an = sin n n (b) bn = n2 en(7 + sin n ´ 5 cos n) (c) cn = (´n)´n Q: Below is a list of sequences, and a list of functions.
(a) Match each sequence ␣ an (8 n=1 to any and all functions f (x) such that f (n) = an for all positive whole numbers n.
(b) Match each sequence ␣ an (8 n=1 to any and all functions f (x) such that lim nÑ8 an = lim xÑ8 f (x).
an = 1 + 1 n f (x) = cos(πx) bn = 1 + 1 |n| g(x) = cos(πx) x cn = e´n h(x) = # x+1 x x is a whole number 1 else dn = (´1)n i(x) = # x+1 x x is a whole number 0 else en = (´1)n n j(x) = 1 ex 93 SEQUENCES AND SERIES 3.1 SEQUENCES Q: Let tanu8 n=1 be a sequence defined by an = cos n.
(a) Give three different whole numbers n that are within 0.1 of an odd integer multiple of π, and find the corresponding values of an.
(b) Give three different whole numbers n such that an is close to 0. Justify your answers.
(c) Give three different whole numbers n such that an is close to 1. Justify your answers.
Remark: this demonstrates intuitively, though not rigorously, why lim nÑ8 cos n is undefined.
We consistently find terms in the sequence that are close to ´1, and also consistently find terms in the sequence that are close to 0. Contrast this to a sequence like ␣ cos(2πn) ( , whose terms are always 1, and whose limit therefore is 1. It is possible to turn the ideas of this question into a rigorous proof that lim nÑ8 cos n is undefined. See the solution.
§ § Stage 2 Q: Determine the limits of the following sequences.
(a) an = 3n2 ´ 2n + 5 4n + 3 (b) bn = 3n2 ´ 2n + 5 4n2 + 3 (c) cn = 3n2 ´ 2n + 5 4n3 + 3 Q: Determine the limit of the sequence an = 4n3 ´ 21 ne + 1 n .
Q: Determine the limit of the sequence bn = 4 ?n + 1 ?
9n + 3.
Q: Determine the limit of the sequence cn = cos(n + n2) n .
Q: Determine the limit of the sequence an = nsin n n2 .
Q: Determine the limit of the sequence dn = e´1/n.
Q: Determine the limit of the sequence an = 1 + 3 sin(n2) ´ 2 sin n n .
Q: Determine the limit of the sequence bn = en 2n + n2.
94 SEQUENCES AND SERIES 3.1 SEQUENCES Q18: Find the limit, if it exists, of the sequence ␣ ak ( , where ak = k! sin3 k (k + 1)!
Q19: Consider the sequence !
(´1)n sin 1 n ) . State whether this sequence converges or diverges, and if it converges give its limit.
Q20: Evaluate lim nÑ8 6n2 + 5n n2 + 1 + 3 cos(1/n2) .
§ § Stage 3 Q21: Find the limit of the sequence " log sin 1 n + log(2n) .
Q: Evaluate lim nÑ8 ha n2 + 5n ´ a n2 ´ 5n i .
Q: Evaluate lim nÑ8 ha n2 + 5n ´ a 2n2 ´ 5 i .
Q: Evaluate the limit of the sequence " n 2 + 1 n 100 ´ 2100 8 n=1 .
Q: Write a sequence tanu8 n=1 whose limit is f 1(a) for a function f (x) that is differentiable at the point a.
Your answer will depend on f and a.
Q: Let tAnu8 n=3 be the area of a regular polygon with n sides, with the distance from the centroid of the polygon to each corner equal to 1.
1 A(3) = 3 ?
3 4 1 A(4) = 2 1 A(5) = 2.5 sin(0.4π) (a) By dividing the polygon into n triangles, give a formula for An.
95 SEQUENCES AND SERIES 3.1 SEQUENCES (b) What is lim nÑ8 An?
Q: Suppose we define a sequence tfnu, which depends on some constant x, as the following: fn(x) = # 1 n ď x ă n + 1 0 else For a fixed constant x ě 1, tfnu is the sequence t0, 0, 0, . . . , 0, 1, 0, . . . , 0, 0, 0, . . .u. The sole nonzero element comes in position k, where k is what we get when we round x down to a whole number. If x ă 1, then the sequence consists of all zeroes.
Since we can plug in different values of x, we can think of fn(x) as a function of sequences: a different x gives you a different sequence. On the other hand, if we imagine fixing n, then fn(x) is just a function, where fn(x) gives the nth term in the sequence corresponding to x.
(a) Sketch the curve y = f2(x).
(b) Sketch the curve y = f3(x).
(c) Define An = ş8 0 fn(x) dx. Give a simple description of the sequence tAnu8 n=1.
(d) Evaluate lim nÑ8 An.
(e) Evaluate lim nÑ8 fn(x) for a constant x, and call the result g(x).
(f) Evaluate ż 8 0 g(x) dx.
Q: Determine the limit of the sequence bn = 1 + 3 n + 5 n2 n .
Q: A sequence ␣ an (8 n=1 of real numbers satisfies the recursion relation an+1 = an + 8 3 for n ě 1.
(a) Suppose a1 = 4. What is lim nÑ8 an?
(b) Find x if x = x + 8 3 .
(c) Suppose a1 = 1. Show that lim nÑ8 an = L, where L is the solution to equation above.
Q: Zipf’s Law applied to word frequency can be phrased as follows: The most-used word in a language is used n times as frequently as the n-th most word used in a language.
(a) Suppose the sequence tw1, w2, w3, . . .u is a list of all words in a language, where wn is the word that is the nth most frequently used. Let fn be the frequency of word wn. Is tf1, f2, f3, . . .u an increasing sequence or a decreasing sequence?
(b) Give a general formula for fn, treating f1 as a constant.
96 SEQUENCES AND SERIES 3.2 SERIES (c) Suppose in a language, w1 (the most frequently used word) has frequency 6%. If the language follows Zipf’s Law, then what frequency does w3 have?
(d) Suppose f6 = 0.3% for a language following Zipf’s law. What is f10?
(e) The word “the” is the most-used word in contemporary American English. In a collection of about 450 million words, “the” appeared 22,038,615 times. The second-most used word is “be,” followed by “and.” About how many usages of these words do you expect in the same collection of 450 million words?
Sources: Zipf’s word frequency law in natural language: A critical review and future directions, Steven T. Piantadosi. Psychon Bull Rev. 2014 Oct; 21(5): 1112–1130. Accessed online 11 October 2017 Word Frequency Data, Accessed online 11 October 2017 3.2IJ Series Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Write out the first five partial sums corresponding to the series 8 ÿ n=1 1 n.
You don’t need to simplify the terms.
Q: Every student who comes to class brings their instructor cookies, and leaves them on the instructor’s desk. Let Ck be the total number of cookies on the instructor’s desk after the kth student comes.
If C11 = 20, and C10 = 17, how many cookies did the 11th student bring to class?
Q: Suppose the sequence of partial sums of the series 8 ÿ n=1 an is tSNu = " N N + 1 .
(a) What is tanu?
(b) What is lim nÑ8 an?
(c) Evaluate 8 ÿ n=1 an.
Q: Suppose the sequence of partial sums of the series 8 ÿ n=1 an is tSNu = " (´1)N + 1 N .
What is tanu?
97 SEQUENCES AND SERIES 3.2 SERIES Q: Let f (N) be a formula for the Nth partial sum of 8 ÿ n=1 an. (That is, f (N) = SN.) If f 1(N) ă 0 for all N ą 1, what does that say about an?
Questions 6 through 8 invite you to explore geometric sums in a geometric way. This is complementary to than the algebraic method discussed in the text.
Q: Suppose the triangle outlined in red in the picture below has area one.
(a) Express the combined area of the black triangles as a series, assuming the pattern continues forever.
(b) Evaluate the series using the picture (not the formula from your book).
Q: Suppose the square outlined in red in the picture below has area one.
(a) Express the combined area of the black squares as a series, assuming the pattern continues forever.
(b) Evaluate the series using the picture (not the formula from your book).
98 SEQUENCES AND SERIES 3.2 SERIES Q: In the style of Questions 6 and 7, draw a picture that represents 8 ÿ n=1 1 3n as an area.
Q: Evaluate 100 ÿ n=0 1 5n .
Q: Every student who comes to class brings their instructor cookies, and leaves them on the instructor’s desk. Let Ck be the total number of cookies on the instructor’s desk after the kth student comes.
If C20 = 53, and C10 = 17, what does C20 ´ C10 = 36 represent?
Q: Evaluate 100 ÿ n=50 1 5n . (Note the starting index.) Q: (a) Starting on day d = 1, every day you give your friend $ 1 d+1, and they give $1 d back to you. After a long time, how much money have you gained by this arrangement?
(b) Evaluate 8 ÿ d=1 1 d ´ 1 (d + 1) .
(c) Starting on day d = 1, every day your friend gives you $(d + 1), and they take $(d + 2) from you. After a long time, how much money have you gained by this arrangement?
(d) Evaluate 8 ÿ d=1 ((d + 1) ´ (d + 2)).
Q: Suppose 8 ÿ n=1 an = A, 8 ÿ n=1 bn = B, and 8 ÿ n=1 cn = C.
Evaluate 8 ÿ n=1 (an + bn + cn+1).
Q: Suppose 8 ÿ n=1 an = A, 8 ÿ n=1 bn = B ‰ 0, and 8 ÿ n=1 cn = C.
True or false: 8 ÿ n=1 an bn + cn = A B + C.
99 SEQUENCES AND SERIES 3.2 SERIES § § Stage 2 Q15: To what value does the series 1 + 1 3 + 1 9 + 1 27 + 1 81 + 1 243 + ¨ ¨ ¨ converge?
Q16: Evaluate 8 ÿ k=7 1 8k Q17: Show that the series 8 ÿ k=1 6 k2 ´ 6 (k + 1)2 converges and find its limit.
Q18: Find the sum of the convergent series 8 ÿ n=3 cos π n ´ cos π n + 1 .
Q19: The nth partial sum of a series 8 ÿ n=1 an is known to have the formula sn = 1 + 3n 5 + 4n.
(a) Find an expression for an, valid for n ě 2.
(b) Show that the series 8 ÿ n=1 an converges and find its value.
Q20: Find the sum of the series 8 ÿ n=2 3 ¨ 4n+1 8 ¨ 5n . Simplify your answer completely.
Q21: Relate the number 0.2¯ 3 = 0.233333 . . . to the sum of a geometric series, and use that to represent it as a rational number (a fraction or combination of fractions, with no decimals).
Q22: Express 2.656565 . . . as a rational number, i.e. in the form p/q where p and q are integers.
Q23: Express the decimal 0.321 = 0.321321321 . . . as a fraction.
Q24: Find the value of the convergent series 8 ÿ n=2 2n+1 3n + 1 2n ´ 1 ´ 1 2n + 1 Simplify your answer completely.
100 SEQUENCES AND SERIES 3.2 SERIES Q25: Evaluate 8 ÿ n=1 1 3 n + ´ 2 5 n´1 Q26: Find the sum of the series 8 ÿ n=0 1 + 3n+1 4n .
Q: Evaluate 8 ÿ n=5 log n ´ 3 n .
Q: Evaluate 8 ÿ n=2 2 n ´ 1 n + 1 ´ 1 n ´ 1 .
§ § Stage 3 Q: An infinitely long, flat cliff has stones hanging off it, attached to thin wire of negligible mass. Starting at position x = 1, every metre (at position x, where x is some whole number) the stone has mass 1 4x kg and is hanging 2x metres below the top of the cliff.
1 2 3 4 5 6 How much work in joules does it take to pull up all the stones to the top of the cliff?
You may use g = 9.8 m/sec2.
101 SEQUENCES AND SERIES 3.2 SERIES Q: Find the combined volume of an infinite collection of spheres, where for each whole number n = 1, 2, 3, . . . there is exactly one sphere of radius 1 πn .
Q: Evaluate 8 ÿ n=3 sin2 n 2n + cos2(n + 1) 2n+1 !
.
Q: Suppose a series 8 ÿ n=1 an has sequence of partial sums tSNu, and the series 8 ÿ N=1 SN has sequence of partial sums tSMu = # M ÿ N=1 SN + .
If SM = M + 1 M , what is an?
Q: Create a bullseye using the following method: Starting with a red circle of area 1, divide the radius into thirds, creating two rings and a circle. Colour the middle ring blue.
Continue the pattern with the inside circle: divide its radius into thirds, and colour the middle ring blue.
Step 1 Step 2 Continue in this way indefinitely: dividing the radius of the innermost circle into thirds, creating two rings and another circle, and colouring the middle ring blue.
102 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS What is the area of the red portion?
3.3IJ Convergence Tests Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Select the series below that diverge by the divergence test.
(A) 8 ÿ n=1 1 n (B) 8 ÿ n=1 n2 n + 1 (C) 8 ÿ n=1 sin n (D) 8 ÿ n=1 sin(πn) Q: Select the series below whose terms satisfy the conditions to apply the integral test.
(A) 8 ÿ n=1 1 n (B) 8 ÿ n=1 n2 n + 1 (C) 8 ÿ n=1 sin n (D) 8 ÿ n=1 sin n + 1 n2 Q: Suppose there is some threshold after which a person is considered old, and before which they are young.
Let Olaf be an old person, and let Yuan be a young person.
103 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS (a) Suppose I am older than Olaf. Am I old?
(b) Suppose I am younger than Olaf. Am I old?
(c) Suppose I am older than Yuan. Am I young?
(d) Suppose I am younger than Yuan. Am I young?
Q: Below are graphs of two sequences with positive terms. Assume the sequences continue as shown. Fill in the table with conclusions that can be made from the direct comparison test, if any.
x y n if ř an converges if ř an diverges and if tanu is the red series then ř bn then ř bn and if tanu is the blue series then ř bn then ř bn Q: For each pair of series below, decide whether the second series is a valid comparison series to determine the convergence of the first series, using the direct comparison test and/or the limit comparison test.
(a) 8 ÿ n=10 1 n ´ 1, compared to the divergent series 8 ÿ n=10 1 n.
(b) 8 ÿ n=1 sin n n2 + 1, compared to the convergent series 8 ÿ n=1 1 n2.
(c) 8 ÿ n=5 n3 + 5n + 1 n6 ´ 2 , compared to the convergent series 8 ÿ n=5 1 n3.
(d) 8 ÿ n=5 1 ?n, compared to the divergent series 8 ÿ n=5 1 4 ?n.
Q: Suppose an is a sequence with lim nÑ8 an = 1 2. Does 8 ÿ n=7 an converge or diverge, or is it not possible to determine this from the information given? Why?
104 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS Q: What flaw renders the following reasoning invalid?
Q: Determine whether 8 ÿ n=1 sin n n converges or diverges.
A: First, we will evaluate lim nÑ8 sin n n .
• Note ´1 n ď sin n n ď 1 n for n ě 1.
• Note also that lim nÑ8 ´1 n = lim nÑ8 1 n = 0.
• Therefore, by the Squeeze Theorem, lim nÑ8 sin n n = 0 as well.
So, by the divergence test, 8 ÿ n=1 sin n n converges.
Q: What flaw renders the following reasoning invalid?
Q: Determine whether 8 ÿ n=1 (sin(πn) + 2) converges or diverges.
A: We use the integral test. Let f (x) = sin(πx) + 2. Note f (x) is always positive, since sin(x) + 2 ě ´1 + 2 = 1. Also, f (x) is continuous.
ż 8 1 [sin(πx) + 2]dx = lim bÑ8 ż b 1 [sin(πx) + 2]dx = lim bÑ8 " ´ 1 π cos(πx) + 2x ˇ ˇ ˇ ˇ b 1 # = lim bÑ8 ´ 1 π cos(πb) + 2b + 1 π(´1) ´ 2 = 8 By the integral test, since the integral diverges, also 8 ÿ n=1 (sin(πn) + 2) diverges.
Q: What flaw renders the following reasoning invalid?
Q: Determine whether the series 8 ÿ n=1 2n+1n2 en + 2n converges or diverges.
A: We want to compare this series to the series 8 ÿ n=1 2n+1 en . Note both this series and the series in the question have positive terms.
First, we find that 2n+1n2 en + 2n ą 2n+1 en when n is sufficiently large. The justification for this claim is as follows: • We note that en(n2 ´ 1) ą n2 ´ 1 ą 2n for n sufficiently large.
105 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS • Therefore, en ¨ n2 ą en + 2n • Therefore, 2n+1 ¨ en ¨ n2 ą 2n+1(en + 2n) • Since en + 2n and en are both expressions that work out to be positive for the values of n under consideration, we can divide both sides of the inequality by these terms without having to flip the inequality. So, 2n+1n2 en + 2n ą 2n+1 en .
Now, we claim 8 ÿ n=1 2n+1 en converges.
Note 8 ÿ n=1 2n+1 en = 2 8 ÿ n=1 2n en = 2 8 ÿ n=1 2 e n . This is a geometric series with r = 2 e.
Since 2/e ă 1, the series converges.
Now, by the Direct Comparison Test, we conclude that 8 ÿ n=1 2n+1n2 en + 2n converges.
Q: Which of the series below are alternating?
(A) 8 ÿ n=1 sin n (B) 8 ÿ n=1 cos(πn) n3 (C) 8 ÿ n=1 7 (´n)2n (D) 8 ÿ n=1 (´2)n 3n+1 Q: Give an example of a convergent series for which the ratio test is inconclusive.
Q: Imagine you’re taking an exam, and you momentarily forget exactly how the inequality in the ratio test works. You remember there’s a ratio, but you don’t remember which term goes on top; you remember there’s something about the limit being greater than or less than one, but you don’t remember which way implies convergence.
Explain why lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ ą 1 or, equivalently, lim nÑ8 ˇ ˇ ˇ ˇ an an+1 ˇ ˇ ˇ ˇ ă 1 should mean that the sum 8 ř n=1 an diverges (rather than converging).
Q: Give an example of a series 8 ÿ n=a an, with a function f (x) such that f (n) = an for all whole numbers n, such that: • ż 8 a f (x) dx diverges, while 106 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS • 8 ÿ n=a an converges.
Q14: Suppose that you want to use the Limit Comparison Test on the series 8 ÿ n=0 an where an = 2n + n 3n + 1. Write down a sequence tbnu such that lim nÑ8 an bn exists and is nonzero. (You don’t have to carry out the Limit Comparison Test) Q15: Decide whether each of the following statements is true or false. If false, provide a counterexample. If true provide a brief justification.
(a) If lim nÑ8 an = 0, then 8 ř n=1 an converges.
(b) If lim nÑ8 an = 0, then 8 ř n=1 (´1)nan converges.
(c) If 0 ď an ď bn and 8 ř n=1 bn diverges, then 8 ř n=1 an diverges.
§ § Stage 2 Q16: Does the series 8 ÿ n=2 n2 3n2 + ?n converge?
Q17: Determine, with explanation, whether the series 8 ÿ n=1 5k 4k + 3k converges or di-verges.
Q18: Determine whether the series 8 ÿ n=0 1 n + 1 2 is convergent or divergent. If it is convergent, find its value.
Q: Does the following series converge or diverge?
8 ÿ k=1 1 ?
k ?
k + 1 Q: Evaluate the following series, or show that it diverges: 8 ÿ k=30 3(1.001)k.
Q: Evaluate the following series, or show that it diverges: 8 ÿ n=3 ´1 5 n .
107 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS Q: Does the following series converge or diverge?
8 ÿ n=7 sin(πn) Q: Does the following series converge or diverge?
8 ÿ n=7 cos(πn) Q: Does the following series converge or diverge?
8 ÿ k=1 ek k!.
Q: Evaluate the following series, or show that it diverges: 8 ÿ k=0 2k 3k+2.
Q: Does the following series converge or diverge?
8 ÿ n=1 n!n!
(2n)!.
Q: Does the following series converge or diverge?
8 ÿ n=1 n2 + 1 2n4 + n.
Q28: Show that the series 8 ÿ n=3 5 n(log n)3/2 converges.
Q29: Find the values of p for which the series 8 ÿ n=2 1 n(log n)p converges.
Q30: Does 8 ÿ n=1 e´?n ?n converge or diverge?
Q31: Use the comparison test (not the limit comparison test) to show whether the series 8 ÿ n=2 ?
3n2 ´ 7 n3 converges or diverges.
Q32: Determine whether the series 8 ÿ k=1 3 ?
k4 + 1 ?
k5 + 9 converges.
Q33: Does 8 ÿ n=1 n42n/3 (2n + 7)4 converge or diverge?
Q34: Determine, with explanation, whether each of the following series converge or 108 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS diverge.
(a) 8 ÿ n=1 1 ?
n2 + 1 (b) 8 ÿ n=1 n cos(nπ) 2n Q35: Determine whether the series 8 ÿ k=1 k4 ´ 2k3 + 2 k5 + k2 + k converges or diverges.
Q36: Determine whether each of the following series converge or diverge.
(a) 8 ÿ n=2 n2 + n + 1 n5 ´ n (b) 8 ÿ m=1 3m + sin ?m m2 Q: Evaluate the following series, or show that it diverges: 8 ÿ n=5 1 en .
Q38: Determine whether the series 8 ÿ n=2 6 7n is convergent or divergent. If it is conver-gent, find its value.
Q39: Determine, with explanation, whether each of the following series converge or diverge.
(a) 1 + 1 3 + 1 5 + 1 7 + 1 9 + ¨ ¨ ¨ .
(b) 8 ÿ n=1 2n + 1 22n+1 Q40: Determine, with explanation, whether each of the following series converges or diverges.
(a) 8 ÿ k=2 3 ?
k k2 ´ k.
(b) 8 ÿ k=1 k1010k(k!)2 (2k)!
.
(c) 8 ÿ k=3 1 k(log k)(log log k).
109 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS Q41: Determine whether the series 8 ÿ n=1 n3 ´ 4 2n5 ´ 6n is convergent or divergent.
Q42: What is the smallest value of N such that the partial sum N ÿ n=1 (´1)n n ¨ 10n approxi-mates 8 ÿ n=1 (´1)n n ¨ 10n within an accuracy of 10´6?
Q43: It is known that 8 ÿ n=1 (´1)n´1 n2 = π2 12 (you don’t have to show this). Find N so that SN, the Nth partial sum of the series, satisfies |π2 12 ´ SN| ď 10´6. Be sure to say why your method can be applied to this particular series.
Q44: The series 8 ÿ n=1 (´1)n+1 (2n + 1)2 converges to some number S (you don’t have to prove this). According to the Alternating Series Estimation Theorem, what is the smallest value of N for which the Nth partial sum of the series is at most 1 100 away from S? For this value of N, write out the Nth partial sum of the series.
§ § Stage 3 Q45: Determine, with explanation, whether the following series converge or diverge.
(a) 8 ÿ n=1 nn 9nn!
(b) 8 ÿ n=1 1 nlog n Q46: (a) Prove that ż 8 2 x + sin x 1 + x2 dx diverges.
(b) Explain why you cannot conclude that 8 ÿ n=1 n + sin n 1 + n2 diverges from part (a) and the Integral Test.
(c) Determine, with explanation, whether 8 ÿ n=1 n + sin n 1 + n2 converges or diverges.
Q47: Show that 8 ÿ n=1 e´?n ?n converges and find an interval of length 0.05 or less that contains its exact value.
Q48: Suppose that the series 8 ÿ n=1 an converges and that 1 ą an ě 0 for all n. Prove that 110 SEQUENCES AND SERIES 3.3 CONVERGENCE TESTS the series 8 ÿ n=1 an 1 ´ an also converges.
Q49: Suppose that the series 8 ř n=0 (1 ´ an) converges, where an ą 0 for n = 0, 1, 2, 3, ¨ ¨ ¨ .
Determine whether the series 8 ř n=0 2nan converges or diverges.
Q50: Assume that the series 8 ÿ n=1 nan ´ 2n + 1 n + 1 converges, where an ą 0 for n = 1, 2, ¨ ¨ ¨ . Is the following series ´ log a1 + 8 ÿ n=1 log an an+1 convergent? If your answer is NO, justify your answer. If your answer is YES, evaluate the sum of the series ´ log a1 + 8 ř n=1 log an an+1 .
Q51: Prove that if an ě 0 for all n and if the series 8 ÿ n=1 an converges, then the series 8 ÿ n=1 a2 n also converges.
A number of phenomena roughly follow a distribution called Zipf’s law. We discuss some of these in Ques-tions 52 and 53.
Q: Suppose the frequency of word use in a language has the following pattern: The n-th most frequently used word accounts for α n percent of the total words used.
So, in a text of 100 words, we expect the most frequently used word to appear α times, while the second-most-frequently used word should appear about α 2 times, and so on.
If books written in this language use 20, 000 distinct words, then the most commonly used word accounts for roughly what percentage of total words used?
Q: Suppose the sizes of cities in a country adhere to the following pattern: if the largest city has population α, then the n-th largest city has population α n.
If the largest city in this country has 2 million people and the smallest city has 1 person, then the population of the entire country is ř2ˆ106 n=1 2ˆ106 n . (For many n’s in this sum 2ˆ106 n is not an integer. Ignore that.) Evaluate this sum approximately, with an error of no more than 1 million people.
111 SEQUENCES AND SERIES 3.4 ABSOLUTE AND CONDITIONAL CONVERGENCE 3.4IJ Absolute and Conditional Convergence Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q1: Decide whether the following statement is true or false. If false, provide a counterexample. If true provide a brief justification.
If 8 ÿ n=1 (´1)n+1bn converges, then 8 ÿ n=1 bn also converges.
Q: Describe the series 8 ÿ n=1 an based on whether 8 ÿ n=1 an and 8 ÿ n=1 |an| converge or diverge, using vocabulary from this section where possible.
ř an converges ř an diverges ř |an| converges ř |an| diverges § § Stage 2 Q3: Determine whether the series 8 ÿ n=1 (´1)n 9n + 5 is absolutely convergent, conditionally convergent, or divergent; justify your answer.
Q4: Determine whether the series 8 ÿ n=1 (´1)2n+1 1 + n is absolutely convergent, conditionally convergent, or divergent.
Q5: The series 8 ÿ n=1 (´1)n´1 1 + 4n 3 + 22n either: converges absolutely; converges conditionally; diverges; or none of the above. Determine which is correct.
112 SEQUENCES AND SERIES 3.4 ABSOLUTE AND CONDITIONAL CONVERGENCE Q6: Does the series 8 ÿ n=5 ?n cos n n2 ´ 1 converge conditionally, converge absolutely, or diverge?
Q7: Determine (with justification!) whether the series 8 ÿ n=1 n2 ´ sin n n6 + n2 converges absolutely, converges but not absolutely, or diverges.
Q8: Determine (with justification!) whether the series 8 ÿ n=0 (´1)n(2n)!
(n2 + 1)(n!)2 converges absolutely, converges but not absolutely, or diverges.
Q9: Determine (with justification!) whether the series 8 ÿ n=2 (´1)n n(log n)101 converges absolutely, converges but not absolutely, or diverges.
Q: Show that the series 8 ÿ n=1 sin n n2 converges.
Q: Show that the series 8 ÿ n=1 sin n 4 ´ 1 8 n converges.
Q: Show that the series 8 ÿ n=1 sin2 n ´ cos2 n + 1 2 2n converges.
§ § Stage 3 Q13: Both parts of this question concern the series S = 8 ÿ n=1 (´1)n´124n2e´n3.
(a) Show that the series S converges absolutely.
(b) Suppose that you approximate the series S by its fifth partial sum S5. Give an upper bound for the error resulting from this approximation.
113 SEQUENCES AND SERIES 3.5 POWER SERIES Q: You may assume without proof the following: 8 ÿ n=0 (´1)n (2n)! = cos(1) Using this fact, approximate cos 1 as a rational number, accurate to within 1 1000.
Check your answer against a calculator’s approximation of cos(1): what was your actual error?
Q: Let an be defined as an = # ´en/2 if n is prime n2 if n is not prime Show that the series 8 ÿ n=1 an en converges.
3.5IJ Power Series Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Suppose f (x) = 8 ÿ n=0 3 ´ x 4 n . What is f (1)?
Q: Suppose f (x) = 8 ÿ n=1 (x ´ 5)n n! + 2 . Give a power series representation of f 1(x).
Q: Let f (x) = 8 ÿ n=a An(x ´ c)n for some positive constants a and c, and some sequence of constants tAnu. For which values of x does f (x) definitely converge?
Q: Let f (x) be a power series centred at c = 5. If f (x) converges at x = ´1, and diverges at x = 11, what is the radius of convergence of f (x)?
114 SEQUENCES AND SERIES 3.5 POWER SERIES § § Stage 2 Q5: (a) Find the radius of convergence of the series 8 ÿ k=0 (´1)k2k+1xk (b) You are given the formula for the sum of a geometric series, namely: 1 + r + r2 + ¨ ¨ ¨ = 1 1 ´ r, |r| ă 1 Use this fact to evaluate the series in part (a).
Q6: Find the radius of convergence for the power series 8 ÿ k=0 xk 10k+1(k + 1)!
Q7: Find the radius of convergence for the power series 8 ÿ n=0 (x ´ 2)n n2 + 1 .
Q8: Consider the power series 8 ÿ n=1 (´1)n(x + 2)n ?n , where x is a real number. Find the interval of convergence of this series.
Q9: Find the radius of convergence and interval of convergence of the series 8 ÿ n=0 (´1)n n + 1 x + 1 3 n Q10: Find the interval of convergence for the power series 8 ÿ n=1 (x ´ 2)n n4/5(5n ´ 4).
Q11: Find all values x for which the series 8 ÿ n=1 (x + 2)n n2 converges.
Q12: Find the interval of convergence for the following series.
(a) 8 ÿ n=1 4n n (x ´ 1)n.
115 SEQUENCES AND SERIES 3.5 POWER SERIES (b) 8 ÿ n=1 4n n (2x + 1)n.
(c) 8 ÿ n=1 4n n (2x + 1)2n.
Q13: Find, with explanation, the radius of convergence and the interval of convergence of the power series 8 ÿ n=0 (´1)n (x ´ 1)n 2n(n + 2) Q14: Find the interval of convergence for the series 8 ÿ n=1 (´1)nn2(x ´ a)2n where a is a constant.
Q15: Find the interval of convergence of the following series: (a) 8 ÿ k=1 (x + 1)k k29k .
(b) 8 ÿ k=1 ak(x ´ 1)k, where ak ą 0 for k = 1, 2, ¨ ¨ ¨ and 8 ÿ k=1 ak ak+1 ´ ak+1 ak+2 = a1 a2 .
Q16: Find a power series representation for x3 1 ´ x.
Q: Suppose f 1(x) = 8 ÿ n=0 (x ´ 1)n n + 2 , and ż x 5 f (t)dt = 3x + 8 ÿ n=1 (x ´ 1)n+1 n(n + 1)2 .
Give a power series representation of f (x).
§ § Stage 3 Q18: Determine the values of x for which the series 8 ÿ n=2 xn 32n log n converges absolutely, converges conditionally, or diverges.
116 SEQUENCES AND SERIES 3.5 POWER SERIES Q19: (a) Find the power–series representation for ż 1 1 + x3 dx centred at 0 (i.e. in powers of x).
(b) The power series above is used to approximate ż 1/4 0 1 1 + x3 dx. How many terms are required to guarantee that the resulting approximation is within 10´5 of the exact value?
Justify your answer.
Q20: (a) Show that 8 ÿ n=0 nxn = x (1 ´ x)2 for ´1 ă x ă 1.
(b) Express 8 ÿ n=0 n2xn as a ratio of polynomials. For which x does this series converge?
Q21: Suppose that you have a sequence tbnu such that the series ř8 n=0(1 ´ bn) converges. Using the tests we’ve learned in class, prove that the radius of convergence of the power series 8 ÿ n=0 bnxn is equal to 1.
Q22: Assume ␣ an ( is a sequence such that nan decreases to C as n Ñ 8 for some real number C ą 0 (a) Find the radius of convergence of 8 ÿ n=1 anxn . Justify your answer carefully.
(b) Find the interval of convergence of the above power series, that is, find all x for which the power series in (a) converges. Justify your answer carefully.
Q: An infinitely long, straight rod of negligible mass has the following weights: • At every whole number n, a mass of weight 1 2n at position n, and • a mass of weight 1 3n at position ´n.
At what position is the centre of mass of the rod?
0 1 ´1 1 21 1 31 2 ´2 1 22 1 32 3 ´3 1 23 1 33 117 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q: Let f (x) = 8 ÿ n=0 An(x ´ c)n, for some constant c and a sequence of constants tAnu.
Further, let f (x) have a positive radius of convergence.
If A1 = 0, show that y = f (x) has a critical point at x = c. What is the relationship between the behaviour of the graph at that point and the value of A2?
Q: Evaluate 8 ÿ n=3 n 5n´1.
Q: Find a polynomial that approximates f (x) = log(1 + x) to within an error of 10´5 for all values of x in 0, 1 10 .
Then, use your polynomial to approximate log(1.05) as a rational number.
Q: Find a polynomial that approximates f (x) = arctan x to within an error of 10´5 for all values of x in ´1 4, 1 4 .
3.6IJ Taylor Series Exercises Jump to HINTS, ANSWERS, SOLUTIONS or TABLE OF CONTENTS.
§ § Stage 1 Q: Below is a graph of y = f (x), along with the constant approximation, linear approximation, and quadratic approximation centred at a = 2. Which is which?
x y 2 y = f (x) A B C Q: Suppose T(x) is the Taylor series for f (x) = arctan3 (ex + 7) centred at a = 5. What is T(5)?
118 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q: Below are a list of common functions, and their Taylor series representations.
Match the function to the Taylor series and give the radius of convergence of the series.
function series A.
1 1 ´ x I.
8 ÿ n=0 (´1)n xn+1 n + 1 B. log(1 + x) II.
8 ÿ n=0 (´1)n x2n+1 (2n + 1)!
C. arctan x III.
8 ÿ n=0 (´1)n x2n (2n)!
D. ex IV.
8 ÿ n=0 (´1)n x2n+1 2n + 1 E. sin x V.
8 ÿ n=0 xn F. cos x VI.
8 ÿ n=0 xn n!
Q: (a) Suppose f (x) = 8 ÿ n=0 n2 (n! + 1)(x ´ 3)n for all real x. What is f (20)(3) (the twentieth derivative of f (x) at x = 3)?
(b) Suppose g(x) = 8 ÿ n=0 n2 (n! + 1)(x ´ 3)2n for all real x. What is g(20)(3)?
(c) If h(x) = arctan(5x2) x2 , what is h(20)(0)? What is h(22)(0)?
§ § Stage 2 In Questions 5 through 8, you will create Taylor series from scratch. In practice, it is often preferable to modify an existing series, rather than creating a new one, but you should understand both ways.
Q: Using the definition of a Taylor series, find the Taylor series for f (x) = log(x) centred at x = 1.
Q: Find the Taylor series for f (x) = sin x centred at a = π.
Q: Using the definition of a Taylor series, find the Taylor series for g(x) = 1 x centred at x = 10. What is the interval of convergence of the resulting series?
119 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q: Using the definition of a Taylor series, find the Taylor series for h(x) = e3x centred at x = a, where a is some constant. What is the radius of convergence of the resulting series?
In Questions 9 through 16, practice creating new Taylor series by modifying known Taylor series, rather than creating your series from scratch.
Q9: Find the Maclaurin series for f (x) = 1 2x ´ 1.
Q10: Let 8 ÿ n=0 bnxn be the Maclaurin series for f (x) = 3 x + 1 ´ 1 2x ´ 1, i.e.
8 ÿ n=0 bnxn = 3 x + 1 ´ 1 2x ´ 1. Find bn.
Q11: Find the coefficient c5 of the fifth degree term in the Maclaurin series 8 ÿ n=0 cnxn for e3x.
Q12: Express the Taylor series of the function f (x) = log(1 + 2x) about x = 0 in summation notation.
Q13: The first two terms in the Maclaurin series for x2 sin(x3) are ax5 + bx11 , where a and b are constants. Find the values of a and b.
Q14: Give the first two nonzero terms in the Maclaurin series for ż e´x2 ´ 1 x dx.
Q15: Find the Maclaurin series for ż x4 arctan(2x) dx.
Q16: Suppose that df dx = x 1 + 3x3 and f (0) = 1. Find the Maclaurin series for f (x).
In past chapters, we were only able to exactly evaluate very specific types of series: geometric and telescoping.
In Questions 17 through 25, we expand our range by relating given series to Taylor series.
120 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q17: The Maclaurin series for arctan x is given by arctan x = 8 ÿ n=0 (´1)n x2n+1 2n + 1 which has radius of convergence equal to 1. Use this fact to compute the exact value of the series below: 8 ÿ n=0 (´1)n (2n + 1)3n Q18: Evaluate 8 ÿ n=0 (´1)n n!
.
Q19: Evaluate 8 ÿ k=0 1 ekk! .
Q20: Evaluate the sum of the convergent series 8 ÿ k=1 1 πkk! .
Q21: Evaluate 8 ÿ n=1 (´1)n´1 n 2n .
Q22: Evaluate 8 ÿ n=1 n + 2 n!
en .
Q: Evaluate 8 ÿ n=1 2n n , or show that it diverges.
Q: Evaluate 8 ÿ n=0 (´1)n (2n + 1)!
π 4 2n+1 1 + 22n+1 or show that it diverges.
Q25: (a) Show that the power series 8 ÿ n=0 x2n (2n)! converges absolutely for all real numbers x.
(b) Evaluate 8 ÿ n=0 1 (2n)!.
121 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q: (a) Using the fact that arctan(1) = π 4 , how many terms of the Taylor series for arctangent would you have to add up to approximate π with an error of at most 4 ˆ 10´5?
(b) Example 3.6.13 in the CLP–2 text mentions the formula π = 16 arctan 1 5 ´ 4 arctan 1 239 Using the Taylor series for arctangent, how many terms would you have to add up to approximate π with an error of at most 4 ˆ 10´5?
(c) Assume without proof the following: arctan 1 2 + arctan 1 3 = arctan 3 + 2 2 ¨ 3 ´ 1 Using the Taylor series for arctangent, how many terms would you have to add up to approximate π with an error of at most 4 ˆ 10´5?
Q: Suppose you wanted to approximate the number log(1.5) as a rational number using the Taylor expansion of log(1 + x). How many terms would you need to add to get 10 decimal places of accuracy? (That is, an absolute error less than 5 ˆ 10´11.) Q: Suppose you wanted to approximate the number e as a rational number using the Maclaurin expansion of ex. How many terms would you need to add to get 10 decimal places of accuracy? (That is, an absolute error less than 5 ˆ 10´11.) You may assume without proof that 2 ă e ă 3.
Q: Suppose you wanted to approximate the number log(0.9) as a rational number using the Taylor expansion of log(1 ´ x). Which partial sum should you use to get 10 decimal places of accuracy? (That is, an absolute error less than 5 ˆ 10´11.) Q: Define the hyperbolic sine function as sinh x = ex ´ e´x 2 .
Suppose you wanted to approximate the number sinh(b) using the Maclaurin series of sinh x, where b is some number in (´2, 1). Which partial sum should you use to guarantee 10 decimal places of accuracy? (That is, an absolute error less than 5 ˆ 10´11.) You may assume without proof that 2 ă e ă 3.
Q: Let f (x) be a function with f (n)(x) = (n ´ 1)!
2 h (1 ´ x)´n + (´1)n´1(1 + x)´ni for all n ě 1.
122 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Give reasonable bounds (both upper and lower) on the error involved in approximating f ´1 3 using the partial sum S6 of the Taylor series for f (x) centred at a = 1 2.
Remark: One function with this quality is the inverse hyperbolic tangent function.1 § § Stage 3 Q32: Use series to evaluate lim xÑ0 1 ´ cos x 1 + x ´ ex .
Q33: Evaluate lim xÑ0 sin x ´ x + x3 6 x5 .
Q: Evaluate lim xÑ0 1 + x + x22/x using a Taylor series for the natural logarithm.
Q: Use series to evaluate lim xÑ8 1 + 1 2x x Q: Evaluate the series 8 ÿ n=0 (n + 1)(n + 2) 7n or show that it diverges.
Q: Write the series f (x) = 8 ÿ n=0 (´1)nx2n+4 (2n + 1)(2n + 2) as a combination of familiar functions.
Q: (a) Find the Maclaurin series for f (x) = (1 ´ x)´1/2. What is its radius of convergence?
(b) Manipulate the series you just found to find the Maclaurin series for g(x) = arcsin x.
What is its radius of convergence?
Q39: Find the Taylor series for f (x) = log(x) centred at a = 2. Find the interval of convergence for this series.
Q40: Let I(x) = ż x 0 1 1 + t4 dt.
(a) Find the Maclaurin series for I(x).
(b) Approximate I(1/2) to within ˘0.0001.
(c) Is your approximation in (b) larger or smaller than the true value of I(1/2)? Explain.
1 Of course it is! Actually, hyperbolic tangent is tanh(x) = ex ´ e´x ex + e´x , and inverse hyperbolic tangent is its functional inverse.
123 SEQUENCES AND SERIES 3.6 TAYLOR SERIES Q41: Using a Maclaurin series, the number a = 1/5 ´ 1/7 + 1/18 is found to be an approximation for I = ż 1 0 x4e´x2 dx. Give the best upper bound you can for |I ´ a|.
Q42: Find an interval of length 0.0002 or less that contains the number I = ż 1 2 0 x2e´x2 dx Q43: Let I(x) = ż x 0 e´t ´ 1 t dt.
(a) Find the Maclaurin series for I(x).
(b) Approximate I(1) to within ˘0.01.
(c) Explain why your answer to part (b) has the desired accuracy.
Q44: The function Σ(x) is defined by Σ(x) = ż x 0 sin t t dt.
(a) Find the Maclaurin series for Σ(x).
(b) It can be shown that Σ(x) has an absolute maximum which occurs at its smallest positive critical point (see the graph of Σ(x) below). Find this critical point.
(c) Use the previous information to find the maximum value of Σ(x) to within ˘0.01.
x y Q45: Let I(x) = ż x 0 cos t ´ 1 t2 dt.
(a) Find the Maclaurin series for I(x).
(b) Use this series to approximate I(1) to within ˘0.01 (c) Is your estimate in (b) greater than I(1)? Explain.
Q46: Let I(x) = ż x 0 cos t + t sin t ´ 1 t2 dt (a) Find the Maclaurin series for I(x).
(b) Use this series to approximate I(1) to within ˘0.001 (c) Is your estimate in (b) greater than or less than I(1)?
Q47: Define f (x) = ż x 0 1 ´ e´t t dt.
(a) Show that the Maclaurin series for f (x) is 8 ÿ n=1 (´1)n´1 n ¨ n!
xn.
124 SEQUENCES AND SERIES 3.6 TAYLOR SERIES (b) Use the ratio test to determine the values of x for which the Maclaurin series 8 ÿ n=1 (´1)n´1 n ¨ n!
xn converges.
Q48: Show that ż 1 0 x3 ex ´ 1 dx ď 1 3.
Q49: Let cosh(x) = ex + e´x 2 .
(a) Find the power series expansion of cosh(x) about x0 = 0 and determine its interval of convergence.
(b) Show that 32 3 ď cosh(2) ď 32 3 + 0.1.
(c) Show that cosh(t) ď e 1 2t2 for all t.
Q: The law of the instrument says “If you have a hammer then everything looks like a nail” — it is really a description of the “tendency of jobs to be adapted to tools rather than adapting tools to jobs.” Anyway, this is a long way of saying that just because we know how to compute things using Taylor series doesn’t mean we should neglect other techniques.
(a) Using Newton’s method, approximate the constant 3 ?
2 as a root of the function g(x) = x3 ´ 2. Using a calculator, make your estimation accurate to within 0.01.
(b) You may assume without proof that 3 ?x = 1 + 1 6(x ´ 1) + 8 ÿ n=2 (´1)n´1(2)(5)(8) ¨ ¨ ¨ (3n ´ 4) 3n n!
(x ´ 1)n.
for all real numbers x. Using the fact that this is an alternating series, how many terms would you have to add for the partial sum to estimate 3 ?
2 with an error less than 0.01?
Quote from Silvan Tomkins’s Computer Simulation of Personality: Frontier of Psychological Theory. See also Birmingham screwdrivers.
Q: Let f (x) = arctan(x3). Write f (10) 1 5 as a sum of rational numbers with an error less than 10´6 using the Maclaurin series for arctangent.
Q: Consider the following function: f (x) = # e´1/x2 x ‰ 0 0 x = 0 (a) Sketch y = f (x).
(b) Assume (without proof) that f (n)(0) = 0 for all whole numbers n. Find the Maclaurin series for f (x).
125 SEQUENCES AND SERIES 3.6 TAYLOR SERIES (c) Where does the Maclaurin series for f (x) converge?
(d) For which values of x is f (x) equal to its Maclaurin series?
Q: Suppose f (x) is an odd function, and f (x) = 8 ÿ n=0 f (n)(0) n!
xn. Simplify 8 ÿ n=0 f (2n)(0) (2n)! x2n.
126 HINTS TO PROBLEMS Part II 127 Hints for Exercises 1.1. — Jump to TABLE OF CONTENTS.
H-1: Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle.
H-2: We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region.
H-3: Four rectangles suffice.
H-4: Try drawing a picture.
H-5: Try an oscillating function.
H-6: The ordering of the parts is intentional: each sum can be written by changing some small part of the sum before it.
H-7: If we raise ´1 to an even power, we get +1, and if we raise it to an odd power, we get ´1.
H-8: Sometimes a little anti-simplification can make the pattern more clear.
(a) Re-write as 1 3 + 3 9 + 5 27 + 7 81 + 9 243.
(b) Compare to the sum in the hint for (a).
(c) Re-write as 1 ¨ 1000 + 2 ¨ 100 + 3 ¨ 10 + 4 1 + 5 10 + 6 100 + 7 1000.
H-9: (a), (b) These are geometric sums.
(c) You can write this as three separate sums.
(d) You can write this as two separate sums. Remember that e is a constant. Don’t be thrown off by the index being n instead of i.
H-10: (a) Write out the terms of the two sums.
(b) A change of index is an easier option than expanding the cubic.
(c) Which terms cancel?
(d) Remember 2n + 1 is odd for every integer n. The index starts at n = 2, not n = 1.
H-11: Since the sum adds four pieces, there will be four rectangles. However, one might be extremely small.
H-12: Write out the general formula for the left Riemann sum from Definition 1.1.11 in the CLP-2 text and choose a, b and n to make it match the given sum.
H-13: Since the sum runs from 1 to 3, there are three intervals. Suppose 2 = ∆x = b´a n .
You may assume the sum given is a right Riemann sum (as opposed to left or midpoint).
H-14: Let ∆x = π 20. Then what is b ´ a?
128 H-15: Notice that the index starts at k = 0, instead of k = 1. Write out the given sum explicitly without using summation notation, and sketch where the rectangles would fall on a graph of y = f (x). Then try to identify b ´ a, and n, followed by “right”, “left”, or “midpoint”, and finally a.
H-16: The area is a triangle.
H-17: There is one triangle of positive area, and one of negative area.
H-18: Review Definition 1.1.11 in the CLP-2 text.
H-20: You’ll want the limit as n goes to infinity of a sum with n terms. If you’re having a hard time coming up with the sum in terms of n, try writing a sum with a finite number of terms of your choosing. Then, think about how that sum would change if it had n terms.
H-21: The main step is to express the given sum as the right Riemann sum, n ÿ i=1 f (a + i∆x)∆x.
Don’t be afraid to guess ∆x and f (x) (review Definition 1.1.11 in the CLP-2 text). Then write out explicitly n ř i=1 f (a + i∆x)∆x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
H-22: The main step is to express the given sum as the right Riemann sum n ř k=1 f (a + k∆x)∆x. Don’t be afraid to guess ∆x and f (x) (review Definition 1.1.11 in the CLP-2 text). Then write out explicitly n ř k=1 f (a + k∆x)∆x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
H-23: The main step is to express the given sum in the form řn i=1 f (x˚ i )∆x. Don’t be afraid to guess ∆x, x˚ i (for either a left or a right or a midpoint sum — review Definition 1.1.11 in the CLP-2 text) and f (x). Then write out explicitly řn i=1 f (x˚ i )∆x with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
H-24: The main step is to express the given sum in the form n ř i=1 f (x˚ i )∆x. Don’t be afraid to guess ∆x, x˚ i (probably, based on the symbol Rn, assuming we have a right Riemann sum — review Definition 1.1.11 in the CLP-2 text) and f (x). Then write out explicitly n ř i=1 f (x˚ i )∆x with your guess substituted in, and compare the result with the given sum.
Adjust your guess if they don’t match.
H-25: Try several different choices of ∆x and x˚ i .
H-26: Let x = r3, and re–write the sum in terms of x.
H-27: Note the sum does not start at r0 = 1.
129 H-28: Draw a picture. See Example 1.1.15 in the CLP-2 text.
H-29: Draw a picture. Remember |x| = " x x ě 0 ´x x ă 0 .
H-30: Draw a picture: the area we want is a trapezoid. If you don’t remember a formula for the area of a trapezoid, think of it as the difference of two triangles.
H-31: You can draw a very similar picture to Question 30, but remember the areas are negative.
H-32: If y = ?
16 ´ x2, then y is nonnegative, and y2 + x2 = 16.
H-33: Sketch the graph of f (x).
H-34: At which time in the interval, for example, 0 ď t ď 0.5, is the car moving the fastest?
H-35: What are the possible speeds the car could have reached at time t = 0.25?
H-36: You need to know the speed of the plane at the midpoints of your intervals, so (for example) noon to 1pm is not one of your intervals.
H-37: Sure looks like a Riemann sum.
H-38: For part (b): don’t panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given identity.
The third step is to evaluate the limit n Ñ 8.
H-39: The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit as n Ñ 8.
H-40: The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit n Ñ 8.
H-41: You’ve probably seen this hint before. It is worth repeating. Don’t panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formula. The third step is to evaluate the limit n Ñ 8.
H-42: Using the definition of a right Riemann sum, we can come up with an expression for f (´5 + 10i). In order to find f (x), set x = ´5 + 10i.
H-43: Recall that for a positive constant a, d dx taxu = ax log a, where log a is the natural logarithm (base e) of a.
H-44: Part (a) follows the same pattern as Question 43–there’s just a little more algebra involved, since our lower limit of integration is not 0.
H-45: Your area can be divided into a section of a circle and a triangle. Then you can use geometry to find the area of each piece.
H-46: (a) The difference between the upper and lower bounds is the area that is outside of the smaller rectangles but inside the larger rectangles. Drawing both sets of rectangles on 130 one picture might make things clearer. Look for an easy way to compute the area you want.
(b) Use your answer from Part (a). Your answer will depend on f, a, and b.
H-47: Since f (x) is linear, there exist real numbers m and c such that f (x) = mx + c. It’s a little easier to first look at a single triangle from each sum, rather than the sums in their entirety.
Hints for Exercises 1.2. — Jump to TABLE OF CONTENTS.
H-1: (a) What is the length of this figure?
(b) Think about cutting the area into two pieces vertically.
(c) Think about cutting the area into two pieces another way.
H-2: Use the identity b ş a f (x) dx = c ş a f (x) dx + b ş c f (x) dx.
H-4: Note that the limits of the integral given are in the opposite order from what we might expect: the smaller number is the top limit of integration.
Recall ∆x = b´a n .
H-5: Split the “target integral” up into pieces that can be evaluated using the given integrals.
H-6: Split the “target integral” up into pieces that can be evaluated using the given integrals.
H-7: Split the “target integral” up into pieces that can be evaluated using the given integrals.
H-8: For part (a), use the symmetry of the integrand. For part (b), the area 1 ş 0 ?
1 ´ x2 dx is easy to find–how is this useful to you?
H-9: The evaluation of this integral was also the subject of Question 9 in Section 1.1. This time try using the method of Example 1.2.7 in the CLP-2 text.
H-10: Use symmetry.
H-11: Check Theorem 1.2.12 in the CLP-2 text.
H-12: Split the integral into a sum of two integrals. Interpret each geometrically.
H-13: Hmmmm. Looks like a complicated integral. It’s probably a trick question. Check for symmetries.
H-14: Check for symmetries again.
H-15: What does the integrand look like to the left and right of x = 3?
131 H-16: In part (b), you’ll have to factor a constant out through a square root. Remember the upper half of a circle looks like ?
r2 ´ x2.
H-17: For two functions f (x) and g(x), define h(x) = f (x) ¨ g(x). If h(´x) = h(x), then the product is even; if h(´x) = ´h(x), then the product is odd.
The table will not be the same as if we were multiplying even and odd numbers.
H-18: Note f (0) = f (´0).
H-19: If f (x) is even and odd, then f (x) = ´f (x) for every x.
H-20: Think about mirroring a function across an axis. What does this do to the slope?
Hints for Exercises 1.3. — Jump to TABLE OF CONTENTS.
H-2: First find the general antiderivative by guessing and checking.
H-3: Be careful. Two of these make no sense at all.
H-4: Check by differentiating.
H-5: Check by differentiating.
H-6: Use the Fundamental Theorem of Calculus Part 1.
H-7: Use the Fundamental Theorem of Calculus, Part 1.
H-8: You already know that F(x) is an antiderivative of f (x).
H-9: (a) Recall d dxtarccos xu = ´1 ?
1´x2.
(b) All antiderivatives of ?
1 ´ x2 differ from one another by a constant. You already know one antiderivative.
H-10: In order to apply the Fundamental Theorem of Calculus Part 2, the integrand must be continuous over the interval of integration.
H-11: Use the definition of F(x) as an area.
H-12: F(x) represents net signed area.
H-13: Note G(x) = ´F(x), when F(x) is defined as in Question 12.
H-14: Using the definition of the derivative, F1(x) = lim hÑ0 F(x + h) ´ F(x) h .
The area of a trapezoid with base b and heights h1 and h2 is 1 2b(h1 + h2).
H-15: There is only one!
H-16: If d dxtF(x)u = f (x), that tells us ş f (x) dx = F(x) + C.
H-17: When you’re differentiating, you can leave the ex factored out.
H-18: After differentiation, you can simplify pretty far. Keep at it!
132 H-19: This derivative also simplifies considerably. You might need to add fractions by finding a common denominator.
H-20: Guess a function whose derivative is the integrand, then use the Fundamental Theorem of Calculus Part 2.
H-21: Split the given integral up into two integrals.
H-22: The integrand is similar to 1 1 + x2, so something with arctangent seems in order.
H-23: The integrand is similar to 1 ?
1 ´ x2, so factoring out ?
2 from the denominator will make it look like some flavour of arcsine.
H-24: We know how to antidifferentiate sec2 x, and there is an identity linking sec2 x with tan2 x.
H-25: Recall 2 sin x cos x = sin(2x).
H-26: cos2 x = 1 + cos(2x) 2 H-28: There is a good way to test where a function is increasing, decreasing, or constant, that also has something to do with topic of this section.
H-29: See Example 1.3.5 in the CLP-2 text.
H-30: See Example 1.3.5 in the CLP-2 text.
H-31: See Example 1.3.5 in the CLP-2 text.
H-32: See Example 1.3.5 in the CLP-2 text.
H-33: See Example 1.3.6 in the CLP-2 text.
H-34: Apply d dx to both sides.
H-35: What is the title of this section?
H-36: See Example 1.3.6 in the CLP-2 text.
H-37: See Example 1.3.6 in the CLP-2 text.
H-38: See Example 1.3.6 in the CLP-2 text.
H-39: See Example 1.3.6 in the CLP-2 text.
H-40: Split up the domain of integration.
H-41: It is possible to guess an antiderivative for f 1(x) f 2(x) that is expressed in terms of f 1(x).
H-42: When does the car stop? What is the relation between velocity and distance travelled?
H-43: See Example 1.3.5 in the CLP-2 text. For the absolute maximum part of the question, study the sign of f 1(x).
133 H-44: See Example 1.3.5 in the CLP-2 text. For the “minimum value” part of the question, study the sign of f 1(x).
H-45: See Example 1.3.5 in the CLP-2 text. For the “maximum” part of the question, study the sign of F1(x).
H-46: Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.
H-47: Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.
H-48: Carefully check the Fundamental Theorem of Calculus: as written, it only applies directly to F(x) when x ě 0.
Is F(x) even or odd?
H-49: In general, the equation of the tangent line to the graph of y = f (x) at x = a is y = f (a) + f 1(a) (x ´ a).
H-50: Recall tan2 x + 1 = sec2 x.
H-51: Since the integration is with respect to t, the x3 term can be moved outside the integral.
H-52: Remember that antiderivatives may have a constant term.
Hints for Exercises 1.4. — Jump to TABLE OF CONTENTS.
H-1: One is true, the other false.
H-2: You can check whether the final answer is correct by differentiating.
H-3: Check the limits.
H-4: Check every step. Do they all make sense?
H-6: What is d dxtf (g(x))u?
H-7: What is the derivative of the argument of the cosine?
H-8: What is the title of the current section?
H-9: What is the derivative of x3 + 1?
H-10: What is the derivative of log x?
H-11: What is the derivative of 1 + sin x?
H-12: cos x is the derivative of what?
H-13: What is the derivative of the exponent?
H-14: What is the derivative of the argument of the square root?
H-15: What is d dx ␣a log x ( ?
134 H-16: There is a short, slightly sneaky method — guess an antiderivative — and a really short, still-more-sneaky method.
H-17: Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.
H-18: If w = u2 + 1, then u2 = w ´ 1.
H-19: Using a trigonometric identity, this is similar (though not identical) to ş tan θ ¨ sec2 θ dθ.
H-20: If you multiply the top and the bottom by ex, what does this look like the antiderivative of?
H-21: You know methods other than substitution to evaluate definite integrals.
H-22: tan x = sin x cos x H-23: Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.
H-24: Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.
H-25: Find the right Riemann sum for both definite integrals.
Hints for Exercises 1.5. — Jump to TABLE OF CONTENTS.
H-1: When we say “area between,” we want positive area, not signed area.
H-2: We’re taking rectangles that reach from one function to the other.
H-3: Draw a sketch first.
H-4: Draw a sketch first.
H-5: You can probably find the intersections by inspection.
H-6: To find the intersection, plug x = 4y2 into the equation x + 12y + 5 = 0.
H-7: If the bottom function is the x-axis, this is a familiar question.
H-8: Part of the job is to determine whether y = x lies above or below y = 3x ´ x2.
H-9: Guess the intersection points by trying small integers.
H-10: Draw a sketch first. You can also exploit a symmetry of the region to simplify your solution.
H-11: Figure out where the two curves cross. To determine which curve is above the other, try evaluating f (x) and g(x) for some simple value of x. Alternatively, consider x very close to zero.
H-12: Think about whether it will easier to use vertical strips or horizontal strips.
H-13: Writing an integral for this is nasty. How can you avoid it?
135 H-14: You are asked for the area, not the signed area. Be very careful about signs.
H-15: You are asked for the area, not the signed area. Draw a sketch of the region and be very careful about signs.
H-16: You have to determine whether • the curve y = f (x) = x ?
25 ´ x2 lies above the line y = g(x) = 3x for all 0 ď x ď 4 or • the curve y = f (x) lies below the line y = g(x) for all 0 ď x ď 4 or • y = f (x) and y = g(x) cross somewhere between x = 0 and x = 4.
One way to do so is to study the sign of f (x) ´ g(x) = x ?
25 ´ x2 ´ 3 .
H-17: Flex those geometry muscles.
H-18: These two functions have three points of intersection. This question is slightly messy, but uses the same concepts we’ve been practicing so far.
Hints for Exercises 1.6. — Jump to TABLE OF CONTENTS.
H-1: The horizontal cross-sections were discussed in Example 1.6.1 of the CLP-2 text.
H-2: What are the dimensions of the cross-sections?
H-3: There are two different kinds of washers.
H-4: Draw sketches. The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which is in the optional section 1.6 of the CLP-2 text. The method of washers also works, but requires you to have more patience and also to have a good idea what the specified region looks like. Look at your sketch very careful when identifying the ends of your horizontal strips.
H-5: Draw sketchs.
H-6: Draw a sketch.
H-7: If you take horizontal slices (parallel to one face), they will all be equilateral triangles.
Be careful not to confuse the height of a triangle with the height of the tetrahedron.
H-8: Sketch the region.
H-9: Sketch the region first.
H-10: You can save yourself quite a bit of work by interpreting the integral as the area of a known geometric figure.
H-11: See Example 1.6.3 in the CLP-2 text.
H-12: See Example 1.6.5 in the CLP-2 text.
136 H-13: Sketch the region. To find where the curves intersect, look at where cos( x 2) and x2 ´ π2 both have roots.
H-14: See Example 1.6.6 in the CLP-2 text.
H-15: See Example 1.6.6 in the CLP-2 text. Imagine cross-sections with shadow parallel to the y-axis, sticking straight out of the xy-plane.
H-16: See Example 1.6.1 in the CLP-2 text.
H-17: (a) Don’t be put off by phrases like “rotating an ellipse about its minor axis.” This is the same kind of volume you’ve been calculating all section.
(b) Hopefully, you sketched the ellipse in part (a). What was its smallest radius? Its largest? These correspond to the polar and equitorial radii, respectively.
(c) Combine your answers from (a) and (b).
(d) Remember that the absolute error is the absolute difference of your two results–that is, you subtract them and take the absolute value. The relative error is the absolute error divided by the actual value (which we’re taking, for our purposes, to be your answer from (c)). When you take the relative error, lots of terms will cancel, so it’s easiest to not use a calculator till the end.
H-18: To find the points of intersection, set 4 ´ (x ´ 1)2 = x + 1.
H-19: You can somewhat simplify your calculations in part (a) (but not part (b)) by using the fact that R is symmetric about the line y = x.
When you’re solving an equation for x, be careful about your signs: x ´ 1 is negative.
H-20: The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what R looks like. So it is crucial to first sketch R. Then be very careful in identifying the left end of your horizontal strips.
H-21: Note that the curves cross. The area of this region was found in Problem 14 of Section 1.5. It would be useful to review that problem.
H-22: You can use ideas from this section to answer the question. If you take a very thin slice of the column, the density is almost constant, so you can find the mass. Then you can add up all your little slices. It’s the same idea as volume, only applied to mass.
Do be careful about units: in the problem statement, some are given in metres, others in kilometres.
If you’re having a hard time with the antiderivative, try writing the exponential function with base e. Remember 2 = elog 2.
Hints for Exercises 1.7. — Jump to TABLE OF CONTENTS.
H-1: Read back over Sections 1.4 and 1.7 of the CLP-2 text. When these methods are introduced, they are justified using the corresponding differentiation rules.
137 H-2: Remember our rule: ş udv = uv ´ ş vdu. So, we take u and use it to make du, and we take dv and use it to make v.
H-3: According to the quotient rule, d dx " f (x) g(x) = g(x) f 1(x) ´ f (x)g1(x) g2(x) .
Antidifferentiate both sides of the equation, then solve for the expression in the question.
H-4: Remember all the antiderivatives differ only by a constant, so you can write them all as v(x) + C for some C.
H-5: What integral do you have to evaluate, after you plug in your choices to the integration by parts formula?
H-6: You’ll probably want to use integration by parts. (It’s the title of the section, after all). You’ll break the integrand into two parts, integrate one, and differentiate the other.
Would you rather integrate log x, or differentiate it?
H-7: This problem is similar to Question 6.
H-8: Example 1.7.5 in the CLP-2 text shows you how to find the antiderivative. Then the Fundamental Theorem of Calculus Part 2 gives you the definite integral.
H-9: Compare to Question 8. Try to do this one all the way through without peeking at another solution!
H-10: If at first you don’t succeed, try using integration by parts a few times in a row.
Eventually, one part will go away.
H-11: Similarly to Question 10, look for a way to use integration by parts a few times to simplify the integrand until it is antidifferentiatable.
H-12: Use integration by parts twice to get an integral with only a trigonometric function in it.
H-13: If you let u = log t in the integration by parts, then du works quite nicely with the rest of the integrand.
H-14: Those square roots are a little disconcerting– get rid of them with a substitution.
H-15: This can be solved using the same ideas as Example 1.7.8 in the CLP-2 text.
H-16: Not every integral should be evaluated using integration by parts.
H-17: You know, or can easily look up, the derivative of arccosine. You can use a similar trick as the book did when antidifferentiating other inverse trigonometric functions in Example 1.7.9 of the CLP-2 text.
H-18: After integrating by parts, do some algebraic manipulation to the integral until it’s clear how to evaluate it.
H-19: After integration by parts, use a substitution.
138 H-20: This example is similar to Example 1.7.10 in the CLP-2 text. The functions ex/2 and cos(2x) both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we’ll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually integrating.
H-21: This looks a bit like a substitution problem, because we have an “inside function.” It might help to review Example 1.7.11 in the CLP-2 text.
H-22: Start by simplifying.
H-23: sin(2x) = 2 sin x cos x H-24: What is the derivative of xe´x?
H-25: You’ll want to do an integration by parts for (a)–check the end result to get a guess as to what your parts should be. A trig identity and some amount of algebraic manipulation will be necessary to get the final form.
H-26: See Examples 1.7.9 and 1.6.5 in the CLP-2 text for refreshers on integrating arctangent, and using washers.
Remember tan2 x + 1 = sec2 x, and sec2 x is easy to integrate.
H-27: Your integral can be broken into two integrals, which yield to two different integration methods.
H-28: Think, first, about how to get rid of the square root in the argument of f 2, and, second, how to convert f 2 into f 1. Note that you are told that f 1(2) = 4 and f (0) = 1, f (2) = 3.
H-29: Interpret the limit as a right Riemann sum.
Hints for Exercises 1.8. — Jump to TABLE OF CONTENTS.
H-1: Go ahead and try it!
H-2: Use the substitution u = sec x.
H-3: Divide both sides of the second identity by cos2 x.
H-4: See Example 1.8.6 in the CLP-2 text. Note that the power of cosine is odd, and the power of sine is even (it’s zero).
H-5: See Example 1.8.7 in the CLP-2 text. All you need is a helpful trig identity.
H-6: The power of cosine is odd, so we can reserve one cosine for du, and turn the rest into sines using the identity sin2 x + cos2 x = 1.
H-7: Since the power of sine is odd (and positive), we can reserve one sine for du, and turn the rest into cosines using the identity sin2 + cos2 x = 1.
H-8: When we have even powers of sine and cosine both, we use the identities in the last two lines of Equation 1.8.3 in the CLP-2 text.
139 H-9: Since the power of sine is odd, you can use the substitution u = cos x.
H-10: Which substitution will work better: u = sin x, or u = cos x?
H-11: Try a substitution.
H-12: For practice, try doing this in two ways, with different substitutions.
H-13: A substitution will work. See Example 1.8.14 in the CLP-2 text for a template for integrands with even powers of secant.
H-14: Try the substitution u = sec x.
H-15: Compare to Question 14.
H-16: What is the derivative of tangent?
H-17: Don’t be scared off by the non-integer power of secant. You can still use the strategies in the notes for an odd power of tangent.
H-18: Since there are no secants in the problem, it’s difficult to use the substitution u = sec x that we’ve enjoyed in the past. Example 1.8.12 in the CLP-2 text provides a template for antidifferentiating an odd power of tangent.
H-19: Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. You’ll want to use the identity tan2 x = sec2 x ´ 1, then use the substitution u = tan x, du = sec2 x dx on (perhaps only a part of) the resulting integral.
Example 1.8.16 in the CLP-2 text show you how this can be accomplished.
H-20: Since there is an even power of secant in the integrand, we can use the substitution u = tan x.
H-21: How have we handled integration in the past that involved an odd power of tangent?
H-22: Remember e is some constant. What are our strategies when the power of secant is even and positive? We’ve seen one such substitution in Example 1.8.15 of the CLP-2 text.
H-23: See Example 1.8.16 in the CLP-2 text for a strategy for integrating powers of tangent.
H-24: Write tan x = sin x cos x.
H-25: 1 cos θ = sec θ H-26: cot x = cos x sin x H-27: Try substituting.
H-28: To deal with the “inside function,” start with a substitution.
H-29: Try an integration by parts.
140 Hints for Exercises 1.9. — Jump to TABLE OF CONTENTS.
H-1: The beginning of this section has a template for choosing a substitution. Your goal is to use a trig identity to turn the argument of the square root into a perfect square, so you can cancel a (something)2 = |something|.
H-2: You want to do the same thing you did in Question 1, but you’ll have to complete the square first.
H-3: Since θ is acute, you can draw it as an angle of a right triangle. The given information will let you label two sides of the triangle, and the Pythagorean Theorem will lead you to the third.
H-4: You can draw a right triangle with angle θ, and use the given information to label two of the sides. The Pythagorean Theorem gives you the third side.
H-5: As in Question 1, choose an appropriate substitution. Your answer should be in terms of your original variable, x, which can be achieved using the methods of Question 3.
H-6: As in Question 1, choose an appropriate substitution. Your answer will be a number, so as long as you change your limits of integration when you substitute, you don’t need to bother changing the antiderivative back into the original variable x. However, you might want to use the techniques of Question 4 to simplify your final answer.
H-7: Question 1 guides the way to finding the appropriate substitution. Since the integral is definite, your final answer will be a number. Your limits of integration should be common reference angles.
H-8: Question 1 guides the way to finding the appropriate substitution. Since you have in indefinite integral, make sure to get your answer back in terms of the original variable, x. Question 3 gives a reliable method for this.
H-9: A trig substitution is not the easiest path.
H-10: To antidifferentiate, change your trig functions into sines and cosines.
H-11: The integrand should simplify quite far after your substitution.
H-12: In part (a) you are asked to integrate an even power of cos x. For part (b) you can use a trigonometric substitution to reduce the integral of part (b) almost to the integral of part (a).
H-13: What is the symmetry of the integrand?
H-14: See Example 1.9.3 in the CLP-2 text.
H-15: To integrate an even power of tangent, use the identity tan2 x = sec2 x ´ 1.
H-16: A trig substitution is not the easiest path.
H-17: Complete the square. Your final answer will have an inverse trig function in it.
H-18: To antidifferentiate even powers of cosine, use the formula cos2 θ = 1 2(1 + cos(2θ)).
Then, remember sin(2θ) = 2 sin θ cos θ.
141 H-19: After substituting, use the identity tan2 x = sec2 x ´ 1 more than once.
Remember ż sec xdx = log ˇ ˇ sec x + tan x ˇ ˇ + C.
H-20: There’s no square root, but we can still make use of the substitution x = tan θ.
H-21: You’ll probably want to use the identity tan2 θ + 1 = sec2 θ more than once.
H-22: Complete the square — refer to Question 2 if you want a refresher. The constants aren’t pretty, but don’t let them scare you.
H-23: After substituting, use the identity sec2 u = tan2 u + 1. It might help to break the integral into a few pieces.
H-24: Make use of symmetry, and integrate with respect to y (rather than x).
H-25: Use the symmetry of the function to re-write your integrals without an absolute value.
H-26: Think of ex as ex/22, and use a trig substitution. Then, use the identity sec2 θ = tan2 θ + 1.
H-27: (a) Use logarithm rules to simplify first.
(b) Think about domains.
(c) What went wrong in part (b)? At what point in the work was that problem introduced?
There is a subtle but important point mentioned in the introductory text to Section 1.9 of the CLP-2 text that may help you make sense of things.
H-28: Consider the ranges of the inverse trigonometric functions. For (c), also consider the domain of ?
x2 ´ a2.
Hints for Exercises 1.10. — Jump to TABLE OF CONTENTS.
H-1: If a quadratic function can be factored as (ax + b)(cx + d) for some constants a, b, c, d, then it has roots ´b a and ´d c.
H-2: Review Equations 1.10.7 through 1.10.11 of the CLP-2 text. Be careful to fully factor the denominator.
H-3: Review Example 1.10.1 in the CLP-2 text. Is the “Algebraic Method” or the “Sneaky Method” going to be easier?
H-4: For each part, use long division as in Example 1.10.4 of the CLP-2 text.
H-5: (a) Look for a pattern you can exploit to factor out a linear term.
(b) If you set y = x2, this is quadratic. Remember (x2 ´ a) = (x + ?a)(x ´ ?a) as long as a is positive.
(c),(d) Look for integer roots, then use long division.
142 H-6: Why do we do partial fraction decomposition at all?
H-7: What is the title of this section?
H-8: You can save yourself some work in developing your partial fraction decomposition by renaming x2 to y and comparing the result with Question 7.
H-9: Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3 in the CLP-2 text.
H-10: Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3 in the CLP-2 text. Remember d dxtarctan xu = 1 1+x2.
H-11: Fill in the blank: the integrand is a function.
H-12: The integrand is yet another function.
H-13: Since the degree of the numerator is the same as the degree of the denominator, we can’t do our partial fraction decomposition before we simplify the integrand.
H-14: The degree of the numerator is not smaller than the degree of the denominator.
Your final answer will have an arctangent in it.
H-15: In the partial fraction decomposition, several constants turn out to be 0.
H-16: Factor (2x ´ 1) out of the denominator to get started. You don’t need long division for this step.
H-17: When it comes time to integrate, look for a convenient substitution.
H-18: csc x = 1 sin x = sin x sin2 x H-19: Use the partial fraction decomposition from Queston 18 to save yourself some time.
H-20: In the final integration, complete the square to make a piece of the integrand look more like the derivative of arctangent.
H-21: Review Question 20 in Section 1.9 for antidifferentiation tips.
H-22: Partial fraction decomposition won’t simplify this any more. Use a trig substitution.
H-23: To evaluate the antiderivative, break one of the fractions into two fractions.
H-24: cos2 θ = 1 ´ sin2 θ H-25: If you’re having a hard time making the substitution, multiply the numerator and the denominator by ex.
H-26: Try the substitution u = ?
1 + ex. You’ll need to do long division before you can use partial fraction decomposition.
H-27: The mechanically easiest way to answer part (c) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what R looks like. So look at the 143 sketch in part (a) very carefully when identifying the left endpoints of your horizontal strips.
H-28: You’ll need to use two regions, because the curves cross.
H-29: For (b), use the Fundamental Theorem of Calculus Part 1.
Hints for Exercises 1.11. — Jump to TABLE OF CONTENTS.
H-1: The absolute error is the difference of the two values; the relative error is the absolute error divided by the exact value; the percent error is one hundred times the relative error.
H-2: You should have four rectangles in one drawing, and four trapezoids in another.
H-3: Sketch the second derivative–it’s quadratic.
H-4: You don’t have to find the actual, exact maximum the second derivative achieves–you only have to give a reasonable “ceiling” that it never breaks through.
H-5: To compute the upper bound on the error, find an upper bound on the fourth derivative of cosine, then use Theorem 1.11.12 in the CLP-2 text.
To find the actual error, you need to find the actual value of A.
H-6: Find a function with f 2(x) = 3 for all x in [0, 1].
H-7: You’re allowed to use common sense for this one.
H-8: For part (b), consider Question 7.
H-9: Draw a sketch.
H-10: The error bound for the approximation is given in Theorem 1.11.12 in the CLP-2 text. You want this bound to be zero.
H-11: Follow the formulas in Equations 1.11.2, 1.11.6, and 1.11.9 in the CLP-2 text.
H-12: See Section 1.11.1 in the CLP-2 text. You should be able to simplify your answer to an exact value (in terms of π).
H-13: See Section 1.11.2 in the CLP-2 text. To set up the volume integral, see Example 1.6.6 in the CLP-2 text. Note the dimensions given for the cross sections are diameters, not radii.
H-14: See Section 1.11.3 in the CLP-2 text, and compare to Question 13. Note the table gives diameters, not radii.
H-15: See §1.11.3 in the CLP-2 text. To set up the volume integral, see Example 1.6.6 in the CLP-2 text, or Question 14.
Note that the table gives the circumference, not radius, of the tree at a given height.
H-18: The main step is to find an appropriate value of M. It is not necessary to find the smallest possible M.
144 H-19: The main step is to find M. This question is unusual in that its wording requires you to find the smallest possible allowed M.
H-20: The main steps in part (b) are to find the smallest possible values of M and L.
H-21: As usual, the biggest part of this problem is finding L. Don’t be thrown off by the error bound being given slightly differently from Theorem 1.11.12 in the CLP-2 text: these expressions are equivalent, since ∆x = b´a n .
H-22: The function e´2x = 1 e2x is positive and decreasing, so its maximum occurs when x is as small as possible.
H-23: H-24: The “best ... approximations that you can” means using the maximum number of intervals, given the information available.
The final sentence in part (b) is just a re-statement of the error bounds we’re familiar with from Theorem 1.11.12 in the CLP-2 text. The information ˇ ˇs(k)(x) ˇ ˇ ď k 1000 gives you values of M and L when you set k = 2 and k = 4, respectively.
H-25: Set the error bound to be less than 0.001, then solve for n.
H-26: See Section 1.11.3 in the CLP-2 text. To set up the volume integral, see Example 1.6.2 in the CLP-2 text.
Since the cross-sections of the pool are semi-circular disks, a section that is d metres across will have area 1 2π d 2 2 square feet. Based on the drawing, you may assume the very ends of the pool have distance 0 feet across.
H-27: See Example 1.11.14 in the CLP-2 text.
Don’t get caught up in the interpretation of the integral. It’s nice to see how integrals can be used, but for this problem, you’re still just approximating the integral given, and bounding the error.
When you find the second derivative to bound your error, pay attention to the difference between the integrand and g(r).
H-28: See Example 1.11.15 in the CLP-2 text. You’ll want to use a calculator for the approximation in (a), and for finding the appropriate number of intervals in (b).
Remember that Simpson’s rule requires an even number of intervals.
H-29: See Example 1.11.15 in the CLP-2 text.
Rather than calculating the fourth derivative of the integrand, use the graph to find the largest absolute value it attains over our interval.
H-30: See Example 1.11.14 in the CLP-2 text.
You’ll have to differentiate f (x). To that end, you may also want to review the fundamental theorem of calculus and, in particular, Example 1.3.5 in the CLP-2 text.
145 You don’t have to find the best possible value for M. A reasonable upper bound on |f 2(x)| will do.
To have five decimal places of accuracy, your error must be less than 0.000005. This ensures that, if you round your approximation to five decimal places, they will all be correct.
H-31: To find the maximum value of |f 2(x)|, check its critical points and endpoints.
H-32: In using Simpson’s rule to approximate ż x 1 1 t dt with n intervals, a = 1, b = x, and ∆x = x ´ 1 n .
H-33: • ş2 1 1 1+x2 dx = arctan(2) ´ π 4 , so arctan(2) = π 4 + ş2 1 1 1+x2 dx • If an approximation A of the integral ş2 1 1 1+x2 dx has error at most ε, then A ´ ε ď ş2 1 1 1+x2 dx ď A + ε.
• Looking at our target interval will tell you how small ε needs to be, which in turn will tell you how many intervals you need to use.
• You can show, by considering the numerator and denominator separately, that |f (4)(x)| ď 30.75 for every x in [1, 2].
• If you use Simpson’s rule to approximate ş2 1 1 1+x2 dx, you won’t need very many intervals to get the requisite accuracy.
Hints for Exercises 1.12. — Jump to TABLE OF CONTENTS.
H-1: There are two kinds of impropreity in an integral: an infinite discontinuity in the integrand, and an infinite limit of integration.
H-2: The integrand is continuous for all x.
H-3: What matters is which function is bigger for large values of x, not near the origin.
H-4: Read both the question and Theorem 1.12.17 in the CLP-2 text very carefully.
H-5: (a) What if h(x) is negative? What if it’s not?
(b) What if h(x) is very close to f (x) or g(x), rather than right in the middle?
(c) Note |h(x)| ď 2f (x).
H-6: First: is the integrand unbounded, and if so, where?
Second: when evaluating integrals, always check to see if you can use a simple substitution before trying a complicated procedure like partial fractions.
H-7: Is the integrand bounded?
146 H-8: See Example 1.12.21 in the CLP-2 text. Rather than antidifferentiating, you can find a nice comparison.
H-9: Which of the two terms in the denominator is more important when x « 0? Which one is more important when x is very large?
H-10: Remember to break the integral into two pieces.
H-11: Remember to break the integral into two pieces.
H-12: The easiest test in this case is limiting comparison, Theorem 1.12.22 in the CLP-2 text.
H-13: Not all discontinuities cause an integral to be improper–only infinite discontinuities.
H-14: Which of the two terms in the denominator is more important when x is very large?
H-15: Which of the two terms in the denominator is more important when x « 0? Which one is more important when x is very large?
H-16: What are the “problem x’s” for this integral? Get a simple approximation to the integrand near each.
H-17: To find the volume of the solid, cut it into horizontal slices, which are thin circular disks.
The true/false statement is equivalent to saying that the improper integral giving the volume of the solid when a = 0 diverges to infinity.
H-18: Review Example 1.12.8 in the CLP-2 text. Remember the antiderivative of 1 x looks very different from the antiderivative of other powers of x.
H-19: Compare to Example 1.12.14 in the CLP-2 text. You can antidifferentiate with a u-substitution.
H-20: To evaluate the integral, you can factor the denominator.
Recall lim xÑ8 arctan x = π 2 . For the other limits, use logarithm rules, and beware of indeterminate forms.
H-21: Break up the integral. The absolute values give you a nice even function, so you can replace |x ´ a| with x ´ a if you’re careful about the limits of integration.
H-22: Use integration by parts twice to find the antiderivative of e´x sin x, as in Example 1.7.10 of the CLP-2 text. Be careful with your signs — it’s easy to make a mistake with all those negatives.
If you’re having a hard time taking the limit at the end, review the Squeeze Theorem, Theorem 1.4.17 in the CLP-1 text.
H-23: What is the limit of the integrand when x Ñ 0?
H-24: The only “source of impropriety” is the infinite domain of integration. Don’t be afraid to be a little creative to make a comparison work.
147 H-25: There are two things that contribute to your error: using t as the upper bound instead of infinity, and using n intervals for the approximation.
First, find a t so that the error introduced by approximating ş8 0 e´x 1+x dx by şt 0 e´x 1+x dx is at most 1 210´4. Then, find your n.
H-26: Look for a place to use Theorem 1.12.20 of the CLP-2 text.
Examples 1.2.10 and 1.2.11 in the CLP-2 text have nice results about the area under an even/odd curve.
H-27: x should be a real number Hints for Exercises 1.13. — Jump to TABLE OF CONTENTS.
H-1: Each option in each column should be used exactly once.
H-2: The integrand is the product of sines and cosines. See how this was handled with a substitution in Section 1.8.1 of the CLP-2 text.
After your substitution, you should have a polynomial expression in u–but it might take some simplification to get it into a form you can easily integrate.
H-3: We notice that the integrand has a quadratic polynomial under the square root. If that polynomial were a perfect square, we could get rid of the square root: try a trig substitution, as in Section 1.9 of the CLP-2 text.
The identity sin(2θ) = 2 sin θ cos θ might come in handy.
H-4: Notice the integral is improper. When you compute the limit, l’Hˆ opital’s rule might help.
If you’re struggling to think of how to antidifferentiate, try writing x ´ 1 ex = (x ´ 1)e´x.
H-5: Which method usually works for rational functions (the quotient of two polynomials)?
H-6: It would be nice to replace logarithm with its derivative, 1 x.
H-7: The integrand is a rational function, so it is possible to use partial fractions. But there is a much easier way!
H-8: You should prepare your own personal internal list of integration techniques ordered from easiest to hardest. You should have associated to each technique your own personal list of signals that you use to decide when the technique is likely to be useful.
H-9: Despite both containing a trig function, the two integrals are easiest to evaluate using different methods.
H-10: For the integral of secant, see See Section 1.8.3 or Example 1.10.5 in the CLP-2 text.
In (c), notice the denominator is not yet entirely factored.
148 H-11: Part (a) can be done by inspection – use a little highschool geometry! Part (b) is reminiscent of the antiderivative of logarithm–how did we find that one out? Part (c) is an improper integral.
H-12: Use the substitution u = sin θ.
H-13: For (c), try a little algebra to split the integral into pieces that are easy to antidifferentiate.
H-14: If you’re stumped, review Sections 1.8, 1.9, and 1.10 in the CLP-2 text.
H-15: For part (a), see Example 1.7.11 in the CLP-2 text. For part (d), see Example 1.10.4 in the CLP-2 text.
H-16: For part (b), first complete the square in the denominator. You can save some work by first comparing the derivative of the denominator with the numerator. For part (d) use a simple substitution.
H-17: For part (b), complete the square in the denominator. You can save some work by first comparing the derivative of the denominator with the numerator.
H-18: For part (a), the numerator is the derivative of a function that appears in the denominator.
H-19: The integral is improper.
H-20: For part (a), can you convert this into a partial fractions integral? For part (b), start by completing the square inside the square root.
H-21: For part (b), the numerator is the derivative of a function that is embedded in the denominator.
H-22: Try a substitution.
H-23: Note the quadratic function under the square root: you can solve this with trigonometric substitution, as in Section 1.9 of the CLP-2 text.
H-24: Try a substitution, as in Section 1.8.2 of the CLP-2 text.
H-25: What’s the usual trick for evaluating a rational function (quotient of polynomials)?
H-26: If the denominator were x2 + 1, the antiderivative would be arctangent.
H-27: Simplify first.
H-28: x3 + 1 = (x + 1)(x2 ´ x + 1) H-29: You have the product of two quite dissimilar functions in the integrand–try integration by parts.
H-30: Use the identity cos(2x) = 2 cos2 x ´ 1.
H-31: Using logarithm rules can make the integrand simpler.
H-32: What is the derivative of the function in the denominator? How could that be useful to you?
H-33: For part (a), the substitution u = log x gives an integral that you have seen before.
149 H-34: For part (a), split the integral in two. One part may be evaluated by interpreting it geometrically, without doing any integration at all. For part (c), multiply both the numerator and denominator by ex and then make a substitution.
H-35: Let u = ?
1 ´ x.
H-36: Use the substitution u = ex.
H-37: Use integration by parts. If you choose your parts well, the resulting integration will be very simple.
H-38: sin x cos2 x = tan x sec x H-39: The cases n = ´1 and n = ´2 are different from all other values of n.
H-40: x4 + 1 = (x2 + ?
2x + 1)(x2 ´ ?
2x + 1) Hints for Exercises 2.1. — Jump to TABLE OF CONTENTS.
H-1: Watch your units: 1 J = 1 kg¨m2 sec2 , but your mass is not given in kilograms, and your height is not given in metres.
H-2: The force of the rock on the ground is the product of its mass and the acceleration due to gravity.
H-3: Adding or subtracting two quantities of the same units doesn’t change the units.
For example, if I have one metre of rope, and I tie on two more metres of rope, I have 1 + 2 = 3 metres of rope–not 3 centimetres of rope, or 3 kilograms of rope.
Multiplying or dividing quantities of some units gives rise to a quantity with the product or quotient of those units. For example, if I buy ten pounds of salmon for $50, the price of my salmon is 50 dollars 10 pounds = 50 10 dollars pound = 5 dollars pound. (Not 5 pound-dollars, or 5 pounds.) H-4: See Question 3.
H-5: Hooke’s law says that the force required to stretch a spring x units past its natural length is proportional to x; that is, there is some constant k associated with the individual spring such that the force required to stretch it x m past its natural length is kx.
H-6: Definition 2.1.1 in the CLP-2 text tells us the work done by the force from x = 1 to x = b is W(b) = şb 1 F(x) dx, where F(x) is the force on the object at position x. To recover the equation for F(x), use the Fundamental Theorem of Calculus.
H-7: Review Definition 2.1.1 in the CLP-2 text for calculating the work done by a force over a distance.
H-8: For (a), c ℓ´x is meausured in Newtons, while ℓand x are in metres. For (b), notice the similarities and differences between the tube of air and a spring obeying Hooke’s law.
H-9: See Example 2.1.2 in the CLP-2 text. Be careful about your units.
H-10: Be careful about the units.
150 H-11: Suppose that the bucket is a distance y above the ground. How much work is required to raise it an additional height dy?
H-12: Since you’re given the area of the cross-section, it doesn’t matter what shape it has.
However, the density of water is given in cubic centimetres, while the measurements of the tank are given in metres.
H-13: Consider the work done to lift a horizontal plate from 2 m below the ground to a height z. You’ll need to know the mass of the plate, which you can calculate from its volume, since its density is given to you.
dz H-14: You can find the spring constant k from the information about the hanging kilogram.
H-15: Follow the method of Example 2.1.6 in the CLP-2 text and Question 11 in this section.
H-16: Calculating the work done on the rope and the weight separately makes the computation somewhat easier.
H-17: When you pull the box, the force you’re exerting is exactly the same as the frictional force, but in the opposite direction. In (a), that force is constant. In (b), it changes. Check Definition 2.1.1 in the CLP-2 text for how to turn force into work.
H-18: Remember that the work done on an object is equal to the change in its kinetic energy, which is 1 2mv2, where m is the mass of the object and v is its velocity. Hooke’s law will tell you how much work was done stretching the spring.
H-19: As in Question 18 in this section, the change in kinetic energy of the car is equal to the work done by the compressing struts. The only added step is to calculate the spring constant, given that a car with mass 2000 kg compresses the spring 2 cm in Earth’s gravity. You’re not calculating work to find the spring constant: you’re using the fact that when the car is sitting still, the force exerted upward by the struts is equal to the force exerted downward by the mass of the car under gravity.
H-20: To find the radius of a horizontal layer of water, use similar triangles. Be careful with centimetres versus metres.
H-21: See Example 2.1.4 in the CLP-2 text for a basic method for calculating the work done pumping water.
To find the area of a horizontal layer of water, use some geometry. A horizontal cross-section of a sphere is a circle, and its radius will depend on the height of the layer 151 in the tank.
H-22: The basic ideas you’ve used already with “cable problems” still work, you only need to take care that the density of the cable is no longer constant. The mass of a tiny piece of cable, say of length dx, is (density)ˆ(length) = (10 ´ x)dx, where x is the distance of our piece from the bottom of the cable.
If you want more work to reference, Question 22 in Section 1.6 finds the mass of an object of variable density.
H-23: To calculate the force on the entire plunger, first find the force on a horizontal rectangle with height dy at depth y.
Checking units can be a good way to make sure your calculation makes sense.
H-24: When y metres of rope have been hauled up, what is the mass of the water?
H-25: The work you’re asked for is an improper integral, moving the earth and moon infinitely far apart.
H-26: You can formulate a guess by considering the work done on the ball versus the work done on the rope in Question 16, Solution 1. But be careful–the ball in that problem did not have the same mass as the rope.
H-27: There are two things that vary with height: the density of the liquid, and the area of the cross-section of the tank. Make a formula M(h) for the mass of a thin layer of liquid h metres below the top of the tank, using mass=volumeˆdensity. The rest of the problem is similar to other tank-pumping problems in this section.
H-28: You can model the motion, instead of a rotation, as dividing the sand into thin horizontal slices and lifting each of them to their new position.
• In order to calculate the work involved lifting a layer of sand, you need to know the mass of the layer of sand.
• To find the mass of a layer of sand, you need its volume and the density of the sand.
• To find the density of the sand, you need to the volume of the sand: that is, the volume of half the hourglass.
• The hourglass is a solid of rotation: you can find its volume using an integral, as in Section 1.6.
H-29: Theorem 1.11.12 in the CLP-2 text gives error bounds for the standard types of numerical approximations. You won’t need very many intervals to achieve the desired accuracy.
Hints for Exercises 2.2. — Jump to TABLE OF CONTENTS.
H-1: See Definition 2.2.2 in the CLP-2 text and the discussion following it for the link between area under the curve and averages.
152 H-2: Average velocity is discussed in Example 2.2.5 of the CLP-2 text. You don’t need an integral for this.
H-3: Much like Problem 2, you don’t need to do any integration here.
H-4: Part (a) is asking the length of the pieces we’ve cut our interval into. Part (c) should be given in terms of f. Our final answer in (d) will resemble a Riemann sum, but without some extra manipulation it won’t be in exactly the form of a Riemann sum we’re used to.
H-5: For (b), the value of f (0) could be much, much larger than g(0).
H-6: The answer is something very simple.
H-7: Apply the definition of “average value” in Section 2.2 of the CLP-2 text.
H-8: You can antidifferentiate x2 log x using integration by parts.
H-9: You can antidifferentiate an odd power of cosine with a substitution; for an even power of cosine, use the identity cos2 x = 1 2 1 + cos(2x) .
H-10: If you’re not sure how to antidifferentiate, try the substitution u = kx, du = kdx, keeping in mind that k is a constant. Interestingly, your final answer won’t depend on k.
H-11: The method of partial fractions can help you antidifferentiate.
H-12: Try the substitution u = log x, du = 1 x dx.
H-13: Remember cos2 x = 1 2 1 + cos(2x) .
H-14: Notice the term 50 cos t 12π has a period of 24 hours, while the term 200 cos t 4380π has a period of one year.
If n is an approximation of c, then the relative error of n is |n´c| c .
H-15: A cross section of S at location x is a circle with radius x2, so area πx4. Part (a) is asking for the average of this function on [0, 2].
H-16: (a) can be done without calculation H-17: tan2 x = sec2 x ´ 1 H-18: Remember force is the product of the spring constant with the distance it’s stretched past its natural length. The units given in the question are not exactly standard, but they are compatible with each other.
You can find part (b) without any calculation. For (c), remember sin2 x = 1 2 1 ´ cos(2x) .
H-19: The trapezoidal rule is found in Section 1.11.2 of the CLP-2 text.
H-20: To find a definite integral of the absolute value of a function, break up the interval of integration into regions where the function is positive, and intervals where it’s negative.
H-21: This is an application of the ideas in Question 20.
H-22: Slice the solid into circular disks of radius |f (x)| and thickness dx.
153 H-23: The question tells you 1 1´0 ş1 0 f (x) dx = f (0)+ f (1) 2 .
H-24: Set up this question just like Question 23, but with variables for your limits of integration.
Note (s ´ t)2 = s2 ´ 2st + t2.
H-25: What are the graphs of f (x) and f (a + b ´ x) like?
H-26: For (b), express A(x) as an integral, then differentiate.
H-27: For (b), consider the cases that f (x) is always bigger or always smaller than 0.
Then, use the intermediate value theorem, Theorem 1.6.12 in the CLP-1 text.
H-28: Try l’Hˆ opital’s rule.
H-29: Use the result of Question 28.
Hints for Exercises 2.3. — Jump to TABLE OF CONTENTS.
H-1: It might help to know that ´x2 + 2x + 1 = 2 ´ (x ´ 1)2.
H-2: The centroid of a region doesn’t have to be a point in the region.
H-3: Read over the very beginning of Section 2.3 in the CLP-2 text, specifically Equation 2.3.1.
H-4: Use Equation 2.3.1 in the CLP-2 text.
H-5: Imagine cutting out the shape and setting it on top of a pencil, so that the pencil lines up with the vertical line x = a. Will the figure balance, or fall to one side? Which side?
H-6: You can find the heights of the centres of mass using symmetry.
H-7: Think about whether your answers should have repetition.
H-8: The definition of a definite integral (Definition 1.1.9 in the CLP-2 text) will tell you how to convert your limits of sums into integrals.
H-9: In (a), the slices all have the same width, so the area of the slices is larger (and hence the density of R is higher) where T(x) ´ B(x) is larger.
H-10: Part (a) is a significantly different model from the last question.
H-11: Which method involves more work: horizontal strips or vertical strips?
H-12: This is a straightforward application of Equation 2.3.2 in the CLP-2 text.
H-13: Remember the derivative of arctangent is 1 1+x2 H-14: This is a straightforward application of Equations 2.3.3 and 2.3.4 in the CLP-2 text.
Note that you’re only asked for the y-coordinate of the centroid.
H-15: You can use a trigonometric substitution to find the area, then a partial fraction decomposition to find the y-coordinate of the centroid. Remember sin(1/2) = π/6.
154 H-16: Vertical slices will be easier than horizontal. An integration by parts might be helpful to find ¯ x, while trigonometric identities are important to finding ¯ y.
H-17: No trigonometric substitution is necessary if you’re clever with your u-substitutions, and remember the derivative of arctangent.
H-18: In R, the top function is x ´ x2, and the bottom function is x2 ´ 3x.
H-19: Remember d dxtarctan xu = 1 1+x2.
H-20: You can save quite a bit of work by, firstly, exploiting symmetry and, secondly, thinking about whether it is more efficient to use vertical strips or horizontal strips.
H-21: Sketch the region, being careful the domain of ?
9 ´ 4x2. You can save quite a bit of work by exploiting symmetry.
H-22: Horizontal slices will be easier than vertical.
H-23: Start with a picture: whether you use vertical slices or horizontal, you’ll need to break your integral into multiple pieces.
H-24: For practice, do the computation twice — once with horizontal strips and once with vertical strips. Watch for improper integrals.
H-25: Draw a sketch. In part (b) be careful about the equation of the right hand boundary of A.
H-26: Draw a sketch. Rotating about a horizontal line is similar to rotating about the x-axis, but for the radius of a slice, you’ll need to know |y ´ (´1)|: the distance from the outer edge of the region (the boundary function’s y-value) to y = ´1.
H-27: Go back to the derivation of Equation 2.3.3 in the CLP-2 text (centroid for a region) to figure out what to do when your surface does not have uniform density. We will consider a rod R that reaches from x = 0 to x = 4, and the mass of the section of the rod along [a, b] is equal to the mass of the strip of our rectangle along [a, b].
H-28: Horizontal slices will help you, where symmetry doesn’t, to set up a rod R whose centre of mass is the same as one coordinate of the centre of mass of the circle. When you’re integrating, trigonometric substitutions are sometimes the easiest way, and sometimes not.
The equation of a circle of radius 3, centred at (0, 3), is x2 + (y ´ 3)2 = 9.
H-29: The model in the question gives you the setup to solve this problem. You know how to find the centre of mass of a rod–that’s Equation 2.3.2 in the CLP-2 text — so all you need to find is ρ(y), the density of the rod at position y. To find this, consider a thin slice of the cone at position y with thickness dy. Its volume V(y) is the same as the mass of the small section of the rod at position y with thickness dy. So, the density of the rod at position y is ρ(y) = V(y) dy .
H-30: Use similar triangles to show that the shape of the lower (also upper) half of the hourglass is a truncated cone, where the untruncated cone would have had a height 10 cm.
155 To calculate the centre of mass of the upturned sand using the result of Question 29, you should find h = 9.8 (not h = 10–think carefully about our model from Question 29) and k = 8.8. For the centre of mass of the sand before turning, h = 10 and k = 6.
H-31: The techniques of Section 2.1 get pretty complicated here, so it’s easiest to use the techniques we developed in Questions 6, 29, and 30 in this section. That is, (1) find the height of the centre of mass of the water in its starting and ending positions, and then (2) model the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second.
The height change of the centre of mass is all that matters to calculate the work done against gravity, so you only have to worry about the height of the centres of mass.
H-32: The area of R is precisely one, so the error in your approximation is the error involved in approximating ş?
π/2 0 2x2 sin(x2) dx.
Hints for Exercises 2.4. — Jump to TABLE OF CONTENTS.
H-1: You don’t need to solve the differential equation from scratch, only verify whether the given function y = f (x) makes it true. Find dy dx and plug it into the differential equation.
H-2: For (d), note the equation given is quadratic in the variable dy dx.
H-3: The step ż 1 g(y) dy = ż f (x) dx shows up whether we’re using our mnemonic or not.
H-4: Note d dxtf (x)u = d dxtf (x) + Cu. Plug in y = f (x) + C to the equation dy dx = xy to see whether it makes the equation is true.
H-5: If a function is differentiable at a point, it is also continuous at that point.
H-6: Let Q(t) be the quantity of morphine in a patient’s bloodstream at time t, where t is measured in minutes.
Using the definition of a derivative, dQ dt = lim hÑ0 Q(t + h) ´ Q(t) h « Q(t + 1) ´ Q(t) 1 So, dQ dt is roughly the change in the amount of morphine in one minute, from t to t + 1.
H-7: If p(t) is the proportion of the new form, then 1 ´ p(t) is the proportion of the old form.
When we say two quantities are proportional, we mean that one is a constant multiple of the other.
H-8: The red marks show the slope y(x) would have at a point if it crosses that point. So, pick a value of y(0); based on the red marks, you can see how fast y(x) is increasing or decreasing at that point, which leads you roughly to a value of y(1); again, the red marks 156 tell you how fast y(x) is increasing or decreasing, which leads you to a value of y(2), etc (unless you’re already off the graph).
H-9: To draw the sketch similar to Question 8(d), don’t actually calculate every single slope; find a few (for instance, where the slope is zero, or where it’s negative), and use a pattern (for instance, the slope increases as y increases) to approximate most of the points.
H-10: Start by multiplying both sides of the equation by ey and dx, pretending that dy dx is a fraction, according to our mnemonic.
H-11: You need to solve for your function y(x) explicitly. Be careful with absolute values: if |y| = F, then y = F or y = ´F. However, y = ˘F is not a function. You have to choose one: y = F or y = ´F.
H-12: If your answer doesn’t quite look like the answer given, try manipulating it with logarithm rules: log a + log b = log(ab), and a log b = log(ba).
H-13: Simplify the equation.
H-14: Be careful with the arbitrary constant.
H-15: Start by cross-multiplying.
H-16: Be careful about signs. If y2 = F, then possibly y = ?
F, and possibly y = ´ ?
F.
However, y = ˘ ?
F is not a function.
H-17: Be careful about signs.
H-18: Be careful about signs. If log |y| = F, then |y| = eF. Since you should give your answer as an explicit function y(x), you need to decide whether y = eF or y = ´eF.
H-19: Move the y from the left hand side to the right hand side, then use partial fractions to integrate.
Be careful about the signs. Remember that we need y = ´1 when x = 1. This suggests how to deal with absolute values.
H-20: The unknown function f (x) satisfies an equation that involves the derivative of f.
H-21: Try guessing the partial fractions expansion of 1 x(x + 1).
Since x = 1 is in the domain and x = 0 is not, you may assume x ą 0 for all x in the domain.
H-22: d dxtsec xu = sec x tan x H-23: The general solution to the differential equation will contain the constant k and one other constant. They are determined by the data given in the question.
H-24: • When you’re solving the differential equation, you should have an integral that you can massage to look something like arctangent.
157 • What is the velocity of the object at its highest point?
• Your final answer will depend on the (unspecified) constants v0, m, g and k.
H-25: The general solution to the differential equation will contain the constant k and one other constant. They are determined by the data given in the question.
H-26: The method of partial fractions will help you integrate.
To solve x´a x´b = Y for x, move the terms containing x out of the denominator, then gather them on one side of the equals sign and factor out the x.
x ´ a x ´ b = Y x ´ a = Y(x ´ b) = Yx ´ Yb x ´ Yx = a ´ Yb x(1 ´ Y) = a ´ Yb x = a ´ Yb 1 ´ Y To find the limit, you can avoid l’Hˆ opital’s rule using some clever algebra–but you can also just use l’Hˆ opital’s rule.
H-27: Be careful about signs.
Part (a) has some algebraic similarities to Question 26.
H-28: The general solution to the differential equation will contain a constant of proportionality and one other constant. They are determined by the data given in the question.
H-29: You do not need to know anything about investing or continuous compounding to do this problem. You are given the differential equation explicitly. The whole first sentence is just window dressing.
H-30: Again, you do not need to know anything about investing to do this problem. You are given the differential equation explicitly.
H-31: Differentiate the given integral equation. Plugging in x = 0 gives you y(0).
H-32: Suppose that in a very short time interval dt, the height of water in the tank changes by dh (which is negative). Express in two different ways the volume of water that has escaped during this time interval. Equating the two gives the needed differential equation.
As the water escapes, it forms a cylinder of radius 1 cm.
H-33: Sketch the mercury in the tank at time t, when it has height h, and also at time t + dt, when it has height h + dh (with dh ă 0). The difference between those two volumes is the volume of (essentially) a disk of thickness ´dh. Figure out the radius and then the volume of that disk. This volume has to be the same as the volume of mercury that left through the hole in the bottom of the sphere, which runs out in the shape of a cylinder. Toricelli’s law tells you what the length of that cylinder is, and from there you 158 can find its volume. Setting the two volumes equal to each other gives the differential equation that determines h(t).
H-34: The fundamental theorem of calculus will be useful in part (b).
H-35: For any p ą 0, determine first y(t) (in terms of p and c) and then the times (also depending on p and c) at which y = 2, y = 1 and y = 0. The condition that “the top half takes exactly the same amount of time to drain as the bottom half” then gives an equation that determines p.
H-36: For (a), think of a very simple function.
The equation in the question statement is equivalent to the equation 1 ?x ´ a ż x a f (t) d(t) = dż x a f 2(t) dt which is, in some cases, easier to use.
For (d), you’ll want to let Y(x) = şx a f (t) dt, and use the quadratic equation.
H-37: Start by antidifferentiating both sides of the equation with respect to x.
Hints for Exercises 3.1. — Jump to TABLE OF CONTENTS.
H-1: Not every limit exists.
H-2: 100 isn’t all that big when you’re contemplating infinity. (Neither is any other number.) H-3: lim nÑ8 a2n+5 = lim nÑ8 an H-4: The sequence might be defined by different functions when n is large than when n is small.
H-5: Recall (´1)n is positive when n is even, and negative when n is odd.
H-6: Modify your answer from Question 5, but make the terms approach zero.
H-7: (´n)´n = (´1)n nn H-8: What might cause your answers in (a) and (b) to differ? Carefully read Theorem 3.1.6 in the CLP-2 text about convergent functions and their corresponding sequences.
H-9: You can use the fact that π is somewhat close to 22 7 , or you can use trial and error.
H-10: You can compare the leading terms, or factor a high power of n from the numerator and denominator.
H-11: This isn’t a rational expression, but you can treat it in a similar way. Recall e ă 3.
H-12: The techniques of evaluating limits of rational sequences are again useful here.
159 H-13: Use the squeeze theorem.
H-14: 1 n ď nsin n ď n H-15: e´1/n = 1 e1/n ; what happens to 1 n as n grows?
H-16: Use the squeeze theorem.
H-17: L’Hˆ opital’s rule might help you decide what happens if you are unsure.
H-18: Simplify ak.
H-19: What happens to 1 n as n gets very big?
H-20: cos 0 = 1 H-21: This is trickier than it looks. Write 1 n = x and look at the limit as x Ñ 0.
H-22: Multiply and divide by the conjugate.
H-23: Compared to Question 22, there’s an easier path.
H-24: Consider f 1(x), when f (x) = x100.
H-25: Look to Question 24 for inspiration.
H-26: The area of an isosceles triangle with two sides of length 1, meeting at an angle θ, is 1 2 sin θ.
θ sin θ 1 1 H-27: Every term of An is the same, and g(x) is a constant function.
H-28: You’ll need to use a logarithm before you can apply l’Hˆ opital’s rule.
H-29: (a) Write out the first few terms of the sequence.
(c) Consider how an+1 ´ L relates to an ´ L. What should happen to these numbers if an converges to L?
H-30: Your answer from (b) will help you a lot with the subsequent parts.
Hints for Exercises 3.2. — Jump to TABLE OF CONTENTS.
H-1: SN is the sum of the terms corresponding to n = 1 through n = N.
H-2: Note Ck is the cumulative number of cookies.
160 H-3: How is (a) related to Question 2?
H-4: You’ll have to calculate a1 separately from the other terms.
H-5: When does adding a number decrease the total sum?
H-6: For (b), imagine cutting up the triangle into its black and white parts, then sharing it equally among a certain number of friends. What is the easiest number of friends to share with, making sure each has the same area in their pile?
H-7: Compare to Question 6.
H-8: Iteratively divide a shape into thirds.
H-9: Equation 3.2.1 in the CLP-2 text tells us N ÿ n=0 arn = a1 ´ rN+1 1 ´ r , for r ‰ 1.
H-10: Note Ck is the cumulative number of cookies.
H-11: To adjust the starting index, either factor out the first term in the series, or subtract two series. For the subtraction option, consider Question 10.
H-12: Express your gains in (a) and (c) as series.
H-13: To find the difference between 8 ÿ n=1 cn and 8 ÿ n=1 cn+1, try writing out the first few terms.
H-14: You might want to first consider a simpler true or false: 8 ÿ n=1 an bn ?
= A B .
H-15: What kind of a series is this?
H-16: This is a special kind of series, that you should recognize.
H-17: When you see ÿ k ¨ ¨ ¨ k ¨ ¨ ¨ ´ ¨ ¨ ¨ k + 1 ¨ ¨ ¨ , you should think “telescoping series.” H-18: When you see ÿ n ¨ ¨ ¨ n ¨ ¨ ¨ ´ ¨ ¨ ¨ n + 1 ¨ ¨ ¨ , you should immediately think “telescoping series”. But be careful not to jump to conclusions — evaluate the nth partial sum explicitly.
H-19: Review Definition 3.2.3 in the CLP-2 text.
H-20: This is a special case of a general series whose sum we know.
H-21: Review Example 3.2.5 in the CLP-2 text. To write the number as a geometric series, the first few terms might not fit the pattern of the rest of the terms.
H-22: Start by writing it as a geometric series.
H-23: Review Example 3.2.5 in the CLP-2 text. Since the pattern repeats every three decimals, your common ratio r will be 1 103.
161 H-24: Split the series into two parts.
H-25: Split the series into two parts.
H-26: Split the series into two parts.
H-27: Use logarithm rules to turn this into a more obvious telescoping series.
H-28: This is a telescoping series.
H-29: The stone at position x has mass 1 4x kg, and we have to pull it a distance of 2x metres. From this, you can find the work involved in pulling up a single stone. Then, add up the work involved in pulling up all the stones.
H-30: The volume of a sphere of radius r is 4 3πr3.
H-31: Use the properties of a telescoping series to simplify the terms.
Recall sin2 θ + cos2 θ = 1.
H-32: Review Question 3 for using the sequence of partial sums.
H-33: What is the ratio of areas between the outermost (red) ring and the next (blue) ring?
Hints for Exercises 3.3. — Jump to TABLE OF CONTENTS.
H-1: That is, which series have terms whose limit is not zero?
H-2: That is, if f (x) is a function with f (n) = an for all whole numbers n, is f (x) nonnegative and decreasing?
H-3: This isn’t a trick. It’s meant to give you intuition to the direct comparison test.
H-4: The comparison test is Theorem 3.3.8 in the CLP-2 text. However, rather than trying to memorize which way the inequalities go in all cases, you can use the same reasoning as Question 3.
H-5: Think about Question 4 to remind yourself which way the inequalities have to go for direct comparison.
Note that all the comparison series have positive terms, so we don’t need to worry about that part of the limit comparison test.
H-6: The divergence test is Theorem 3.3.1 in the CLP-2 text.
H-7: The limit is calculated correctly.
H-8: It is true that f (x) is positive. What else has to be true of f (x) for the integral test to apply?
H-9: Refer to Question 4.
H-10: The definition of an alternating series is given in the start of Section 3.3.4 in the CLP-2 text.
162 H-11: For the ratio test to be inconclusive, lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ should be 1 or nonexistent.
H-12: By the divergence test, for a series ř an to converge, we need lim nÑ8 an = 0. That is, the magnitude (absolute value) of the terms needs to be getting smaller.
H-13: If f (x) is positive and decreasing, then the integral test tells you that the integral and the series either both increase or both decrease. So, in order to find an example with the properties required in the question, you need f (x) to not be both positive and decreasing.
H-14: Review Theorem 3.3.11 and Example 3.3.12 in the CLP-2 text.
H-15: Don’t jump to conclusions about properties of the an’s.
H-16: Always try the divergence test first (in your head).
H-17: Which test should you always try first (in your head)?
H-18: Review the integral test, which is Theorem 3.3.5 in the CLP-2 text.
H-19: A comparison might be helpful–try some algebraic manipulation to find a likely series to compare it to.
H-20: This is a geometric series.
H-21: Notice that the series is geometric, but it doesn’t start at n = 0.
H-22: Note n only takes integer values: what’s sin(πn) when n is an integer?
H-23: Note n only takes integer values: what’s cos(πn) when n is an integer?
H-24: What’s the test that you should always think of when you see a factorial?
H-25: This is a geometric series, but you’ll need to do a little algebra to figure out r.
H-26: Which test fits most often with factorials?
H-27: Try finding a nice comparison.
H-28: With the substitution u = log x, the function 1 x(log x)3/2 is easily integrable.
H-29: Combine the integral test with the results about p-series, Example 3.3.6 in the CLP-2 text.
H-30: Try the substitution u = ?x.
H-31: Review Example 3.3.9 in the CLP-2 text for developing intuition about comparisons, and Example 3.3.10 for an example where finding an appropriate comparison series calls for some creativity.
H-32: What does the summand look like when k is very large?
H-33: What does the summand look like when n is very large?
H-34: cos(nπ) is a sneaky way to write (´1)n.
H-35: What is the behaviour for large k?
163 H-36: When m is large, 3m + sin ?m « 3m.
H-37: This is a geometric series, but it doesn’t start at n = 0.
H-38: The series is geometric.
H-39: The first series can be written as 8 ÿ n=1 1 2n ´ 1 .
H-41: What does the summand look like when n is very large?
H-42: Review the alternating series test, which is given in Theorem 3.3.14 in the CLP-2 text.
H-43: Review the alternating series test, which is given in Theorem 3.3.14 in the CLP-2 text.
H-44: Review the alternating series test, which is given in Theorem 3.3.14 in the CLP-2 text.
H-46: For part (a), see Example 1.12.23 in the CLP-2 text.
For part (b), review Theorem 3.3.5 in the CLP-2 text.
For part (c), see Example 3.3.12 in the CLP-2 text.
H-47: The truncation error arising from the approximation 8 ÿ n=1 e´?n ?n « N ÿ n=1 e´?n ?n is precisely EN = 8 ÿ n=N+1 e´?n ?n . You’ll want to find a bound on this sum using the integral test.
A key observation is that, since f (x) = e´?x ?x is decreasing, we can show that e´?n ?n ď ż n n´1 e´?x ?x dx for every n ě 1.
H-48: What does the fact that the series 8 ř n=0 an converges guarantee about the behavior of an for large n?
H-49: What does the fact that the series 8 ř n=0 (1 ´ an) converges guarantee about the behavior of an for large n?
H-50: What does the fact that the series 8 ÿ n=1 nan ´ 2n + 1 n + 1 converges guarantee about the behavior of an for large n?
H-51: What does the fact that the series ř8 n=1 an converges guarantee about the behavior of an for large n? When is x2 ď x?
164 H-52: If we add together the frequencies of all the words, they should amount to 100%.
We can approximate this sum using ideas from Example 3.3.4 in the CLP-2 text.
H-53: We are approximating a finite sum — not an infinite series. To get greater accuracy, use exact values for the first several terms in the sum, and use an integral to approximate the rest.
Hints for Exercises 3.4. — Jump to TABLE OF CONTENTS.
H-1: What is conditional convergence?
H-2: If ř |an| converges, then ř an is guaranteed to converge as well.
(That’s Theorem 3.4.2 in the CLP-2 text.) So, one of the blank spaces describes an impossible sequence.
H-4: Be careful about the signs.
H-5: Does the alternating series test really apply?
H-6: What does the summand look like when n is very large?
H-7: What does the summand look like when n is very large?
H-8: This is a trick question. Be sure to verify all of the hypotheses of any convergence test you apply.
H-9: Try the substitution u = log x.
H-10: Show that it converges absolutely.
H-11: Use a similar method to Queston 10.
H-12: Show it converges absolutely using a direct comparison test.
H-13: For part (a), replace n by x in the absolute value of the summand. Can you integrate the resulting function?
H-14: You don’t need to add up very many terms for this level of accuracy.
H-15: Use the direct comparison test to show that the series converges absolutely.
Hints for Exercises 3.5. — Jump to TABLE OF CONTENTS.
H-1: f (1) is the sum of a geometric series.
H-2: Calculate d dx "(x ´ 5)n n! + 2 when n is a constant.
H-3: There is only one.
H-4: Use Theorem 3.5.9 in the CLP-2 text.
H-5: Review the discussion immediately following Definition 3.5.1 in the CLP-2 text.
H-6: Review the discussion immediately following Definition 3.5.1 in the CLP-2 text.
165 H-7: Review the discussion immediately following Definition 3.5.1 in the CLP-2 text.
H-8: See Example 3.5.11 in the CLP-2 text.
H-9: See Example 3.5.11 in the CLP-2 text.
H-15: Start part (b) by computing the partial sums of 8 ÿ k=1 ak ak+1 ´ ak+1 ak+2 H-16: You should know a power series representation for 1 1 ´ x. Use it.
H-17: You can safely ignore one of the given equations, but not the other.
H-18: n ě log n for all n ě 1.
H-19: See Example 3.5.21 in the CLP-2 text. For part (b), review §3.3.4 in the CLP-2 text.
H-20: You know the geometric series expansion of 1 1´x. What (calculus) operation(s) can you apply to that geometric series to convert it into the given series?
H-21: First show that the fact that the series ř8 n=0(1 ´ bn) converges guarantees that limnÑ8 bn = 1.
H-22: What does an look like for large n?
H-23: Equation 2.3.1 in the CLP-2 text tells us the centre of mass of a rod with weights tmnu at positions txnu is ¯ x = ř mnxn ř mn .
H-24: Use the second derivative test.
H-25: What function has 8 ÿ n=1 nxn´1 as its power series representation?
H-26: The power series representation in Example 3.5.20 is an alternating series when x is positive.
H-27: The power series representation in Example 3.5.21 is an alternating series when x is nonzero.
Hints for Exercises 3.6. — Jump to TABLE OF CONTENTS.
H-1: Which of the functions are constant, linear, and quadratic?
H-2: You don’t have to actually calculate the entire series T(x) to answer the question.
H-3: If you don’t have these memorized, it’s good to be able to derive them. For instance, log(1 + x) is the antiderivative of 1 1 + x, whose Taylor series can be found by modifying the geometric series ř xn.
H-4: See Example 3.6.16 in the CLP-2 text.
H-5: The series will bear some resemblance to the Maclaurin series for log(1 + x).
166 H-6: The terms f (n)(π) are going to be similar to the terms f (n)(0) that we used in the Maclaurin series for sine.
H-7: The Taylor series will look similar to a geometric series.
H-8: Your answer will depend on a.
H-9: You should know the Maclaurin series for 1 1 ´ x. Use it.
H-10: You should know the Maclaurin series for 1 1 ´ x. Use it.
H-11: You should know the Maclaurin series for ex. Use it.
H-12: Review Example 3.5.20 in the CLP–2 text.
H-13: You should know the Maclaurin series for sin x. Use it.
H-14: You should know the Maclaurin series for ex. Use it.
H-15: You should know the Maclaurin series for arctan(x). Use it.
H-16: You should know the Maclaurin series for 1 1 ´ x. Use it.
H-17: Set (´1)n x2n+1 2n + 1 = C (´1)n (2n + 1)3n , for some constant C. What are x and C?
H-18: There is an important Taylor series, one of the series in Theorem 3.6.5 of the CLP-2 text, that looks a lot like the given series.
H-19: There is an important Taylor series, one of the series in Theorem 3.6.5 of the CLP-2 text, that looks a lot like the given series.
H-20: There is an important Taylor series, one of the series in Theorem 3.6.5 of the CLP-2 text, that looks a lot like the given series. Be careful about the limits of summation.
H-21: There is an important Taylor series, one of the series in Theorem 3.6.5 of the CLP-2 text, that looks a lot like the given series.
H-22: Split the series into a sum of two series. There is an important Taylor series, one of the series in Theorem 3.6.5 of the CLP-2 text, that looks a lot like each of the two series.
H-23: Try the ratio test.
H-24: Write it as the sum of two Taylor series.
H-25: Can you think of a way to eliminate the odd terms from ex = 8 ÿ n=0 xn n! ?
H-26: The series you’re adding up are alternating, so it’s simple to bound the error using a partial sum.
H-27: The Taylor Series is alternating, so bounding the error in a partial-sum approximation is straightforward.
167 H-28: The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.
H-29: The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.
H-30: Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. This theorem requires you to consider values of c between x and x = 0; since x could be anything from ´2 to 1, you should think about values of c between ´2 and 1.
H-31: Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.
To bound the derivative over the appropriate range, remember how to find absolute extrema.
H-32: See Example 3.6.21 in the CLP-2 text H-33: See Example 3.6.21 in the CLP-2 text H-34: Set f (x) = 1 + x + x22/x, and find lim xÑ0 log ( f (x)).
H-35: Use the substitution y = 1 x, and compare to Question 34.
H-36: Start by differentiating 8 ÿ n=0 xn.
H-37: The series bears a resemblance to the Taylor series for arctangent.
H-38: For simplification purposes, note (1)(3)(5)(7) ¨ ¨ ¨ (2n ´ 1) = (2n)!
2n n! .
H-39: You know the Maclaurin series for log(1 + y). Use it! Remember that you are asked for a series expansion in powers of x ´ 2. So you want y to be some constant times x ´ 2.
H-40: See Example 3.5.21 in the CLP-2 text. For parts (b) and (c), review §3.3.4 in the CLP-2 text.
H-41: Look at the signs of successive terms in the series.
H-42: The magic word is “series”.
H-43: See Example 3.6.14 in the CLP-2 text. For parts (b) and (c), review §3.3.4 in the CLP-2 text.
H-44: See Example 3.6.14 in the CLP-2 text. For part (b), review the fundamental theorem of calculus in §1.3 of the CLP-2 text. For part (c), review §3.3.4 in the CLP-2 text.
H-45: See Example 3.6.14 in the CLP-2 text. For parts (b) and (c), review §3.3.4 in the CLP-2 text.
H-46: See Example 3.6.14 in the CLP-2 text. For parts (b) and (c), review §3.3.4 in the CLP-2 text.
168 H-48: Use the Maclaurin series for ex.
H-49: For part (c), compare two power series term-by-term.
H-50: For Newton’s method, recall we approximate a root of the function g(x) in iterations: given an approximation xn, our next approximation is xn+1 = xn ´ g(xn) g1(xn).
To gauge your error, note that from approximation to approximation, the first digits stabilize. Keep refining your approximation until the first two digits stop changing.
H-51: First, modify your known Maclaurin series for arctangent into a Maclaurin series for f (x). This series is not hard to repeatedly differentiate, so use it to find a power series for f (10)(x).
H-52: Remember ex is never negative for any real number x.
H-53: Since f (x) is odd, f (´x) = ´f (x) for all x in its domain. Consider the even-indexed terms and odd-indexed terms of the Taylor series.
169 ANSWERS TO PROBLEMS Part III 170 Answers to Exercises 1.1 — Jump to TABLE OF CONTENTS A-1: The area is between 1.5 and 2.5 square units.
A-2: The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but this is a reasonable estimate, using methods from this chapter.) A-3: The area under the curve is a number in the interval 3 8 h 1 2 + 1 ?
2 i , 3 8 h 1 + 1 ?
2 i .
A-4: left A-5: Many answers are possible. One example is f (x) = sin x, [a, b] = [0, π], n = 1.
Another example is f (x) = sin x, [a, b] = [0, 5π], n = 5.
A-6: Some of the possible answers are given, but more exist.
(a) 7 ÿ i=3 i ; 5 ÿ i=1 (i + 2) (b) 7 ÿ i=3 2i ; 5 ÿ i=1 (2i + 4) (c) 7 ÿ i=3 (2i + 1) ; 5 ÿ i=1 (2i + 5) (d) 8 ÿ i=1 (2i ´ 1) ; 7 ÿ i=0 (2i + 1) A-7: Some answers are below, but others are possible.
(a) 4 ÿ i=1 1 3i ; 4 ÿ i=1 1 3 i (b) 4 ÿ i=1 2 3i ; 4 ÿ i=1 2 1 3 i (c) 4 ÿ i=1 (´1)i 2 3i ; 4 ÿ i=1 2 (´3)i (d) 4 ÿ i=1 (´1)i+1 2 3i ; 4 ÿ i=1 ´ 2 (´3)i A-8: (a) 5 ÿ i=1 2i ´ 1 3i 171 (b) 5 ÿ i=1 1 3i + 2 (c) 7 ÿ i=1 i ¨ 104´i ; 7 ÿ i=1 i 10i´4 A-9: (a) 5 2 " 1 ´ 3 5 101# (b) 5 2 3 5 50 " 1 ´ 3 5 51# (c) 270 (d) 1 ´ 1 e b e ´ 1 + e 4 [b(b + 1)]2 A-10: (a) 50 ¨ 51 = 2550 (b) h 1 2(95)(96) i2 ´ h 1 2(4)(5) i2 = 20, 793, 500 (c) ´1 (d) ´10 A-11: x y b a y = f (x) A-12: n = 4, a = 2, and b = 6 A-13: One answer is below, but other interpretations exist.
172 x y 5 7 9 11 49 81 121 y = x2 A-14: Many interpretations are possible–see the solution to Question 13 for a more thorough discussion–but the most obvious is given below.
x y 2π 20 3π 20 4π 20 5π 20 π 20 y = tan x A-15: Three answers are possible. It is a midpoint Riemann sum for f on the interval [1, 5] with n = 4. It is also a left Riemann sum for f on the interval [1.5, 5.5] with n = 4. It is also a right Riemann sum for f on the interval [0.5, 4.5] with n = 4.
A-16: 25 2 A-17: 21 2 A-18: 50 ř i=1 5 + i ´ 1/2 1 5 8 1 5 173 A-19: 54 A-20: ż 7 ´1 f (x) dx = lim nÑ8 n ÿ i=1 f ´1 + 8i n 8 n A-21: f (x) = sin2(2 + x) and b = 4 A-22: f (x) = x ?
1 ´ x2 A-23: ş3 0 e´x/3 cos(x) dx A-24: ż 1 0 xex dx A-25: Possible answers include: 2 ż 0 e´1´x dx, 3 ż 1 e´x dx, 2 ż 3/2 1/2 e´2x dx, and 2 1 ż 0 e´1´2x dx.
A-26: r3n+3 ´ 1 r3 ´ 1 A-27: r5 r96 ´ 1 r ´ 1 A-28: 5 A-29: 16 A-30: b2 ´ a2 2 A-31: b2 ´ a2 2 A-32: 4π A-33: ż 3 0 f (x) dx = 2.5 A-34: 53 m A-35: true A-36: 3200 km A-37: (a) There are many possible answers. Two are ş0 ´2 ?
4 ´ x2 dx and ş2 0 a 4 ´ (´2 + x)2 dx. (b) π A-38: (a) 30 (b) 411 4 A-39: 56 3 A-40: 6 A-41: 12 174 A-42: f (x) = 3 10 x 5 + 8 2 sin 2x 5 + 2 A-43: 1 log 2 A-44: (a) 1 log 10 10b ´ 10a (b) 1 log c cb ´ ca ; yes, it agrees.
A-45: π 4 ´ 1 2 arccos(a) + 1 2a ?
1 ´ a2 A-46: (a) [ f (b) ´ f (a)] ¨ b ´ a n (b) Choose n to be an integer that is greater than or equal to 100 [ f (b) ´ f (a)] (b ´ a).
A-47: true (but note, for a non-linear function, it is possible that the midpoint Riemann sum is not the average of the other two) Answers to Exercises 1.2 — Jump to TABLE OF CONTENTS A-1: Possible drawings: x y a y = f (x) x y a c b y = f (x) x y a b y = f (x) y = f (x) + g(x) A-2: sin b ´ sin a A-3: (a) False. For example, the function f (x) = # 0 for x ă 0 1 for x ě 0 provides a counterexample.
(b) False. For example, the function f (x) = x provides a counterexample.
(c) False. For example, the functions f (x) = # 0 for x ă 1 2 1 for x ě 1 2 and g(x) = # 0 for x ě 1 2 1 for x ă 1 2 175 provide a counterexample.
A-4: (a) ´ 1 20 (b) positive (c) negative (d) positive A-5: ´21 A-6: ´6 A-7: 20 A-8: (a) π 4 ´ 1 2 arccos(´a) ´ 1 2a ?
1 ´ a2 = ´π 4 + 1 2 arccos(a) ´ 1 2a ?
1 ´ a2 (b) 1 2 arccos(a) ´ 1 2a ?
1 ´ a2 A-9: 5 A-10: 0 A-11: 5 A-12: 20 + 2π A-13: 0 A-14: 0 A-15: 0 A-16: (a) y = 1 b a 1 ´ (ax)2 (b) a b ż 1 a ´ 1 a c 1 a2 ´ x2 dx (c) π ab A-17: ˆ even odd even even odd odd odd even A-18: f (0) = 0; g(0) can be any real number A-19: f (x) = 0 for every x A-20: The derivative of an even function is odd, and the derivative of an odd function is even.
Answers to Exercises 1.3 — Jump to TABLE OF CONTENTS A-1: e2 ´ e´2 A-2: F(x) = x4 4 + 1 2 cos 2x + 1 2.
A-3: (a) True (b) False (c) False, unless şb a f (x) dx = şb a x f (x) dx = 0.
A-4: false 176 A-5: false A-6: sin(x2) A-7: 3 ?e A-8: For any constant C, F(x) + C is an antiderivative of f (x). So, for example, F(x) and F(x) + 1 are both antiderivatives of f (x).
A-9: (a) We differentiate with respect to a. Recall d dxtarccos xu = ´1 ?
1´x2. To differentiate 1 2a ?
1 ´ a2, we use the product and chain rules.
d da "π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2 = 0 ´ 1 2 ¨ ´1 ?
1 ´ a2 + 1 2a ¨ ´2a 2 ?
1 ´ a2 + 1 2 a 1 ´ a2 = 1 2 ?
1 ´ a2 ´ a2 2 ?
1 ´ a2 + 1 ´ a2 2 ?
1 ´ a2 = 1 ´ a2 + 1 ´ a2 2 ?
1 ´ a2 = 2(1 ´ a2) 2 ?
1 ´ a2 = a 1 ´ a2 (b) F(x) = 5π 4 ´ 1 2 arccos(x) + 1 2x ?
1 ´ x2 A-10: (a) 0 (b),(c) The FTC does not apply, because the integrand is not continuous over the interval of integration.
A-11: t y a x x + h y = f (t) A-12: (a) zero (b) increasing when 0 ă x ă 1 and 3 ă x ă 4; decreasing when 1 ă x ă 3 A-13: (a) zero (b) G(x) is increasing when 1 ă x ă 3, and it is decreasing when 0 ă x ă 1 and when 3 ă x ă 4.
177 A-14: Using the definition of the derivative, F1(x) = lim hÑ0 F(x + h) ´ F(x) h = lim hÑ0 şx+h a t dt ´ şx a t dt h = lim hÑ0 şx+h x t dt h The numerator describes the area of a trapezoid with base h and heights x and x + h.
= lim hÑ0 1 2h(x + x + h) h = lim hÑ0 x + 1 2h = x t y x x + h x x + h y = t şx+h x t dt So, F1(x) = x.
A-15: f (t) = 0 A-16: ş log(ax) dx = x log(ax) ´ x + C, where a is a given constant, and C is any constant.
A-17: ş x3ex dx = ex x3 ´ 3x2 + 6x ´ 6 + C A-18: ż 1 ?
x2 + a2 dx = log ˇ ˇ ˇx + a x2 + a2 ˇ ˇ ˇ + C when a is a given constant. As usual, C is an arbitrary constant.
A-19: ż x a x(a + x) dx = b x(a + x) ´ a log ?x + ?
a + x + C A-20: 5 ´ cos 2 A-21: 2 A-22: 1 5 arctan(5x) + C A-23: arcsin x ?
2 + C 178 A-24: tan x ´ x + C A-25: ´3 4 cos(2x) + C, or equivalently, 3 2 sin2 x + C A-26: 1 2x + 1 4 sin(2x) + C A-27: F1 π 2 = log(3) G1 π 2 = ´ log(3) A-28: f (x) is increasing when ´8 ă x ă 1 and when 2 ă x ă 8.
A-29: F1(x) = ´ sin x cos3 x + 6 A-30: 4x3e(1+x4)2 A-31: sin6 x + 8) cos x A-32: F1(1) = 3e´1 A-33: sin u 1 + cos3 u A-34: f (x) = 2x A-35: f (4) = 4π A-36: (a) (2x + 1)e´x2 (b) x = ´1/2 A-37: esin x ´ esin(x4´x3) 4x3 ´ 3x2 A-38: ´2x cos e´x2 ´ 5x4 cos ex5 A-39: exa sin(ex) ´ a sin(x) A-40: 14 A-41: 5 2 A-42: 45 m A-43: f 1(x) = (2 ´ 2x) log 1 + e2x´x2 and f (x) achieves its absolute maximum at x = 1, because f (x) is increasing for x ă 1 and decreasing for x ą 1.
A-44: The minimum is ş´1 0 dt 1+t4. As x runs from ´8 to 8, the function f (x) = şx2´2x 0 dt 1+t4 decreases until x reaches 1 and then increases all x ą 1. So the minimum is achieved for x = 1. At x = 1, x2 ´ 2x = ´1.
A-45: F achieves its maximum value at x = π.
A-46: 2 A-47: log 2 179 A-48: In the sketch below, open dots denote inflection points, and closed dots denote extrema.
x y y = F(x) y = f (x) ´5 ´3 ´1 1 3 5 A-49: (a) 3x2 ż x3+1 0 et3dt + 3x5e(x3+1)3 (b) y = ´3(x + 1) A-50: Both students.
A-51: (a) 27(1 ´ cos 3) (b) x3 sin(x) + 3x2[1 ´ cos(x)] A-52: If f (x) = 0 for all x, then F(x) is even and possibly also odd.
If f (x) ‰ 0 for some x, then F(x) is not even. It might be odd, and it might be neither even nor odd.
(Perhaps surprisingly, every antiderivative of an odd function is even.) Answers to Exercises 1.4 — Jump to TABLE OF CONTENTS A-1: (a) true (b) false A-2: The reasoning is not sound: when we do a substitution, we need to take care of the differential (dx). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here’s the way to do it: Problem: Evaluate ż (2x + 1)2dx.
Work: We use the substitution u = 2x + 1. Then du = 2dx, so dx = 1 2du: ż (2x + 1)2dx = ż u2 ¨ 1 2 du = 1 6u3 + C = 1 6 (2x + 1)3 + C 180 A-3: The problem is with the limits of integration, as in Question 1. Here’s how it ought to go: Problem: Evaluate ż π 1 cos(log t) t dt.
Work: We use the substitution u = log t, so du = 1 t dt. When t = 1, we have u = log 1 = 0 and when t = π, we have u = log(π). Then: ż π 1 cos(log t) t dt = ż log(π) log 1 cos(u)du = ż log(π) 0 cos(u)du = sin(log(π)) ´ sin(0) = sin(log(π)).
A-4: This one is OK.
A-5: ż 1 0 f (u) ?
1 ´ u2 du. Because the denominator ?
1 ´ u2 vanishes when u = 1, this is what is known as an improper integral. Improper integrals will be discussed in § 1.12 of the CLP-2 text.
A-6: some constant C A-7: 1 2 sin(e) ´ sin(1) A-8: 1 3 A-9: ´ 1 300(x3 + 1)100 + C A-10: log 4 A-11: log 2 A-12: 4 3 A-13: e6 ´ 1 A-14: 1 3(4 ´ x2)3/2 + C A-15: e ?
log x + C A-16: 0 A-17: 1 2[cos 1 ´ cos 2] « 0.478 A-18: 1 2 ´ 1 2 log 2 A-19: 1 2 tan2 θ ´ log | sec θ| + C 181 A-20: arctan(ex) + C A-21: π 4 ´ 2 3 A-22: ´1 2 (log(cos x))2 + C A-23: 1 2 sin(1) A-24: 1 3[2 ?
2 ´ 1] « 0.609 A-25: Using the definition of a definite integral with right Riemann sums: ż b a 2f (2x)dx = lim nÑ8 n ÿ i=1 ∆x ¨ 2f (2(a + i∆x)) ∆x = b ´ a n = lim nÑ8 n ÿ i=1 b ´ a n ¨ 2f 2 a + i b ´ a n = lim nÑ8 n ÿ i=1 2b ´ 2a n ¨ f 2a + i 2b ´ 2a n ż 2b 2a f (x)dx = lim nÑ8 n ÿ i=1 ∆x ¨ f (2a + i∆x) ∆x = 2b ´ 2a n = lim nÑ8 n ÿ i=1 2b ´ 2a n ¨ f 2a + i 2b ´ 2a n Since the Riemann sums are exactly the same, ż b a 2f (2x)dx = ż 2b 2a f (x)dx Answers to Exercises 1.5 — Jump to TABLE OF CONTENTS A-1: Area between curves « π 4 2 + ?
2 x y π π 2 π 4 3π 4 y = cos x y = sin x A-2: (a) Vertical rectangles: 182 x y π 10 π 5 3π 10 2π 5 π 2 y = arcsin 2x π y = b xπ 2 (b) One possible answer: x y π 10 π 5 3π 10 2π 5 π 2 x = π 2 sin y x = 2 πy2 A-3: ż ?
2 0 2x ´ x3 dx A-4: ż 4 ´3/2 4 5(6 ´ y2) + 2y dy A-5: ż 4a 0 ?
4ax ´ x2 4a dx A-6: ż 25 1 ´ 1 12(x + 5) + 1 2 ?x dx 183 A-7: 1 8 A-8: 4 3 A-9: 5 3 ´ 1 log 2 A-10: 8 π ´ 1 A-11: 20 9 A-12: 1 6 A-13: 2π A-14: 2 h π ´ 1 4π2i A-15: 31 6 A-16: 26 3 A-17: 7π 8 ´ 1 2 A-18: 12 ?
2 ´ 13 4 Answers to Exercises 1.6 — Jump to TABLE OF CONTENTS A-1: The horizontal cross-sections are circles, but the vertical cross-sections are not.
A-2: The columns have the same volume.
A-3: Washers when 1 ă y ď 6: If y ą 1, then our washer has inner radius 2 + 2 3y, outer radius 6 ´ 2 3y, and height dy.
184 y thickness: dy R = 6 ´ 2 3y r = 2 + 2 3y Washers when 0 ď y ă 1: When 0 ď y ă 1, we have a “double washer,” two concentric rings. The inner washer has inner radius r1 = y and outer radius R1 = 2 ´ y. The outer washer has inner radius r2 = 2 + 2 3y and outer radius R2 = 6 ´ 2 3y. The thickness of the washers is dy.
y thickness: dy R1 = 2 ´ y R2 = 6 ´ 2 3y r1 = y r2 = 2 + 2 3y 185 A-4: (a) π ż 3 0 xe2x2 dx (b) ż 1 0 π 3 + ?y 2 ´ 3 ´ ?y 2 dy + ż 4 1 π 5 ´ y 2 ´ 3 ´ ?y 2 dy A-5: (a) ż 1 ´1 π (5 ´ 4x2) 2 ´ (2 ´ x2) 2 dx (b) ż 0 ´1 π 5 + a y + 1 2 ´ 5 ´ a y + 1 2 dy A-6: π ż 2 ´2 (9 ´ x2) 2 ´ (x2 + 1) 2 dx A-7: ?
2 12 ℓ3 A-8: π 4 e2a2 ´ 1 A-9: π 38 3 ´ 514 34 = π512 81 A-10: (a) 8π ş1 ´1 ?
1 ´ x2 dx (b) 4π2 A-11: (a) The region R is the region between the blue and red curves, with 3 ď x ď 5, in the figures below.
(3, √ 24) (5, √ 40) R y2 = 8x y2 = x2 + 15 x y (3, √ 24) (5, √ 40) R (b) 4 3π « 4.19 A-12: (a) The region R is sketched below.
x y y = log x x = 2 (b) π h 4 log 2 ´ 3 2 i « 3.998 A-13: π2 + 8π3 + 8π6 5 186 A-14: 8 3 A-15: 256 ˆ 8 15 = 136.5˙ 3 A-16: 28 3 πh A-17: (a) 4π 3b2a cubic units (b) a = 1 6356.752 and b = 1 6378.137 (c) Approximately 1.08321 ˆ 1012 km3, or 1.08321 ˆ 1021 m3 (d) Absolute error is about 3.64 ˆ 109 km3, and relative error is about 0.00336, or 0.336%.
A-18: (a) 9 2 (b) π ż 2 ´1 4 ´ x 2 ´ 1 + (x ´ 1)22 dx A-19: (a) π 2 ´ 1 (b) π2 2 ´ π « 1.793 A-20: (a) V1 = 4 3πc2 (b) V2 = π c 3 4 ?
2 ´ 2 (c) c = 0 or c = ?
2 ´ 1 2 A-21: ż π π/2 π (5 + π sin x)2 ´ (5 + 2π ´ 2x)2 dx + ż 3π/2 π π (5 + 2π ´ 2x)2 ´ (5 + π sin x)2 dx A-22: (a) 6000cπ log 2 1 ´ 1 210 , which is close to 6000cπ log 2 .
(b) 6km: that is, there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.
Answers to Exercises 1.7 — Jump to TABLE OF CONTENTS A-1: chain; product A-2: The part chosen as u will be differentiated. The part chosen as dv will be antidifferentiated.
A-3: ż f 1(x) g(x) dx = f (x) g(x) + ż f (x)g1(x) g2(x) dx A-4: All the antiderivatives differ only by a constant, so we can write them all as 187 v(x) + C for some C. Then, using the formula for integration by parts, ż u(x) ¨ v1(x)dx = u(x) lo omo on u v(x) + C looooomooooon v ´ ż v(x) + C looooomooooon v u1(x)dx looomooon du = u(x)v(x) + Cu(x) ´ ż v(x)u1(x)dx ´ ż Cu1(x)dx = u(x)v(x) + Cu(x) ´ ż v(x)u1(x)dx ´ Cu(x) + D = u(x)v(x) ´ ż v(x)u1(x)dx + D where D is any constant.
Since the terms with C cancel out, it didn’t matter what we chose for C–all choices end up the same.
A-5: Suppose we choose dv = f (x)dx, u = 1. Then v = ż f (x)dx, and du = dx. So, our integral becomes: ż (1) lo omo on u f (x)dx loomoon dv = (1) lo omo on u ż f (x)dx loooomoooon v ´ ż ż f (x)dx loooooomoooooon v dx lo omo on du In order to figure out the first product (and the second integrand), you need to know the antiderivative of f (x)–but that’s exactly what you’re trying to figure out!
A-6: x2 log x 2 ´ x2 4 + C A-7: ´log x 6x6 ´ 1 36x6 + C A-8: π A-9: π 2 ´ 1 A-10: ex x3 ´ 3x2 + 6x ´ 6 + C A-11: x2 2 log3 x ´ 3x2 4 log2 x + 3x2 4 log x ´ 3x2 8 + C A-12: (2 ´ x2) cos x + 2x sin x + C A-13: t3 ´ 5 2t2 + 6t log t ´ 1 3t3 + 5 4t2 ´ 6t + C A-14: e ?s (2s ´ 4?s + 4) + C A-15: x log2 x ´ 2x log x + 2x + C A-16: ex2+1 + C A-17: y arccos y ´ a 1 ´ y2 + C 188 A-18: 2y2 arctan(2y) ´ y + 1 2 arctan(2y) + C A-19: x3 3 arctan x ´ 1 6(1 + x2) + 1 6 log(1 + x2) + C A-20: 2 17ex/2 cos(2x) + 8 17ex/2 sin(2x) + C A-21: x 2 sin(log x) ´ cos(log x) + C A-22: 2x log 2 x ´ 1 log 2 + C A-23: 2ecos x[1 ´ cos x] + C A-24: ż xe´x (1 ´ x)2dx = xe´x 1 ´ x + e´x + C = e´x 1 ´ x + C A-25: (a) We integrate by parts with u = sinn´1 x and dv = sin x dx, so that du = (n ´ 1) sinn´2 x cos x and v = ´ cos x.
ż sinn x dx = ´ sinn´1 x cos x looooooooomooooooooon uv + (n ´ 1) ż cos2 x sinn´2 x dx looooooooooooooooomooooooooooooooooon ´ ş vdu Using the identity sin2 x + cos2 x = 1, = ´ sinn´1 x cos x + (n ´ 1) ż (1 ´ sin2 x) sinn´2 x dx = ´ sinn´1 x cos x + (n ´ 1) ż sinn´2 x dx ´ (n ´ 1) ż sinn x dx Moving the last term on the right hand side to the left hand side gives n ż sinn x dx = ´ sinn´1 x cos x + (n ´ 1) ż sinn´2 x dx Dividing across by n gives the desired reduction formula.
(b) 35 256π « 0.4295 A-26: (a) Area: π 4 ´ log 2 2 x = 1 y = tan−1 x x y (b) Volume: π2 2 ´ π 189 A-27: π 17e18 ´ 4373 36 A-28: 12 A-29: 2 e Answers to Exercises 1.8 — Jump to TABLE OF CONTENTS A-1: (e) A-2: 1 n secn x + C A-3: We divide both sides by cos2 x, and simplify.
sin2 x + cos2 x = 1 sin2 x + cos2 x cos2 x = 1 cos2 x sin2 x cos2 x + 1 = sec2 x tan2 x + 1 = sec2 x A-4: sin x ´ sin3 x 3 + C A-5: π 2 A-6: sin37 t 37 ´ sin39 t 39 + C A-7: 1 3 cos3 x ´ 1 cos x + C A-8: π 8 ´ 9 ?
3 64 A-9: ´ cos x + 2 3 cos3 x ´ 1 5 cos5 x + C A-10: 1 2.2 sin2.2 x + C A-11: 1 2 tan2 x + C, or equivalently, 1 2 sec2 +C A-12: 1 7 sec7 x ´ 1 5 sec5 x + C A-13: tan49 x 49 + tan47 x 47 + C A-14: 1 3.5 sec3.5 x ´ 1 1.5 sec1.5 x + C 190 A-15: 1 4 sec4 x ´ 1 2 sec2 x + C or 1 4 tan4 x + C A-16: 1 5 tan5 x + C A-17: 1 1.3 sec1.3 x + 1 0.7 cos0.7 x + C A-18: = 1 4 sec4 x ´ sec2 x + log | sec x| + C A-19: 41 45 ?
3 ´ π 6 A-20: 1 11 + 1 9 A-21: 2?sec x + C A-22: tane+1 θ tan6 θ 7 + e + 3 tan4 θ 5 + e + 3 tan2 θ 3 + e + 1 1 + e + C A-23: (a) Using the trig identity tan2 x = sec2 x ´ 1 and the substitution y = tan x, dy = sec2 x dx, ż tann x dx = ż tann´2 x tan2 x dx = ż tann´2 x sec2 x dx ´ ż tann´2 x dx = ż yn´2 dy ´ ż tann´2 x dx = yn´1 n ´ 1 ´ ż tann´2 x dx = tann´1 x n ´ 1 ´ ż tann´2 x dx (b) 13 15 ´ π 4 « 0.0813 A-24: 1 2 cos2 x + 2 log | cos x| ´ 1 2 cos2 x + C A-25: tan θ + C A-26: log | sin x| + C A-27: 1 2 sin2(ex) + C A-28: (sin2 x + 2) cos(cos x) + 2 cos x sin(cos x) + C A-29: x 2 sin2 x ´ x 4 + 1 4 sin x cos x + C Answers to Exercises 1.9 — Jump to TABLE OF CONTENTS A-1: (a) x = 4 3 sec θ (b) x = 1 2 sin θ (c) x = 5 tan θ 191 A-2: (a) x ´ 2 = ?
3 sec u (b) x ´ 1 = ?
5 sin u (c) 2x + 3 2 = ?
31 2 tan u (d) x ´ 1 2 = 1 2 sec u A-3: (a) ?
399 20 (b) 5 ?
2 7 (c) ?
x ´ 5 2 A-4: (a) ?
4 ´ x2 2 (b) 1 2 (c) 1 ?
1 ´ x A-5: 1 4 ¨ x ?
x2 + 4 + C A-6: 1 2 ?
5 A-7: π 6 A-8: log ˇ ˇ ˇ ˇ ˇ c 1 + x2 25 + x 5 ˇ ˇ ˇ ˇ ˇ + C A-9: 1 2 ?
2x2 + 4x + C A-10: ´ 1 16 ?
x2 + 16 x + C A-11: ?
x2 ´ 9 9x + C A-12: (a) We’ll use the trig identity cos 2θ = 2 cos2 θ ´ 1. It implies that cos2 θ = cos 2θ + 1 2 ù ñ cos4 θ = 1 4 cos2 2θ + 2 cos 2θ + 1 = 1 4 hcos 4θ + 1 2 + 2 cos 2θ + 1 i = cos 4θ 8 + cos 2θ 2 + 3 8 So, ż π/4 0 cos4 θ dθ = ż π/4 0 cos 4θ 8 + cos 2θ 2 + 3 8 dθ = sin 4θ 32 + sin 2θ 4 + 3 8θ π/4 0 = 1 4 + 3 8 ¨ π 4 = 8 + 3π 32 as required.
(b) 8 + 3π 16 192 A-13: 0 A-14: 2 arcsin x 2 + x 2 a 4 ´ x2 + C A-15: ?
25x2 ´ 4 ´ 2 arcsec 5x 2 + C A-16: 40 3 A-17: arcsin x + 1 2 + C A-18: 1 4 arccos 1 2x ´ 3 + ?
4x2 ´ 12x + 8 (2x ´ 3)2 !
+ C, or equivalently, 1 4 arcsec (2x ´ 3) + ?
4x2 ´ 12x + 8 (2x ´ 3)2 !
+ C A-19: log(1 + ?
2) ´ 1 ?
2 A-20: 1 2 arctan x + x x2 + 1 + C A-21: 3 + x 2 ?
x2 ´ 2x + 2 + 1 2 log ˇ ˇ ˇ ?
x2 ´ 2x + 2 + x ´ 1 ˇ ˇ ˇ + C A-22: 1 ?
3 log ˇ ˇ ˇ ˇ 6 5x + 1 + 2 5 a 9x2 + 15x ˇ ˇ ˇ ˇ + C A-23: 1 3 ?
1 + x2(4 + x2) + log ˇ ˇ ˇ ˇ ˇ 1 ´ ?
1 + x2 x ˇ ˇ ˇ ˇ ˇ + C A-24: 8π 3 + 4 ?
3 A-25: Area: 4 3 ´ 4 c 4 3 Volume: π2 6 ´ ?
3π 4 A-26: 2 ?
1 + ex + 2 log ˇ ˇ1 ´ ?
1 + exˇ ˇ ´ x + C A-27: (a) 1 1 ´ x2 (b) False (c) The work in the question is not correct. The most salient problem is that when we make the substitution x = sin θ, we restrict the possible values of x to [´1, 1], since this is the range of the sine function. However, the original integral had no such restriction.
How can we be sure we avoid this problem in the future? In the introductory text to Section 1.9 (before Example 1.9.1), the CLP-2 text tells us that we are allowed to write our old variable as a function of a new variable (say x = s(u)) as long as that function 193 is invertible to recover our original variable x. There is one very obvious reason why invertibility is necessary: after we antidifferentiate using our new variable u, we need to get it back in terms of our original variable, so we need to be able to recover x. Moreover, invertibility reconciles potential problems with domains: if an inverse function u = s´1(x) exists, then for any x, there exists a u with s(u) = x. (This was not the case in the work for the question, because we chose x = sin θ, but if x = 2, there is no corresponding θ. Note, however, that x = sin θ is invertible over [´1, 1], so the work is correct if we restrict x to those values.) A-28: (a), (b): None.
(c): x ă ´a Answers to Exercises 1.10 — Jump to TABLE OF CONTENTS A-1: (a) (iii) (b) (ii) (c) (ii) (d) (i) A-2: A x ´ 1 + B (x ´ 1)2 + C x + 1 + D (x + 1)2 + Ex + F x2 + 1 A-3: 3 A-4: (a) x3 + 2x + 2 x2 + 1 = x + x + 2 x2 + 1 (b) 15x4 + 6x3 + 34x2 + 4x + 20 5x2 + 2x + 8 = 3x2 + 2 + 4 5x2 + 2x + 8 (c) 2x5 + 9x3 + 12x2 + 10x + 30 2x2 + 5 = x3 + 2x + 6 A-5: (a) 5x3 ´ 3x2 ´ 10x + 6 = (x + ?
2)(x ´ ?
2)(5x ´ 3) (b) x4 ´ 3x2 ´ 5 = x + d 3 + ?
29 2 x ´ d 3 + ?
29 2 x2 + ?
29 ´ 3 2 (c) x4 ´ 4x3 ´ 10x2 ´ 11x ´ 6 = (x + 1)(x ´ 6)(x2 + x + 1) (d) 2x4 + 12x3 ´ x2 ´ 52x + 15 = (x + 3)(x + 5) x ´ 1 + ?
2 2 x ´ 1 ´ ?
2 2 A-6: The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution u = (ax + b). It’s less clear how to find the antiderivative of (2).
A-7: log 4 3 A-8: ´1 x ´ arctan x + C A-9: 4 log |x ´ 3| ´ 2 log(x2 + 1) + C A-10: F(x) = log |x ´ 2| + log(x2 + 4) + 2 arctan(x/2) + D A-11: ´2 log |x ´ 3| + 3 log |x + 2| + C 194 A-12: ´9 log |x + 2| + 14 log |x + 3| + C A-13: 5x + 1 2 log |x ´ 1| ´ 7 2 log |x + 1| + C A-14: x ´ 2 x + 5 2 arctan(2x) + C A-15: 1 x ´ 2 x ´ 1 + C A-16: ´1 2 log |x ´ 2| + 1 2 log |x + 2| + 3 2 log |2x ´ 1| + C A-17: log 4 ¨ 63 53 A-18: 1 2 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ + C A-19: ´ cos x 2 sin2 x + 1 4 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ + C A-20: 3 log 2 + 1 2 + 2 ?
15 arctan 7 ?
15 ´ arctan 9 ?
15 A-21: = 9 4 ?
2 arctan x ?
2 ´ 2 + 3x 4(x2 + 2) + C A-22: 3 8 arctan x + 3x3 + 5x 8(1 + x2)2 + C A-23: 3 2x2 + 1 ?
5 arctan x ?
5 + 3 2 log |x2 + 5| ´ 3 2x2 + 10 + C A-24: log ˇ ˇ ˇ ˇ sin θ ´ 1 sin θ ´ 2 ˇ ˇ ˇ ˇ + C A-25: t ´ 1 2 log |e2t + et + 1| ´ 1 ?
3 arctan 2et + 1 ?
3 + C A-26: 2 ?
1 + ex + log ˇ ˇ ˇ ˇ ?
1 + ex ´ 1 ?
1 + ex + 1 ˇ ˇ ˇ ˇ + C A-27: (a) The region R is 3 4 y = 10 √ 25−x2 x y 195 (b) 10π log 9 4 = 20π log 3 2 (c) 20π A-28: 2 log 5 3 + 4 ?
3 arctan 1 4 ?
3 A-29: (a) 1 6 log ˇ ˇ ˇ ˇ2 ¨ x ´ 3 x + 3 ˇ ˇ ˇ ˇ (b) F1(x) = 1 x2´9 Answers to Exercises 1.11 — Jump to TABLE OF CONTENTS A-1: Relative error: « 0.08147; absolute error: 0.113; percent error: « 8.147%.
A-2: Midpoint rule: x y 2 10 Trapezoidal rule: x y 2 10 A-3: M = 6.25, L = 2 A-4: One reasonable answer is M = 3.
A-5: (a) π5 180 ¨ 8 (b) 0 (c) 0 A-6: Possible answers: f (x) = 3 2x2 + Cx + D for any constants C, D.
A-7: my mother A-8: (a) true (b) false 196 A-9: True. Because f (x) is positive and concave up, the graph of f (x) is always below the top edges of the trapezoids used in the trapezoidal rule.
x y y = f (x) A-10: Any polynomial of degree at most 3 will do. For example, f (x) = 5x3 ´ 27, or f (x) = x2.
A-11: Midpoint: ż 30 0 1 x3 + 1 dx « 1 (2.5)3+1 + 1 (7.5)3+1 + 1 (12.5)3+1 + 1 (17.5)3+1 + 1 (22.5)3+1 + 1 (27.5)3+1 5 Trapezoidal: ż 30 0 1 x3 + 1 dx « 1/2 03 + 1 + 1 53 + 1 + 1 103 + 1 + 1 153 + 1 + 1 203 + 1 + 1 253 + 1 + 1/2 303 + 1 5 Simpson’s: ż 30 0 1 x3 + 1 dx « h 1 03 + 1+ 4 53 + 1+ 2 103 + 1+ 4 153 + 1+ 2 203 + 1+ 4 253 + 1+ 1 303 + 1 i5 3 A-12: 2π 3 A-13: 1720π « 5403.5 cm3 A-14: π 12(16.72) « 4.377 m3 A-15: 12.94 6π « 0.6865 m3 A-16: (a) 363,500 (b) 367,000 A-17: (a) 49 2 (b) 77 3 A-18: Let f (x) = sin(x2). Then f 1(x) = 2x cos(x2) and f 2(x) = 2 cos(x2) ´ 4x2 sin(x2).
Since |x2| ď 1 when |x| ď 1, and |sin θ| ď 1 and |cos θ| ď 1 for all θ, we have ˇ ˇ ˇ2 cos(x2) ´ 4x2 sin(x2) ˇ ˇ ˇ ď 2| cos(x2)| + 4x2| sin(x2)| ď 2 ˆ 1 + 4 ˆ 1 ˆ 1 = 2 + 4 = 6 197 We can therefore choose M = 6, and it follows that the error is at most M[b ´ a]3 24n2 ď 6 ¨ [1 ´ (´1)]3 24 ¨ 10002 = 2 106 = 2 ¨ 10´6 A-19: 3 100 A-20: (a) 1/3 3 (´3)5 + 4 1 3 ´ 3 5 + 2 2 3 ´ 3 5 + 4(´2)5 + 2 4 3 ´ 3 5 + 4 5 3 ´ 3 5 + (´1)5 (b) Simpson’s Rule results in a smaller error bound.
A-21: 8 15 A-22: 1 180 ˆ 34 = 1 14580 A-23: (a) T4 = 1 4 1 2 ˆ 1 + 4 5 + 2 3 + 4 7 + 1 2 ˆ 1 2 , (b) S4 = 1 12 1 + 4 ˆ 4 5 + 2 ˆ 2 3 + 4 ˆ 4 7 + 1 2 (c) ˇ ˇ ˇI ´ S4 ˇ ˇ ˇ ď 24 180 ˆ 44 = 1 1920 A-24: (a) T4 = 8.03515, S4 « 8.03509 (b) ˇ ˇ ˇ ż b a f (x) dx ´ Tn ˇ ˇ ˇ ď 2 1000 83 12(4)2 ď 0.00533, ˇ ˇ ˇ ż b a f (x) dx ´ Sn ˇ ˇ ˇ ď 4 1000 85 180(4)4 ď 0.00284 A-25: Any n ě 68 works.
A-26: 472 3 « 494 ft3 A-27: (a) 0.025635 (b) 1.8 ˆ 10´5 A-28: (a) « 0.6931698 (b) n ě 12 with n even A-29: (a) 0.01345 (b) n ě 28 with n even A-30: n ě 259 A-31: (a) When 0 ď x ď 1, then x2 ď 1 and x + 1 ě 1, so |f 2(x)| = x2 |x + 1| ď 1 1 = 1.
(b) 1 2 (c) n ě 65 (d) n ě 46 A-32: x ´ 1 12 1 + 16 x + 3 + 4 x + 1 + 16 3x + 1 + 1 x 198 A-33: Note: for more detail, see the solutions.
First, we use Simpson’s rule with n = 4 to approximate ş2 1 1 1+x2 dx. The choice of this method (what we’re approximating, why n = 4, etc.) is explained in the solutions–here, we only show that it works.
ż 2 1 1 1 + x2 dx « 1 12 1 2 + 64 41 + 8 13 + 64 65 + 1 5 « 0.321748 For ease of notation, define A = 0.321748.
Now, we bound the error associated with this approximation. Define N(x) = 24(5x4 ´ 10x2 + 1) and D(x) = (x2 + 1)5, so N(x)/D(x) gives the fourth derivative of 1 1+x2. When 1 ď x ď 2, |N(x)| ď N(2) = 984 (because N(x) is increasing over that interval) and |D(x)| ě D(1) = 25 (because D(x) is also increasing over that interval), so ˇ ˇ ˇ d4 dx4 !
1 1+x2 )ˇ ˇ ˇ = ˇ ˇ ˇ N(x) D(x) ˇ ˇ ˇ ď 984 25 = 30.75. Now we find the error bound for Simpson’s rule with L = 30.75, b = 2, a = 1, and n = 4.
ˇ ˇ ˇ ˇ ˇ ż 2 1 1 1 + x2 dx ´ A ˇ ˇ ˇ ˇ ˇ = |error| ď L(b ´ a)5 180 ¨ n4 = 30.75 180 ¨ 44 ă 0.00067 So, ´0.00067 ă ż 2 1 1 1 + x2 dx ´ A ă 0.00067 A ´ 0.0067 ă ż 2 1 1 1 + x2 dx ă A + 0.00067 A ´ 0.00067 ă arctan(2) ´ arctan(1) ă A + 0.00067 A ´ 0.00067 ă arctan(2) ´ π 4 ă A + 0.00067 π 4 + A ´ 0.00067 ă arctan(2) ă π 4 + A + 0.00067 π 4 + 0.321748 ´ 0.00067 ă arctan(2) ă π 4 + 0.321748 + 0.00067 π 4 + 0.321078 ă arctan(2) ă π 4 + 0.322418 π 4 + 0.321 ă arctan(2) ă π 4 + 0.323 This was the desired bound.
Answers to Exercises 1.12 — Jump to TABLE OF CONTENTS A-1: Any real number in [1, 8) or (´8, ´1], and b = ˘8.
A-2: b = ˘8 A-3: The red function is f (x), and the blue function is g(x).
199 A-4: False. For example, the functions f (x) = e´x and g(x) = 1 provide a counterexample.
A-5: (a) Not enough information to decide. For example, consider h(x) = 0 versus h(x) = ´1.
(b) Not enough information to decide. For example, consider h(x) = f (x) versus h(x) = g(x).
(c) ż 8 0 h(x) dx converges by the comparison test, since |h(x)| ď 2f (x) and ż 8 0 2f (x) dx converges.
A-6: The integral diverges.
A-7: The integral diverges.
A-8: The integral does not converge.
A-9: The integral converges.
A-10: The integral diverges.
A-11: The integral diverges.
A-12: The integral diverges.
A-13: The integral diverges.
A-14: The integral diverges.
A-15: The integral converges.
A-16: The integral converges.
A-17: false A-18: q = 1 5 A-19: p ą 1 A-20: log 3 ´ π 4 + 1 2 arctan 2 A-21: The integral converges.
A-22: 1 2 A-23: The integral converges.
A-24: The integral converges.
A-25: t = 10 and n = 2042 will do the job. There are many other correct answers.
A-26: (a) The integral converges.
(b) The interval converges.
A-27: false 200 Answers to Exercises 1.13 — Jump to TABLE OF CONTENTS A-1: (A)–(I), (B)–(IV), (C)–(II), (D)–(III) A-2: 1 5 ´ 2 7 + 1 9 = 8 315 A-3: 3 2 ?
5 arcsin x c 5 3 !
+ x 2 ?
3 ´ 5x2 + C A-4: 0 A-5: log ˇ ˇ ˇ ˇ x + 1 3x + 1 ˇ ˇ ˇ ˇ + C A-6: 8 3 log 2 ´ 7 9 A-7: 1 2 log ˇ ˇx2 ´ 3 ˇ ˇ + C A-8: (a) 2 (b) 2 15 (c) 3e4 16 + 1 16 A-9: (a) 1 (b) 8 15 A-10: (a) e2 + 1 (b) log( ?
2 + 1) (c) log 15 13 « 0.1431 A-11: (a) 9 4π (b) log 2 ´ 2 + π 2 « 0.264 (b) 2 log 2 ´ 1 2 « 0.886 A-12: 1 3 sin3 θ ´ 2 sin θ + 12 log ˇ ˇ ˇ ˇ sin θ ´ 3 sin θ ´ 2 ˇ ˇ ˇ ˇ + C A-13: (a) 1 15 (b) 1 9 ¨ x ?
x2 + 9 + C (c) 1 2 log |x ´ 1| ´ 1 4 log(x2 + 1) ´ 1 2 arctan x + C (d) 1 2 x2 arctan x ´ x + arctan x + C A-14: (a) 1 12 (b) 2 sin´1 x 2 + x c 1 ´ x2 4 + C (c) ´2 log |x| + 1 x + 2 log |x ´ 1| + C A-15: (a) 2 5 (b) 1 2 ?
2 (c) log 2 ´ 1 2 « 0.193 (d) log 2 ´ 1 2 « 0.193 A-16: (a) 1 2x2 log x ´ 1 4x2 + C (b) 1 2 log[x2 + 4x + 5] ´ 3 arctan(x + 2) + C (c) 1 2 log |x ´ 3| ´ 1 2 log |x ´ 1| + C (d) 1 3 arctan x3 + C A-17: (a) π 4 ´ 1 2 log 2 (b) log |x2 ´ 2x + 5| + 1 2 arctan x ´ 1 2 + C 201 A-18: (a) ´ 1 300(x3 + 1)100 + C (b) sin5x 5 ´ sin7x 7 + C A-19: -2 A-20: (a) ´1 4 log |ex + 1| + 1 4 log |ex ´ 3| + C (b) 4π 3 ´ 2 ?
3 A-21: (a) 1 2 sec2 x + log | cos x| + C (b) 1 10 arctan 8 « 0.1446 A-22: 2 5(x ´ 1)5/2 + 2 3(x ´ 1)3/2 + C A-23: log ˇ ˇ ˇx + ?
x2 ´ 2 ˇ ˇ ˇ ´ ?
x2 ´ 2 x + C A-24: 7 24 A-25: 3 log |x + 1| + 2 x + 1 ´ 5 2(x + 1)2 + C A-26: 2 ?
3 arctan 2 ?
3x + 1 ?
3 + C A-27: 1 2 (x ´ sin x cos x) + C A-28: 1 3 log |x + 1| ´ 1 6 log |x2 + x + 1| + 1 ?
3 arctan 2x ´ 1 ?
3 + C A-29: 3x3 arcsin x + 3 ?
1 ´ x2 ´ (1 ´ x2)3/2 + C A-30: 2 A-31: 1 4 A-32: log log(cos(0.1)) log(cos(0.2)) A-33: (a) 1 2x sin(log x) ´ cos(log x) + C (b) 2 log 2 ´ log 3 = log 4 3 A-34: (a) 9 4π + 9 (b) 2 log |x ´ 2| ´ log(x2 + 4) + C (c) π 2 A-35: ´ arcsin( ?
1 ´ x) ´ ?
1 ´ x?x + C A-36: ee(e ´ 1) A-37: ex x + 1 + C A-38: x sec x ´ log | sec x + tan x| + C 202 A-39: ż x(x + a)n dx = $ ’ & ’ % (x+a)(n+2) n+2 ´ a (x+a)n+1 n+1 + C if n ‰ ´1, ´2 (x + a) ´ a log |x + a| + C if n = ´1 log |x + a| + a x+a + C if n = ´2 A-40: x arctan(x2) ´ 1 ?
2 1 2 log ˇ ˇ ˇ x2´ ?
2x+1 x2+ ?
2x+1 ˇ ˇ ˇ + arctan ?
2x + 1 + arctan ?
2x ´ 1 !
+ C Answers to Exercises 2.1 — Jump to TABLE OF CONTENTS A-1: 0.00294 J A-2: The rock has mass 1 9.8 kg (about 102 grams); lifting it one metre takes 1 J of work.
A-3: (a) metres (b) newtons (c) joules A-4: smoot ¨ barn megaFonzie (smoot-barns per megaFonzie) A-5: 10 cm below the bottom of the unloaded spring A-6: x = 2 A-7: a = 3 A-8: (a) joules (b) c log ℓ´ 1 ℓ´ 1.5 J A-9: 1 4 J A-10: 25 J A-11: 196 J A-12: 14700 J A-13: ż 3 0 (9.8)(8000)(2 + z)(3 ´ z)2 dz joules A-14: 0.2352 J A-15: 20 49 kg, or about 408 grams A-16: 294 J A-17: (a) 117.6 J (b) 3.92 30 ´ 2 ?
3 « 104 J A-18: 1 2 ?
5 m/sec, or about 22.36 cm/sec A-19: yes (at least, the car won’t scrape the ground) A-20: « 0.144 J A-21: 904,050π J 203 A-22: 10205 6 J A-23: (a) 4900 N (b) 44100 x2 N (c) 29 400 J A-24: 220.5 J A-25: About 7 ˆ 1028 J A-26: true A-27: 92555 9 J A-28: 7 40 = 0.175 J A-29: One possible answer: 1 4 d 1 ´ 1 8 4 + d 1 ´ 3 8 4 Answers to Exercises 2.2 — Jump to TABLE OF CONTENTS A-1: The most straightforward of many possible answers is shown.
x y 5 A A-2: 500 km A-3: W b ´ a N A-4: (a) b ´ a n (b) a + 3b ´ a n (c) f a + 3b ´ a n (d) 1 n n ř i=1 f a + (i ´ 1) b´a n A-5: (a) yes (b) not enough information A-6: 0 A-7: 1 A-8: 1 e ´ 1 h2 9e3 + 1 9 i A-9: 4 π + 1 204 A-10: 2 π A-11: 10 3 log 7 degrees Celsius A-12: 1 2(e ´ 1) A-13: 1 2 A-14: (a) 400 ppm (b) « 599.99 ppm (c) 0.125, or 12.5% A-15: (a) 16π 5 (b) 32π 5 (c) 32π 5 A-16: (a) 0 (b) ?
3 A-17: c 4 π ´ 1 « 0.52 A-18: (a) F(t) = 3f (t) = 3 sin (tπ) N (b) 0 (c) 3 ?
2 « 2.12 A-19: (a) 130 km (b) 65 km/hr A-20: (a) A = e ´ 1 (b) 0 (c) 4 ´ 2e + 2(e ´ 1) log(e ´ 1) « 0.42 A-21: (a) neither–both are zero (b) |f (x) ´ A| has the larger average on [0, 4] A-22: (b ´ a)πR2 A-23: 0 A-24: Yes, but if a ‰ 0, then s = t.
A-25: A A-26: (a) bA(b) ´ aA(a) b ´ a (b) f (t) = A(t) + tA1(t) A-27: (a) One of many possible answers: f (x) = # ´1 if x ď 0 1 if x ą 0.
(b) No such function exists.
Note 1: Suppose f (x) ą 0 for all x in [´1, 1]. Then 1 2 ş1 ´1 f (x) dx ą 1 2 ş1 ´1 0 dx = 0.
That is, the average value of f (x) on the interval [´1, 1] is not zero–it’s something greater than zero.
Note 2: Suppose f (x) ă 0 for all x in [´1, 1]. Then 1 2 ş1 ´1 f (x) dx ă 1 2 ş1 ´1 0 dx = 0.
That is, the average value of f (x) on the interval [´1, 1] is not zero–it’s something less than zero.
So, if the average value of f (x) is zero, then f (x) ě 0 for some x in [´1, 1], and f (y) ď 0 for some y P [´1, 1]. Since f is a continuous function, and 0 is between f (x) 205 and f (y), by the intermediate value theorem (Theorem 1.6.12 in the in the CLP-1 text) there is some value c between x and y such that f (c) = 0. Since x and y are both in [´1, 1], then c is as well. Therefore, no function exists as described in the question.
A-28: true A-29: 0 Answers to Exercises 2.3 — Jump to TABLE OF CONTENTS A-1: (1, 1) A-2: (0, 0) A-3: In general, false.
A-4: 3.5 metres from the left end A-5: (a) to the left (b) to the left (c) not enough information (d) along the line x = a (e) to the right A-6: 39200π 9 (12 ´ π) « 121, 212 J A-7: (a), (b) 1 x dx (c), (d) log 3 (e), (f) 2 log 3 A-8: (a) n ř i=1 h b´a n ρ a + i ´ 1 2 ( b´a n ) ˆ a + (i ´ 1 2) b´a n i n ř i=1 b´a n ρ a + (i ´ 1 2) b´a n (b) ¯ x = şb a xρ(x) dx şb a ρ(x) dx A-9: (a) x y T(x) B(x) a b a1 b1 R (b) (T(x) ´ B(x)) dx (c) T(x) ´ B(x) (d) ¯ x = şb a x(T(x) ´ B(x)) dx şb a(T(x) ´ B(x)) dx A-10: (a) The strips between x = a and x = a1 at the left end of the figure all have the same centre of mass, which is the y-value where T(x) = B(x), x ă 0. So, there should be multiple weights of different mass piled up at that y-value.
206 Similarly, the strips between x = b1 and x = b at the right end of the figure all have the same centre of mass, which is the y-value where T(x) = B(x), x ą 0. So, there should be a second pile of weights of different mass, at that (higher) y-value.
Between these two piles, there are a collection of weights with identical mass distributed fairly evenly. The top and bottom ends of R (above the uppermost pile, and below the lowermost pile) have no weights.
One possible answer (using twelve slices): x y T(x) B(x) a b R (b) The area of the strip is (T(x) ´ B(x)) dx, and its centre of mass is at height T(x) + B(x) 2 .
(c) ¯ y = şb a T(x)2 ´ B(x)2 dx 2 şb a T(x) ´ B(x) dx A-11: ¯ x = ´1 3 ż 0 ´1 6x2 dx A-12: ¯ x = 14 3 A-13: ¯ x = log 10.1 2(arctan 10 + arctan(3)) « 0.43 A-14: ¯ y = 3 4e ´ e 4 A-15: (a) 207 x = 2 y = 1 √ 16−x2 x y (b) 3 log 3 8π A-16: ¯ x = π 4 ?
2 ´ 1 ?
2 ´ 1 and ¯ y = 1 4( ?
2 ´ 1) A-17: (a) ¯ x = k A ?
2 ´ 1 , ¯ y = k2π 8A (b) k = 8 π ?
2 ´ 1 A-18: (a) y=x−x2 y=x2−3x x y (2, −2) (b) 8 3 (c) 1 A-19: 2 π log 2 « 0.44127 A-20: ¯ x = 0 and ¯ y = 12 24 + 9π A-21: (a) 9 4π (b) ¯ x = 0 and ¯ y = 4 π A-22: ( ¯ x, ¯ y) = 1, ´ 2 π A-23: e2 ´ 3/2 e2 ´ 5/2, e4 ´ 7 4e2 ´ 10 « (1.2, 2.4) A-24: ¯ y = 8 5 A-25: (a) ¯ x = 8 11, ¯ y = 166 55 (b) π ż 4 0 y dy + π ż 6 4 (6 ´ y)2 dy A-26: (a) ¯ y = e 4 ´ 3 4e (b) π e2 2 + 2e ´ 3 2 A-27: (3, 1.5) 208 A-28: (0, 3.45) A-29: (a) h 4 (b) 1 2h2k ´ 2 3hk2 + 1 4k3 h2 ´ hk + 1 3k2 A-30: about 0.833 N A-31: (a) 17,150π J (b) 2450 9 π (8π ´ 9) « 13, 797 J (c) about 74% A-32: ¯ x = π 162 c π 2 sin π 72 + 2 sin π 18 + 9 sin π 8 + 8 sin 2π 9 + 25 sin 25π 72 + 9 « 0.976 Answers to Exercises 2.4 — Jump to TABLE OF CONTENTS A-1: (a) yes (b) yes (c) no A-2: (a) One possible answer: f (x) = x, g(y) = sin y 3y .
(b) One possible answer: f (x) = ex, g(y) = ey.
(c) One possible answer: f (x) = x ´ 1, g(y) = 1.
(d) The given equation is equivalent to the equation dy dx = x, which fits the form of a separable equation with f (x) = x, g(y) = 1.
A-3: The mnemonic allows us to skip from the separable differential equation we want to solve (very first line) to the equation ż 1 g(y) dy = ż f (x) dx A-4: false A-5: (a) [0, 8) (b) No such function exists. If |f (x)| = Cx and f (x) switches from f (x) = Cx to f (x) = ´Cx at some point, then that point is a jump discontinuity. Where f (x) contains a discontinuity, dy dx does not exist.
A-6: dQ dt = ´0.003Q(t) A-7: dp dt = αp(t) 1 ´ p(t) , for some constant α.
A-8: (a) ´1 (b) 0 (c) 0.5 (d) Two possible answers are shown below: 209 x y 1 1 x y 1 1 Another possible answer is the constant function y = 2.
A-9: (a) ´1 2 (b) 3 2 (c) ´5 2 (d) Your sketch should look something like this: x y 1 1 (e) There are lots of possible answers. Several are shown below.
210 x y 1 1 x y 1 1 x y 1 1 x y 1 1 A-10: y = log(x2 + 2) A-11: y(x) = 3 ?
1 + x2 A-12: y(t) = 3 log ´3 C + sin t A-13: y = 3 b 3 2ex2 + C.
A-14: y = ´ log C ´ x2 2 The solution only exists for C ´ x2 2 ą 0, i.e. C ą 0 and the function has domain ␣ x : |x| ă ?
2C ( .
A-15: y = (3ex ´ 3x2 + 24)1/3 211 A-16: y = f (x) = ´ 1 ?
x2 + 16 A-17: y = ?
10x3 + 4x2 + 6x ´ 4 A-18: y(x) = ex4/4 A-19: y = 1 1´2x A-20: f (x) = e ¨ ex2/2 A-21: y(x) = b 4 + 2 log 2x x+1. Note that, to satisfy y(1) = 2, we need the positive square root.
A-22: y2 + 2 3(y2 ´ 4)3/2 = 2 sec x + 2 A-23: 12 weeks A-24: t = c m kg arctan d k mg v0 !
A-25: (a) k = 1 400 (b) t = 70sec A-26: (a) x(t) = 3 ´ 4ekt 1 ´ 2ekt (b) As t Ñ 8, x Ñ 2.
A-27: (a) P = 4 1 + e´4t (b) At t = 1 2, P « 3.523. As t Ñ 8, P Ñ 4.
A-28: (a) dv dt = ´kv2 (b) v = 400 t + 1 (c) t = 7 A-29: (a) B(t) = C e0.06t´0.02 cos t with the arbitrary constant C ě 0.
(b) $1159.89 A-30: (a) B(t) = t30000 ´ 50mu et/50 + 50m (b) $600 A-31: y(x) = 4 ´ e1´cos x 2 ´ e1´cos x . The largest allowed interval is ´ arccos(1 ´ log 2) ă x ă arccos(1 ´ log 2) or, roughly, ´1.259 ă x ă 1.259.
A-32: 180, 000 b 3 g « 99, 591 sec « 27.66 hr A-33: t = 4 ˆ 144 15 d 125 2g « 2, 394 sec « 0.665 hr A-34: (a) 3 (b) y1 = (y ´ 1)(y ´ 2) (c) f (x) = 4 ´ ex 2 ´ ex A-35: p = 1 4 A-36: 212 (a) One possible answer: f (t) = 0 (b) 1 ?x ´ a f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) 2 bşx a f 2(t) dt (c) 2 x ´ a ż x a f (t) dt f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) (d) Y(x) = D(x ´ a), where D is any constant (e) f (t) = D, for any nonnegative constant D A-37: x = 1 4 y ´ 1 + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ Answers to Exercises 3.1 — Jump to TABLE OF CONTENTS A-1: (a) ´2 (b) 0 (c) the limit does not exist A-2: true A-3: (a) A ´ B C (b) 0 (c) A B A-4: Two possible answers, of many: • an = # 3000 ´ n if n ď 1000 ´2 + 1 n if n ą 1000 • an = 1, 002, 001 n ´ 2 A-5: One possible answer is an = (´1)n = t´1, 1, ´1, 1, ´1, 1, ´1, . . .u.
Another is an = n(´1)n = t´1, 2, ´3, 4, ´5, 6, ´7, . . .u.
A-6: One sequence of many possible is an = (´1)n n = " ´1, 1 2, ´1 3, 1 4, ´1 5, 1 6, . . .
.
A-7: Some possible answers: (a) ´1 n ď sin n n ď 1 n (b) n2 13en ď n2 en(7 + sin n ´ 5 cos n) ď n2 en or 0 ď n2 en(7 + sin n ´ 5 cos n) ď n2 en (c) ´1 nn ď (´n)´n ď 1 nn A-8: (a) an = bn = h(n) = i(n), cn = j(n), dn = f (n), en = g(n) (b) lim nÑ8 an = lim nÑ8 bn = lim xÑ8 h(x) = 1, lim nÑ8 cn = lim nÑ8 en = lim xÑ8 g(x) = lim xÑ8 j(x) = 0, lim nÑ8 dn, lim xÑ8 f (x) and lim xÑ8 i(x) do not exist.
A-9: (a) Some possible answers: a22 « ´0.99996, a66 « ´0.99965, and a110 « ´0.99902.
213 (b) Some possible answers: a11 « 0.0044, a33 « ´0.0133, and a55 « 0.0221.
The integers 11, 33, and 55 were found by approximating π by 22 7 and finding when an odd multiple of 11 7 (which is the corresponding approximation of π 2 ) is an integer.
(c) Some possible answers: a44 « 0.9998, a132 « 0.9986 and a220 « 0.09961.
See the solution for how we found them.
A-10: (a) 8 (b) 3 4 (c) 0 A-11: 8 A-12: 0 A-13: 0 A-14: 0 A-15: 1 A-16: 0 A-17: 8 A-18: lim kÑ8 ak = 0.
A-19: The sequence converges to 0.
A-20: 9 A-21: log 2 A-22: 5 A-23: ´8 A-24: 100 ¨ 299.
A-25: Possible answers are tanu = " n f a + 1 n ´ f (a) or tanu = " n f (a) ´ f a ´ 1 n .
A-26: (a) An = n 2 sin 2π n (b) π A-27: (a) x y 1 2 3 214 (b) x y 1 4 3 (c) An = 1 for all n (d) lim nÑ8 An = 1.
(e) g(x) = 0 (f) ż 8 0 g(x) dx = 0.
A-28: e3 A-29: (a) 4 (b) x = 4 (c) see solution A-30: (a) decreasing (b)fn = 1 n f1 (c) 2% (d) 0.18% (e) “be”: 11,019,308; “and”: 7,346,205 Answers to Exercises 3.2 — Jump to TABLE OF CONTENTS A-1: N SN 1 1 2 1 + 1 2 3 1 + 1 2 + 1 3 4 1 + 1 2 + 1 3 + 1 4 5 1 + 1 2 + 1 3 + 1 4 + 1 5 A-2: 3 A-3: (a) an = $ & % 1 2 if n = 1 1 n(n + 1) else (b) 0 (c) 1 A-4: an = # 0 if n = 1 2(´1)n ´ 1 n(n´1) else A-5: an ă 0 for all n ě 2 A-6: (a) 8 ÿ n=1 2 4n (b) 2 3 215 A-7: (a) 8 ÿ n=1 1 9n (b) 1 8 A-8: Two possible pictures: A-9: 5101 ´ 1 4 ¨ 5100 A-10: All together, there were 36 cookies brought by Student 11 through Student 20.
A-11: 551 ´ 1 4 ¨ 5100 A-12: (a) As time passes, your gains increase, approaching $1.
(b) 1 (c) As time passes, you lose more and more money, without bound.
(d) ´8 A-13: A + B + C ´ c1 A-14: in general, false A-15: 3 2 A-16: 1 7 ˆ 86 A-17: 6 216 A-18: cos π 3 ´ cos(0) = ´1 2 A-19: (a) an = 11 16n2 + 24n + 5 (b) 3 4 A-20: 24 5 A-21: 7 30 A-22: 263 99 A-23: 321 999 = 107 333 A-24: 3 A-25: 1 2 + 5 7 = 17 14 A-26: 40 3 A-27: The series diverges to ´8.
A-28: ´1 2 A-29: 9.8 J A-30: 4π 3 (π3 ´ 1) A-31: sin2 3 8 + 32 « 32.0025 A-32: an = $ ’ & ’ % 2 n(n´1)(n´2) if n ě 3, ´5 2 if n = 2, 2 if n = 1 A-33: 5 8 Answers to Exercises 3.3 — Jump to TABLE OF CONTENTS A-1: (B), (C) A-2: (A) A-3: (a) I am old (b) not enough information to tell (c) not enough information to tell (d) I am young A-4: 217 if ř an converges if ř an diverges and if tanu is the red series then ř bn CONVERGES inconclusive and if tanu is the blue series inconclusive then ř bn DIVERGES A-5: (a) both direct comparison and limit comparison (b) direct comparison (c) limit comparison (d) neither A-6: It diverges by the divergence test, because lim nÑ8 an ‰ 0.
A-7: We cannot use the divergence test to show that a series converges. It is inconclusive in this case.
A-8: The integral test does not apply because f (x) is not decreasing.
A-9: The inequality goes the wrong way, so the direct comparison test (with this comparison series) is inconclusive.
A-10: (B), (D) A-11: One possible answer: 8 ÿ n=1 1 n2.
A-12: By the divergence test, for a series ř an to converge, we need lim nÑ8 an = 0. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If lim nÑ8 ˇ ˇ ˇ ˇ an an+1 ˇ ˇ ˇ ˇ ă 1 or (equivalently) lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ ą 1, then |an+1| ą |an| for sufficiently large n, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.
A-13: One possible answer: f (x) = sin(πx), an = 0 for every n.
By the integral test, any answer will use a function f (x) that is not both positive and decreasing.
A-14: One possible answer: bn = 2n 3n A-15: (a) In general false. The harmonic series 8 ř n=1 1 n provides a counterexample.
(b) In general false. If an = (´1)n 1 n, then 8 ř n=1 (´1)nan is again the harmonic series 8 ř n=1 1 n, which diverges.
(c) In general false. Take, for example, an = 0 and bn = 1.
A-16: No. It diverges.
A-17: It diverges.
218 A-18: The series diverges.
A-19: It diverges.
A-20: This is a geometric series with r = 1.001. Since |r| ą 1, it is divergent.
A-21: The series converges to ´ 1 150.
A-22: The series converges.
A-23: It diverges.
A-24: The series converges.
A-25: The series converges to 1 3.
A-26: The series converges.
A-27: It converges.
A-28: Let f (x) = 5 x(log x)3/2. Then f (x) is positive and decreases as x increases. So the sum 8 ÿ 3 f (n) and the integral ż 8 3 f (x) dx either both converge or both diverge, by the integral test, which is Theorem 3.3.5 in the CLP-2 text. For the integral, we use the substitution u = log x, du = dx x to get ż 8 3 5 dx x(log x)3/2 = ż 8 log 3 5 du u3/2 which converges by the p–test (which is Example 1.12.8 in the CLP-2 text) with p = 3 2 ą 1.
A-29: p ą 1 A-30: It converges.
A-31: The series 8 ÿ n=2 ?
3 n2 converges by the p–test with p = 2.
Note that 0 ă an = ?
3n2 ´ 7 n3 ă ?
3n2 n3 = ?
3 n2 for all n ě 2. As the series 8 ř n=2 ?
3 n2 converges, the comparison test says that 8 ř n=2 ?
3n2´7 n3 converges too.
A-32: The series converges.
A-33: It diverges.
A-34: (a) diverges (b) converges A-35: The series diverges.
219 A-36: (a) converges (b) diverges A-37: 1 e5 ´ e4 A-38: 1 7 A-39: (a) diverges by limit comparison with the harmonic series (b) converges by the ratio test A-40: (a) Converges by the limit comparison test with b = 1 k5/3.
(b) Diverges by the ratio test.
(c) Diverges by the integral test.
A-41: It converges.
A-42: N = 5 A-43: N ě 999 A-44: We need N = 4 and then S4 = 1 32 ´ 1 52 + 1 72 ´ 1 92 A-45: (a) converges (b) converges A-46: (a) See the solution.
(b) f (x) = x + sin x 1 + x2 is not a decreasing function.
(c) See the solution.
A-47: The sum is between 0.9035 and 0.9535.
A-48: Since lim nÑ8 an = 0, there must be some integer N such that 1 2 ą an ě 0 for all n ą N.
Then, for n ą N, an 1 ´ an ď an 1 ´ 1/2 = 2an From the information in the problem statement, we know 8 ÿ n=N+1 2an = 2 8 ÿ n=N+1 an converges.
So, by the direct comparison test, 8 ÿ n=N+1 an 1 ´ an converges as well.
Since the convergence of a series is not affected by its first N terms, as long as N is finite, we conclude 8 ÿ n=1 an 1 ´ an converges.
220 A-49: It diverges.
A-50: It converges to ´ log 2 = log 1 2, A-51: See the solution.
A-52: About 9% to 10% A-53: The total population is between 29,820,091 and 30,631,021 people.
Answers to Exercises 3.4 — Jump to TABLE OF CONTENTS A-1: False. For example, bn = 1 n provides a counterexample.
A-2: ř an converges ř an diverges ř |an| converges converges absolutely not possible ř |an| diverges converges conditionally diverges A-3: conditionally convergent A-4: The series diverges.
A-5: It diverges.
A-6: It converges absolutely.
A-7: It converges absolutely.
A-8: It diverges.
A-9: It converges absolutely.
A-10: See solution.
A-11: See solution.
A-12: See solution.
A-13: (a) See the solution.
(b) |S ´ S5| ď 24 ˆ 36e´63 A-14: cos 1 « 389 720; the actual associated error (using a calculator) is about 0.000025.
A-15: See solution.
Answers to Exercises 3.5 — Jump to TABLE OF CONTENTS A-1: 2 A-2: f (x) = 8 ÿ n=1 n(x ´ 5)n´1 n! + 2 221 A-3: only x = c A-4: R = 6 A-5: (a) R = 1 2 (b) 2 1 + 2x for all |x| ă 1 2 A-6: R = 8 A-7: 1 A-8: The interval of convergence is ´1 ă x + 2 ď 1 or (´3, ´1].
A-9: The radius of convergence is 3. The interval of convergence is ´4 ă x ď 2, or simply (´4, 2].
A-10: ´3 ď x ă 7 or [´3, 7) A-11: The given series converges if and only if ´3 ď x ď ´1. Equivalently, the series has interval of convergence [´3, ´1].
A-12: (a) 3 4 ď x ă 5 4 or 3 4, 5 4 (b) ´5 8 ď x ă ´3 8 or ´ 5 8, ´3 8 (c) ´3 4 ă x ă ´1 4, or ´ 3 4, ´1 4 A-13: The radius of convergence is 2. The interval of convergence is ´1 ă x ď 3, or ´ 1, 3 .
A-14: The interval of convergence is a ´ 1 ă x ă a + 1, or a ´ 1, a + 1 .
A-15: (a) |x + 1| ď 9 or ´10 ď x ď 8 or [´10, 8] (b) This series converges only for x = 1.
A-16: 8 ÿ n=0 xn+3 = 8 ÿ n=3 xn A-17: f (x) = 3 + 8 ÿ n=1 (x ´ 1)n n(n + 1) A-18: The series converges absolutely for |x| ă 9, converges conditionally for x = ´9 and diverges otherwise.
A-19: (a) 8 ÿ n=0 (´1)n x3n+1 3n + 1 + C (b) We need to keep two terms (the n = 0 and n = 1 terms).
A-20: (a) See the solution.
(b) 8 ÿ n=0 n2xn = x(1 + x) (1 ´ x)3 . The series converges for ´1 ă x ă 1.
A-21: See the solution.
A-22: (a) 1.
(b) The series converges for ´1 ď x ă 1, i.e. for the interval [´1, 1) A-23: 5 6 222 A-24: The point x = c corresponds to a local maximum if A2 ă 0 and a local minimum if A2 ą 0.
A-25: 13 80 A-26: x ´ x2 2 + x3 3 ´ x4 4 A-27: x ´ x3 3 + x5 5 Answers to Exercises 3.6 — Jump to TABLE OF CONTENTS A-1: A: linear B: constant C: quadratic A-2: T(5) = arctan3 e5 + 7 A-3: A - V, radius= 1 B - I, radius= 1 C - IV, radius= 1 D - VI, radius= +8 E - II, radius= +8 F - III, radius= +8 A-4: (a) f (20)(3) = 202 20!
20! + 1 (b) g(20)(3) = 102 20!
10! + 1 (c) h(20)(0) = ´20! ¨ 511 11 ; h(22)(0) = 0 A-5: 8 ÿ n=1 (´1)n´1 n (x ´ 1)n A-6: 8 ÿ n=0 (´1)n+1 (2n + 1)!(x ´ π)2n+1 A-7: 1 10 8 ÿ n=0 10 ´ x 10 n with interval of convergence (0, 20).
A-8: 8 ÿ n=0 3ne3a n! (x ´ a)n, with infinite radius of convergence A-9: ´ 8 ÿ n=0 2nxn A-10: bn = 3(´1)n + 2n A-11: c5 = 35 5!
A-12: 8 ÿ n=0 (´1)n 2n+1xn+1 n + 1 for all |x| ă 1 2 223 A-13: a = 1, b = ´ 1 3! = ´1 6.
A-14: ż e´x2 ´ 1 x dx = C ´ x2 2 + x4 8 + ¨ ¨ ¨ .
It is not clear from the wording of the question whether or not the arbitrary constant C is to be counted as one of the “first two nonzero terms”.
A-15: 8 ÿ n=0 (´1)n 22n+1x2n+6 (2n + 1)(2n + 6) + C = 8 ÿ n=0 (´1)n 22nx2n+6 (2n + 1)(n + 3) + C A-16: f (x) = 1 + 8 ÿ n=0 (´1)n 3n 3n + 2x3n+2 A-17: π 2 ?
3 A-18: 1 e A-19: e1/e A-20: e1/π ´ 1 A-21: log(3/2) A-22: (e + 2)ee ´ 2 A-23: The sum diverges–see the solution.
A-24: 1 + ?
2 ?
2 A-25: (a) See the solution.
(b) 1 2 e + 1 e A-26: (a) 50,000 (b) three terms (n = 0 to n = 2) (c) six terms (n = 0 to n = 5) A-27: 29 A-28: S13 or higher A-29: S9 or higher A-30: S18 or higher A-31: The error is in the interval ´57 14 ¨ 37 1 + 1 37 , ´57 7 ¨ 67 « (´0.199, ´0.040) A-32: ´1 A-33: 1 5! = 1 120 A-34: e2 A-35: ?e 224 A-36: 2 (6/7)3 = 343 108 A-37: 8 ÿ n=0 (´1)n x2n+4 (2n + 1)(2n + 2) = x3 arctan x ´ x2 2 log(1 + x2) A-38: (a) the Maclaurin series for f (x) is 8 ÿ n=0 (2n)!
22n (n!)2 xn, and its radius of convergence is R = 1.
(b) the Maclaurin series for arcsin x is 8 ÿ n=0 (2n)!
22n (n!)2(2n + 1)x2n+1, and its radius of convergence is R = 1.
A-39: log(x) = log 2 + 8 ÿ n=1 (´1)n´1 n 2n (x ´ 2)n. It converges when 0 ă x ď 4.
A-40: (a) 8 ÿ n=0 (´1)n x4n+1 4n + 1 (b) 0.493967 (c) The approximate value of part (b) is larger than the true value of I(1/2) A-41: 1 66 A-42: Any interval of length 0.0002 that contains 0.03592 and 0.03600 is fine.
A-43: (a) 8 ÿ n=1 (´1)n xn n n!
(b) ´0.80 (c) See the solution.
A-44: (a) Σ(x) = 8 ÿ n=0 (´1)n x2n+1 (2n + 1)(2n + 1)!
(b) x = π (c) 1.8525 A-45: (a) I(x) = 8 ÿ n=1 (´1)n x2n´1 (2n)!(2n ´ 1) (b) I(1) = ´1 2 + 1 4!3 ˘ 1 6!5 = ´0.486 ˘ 0.001 (c) I(1) ă ´1 2 + 1 4!3 A-46: (a) I(x) = 8 ÿ n=1 (´1)n+1 x2n´1 (2n)! = 1 2!x ´ 1 4!x3 + 1 6!x5 ´ 1 8!x8 + ¨ ¨ ¨ (b) 0.460 (c) I(1) ă 1 2! ´ 1 4! + 1 6! ă 0.460 A-47: (a) See the solution.
(b) The series converges for all x.
A-48: See the solution.
A-49: (a) cosh(x) = 8 ÿ n=0 n even xn n! = 8 ÿ n=0 x2n (2n)! for all x.
225 A-50: (a) 3 ?
3 « 1.26 (b) 12 terms (S11) A-51: 15!
5! ¨ 56 ´ 21!
7! ¨ 11! ¨ 511 + 27!
9! ¨ 17! ¨ 517 ´ 33!
11! ¨ 23! ¨ 523 A-52: (a) x y 1 ´ ?
2/3 ?
2/3 y = f (x) (b) the constant function 0 (c) everywhere (d) only at x = 0 A-53: 0 226 SOLUTIONS TO PROBLEMS Part IV 227 Solutions to Exercises 1.1 — Jump to TABLE OF CONTENTS S-1: x y 1 3 0.75 1.25 x y 1 3 0.75 1.25 The diagram on the left shows a rectangle with area 2 ˆ 1.25 = 2.5 square units. Since the blue-shaded region is entirely inside this rectangle, the area of the blue-shaded region is no more than 2.5 square units.
The diagram on the right shows a rectangle with area 2 ˆ 0.75 = 1.5 square units. Since the blue-shaded region contains this entire rectangle, the area of the blue region is no less than 1.5 square units.
So, the area of the blue-shaded region is between 1.5 and 2.5 square units.
Remark: we could also give an obvious range, like “the shaded area is between zero and one million square units.” This would be true, but not very useful or interesting.
S-2: Solution 1: One naive way to solve this is to simply use the same method as Question 1.
x y 1 2 3 4 0.75 1.25 0.25 2.25 1.75 x y 1 2 3 4 0.75 1.25 0.25 2.25 1.75 The rectangle on the left has area 3 ˆ 2.25 = 6.75 square units, and encompasses the entire shaded region. The rectangle on the right has area 3 ˆ 0.25 = 0.75 square units, and is entirely contained inside the blue-shaded region. So, the area of the blue-shaded region is between 0.75 and 6.75 square units.
This is a legitimate approximation, but we can easily do much better. The shape of this graph suggests that using the areas of three rectangles would be a natural way to improve our estimate.
Solution 2: Let’s use these rectangles instead: 228 x y 1 2 3 4 0.75 1.25 0.25 2.25 1.75 x y 1 2 3 4 0.75 1.25 0.25 2.25 1.75 In the left picture, the red area is (1 ˆ 1.25) + (1 ˆ 2.25) + (1 ˆ 0.75) = 4.25 square units. In the right picture, the red area is (1 ˆ 0.75) + (1 ˆ 1.75) + (1 ˆ 0.25) = 2.75 square units. So, the blue shaded area is between 2.75 and 4.25 square units.
S-3: Remark: in the solution below, we find the appropriate approximation using trial and error. In Question 46, we take a more systematic approach.
Try 1: First, we can try by using a single rectangle as an overestimate, and a single rectangle as an underestimate.
x y y = 1 2x 1 3 1/2 1/8 x y y = 1 2x 1 3 1/2 1/8 The area under the curve is less than the area of the rectangle on the left (2 ˆ 1 2 = 1) and greater than the area of the rectangle on the right (2 ˆ 1 8 = 1 4). So, the area is in the range 1 4, 1 . Unfortunately, this range is too big–we need our range to have length at most 0.2. So, we refine our approximation by using more rectangles.
Try 2: Let’s try using two rectangles each for the upper and lower bounds.
229 x y y = 1 2x 1 2 3 1/2 1/4 1/8 x y y = 1 2x 1 3 2 1/2 1/4 1/8 The rectangles in the left picture have area 1 ˆ 1 2 + 1 ˆ 1 4 = 3 4, and the rectangles in the right picture have area 1 ˆ 1 4 + 1 ˆ 1 8 = 3 8. So, the area under the curve is in the interval 3 8, 3 4 . The length of this interval is 3 8, and 3 8 ą 3 15 = 1 5 = 0.2. (Indeed, 3 8 = 0.375 ą 0.2.) Since the length of our interval is still bigger than 0.2, we need even more rectangles.
Try 3: Let’s go ahead and try four rectangles each for the upper and lower estimates.
x y y = 1 2x 1 2 3 1/2 1/4 1/(4 ?
2) 1/(2 ?
2) 1/8 x y y = 1 2x 1 2 3 1/2 1/4 1/(4 ?
2) 1/(2 ?
2) 1/8 The area of the rectangles on the left is: 1 2 ˆ 1 2 + 1 2 ˆ 1 2 ?
2 + 1 2 ˆ 1 4 + 1 2 ˆ 1 4 ?
2 = 3 8 1 + 1 ?
2 , and the area of the rectangles on the right is: 1 2 ˆ 1 2 ?
2 + 1 2 ˆ 1 4 + 1 2 ˆ 1 4 ?
2 + 1 2 ˆ 1 8 = 3 8 1 2 + 1 ?
2 .
So, the area under the curve is in the interval 3 8 h 1 2 + 1 ?
2 i , 3 8 h 1 + 1 ?
2 i . The length of this interval is 3 16, and 3 16 ă 3 15 = 1 5 = 0.2, as desired. (Indeed, 3 16 = 0.1875 ă 0.2.) Note, if we choose any value in the interval 3 8 h 1 2 + 1 ?
2 i , 3 8 h 1 + 1 ?
2 i as an approximation for the area under the curve, our error is no more than 0.2.
230 S-4: Since f (x) is decreasing, it is larger on the left endpoint of an interval than on the right endpoint of an interval. So, a left Riemann sum gives a larger approximation.
Notice this does not depend on n.
Furthermore, the actual area ż 5 0 f (x)dx is larger than its right Riemann sum, and smaller than its left Riemann sum.
x y left Riemann sum x y right Riemann sum S-5: If f (x) is always increasing or always decreasing, then the midpoint Riemann sum will be between the left and right Riemann sums. So, we need a function that goes up and down. Many examples are possible, but let’s work with a familiar one: sin x.
If our intervals have endpoints that are integer multiples of π, then the left and right Riemann sums will be 0, since sin(0) = sin(π) = sin(2π) = ¨ ¨ ¨ = 0. The midpoints of these intervals will give y-values of 1 and -1. So, for example, we can let f (x) = sin x, [a, b] = [0, π], and n = 1. Then the right and left Riemann sums are 0, while the midpoint Riemann sum is π.
We can extend the example of f (x) = sin x to have more intervals. As long as we have more positive terms than negative, the midpoint approximation will be a positive number, and so it will be larger than both the left and right Riemann sums. So, for example, we can let f (x) = sin x, [a, b] = [0, 5π], and n = 5. Then the midpoint Riemann sum is π ´ π + π ´ π + π = π, which is strictly larger than 0 and so it is larger than both the left and right Riemann sums.
x y 5π 231 S-6: (a) Two possible answers are 7 ÿ i=3 i and 5 ÿ i=1 (i + 2). The first has simpler terms (i versus i + 2), while the second has simpler indices (we often like to start at i = 1). Neither is objectively better than the other, but depending on your purposes you might find one more useful.
(b) The terms of this sum are each double the terms of the sum from part (a), so two possible answers are 7 ÿ i=3 2i and 5 ÿ i=1 (2i + 4).
We often want to write a sum that involves even numbers: it will be useful for you to remember that the term 2i (with index i) generates evens.
(c) The terms of this sum are each one more than the terms of the sum from part (b), so two possible answers are 7 ÿ i=3 (2i + 1) and 5 ÿ i=1 (2i + 5).
In the last part, we used the expression 2i to generate even numbers; 2i + 1 will generate odds. So will the index 2i + 5, and indeed, 2i + k for any odd number k. The choice of what you add will depend on the limits of i.
(d) This sum adds up the odd numbers from 1 to 15. From Part (c), we know that the formula 2i + 1 is a simple way of generating odd numbers. Since our first term should be 1 and our last term should be 15, if we use ř(2i + 1), then i should run from 0 to 7. So, one way of expressing our sum in sigma notation is 7 ÿ i=0 (2i + 1).
Sometimes we like our sum to start at i = 1 instead of i = 0. If this is our desire, we can use 2i ´ 1 as our terms, and let i run from 1 to 8. This gives us another way of expressing our sum: 8 ÿ i=1 (2i ´ 1).
S-7: (a) The denominators are successive powers of three, so one way of writing this is 4 ÿ i=1 1 3i .
Equivalently, the terms we’re adding are powers of 1/3, so we can also write 4 ÿ i=1 1 3 i .
(b) This sum is obtained from the sum in (a) by multiplying each term by two, so we can write 4 ÿ i=1 2 3i or 4 ÿ i=1 2 1 3 i .
232 (c) The difference between this sum and the previous sum is its alternating sign, minus-plus-minus-plus. This behaviour appears when we raise a negative number to successive powers. We can multiply each term by (´1)i, or we can slip a negative into the number that is already raised to the power i: 4 ÿ i=1 (´1)i 2 3i , or 4 ÿ i=1 2 (´3)i .
(d) This sum is the negative of the sum in part (c), so we can simply multiply each term by negative one: 4 ÿ i=1 (´1)i+1 2 3i , or 4 ÿ i=1 ´ 2 (´3)i .
Be careful with the second form: a common mistake is to think that ´ 2 (´3)i = 2 3i , but these are not the same.
S-8: (a) If we re-write the second term as 3 9 instead of 1 3, our sum becomes: 1 3 + 3 9 + 5 27 + 7 81 + 9 243 The numerators are the first five odd numbers, and the denominators are the first five positive powers of 3. We learned how to generate odd numbers in Question 6, and we learned how to generate powers of three in Question 7. Combining these, we can write our sum as 5 ÿ i=1 2i ´ 1 3i .
(b) The denominators of these terms differ from the denominators of part (a) by precisely two, while the numerators are simply 1. So, we can modify our previous answer: 5 ÿ i=1 1 3i + 2 .
(c) Let’s re-write the sum to make the pattern clearer.
1000 + 200 + 30 + 4 + 1 2 + 3 50 + 7 1000 = 1 ¨ 1000 + 2 ¨ 100 + 3 ¨ 10 + 4 1 + 5 10 + 6 100 + 7 1000 = 1 ¨ 103 + 2 ¨ 102 + 3 ¨ 101 + 4 ¨ 100 + 5 ¨ 10´1 + 6 ¨ 10´2 + 7 ¨ 10´3 = 1 ¨ 104´1 + 2 ¨ 104´2 + 3 ¨ 104´3 + 4 ¨ 104´4 + 5 ¨ 104´5 + 6 ¨ 104´6 + 7 ¨ 104´7 If we let the red numbers be our index i, this gives us the expression 7 ÿ i=1 i ¨ 104´i .
Equivalently, we can write 7 ÿ i=1 i 10i´4 .
S-9: 233 (a) Using Theorem 1.1.6.a in the CLP-2 text, with a = 1, r = 3 5 and n = 100: 100 ÿ i=0 3 5 i = 1 ´ 3 5 101 1 ´ 3 5 = 5 2 " 1 ´ 3 5 101# (b) We want to use Theorem 1.1.6, part (a) again, but our sum doesn’t start at 3 5 0 = 1.
We have two options: factor out the leading term, or use the difference of two sums that start where we want them to.
Solution 1: In this solution, we’ll make our sum start at 1 by factoring out the leading term. We wrote our work out the long way (expanding the sigma into “dot-dot-dot” notation) for clarity, but it’s faster to do the algebra in sigma notation all the way through.
100 ÿ i=50 3 5 i = 3 5 50 + 3 5 51 + 3 5 52 + ¨ ¨ ¨ + 3 5 100 = 3 5 50 " 1 + 3 5 + 3 5 2 + ¨ ¨ ¨ + 3 5 50# = 3 5 50 1 ´ 3 5 51 1 ´ 3 5 = 5 2 3 5 50 " 1 ´ 3 5 51# .
Solution 2: In this solution, we write our given expression as the difference of two sums, both starting at i = 0.
100 ÿ i=50 3 5 i = 100 ÿ i=0 3 5 i ´ 49 ÿ i=0 3 5 i = 1 ´ 3 5 101 1 ´ 3 5 ´ 1 ´ 3 5 50 1 ´ 3 5 = 5 2 "3 5 50 ´ 3 5 101# = 5 2 3 5 50 " 1 ´ 3 5 51# .
(c) Before we can use the equations in Theorem 1.1.6, we’ll need to do a little 234 simplification.
10 ÿ i=1 i2 ´ 3i + 5 = 10 ÿ i=1 i2 + 10 ÿ i=1 ´3i + 10 ÿ i=1 5 = 10 ÿ i=1 i2 ´ 3 10 ÿ i=1 i + 5 10 ÿ i=1 1 = 1 6(10)(11)(21) ´ 3 1 2(10 ¨ 11) + 5 ¨ 10 = 270 (d) As in part (c), we’ll simplify first. The first part (shown here in red) is a geometric sum, but it does not start at 1 = 1 e 0 .
b ÿ n=1 1 e n + en3 = b ÿ n=1 1 e n + b ÿ n=1 en3 = b ÿ n=0 1 e n ´ 1 + e b ÿ n=1 n3 = 1 ´ 1 e b+1 1 ´ 1 e ´ 1 + e 1 2b(b + 1) 2 = 1 e ´ 1 e b+1 1 ´ 1 e + e 1 2b(b + 1) 2 = 1 ´ 1 e b e ´ 1 + e 4 [b(b + 1)]2 S-10: (a) The two pieces are very similar, which we can see by changing the index, or expanding them out: 100 ÿ i=50 (i ´ 50) + 50 ÿ i=0 i = (0 + 1 + 2 + ¨ ¨ ¨ + 50) + (0 + 1 + 2 + ¨ ¨ ¨ + 50) = (1 + 2 + ¨ ¨ ¨ + 50) + (1 + 2 + ¨ ¨ ¨ + 50) = 2 (1 + 2 + ¨ ¨ ¨ + 50) = 2 50 ÿ i=1 i = 2 50 ¨ 51 2 = 50 ¨ 51 = 2550 235 (b) If we expand (i ´ 5)3 = i3 ´ 15i2 + 75i ´ 125, we can break the sum into four parts, and evaluate each separately. However, it is much simpler to change the index and make the term (i ´ 5)3 into i3.
100 ÿ i=10 (i ´ 5)3 = 53 + 63 + 73 + ¨ ¨ ¨ + 953 We have a formula to evaluate the sum of cubes if they start at 1, so we turn our expression into the difference of two sums starting at 1: = h 13 + 23 + 33 + 43 + 53 + 63 + 73 + ¨ ¨ ¨ + 953i ´ h 13 + 23 + 33 + 43i = 95 ÿ i=1 i3 ´ 4 ÿ i=1 i3 = 1 2(95)(96) 2 ´ 1 2(4)(5) 2 = 20, 793, 500 .
(c) Notice every two terms cancel with each other, since the sum is (´1) + (+1), etc.
Then the terms n = 1 through n = 10 cancel, and we’re left only with the final term, (´1)11 = ´1.
Written out more explicitly: 11 ÿ n=1 (´1)n = ´1 + 1 ´ 1 + 1 ´ 1 + 1 ´ 1 + 1 ´ 1 + 1 ´ 1 = [´1 + 1] + [´1 + 1] + [´1 + 1] + [´1 + 1] + [´1 + 1] ´ 1 = 0 + 0 + 0 + 0 + 0 ´ 1 = ´1.
(d) For every integer n, 2n + 1 is odd, so (´1)2n+1 = ´1. Then 11 ÿ n=2 (´1)2n+1 = 11 ÿ n=2 ´1 = ´10.
S-11: The index of the sum runs from 1 to 4: the first, second, third, and fourth rectangles.
So, we have four rectangles in our Riemann sum. Let’s start by drawing in the intervals along the x-axis taken up by these four rectangles. Note each has the same width: b ´ a 4 .
236 x y b a y = f (x) Since this is a midpoint Riemann sum, the height of each rectangle is given by the y-value of the function in the midpoint of the interval. So, now let’s find the height of the function at the midpoints of each of the four intervals.
x y b a y = f (x) The left-most interval has a height of about 0, so it gives a “trivial” rectangle with no height and no area. The middle two intervals have rectangles of about the same height, and the right-most interval has the highest rectangle.
237 x y b a y = f (x) S-12: In general, the left Riemann sum for the integral şb a f (x) dx is of the form n ÿ k=1 f a + (k ´ 1)b ´ a n b ´ a n • To get the limits of summation to match the given sum, we need n = 4.
• Then to get the factor multiplying f to match that in the given sum, we need b´a n = 1, so b ´ a = 4.
• Finally, to get the argument of f to match that in the given sum, we need a + (k ´ 1)b ´ a n = a ´ b ´ a n + kb ´ a n = 1 + k Subbing in n = 4 and b ´ a = 4 gives a ´ 1 + k = 1 + k, so a = 2 and b = 6.
S-13: The general form of a Riemann sum is n ÿ i=1 ∆x ¨ f (x˚ i ), where ∆x = b´a n is the width of each rectangle, and f (x˚ i ) is the height.
There are different ways to interpret the given sum as a Riemann sum. The most obvious is given in Solution 1. You may notice that we make some convenient assumptions in this solution about values for ∆x and a, and we assume the sum is a right Riemann sum.
Other visualizations of the sum arise from making more exotic choices. Some of these are explored in Solutions 2-4.
All cases have three rectangles, and the three rectangles will have the same areas: 98, 162, and 242 square units, respectively. This is because the terms of the given sum simplify to 98 + 162 + 242.
Solution 1: • Because the index runs from 1 to 3, there are three intervals: n = 3.
238 • Looking at our sum, it seems reasonable to interpret ∆x = 2. Then, since n = 3, we conclude b´a 3 = 2, hence b ´ a = 6.
• If ∆x = 2, then f (x˚ i ) = (5 + 2i)2. Recall that x˚ i is the x-coordinate we use to decide the height of the ith rectangle. In a right Riemann sum, x˚ i = a + i ¨ ∆x.
So, using 2 = ∆x, we can let f (x˚ i ) = f (a + 2i) = (5 + 2i)2. This fits with the function f (x) = x2, and a = 5.
• Since b ´ a = 6, and a = 5, this tells us b = 11 To sum up, we can interpret the Riemann sum as a right Riemann sum, with three intervals, of the function f (x) = x2 from x = 5 to x = 11.
x y 5 7 9 11 49 81 121 y = x2 Solution 2: We could have chosen a different value for ∆x.
• The index of the sum runs from 1 to 3, so we have n = 3.
• We didn’t have to interpret ∆x as 2–that was just the path of least resistance.
We could have chosen it to be any other number–for the sake of argument, let’s say ∆x = 10. (Positive numbers are easiest to interpret, but negatives are technically allowed as well.) • Then 10 = b´a n = b´a 3 , so b ´ a = 30.
• Let’s use the paradigm of a right Riemann sum, and match up the terms of the 239 sum given in the problem to the terms in the definition: ∆x ¨ f (a + i ¨ ∆x) = 2 ¨ (5 + 2i)2 10 ¨ f (a + 10i) = 2 ¨ (5 + 2i)2 f (a + 10i) = 1 5 ¨ (5 + 2i)2 f (a + 10i) = 1 5 ¨ 5 + 1 5 ¨ 10i 2 • The easiest value of a in this case is a = 0. Then f (10i) = 1 5 ¨ 5 + 1 5 ¨ 10i 2 , so f (x) = 1 5 ¨ 5 + 1 5 ¨ x 2 .
• If a = 0 and b ´ a = 30, then b = 30.
• To sum up: n = 3, a = 0, b = 30, ∆x = 10, and f (x) = 1 5 ¨ 5 + x 5 2.
x y 10 20 30 49 5 81 5 121 5 y = 1 5 5 + x 5 2 By changing ∆x, we changed the widths of the rectangles. The rectangles in this picture are wider and shorter than the rectangles in Solution 1. Their areas are the same: 98, 162, and 242.
Solution 3: We could have chosen a different value of a.
• Suppose ∆x = 2, and we interpret our sum as a right Riemann sum, but we didn’t assume a = 5. We could have chosen a to be any number–say, a = 1.
• Let’s match up what we’re given in the problem to what we’re given as a definition: ∆x ¨ f (a + i ¨ ∆x) = 2 ¨ (5 + 2i)2 2 ¨ f (1 + 2i) = 2 ¨ (5 + 2i)2 f (1 + 2i) = (5 + 2i)2 f (1 + 2i) = (4 + 1 + 2i)2 • Since f (1 + 2i) = (4 + 1 + 2i)2, we have f (x) = (4 + x)2 240 • Since a = 1 and b´a 3 = 2, in this case b = 7.
• To sum up: n = 3, a = 1, b = 7, ∆x = 2, and f (x) = (4 + x)2.
x y 1 3 5 7 49 81 121 y = (4 + x)2 This picture is a lot like the picture in Solution 1, but shifted to the left. By changing a, we changed the left endpoint of our region.
Solution 4: We could have chosen a different kind of Riemann sum.
• We didn’t have to assume that we were dealing with a right Riemann sum.
Suppose ∆x = 2, and we have a midpoint Riemann sum.
• Let’s match up what we’re given in the problem with what we’re given in the definition: ∆x ¨ f a + i ´ 1 2 ∆x = 2 ¨ (5 + 2i)2 2 ¨ f a + i ´ 1 2 2 = 2 ¨ (5 + 2i)2 f a + i ´ 1 2 2 = (5 + 2i)2 f (a + 2i ´ 1) = (5 + 2i)2 f ((a ´ 1) + 2i) = (5 + 2i)2 • It is now convenient to set a ´ 1 = 5, hence a = 6.
• Then f (5 + 2i) = (5 + 2i)2, so f (x) = x2 • Since 2 = b´a 3 and a = 6, we see b = 12.
241 • To sum up: n = 3, a = 6, b = 12, ∆x = 2, and f (x) = x2.
x y 6 8 10 12 49 81 121 y = x2 By choosing to interpret our sum as a midpoint Riemann sum instead of a right Riemann sum, we changed where our rectangles intersect the graph y = f (x): instead of the graph hitting the right corner of the rectangle, it hits in the middle.
S-14: Many interpretations are possible–see the solution to Question 13 for a more thorough discussion–but the most obvious is given below. Recall the definition of a left Riemann sum: n ÿ i=1 ∆x ¨ f (a + (i ´ 1)∆x) We chose a left Riemann sum instead of right or midpoint because our given sum has (i ´ 1) in it, rather than (i ´ 1 2) or simply i.
• Since the sum has five terms (i runs from 1 to 5), there are 5 rectangles. That is, n = 5.
• In the definition of the Riemann sum, note that the term ∆x appears twice: once multiplied by the entire term, and once multiplied by i ´ 1. So, a convenient choice for ∆x is π 20, because this is the constant that is both multiplied at the start of the term, and multiplied by i ´ 1.
• Since π 20 = ∆x = b ´ a n = b ´ a 5 , we see b ´ a = 5π 20 = π 4 .
• We match the terms in the definition with the terms in the problem: f (a + (i ´ 1)∆x) = tan π(i ´ 1) 20 f a + (i ´ 1) π 20 = tan (i ´ 1) π 20 242 So, we choose a = 0 and f (x) = tan x.
• Since a = 0 and b ´ a = π 4 , we see b = π 4 .
x y 2π 20 3π 20 4π 20 5π 20 π 20 y = tan x We note that the first rectangle of the five is a “trivial” rectangle, with height (and area) 0.
S-15: Since there are four terms in the sum, n = 4. (Note the sum starts at k = 0, instead of k = 1.) Since the function is multiplied by 1, 1 = ∆x = b ´ a n = b ´ a 4 , hence b ´ a = 4.
We can choose to view the given sum as a left, right, or midpoint Riemann sum. The choice we make determines the interval. Note that the heights of the rectangles are determined when x = 1.5, 2.5, 3.5, and 4.5.
x y f (1.5) f (2.5) f (3.5) f (4.5) Option 1: right Riemann sum If our sum is a right Riemann sum, then we take the heights of the rectangles from the right endpoint of each interval.
243 x y f (1.5) 1.5 f (2.5) 2.5 f (3.5) 3.5 f (4.5) 4.5 Then a = 0.5 and b = 4.5. Therefore: 3 ř k=0 f (1.5 + k) ¨ 1 is a right Riemann sum on the interval [0.5, 4.5] with n = 4.
Option 2: left Riemann sum If our sum is a left Riemann sum, then we take the heights of the rectangles from the left endpoint of each interval.
x y f (1.5) 1.5 f (2.5) 2.5 f (3.5) 3.5 f (4.5) 4.5 Then a = 1.5 and b = 5.5. Therefore: 3 ř k=0 f (1.5 + k) ¨ 1 is a left Riemann sum on the interval [1.5, 5.5] with n = 4.
Option 3: midpoint Riemann sum If our sum is a midpoint Riemann sum, then we take the heights of the rectangles from the midpoint of each interval.
244 x y f (1.5) 1.5 f (2.5) 2.5 f (3.5) 3.5 f (4.5) 4.5 Then a = 1 and b = 5. Therefore: 3 ř k=0 f (1.5 + k) ¨ 1 is a midpoint Riemann sum on the interval [1, 5] with n = 4.
S-16: The area in question is a triangle with base 5 and height 5, so its area is 25 2 .
x y y = x 5 5 5 S-17: There is a positive and a negative portion of this area. The positive area is a triangle with base 5 and height 5, so area 25 2 square units. The negative area is a triangle with base 2 and height 2, so negative area 4 2 = 2 square units. So, the net area is 25 2 ´ 4 2 = 21 2 square units.
245 x y y = x 5 5 5 2 2 S-18: In general, the midpoint Riemann sum is given by n ÿ i=1 f a + i ´ 1/2 ∆x ∆x , where ∆x = b ´ a n .
In this problem we are told that f (x) = x8, a = 5, b = 15 and n = 50, so that ∆x = b´a n = 1 5 and the desired Riemann sum is: 50 ÿ i=1 5 + i ´ 1/2 1 5 8 1 5 S-19: The given integral has interval of integration going from a = ´1 to b = 5. So when we use three approximating rectangles, all of the same width, the common width is ∆x = b´a n = 2. The first rectangle has left endpoint x0 = a = ´1, the second has left hand endpoint x1 = a + ∆x = 1, and the third has left hand end point x2 = a + 2∆x = 3. So ż 5 ´1 x3 dx « f (x0) + f (x1) + f (x2) ∆x = (´1)3 + 13 + 33 ˆ 2 = 54 S-20: In the given integral, the domain of integration runs from a = ´1 to b = 7. So, we have ∆x = (b´a) n = (7´(´1)) n = 8 n. The left-hand end of the first subinterval is at x0 = a = ´1. So, the right-hand end of the ith interval is at x˚ i = ´1 + 8i n . So: ż 7 ´1 f (x) dx = lim nÑ8 n ÿ i=1 f ´1 + 8i n 8 n 246 S-21: We identify the given sum as the right Riemann sum n ř i=1 f (a + i∆x)∆x, with a = 0 (that’s specified in the statement of the question). Since 4 n is multiplied in every term, and is also multiplied by i, we let ∆x = 4 n. Then x˚ i = a + i∆x = 4i n and f (x) = sin2(2 + x). So, b = a + n∆x = 0 + n ¨ 4 n = 4.
S-22: The given sum is of the form lim nÑ8 n ÿ k=1 k n2 c 1 ´ k2 n2 = lim nÑ8 n ÿ k=1 1 n k n d 1 ´ k n 2 = lim nÑ8 n ÿ k=1 ∆x f (x˚ k) with ∆x = 1 n, a = 0, x˚ k = k n = a + k∆x and f (x) = x ?
1 ´ x2. Since x˚ 0 = 0 and x˚ n = 1, the right hand side is the definition (using the right Riemann sum) of ş1 0 f (x) dx.
S-23: As i ranges from 1 to n, 3i/n range from 3/n to 3 with jumps of ∆x = 3/n, so this is lim nÑ8 n ÿ i=1 3 ne´i/n cos(3i/n) = lim nÑ8 n ÿ i=1 f (x˚ i )∆x = ż b a f (x) dx where x˚ i = 3i/n, f (x) = e´x/3 cos(x), a = x0 = 0 and b = xn = 3. Thus lim nÑ8 n ÿ i=1 3 ne´i/n cos(3i/n) = ż 3 0 e´x/3 cos(x) dx S-24: As i ranges from 1 to n, the exponent i n ranges from 1 n to 1 with jumps of ∆x = 1 n. So let’s try x˚ i = i n, ∆x = 1 n. Then: Rn = n ÿ i=1 iei/n n2 = n ÿ i=1 i nei/n 1 n = n ÿ i=1 x˚ i ex˚ i ∆x = n ÿ i=1 f (x˚ i )∆x with f (x) = xex, and the limit lim nÑ8 Rn = lim nÑ8 n ÿ i=1 f (x˚ i )∆x = ż b a f (x) dx Since we chose x˚ i = i n = 0 + i∆x, we let a = 0. Then 1 n = ∆x = b´a n = b n tells us b = 1.
Thus, lim nÑ8 Rn = ż 1 0 xex dx .
S-25: 247 Choice #1: If we set ∆x = 2 n and x˚ i = 2i n , i.e. x˚ i = a + i∆x with a = 0, then lim nÑ8 n ÿ i=1 e´1´2i/n ¨ 2 n = lim nÑ8 n ÿ i=1 e´1´x˚ i ∆x = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = e´1´x = ż b a f (x) dx with a = x0 = 0 and b = xn = 2 = ż 2 0 e´1´x dx Choice #2: If we set ∆x = 2 n and x˚ i = 1 + 2i n , i.e. x˚ i = a + i∆x with a = 1, then lim nÑ8 n ÿ i=1 e´1´2i/n ¨ 2 n = lim nÑ8 n ÿ i=1 e´x˚ i ∆x = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = e´x = ż b a f (x) dx with a = x0 = 1 and b = xn = 3 = ż 3 1 e´x dx Choice #3: If we set ∆x = 1 n and x˚ i = i n, i.e. x˚ i = a + i∆x with a = 0, then lim nÑ8 n ÿ i=1 e´1´2i/n ¨ 2 n = lim nÑ8 n ÿ i=1 e´1´2x˚ i 2∆x = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = 2e´1´2x = ż b a f (x) dx with a = x0 = 0 and b = xn = 1 = 2 ż 1 0 e´1´2x dx 248 Choice #4: If we set ∆x = 1 n and x˚ i = 1 2 + i n, i.e. xi = a + i∆x with a = 1 2, then lim nÑ8 n ÿ i=1 e´1´2i/n ¨ 2 n = lim nÑ8 n ÿ i=1 e´2xi 2∆x = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = 2e´2x = ż b a f (x) dx with a = x0 = 1 2 and b = xn = 3 2 = 2 ż 3/2 1/2 e´2x dx S-26: This is similar to the familiar form of a geometric sum, but the powers go up by threes. So, we make a subsitution. If x = r3, then: 1 + r3 + r6 + r9 + ¨ ¨ ¨ + r3n = 1 + x + x2 + x3 + ¨ ¨ ¨ + xn Now, using Equation 1.1.3 in the CLP-2 text, 1 + x + x2 + x3 + ¨ ¨ ¨ + xn = xn+1 ´ 1 x ´ 1 Substituting back in x = r3, we find our sum is equal to (r3)n+1 ´ 1 r3 ´ 1 , or r3n+3 ´ 1 r3 ´ 1 .
S-27: The sum does not start at 1, so we need to do some algebra. We can either factor out the first term, or subtract off the initial terms that are missing.
Solution 1: If we factor out r5, then what’s left fits the form of Equation 1.1.3 in the CLP-2 text: r5 + r6 + r7 + ¨ ¨ ¨ + r100 = r5 h 1 + r + r2 + ¨ ¨ ¨ + r95i = r5 r96 ´ 1 r ´ 1 .
Solution 2: We know how to evaluate sums of this form if they start at 1, so we re-write our sum as follows: r5 + r6 + r7 + ¨ ¨ ¨ + r100 = 1 + r + r2 + r3 + r4 + r5 + ¨ ¨ ¨ + r100 ´ 1 + r + r2 + r3 + r4 = r101 ´ 1 r ´ 1 ´ r5 ´ 1 r ´ 1 = r101 ´ 1 ´ r5 + 1 r ´ 1 = r101 ´ r5 r ´ 1 = r5 r96 ´ 1 r ´ 1 .
249 S-28: Recall that |x| = # ´x if x ď 0 x if x ě 0 so that |2x| = # ´2x if x ď 0 2x if x ě 0 To picture the geometric figure whose area the integral represents observe that • at the left hand end of the domain of integration x = ´1 and the integrand |2x| = | ´ 2| = 2 and • as x increases from ´1 towards 0, the integrand |2x| = ´2x decreases linearly, until • when x hits 0 the integrand hits |2x| = |0| = 0 and then • as x increases from 0, the integrand |2x| = 2x increases linearly, until • when x hits +2, the right hand end of the domain of integration, the integrand hits |2x| = |4| = 4.
So the integral ş2 ´1 |2x| dx is the area of the union of the two shaded triangles (one of base 1 and of height 2 and the other of base 2 and height 4) in the figure on the right below and ż 2 ´1 |2x| dx = 1 2 ˆ 1 ˆ 2 + 1 2 ˆ 2 ˆ 4 = 5 x y −1 2 2 4 S-29: The area we want is two triangles, both above the x-axis. Each triangle has base 4 and height 4, so the total area is 2 ¨ 4 ¨ 4 2 = 16.
x y y = |t ´ 1| 5 1 ´3 4 250 If you had a hard time sketching the function, recall that the absolute value of a number leaves it unchanged if it is positive or zero, and flips the sign if it is negative. So, when t ´ 1 ě 0 (that is, when t ě 1), our function is simply f (t) = |t ´ 1| = t ´ 1. On the other hand, when t = 1 is negative (that is, when t ă 1), the absolute value changes the sign, so f (t) = |t ´ 1| = ´(t ´ 1) = ´t + 1.
S-30: The area we want is a trapezoid with base (b ´ a) and heights a and b, so its area is (b ´ a)(b + a) 2 = b2 ´ a2 2 .
x y y = x b b a a Instead of using a formula for the area of a trapezoid, you can find the blue area as the area of a triangle with base and height b, minus the area of a triangle with base and height a.
S-31: The area is negative. The shape is a trapezoid with base length (b ´ a) and heights 0 ´ a = ´a and 0 ´ b = ´b (note: those are nonnegative numbers), so its area is (b ´ a)(´b ´ a) 2 = ´b2 + a2 2 . Since the shape is below the x-axis, we change its sign.
Thus, the integral evaluates to b2 ´ a2 2 .
x y y = x a a b b The signs can be a little hard to keep track of. The base of our trapezoid is |a ´ b|; since b ą a, this is b ´ a. The heights of the trapezoid are |a| and |b|; since these are both negative, |a| = ´a and |b| = ´b.
251 We note that this is the same result as in Question 30.
S-32: If y = ?
16 ´ x2, then y is nonnegative, and y2 + x2 = 16. So, the graph y = ?
16 ´ x2 is the upper half of a circle of radius 4. Since x only runs from 0 to 4, we have a quarter of a circle of radius 4. Then the area under the curve is 1 4 π ¨ 42 = 4π.
x y y = ?
16 ´ x2 S-33: Here is a sketch the graph of f (x).
x y 1 3 1 y = f(x) There is a linear increase from x = 0 to x = 1, followed by a constant. Using the interpretation of ş3 0 f (x) dx as the area between y = f (x) and the x–axis with x between 0 and 3, we can break this area into: • ş1 0 f (x) dx: a right-angled triangle of height 1 and base 1 and hence area 0.5.
• ş3 1 f (x) dx: a rectangle of height 1 and base 2 and hence area 2.
Summing up: ş3 0 f (x) dx = 2.5.
S-34: The car’s speed increases with time. So its highest speed on any time interval occurs at the right hand end of the interval and the best possible upper estimate for the distance traveled is given by the right Riemann sum with ∆x = 0.5, which is v(0.5) + v(1.0) + v(1.5) + v(2.0) ˆ 0.5 = 14 + 22 + 30 + 40 ˆ 0.5 = 53 m S-35: There is a key detail in the statement of Question 34: namely, that the car is continuously accelerating. So, although we don’t know exactly what’s going on in 252 between our brief snippets of information, we know that the car is not going any faster during an interval than at the end of that interval. Therefore, the car certainly travelled no farther than our estimation.
We ask this question in order to point out an important detail. If we did not have the information that the car was continuously accelerating, we would not be able to give a certain upper bound on its distance travelled. It would be possible that, when the car is not being observed (for example, when t = 0.25), it is going much faster than when it is being observed.
S-36: First, note that the distance travelled by the plane is equal to the area under the graph of its speed.
We need to know the speed of the plane at the midpoints of our intervals. So (for example) noon to 1pm is not one of your intervals–we don’t know the speed at 12:30. (A common idea is to average the two end values, 700 and 800. This is a fine approximation, but it is not a Riemann sum.) So, we use the two intervals 12:00 to 2:00, and 2:00 to 4:00.
Then our intervals have length 2 hours, and at the midpoints of the intervals the speed of the plane is 700 kph and 900 kph, respectively. So, our midpoint Riemann sum gives us: 700(2) + 900(2) = 3200 an approximation of 3200 km travelled by the plane from noon to 4:00 pm.
Remark: if we had been asked to approximate the distance travelled from 11:30 am to 4:30 pm, then we could have used the midpoint rule with five intervals and made use of every entry in the data table. With the question as stated, however, we ignore three out of five entries in the table because they are not the midpoints of our intervals.
S-37: Solution #1: Set x˚ i = ´2 + 2i n . Then a = x0 = ´2 and b = xn = 0 and ∆x = 2 n. So lim nÑ8 n ÿ i=1 2 n d 4 ´ ´2 + 2i n 2 = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = a 4 ´ x2 and ∆x = 2 n = ż 0 ´2 a 4 ´ x2 dx For the integral ş0 ´2 ?
4 ´ x2 dx, y = ?
4 ´ x2 is equivalent to x2 + y2 = 4, y ě 0. So the integral represents the area between the upper half of the circle x2 + y2 = 4 (which has radius 2) and the x-axis with ´2 ď x ď 0, which is a quarter circle with area 1 4 ¨ π 22 = π.
253 x y y = ?
4 ´ x2 ´2 Solution #2: Set x˚ i = 2i n . Then a = x0 = 0 and b = xn = 2 and ∆x = 2 n. So lim nÑ8 n ÿ i=1 2 n d 4 ´ ´2 + 2i n 2 = lim nÑ8 n ÿ i=1 f (x˚ i )∆x with f (x) = b 4 ´ (´2 + x)2, ∆x = 2 n = ż 2 0 b 4 ´ (´2 + x)2 dx For the integral ş2 0 a 4 ´ (´2 + x)2 dx , y = a 4 ´ (x ´ 2)2 is equivalent to (x ´ 2)2 + y2 = 4, y ě 0. So the integral represents the area between the upper half of the circle (x ´ 2)2 + y2 = 4 (which is centered at (2, 0) and has radius 2) and the x-axis with 0 ď x ď 2, which is a quarter circle with area 1 4 ¨ π 22 = π.
x y y = a 4 ´ (x ´ 2)2 2 S-38: (a) The left Riemann sum is defined as Ln = n ÿ i=1 f (xi´1)∆x with xi = a + i∆x We subdivide into n = 3 intervals, so that ∆x = b´a n = 3´0 3 = 1, x0 = 0, x1 = 1 and x2 = 2. The function f (x) = 7 + x3 has the values f (x0) = 7 + 03 = 7, f (x1) = 7 + 13 = 8, and f (x2) = 7 + 23 = 15, from which we evaluate L3 = f (x0) + f (x1) + f (x2) ∆x = 7 + 8 + 15 ˆ 1 = 30 254 (b) We divide into n intervals so that ∆x = b´a n = 3 n and xi = a + i∆x = 3i n . The right Riemann sum is therefore: Rn = n ÿ i=1 f (xi)∆x = n ÿ i=1 7 + (3i)3 n3 3 n = n ÿ i=1 21 n + 81 i3 n4 To calculate the sum: Rn = 21 n n ÿ i=1 1 !
+ 81 n4 n ÿ i=1 i3 !
= 21 n ˆ n + 81 n4 ˆ n4 + 2n3 + n2 4 = 21 + 81 4 (1 + 2/n + 1/n2) To evaluate the limit exactly, we take n Ñ 8. The expressions involving 1/n vanish leaving: ż 3 0 (7 + x3) dx = lim nÑ8 Rn = 21 + 81 4 = 411 4 S-39: In general, the right–endpoint Riemann sum approximation to the integral şb a f (x) dx using n rectangles is n ÿ i=1 f (a + i∆x)∆x where ∆x = b´a n . In this problem, a = 2, b = 4, and f (x) = x2, so that ∆x = 2 n and the right–endpoint Riemann sum approximation becomes n ÿ i=1 f 2 + 2i n 2 n = n ÿ i=1 2 + 2i n 2 2 n = n ÿ i=1 4 + 8i n + 4i2 n2 2 n = n ÿ i=1 8 n + 16i n2 + 8i2 n3 = n ÿ i=1 8 n + n ÿ i=1 16i n2 + n ÿ i=1 8i2 n3 = 8 n n ÿ i=1 1 + 16 n2 n ÿ i=1 i + 8 n3 n ÿ i=1 i2 = 8 nn + 16 n2 ¨ n(n + 1) 2 + 8 n3 ¨ n(n + 1)(2n + 1) 6 = 8 + 8 1 + 1 n + 4 3 1 + 1 n 2 + 1 n 255 So ż 4 2 x2 dx = lim nÑ8 h 8 + 8 1 + 1 n + 4 3 1 + 1 n 2 + 1 n i = 8 + 8 + 4 3 ˆ 2 = 56 3 S-40: We’ll use right Riemann sums with a = 0 and b = 2. When there are n rectangles, ∆x = b´a n = 2 n and xi = a + i∆x = 2i/n. So we need to evaluate lim nÑ8 n ÿ i=1 f (xi)∆x = lim nÑ8 n ÿ i=1 (xi)3 + xi ∆x = lim nÑ8 n ÿ i=1 2i n 3 + 2i n !
2 n = lim nÑ8 2 n n ÿ i=1 8i3 n3 + 2i n = lim nÑ8 16 n4 n ÿ i=1 i3 + 4 n2 n ÿ i=1 i !
= lim nÑ8 16(n4 + 2n3 + n2) n4 ¨ 4 + 4(n2 + n) n2 ¨ 2 = lim nÑ8 16 4 1 + 2 n + 1 n2 + 4 2 1 + 1 n = 16 4 + 4 2 = 6.
S-41: We’ll use right Riemann sums with a = 1, b = 4 and f (x) = 2x ´ 1. When there are n rectangles, ∆x = b´a n = 3 n and xi = a + i∆x = 1 + 3i/n. So we need to evaluate lim nÑ8 n ÿ i=1 f (xi)∆x = lim nÑ8 n ÿ i=1 (2xi ´ 1) ∆x = lim nÑ8 n ÿ i=1 2 + 6i n ´ 1 3 n = lim nÑ8 3 n n ÿ i=1 6i n + 1 = lim nÑ8 18 n2 n ÿ i=1 i + 3 n n ÿ i=1 1 !
= lim nÑ8 18 ¨ n(n + 1) n2 ¨ 2 + 3 nn = lim nÑ8 9 1 + 1 n + 3 = 9 + 3 = 12.
256 S-42: Using the definition of a right Riemann sum, 10 ÿ i=1 3(7 + 2i)2 sin(4i) = 10 ÿ i=1 ∆x f (a + i∆x) Since ∆x = 10 and a = ´5, 10 ÿ i=1 3(7 + 2i)2 sin(4i) = 10 ÿ i=1 10f (´5 + 10i) Dividing both expressions by 10, 10 ÿ i=1 3 10(7 + 2i)2 sin(4i) = 10 ÿ i=1 f (´5 + 10i) So, we have an expression for f (´5 + 10i): f (´5 + 10i) = 3 10(7 + 2i)2 sin(4i) In order to find f (x), let x = ´5 + 10i. Then i = x 10 + 1 2.
f (x) = 3 10 7 + 2 x 10 + 1 2 2 sin 4 x 10 + 1 2 = 3 10 x 5 + 8 2 sin 2x 5 + 2 .
S-43: As in the text, we’ll set up a Riemann sum for the given integral. Right Riemann sums have the simplest form, so we use a right Riemann sum, but we could equally well use left or midpoint.
ż 1 0 2xdx = lim nÑ8 n ÿ i=1 ∆x f (a + i∆x) = lim nÑ8 n ÿ i=1 1 n f i n = lim nÑ8 n ÿ i=1 1 n ¨ 2i/n = lim nÑ8 1 n 21/n + 22/n + 23/n + ¨ ¨ ¨ + 2n/n = lim nÑ8 21/n n 1 + 21/n + 22/n + ¨ ¨ ¨ + 2 n´1 n = lim nÑ8 21/n n 1 + 21/n + 21/n2 + ¨ ¨ ¨ + 21/nn´1 257 The sum in parenthesis has the form of a geometric sum, with r = 21/n: = lim nÑ8 21/n n 21/nn ´ 1 21/n ´ 1 !
= lim nÑ8 21/n n 2 ´ 1 21/n ´ 1 = lim nÑ8 21/n n(21/n ´ 1) Note as n Ñ 8, 1/n Ñ 0, so the numerator has limit 1, while the denominator has indeterminate form 8 ¨ 0. So, we’ll do a little algebra to get this into a l’Hˆ opital-style indeterminate form: = lim nÑ8 1 n ¨ 21/n 21/n ´ 1 = lim nÑ8 1 n 1 ´ 2´1/n loooomoooon numÑ0 denÑ0 Now we can use l’Hˆ opital’s rule. Recall d dx t2xu = 2x log x, where log x is the natural logarithm of x, also sometimes written ln x. We’ll need to use the chain rule when we differentiate the denominator.
= lim nÑ8 ´1 n2 ´2´1/n log 2 ¨ 1 n2 = lim nÑ8 21/n log 2 = 1 log 2 Using a calculator, we see this is about 1.44 square units.
S-44: As in the text, we’ll set up a Riemann sum for the given integral. Right Riemann 258 sums have the simplest form: ż b a 10xdx = lim nÑ8 n ÿ i=1 ∆x f (a + i∆x) = lim nÑ8 n ÿ i=1 b ´ a n f a + ib ´ a n = lim nÑ8 n ÿ i=1 b ´ a n ¨ 10a+i b´a n = lim nÑ8 n ÿ i=1 b ´ a n ¨ 10a ¨ 10 b´a n i = lim nÑ8 b ´ a n ¨ 10a 10 b´a n 1 + 10 b´a n 2 + 10 b´a n 3 + ¨ ¨ ¨ + 10 b´a n n = lim nÑ8 b ´ a n ¨ 10a ¨ 10 b´a n 1 + 10 b´a n + 10 b´a n 2 + ¨ ¨ ¨ + 10 b´a n n´1 Now the sum in parentheses has the form of a geometric sum, with r = 10 b´a n : = lim nÑ8 b ´ a n ¨ 10a ¨ 10 b´a n 10 b´a n n ´ 1 10 b´a n ´ 1 = lim nÑ8 b ´ a n ¨ 10a ¨ 10 b´a n 10b´a ´ 1 10 b´a n ´ 1 !
The coloured parts do not depend on n, so for simplicity we can move them outside the limit.
= (b ´ a) ¨ 10a 10b´a ´ 1 lim nÑ8 1 n ¨ 10 b´a n 10 b´a n ´ 1 !
= (b ´ a) ¨ 10b ´ 10a lim nÑ8 1/n 1 ´ 10´ b´a n loooooooomoooooooon numÑ0 denÑ0 259 Now we can use l’Hˆ opital’s rule. Recall d dx t10xu = 10x log x, where log x is the natural logarithm of x, also sometimes written ln x. For the denominator, we will have to use the chain rule.
= (b ´ a) ¨ 10b ´ 10a lim nÑ8 ´1/n2 ´10´ b´a n ¨ log 10 ¨ b´a n2 = (b ´ a) ¨ 10b ´ 10a lim nÑ8 1 10´ b´a n ¨ log 10 ¨ (b ´ a) !
= (b ´ a) ¨ 10b ´ 10a 1 log 10 ¨ (b ´ a) = 1 log 10 10b ´ 10a For part (b), we can guess that if 10 were changed to c, our answer would be ż b a cx dx = 1 log c cb ´ ca In Question 43, we had a = 0, b = 1, and c = 2. In this case, the formula we guessed above gives ż 1 0 2x dx = 1 log 2 21 ´ 20 = 1 log 2 This does indeed match the answer we calculated.
(In fact, we can directly show ż b a cx dx = 1 log c cb ´ ca using the method of this problem.) S-45: First, we note y = ?
1 ´ x2 is the upper half of a circle of radius 1, centred at the origin. We’re taking the area under the curve from 0 to a, so the area in question is as shown in the picture below.
x y ´1 1 a In order to use geometry to find this area, we break it up into two pieces: a sector of a circle, and a triangle, shown below.
260 x y ´1 1 a θ Area of sector: The sector is a portion of a circle with radius 1, with inner angle θ. So, its area is θ 2π (area of circle) = θ 2π (π) = θ 2.
Our job now is to find θ in terms of a. Note π 2 ´ θ is the inner angle of the red triangle, which lies in the unit circle. So, cos π 2 ´ θ = a. Then π 2 ´ θ = arccos(a), and so θ = π 2 ´ arccos(a).
Then the area of the sector is π 4 ´ 1 2 arccos(a) square units.
Area of triangle: The triangle has base a. Its height is the y-value of the function when x = a, so its height is ?
1 ´ a2. Then the area of the triangle is 1 2a ?
1 ´ a2.
We conclude ż a 0 a 1 ´ x2 dx = π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2.
S-46: (a) The difference between our upper and lower bounds is the difference in areas between the larger set of rectangles and the smaller set of rectangles. Drawing them on a single picture makes this a little clearer.
x y y = f (x) a b Each of the rectangles has width b´a n , since we took a segment of the x-axis with length b ´ a and chopped it into n pieces. We could calculate the height of each rectangle, but it would be a little complicated, since it differs for each of them. An 261 easier method is to notice that the area we want to calculate can be imagined as a single rectangle: x y y = f (x) a b f (a) f (b) The rectangle has base b´a n . Its highest coordinate is f (a), and its lowest is f (b), so its height is f (b) ´ f (a). Therefore, the difference in area between our lower bound and our upper bound is: [ f (b) ´ f (a)] ¨ b ´ a n (b) We want to give a range with length at most 0.01, and guarantee that the area under the curve y = f (x) is inside that range. In the previous part, we figured out that when we use n rectangles, the length of our range is [ f (b) ´ f (a)] ¨ b´a n . So, all we have to do is set this to be less than or equal to 0.01, and solve for n: [ f (b) ´ f (a)] ¨ b ´ a n ď 0.01 100 [ f (b) ´ f (a)] ¨ (b ´ a) ď n We can choose n to be an integer that is greater than or equal to 100 [ f (b) ´ f (a)] ¨ (b ´ a). Using that many rectangles, we find an upper and lower bound for the area under the curve. If we choose any number between our upper and lower bound as an approximation for the area under the curve, our error is no more than 0.01.
Remark: this question depends on the fact that f is decreasing and positive from a to b.
In general, bounding errors on approximations like this is not so straightforward.
S-47: Since f (x) is linear, there exist real numbers m and c such that f (x) = mx + c. Now we can do some calculations. Suppose we have a rectangle in our Riemann sum that takes up the interval [x, x + w].
• If we are using a left Riemann sum, our rectangle has height f (x) = mx + c. Then it has area w(mx + c).
• If we are using a right Riemann sum, our rectangle has height f (x + w) = m(x + w) + c = mx + c + mw. Then it has area w(mx + c + mw).
262 • If we are using a midpoint Riemann sum, our rectangle has height f (x + 1 2w) = m(x + 1 2w) + c = mx + c + 1 2mw. Then it has area w mx + c + 1 2w .
So, for each rectangle in our sums, the midpoint rectangle has the same area as the average of the left and right rectangles: w mx + c + 1 2mw = w(mx + c) + w(mx + c + mw) 2 It follows that the midpoint Riemann sum has a value equal to the average of the values of the left and right Riemann sums. To see this, let the rectangles in the midpoint Riemann sum have areas M1, M2, . . . , Mn, let the rectangles in the left Riemann sum have areas L1, L2, . . . , Ln, and let the rectangles in the right Riemann sum have areas R1, R2, . . . , Rn. Then the midpoint Riemann sum evaluates to M1 + M2 + ¨ ¨ ¨ + Mn, and: [L1 + L2 + . . . + Ln] + [R1 + R2 + . . . + Rn] 2 = L1 + R1 2 + L2 + R2 2 + ¨ ¨ ¨ + Ln + Rn 2 = M1 + M2 + ¨ ¨ ¨ + Mn So, the statement is true.
(Note, however, it is false for many non-linear functions f (x).) Solutions to Exercises 1.2 — Jump to TABLE OF CONTENTS S-1: (a) ż a a f (x) dx = 0 x y a y = f (x) The area under the curve is zero, because it’s a region with no width.
(b) ż b a f (x) dx = ż c a f (x) dx + ż b c f (x)dx x y a c b y = f (x) 263 If we assume a ď c ď b, then this identity simply tells us that if we add up the area under the curve from a to c, and from c to b, then we get the whole area under the curve from a to b.
(The situation is slightly more complicated when c is not between a and b, but it still works out.) (c) ż b a ( f (x) + g(x)) dx = ż b a f (x) dx + ż b a g(x) dx x y a b y = f (x) y = f (x) + g(x) The blue-shaded area in the picture above is ż b a f (x) dx. The area under the curve f (x) + g(x) but above the curve f (x) (shown in red) is ż b a g(x) dx.
S-2: Using the identity b ż a f (x) dx = c ż a f (x) dx + b ż c f (x) dx , we see b ż a cos x dx = 0 ż a cos x dx + b ż 0 cos x dx = ´ a ż 0 cos x dx + b ż 0 cos x dx = ´ sin a + sin b = sin b ´ sin a S-3: (a) False. For example if f (x) = # 0 for x ă 0 1 for x ě 0 then ş´2 ´3 f (x)dx = 0 and ´ ş2 3 f (x)dx = ş3 2 f (x)dx = 1.
264 x y ´3 ´2 3 2 (b) False. For example, if f (x) = x, then ş´2 ´3 f (x) dx is negative while ş3 2 f (x) dx is positive, so they cannot be the same.
x y ´3 ´2 3 2 (c) False. For example, consider the functions f (x) = # 0 for x ă 1 2 1 for x ě 1 2 and g(x) = # 0 for x ě 1 2 1 for x ă 1 2 Then f (x) ¨ g(x) = 0 for all x, so ş1 0 f (x) ¨ g(x)dx = 0. However, ş1 0 f (x)dx = 1 2 and ş1 0 g(x)dx = 1 2, so ş1 0 f (x)dx ¨ ş1 0 g(x)dx = 1 4.
1 2 1 1 x y f (x) 1 2 1 1 x y g(x) S-4: (a) ∆x = b ´ a n = 0 ´ 5 100 = ´ 1 20 Note: if we were to use the Riemann-sum definition of a definite integral, this is how we would justify the identity b ş a f (x)dx = ´ a ş b f (x)dx.
(b) The heights of the rectangles are given by f (xi), where xi = a + i∆x = 5 ´ i 20. Since f (x) only gives positive values, f (xi) ą 0, so the heights of the rectangles are positive.
265 (c) Our Riemann sum is the sum of the signed areas of individual rectangles. Each rectangle has a negative base (∆x) and a positive height (f (xi)). So, each term of our sum is negative. If we add up negative numbers, the sum is negative. So, the Riemann sum is negative.
(d) Since f (x) is always above the x-axis, 5 ş 0 f (x)dx is positive.
S-5: The operation of integration is linear (that’s part (d) of the “arithmetic of integration” Theorem 1.2.1 in the CLP-2 text), so that: ż 3 2 [6f (x) ´ 3g(x)] dx = ż 3 2 6f (x) dx ´ ż 3 2 3g(x) dx = 6 ż 3 2 f (x) dx ´ 3 ż 3 2 g(x) dx = (6 ˆ (´1)) ´ (3 ˆ 5) = ´21 S-6: The operation of integration is linear (that’s part (d) of the “arithmetic of integration” Theorem 1.2.1 in the CLP-2 text), so that: ż 2 0 [2f (x) + 3g(x)] dx = ż 2 0 2f (x) dx + ż 2 0 3g(x) dx = 2 ż 2 0 f (x) dx + 3 ż 2 0 g(x) dx = (2 ˆ 3) + (3 ˆ (´4)) = ´6 S-7: Using part (d) of the “arithmetic of integration” Theorem 1.2.1, followed by parts (c) and (b) of the “arithmetic for the domain of integration” Theorem 1.2.3 in the in the CLP-2 text, ż 2 ´1 3g(x) ´ f (x) dx = 3 ż 2 ´1 g(x) dx ´ ż 2 ´1 f (x) dx = 3 ż 0 ´1 g(x) dx + 3 ż 2 0 g(x) dx ´ ż 0 ´1 f (x) dx ´ ż 2 0 f (x) dx = 3 ż 0 ´1 g(x) dx + 3 ż 2 0 g(x) dx + ż ´1 0 f (x) dx ´ ż 2 0 f (x) dx = 3 ˆ 3 + 3 ˆ 4 + 1 ´ 2 = 20 S-8: (a) Since ?
1 ´ x2 is an even function, ż 0 a a 1 ´ x2 dx = ż |a| 0 a 1 ´ x2 dx = π 4 ´ 1 2 arccos(|a|) + 1 2|a| b 1 ´ |a|2 = π 4 ´ 1 2 arccos(´a) ´ 1 2a a 1 ´ a2 266 Alternatively, since arccos(´a) = π ´ arccos(a) we also have ż 0 a a 1 ´ x2 dx = ´π 4 + 1 2 arccos(a) ´ 1 2a a 1 ´ a2 (b) Note ż 1 0 a 1 ´ x2 dx = π 4 , since the area under the curve represents one-quarter of the unit circle. Then, ż 1 a a 1 ´ x2 dx = ż 1 0 a 1 ´ x2 dx ´ ż a 0 a 1 ´ x2 dx = π 4 ´ π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2 = 1 2 arccos(a) ´ 1 2a a 1 ´ a2 S-9: Recall that |x| = # ´x if x ď 0 x if x ě 0 so that |2x| = # ´2x if x ď 0 2x if x ě 0 Also recall, from Example 1.2.6 in the CLP-2 text that ż b a x dx = b2 ´ a2 2 So ż 2 ´1 |2x| dx = ż 0 ´1 |2x| dx + ż 2 0 |2x| dx = ż 0 ´1 (´2x) dx + ż 2 0 2x dx = ´2 ż 0 ´1 x dx + 2 ż 2 0 x dx = ´2 ¨ 02 ´ (´1)2 2 + 2 ¨ 22 ´ 02 2 = 1 + 4 = 5 S-10: We note that the integrand f (x) = x|x| is an odd function, because f (´x) = ´x| ´ x| = ´x|x| = ´f (x). Then, by Theorem 1.2.12.b in the CLP-2 text, ż 5 ´5 x|x| dx = 0.
267 S-11: Using Theorem 1.2.12.a in the CLP-2 text, 10 = ż 2 ´2 f (x)dx = 2 ż 2 0 f (x)dx 5 = ż 2 0 f (x)dx Also, ż 2 ´2 f (x)dx = ż 0 ´2 f (x)dx + ż 2 0 f (x)dx So, ż 0 ´2 f (x)dx = ż 2 ´2 f (x)dx ´ ż 2 0 f (x)dx = 10 ´ 5 = 5 Indeed, for any even function f (x), 0 ş ´a f (x)dx = a ş 0 f (x)dx.
S-12: We first use additivity: ż 2 ´2 5 + a 4 ´ x2 dx = ż 2 ´2 5 dx + ż 2 ´2 a 4 ´ x2 dx The first integral represents the area of a rectangle of height 5 and width 4 and so equals 20. The second integral represents the area above the x–axis and below the curve y = ?
4 ´ x2 or x2 + y2 = 4. That is a semicircle of radius 2, which has area 1 2π22. So ż 2 ´2 5 + a 4 ´ x2 dx = 20 + 2π x y 2 ´2 y = 5 2 ş ´2 5dx = 5 x y 2 ´2 y = ?
4 ´ x2 2 ş ´2 ?
4 ´ x2dx = 2π 268 S-13: Note that the integrand f (x) = sin x log(3+x2) is an odd function, because: f (´x) = sin(´x) log(3 + (´x)2) = ´ sin x log(3 + x2) = ´f (x) The domain of integration ´2012 ď x ď 2012 is symmetric about x = 0. So, by Theorem 1.2.12 of the CLP-2 text, ż +2012 ´2012 sin x log(3 + x2)dx = 0 S-14: Note that the integrand f (x) = x1/3 cos x is an odd function, because: f (´x) = (´x)1/3 cos(´x) = ´x1/3 cos x = ´f (x) The domain of integration ´2012 ď x ď 2012 is symmetric about x = 0. So, by Theorem 1.2.12 of the CLP-2 text, ż +2012 ´2012 x1/3 cos x dx = 0 S-15: Our integrand f (x) = (x ´ 3)3 is neither even nor odd. However, it does have a similar symmetry. Namely, f (3 + x) = ´f (3 ´ x). So, f is “negatively symmetric” across the line x = 3. This suggests that the integral should be 0: the positive area to the right of x = 3 will be the same as the negative area to the left of x = 3.
Another way to see this is to notice that the graph of f (x) = (x ´ 3)3 is equivalent to the graph of g(x) = x3 shifted three units to the right, and g(x) is an odd function. So, ż 6 0 (x ´ 3)3 dx = ż 3 ´3 x3 dx = 0 x y ´3 3 6 y = x3 y = (x ´ 3)3 269 S-16: (a) (ax)2 + (by)2 = 1 by = b 1 ´ (ax)2 y = 1 b b 1 ´ (ax)2 (b) The values of x in the domain of the function above are those that satisfy 1 ´ (ax)2 ě 0. That is, ´1 a ď x ď 1 a. Therefore, the upper half of the ellipse has area 1 b ż 1 a ´ 1 a b 1 ´ (ax)2 dx The upper half of a circle has equation y = ?
r2 ´ x2.
= 1 b ż 1 a ´ 1 a d a2 1 a2 ´ x2 dx = 1 b ż 1 a ´ 1 a a c 1 a2 ´ x2 dx = a b ż 1 a ´ 1 a c 1 a2 ´ x2 dx (c) The function y = c 1 a2 ´ x2 is the upper-half of the circle centred at the origin with radius 1 a. So, the expression from (b) evaluates to a b π 2a2 = π 2ab.
The expression from (b) was half of the ellipse, so the area of the ellipse is π ab.
Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this: • The area of the unit circle x2 + y2 = 1 is π .
• The ellipse (ax)2 + y2 = 1 is obtained by shrinking the unit circle horizontally by a factor of a. So, its area is π a .
• Further, the ellipse (ax)2 + (by)2 = 1 is obtained from the previous ellipse by shrinking it vertically by a factor of b. So, its area is π ab .
270 S-17: Let’s recall the definitions of even and odd functions: f (x) is even if f (´x) = f (x) for every x in its domain, and f (x) is odd if f (´x) = ´f (x) for every x in its domain.
Let h(x) = f (x) ¨ g(x).
even ˆ even: If f and g are both even, then h(´x) = f (´x) ¨ g(´x) = f (x) ¨ g(x) = h(x), so their product is even.
odd ˆ odd: If f and g are both odd, then h(´x) = f (´x) ¨ g(´x) = [´f (x)] ¨ [´g(x)] = f (x) ¨ g(x) = h(x), so their product is even.
even ˆ odd: If f is even and g is odd, then h(´x) = f (´x) ¨ g(´x) = f (x) ¨ [´g(x)] = ´[ f (x) ¨ g(x)] = ´h(x), so their product is odd. Because multiplication is commutative, the order we multiply the functions in doesn’t matter.
We note that the table would be the same as if we were adding (not multiplying) even and odd numbers (not functions).
S-18: Since f (x) is odd, f (0) = ´f (´0) = ´f (0). So, f (0) = 0.
However, this restriction does not apply to g(x). For example, for any constant c, let g(x) = c. Then g(x) is even and g(0) = c. So, g(0) can be any real number.
S-19: Let x be any real number.
• f (x) = f (´x) (since f (x) is even), and • f (x) = ´f (´x) (since f (x) is odd).
• So, f (x) = ´f (x).
• Then (adding f (x) to both sides) we see 2f (x) = 0, so f (x) = 0.
So, f (x) = 0 for every x.
S-20: Solution 1: Suppose f (x) is an odd function. We investigate f 1(x) using the chain rule: f (´x) = ´f (x) (odd function) d dxtf (´x)u = d dxt´f (x)u ´f 1(´x) = ´f 1(x) (chain rule) f 1(´x) = f 1(x) So, when f (x) is odd, f 1(x) is even.
Similarly, suppose f (x) is even.
271 f (´x) = f (x) (even function) d dxtf (´x)u = d dxtf (x)u ´f 1(´x) = f 1(x) (chain rule) f 1(´x) = ´f 1(x) So, when f (x) is even, f 1(x) is odd.
Solution 2: Another way to think about this problem is to notice that “mirroring” a function changes the sign of its derivative. Then since an even function is “mirrored once” (across the y-axis), it should have f 1(x) = ´f 1(´x), and so the derivative of an even function should be an odd function. Since an odd function is “mirrored twice” (across the y-axis and across the x-axis), it should have f 1(x) = ´(´f 1(´x)) = f 1(´x). So the derivative of an odd function should be even. These ideas are presented in more detail below.
First, we consider the case where f (x) is even, and investigate f 1(x).
x y y = f (x) a1 ´a1 a2 ´a2 a3 ´a3 The whole function has a mirror-like symmetry across the y-axis. So, at x and ´x, the function will have the same “steepness,” but if one is increasing then the other is decreasing. That is, f 1(´x) = ´f 1(x). (In the picture above, compare the slope at some point ai with its corresponding point ´ai.) So, f 1(x) is odd when f (x) is even.
Second, let’s consider the case where f (x) is odd, and investigate f 1(x). Suppose the blue graph below is y = f (x). If f (x) were even, then to the left of the y-axis, it would look like the orange graph, which we’ll call y = g(x).
272 x y y = g(x) y = f (x) From our work above, we know that, for every x ą 0, ´f 1(x) = g1(´x). When x ă 0, f (x) = ´g(x). So, if x ą 0, then ´f 1(x) = g1(´x) = ´f 1(´x). In other words, f 1(x) = f 1(´x). Similarly, if x ă 0, then f 1(x) = ´g1(x) = f 1(´x). Therefore f 1(x) is even. (In the graph below, you can anecdotally verify that f 1(ai) = f 1(´ai).) x y y = g(x) y = f (x) a1 ´a1 a2 ´a2 a3 ´a3 Solutions to Exercises 1.3 — Jump to TABLE OF CONTENTS S-1: The Fundamental Theorem of Calculus Part 2 (Theorem 1.3.1 in the CLP-2 text) tells us that ż ?
5 1 f (x) dx = F( ?
5) ´ F(1) = e( ?
52´3) + 1 ´ e(12´3) + 1 = e5´3 ´ e1´3 = e2 ´ e´2 S-2: First, let’s find a general antiderivative of x3 ´ sin(2x).
273 • One function with derivative x3 is x4 4 .
• To find an antiderivative of sin(2x), we might first guess cos(2x); checking, we see d dxtcos(2x)u = ´2 sin(2x). So, we only need to multiply by ´1 2: d dx " ´1 2 cos 2x = sin(2x).
So, the general antiderivative of f (x) is x4 4 + 1 2 cos 2x + C. To satisfy F(0) = 1, we need2 hx4 4 + 1 2 cos 2x + C i x=0 = 1 ð ñ 1 2 + C = 1 ð ñ C = 1 2 So F(x) = x4 4 + 1 2 cos 2x + 1 2.
S-3: (a) This is true, by part 2 of the Fundamental Theorem of Calculus, Thereom 1.3.1 in the CLP-2 text with G(x) = f (x) and f (x) replaced by f 1(x).
(b) This is not only false, but it makes no sense at all. The integrand is strictly positive so the integral has to be strictly positive. In fact it’s +8. The Fundamental Theorem of Calculus does not apply because the integrand has an infinite discontinuity at x = 0.
x y ´1 1 y = 1 x2 (c) This is not only false, but it makes no sense at all, unless şb a f (x) dx = şb a x f (x) dx = 0.
The left hand side is a number. The right hand side is a number times x.
ż b a x f (x) dx loooooomoooooon area vs x lo omo on variable ¨ ż b a f (x) dx looooomooooon area For example, if a = 0, b = 1 and f (x) = 1, then the left hand side is ş1 0 x dx = 1 2 and the 2 The symbol ð ñ is read “if and only if”. This is used in mathematics to express the logical equivalence of two statements. To be more precise, the statement P ð ñ Q tells us that P is true whenever Q is true and Q is true whenever P is true.
274 right hand side is x ş1 0 dx = x.
S-4: This is a tempting thought: ż 1 x dx = log |x| + C so perhaps similarly ż 1 x2 dx ?
= log |x2| + C = log(x2) + C We check by differentiating: d dxtlog(x2)u = d dxt2 log xu = 2 x ‰ 1 x2 So, it wasn’t so easy: false.
When we’re guessing antiderivatives, we often need to adjust our original guesses a little. Changing constants works well; changing functions usually does not.
S-5: This is tempting: d dxtsin(ex)u = ex cos(ex) so perhaps d dx "sin(ex) ex ?
= cos(ex) We check by differentiating: d dx "sin(ex) ex = ex (cos(ex) ¨ ex) ´ sin(ex)ex e2x (quotient rule) = cos(ex) ´ sin(ex) ex ‰ cos(ex) So, the statement is false.
When we’re guessing antiderivatives, we often need to adjust our original guesses a little. Dividing by constants works well; dividing by functions usually does not.
S-6: “The instantaneous rate of change of F(x) with respect to x” is another way of saying “F1(x)”. From the Fundamental Theorem of Calculus Part 1, we know this is sin(x2).
S-7: The slope of the tangent line to y = F(x) when x = 3 is exactly F1(3). By the Fundamental Theorem of Calculus Part 1, F1(x) = e1/x. Then F1(3) = e1/3 = 3 ?e.
275 S-8: For any constant C, F(x) + C is an antiderivative of f (x), because d dxtF(x) + Cu = d dxtF(x)u = f (x). So, for example, F(x) and F(x) + 1 are both antiderivatives of f (x).
S-9: (a) We differentiate with respect to a. Recall d dxtarccos xu = ´1 ?
1´x2. To differentiate 1 2a ?
1 ´ a2, we use the product and chain rules.
d da "π 4 ´ 1 2 arccos(a) + 1 2a a 1 ´ a2 = 0 ´ 1 2 ¨ ´1 ?
1 ´ a2 + 1 2a ¨ ´2a 2 ?
1 ´ a2 + 1 2 a 1 ´ a2 = 1 2 ?
1 ´ a2 ´ a2 2 ?
1 ´ a2 + 1 ´ a2 2 ?
1 ´ a2 = 1 ´ a2 + 1 ´ a2 2 ?
1 ´ a2 = 2(1 ´ a2) 2 ?
1 ´ a2 = a 1 ´ a2 (b) Let G(x) = π 4 ´ 1 2 arccos(x) + 1 2x ?
1 ´ x2. We showed in part (a) that G(x) is an antiderivative of ?
1 ´ x2. Since F(x) is also an antiderivative of ?
1 ´ x2, F(x) = G(x) + C for some constant C (this is Lemma 1.3.8 in the CLP-2 text).
Note G(0) = ż 0 0 a 1 ´ x2 dx = 0, so if F(0) = π, then F(x) = G(x) + π. That is, F(x) = 5π 4 ´ 1 2 arccos(x) + 1 2x a 1 ´ x2 .
S-10: (a) The antiderivative of cos x is sin x, and cos x is continuous everywhere, so ż π ´π cos x dx = sin(π) ´ sin(´π) = 0.
(b) Since sec2 x is discontinuous at x = ˘π 2 , the Fundamental Theorem of Calculus Part 2 does not apply to ż π ´π sec2 x dx.
(c) Since 1 x+1 is discontinuous at x = ´1, the Fundamental Theorem of Calculus Part 2 does not apply to ż 0 ´2 1 x + 1 dx.
S-11: Using the definition of F, F(x) is the area under the curve from a to x, and F(x + h) is the area under the curve from a to x + h. These are shown on the same diagram, below.
276 t y a x x + h y = f (t) Then the area represented by F(x + h) ´ F(x) is the area that is outside the red, but inside the blue. Equivalently, it is x+h ş x f (t) dt.
t y a x x + h y = f (t) S-12: We evaluate F(0) using the definition: F(0) = ş0 0 f (t) dt = 0. Although f (0) ą 0, the area from t = 0 to t = 0 is zero.
As x moves along, F(x) adds bits of signed area. If it’s adding positive area, it’s increasing, and if it’s adding negative area, it’s decreasing. So, F(x) is increasing when 0 ă x ă 1 and 3 ă x ă 4, and F(x) is decreasing when 1 ă x ă 3.
S-13: This question is nearly identical to Question 12, with G(x) = ż 0 x f (t) dt = ´ ż x 0 f (t) dt = ´F(x).
So, G(x) increases when F(x) decreases, and vice-versa. Therefore: G(0) = 0, G(x) is increasing when 1 ă x ă 3, and G(x) is decreasing when 0 ă x ă 1 and when 3 ă x ă 4.
S-14: Using the definition of the derivative, F1(x) = lim hÑ0 F(x + h) ´ F(x) h = lim hÑ0 şx+h a t dt ´ şx a t dt h = lim hÑ0 şx+h x t dt h 277 The numerator describes the area of a trapezoid with base h and heights x and x + h.
= lim hÑ0 1 2h(x + x + h) h = lim hÑ0 x + 1 2h = x t y x x + h x x + h y = t şx+h x t dt So, F1(x) = x.
S-15: If F(x) is constant, then F1(x) = 0. By the Fundamental Theorem of Calculus Part 1, F1(x) = f (x). So, the only possible continuous function fitting the question is f (x) = 0.
This makes intuitive sense: if moving x doesn’t add or subtract area under the curve, then there must not be any area under the curve–the curve should be the same as the x-axis.
As an aside, we mention that there are other, non-continuous functions f (t) such that şx 0 f (t) dt = 0 for all x. For example, f (t) = " 0 x ‰ 0 1 x = 0 . These kinds of removable discontinuities will not factor heavily in our discussion of integrals.
S-16: d dxtx log(ax) ´ xu = x a ax + log(ax) ´ 1 (product rule, chain rule) = log(ax) So, we know ż log(ax) dx = x log(ax) ´ x + C where a is a given constant, and C is any constant.
Remark: ş log(ax) dx can be calculated using the method of Integration by Parts, which you will learn in Section 1.7 of the CLP-2 text.
278 S-17: d dx !
ex x3 ´ 3x2 + 6x ´ 6 ) = ex 3x2 ´ 6x + 6 + ex x3 ´ 3x2 + 6x ´ 6 (product rule) = ex 3x2 ´ 6x + 6 + x3 ´ 3x2 + 6x ´ 6 = x3ex So, ż x3ex dx = ex x3 ´ 3x2 + 6x ´ 6 + C Remark: ş x3ex dx can be calculated using the method of Integration by Parts, which you will learn in Section 1.7 of the CLP-2 text.
S-18: d dx !
log ˇ ˇ ˇx + a x2 + a2 ˇ ˇ ˇ ) = 1 x + ?
x2 + a2 ¨ 1 + 1 2 ?
x2 + a2 ¨ 2x (chain rule) = 1 + x ?
x2+a2 x + ?
x2 + a2 = ?
x2+a2+x ?
x2+a2 x + ?
x2 + a2 = 1 ?
x2 + a2 So, ż 1 ?
x2 + a2 dx = log ˇ ˇ ˇx + a x2 + a2 ˇ ˇ ˇ + C Remark: ş 1 ?
x2+a2 dx can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9 of the CLP-2 text.
S-19: Using the chain rule: d dx "b x(a + x) ´ a log ?x + ?
a + x = x + (a + x) 2 a x(a + x) ´ a 1 ?x + ?a + x ¨ 1 2?x + 1 2?a + x = 2x + a 2 a x(a + x) ´ a 1 ?x + ?a + x ¨ ?a + x + ?x 2 a x(a + x) !!
= 2x + a 2 a x(a + x) ´ a 1 2 a x(a + x) !
= 2x 2 a x(a + x) = x a x(a + x) 279 So, ż x a x(a + x) dx = b x(a + x) ´ a log ?x + ?
a + x + C Remark: ş x ?
x(a+x) dx can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9 of the CLP-2 text.
S-20: By the Fundamental Theorem of Calculus, ż 2 0 x3 + sin x) dx = x4 4 ´ cos x 2 0 = 24 4 ´ cos 2 ´ (0 ´ cos 0) = 4 ´ cos 2 + 1 = 5 ´ cos 2.
S-21: By part (d) of our “Arithmetic of Integration” theorem, Theorem 1.2.1 in the CLP-2 text, ż 2 1 x2 + 2 x2 dx = ż 2 1 h 1 + 2 x2 i dx = ż 2 1 dx + 2 ż 2 1 1 x2 dx Then by the Fundamental Theorem of Calculus Part 2, ż 2 1 dx + 2 ż 2 1 1 x2 dx = h x i2 1 + 2 h ´ 1 x i2 1 = 2 ´ 1 + 2 h ´ 1 2 + 1 i = 2 S-22: The integrand is similar to 1 1 + x2, which is the derivative of arctangent. Indeed, we have ż 1 1 + 25x2dx = ż 1 1 + (5x)2dx.
So, a reasonable first guess for the antiderivative might be F(x) ?
= arctan(5x).
However, because of the chain rule, F1(x) = 5 1 + (5x)2.
In order to “fix” the numerator, we make a second guess: F(x) = 1 5 arctan(5x) F1(x) = 1 5 5 1 + (5x)2 = 1 1 + 25x2 So, ż 1 1 + 25x2dx = 1 5 arctan(5x) + C.
280 S-23: The integrand is similar to 1 ?
1 ´ x2. In order to formulate a guess for the antiderivative, let’s factor out ?
2 from the denominator: ż 1 ?
2 ´ x2dx = ż 1 c 2 1 ´ x2 2 dx = ż 1 ?
2 b 1 ´ x2 2 dx = ż 1 ?
2 ¨ 1 d 1 ´ x ?
2 2dx At this point, we might guess that our antiderivative is something like F(x) = arcsin x ?
2 . To explore this possibility, we can differentiate, and see what we get.
d dx " arcsin x ?
2 = 1 ?
2 ¨ 1 d 1 ´ x ?
2 2 This is exactly what we want! So, ż 1 ?
2 ´ x2dx = arcsin x ?
2 + C S-24: We know that ş sec2 x dx = tan x + C, and sec2 x = tan2 x + 1, so ż tan2 x dx = ż sec2 x ´ 1 dx = ż sec2 x dx ´ ż 1 dx = tan x ´ x + C S-25: Solution 1: This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: 2 sin x cos x = sin(2x).
ż 3 sin x cos x dx = ż 3 2 ¨ 2 sin x cos x dx = ż 3 2 sin(2x) dx 281 So, we might guess that the antiderivative is something like ´ cos(2x). We only need to figure out the constants.
d dxt´ cos(2x)u = 2 sin(2x) So, d dx " ´3 4 cos(2x) = 3 2 sin(2x) Therefore, ż 3 sin x cos x dx = ´3 4 cos(2x) + C Solution 2: You might notice that the integrand looks like it came from the chain rule, since cos x is the derivative of sin x. Using this observation, we can work out the antideriative: d dx !
sin2 x ) = 2 sin x cos x d dx "3 2 sin2 x = 3 sin x cos x So, ż 3 sin x cos x dx = 3 2 sin2 x + C These two answers look different. Using the identity cos(2x) = 1 ´ 2 sin2(x), we reconcile them: ´3 4 cos(2x) + C = ´3 4 1 ´ 2 sin2 x + C = 3 2 sin2 x + C ´ 3 4 The 3 4 here is not significant. Remember that C is used to designate a constant that can take any value between ´8 and +8. So C ´ 3 4 is also just a constant that can take any value between ´8 and +8. As the two answers we found differ by a constant, they are equivalent.
S-26: It’s not immediately obvious which function has cos2 x as its derivative, but we can make the situation a little clearer by using the identity cos2 x = 1 + cos(2x) 2 : ż cos2 x dx = ż 1 2 ¨ (1 + cos(2x)) dx = ż 1 2 dx + ż 1 2 cos(2x) dx = 1 2x + C + ż 1 2 cos(2x) dx 282 For the remaining integral, we might guess something like F(x) = sin(2x). Let’s figure out the appropriate constant: d dx tsin(2x)u = 2 cos(2x) d dx "1 4 sin(2x) = 1 2 cos(2x) So, ż 1 2 cos(2x) dx = 1 4 sin(2x) + C Therefore, ż cos2 x dx = 1 2x + 1 4 sin(2x) + C S-27: By the Fundamental Theorem of Calculus Part 1, F1(x) = d dx ż x 0 log(2 + sin t) dt = log(2 + sin x) G1(y) = d dy ´ ż y 0 log(2 + sin t) dt = ´ log(2 + sin y) So, F1π 2 = log 3 G1π 2 = ´ log(3) S-28: By the Fundamental Theorem of Calculus Part 1, f 1(x) = 100(x2 ´ 3x + 2)e´x2 = 100(x ´ 1)(x ´ 2)e´x2 As f (x) is increasing whenever f 1(x) ą 0 and 100e´x2 is always strictly bigger than 0, we have f (x) increasing if and only if (x ´ 1)(x ´ 2) ą 0, which is the case if and only if (x ´ 1) and (x ´ 2) are of the same sign. Both are positive when x ą 2 and both are negative when x ă 1. So f (x) is increasing when ´8 ă x ă 1 and when 2 ă x ă 8.
Remark: even without the Fundamental Theorem of Calculus, since f (x) is the area under a curve from 1 to x, f (x) is increasing when the curve is above the x-axis (because we’re adding positive area), and it’s decreasing when the curve is below the x-axis (because we’re adding negative area).
S-29: Write G(x) = ż x 0 1 t3 + 6 dt. By the Fundamental Theorem of Calculus Part 1, G1(x) = 1 x3 + 6. Since F(x) = G(cos x), the chain rule gives us F1(x) = G1(cos x) ¨ (´ sin x) = ´ sin x cos3 x + 6 283 S-30: Define g(x) = ż x 0 et2dt. By the Fundamental Theorem of Calculus Part 1, g1(x) = ex2. As f (x) = g(1 + x4) the chain rule gives us f 1(x) = 4x3g1(1 + x4) = 4x3e(1+x4)2 S-31: Define g(x) = şx 0(t6 + 8)dt. By the fundamental theorem of calculus, g1(x) = x6 + 8.
We are to compute the derivative of f (x) = g(sin x). The chain rule gives d dx #ż sin x 0 (t6 + 8)dt + = g1(sin x) ¨ cos x = sin6 x + 8 cos x S-32: Let G(x) = ż x 0 e´t sin πt 2 dt. By the Fundamental Theorem of Calculus Part 1, G1(x) = e´x sin πx 2 and, since F(x) = G(x3), F1(x) = 3x2G1(x3) = 3x2e´x3 sin πx3 2 .
Then F1(1) = 3e´1 sin π 2 = 3e´1.
S-33: Define G(x) = ż 0 x dt 1 + t3 = ´ ż x 0 1 1 + t3 dt, so that G1(x) = ´ 1 1 + x3 by the Fundamental Theorem of Calculus Part 1. Then by the chain rule, d du #ż 0 cos u dt 1 + t3 + = d duG(cos u) = G1(cos u) ¨ d du cos u = ´ 1 1 + cos3 u ¨ (´ sin u).
S-34: Applying d dx to both sides of x2 = 1 + şx 1 f (t) dt gives, by the Fundamental Theorem of Calculus Part 1, 2x = f (x).
S-35: Apply d dx to both sides of x sin(πx) = şx 0 f (t) dt. Then, by the Fundamental Theorem of Calculus Part 1, f (x) = d dx ż x 0 f (t) dt = d dx ␣ x sin(πx) ( ù ñ f (x) = d dx ␣ x sin(πx) ( = sin(πx) + πx cos(πx) ù ñ f (4) = sin(4π) + 4π cos(4π) = 4π S-36: (a) Write F(x) = G(x2) ´ H(´x) with G(y) = ż y 0 e´t dt, H(y) = ż y 0 e´t2 dt 284 By the Fundamental Theorem of Calculus Part 1, G1(y) = e´y , H1(y) = e´y2 Hence, by the chain rule, F1(x) = 2xG1(x2) ´ (´1)H1(´x) = 2xe´(x2) + e´(´x)2 = (2x + 1)e´x2 (b) Observe that F1(x) ă 0 for x ă ´1/2 and F1(x) ą 0 for x ą ´1/2. Hence F(x) is decreasing for x ă ´1/2 and increasing for x ą ´1/2, and F(x) must take its minimum value when x = ´1/2.
S-37: Define G(y) = ż y 0 esin t dt. Then: F(x) = ż x 0 esin t dt + ż 0 x4´x3 esin t dt = ż x 0 esin t dt ´ ż x4´x3 0 esin t dt = G(x) ´ G(x4 ´ x3) By the Fundamental Theorem of Calculus Part 1, G1(y) = esin y Hence, by the chain rule, F1(x) = G1(x) ´ G1(x4 ´ x3) d dx ␣ x4 ´ x3( = G1(x) ´ G1(x4 ´ x3) (4x3 ´ 3x2) = esin x ´ esin(x4´x3) 4x3 ´ 3x2 S-38: Define with G(y) = ż y 0 cos et dt. Then: F(x) = ż ´x2 x5 cos et dt = ż ´x2 0 cos et dt + ż 0 x5 cos et dt = ż ´x2 0 cos et dt ´ ż x5 0 cos et dt = G(´x2) ´ G(x5) By the Fundamental Theorem of Calculus, G1(y) = cos ey 285 Hence, by the chain rule, F1(x) = G1(´x2) d dx ␣ ´ x2( ´ G1(x5) d dx ␣ x5( = G1(´x2) (´2x) ´ G1(x5) (5x4) = ´2x cos e´x2 ´ 5x4 cos ex5 S-39: Define with G(y) = ż y 0 ?
sin t dt. Then: F(x) = ż ex x ?
sin t dt = ż ex 0 ?
sin t dt + ż 0 x ?
sin t dt = ż ex 0 ?
sin t dt ´ ż x 0 ?
sin t dt = G(ex) ´ G(x) By the Fundamental Theorem of Calculus Part 1, G1(y) = a sin y Hence, by the chain rule, F1(x) = G1(ex) d dx ␣ ex( ´ G1(x) = exG1(ex) ´ G1(x) = ex b sin(ex) ´ b sin(x) S-40: Splitting up the domain of integration, ż 5 1 f (x) dx = ż 3 1 f (x) dx + ż 5 3 f (x) dx = ż 3 1 3 dx + ż 5 3 x dx = 3x ˇ ˇ ˇ ˇ x=3 x=1 + x2 2 ˇ ˇ ˇ ˇ x=5 x=3 = 14 x y 1 3 5 3 y = f (x) 286 S-41: By the chain rule, d dx ␣( f 1(x))2 ( = 2f 1(x) f 2(x) so 1 2 f 1(x)2 is an antiderivative for f 1(x) f 2(x) and, by the Fundamental Theorem of Calculus Part 2, ż 2 1 f 1(x) f 2(x) dx = 1 2 ( f 1(x))2 x=2 x=1 = 1 2 f 1(2)2 ´ 1 2 f 1(1)2 = 5 2 Remark: evaluating antiderivatives of this type will occupy the next section, Section 1.4 of the CLP-2 text.
S-42: The car stops when v(t) = 30 ´ 10t = 0, which occurs at time t = 3. The distance covered up to that time is ż 3 0 v(t) dt = (30t ´ 5t2) ˇ ˇ ˇ 3 0 = (90 ´ 45) ´ 0 = 45 m.
S-43: Define g(x) = ż x 0 log 1 + et dt. By the Fundamental Theorem of Calculus Part 1, g1(x) = log 1 + ex . But f (x) = g(2x ´ x2), so by the chain rule, f 1(x) = g1(2x ´ x2) ¨ d dxt2x ´ x2u = (2 ´ 2x) ¨ log 1 + e2x´x2 Observe that e2x´x2 ą 0 for all x so that 1 + e2x´x2 ą 1 for all x and log 1 + e2x´x2 ą 0 for all x. Since 2 ´ 2x is positive for x ă 1 and negative for x ą 1, f 1(x) is also positive for x ă 1 and negative for x ą 1. That is, f (x) is increasing for x ă 1 and decreasing for x ą 1. So f (x) achieves its absolute maximum at x = 1.
S-44: Let f (x) = şx2´2x 0 dt 1+t4 and g(x) = şx 0 dt 1+t4. Then g1(x) = 1 1+x4 and, since f (x) = g(x2 ´ 2x), f 1(x) = (2x ´ 2)g1(x2 ´ 2x) = 2 x´1 1+(x2´2x)4. This is zero for x = 1, negative for x ă 1 and positive for x ą 1. Thus as x runs from ´8 to 8, f (x) decreases until x reaches 1 and then increases all x ą 1. So the minimum of f (x) is achieved for x = 1. At x = 1, x2 ´ 2x = ´1 and f (1) = ş´1 0 dt 1+t4.
S-45: Define G(x) = ż x 0 sin( ?
t) dt. By the Fundamental Theorem of Calculus Part 1, G1(x) = sin(?x). Since F(x) = G(x2), and since x ą 0, we have F1(x) = 2xG1(x2) = 2x sin |x| = 2x sin x.
287 Thus F increases as x runs from to 0 to π (since F1(x) ą 0 there) and decreases as x runs from π to 4 (since F1(x) ă 0 there). Thus F achieves its maximum value at x = π.
S-46: The given sum is of the form lim nÑ8 n ÿ j=1 π n sin jπ n = lim nÑ8 n ÿ j=1 f (x˚ j )∆x with ∆x = π n , x˚ j = jπ n and f (x) = sin(x). Since x˚ 0 = 0 and x˚ n = π, the right hand side is the definition (using the right Riemann sum) of ż π 0 f (x) dx = ż π 0 sin(x) dx = [´ cos(x)]π 0 = 2 where we evaluate the definite integral using the Fundamental Theorem of Calculus Part 2.
S-47: The given sum is of the form lim nÑ8 1 n n ÿ j=1 1 1 + j n = lim nÑ8 n ÿ j=1 f (xj)∆x with ∆x = 1 n, xj = j n and f (x) = 1 1+x. The right hand side is the definition (using the right Riemann sum) of ż 1 0 f (x) dx = ż 1 0 1 1 + x dx = log |1 + x| ˇ ˇ ˇ 1 0 = log 2 S-48: F(x), x ě 0 We learned quite a lot last semester about curve sketching. We can use those techniques here. We have to be quite careful about the sign of x, though. We can only directly apply the Fundamental Theorem of Calculus Part 1 (as it’s written in your text) when x ě 0. So first, let’s graph the right-hand portion. Notice f (x) has even symmetry–so, if we know one half of F(x), we should be able to figure out the other half with relative ease.
• F(0) = ż 0 0 f (t) dt = 0 (so, F(x) passes through the origin) • Using the Fundamental Theorem of Calculus Part 1, F1(x) ą 0 when 0 ă x ă 1 and when 3 ă x ă 5; F1(x) ă 0 when 1 ă x ă 3. So, F(x) is decreasing from 1 to 3, and increasing from 0 to 1 and also from 3 to 5. That gives us a skeleton to work with.
288 x y 1 3 5 We get the relative sizes of the maxes and mins by eyeballing the area under y = f (t). The first lobe (from x = 0 to x = 1 has a small positive area, so F(1) is a small positive number. The next lobe (from x = 1 to x = 3) has a larger absolute area than the first, so F(3) is negative. Indeed, the second lobe seems to have more than twice the area of the first, so |F(3)| should be larger than F(1). The third lobe is larger still, and even after subtracting the area of the second lobe it looks much larger than the first or second lobe, so |F(3)| ă F(5).
• We can use F2(x) to get the concavity of F(x). Note F2(x) = f 1(x). We observe f (x) is decreasing on (roughly) (0, 2.5) and (4, 5), so F(x) is concave down on those intervals. Further, f (x) is increasing on (roughly) (2.5, 4), so F(x) is concave up there, and has inflection points at about x = 2.5 and x = 4.
x y y = F(x) y = f (x) ´5 ´3 ´1 1 3 5 In the sketch above, closed dots are extrema, and open dots are inflection points.
F(x), x ă 0 Now we can consider the left half of the graph. If you stare at it long enough, you might convince yourself that F(x) is an odd function. We can also show this 289 with the following calculation: F(´x) = ż ´x 0 f (t) dt As in Example 1.2.10 of the CLP-2 text, since f (t) is even, = ż 0 x f (t) dt = ´ ż x 0 f (t) dt = ´F(x) Knowing that F(x) is odd allows us to finish our sketch.
x y y = F(x) y = f (x) ´5 ´3 ´1 1 3 5 S-49: (a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1, f 1(x) = 3x2 ż x3+1 0 et3dt + x3 d dx ż x3+1 0 et3dt = 3x2 ż x3+1 0 et3dt + x3 3x2 " d dy ż y 0 et3dt # y=x3+1 = 3x2 ż x3+1 0 et3dt + x3 3x2h ey3i y=x3+1 = 3x2 ż x3+1 0 et3dt + x3 3x2 e(x3+1)3 = 3x2 ż x3+1 0 et3dt + 3x5e(x3+1)3 (b) In general, the equation of the tangent line to the graph of y = f (x) at x = a is y = f (a) + f 1(a) (x ´ a) 290 Substituting in the given f (x) and a = ´1: f (a) = f (´1) = (´1)3 ż 0 0 et3 dt = 0 f 1(a) = f 1(´1) = 3(´1)2 ż 0 0 et3 dt + 3(´1)5e0 = 0 ´ 3 = ´3 (x ´ a) = x ´ (´1) = x + 1 So, the equation of the tangent line is y = ´3(x + 1) .
S-50: Recall that “+C” means that we can add any constant to the function. Since tan2 x = sec2 x ´ 1, Students A and B have equivalent answers: they only differ by a constant.
So, if one is right, both are right; if one is wrong, both are wrong. We check Student A’s work: d dxttan2 x + x + Cu = d dxttan2 xu + 1 + 0 = f (x) ´ 1 + 1 = f (x) So, Student A’s answer is indeed an anditerivative of f (x). Therefore, both students ended up with the correct answer.
Remark: it is a frequent occurrence that equivalent answers might look quite different.
As you are comparing your work to others’, this is a good thing to keep in mind!
S-51: (a) When x = 3, F(3) = ż 3 0 33 sin(t) dt = 27 ż 3 0 sin t dt Using the Fundamental Theorem of Calculus Part 2, = 27 [´ cos t]t=3 t=0 = 27 [´ cos 3 ´ (´ cos 0)] = 27(1 ´ cos 3) (b) Since the integration is with respect to t, the x3 term can be moved outside the integral. That is: for the purposes of the integral, x3 is a constant (although for the purposes of the derivative, it certainly is not).
F(x) = ż x 0 x3 sin(t) dt = x3 ż x 0 sin(t) dt 291 Using the product rule and the Fundamental Theorem of Calculus Part 1, F1(x) = x3 ¨ sin(x) + 3x2 ż x 0 sin(t) dt = x3 sin(x) + 3x2 [´ cos(t)]t=x t=0 = x3 sin(x) + 3x2[´ cos(x) ´ (´ cos(0))] = x3 sin(x) + 3x2[1 ´ cos(x)] Remark: Since x and t play different roles in our problem, it’s crucial that they have different names. This is one reason why we should avoid the common mistake of writing şx a f (x)dx when we mean şx a f (t)dt.
S-52: If F(x) is even, then f (x) is odd (by the result of Question 20 in Section 1.2). So, F(x) can only be even if f (x) is both even and odd. By the result in Question 19, Section 1.2, this means F(x) is only even if f (x) = 0 for all x. Note if f (x) = 0, then F(x) is a constant function. So, it is certainly even, and it might be odd as well if F(x) = f (x) = 0.
Therefore, if f (x) ‰ 0 for some x, then F(x) is not even. It could be odd, or it could be neither even nor odd. We can come up with examples of both types: if f (x) = 1, then F(x) = x is an odd antiderivative, and F(x) = x + 1 is an antiderivative that is neither even nor odd.
Interestingly, the antiderivative of an odd function is always even. The proof is a little beyond what we might ask you, but is given below for completeness. The proof goes like this: First, we’ll show that if g(x) is odd, then there is some antiderivative of g(x) that is even. Then, we’ll show that every antiderivative of g(x) is even.
So, suppose g(x) is odd and define G(x) = ż x 0 g(t)dt. By the Fundamental Theorem of Calculus Part 1, G1(x) = g(x), so G(x) is an antiderivative of g(x). Since g(x) is odd, for any x ě 0, the net signed area under the curve along [0, x] is the negative of the net signed area under the curve along [´x, 0]. So, ż x 0 g(t) dt = ´ ż 0 ´x g(t) dt (See Example 1.2.11 in the CLP-2 text) = ż ´x 0 g(t) dt By the definition of G(x), G(x) = G(´x) That is, G(x) is even. We’ve shown that there exists some antiderivative of g(x) that is even; it remains to show that all of them are even.
Recall that every antiderivative of g(x) differs from G(x) by some constant. So, any antiderivative of g(x) can be written as G(x) + C, and G(´x) + C = G(x) + C. So, every antiderivative of an odd function is even.
292 Solutions to Exercises 1.4 — Jump to TABLE OF CONTENTS S-1: (a) This is true: it is an application of Theorem 1.4.2 in the CLP-2 text with f (x) = sin x and u(x) = ex.
(b) This is false: the upper limit of integration is incorrect. Using Theorem 1.4.6 in the CLP-2 text, the correct form is ż 1 0 sin(ex) ¨ ex dx = ż e 1 sin(u) du = ´ cos(e) + cos(1) = cos(1) ´ cos(e).
Alternately, we can use the Fundamental Theorem of Calculus Part 2, and our answer from (a): ż 1 0 sin(ex) ¨ ex dx = [´ cos(ex) + C]1 0 = cos(1) ´ cos(e) .
S-2: The reasoning is not sound: when we do a substitution, we need to take care of the differential (dx). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here’s the way to do it: Problem: Evaluate ż (2x + 1)2dx.
Work: We use the substitution u = 2x + 1. Then du = 2dx, so dx = 1 2du: ż (2x + 1)2dx = ż u2 ¨ 1 2 du = 1 6u3 + C = 1 6 (2x + 1)3 + C S-3: The problem is with the limits of integration, as in Question 1. Here’s how it ought to go: Problem: Evaluate ż π 1 cos(log t) t dt.
Work: We use the substitution u = log t, so du = 1 t dt. When t = 1, we have u = log 1 = 0 and when t = π, we have u = log(π). Then: ż π 1 cos(log t) t dt = ż log(π) log 1 cos(u)du = ż log(π) 0 cos(u)du = sin(log(π)) ´ sin(0) = sin(log(π)).
293 S-4: Perhaps shorter ways exist, but the reasoning here is valid.
Problem: Evaluate ż π/4 0 x tan(x2) dx.
Work: We begin with the substitution u = x2, du = 2xdx: If u = x2, then du dx = 2x, so indeed du = 2xdx.
ż π/4 0 x tan(x2) dx = ż π/4 0 1 2 tan(x2) ¨ 2xdx algebra = ż π2/16 0 1 2 tan u du Every piece is changed from x to u: integrand, differential, limits.
= 1 2 ż π2/16 0 sin u cos udu tan u = sin u cos u Now we use the substitution v = cos u, dv = ´ sin u du: = 1 2 ż cos(π2/16) cos 0 ´1 vdv Every piece is changed from u to v: integrand, differential, limits.
= ´1 2 ż cos(π2/16) 1 1 vdv cos(0) = 1 = ´1 2 log |v| cos(π2/16) 1 FTC Part 2 = ´1 2 log cos(π2/16) ´ log(1) = ´1 2 log cos(π2/16) log(1) = 0 S-5: We substitute: u = sin x, du = cos x dx, cos x = a 1 ´ sin2 x = a 1 ´ u2, dx = du cos x = du ?
1 ´ u2 u(0) = sin 0 = 0 u π 2 = sin π 2 = 1 294 So, ż x=π/2 x=0 f (sin x) dx = ż u=1 u=0 f (u) du ?
1 ´ u2 Because the denominator ?
1 ´ u2 vanishes when u = 1, this is what is known as an improper integral. Improper integrals will be discussed in § 1.12 of the CLP-2 text.
S-6: Using the chain rule, we see that d dxtf (g(x))u = f 1(g(x))g1(x) So, f (g(x)) is an antiderivative of f 1(g(x))g1(x). All antiderivatives of f 1(g(x))g1(x) differ by only a constant, so: ż f 1(g(x))g1(x) dx ´ f (g(x)) = f (g(x)) + C ´ f (g(x)) = C That is, our expression simplifies to some constant C.
Remark: since ż f 1(g(x))g1(x) dt ´ f (g(x)) = C we conclude ż f 1(g(x))g1(x) dt = f (g(x)) + C which is precisely how we perform substitution on integrals.
S-7: We write u(x) = ex2 and find du = u1(x) dx = 2xex2dx. Note that u(1) = e12 = e when x = 1, and u(0) = e02 = 1 when x = 0. Therefore: ż 1 0 xex2 cos(ex2) dx = 1 2 ż x=1 x=0 cos(u(x))u1(x) dx = 1 2 ż u=e u=1 cos(u) du = 1 2 sin(u) e 1 = 1 2 sin(e) ´ sin(1) .
S-8: Substituting y = x3, dy = 3x2 dx : ż 2 1 x2 f (x3) dx = 1 3 ż 8 1 f (y) dy = 1 3 295 S-9: Setting u = x3 + 1, we have du = 3x2 dx and so ż x2 dx (x3 + 1)101 = ż du/3 u101 = 1 3 ż u´101 du = 1 3 ¨ u´100 ´100 = ´ 1 3 ˆ 100u100 + C = ´ 1 300(x3 + 1)100 + C S-10: Setting u = log x, we have du = 1 x dx and so ż e4 e dx x ¨ log x = ż x=e4 x=e 1 log x ¨ 1 x dx = ż u=4 u=1 1 u du, since u = log(e) = 1 when x = e and u = log(e4) = 4 when x = e4. Then, by the Fundamental Theorem of Calculus Part 2, ż 4 1 1 u du = h log |u| i4 1 = log 4 ´ log 1 = log 4.
S-11: Setting u = 1 + sin x, we have du = cos x dx and so ż π/2 0 cos x 1 + sin xdx = ż x=π/2 x=0 1 1 + sin x cos x dx = ż u=2 u=1 du u since u = 1 + sin 0 = 1 when x = 0 and u = 1 + sin(π/2) = 2 when x = π/2. Then, by the Fundamental Theorem of Calculus Part 2, ż u=2 u=1 du u = h log |u| i2 1 = log 2 S-12: Setting u = sin x, we have du = cos x dx and so ż π/2 0 cos x ¨ (1 + sin2x)dx = ż x=π/2 x=0 (1 + sin2x) ¨ cos x dx = ż u=1 u=0 (1 + u2) du, since u = sin 0 = 0 when x = 0 and u = sin(π/2) = 1 when x = π/2. Then, by the Fundamental Theorem of Calculus Part 2, ż 1 0 (1 + u2) du = u + u3 3 1 0 = 1 + 1 3 ´ 0 = 4 3.
296 S-13: Substituting t = x2 ´ x, dt = (2x ´ 1) dx and noting that t = 0 when x = 1 and t = 6 when x = 3, ż 3 1 (2x ´ 1)ex2´xdx = ż 6 0 et dt = et6 0 = e6 ´ 1 S-14: We use the substitution u = 4 ´ x2, for which du = ´2x dx : ż x2 ´ 4 ?
4 ´ x2 x dx = ż 1 2 ¨ 4 ´ x2 ?
4 ´ x2(´2x) dx = 1 2 ż u ?u du = 1 2 ż ?u du = 1 2 u3/2 3/2 + C = 1 3(4 ´ x2)3/2 + C S-15: Solution 1: If we let u = a log x, then du = 1 2x a log x dx, and: ż e ?
log x 2x a log x dx = ż eu du = eu + C = e ?
log x + C Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it’s not unnatural to think that u = log x, du = 1 x dx would be a good choice. In that case: ż e ?
log x 2x a log x dx = ż e ?u 2?udu Now, we should be able to see that w = ?u, dw = 1 2?u du is a good choice: ż e ?u 2?u du = ż ew dw = e ?u + C = e ?
log x + C 297 S-16: The straightforward method: We use the substitution u = x2, for which du = 2x dx, and note that u = 4 for both x = 2 and x = ´2: ż 2 ´2 xex2 dx = ż 2 ´2 1 2ex2 2xdx = ż 4 4 1 2eu du = 0 The slightly sneaky method: We note that d dx !
ex2) = 2x ex2, so that 1 2ex2 is a antiderivative for the integrand xex2. So ż 2 ´2 xex2 dx = 1 2ex22 ´2 = 1 2e4 ´ 1 2e4 = 0 The really sneaky method: The integrand f (x) = xex2 is an odd function (meaning that f (´x) = ´f (x)). So by Theorem 1.2.12 in the CLP-2 text every integral of the form şa ´a xex2 dx is zero.
S-17: The given sum is of the form lim nÑ8 n ÿ j=1 j n2 sin 1 + j2 n2 = lim nÑ8 n ÿ j=1 f (x˚ j )∆x with ∆x = 1 n, x˚ j = j n and f (x) = x sin(1 + x2). Since x˚ 0 = 0 and x˚ n = 1, the right hand side is the definition (using the right Riemann sum) of ż 1 0 f (x) dx = ż 1 0 x sin(1 + x2) dx = 1 2 ż 2 1 sin(y) dy with y = 1 + x2, dy = 2x dx = 1 2 h ´ cos(y) iy=2 y=1 = 1 2[cos 1 ´ cos 2] Using a calculator, we see this is close to 0.478.
S-18: Often, the denominator of a function is a good guess for the substitution. So, let’s try setting w = u2 + 1. Then dw = 2u du: ż 1 0 u3 u2 + 1 du = 1 2 ż 1 0 u2 u2 + 1 2u du 298 The numerator now is u2, and looking at our substitution, we see u2 = w ´ 1: = 1 2 ż 2 1 w ´ 1 w dw = 1 2 ż 2 1 1 ´ 1 w dw = 1 2 [w ´ log |w|]w=2 w=1 = 1 2 (2 ´ log 2 ´ 1) = 1 2 ´ 1 2 log 2 S-19: The only thing we really have to work with is a tangent, so it’s worth considering what would happen if we substituted u = tan θ. Then du = sec2 θ dθ. This doesn’t show up in the integrand as it’s written, but we can try and bring it out by using the identity tan2 θ = sec2 θ ´ 1: ż tan3 θ dθ = ż tan θ ¨ tan2 θdθ = ż tan θ ¨ sec2 θ ´ 1 dθ = ż tan θ ¨ sec2 θdθ ´ ż tan θ dθ In Example 1.4.17 of the CLP-2 text, we learned ş tan θ dθ = log | sec θ| + C = ż u du ´ log | sec θ| + C = 1 2u2 ´ log | sec θ| + C = 1 2 tan2 θ ´ log | sec θ| + C S-20: At first glance, it’s not clear what substitution to use. If we try the denominator, u = ex + e´x, then du = (ex ´ e´x) dx, but it’s not clear how to make this work with our integral. So, we can try something else.
If we want to tidy things up, we might think to take u = ex as a substitution. Then du = ex dx, so we need an ex in the numerator. That can be arranged.
ż 1 ex + e´x ¨ ex ex dx = ż ex (ex)2 + 1 dx = ż 1 u2 + 1 du = arctan(u) + C = arctan(ex) + C 299 S-21: We often like to take the “inside” function as our substitution, in this case u = 1 ´ x2, so du = ´2x dx. This takes care of part of the integral: ż 1 0 (1 ´ 2x) a 1 ´ x2 dx = ż 1 0 a 1 ´ x2 dx + ż 1 0 (´2x) a 1 ´ x2 dx The left integral is tough to solve with substitution, but luckily we don’t have to–it’s the area of a quarter of a circle of radius 1.
= π 4 + ż 0 1 ?u du = π 4 + 2 3u3/2 u=0 u=1 = π 4 + 0 ´ 2 3 = π 4 ´ 2 3 S-22: Solution 1: We often find it useful to take “inside” functions as our substitutions, so let’s try u = cos x, du = ´ sin x dx. In order to dig up a sine, we use the identity tan x = sin x cos x : ż tan x ¨ log (cos x) dx = ´ ż ´ sin x cos x ¨ log (cos x) dx = ´ ż 1 u log(u) du Now, it is convenient to let w = log u, dw = 1 u du : ´ ż 1 ulog(u) du = ´ ż w dw = ´1 2w2 + C = ´1 2 (log u)2 + C = ´1 2 (log(cos x))2 + C Solution 2: We might guess that it’s useful to have u = log(cos x), 300 du = ´ sin x cos x dx = ´ tan x dx: ż tan x ¨ log (cos x)dx = ´ ż ´ tan x ¨ log (cos x)dx = ´ ż u du = ´1 2u2 + C = ´1 2 (log(cos x))2 + C S-23: The given sum is of the form lim nÑ8 n ÿ j=1 j n2 cos j2 n2 = lim nÑ8 n ÿ j=1 f (x˚ j )∆x with ∆x = 1 n, x˚ j = j n and f (x) = x cos(x2). Since x˚ 0 = 0 and x˚ n = 1, the right hand side is the definition (using the right Riemann sum) of ż 1 0 f (x) dx = ż 1 0 x cos(x2) dx = 1 2 ż 1 0 cos(y) dy with y = x2, dy = 2x dx = 1 2 h sin(y) i1 0 = 1 2 sin 1 S-24: The given sum is of the form lim nÑ8 n ÿ j=1 j n2 c 1 + j2 n2 = lim nÑ8 n ÿ j=1 f (x˚ j )∆x with ∆x = 1 n, x˚ j = j n and f (x) = x ?
1 + x2. Since x˚ 0 = 0 and x˚ n = 1, the right hand side is the definition (using the right Riemann sum) of ż 1 0 f (x) dx = ż 1 0 x a 1 + x2 dx = 1 2 ż 2 1 ?y dy with y = 1 + x2, dy = 2x dx = 1 2 2 3y3/2 y=2 y=1 = 1 3[2 ?
2 ´ 1] 301 Using a calculator, we see this is approximately 0.609.
S-25: Using the definition of a definite integral with right Riemann sums: ż b a 2f (2x)dx = lim nÑ8 n ÿ i=1 ∆x ¨ 2f (2(a + i∆x)) ∆x = b ´ a n = lim nÑ8 n ÿ i=1 b ´ a n ¨ 2f 2 a + i b ´ a n = lim nÑ8 n ÿ i=1 2b ´ 2a n ¨ f 2a + i 2b ´ 2a n ż 2b 2a f (x)dx = lim nÑ8 n ÿ i=1 ∆x ¨ f (2a + i∆x) ∆x = 2b ´ 2a n = lim nÑ8 n ÿ i=1 2b ´ 2a n ¨ f 2a + i 2b ´ 2a n Since the Riemann sums are exactly the same, ż b a 2f (2x)dx = ż 2b 2a f (x)dx Looking at the Riemann sum in this way is instructive, because it is very clear why the two integrals should be equal (without using substitution). The rectangles in the first Riemann sum are half as wide, but twice as tall, as the rectangles in the second Riemann sum. So, the two Riemann sums have rectangles of the same area.
b´a n 2f (2x˚ i ) 2b´a n f (x˚ i ) In the integral on the left, the variable is red x and in the integral on the right, the variable is blue x. Red x and blue x are not the same. In fact 2x˚ i = x˚ i . (Not every substitution corresponds to such a simple picture.) Solutions to Exercises 1.5 — Jump to TABLE OF CONTENTS 302 S-1: x y π π 2 π 4 3π 4 y = cos x y = sin x The intervals of our rectangles are [0, π 4 ], [ π 4 , π 2 ], [ π 2 , 3π 4 ], and [ 3π 4 , π]. Since we’re taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.
x = 0: The distance from cos 0 to sin 0 is 1, so our first rectangle has height 1.
x = π 4 : The distance from cos π 4 to sin π 4 is 0, so our second rectangle has height 0.
x = π 2 : The distance from cos π 2 to sin π 2 is 1, so our third rectangle has height 1.
x = 3π 4 : The distance from cos 3π 4 to sin 3π 4 is sin(3π/4) ´ cos(3π/4) = 1 ?
2 ´ ´ 1 ?
2 = ?
2, so our fourth rectangle has height ?
2.
So, our approximation for the area between the two curves is π 4 1 + 0 + 1 + ?
2 = π 4 2 + ?
2 S-2: (a) We are finding the area in the interval from x = 0 to x = π 2 . Since we’re taking n = 5 rectangles, our rectangles cover the following intervals: h 0, π 10 i , h π 10, π 5 i , π 5 , 3π 10 , 3π 10 , 2π 5 , 2π 5 , π 2 .
303 x y π 10 π 5 3π 10 2π 5 π 2 y = arcsin 2x π y = b xπ 2 (b) We are finding the area in the interval from y = 0 to y = π 2 . (In general, when we switch from horizontal rectangles to vertical, the limits of integration will change–it’s only coincidence that they are the same in this example.) Since we’re taking n = 5 rectangles, these rectangles cover the following intervals of the y-axis: h 0, π 10 i , h π 10, π 5 i , π 5 , 3π 10 , 3π 10 , 2π 5 , 2π 5 , π 2 .
The question doesn’t specify which endpoints we’re using. Let’s use upper endpoints, to match part (a).
x y π 10 π 5 3π 10 2π 5 π 2 x = π 2 sin y x = 2 πy2 304 S-3: The curves intersect when y = x and y = x3 ´ x. To find these points, we set: x = x3 ´ x 0 = x3 ´ 2x 0 = x(x2 ´ 2) 0 = x or 0 = x2 ´ 2 For x ě 0, the curves intersect at (0, 0) and ( ?
2, ?
2).
A handy observation is that, since both curves are continuous and they do not meet each other between x = 0 and x = ?
2, we don’t have to worry about dividing our area into two regions: one of the functions is always on the top, and the other is always on the bottom.
Using vertical strips: x y y = x3 −x y = x ( √ 2, √ 2) The top and bottom boundaries of the specified region are y = T(x) = x and y = B(x) = x3 ´ x, respectively. So, Area = ż ?
2 0 T(x) ´ B(x) dx = ż ?
2 0 x ´ (x3 ´ x) dx = ż ?
2 0 2x ´ x3 dx S-4: We need to find where the curves intersect.
x2 4 = y2 = 6 ´ 5x 4 1 4x2 + 5 4x ´ 6 = 0 x2 + 5x ´ 24 = 0 (x + 8)(x ´ 3) = 0 x = ´8, x = 3 The curves intersect at (´8, 4) and (3, ´3 2). Using horizontal strips: 305 x y y = −x/2 or x = −2y y2 = 6 −5 4x or x = 4 5(6 −y2) (−8, 4) (3, −3/2) we have Area = ż 4 ´3/2 h4 5(6 ´ y2) + 2y i dy S-5: If the curves intersect at (x, y), then x22 = (4a)2 y2 = (4a)24ax x4 = (4a)3x x4 ´ (4a)3x = 0 x(x3 ´ (4a)3) = 0 x = 0 or x3 = (4a)3 The curves intersect at (0, 0) and (4a, 4a). (It is also possible to find these points by inspection.) Using vertical strips: x y x2 = 4ay y2 = 4ax (4a, 4a) We want the y-values of the functions. We write the top function as y = ?
4ax (we care about the positive square root, not the negative one) and we write the bottom function as y = x2 4a. Then we have Area = ż 4a 0 ?
4ax ´ x2 4a dx S-6: The curves intersect when x = 4y2 and 0 = 4y2 + 12y + 5 = (2y + 5)(2y + 1). So, the curves intersect at (1, ´1 2) and (25, ´5 2). Using vertical strips: 306 x y x + 12y + 5 = 0 or y = −1 12(x + 5) x = 4y2 or y = ±√x/2 (25, −5 2) (1, −1 2) we have Area = ż 25 1 ´ 1 12(x + 5) + 1 2 ?x dx S-7: x y 1 y = 1 (2x´4)2 The area between the curve y = 1 (2x´4)2 and the x-axis, with x running from a = 0 to b = 1, is exactly the definite integral of 1 (2x´4)2 with limits 0 and 1.
Area = ż 1 0 dx (2x ´ 4)2 u = 2x ´ 4, du = 2 dx = 1 2 ż ´2 ´4 1 u2du = 1 2 ´1 u u=´2 u=´4 = 1 2 h1 2 ´ 1 4 i = 1 8 S-8: If the curves y = f (x) = x and y = g(x) = 3x ´ x2 intersect at (x, y), then 3x ´ x2 = y = x x2 ´ 2x = 0 x(x ´ 2) = 0 x = 0 or x = 2 Furthermore, g(x) ´ f (x) = 2x ´ x2 = x(2 ´ x) is positive for all 0 ď x ď 2. That is, the curve y = 3x ´ x2 lies above the line y = x for all 0 ď x ď 2.
307 x y y = x y = 3x ´ x2 2 We therefore evaluate the integral: ż 2 0 (3x ´ x2) ´ x dx = ż 2 0 [2x ´ x2] dx = x2 ´ x3 3 2 0 = 4 ´ 8 3 ´ 0 = 4 3 S-9: The following sketch contains the graphs of y = 2x and y = ?x + 1.
x y y = ?x + 1 y = 2x 1 From the sketch, it looks like the two curves cross when x = 0 and when x = 1 and nowhere3 else. Indeed, when x = 0 we have 2x = ?x + 1 = 1 and when x = 1 we have 2x = ?x + 1 = 2.
To antidifferentiate 2x, we write 2x = (elog 2) x = ex log 2.
Area = ż 1 0 (?x + 1) ´ ex log 2 dx = 2 3x3/2 + x ´ 1 log 22x 1 0 = 2 3 + 1 ´ 1 log 2[2 ´ 1] = 5 3 ´ 1 log 2 S-10: Here is a sketch of the specified region.
3 To verify analytically that the curves have no other crossings, write f (x) = ?x + 1 ´ 2x and compute f 1(x) = 1 2?x ´ (log 2)2x. Notice that f 1(x) decreases as x increases and so can take the value 0 for at most a single value of x. Then, by the mean value theorem (or Rolle’s theorem, which is Theorem 2.13.1 in the CLP-1 text), f (x) can take the value 0 for at most two distinct values of x.
308 y = √ 2 cos(πx/4) y = |x| y = |x| (1, 1) (−1, 1) x y Both functions are even, so the region is symmetric about the y–axis. So, we will compute the area of the part with x ě 0 and multiply by 2. The curves y = ?
2 cos(πx/4) and y = x intersect when x = ?
2 cos(πx/4) or cos(πx/4) = x ?
2, which is the case4 when x = 1. So, using vertical strips as in the figure above, the area (including the multiplication by 2) is 2 ż 1 0 ?
2 cos(πx/4) ´ x dx = 2 ?
2 4 π sin(πx/4) ´ x2 2 1 0 = 2 4 π ´ 1 2 = 8 π ´ 1 S-11: For our computation, we will need an antiderivative of x2?
x3 + 1, which can be found using the substitution u = x3 + 1, du = 3x2 dx: ż x2a x3 + 1 dx = ż ?u ¨ 1 3 du = 1 3 ż u1/2 du = 1 3 ¨ u3/2 3/2 + C = 2 9(x3 + 1)3/2 + C.
The two functions f (x) and g(x) are clearly equal at x = 0. If x ‰ 0, then the functions are equal when 3x2 = x2a x3 + 1 3 = a x3 + 1 9 = x3 + 1 8 = x3 2 = x.
The function g(x) = 3x2 is the larger of the two on the interval [0, 2], as can be seen by plugging in x = 1, say, or by observing that when x is very small f (x) = x2?
x3 + 1 « x2 and g(x) = 3x2.
4 The solution x = 1 was found by guessing. To guess a solution to cos(πx/4) = x ?
2 just ask yourself what simple angle has a cosine that involves ?
2. This guessing strategy is essentially useless in the real world, but works great on problem sets and exams.
309 x y y = 3x2 y = x2√ x3 + 1 (2, 12) The area in question is therefore: ż 2 0 3x2 ´ x2a x3 + 1 dx = x3 ´ 2 9(x3 + 1)3/2 ˇ ˇ ˇ ˇ 2 0 = 23 ´ 2 9(23 + 1)3/2 ´ 03 ´ 2 9(03 + 1)3/2 = 8 ´ 6 ´ 0 ´ 2 9 = 20 9 .
S-12: First, let’s figure out what our curve x = y2 + y = y(y + 1) looks like.
• The curve intercepts the y-axis when y = 0 and y = ´1.
• The x-values of the curve are negative when ´1 ă y ă 0, and positive elsewhere.
This leads to the figure below. We’re evaluating the area from y = ´1 to y = 0. Since y2 + y is negative there, the length of our (horizontal) slices are 0 ´ (y2 + y).
Area = ż 0 ´1 0 ´ (y2 + y) dy = ´ y3 3 + y2 2 0 ´1 = ´1 3 + 1 2 = 1 6 (0, 0) (0, −1) x y x = y + y2 S-13: Let’s begin by sketching our region. Note that y = ?
1 ´ x2 and y = ?
9 ´ x2 are the top halves of circles centred at the origin with radii 1 and 3, respectively.
310 x y y = ?
9 ´ x2 y = |x| Our region is the difference of two quarter-circles, so we find its area using geometry: Area = 1 4 π ¨ 32 ´ 1 4 π ¨ 12 = 2π S-14: We will compute the area by using thin vertical strips, as in the sketch below: y x π 2 π 3π 2 2π 2 4 6 8 y = 4 + π sin(x) y = 4 + 2π −2x By looking at the sketch above, we guess the line y = 4 + 2π ´ 2x intersects the curve y = 4 + π sin x when x = π 2 , x = π, and x = 3π 2 . Let’s make sure these are correct by plugging them into the two equations, and making sure the y-values match: x 4 + 2π ´ 2x 4 + π sin(x) match?
π 2 4 + π 4 + π ✓ π 4 4 ✓ 3π 2 4 ´ π 4 ´ π ✓ Also from the sketch, we see that: • When π 2 ď x ď π, the top of the strip is at y = 4 + π sin x and the bottom of the strip is at y = 4 + 2π ´ 2x. So the strip has height (4 + π sin x) ´ (4 + 2π ´ 2x) and width dx, and hence area (4 + π sin x) ´ (4 + 2π ´ 2x) dx.
• When π ď x ď 3π 2 , the top of the strip is at y = 4 + 2π ´ 2x and the bottom of the strip is at y = 4 + π sin x. So the strip has height (4 + 2π ´ 2x) ´ (4 + π sin x) and width dx, and hence area (4 + 2π ´ 2x) ´ (4 + π sin x) dx.
311 Now we can calculate: Area = ż π π/2 (4 + π sin x) ´ (4 + 2π ´ 2x) dx + ż 3π/2 π (4 + 2π ´ 2x) ´ (4 + π sin x) dx = ż π π/2 π sin x ´ 2π + 2x dx + ż 3π/2 π 2π ´ 2x ´ π sin x dx = h ´ π cos x ´ 2πx + x2iπ π/2 + h 2πx ´ x2 + π cos x i3π/2 π = π ´ π2 + 3 4π2 + π2 ´ 5 4π2 + π = 2 h π ´ 1 4π2i S-15: First, here is a sketch of the region. We are not asked for it, but it is crucial for understanding the question.
2 3 x y y = x + 2 y = x2 The two curves y = x + 2 and y = x2 cross at (2, 4). The area of the part between them with 0 ď x ď 2 is: ż 2 0 x + 2 ´ x2 dx = h1 2x2 + 2x ´ 1 3x3i2 0 = 2 + 4 ´ 8 3 = 10 3 The area of the part between the two curves with 2 ď x ď 3 is: ż 3 2 x2 ´ (x + 2) dx = h1 3x3 ´ 1 2x2 ´ 2x i3 2 = 9 ´ 9 2 ´ 6 ´ 8 3 + 2 + 4 = 11 6 The total area is 10 3 + 11 6 = 31 6 .
S-16: We need to figure out which curve is on top, when. To do this, set h(x) = 3x ´ x ?
25 ´ x2. If h(x) ą 0, then y = 3x is the top curve; if h(x) ă 0, then y = x ?
25 ´ x2 is the top curve.
h(x) = 3x ´ x a 25 ´ x2 = x h 3 ´ a 25 ´ x2 i 312 We only care about values of x in [0, 4], so x is nonnegative. Then h(x) is positive when: 3 ą a 25 ´ x2 9 ą 25 ´ x2 x2 ą 16 x ą 4 That is, h(x) is never positive over the interval [0, 4]. So, y = x ?
25 ´ x2 lies above y = 3x for all 0 ď x ď 4.
The area we need to calculate is therefore: A = ż 4 0 h x a 25 ´ x2 ´ 3x i dx = ż 4 0 x a 25 ´ x2 dx ´ ż 4 0 3x dx = A1 ´ A2.
To evaluate A1, we use the substitution u(x) = 25 ´ x2, for which du = u1(x) dx = ´2x dx; and u(4) = 25 ´ 42 = 9 when x = 4, while u(0) = 25 ´ 02 = 25 when x = 0. Therefore A1 = ż x=4 x=0 x a 25 ´ x2 dx = ´1 2 ż u=9 u=25 ?u du = ´1 3u3/2 9 25 = 125 ´ 27 3 = 98 3 For A2 we use the antiderivative directly: A2 = ż 4 0 3x dx = 3x2 2 4 0 = 24 Therefore the total area is: A = 98 3 ´ 24 = 26 3 S-17: Let’s begin by sketching our region. Note that y = ?
9 ´ x2 is the top half of a circle centred at the origin with radius 3, while y = a 1 ´ (x ´ 1)2 is the top half of a circle of radius 1 centred at (1, 0).
x y 313 Note y = x intersects y = a 1 ´ (x ´ 1)2 at (1, 1), the highest part of the smaller half-circle.
We can easily take the area of triangles and sectors of circles. With that in mind, we cut up our region the following way: x y A1 A2 A3 • The desired area is A3 ´ (A1 + A2).
• A1 is the area of right a triangle with base 1 and height 1, so A1 = 1 2.
• A2 is the area of a quarter circle of radius 1, so A2 = π 4 .
• A3 is the area of an eighth of a circle of radius 3, so A3 = 9π 8 So, the area of our region is 9π 8 ´ 1 2 ´ π 4 = 7π 8 ´ 1 2.
S-18: The first function is a cubic, with intercepts at x = 0, ˘2. The second is a straight line with a positive slope.
We need to figure out what these functions look like in relation to one another, so let’s find their points of intersection.
x(x2 ´ 4) = x ´ 2 x(x + 2)(x ´ 2) = x ´ 2 x ´ 2 = 0 or x(x + 2) = 1 x2 + 2x ´ 1 = 0 x = ´2 ˘ a 4 ´ 4(1)(´1) 2 x = ´1 ˘ ?
2 So, our three points of intersection are when x = 2 and when x = ´1 ˘ ?
2. We note ´1 ´ ?
2 ă ´1 + ?
2 ă ´1 + ?
4 ă 2 .
So, we need to see which function is on top over the two intervals ´1 ´ ?
2, ´1 + ?
2 and ´1 + ?
2, 2 . It suffices to check points in these intervals.
314 x x(x2 ´ 4) x ´ 2 top function: 0 0 ´2 x(x2 ´ 4) 1 -3 ´1 x ´ 2 Since 0 is in the interval ´1 ´ ?
2, ´1 + ?
2 , x(x2 ´ 4) is the top function in that interval.
Since 1 is in the interval ´1 + ?
2, 2 , x ´ 2 is the top function in that interval. Now we can set up the integral to evaluate the area: Area = ż ´1+ ?
2 ´1´ ?
2 h x(x2 ´ 4) ´ (x ´ 2) i dx + ż 2 ´1+ ?
2 h (x ´ 2) ´ x(x2 ´ 4) i dx = ż ´1+ ?
2 ´1´ ?
2 h x3 ´ 5x + 2 i dx + ż 2 ´1+ ?
2 h ´x3 + 5x ´ 2 i dx = 1 4x4 ´ 5 2x2 + 2x ´1+ ?
2 ´1´ ?
2 + ´1 4x4 + 5 2x2 ´ 2x 2 ´1+ ?
2 After some taxing but rudimentary algebra: = 8 ?
2 + 4 ?
2 ´ 13 4 = 12 ?
2 ´ 13 4 Solutions to Exercises 1.6 — Jump to TABLE OF CONTENTS S-1: If we take a horizontal slice of a cone, we get a circle. If we take a vertical cross-section, the base is flat (it’s a chord on the circular base of the cone), so we know right away it isn’t a circle. Indeed, if we slice down through the very centre, we get a triangle. (Other vertical slices have a curvy top, corresponding to a class of curves known as hyperbolas.) S-2: The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, A, and height dy. Then in both cases, the volume of the column is ż h 0 A dy = hA cubic units S-3: Notice f (x) is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the x-values, so below we give the x-values as functions of y.
315 x y y = f (x) 2 1 4 6 1 3 x = y x = 2 ´ y x = 2 + 2 3y x = 6 ´ 2 3y If we imagine rotating the region from the picture about the y-axis, there will be two kinds of washers formed: when y ă 1, we have a “double washer,” two concentric rings.
When y ą 1, we have a single ring.
Washers when 1 ă y ď 6: If y ą 1, then our washer has inner radius 2 + 2 3y, outer radius 6 ´ 2 3y, and height dy.
y thickness: dy R = 6 ´ 2 3y r = 2 + 2 3y Washers when 0 ď y ă 1: When 0 ď y ă 1, we have a “double washer,” two concentric rings corresponding to the two “humps” in the function. The inner washer has inner radius r1 = y and outer radius R1 = 2 ´ y. The outer washer has inner radius r2 = 2 + 2 3y and outer radius R2 = 6 ´ 2 3y. The thickness of the washers is dy.
316 y thickness: dy R1 = 2 ´ y R2 = 6 ´ 2 3y r1 = y r2 = 2 + 2 3y S-4: (a) When the strip shown in the figure x 3 y y = √x ex2 is rotated about the x–axis, it forms a thin disk of radius ?xex2 and thickness dx and hence of cross sectional area πxe2x2 and volume πxe2x2 dx So the volume of the solid is π ż 3 0 xe2x2 dx (b) The curves intersect at (´1, 1) and (2, 4).
317 x y y = x2 or x = ±√y y = x + 2 or x = y −2 x = 3 y = 1 (−1, 1) (2, 4) We’ll use horizontal washers as in Example 1.6.5 of the CLP-2 text.
• We use thin horizontal strips of width dy as in the figure above.
• When we rotate about the line x = 3, each strip sweeps out a thin washer – whose inner radius is rin = 3 ´ ?y, and – whose outer radius is rout = 3 ´ (y ´ 2) = 5 ´ y when y ě 1 (see the red strip in the figure on the right above), and whose outer radius is rout = 3 ´ (´?y) = 3 + ?y when y ď 1 (see the blue strip in the figure on the right above) and – whose thickness is dy and hence – whose volume is π(r2 out ´ r2 in)dy = π 5 ´ y 2 ´ 3 ´ ?y 2 dy when y ě 1 and whose volume is π(r2 out ´ r2 in)dy = π 3 + ?y 2 ´ 3 ´ ?y 2 dy when y ď 1 and • As our bottommost strip is at y = 0 and our topmost strip is at y = 4, the total volume is ż 1 0 π 3 + ?y 2 ´ 3 ´ ?y 2 dy + ż 4 1 π 5 ´ y 2 ´ 3 ´ ?y 2 dy S-5: (a) The curves intersect at (1, 0) and (´1, 0). When the strip shown in the figure x y = −1 y y = 4 −4x2 y = 1 −x2 (−1, 0) (1, 0) is rotated about the line y = ´1, it forms a thin washer with: • inner radius (1 ´ x2) ´ (´1) = 2 ´ x2, 318 • outer radius (4 ´ 4x2) ´ (´1) = 5 ´ 4x2 and • thickness dx ; so, it has • cross sectional area π (5 ´ 4x2)2 ´ (2 ´ x2)2 and • volume π (5 ´ 4x2)2 ´ (2 ´ x2)2 dx.
So the volume of the solid is ż 1 ´1 π (5 ´ 4x2) 2 ´ (2 ´ x2) 2 dx (b) The curve y = x2 ´ 1 intersects y = 0 at (1, 0) and (´1, 0).
x y y = x2 −1 or x = ±√y + 1 x = 5 (−1, 0) (1, 0) (0, −1) We’ll use horizontal washers.
• We use thin horizontal strips of height dy as in the figure above.
• When we rotate about the line x = 5, each strip sweeps out a thin washer – whose inner radius is rin = 5 ´ a y + 1, and – whose outer radius is rout = 5 ´ (´ a y + 1) = 5 + a y + 1 and – whose thickness is dy and hence – whose volume is π(r2 out ´ r2 in) dy = π 5 + a y + 1 2 ´ 5 ´ a y + 1 2 dy • As our topmost strip is at y = 0 and our bottommost strip is at y = ´1 (when x = 0), the total volume is ż 0 ´1 π h 5 + a y + 1 2 ´ 5 ´ a y + 1 2i dy S-6: The curves intersect at (´2, 4) and (2, 4). When the strip shown in the figure x y y = x2 y = 8 −x2 y = −1 (−2, 4) (2, 4) 319 is rotated about the line y = ´1, it forms a thin washer (punctured disc) of • inner radius x2 + 1, • outer radius 9 ´ x2 and • thickness dx and hence of • cross sectional area π (9 ´ x2)2 ´ (x2 + 1)2 and • volume π (9 ´ x2)2 ´ (x2 + 1)2 dx.
So the volume of the solid is π ż 2 ´2 (9 ´ x2) 2 ´ (x2 + 1) 2 dx S-7: We’ll make horizontal slices, parallel to one of the faces of the tetrahedron. Then our slices will be equilateral triangles, of varying sizes.
ℓ For the sake of ease, as in Example 1.6.1 of the CLP-2 text, we picture the tetrahedron perched on a tip, one base horizontal on top.
x y y b 2 3ℓ ℓ 2 ´ ℓ 2 Notice our slice forms the horizontal top of a smaller tetrahedron. The horizontal top of the full tetrahedron has side length ℓ, which is b 3 2 times the height of the full tetrahedron. Our slice is the horizontal top of a tetrahedron of height y and so has side length b 3 2y. An equilateral triangle with side length L has base L and height ?
3 2 L, and 320 hence area ?
3 4 L2. So, the area of our slice with side length b 3 2y is A = ?
3 4 c 3 2y !2 = 3 ?
3 8 y2 So, the volume of a tetrahedron with side length ℓis: Volume = ż b 2 3ℓ 0 3 ?
3 8 y2 dy = ?
3 8 ¨ c 2 3ℓ !3 = ?
2 12 ℓ3 You were given the height of a tetrahedron, but for completeness we calculate it here.
Draw a line starting at one tip, and dropping straight down to the middle of the opposite face. It forms a right triangle with one edge of the tetrahedron, and a line from the middle of the face to the corner.
ℓ c A C We know the length of the hypotenuse of this right triangle (it’s ℓ), so if we know the length of its base (labeled Ac in the diagram), we can figure out its third side, the height of our tetrahedron. Note by using the Pythagorean theorem, we see that the height of an equilateral triangle with edge length ℓis b 3 2ℓ.
Here is a sketch of the base of the pyramid: A C B b c ℓ 321 The triangles ABC and Abc are similar (since b and B are right angles, and also A has the same angle in both). Therefore, Ac Ab = AC AB Ac ℓ/2 = ℓ ?
3ℓ/2 Ac = 1 ?
3ℓ With this in our pocket, we can find the height of the tetrahedron: c ℓ2 ´ 1 ?
3ℓ 2 = b 2 3ℓ.
S-8: Let f (x) = 1 + ?xex2. On the vertical slice a distance x from the y-axis, sketched in the figure below, y runs from 1 to f (x). Upon rotation about the line y = 1, this thin slice sweeps out a thin disk of thickness dx and radius f (x) ´ 1 and hence of volume π[ f (x) ´ 1]2 dx. The full volume generated (for any fixed a ą 0) is ż a 0 π[ f (x) ´ 1]2 dx = π ż a 0 xe2x2 dx.
Using the substitution u = 2x2, so that du = 4x dx: Volume = π ż 2a2 0 eu du 4 = π 4 euˇ ˇ ˇ 2a2 0 = π 4 e2a2 ´ 1 x x = a y y = 1 + √xex2 y = 1 Remark: we spent a good deal of time last semester developing highly accurate but time-consuming methods for sketching common functions. For the purposes of questions like this, we don’t need a detailed picture of a function–broad outlines suffice.
Notice that ?x ą 0 whenever x ą 0, and ex2 ą 0 for all x. Therefore, ?xex2 is nonnegative over its entire domain, and so the graph y = 1 + ?xex2 is always the top function, above the bottom function y = 1. That is the only information we needed to perform our calculation.
S-9: The curves y = 1/x and 3x + 3y = 10, i.e. y = 10 3 ´ x intersect when 1 x = 10 3 ´ x ð ñ 3 = 10x ´ 3x2 ð ñ 3x2 ´ 10x + 3 = 0 ð ñ (3x ´ 1)(x ´ 3) = 0 ð ñ x = 3 , 1 3 322 y = 0 y (3, 1/3) (1/3, 3) y = T(x) y = B(x) When the region is rotated about the x–axis, the vertical strip in the figure above sweeps out a washer with thickness dx, outer radius T(x) = 10 3 ´ x and inner radius B(x) = 1 x.
This washer has volume π T(x)2 ´ B(x)2 dx = π 100 9 ´ 20 3 x + x2 ´ 1 x2 dx Hence the volume of the solid is π ż 3 1/3 100 9 ´ 20 3 x + x2 ´ 1 x2 dx = π h100x 9 ´ 10 3 x2 + 1 3x3 + 1 x i3 1/3 = π h38 3 ´ 514 34 i = π512 81 S-10: (a) The top and the bottom of the circle have equations y = T(x) = 2 + ?
1 ´ x2 and y = B(x) = 2 ´ ?
1 ´ x2, respectively.
y = 0 x = −1 x = 1 y = T(x) y = B(x) When R is rotated about the x–axis, the vertical strip of R in the figure above sweeps out a washer with thickness dx, outer radius T(x) and inner radius B(x). This washer has volume π T(x)2 ´ B(x)2 dx = π T(x) + B(x) T(x) ´ B(x) dx = π ˆ 4 ˆ 2 a 1 ´ x2 dx Hence the volume of the solid is 8π ż 1 ´1 a 1 ´ x2 dx 323 (b) Since y = ?
1 ´ x2 is equivalent to x2 + y2 = 1, y ě 0, the integral is 8π times the area of the upper half of the circle x2 + y2 = 1 and hence is 8π ˆ 1 2π12 = 4π2.
S-11: (a) The two curves intersect when x obeys 8x = x2 + 15 or x2 ´ 8x + 15 = (x ´ 5)(x ´ 3) = 0. The points of intersection, in the first quadrant, are (3, ?
24) and (5, ?
40). The region R is the region between the blue and red curves, with 3 ď x ď 5, in the figures below.
(3, √ 24) (5, √ 40) R y2 = 8x y2 = x2 + 15 x y (3, √ 24) (5, √ 40) R (b) The part of the solid with x coordinate between x and x + dx is a “washer” shaped region with inner radius ?
x2 + 15, outer radius ?
8x and thickness dx. The surface area of the washer is π( ?
8x)2 ´ π( ?
x2 + 15)2 = π(8x ´ x2 ´ 15) and its volume is π(8x ´ x2 ´ 15) dx. The total volume is ż 5 3 π(8x ´ x2 ´ 15) dx = π h 4x2 ´ 1 3x3 ´ 15x i5 3 = π h 100 ´ 125 3 ´ 75 ´ 36 + 9 + 45 i = 4 3π « 4.19 S-12: (a) The region R is sketched in the figure on the left below. (The bound y = 0 renders the bound x = 1 unnecessary, since the graph y = log x hits the x-axis when x = 1.) x y y = log x x = 2 x y x = ey x = 2 (b) We’ll use horizontal washers as in Example 1.6.5 of the CLP-2 text.
• We cut R into thin horizontal strips of height dy as in the figure on the right above.
• When we rotate R about the y–axis, i.e. about the line x = 0, each strip sweeps out a thin washer – whose inner radius is rin = ey and outer radius is rout = 2, and – whose thickness is dy and hence 324 – whose volume π(r2 out ´ r2 in)dy = π 4 ´ e2y dy.
• As our bottommost strip is at y = 0 and our topmost strip is at y = log 2 (since at the top x = 2 and x = ey), the total Volume = ż log 2 0 π 4 ´ e2y dy = π 4y ´ e2y/2 log 2 0 = π h 4 log 2 ´ 2 + 1 2 i = π h 4 log 2 ´ 3 2 i Using a calculator, we see this is approximately 3.998.
S-13: Here is a sketch of the curves y = cos( x 2) and y = x2 ´ π2.
y y = −π2 (−π, 0) (π, 0) y = cos( x 2) y = x2 −π2 By inspection, the curves meet at x = ˘π where both cos( x 2) and x2 ´ π2 take the value zero. We’ll use vertical washers as specified in the question.
• We cut the specified region into thin vertical strips of width dx as in the figure above.
• When we rotate about the line y = ´π2, each strip sweeps out a thin washer – whose inner radius is rin = (x2 ´ π2) ´ (´π2) = x2 and outer radius is rout = cos( x 2) ´ (´π2) = cos( x 2) + π2, and – whose thickness is dx and hence – whose volume π(r2 out ´ r2 in)dx = π (cos( x 2) + π2)2 ´ (x2)2 dx.
• As our leftmost strip is at x = ´π and our rightmost strip is at x = π, the total volume is π ż π ´π cos2( x 2) + 2π2 cos( x 2) + π4 ´ x4 dx = π ż π ´π 1 + cos(x) 2 + 2π2 cos( x 2) + π4 ´ x4 dx 325 Because the integrand is even, = 2π ż π 0 1 + cos(x) 2 + 2π2 cos( x 2) + π4 ´ x4 dx = 2π 1 2x + 1 2 sin(x) + 4π2 sin( x 2) + π4x ´ 1 5x5 π 0 = 2π π 2 + 0 + 4π2 + π5 ´ π5 5 = π2 + 8π3 + 8π6 5 We used the fact that the integrand is an even function and the interval of integration [´π, π] is symmetric, but one can also compute directly.
S-14: As in Example 1.6.6 of the CLP-2 text notes, we slice V into thin horizontal “square pancakes”.
• We are told that the pancake at height x is a square of side 2 1+x and so • has cross-sectional area 2 1+x 2 and thickness dx and hence • has volume 2 1+x 2dx.
Hence the volume of V is ż 2 0 h 2 1 + x i2 dx = ż 3 1 4 u2 du = 4u´1 ´1 ˇ ˇ ˇ ˇ 3 1 = ´4 h1 3 ´ 1 i = 8 3 We made the change of variables u = 1 + x, du = dx.
S-15: Here is a sketch of the base region.
x y y = 8 −x2 y = x2 Consider the thin vertical cross–section resting on the heavy red line in the figure above.
It has thickness dx. Its face is a square whose side runs from y = x2 to y = 8 ´ x2, a distance of 8 ´ 2x2. So the face has area (8 ´ 2x2)2 and the slice has volume (8 ´ 2x2)2 dx.
The two curves cross when x2 = 8 ´ x2, i.e. when x2 = 4 or x = ˘2. So x runs from ´2 to 2 and the total volume is ż 2 ´2 (8 ´ 2x2) 2 dx = 2 ż 2 0 4(4 ´ x2) 2 dx = 8 ż 2 0 16 ´ 8x2 + x4 dx = 8 h 16 ˆ 2 ´ 8 323 + 1 525i = 256 ˆ 8 15 = 136.5˙ 3 326 In the first simplification step, we used the fact that our integrand was even, but we also could have finished our computation without this step.
S-16: Slice the frustum into horizontal discs. When the disc is a distance t from the top of the frustum it has radius 2 + 2t/h. Note that as t runs from 0 (the top of the frustum) to t = h (the bottom of the frustum) the radius 2 + 2t/h increases linearly from 2 to 4.
h 2 4 t Thus the disk has volume π 2 + 2t/h 2dt. The total volume of the frustum is π ż h 0 2 + 2t/h 2dt = 4π ż h 0 1 + t/h 2dt = 4π (1 + t/h)3 3/h h 0 = 4 3πh ˆ 7 = 28 3 πh Remark: we could also solve this problem using the formula for the volume of a cone.
Using similar triangles, the frustum in question is shaped like a right circular cone of height 2h and base radius 4 (and hence of volume 1 3π(42)(2h)), but missing its top, which is a right circular cone of height h and base radius 2 (and hence volume 1 3π(22)h).
So, the volume of the frustum is 1 3π(42)(2h) ´ 1 3π(22)h = 28 3 πh.
S-17: (a) We’ll want to start by graphing the upper half of the ellipse (ax)2 + (by)2 = 1. Its intercepts will be enough to get us an idea: (0, 1 b) and (˘1 a, 0): x y 1 a ´1 a 1 b y = 1 b a 1 ´ (ax)2 327 We note a few things at the outset: first, since a ě b, then 1 a ď 1 b, so indeed the x-axis is the minor axis. That is, we’re rotating about the proper axis to create an oblate spheroid.
Second, if we solve our equation for y, we get y = 1 b a 1 ´ (ax)2. (Since we only want the upper half of the ellipse, we only need to consider the positive square root.) Now, we have a standard volume-of-revolution problem. We make vertical slices, of width dx and height y = 1 b a 1 ´ (ax)2. When we rotate these slices about the x-axis, they form thin disks of volume π h 1 b a 1 ´ (ax)2 i2 dx . Since x runs from ´1 a to 1 a, the volume of our oblate spheroid is: Volume = ż 1 a ´ 1 a π 1 b b 1 ´ (ax)2 2 dx = π b2 ż 1 a ´ 1 a 1 ´ (ax)2 dx = 2π b2 ż 1 a 0 1 ´ (ax)2 dx (even function) = 2π b2 x ´ a2x3 3 1 a 0 = 2π b2 1 a ´ 1 3a = 4π 3b2a (b) As we saw in the sketch from part (a), the shortest radius of the ellipse is 1 a, while the largest is 1 b. So, 1 a = 6356.752, and 1 b = 6378.137. That is, a = 1 6356.752 and b = 1 6378.137.
Note a ě b, as specified in part (a).
(c) Combining our answers from (a) and (b), the volume of an oblate spheroid with approximately the same dimensions as the earth is: 4π 3b2a = 4π 3 1 b 2 1 a = 4π 3 (6378.137)2 (6356.752) « 1.08321 ˆ 1012 km3 « 1.08321 ˆ 1021 m3 (d) A sphere of radius 6378.137 has volume 4 3π (6378.137)3 328 So, our absolute error is: ˇ ˇ ˇ ˇ 4π 3 (6378.137)2 (6356.752) ´ 4 3π (6378.137)3 ˇ ˇ ˇ ˇ =4π 3 (6378.137)2 ˇ ˇ6356.752 ´ 6378.137 ˇ ˇ =4π 3 (6378.137)2 (21.385) « 3.64 ˆ 109 km3 And our relative error is: abs error actual value = 4π 3 (6378.137)2 ˇ ˇ6356.752 ´ 6378.137 ˇ ˇ 4π 3 (6378.137)2 (6356.752) = ˇ ˇ6356.752 ´ 6378.137 ˇ ˇ 6356.752 = 6378.137 6356.752 ´ 1 « 0.00336 That is, about 0.336%, or about one-third of one percent.
S-18: (a) The curve y = 4 ´ (x ´ 1)2 is an “upside down parabola” and line y = x + 1 has slope 1. They intersect at points (x, y) which satisfy both y = x + 1 and y = 4 ´ (x ´ 1)2.
That is, when x obeys x + 1 = 4 ´ (x ´ 1)2 x + 1 = 4 ´ x2 + 2x ´ 1 x2 ´ x ´ 2 = 0 (x ´ 2)(x + 1) = 0 x = ´1 or x = 2 Thus the intersection points are (´1, 0) and (2, 3). Here is a sketch of R: y x (−1, 0) (2, 3) y = 4 −(x −1)2 y = x + 1 The red strip in the sketch above runs from y = x + 1 to y = 4 ´ (x ´ 1)2 and so has area 329 [4 ´ (x ´ 1)2 ´ (x + 1)] dx = [2 + x ´ x2] dx. All together R has Area = ż 2 ´1 2 + x ´ x2 dx = 2x + x2 2 ´ x3 3 2 ´1 = 6 + 3 2 ´ 9 3 = 9 2 (b) We’ll use vertical washers as in Example 1.6.3 of the CLP-2 text. Note that the highest point achieved by y = 4 ´ (x ´ 1)2 is y = 4, so rotating around the line y = 5 causes no unexpected problems.
y x y = 5 (−1, 0) (2, 3) y = 4 −(x −1)2 y = x + 1 • We cut R into thin vertical strips of width dx like the red strip in the figure above.
• When we rotate R about the horizontal line y = 5, each strip sweeps out a thin washer – whose inner radius is rin = 5 ´ [4 ´ (x ´ 1)2] = 1 + (x ´ 1)2, and – whose outer radius is rout = 5 ´ [x + 1] = 4 ´ x and – whose thickness is dx and hence – whose volume is π r2 out ´ r2 in dx = π 4 ´ x 2 ´ 1 + (x ´ 1)22 dx • As our leftmost strip is at x = ´1 and our rightmost strip is at x = 2, the total Volume = π ż 2 ´1 4 ´ x 2 ´ 1 + (x ´ 1)22 dx S-19: (a) The curves (x ´ 1)2 + y2 = 1 and x2 + (y ´ 1)2 = 1 are circles of radius 1 centered on (1, 0) and (0, 1) respectively. Both circles pass through (0, 0) and (1, 1). They are sketched below.
330 x y y = x (1, 1) x2 + (y −1)2 = 1 (x −1)2 + y2 = 1 The region R is symmetric about the line y = x, so the area of R is twice the area of the part of R to the left of the line y = x. The red strip in the sketch above runs from the edge of the lower circle to x = y. So, given a value of y in [0, 1], we need to find the corresponding value of x along the circle. We solve (x ´ 1)2 + y2 = 1 for x, keeping in mind that 0 ď x ď 1: (x ´ 1)2 + y2 = 1 (x ´ 1)2 = 1 ´ y2 |x ´ 1| = b 1 ´ y2 1 ´ x = b 1 ´ y2 x = 1 ´ b 1 ´ y2 Now, we calculate: Area = 2 ż 1 0 y ´ 1 ´ b 1 ´ y2 dy = 2 #ż 1 0 y ´ 1 dy + ż 1 0 b 1 ´ y2 dy + = 2 !hy2 2 ´ y i1 0 + ż 1 0 b 1 ´ y2 dy ) = π 2 ´ 1 Here the integral ş1 0 a 1 ´ y2 dy was evaluated simply as the area of one quarter of a cicular disk of radius 1. It can also be evaluated by substituting y = sin θ, a technique we’ll learn more about in Section 1.9 of the CLP-2 text.
(b) We’ll use horizontal washers as in Example 1.6.5 of the in the CLP-2 text.
• We cut R into thin horizontal strips of width dy like the blue strip in the figure above.
• When we rotate R about the y–axis, each strip sweeps out a thin washer – whose inner radius is rin = 1 ´ a 1 ´ y2, and 331 – whose outer radius is rout = a 1 ´ (y ´ 1)2 and – whose thickness is dy and hence – whose volume is π b 1 ´ (y ´ 1)22 ´ 1 ´ b 1 ´ y2 2 dy =π 1 ´ (y ´ 1)2 ´ 1 + 2 b 1 ´ y2 ´ (1 ´ y2) =2π b 1 ´ y2 + y ´ 1 dy • As our bottommost strip is at y = 0 and our topmost strip is at y = 1, the total Volume = 2π ż 1 0 b 1 ´ y2 + y ´ 1 dy = 2π hπ 4 + 1 2 ´ 1 i = π2 2 ´ π « 1.793 Here, we again used that ş1 0 a 1 ´ y2 dy is the area of a quarter circle of radius one, and we used a calculator to approximate the final answer.
S-20: Before we start, it will be useful to have a reasonable sketch of the graph y = c ?
1 + x2 over the interval [0, 1]. Its endpoints are (0, c) and (1, c ?
2). The function is entirely above the x-axis, which we need to know for part (a). For part (b), we need to know whether it is always increasing or not: when we’re drawing horizontal strips, we need to know their endpoints, and if the function has “humps,” the right endpoint will not be simply the line x = 1.
If you’re comfortable noticing that 1 + x2 increases as x increases because we only consider nonnegative values of x, then you can also be confident that ?
1 + x2 is simply increasing. Alternately, we can consider the derivative: d dx !
c a 1 + x2 ) = c ¨ 1 2 ?
1 + x2 ¨ 2x = cx ?
1 + x2 Since we only consider positive values of x, this derivative is never negative, so the function is never decreasing. This gives us the following basic sketch: x y 1 c 332 The figures in the solution below use a slightly more detailed rendering of our function, but so much accuracy is not necessary.
(a) Let V1 be the solid obtained by revolving R about the x–axis. The portion of V1 with x–coordinate between x and x + dx is obtained by rotating the red vertical strip in the figure on the left below about the x–axis. That portion is a disk of radius c ?
1 + x2 and thickness dx. The volume of this disk is π(c ?
1 + x2)2dx = πc2(1 + x2) dx. So the total volume of V1 is V1 = ż 1 0 πc2(1 + x2) dx = πc2h x + x3 3 i1 0 = 4 3πc2 x = 1 y = c √ 1 + x2 x y c x = 1 x = q y2 c2 −1 x y c (b) We’ll use horizontal washers as in Example 1.6.5 of the in the CLP-2 text.
• We cut R into thin horizontal strips of width dy as in the figure on the right above.
• When we rotate R about the y–axis, i.e. about the line x = 0, each strip sweeps out a thin washer – whose outer radius is rout = 1, and – whose inner radius is rin = b y2 c2 ´ 1 when y ě c ?
1 + 02 = c (see the red strip in the figure on the right above), and whose inner radius is rin = 0 when y ď c (see the blue strip in the figure on the right above) and – whose thickness is dy and hence – whose volume is π(r2 out ´ r2 in)dy = π 2 ´ y2 c2 dy when y ě c and whose volume is π(r2 out ´ r2 in)dy = π dy when y ď c and • As our bottommost strip is at y = 0 and our topmost strip is at y = ?
2 c (since at the top x = 1 and y = c ?
1 + x2), the total V2 = ż ?
2 c c π 2 ´ y2 c2 dy + ż c 0 π dy = π h 2y ´ y3 3c2 i ?
2 c c + πc = π c h4 ?
2 3 ´ 5 3 i + πc = π c 3 4 ?
2 ´ 2 333 (c) We have V1 = V2 if and only if 4 3πc2 = π c 3 4 ?
2 ´ 2 4c2 = c 4 ?
2 ´ 2 4c2 ´ c 4 ?
2 ´ 2 = 0 4c c ´ ?
2 ´ 1 2 = 0 c = 0 or c = ?
2 ´ 1 2 S-21: We will compute the volume by rotating thin vertical strips as in the sketch y x π 2 π 3π 2 2π 2 4 6 8 y = 4 + π sin(x) y = 4 + 2π −2x about the line y = ´1 to generate thin washers. We need to know when the line y = 4 + 2π ´ 2x intersects the curve y = 4 + π sin x. Looking at the graph, it appears to be at π 2 , π, and 3π 2 . By plugging in these values of x to both functions, we see they are indeed the points of intersection.
• When π 2 ď x ď π, the top of the strip is at y = 4 + π sin x and the bottom of the strip is at y = 4 + 2π ´ 2x. When the strip is rotated, we get a thin washer with outer radius R1(x) = 1 + 4 + π sin x = 5 + π sin x and inner radius r1(x) = 1 + 4 + 2π ´ 2x = 5 + 2π ´ 2x.
• When π ď x ď 3π 2 , the top of the strip is at y = 4 + 2π ´ 2x and the bottom of the strip is at y = 4 + π sin x. When the strip is rotated, we get a thin washer with outer radius R2(x) = 1 + 4 + 2π ´ 2x = 5 + 2π ´ 2x and inner radius r2(x) = 1 + 4 + π sin x = 5 + π sin x.
So, the total Volume = ż π π/2 π R1(x)2 ´ r1(x)2 dx + ż 3π/2 π π R2(x)2 ´ r2(x)2 dx = ż π π/2 π (5 + π sin x)2 ´ (5 + 2π ´ 2x)2 dx + ż 3π/2 π π (5 + 2π ´ 2x)2 ´ (5 + π sin x)2 dx 334 S-22: (a) We use the same ideas for volume, and apply them to mass. We want to take slices of the column, approximate their mass, then add them up. To reconcile our units, let k = 1000h, so k is the height in metres. Then the density of air at height k is c2´k/6000 kg m3.
A horizontal slice of the column is a circular disk with height dk and radius 1 m. So, its volume is π dk m3. What we’re interested in, though, is its mass. At height k, its mass is (volume) ˆ (density) = π dk m3 ˆ c2´k/6000 kg m3 = cπ2´k/6000 dk kg Since k runs from 0 to 60, 000, the total mass is given by ż 60000 0 cπ2´k/6000 dk = cπ ż 60000 0 2´k/6000 dk To facilitate integration, we can write our exponential function in terms of e, then use the substitution u = ´ k 6000 log 2, du = ´ 1 6000 log 2 dk.
= cπ ż 60000 0 elog 2´k/6000 dk = cπ ż 60000 0 e´ k 6000 log 2dk = ´6000cπ log 2 ż ´10 log 2 0 eudu = 6000cπ log 2 ż 0 ´10 log 2 eudu = 6000cπ log 2 1 ´ 1 210 We note this is fairly close to 6000cπ log 2 .
We also remark that this is a demonstration of the usefulness of integrals. We wanted to know how much of something there was, but the amount of that something was different everywhere: more in some places, less in others. Integration allowed us to account for this gradient. You’ve seen this behaviour exploited to find distances travelled, areas, volumes, and now mass. In your studies, you will doubtless learn to use it to find still more quantities, and we will discuss other applications in Chapter 2 of the CLP-2 text.
(b) We want to find the value of k that gives a mass of 3000cπ log 2 . By following our reasoning above, the mass of air in the column from the ground to height k is 6000cπ log 2 1 ´ 1 2k/6000 335 So, we set this equal to the mass we want, and solve for k.
6000cπ log 2 1 ´ 1 2k/6000 = 3000cπ log 2 2 1 ´ 1 2k/6000 = 1 1 = 2 2k/6000 2k/6000 = 21 k = 6000 h = 6 This means that there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.
Solutions to Exercises 1.7 — Jump to TABLE OF CONTENTS S-1: Integration by substitution is just using the chain rule, backwards: d dxtf (g(x))u = f 1(g(x))g1(x) ô ż d dxtf (g(x))udx = ż f 1(g(x))g1(x)dx ô f (g(x)) looomooon f (u) +C = ż f 1(g(x)) looomooon f 1(u) g1(x)dx looomooon du Similarly, integration by parts comes from the product rule: d dxtf (x)g(x)u = f 1(x)g(x) + f (x)g1(x) ô ż d dxtf (x)g(x)udx = ż f 1(x)g(x) + f (x)g1(x)dx ô f (x)g(x) + C = ż f 1(x)g(x)dx + ż f (x)g1(x)dx ô ż f (x) lo omo on u g1(x)dx looomooon dv = f (x) lo omo on u g(x) lo omo on v ´ ż g(x) lo omo on v f 1(x)dx looomooon du In the last line, the “+C” has been absorbed into the indefinite integral on the right hand side.
S-2: Remember our rule: ş udv = uv ´ ş vdu. So, we take u and use it to make du–that is, we differentiate it. We take dv and use it to make v–that is, we antidifferentiate it.
336 S-3: We’ll use the same ideas that led to the method of integration by parts. (You can review this in your text, or see the solution to Question 1 in this section.) According to the quotient rule, d dx " f (x) g(x) = g(x) f 1(x) ´ f (x)g1(x) g2(x) .
Antidifferentiating both sides gives us: ż d dx " f (x) g(x) dx = ż g(x) f 1(x) ´ f (x)g1(x) g2(x) dx f (x) g(x) + C = ż f 1(x) g(x) dx ´ ż f (x)g1(x) g2(x) dx ż f 1(x) g(x) dx = f (x) g(x) + ż f (x)g1(x) g2(x) dx In the last line, the “+C” has been absorbed into the indefinite integral on the right hand side.
This is exactly the integration by parts formula for the functions u = 1/g and v = f.
S-4: All the antiderivatives differ only by a constant, so we can write them all as v(x) + C for some C. Then, using the formula for integration by parts, ż u(x) ¨ v1(x)dx = u(x) lo omo on u v(x) + C looooomooooon v ´ ż v(x) + C looooomooooon v u1(x)dx looomooon du = u(x)v(x) + Cu(x) ´ ż v(x)u1(x)dx ´ ż Cu1(x)dx = u(x)v(x) + Cu(x) ´ ż v(x)u1(x)dx ´ Cu(x) + D = u(x)v(x) ´ ż v(x)u1(x)dx + D where D is any constant.
Since the terms with C cancel out, it didn’t matter what we chose for C–all choices end up the same.
S-5: Suppose we choose dv = f (x)dx, u = 1. Then v = ż f (x)dx, and du = dx. So, our integral becomes: ż (1) lo omo on u f (x)dx loomoon dv = (1) lo omo on u ż f (x)dx loooomoooon v ´ ż ż f (x)dx loooooomoooooon v dx lo omo on du In order to figure out the first product (and the second integrand), you need to know the antiderivative of f (x)–but that’s exactly what you’re trying to figure out! So, using integration by parts has not eased your pain.
337 We note here that in certain cases, such as ş log x dx (Example 1.7.8 in the CLP-2 text), it is useful to choose dv = 1dx. This is similar to, but crucially different from, the do-nothing method in this problem.
S-6: For integration by parts, we want to break the integrand into two pieces, multiplied together. There is an obvious choice for how to do this: one piece is x, and the other is log x. Remember that one piece will be integrated, while the other is differentiated. The question is, which choice will be more helpful. After some practice, you’ll get the hang of making the choice. For now, we’ll present both choices–but when you’re writing a solution, you don’t have to write this part down. It’s enough to present your choice, and then a successful computation is justification enough.
Option 1: u = x du = 1 dx dv = log x dx v =??
Option 2: u = log x du = 1 x dx dv = x dx v = 1 2x2 In Example 1.7.8 of the CLP-2 text, we found the antiderivative of logarithm, but it wasn’t trivial. We might reasonably want to avoid using this complicated antiderivative.
Indeed, Option 2 (differentiating logarithm, antidifferentiating x) looks promising–when we multiply the blue equations, we get something easily integrable– so let’s not even bother going deeper into Option 1.
That is, we perform integration by parts with u = log x and dv = x dx, so that du = dx x and v = x2 2 .
ż log x lo omo on u x dx lo omo on dv = x2 log x 2 looomooon uv ´ ż x2 2 lo omo on v dx x lo omo on du = x2 log x 2 ´ 1 2 ż x dx = x2 log x 2 ´ x2 4 + C S-7: Our integrand is the product of two functions, and there’s no clear substitution. So, we might reasonably want to try integration by parts. Again, we have two obvious pieces: log x, and x´7. We’ll consider our options for assigning these to u and dv: 338 Option 1: u = log x du = 1 x dx dv = x´7 dx v = 1 ´6x´6 Option 2: u = x´7 du = ´7x´8 dx dv = log x dx v =??
Again, we remember that logarithm has some antiderivative we found in Example 1.7.8 of the CLP-2 text, but it was something complicated. Luckily, we don’t need to bother with it: when we multiply the red equations in Option 1, we get a perfectly workable integral.
We perform integration by parts with u = log x and dv = x´7 dx, so that du = dx x and v = ´ x´6 6 .
ż log x x7 dx = ´ log x x´6 6 looooomooooon uv + ż x´6 6 lo omo on ´v dx x lo omo on du = ´log x 6x6 + 1 6 ż x´7 dx = ´log x 6x6 ´ 1 36x6 + C S-8: To integrate by parts, we need to decide what to use as u, and what to use as dv. The salient parts of this integrand are x and sin x, so we only need to decide which is u and which dv. Again, this process will soon become familiar, but to help you along we show you both options below.
Option 1: u = x du = 1 dx dv = sin x dx v = ´ cos x Option 2: u = sin x du = cos x dx dv = x dx v = 1 2x2 The derivative and antiderivative of sine are almost the same, but x turns into something simpler when we differentiate it. So, we choose Option 1.
339 We integrate by parts, using u = x, dv = sin x dx so that v = ´ cos x and du = dx: ż π 0 x sin x dx = ´x cos x looomooon uv ˇ ˇ ˇ π 0 ´ ż π 0 (´ cos x) loooomoooon v dx lo omo on du = h ´ x cos x + sin x iπ 0 = ´π(´1) = π S-9: When we have two functions multiplied like this, and there’s no obvious substitution, our minds turn to integration by parts. We hope that our integral will be improved by differentiating one part and antidifferentiating the other. Let’s see what our choices are: Option 1: u = x du = 1 dx dv = cos x dx v = sin x Option 2: u = cos x du = ´ sin x dx dv = x dx v = 1 2x2 Option 1 seems preferable. We integrate by parts, using u = x, dv = cos x dx so that v = sin x and du = dx: ż π 2 0 x cos x dx = x lo omo on u sin x lo omo on v ˇ ˇ ˇ π 2 0 ´ ż π 2 0 sin x lo omo on v dx lo omo on du = h x sin x + cos x i π 2 0 = π 2 ´ 1 S-10: This integrand is the product of two functions, with no obvious substitution. So, let’s try integration by parts, with one part ex and one part x3.
Option 1: u = ex du = ex dx dv = x3 dx v = 1 4x4 Option 2: u = x3 du = 3x2 dx dv = ex dx v = ex At first glance, multiplying the red functions and multiplying the blue functions give largely equivalent integrands to what we started with–none of them with obvious antiderivatives. In previous questions, we were able to choose u = x, and then du = dx, so the “x” in the integrand effectively went away. Here, we see that choosing u = x3 will 340 lead to du = 3x2dx, which has a lower power. If we repeatedly perform integration by parts, choosing u to be the power of x each time, then after a few iterations it should go away, because the third derivative of x3 is a constant.
So, we start with Option 2: u = x3, dv = exdx, du = 3x2dx, and v = ex.
ż x3exdx = x3 lo omo on u ex lo omo on v ´ ż ex lo omo on v ¨ 3x2dx lo omo on du = x3ex ´ 3 ż ex ¨ x2dx Now, we take u = x2 and dv = exdx, so du = 2xdx and v = ex. We’re only using integration by parts on the actual integral–the rest of the function stays the way it is.
= x3ex ´ 3 x2ex lo omo on uv ´ ż ex lo omo on v ¨ 2xdx lo omo on du = x3ex ´ 3x2ex + 6 ż xexdx Continuing, we take u = x and dv = exdx, so du = dx and v = ex. This is the step where the polynomial part of the integrand finally disappears.
= x3ex ´ 3x2ex + 6 xex lo omo on uv ´ ż ex lo omo on v dx lo omo on du = x3ex ´ 3x2ex + 6xex ´ 6ex + C = ex x3 ´ 3x2 + 6x ´ 6 + C Let’s check that this makes sense: the derivative of ex x3 ´ 3x2 + 6x ´ 6 + C should be x3ex. We differentiate using the product rule.
d dx !
ex x3 ´ 3x2 + 6x ´ 6 + C ) = ex x3 ´ 3x2 + 6x ´ 6 + ex 3x2 ´ 6x + 6 = ex x3 ´ 3x2 + 3x2 + 6x ´ 6x ´ 6 + 6 = x3ex Remark: In order to be technically correct in our antidifferentiation, we should add the +C as soon as we do the first integration by parts. However, when we are using integration by parts, we usually end up evaluating an integral at the end, and we add the +C at that point. Since the +C comes up eventually, it is common practice to not clutter our calculations with it until the end.
S-11: Since our integrand is two functions multiplied together, and there isn’t an obvious substitution, let’s try integration by parts. Here are our salient options.
341 Option 1: u = x du = 1 dx dv = log3 x dx v =??
Option 2: u = log3 x du = 3 log2 x ¨ 1 x dx dv = x dx v = 1 2x2 This calls for some strategizing. Using the template of Example 1.7.8 in the CLP-2 text, we could probably figure out the antiderivative of log3 x. Option 1 is tempting, because our x-term goes away. So, there might be a benefit there, but on the other hand, the antiderivative of log3 x is probably pretty complicated.
Now let’s consider Option 2. When we multiply the blue functions together, we get something similar to our original integrand, but the power of logarithm is smaller. If we were to iterate this method (using integration by parts a few times, always choosing u to be the part with a logarithm) then eventually we would end up differentiating logarithm.
This seems like a safer plan: let’s do Option 2.
We use integration by parts with u = log3 x, dv = xdx, du = 3 x log2 xdx, and v = 1 2x2.
ż x log3 x dx = 1 2x2 log3 x loooomoooon uv ´ ż 3 2x log2 xdx loooooomoooooon vdu = 1 2x2 log3 x ´ 3 2 ż x log2 xdx Continuing our quest to differentiate away the logarithm, we use integration by parts with u = log2 x, dv = xdx, du = 2 x log xdx, and v = 1 2x2.
= 1 2x2 log3 x ´ 3 2 1 2x2 log2 x loooomoooon uv ´ ż x log xdx loooomoooon vdu = 1 2x2 log3 x ´ 3 4x2 log2 x + 3 2 ż x log xdx 342 One last integration by parts: u = log x, dv = xdx, du = 1 xdx, and v = 1 2x2.
= 1 2x2 log3 x ´ 3 4x2 log2 x + 3 2 1 2x2 log x loooomoooon uv ´ ż 1 2xdx lo omo on vdu = 1 2x2 log3 x ´ 3 4x2 log2 x + 3 4x2 log x ´ 3 4 ż xdx = 1 2x2 log3 x ´ 3 4x2 log2 x + 3 4x2 log x ´ 3 8x2 + C Once again, technically there is a +C in the work after the first integration by parts, but we follow convention by conveniently suppressing it until the final integration.
S-12: The integrand is the product of two functions, without an obvious substitution, so let’s see what integration by parts can do for us.
Option 1: u = x2 du = 2x dx dv = sin x dx v = ´ cos x Option 2: u = sin x du = cos x dx dv = x2 dx v = 1 3x3 Neither option gives us something immediately integrable, but Option 1 replaces our x2 term with a lower power of x. If we repeatedly apply integration by parts, we can reduce this power to zero. So, we start by choosing u = x2 and dv = sin xdx, so du = 2xdx and v = ´ cos x.
ż x2 sin x dx = ´x2 cos x loooomoooon uv + ż 2x cos xdx loooooomoooooon ´vdu = ´x2 cos x + 2 ż x cos xdx Using integration by parts again, we want to be differentiating (not antidifferentiating) x, so we choose u = x, dv = cos xdx, and then du = dx (x went away!), v = sin x.
= ´x2 cos x + 2 x sin x lo omo on uv ´ ż sin xdx loomoon vdu = ´x2 cos x + 2x sin x + 2 cos x + C = (2 ´ x2) cos x + 2x sin x + C 343 S-13: This problem is similar to Questions 6 and 7: integrating a polynomial multiplied by a logarithm. Just as in these questions, if we use integration by parts with u = log t, then du = 1 t dt, and our new integrand will consist of powers of t–which are easy to antidifferentiate.
So, we use u = log t, dv = 3t2 ´ 5t + 6, du = 1 t dt, and v = t3 ´ 5 2t2 + 6t.
ż (3t2 ´ 5t + 6) log t dt = log t lo omo on u t3 ´ 5 2t2 + 6t loooooomoooooon v ´ ż 1 t t3 ´ 5 2t2 + 6t dt loooooooooooomoooooooooooon vdu = t3 ´ 5 2t2 + 6t log t ´ ż t2 ´ 5 2t + 6 dt = t3 ´ 5 2t2 + 6t log t ´ 1 3t3 + 5 4t2 ´ 6t + C S-14: Before we jump to integration by parts, we notice that the square roots lend themselves to substitution. Let’s take w = ?s. Then dw = 1 2?s ds, so 2w dw = ds.
ż ?se ?sds = ż w ¨ ew ¨ 2wdw = 2 ż w2ewdw Now we have nearly the situation of Question 10. We can repeatedly use integration by parts, with u as the power of w, to get rid of the polynomial part. We’ll start with u = w2, dv = ewdw, du = 2wdw, and v = ew.
= 2 w2ew lo omo on uv ´ ż 2wewdw looomooon vdu = 2w2ew ´ 4 ż wewdw We use integration by parts again, this time with u = w, dv = ewdw, du = dw, and v = ew.
= 2w2ew ´ 4 wew lo omo on uv ´ ż ewdw lo omo on vdu = 2w2ew ´ 4wew + 4ew + C = ew 2w2 ´ 4w + 4 + C = e ?s 2s ´ 4?s + 4 + C 344 S-15: Let’s use integration by parts. What are our parts? We have a few options.
Solution 1: Following Example 1.7.8 in the CLP-2 text, we choose u = log2 x and dv = dx, so that du = 2 x log x dx and v = x.
ż log2 xdx = x log2 x loomoon uv ´ ż 2 log xdx loooomoooon vdu Here we can either use the antiderivative of logarithm from memory, or re-derive it.
We do the latter, using integration by parts with u = log x, dv = 2dx, du = 1 xdx, and v = 2x.
= x log2 x ´ 2x log x looomooon uv ´ ż 2dx lo omo on vdu = x log2 x ´ 2x log x + 2x + C Solution 2: Our integrand is two functions multiplied together: log x and log x. So, we will use integration by parts with u = log x, dv = log x, du = 1 xdx, and (using the antiderivative of logarithm, found in Example 1.7.8 in the CLP-2 text) v = x log x ´ x.
ż log2 x dx = ( log x lo omo on u )(x log x ´ x looooomooooon v ) ´ ż (x log x ´ x looooomooooon v ) 1 xdx lo omo on du = x log2 x ´ x log x ´ ż (log x ´ 1) dx = x log2 x ´ x log x ´ [(x log x ´ x) ´ x] + C = x log2 x ´ 2x log x + 2x + C S-16: This is your friendly reminder that to a person with a hammer, everything looks like a nail. The integral in the problem is a classic example of an integral to solve using substitution. We have an “inside function,” x2 + 1, whose derivative shows up multiplied to the rest of the integrand. We take u = x2 + 1, then du = 2xdx, so ż 2xex2+1dx = ż eudu = eu + C = ex2+1 + C S-17: In Example 1.7.9 of the CLP-2 text, we saw that integration by parts was useful when the integrand has a derivative that works nicely when multiplied by x. We use the 345 same idea here. Let u = arccos y and dv = dy, so that v = y and du = ´ dy ?
1´y2.
ż arccos y dy = y arccos y loooomoooon uv + ż y a 1 ´ y2dy looooomooooon ´vdu Using the substitution u = 1 ´ y2, du = ´2ydy, = y arccos y ´ 1 2 ż u´1/2du = y arccos y ´ u1/2 + C = y arccos y ´ b 1 ´ y2 + C S-18: We integrate by parts, using u = arctan(2y), dv = 4y dy, so that v = 2y2 and du = 2 dy 1+(2y)2: ż 4y arctan(2y) dy = 2y2 arctan(2y) looooooomooooooon uv ´ ż 4y2 (2y)2 + 1 dy loooooomoooooon vdu The integrand 4y2 (2y)2+1 is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that 4y2 4y2+1 = 4y2+1 4y2+1 ´ 1 4y2+1. We therefore have: ż 4y2 4y2 + 1 dy = ż 1 ´ 1 4y2 + 1 dy = y ´ 1 2 arctan(2y) + C The final answer is then ż 4y arctan(2y) dy = 2y2 arctan(2y) ´ y + 1 2 arctan(2y) + C S-19: We’ve got an integrand that consists of two functions multiplied together, and no obvious substitution. So, we think about integration by parts. Let’s consider our options.
Note in Example 1.7.9 of the CLP-2 text, we found that the antiderivative of arctangent is 346 x arctan x ´ 1 2 log(1 + x2) + C.
Option 1: u = arctan x du = 1 1 + x2 dx dv = x2 dx v = 1 3x3 Option 2: u = x2 du = 2x dx dv = arctan x dx v = x arctan x ´ 1 2 log(1 + x2) Option 1 Option 1 seems likelier. Let’s see how it plays out. We use integration by parts with u = arctan x, dv = x2dx, du = dx 1+x2, and v = 1 3x3.
ż x2 arctan x dx = x3 3 arctan x looooomooooon uv ´ ż x3 3(1 + x2)dx loooooomoooooon vdu = x3 3 arctan x ´ 1 3 ż x3 1 + x2dx This is starting to look like a candidate for a substitution! Let’s try the denominator, s = 1 + x2. Then ds = 2xdx, and x2 = s ´ 1.
= x3 3 arctan x ´ 1 6 ż x2 1 + x2 ¨ 2xdx = x3 3 arctan x ´ 1 6 ż s ´ 1 s ds = x3 3 arctan x ´ 1 6 ż 1 ´ 1 s ds = x3 3 arctan x ´ 1 6s + 1 6 log |s| + C = x3 3 arctan x ´ 1 6(1 + x2) + 1 6 log(1 + x2) + C Option 2: What if we had tried the other option? That is, u = x2, du = 2xdx, dv = arctan x, and v = x arctan x ´ 1 2 log(1 + x2). It’s not always the case that both options work, but sometimes they do. (They are almost never of equal difficulty.) This solution takes advantage of two previously hard-won results: the antiderivatives of logarithm and arctangent.
347 ż x2 arctan xdx = x2 lo omo on u x arctan x ´ 1 2 log(1 + x2) loooooooooooooooooomoooooooooooooooooon v ´ ż x arctan x ´ 1 2 log(1 + x2) loooooooooooooooooomoooooooooooooooooon v ¨ 2xdx lo omo on du = x3 arctan x ´ x2 2 log(1 + x2) ´ 2 ż x2 arctan xdx + ż x log(1 + x2)dx Adding 2 ż x2 arctan xdx to both sides: 3 ż x2 arctan xdx = x3 arctan x ´ x2 2 log(1 + x2) + ż x log(1 + x2)dx ż x2 arctan xdx = x3 3 arctan x ´ x2 6 log(1 + x2) + 1 3 ż x log(1 + x2)dx Using the substitution s = 1 + x2, ds = 2xdx: = x3 3 arctan x ´ x2 6 log(1 + x2) + 1 6 ż log sds Using the antiderivative of logarithm found in Example 1.7.8 of the CLP-2 text, = x3 3 arctan x ´ x2 6 log(1 + x2) + 1 6 (s log s ´ s) + C = x3 3 arctan x ´ x2 6 log(1 + x2) + 1 6 (1 + x2) log(1 + x2) ´ (1 + x2) + C = x3 3 arctan x + ´x2 6 + 1 + x2 6 log(1 + x2) ´ 1 6(1 + x2) + C = x3 3 arctan x + 1 6 log(1 + x2) ´ 1 6(1 + x2) + C S-20: This example is similar to Example 1.7.10 in the CLP-2 text. The functions ex/2 and cos(2x) both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we’ll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually antidifferentiating.
Let’s use u = cos(2x) and dv = ex/2dx, so du = ´2 sin(2x)dx and v = 2ex/2.
ż ex/2 cos(2x)dx = 2ex/2 cos(2x) looooooomooooooon uv ´ ż ´4ex/2 sin(2x)dx loooooooooomoooooooooon vdu = 2ex/2 cos(2x) + 4 ż ex/2 sin(2x)dx 348 Similarly to our first integration by parts, we use u = sin(2x), dv = ex/2dx, du = 2 cos(2x)dx, and v = 2ex/2.
= 2ex/2 cos(2x) + 4 2ex/2 sin(2x) loooooomoooooon uv ´ ż 4ex/2 cos(2x)dx looooooooomooooooooon vdu So, we’ve found the equation ż ex/2 cos(2x)dx = 2ex/2 cos(2x) + 8ex/2 sin(2x) ´ 16 ż ex/2 cos(2x)dx + C We add 16 ż ex/2 cos(2x)dx to both sides.
17 ż ex/2 cos(2x)dx = 2ex/2 cos(2x) + 8ex/2 sin(2x) + C ż ex/2 cos(2x)dx = 2 17ex/2 cos(2x) + 8 17ex/2 sin(2x) + C Remark: remember that C is a stand-in for “we can add any real constant”. Since C can be any number in (´8, 8), also C 17 can be any number in (´8, 8). So, rather than write C 17 in the last line, we re-named C 17 to C.
S-21: Solution 1: This question looks like a substitution, since we have an “inside function.” So, let’s see where that leads: let u = log x. Then du = 1 x dx. We don’t see this right away in our function, but we can bring it into the function by multiplying and dividing by x, and noting from our substitution that eu = x.
ż sin(log x)dx = ż x sin(log x) x dx = ż eu sin u du Using the result of Example 1.7.11 in the CLP-2 text: = 1 2eu (sin u ´ cos u) + C = 1 2elog x (sin(log x) ´ cos(log x)) + C = 1 2x (sin(log x) ´ cos(log x)) + C Solution 2: It’s not clear how to antidifferentiate the integrand, but we can certainly differentiate it. So, keeping in mind the method of Example 1.7.11 in the CLP-2 text, we take u = sin(log x) and dv = dx, so du = 1 x cos(log x)dx and v = x.
ż sin(log x)dx = x sin(log x) looooomooooon uv ´ ż cos(log x)dx loooooomoooooon vdu 349 Continuing on, we again use integration by parts, with u = cos(log x), dv = dx, du = ´ 1 x sin(log x)dx, and v = x.
= x sin(log x) ´ x cos(log x) loooooomoooooon uv + ż sin(log x) loooomoooon ´vdu dx That is, we have ż sin(log x)dx = x [sin(log x) ´ cos(log x)] ´ ż sin(log x)dx + C Adding ş sin(log x)dx to both sides, 2 ż sin(log x)dx = x [sin(log x) ´ cos(log x)] + C ż sin(log x)dx = x 2 [sin(log x) ´ cos(log x)] + C Remark: remember that C is a stand-in for “we can add any real constant”. Since C can be any number in (´8, 8), also C 2 can be any number in (´8, 8). So, rather than write C 2 in the last line, we re-named C 2 to C.
S-22: We begin by simplifying the integrand.
ż 2x+log2 xdx = ż 2x ¨ 2log2 xdx = ż 2x ¨ x dx This is similar to the integral ż xexdx, which we saw in Example 1.7.1 of the CLP-2 text.
Let’s write 2 = elog 2 to take advantage of the easy integrability of ex.
= ż x ¨ ex log 2dx We use integration by parts with u = x, dv = ex log 2dx; du = dx, v = 1 log 2ex log 2.
(Remember log 2 is a constant. If you’d prefer, you can do a substitution with s = x log 2 first, to have a simpler exponent of e.) = x log 2ex log 2 looooomooooon uv ´ ż 1 log 2ex log 2dx looooooomooooooon vdu = x log 2ex log 2 ´ 1 (log 2)2 ex log 2 + C = x log 22x ´ 1 (log 2)22x + C 350 S-23: It’s not obvious where to start, but in general it’s nice to have the arguments of our trig functions the same. So, we use the identity sin(2x) = 2 sin x cos x.
ż ecos x sin(2x)dx = 2 ż ecos x cos x sin x dx Now we can use the substitution w = cos x, dw = ´ sin xdx.
= ´2 ż wewdw From here the integral should look more familiar. We can use integration by parts with u = w, dv = ewdw, du = dw, and v = ew.
= ´2 wew lo omo on uv ´ ż ewdw lo omo on vdu = 2ew [1 ´ w] + C = 2ecos x[1 ´ cos x] + C S-24: We’ve got an integrand that consists of several functions multiplied together, and no obvious substitution. So, we think about integration by parts. We know an antiderivative for 1 (1´x)2, because we know d dx 1 1´x = 1 (1´x)2. So let’s try dv = dx (1´x)2 and u = xe´x. Then v = 1 1´x and du = (1 ´ x)e´x dx. So, by integration by parts, ż xe´x lo omo on u dx (1 ´ x)2 looomooon dv = xe´x 1 ´ x lo omo on uv ´ ż 1 1 ´ x lo omo on v (1 ´ x)e´x dx looooooomooooooon du = xe´x 1 ´ x ´ ż e´x dx = xe´x 1 ´ x + e´x + C = e´x 1 ´ x + C S-25: (a) The “parts” in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of sinn´1 x, although not exactly. So, let’s integrate by parts with u = sinn´1 x and dv = sin x dx, so that du = (n ´ 1) sinn´2 x cos x and v = ´ cos x.
ż sinn x dx = ´ sinn´1 x cos x looooooooomooooooooon uv + (n ´ 1) ż cos2 x sinn´2 x dx looooooooooooooooomooooooooooooooooon ´ ş vdu 351 Using the identity sin2 x + cos2 x = 1, = ´ sinn´1 x cos x + (n ´ 1) ż (1 ´ sin2 x) sinn´2 x dx = ´ sinn´1 x cos x + (n ´ 1) ż sinn´2 x dx ´ (n ´ 1) ż sinn x dx Moving the last term on the right hand side to the left hand side gives n ż sinn x dx = ´ sinn´1 x cos x + (n ´ 1) ż sinn´2 x dx Dividing across by n gives the desired reduction formula.
(b) By the reduction formula of part (a), if n ě 2, ż π/2 0 sinn(x) dx = n ´ 1 n ż π/2 0 sinn´2(x) dx since sin 0 = cos π 2 = 0. Applying this reduction formula, with n = 8, 6, 4, 2: ż π/2 0 sin8(x) dx = 7 8 ż π/2 0 sin6(x) dx = 7 8 ¨ 5 6 ż π/2 0 sin4(x) dx = 7 8 ¨ 5 6 ¨ 3 4 ż π/2 0 sin2(x) dx = 7 8 ¨ 5 6 ¨ 3 4 ¨ 1 2 ż π/2 0 dx = 7 8 ¨ 5 6 ¨ 3 4 ¨ 1 2 ¨ π 2 = 35 256π Using a calculator, we see this is approximately 0.4295.
S-26: (a) The sketch is the figure on the left below. By integration by parts with u = arctan x, dv = dx, v = x and du = 1 1+x2 dx, and then the substitution s = 1 + x2, A = ż 1 0 arctan x dx = x arctan x loooomoooon uv ˇ ˇ ˇ 1 0 ´ ż 1 0 x 1 + x2 dx loooomoooon vdu = arctan 1 ´ 1 2 log(1 + x2) ˇ ˇ ˇ 1 0 = π 4 ´ log 2 2 x = 1 y = tan−1 x x y x = 1 x = tan y x y (b) We’ll use horizontal washers as in Example 1.6.5 of the CLP-2 text.
• We cut R into thin horizontal strips of width dy as in the figure on the right above.
• When we rotate R about the y–axis, each strip sweeps out a thin washer 352 – whose inner radius is rin = tan y and outer radius is rout = 1, and – whose thickness is dy and hence – whose volume π(r2 out ´ r2 in)dy = π(1 ´ tan2 y)dy.
• As our bottommost strip is at y = 0 and our topmost strip is at y = π 4 (since at the top x = 1 and x = tan y), the total Volume = ż π/4 0 π(1 ´ tan2 y) dy = ż π/4 0 π(2 ´ sec2 y) dy = π 2y ´ tan y π/4 0 = π2 2 ´ π S-27: For a fixed value of x, if we rotate about the x-axis, we form a washer of inner radius B(x) and outer radius T(x) and hence of area π[T(x)2 ´ B(x)2]. We integrate this function from x = 0 to x = 3 to find the total volume V: V = ż 3 0 π[T(x)2 ´ B(x)2] dx = π ż 3 0 (?xe3x)2 ´ (?x(1 + 2x))2 dx = π ż 3 0 xe6x ´ (x + 4x2 + 4x3) dx = π ż 3 0 xe6x dx ´ π hx2 2 + 4x3 3 + x4i3 0 = π ż 3 0 xe6x dx ´ π h32 2 + 4 ¨ 33 3 + 34i For the first integral, we use integration by parts with u(x) = x, dv = e6xdx, so that du = dx and v(x) = 1 6e6x: ż 3 0 xe6x dx = xe6x 6 lo omo on uv ˇ ˇ ˇ ˇ 3 0 ´ ż 3 0 1 6e6x dx looomooon vdu = 3e18 6 ´ 0 ´ 1 36e6x ˇ ˇ ˇ ˇ 3 0 = e18 2 ´ e18 36 ´ 1 36 .
Therefore, the total volume is V = π e18 2 ´ e18 36 ´ 1 36 ´ π 32 2 + 4 ¨ 33 3 + 34 = π 17e18 ´ 4373 36 .
353 S-28: To get rid of the square root in the argument of f 2, we make the change of variables (also called “substitution”) x = t2, dx = 2t dt.
ż 4 0 f 2 ?x dx = 2 ż 2 0 t f 2(t) dt Then, to convert f 2 into f 1, we integrate by parts with u = t, dv = f 2(t) dt, v = f 1(t).
ż 4 0 f 2 ?x dx = 2 "h t f 1(t) lo omo on uv i2 0 ´ ż 2 0 f 1(t) dt loomoon vdu = 2 h t f 1(t) ´ f (t) i2 0 = 2 2f 1(2) ´ f (2) + f (0) = 2 2 ˆ 4 ´ 3 + 1 = 12 S-29: As we saw in Section 1.1 of the CLP-2 text, there are many different ways to interpret a limit as a Riemann sum. In the absence of instructions that restrain our choices, we go with the most convenient interpretations.
With that in mind, we choose: • that our Riemann sum is a right Riemann sum (because we see i, not i ´ 1 or i ´ 1 2) • ∆x = 2 n (because it is multiplied by the rest of the integrand, and also shows up multiplied by i), • then xi = a + i∆x = 2 ni ´ 1, which leads us to a = ´1 and • f (x) = xex.
• Finally, since ∆x = b´a n = 2 n and a = ´1, we have b = 1.
So, the limit is equal to the definite integral lim nÑ8 n ÿ i=1 2 n 2 ni ´ 1 e 2 ni´1 = ż 1 ´1 xex dx which we evaluate using integration by parts with u = x, dv = exdx, du = dx, and v = ex.
= h xex lo omo on uv i1 ´1 ´ ż 1 ´1 exdx lo omo on vdu = e + 1 e ´ e ´ 1 e = 2 e Solutions to Exercises 1.8 — Jump to TABLE OF CONTENTS 354 S-1: If u = cos x, then du = ´ sin x dx. If n ‰ ´1, then ż π/4 0 sin x cosn x dx = ´ ż 1/ ?
2 1 undu = ´ 1 n + 1un+1 1/ ?
2 1 = 1 n + 1 1 ´ 1 ?
2 n+1 !
If n = ´1, then ż π/4 0 sin x cosn x dx = ´ ż 1/ ?
2 1 undu = ´ ż 1/ ?
2 1 1 udu = ´ log |u| 1/ ?
2 1 = ´ log 1 ?
2 = 1 2 log 2 So, (e) n can be any real number.
S-2: We use the substitution u = sec x, du = sec x tan x dx.
ż secn x tan xdx = ż secn´1 x ¨ sec x tan x dx = ż un´1du Since n is positive, n ´ 1 ‰ ´1, so we antidifferentiate using the power rule.
= un n + C = 1 n secn x + C S-3: We divide both sides by cos2 x, and simplify.
sin2 x + cos2 x = 1 sin2 x + cos2 x cos2 x = 1 cos2 x sin2 x cos2 x + 1 = sec2 x tan2 x + 1 = sec2 x S-4: The power of cosine is odd, and the power of sine is even (zero). Following the strategy in the text, we make the substitution u = sin x, so that du = cos x dx and cos2 x = 1 ´ sin2 x = 1 ´ u2: ż cos3 x dx = ż (1 ´ sin2 x) cos x dx = ż (1 ´ u2) du = u ´ u3 3 + C = sin x ´ sin3 x 3 + C 355 S-5: Using the trig identity cos2 x = 1 + cos(2x) 2 , we have ż cos2 xdx = 1 2 ż π 0 1 + cos(2x) dx= 1 2 h x + 1 2 sin(2x) iπ 0 = π 2 S-6: Since the power of cosine is odd, following the strategies in the text, we make the substitution u = sin t, so that du = cos t dt and cos2 t = 1 ´ sin2 t = 1 ´ u2.
ż sin36 t cos3 t dt = ż sin36 t (1 ´ sin2 t) cos t dt = ż u36(1 ´ u2) du = u37 37 ´ u39 39 + C = sin37 t 37 ´ sin39 t 39 + C S-7: Since the power of sine is odd (and positive), we can reserve one sine for du, and turn the rest into cosines using the identity sin2 + cos2 x = 1. This allows us to use the substitution u = cos x, du = ´ sin x dx, and sin2 x = 1 ´ cos2 x = 1 ´ u2.
ż sin3 x cos4 x dx = ż sin2 x cos4 x sin x dx = ż ´1 ´ u2 u4 du = ż ´ 1 u4 + 1 u2 du = 1 3u3 ´ 1 u + C = 1 3 cos3 x ´ 1 cos x + C S-8: Both sine and cosine have even powers (four and zero, respectively), so we don’t have the option of using a substitution like u = sin x or u = cos x. Instead, we use the identity sin2 θ = 1 ´ cos(2θ) 2 .
ż π/3 0 sin4 x dx = ż π/3 0 sin2 x 2 dx = ż π/3 0 1 ´ cos(2x) 2 2 dx = 1 4 ż π/3 0 1 ´ 2 cos(2x) + cos2(2x) dx = 1 4 ż π/3 0 (1 ´ 2 cos(2x)) dx + 1 4 ż π/3 0 cos2(2x) dx 356 We can antidifferentiate the first integral right away. For the second integral, we use the identity cos2 θ = 1 + cos(2θ) 2 , with θ = 2x.
= 1 4 h x ´ sin(2x) iπ/3 0 + 1 8 ż π/3 0 (1 + cos(4x)) dx = 1 4 π 3 ´ ?
3 2 + 1 8 h x + 1 4 sin(4x) iπ/3 0 = 1 4 π 3 ´ ?
3 2 + 1 8 π 3 ´ ?
3 8 = π 8 ´ 9 ?
3 64 S-9: Since the power of sine is odd, we can reserve one sine for du, and change the remaining four into cosines. This sets us up to use the substitution u = cos x, du = ´ sin x dx.
ż sin5 x dx = ż sin4 x ¨ sin x dx = ż (1 ´ cos2 x)2 sin x dx = ´ ż (1 ´ u2)2 du = ´ ż (1 ´ 2u2 + u4)du = ´u + 2 3u3 ´ 1 5u5 + C = ´ cos x + 2 3 cos3 x ´ 1 5 cos5 x + C S-10: If we use the substitution u = sin x, then du = cos x dx, which very conveniently shows up in the integrand.
ż sin1.2 x cos x dx = ż u1.2du = u2.2 2.2 + C = 1 2.2 sin2.2 x + C Note this is exactly the strategy described in the text when the power of cosine is odd.
The non-integer power of sine doesn’t cause a problem.
S-11: Solution 1: Let’s use the substitution u = tan x, du = sec2 x dx: ż tan x sec2 xdx = ż u du = 1 2u2 + C = 1 2 tan2 x + C Solution 2: We can also use the substitution u = sec x, du = sec x tan x dx: ż tan x sec2 xdx = ż u du = 1 2u2 + C = 1 2 sec2 x + C 357 We note that because tan2 x and sec2 x only differ by a constant, the two answers are equivalent.
S-12: Solution 1: Substituting u = cos x, du = ´ sin x dx, sin2 x = 1 ´ cos2 x = 1 ´ u2, gives ż tan3 x sec5 x dx = ż sin3 x cos8 x dx = ż (1 ´ cos2 x) sin x cos8 x dx = ´ ż 1 ´ u2 u8 du = ´ hu´7 ´7 ´ u´5 ´5 i + C = 1 7 sec7 x ´ 1 5 sec5 x + C Solution 2: Alternatively, substituting u = sec x, du = sec x tan x dx, tan2 x = sec2 x ´ 1 = u2 ´ 1, gives ż tan3 x sec5 x dx = ż tan2 x sec4 x (tan x sec x) dx = ż (u2 ´ 1)u4 du = hu7 7 ´ u5 5 i + C = 1 7 sec7 x ´ 1 5 sec5 x + C S-13: Use the substitution u = tan x, so that du = sec2 x dx: ż sec4 x tan46 x dx = ż (tan2 x + 1) tan46 x sec2 x dx = ż (u2 + 1)u46 du = u49 49 + u47 47 + C = tan49 x 49 + tan47 x 47 + C S-14: We use the substitution u = sec x, du = sec x tan x dx. Then tan2 x = sec2 x ´ 1 = u2 ´ 1.
ż tan3 x sec1.5 x dx = ż tan2 x ¨ sec0.5 x ¨ sec x tan xdx = ż (u2 ´ 1)u0.5 du = ż u2.5 ´ u0.5 du = u3.5 3.5 ´ u1.5 1.5 + C = 1 3.5 sec3.5 x ´ 1 1.5 sec1.5 x + C Note this solution used the same method as Example 1.8.13 in the CLP-2 text for the case that the power of tangent is odd and there is at least one secant.
S-15: We’ll give two solutions.
358 Solution 1: As in Question 14, we have an odd power of tangent and at least one secant.
So, as in strategy (2) of Section 1.8.2 in the CLP-2 text, we can use the substitution u = sec x, du = sec x tan x dx, and tan2 x = sec2 x ´ 1 = u2 ´ 1.
ż tan3 x sec2 x dx = ż tan2 x sec x ¨ sec x tan x dx = ż (u2 ´ 1)u du = ż u3 ´ u du = 1 4u4 ´ 1 2u2 + C = 1 4 sec4 x ´ 1 2 sec2 x + C Solution 2: We have an even, strictly positve, power of sec x. So, as in strategy (3) of Section 1.8.2 in the CLP-2 text, we can use the substitution u = tan x, du = sec2 x dx.
ż tan3 x sec2 x dx = ż tan3 x ¨ sec2 x dx = ż u3 du = 1 4u4 + C = 1 4 tan4 x + C It looks like we have two different answers. But, because tan2 x = sec2 x ´ 1, 1 4 tan4 = 1 4(sec2 x ´ 1) 2 = 1 4 sec4 x ´ 1 2 sec2 x + 1 4 and the two answers are really the same, except that the arbitrary constant C of Solution 1 is 1 4 plus the arbitrary constant C of Solution 2.
S-16: In contrast to Questions 14 and 15, we do not have an odd power of tangent, so we should consider a different substitution. Luckily, if we choose u = tan x, then du = sec2 x dx, and this fits our integrand nicely.
ż tan4 x sec2 x dx = ż u4 du = 1 5u5 + C = 1 5 tan5 x + C S-17: Solution 1: Since the power of tangent is odd, let’s try to use the substitution u = sec x, du = sec x tan x dx, and tan2 x = sec2 x ´ 1 = u2 ´ 1, as in Questions 14 and 15. In 359 order to make this work, we need to see sec x tan x dx in the integrand, so we do a little algebraic manipulation.
ż tan3 x sec´0.7 x dx = ż tan3 x sec0.7x dx = ż tan3 x sec1.7x sec x dx = ż tan2 x sec1.7 x ¨ sec x tan x dx = ż u2 ´ 1 u1.7 du = ż u0.3 ´ u´1.7 du = u1.3 1.3 + 1 0.7u0.7 + C = 1 1.3 sec1.3 x + 1 0.7 sec0.7 x + C = 1 1.3 sec1.3 x + 1 0.7 cos0.7 x + C Solution 2: Let’s convert the secants and tangents to sines and cosines.
ż tan3 x sec´0.7 x dx = ż sin3 x cos3 x ¨ cos0.7 x dx = ż sin3 x cos2.3 x dx = ż sin2 x cos2.3 x ¨ sin x dx Using the substitution u = cos x, du = ´ sin dx, and sin2 x = 1 ´ cos2 x = 1 ´ u2: = ´ ż 1 ´ u2 u2.3 du = ż ´u´2.3 + u´0.3 du = 1 1.3u´1.3 + 1 0.7u0.7 + C = 1 1.3 sec1.3 x + 1 0.7 cos0.7 x + C S-18: We replace tan x with sin x cos x.
ż tan5 x dx = ż sin x cos x 5 dx = ż sin4 x cos5 x ¨ sin x dx Now we use the substitution u = cos x, du = ´ sin x dx, and sin2 x = 1 ´ cos2 x = 1 ´ u2.
= ´ ż (1 ´ u2)2 u5 du = ż ´u´5 + 2u´3 ´ u´1 du = 1 4u´4 ´ u´2 ´ log |u| + C = 1 4 sec4 x ´ sec2 x ´ log | cos x| + C = 1 4 sec4 x ´ sec2 x + log | sec x| + C 360 where in the last line, we used the logarithm rule log(ba) = a log b, with ba = cos x = (sec x)´1.
S-19: Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. For even powers, we use the identity tan2 x = sec2 x ´ 1, then use the substitution u = tan x, du = sec2 x dx on (perhaps only a part of) the resulting integral.
ż π/6 0 tan6 x dx = ż π/6 0 tan4 x(sec2 x ´ 1) dx = ż π/6 0 tan4 x sec2 x loooooomoooooon u4 du ´ tan4 x dx = ż π/6 0 tan4 x sec2 x ´ tan2 x(sec2 x ´ 1) dx = ż π/6 0 tan4 x sec2 x ´ tan2 x sec2 x loooooomoooooon u2 du + tan2 x dx = ż π/6 0 tan4 x sec2 x ´ tan2 x sec2 x + (sec2 x lo omo on du ´1) dx = ż π/6 0 tan4 x ´ tan2 x + 1 sec2 x dx ´ ż π/6 0 1dx Note tan(0) = 0, and tan(π/6) = 1/ ?
3.
= ż 1/ ?
3 0 (u4 ´ u2 + 1) du ´ x π/6 0 = 1 5u5 ´ 1 3u3 + u 1/ ?
3 0 ´ π 6 = 1 5 ?
3 5 ´ 1 3 ?
3 3 + 1 ?
3 ´ π 6 = 41 45 ?
3 ´ π 6 S-20: Since there is an even power of secant in the integrand, we can reserve two secants for du and change the rest to tangents. That sets us up nicely to use the substitution 361 u = tan x, du = sec2 x dx. Note tan(0) = 0 and tan(π/4) = 1.
ż π/4 0 tan8 x sec4 x dx = ż π/4 0 tan8 x (tan2 x + 1) sec2 x dx = ż 1 0 u8 (u2 + 1) du = ż 1 0 u10 + u8 du = 1 11 + 1 9 S-21: Solution 1: Let’s use the substitution u = sec x, du = sec x tan x dx. In order to make this work, we need to see sec x tan x in the integrand, so we start with some algebraic manipulation.
ż tan x?sec x ?sec x ?sec x dx = ż 1 ?sec x sec x tan x dx = ż 1 ?u du = 2?u + C = 2?sec x + C Solution 2: Let’s turn our secants and tangents into sines and cosines.
ż tan x?sec x dx = ż sin x cos x ¨ ?cos x dx = ż sin x cos1.5 x dx We use the substitution u = cos x, du = ´ sin x dx.
= ż ´u´1.5 du = 2 ?u + C = 2?sec x + C S-22: Since the power of secant is even and positive, we can reserve two secants for du, and change the rest into tangents, setting the stage for the substitution u = tan θ, 362 du = sec2 θ dθ.
ż sec8 θ tane θ dθ = ż sec6 θ tane θ sec2 θ dθ = ż (tan2 θ + 1)3 tane θ sec2 θ dθ = ż (u2 + 1)3 ¨ ue du = ż (u6 + 3u4 + 3u2 + 1) ¨ ue du = ż (u6+e + 3u4+e + 3u2+e + ue) du = 1 7 + eu7+e + 3 5 + eu5+e + 3 3 + eu3+e + 1 1 + eu1+e + C = 1 7 + e tan7+e θ + 3 5 + e tan5+e θ + 3 3 + e tan3+e θ + 1 1 + e tan1+e θ + C = tan1+e θ tan6 θ 7 + e + 3 tan4 θ 5 + e + 3 tan2 θ 3 + e + 1 1 + e + C S-23: (a) Using the trig identity tan2 x = sec2 x ´ 1 and the substitution y = tan x, dy = sec2 x dx, ż tann x dx = ż tann´2 x tan2 x dx = ż tann´2 x sec2 x dx ´ ż tann´2 x dx = ż yn´2 dy ´ ż tann´2 x dx = yn´1 n ´ 1 ´ ż tann´2 x dx = tann´1 x n ´ 1 ´ ż tann´2 x dx (b) By the reduction formula of part (a), ż π/4 0 tann(x) dx = tann´1 x n ´ 1 π/4 0 ´ ż π/4 0 tann´2(x) dx = 1 n ´ 1 ´ ż π/4 0 tann´2(x) dx for all integers n ě 2, since tan 0 = 0 and tan π 4 = 1. We apply this reduction formula, with n = 6, 4, 2.
ż π/4 0 tan6(x) dx = 1 5 ´ ż π/4 0 tan4(x) dx = 1 5 ´ 1 3 + ż π/4 0 tan2(x) dx = 1 5 ´ 1 3 + 1 ´ ż π/4 0 dx = 1 5 ´ 1 3 + 1 ´ π 4 = 13 15 ´ π 4 Using a calculator, we see this is approximately 0.0813.
Notice how much faster this was than the method of Question 19.
363 S-24: Recall tan x = sin x cos x.
ż tan5 x cos2 x dx = ż sin5 x cos5 x cos2 x dx = ż sin5 x cos3 x dx Substitute u = cos x, so du = ´ sin x dx and sin2 x = 1 ´ cos2 x = 1 ´ u2.
= ż sin4 x cos3 x sin x dx = ´ ż (1 ´ u2)2 u3 du = ´ ż 1 ´ 2u2 + u4 u3 du = ż ´ 1 u3 + 2 u ´ u du = 1 2u2 + 2 log |u| ´ 1 2u2 + C = 1 2 cos2 x + 2 log | cos x| ´ 1 2 cos2 x + C S-25: We can use the definition of secant to make this integral look more familiar.
ż 1 cos2 θ dθ = ż sec2 θ dθ = tan θ + C S-26: We re-write cot x = cos x sin x , and use the substitution u = sin x, du = cos x dx.
ż cot x dx = ż cos x sin x dx = ż 1 u du = log |u| + C = log | sin x| + C S-27: Solution 1: We begin with the obvious substitution, w = ex, dw = exdw.
ż ex sin(ex) cos(ex) dx = ż sin w cos w dw Now we see another substitution, u = sin w, du = cos w dw.
= ż u du = 1 2u2 + C = 1 2 sin2 w + C = 1 2 sin2(ex) + C Solution 2: Notice that d dxtsin(ex)u = ex cos(ex). This suggests to us the substitution u = sin(ex), du = ex cos(ex) dx.
ż ex sin(ex) cos(ex) dx = ż u du = 1 2u2 + C = 1 2 sin2(ex) + C 364 S-28: Since we have an “inside function,” we start with the substitution s = cos x, so ´ds = sin x dx and sin2 x = 1 ´ cos2 x = 1 ´ s2.
ż sin(cos x) sin3 x dx = ż sin(cos x) ¨ sin2 x ¨ sin xdx = ´ ż sin(s) ¨ (1 ´ s2) ds We use integration by parts with u = (1 ´ s2), dv = sin s ds; du = ´2s ds, and v = ´ cos s.
= ´ ´(1 ´ s2) cos s ´ ż 2s cos s ds = (1 ´ s2) cos s + ż 2s cos s ds We integrate by parts again, with u = 2s, dv = cos s ds; du = 2 ds, and v = sin s.
= (1 ´ s2) cos s + 2s sin s ´ ż 2 sin s ds = (1 ´ s2) cos s + 2s sin s + 2 cos s + C = sin2 x ¨ cos(cos x) + 2 cos x ¨ sin(cos x) + 2 cos(cos x) + C = (sin2 x + 2) cos(cos x) + 2 cos x ¨ sin(cos x) + C S-29: Since the integrand is the product of polynomial and trigonometric functions, we suspect it might yield to integration by parts. There are a number of ways this can be accomplished.
Solution 1: Before we choose parts, let’s use the identity sin(2x) = 2 sin x cos x.
ż x sin x cos xdx = 1 2 ż x sin(2x)dx Now let u = x, dv = sin(2x)dx; du = dx, and v = ´1 2 cos(2x). Using integration by parts: = 1 2 ´x 2 cos(2x) + 1 2 ż cos(2x)dx = ´x 4 cos(2x) + 1 8 sin(2x) + C = ´x 4(1 ´ 2 sin2 x) + 1 4 sin x cos x + C = ´x 4 + x 2 sin2 x + 1 4 sin x cos x + C 365 Solution 2: If we let u = x, then du = dx, and this seems desirable for integration by parts. If u = x, then dv = sin x cos xdx. To find v we can use the substitution u = sin x, du = cos xdx.
v = ż sin x cos xdx = ż udu = 1 2u2 + C = 1 2 sin2 x + C So, we take v = 1 2 sin2 x. Now we can apply integration by parts to our original integral.
ż x sin x cos x dx = x 2 sin2 x ´ ż 1 2 sin2 xdx Apply the identity sin2 x = 1 ´ cos(2x) 2 .
= x 2 sin2 x ´ 1 4 ż 1 ´ cos(2x)dx = x 2 sin2 x ´ x 4 + 1 8 sin(2x) + C = x 2 sin2 x ´ x 4 + 1 4 sin x cos x + C Solution 3: Let u = x sin x and dv = cos xdx; then du = (x cos x + sin x)dx and v = sin x.ż x sin x cos xdx = x sin2 x ´ ż sin x(x cos x + sin x)dx = x sin2 x ´ ż x sin x cos xdx ´ ż sin2 xdx Apply the identity sin2 x = 1 ´ cos(2x) 2 to the second integral.
= x sin2 x ´ ż x sin x cos xdx ´ ż 1 ´ cos(2x) 2 dx = x sin2 x ´ ż x sin x cos xdx ´ x 2 + 1 4 sin(2x) + C So, we have the equation ż x sin x cos xdx = x sin2 x ´ ż x sin x cos xdx ´ x 2 + 1 4 sin(2x) + C 2 ż x sin x cos xdx = x sin2 x ´ x 2 + 1 4 sin(2x) + C ż x sin x cos xdx = x 2 sin2 x ´ x 4 + 1 8 sin(2x) + C 2 = x 2 sin2 x ´ x 4 + 1 4 sin x cos x + C 2 Since C is an arbitrary constant that can take any number in (´8, 8), also C 2 is an arbitrary constant that can take any number in (´8, 8), so we’re free to rename C 2 to C.
366 Solutions to Exercises 1.9 — Jump to TABLE OF CONTENTS S-1: In the text, there is a template for choosing an appropriate substitution, but for this problem we will explain the logic of the choices.
The trig identities that we can use are: 1 ´ sin2 θ = cos2 θ tan2 θ + 1 = sec2 θ sec2 θ ´ 1 = tan2 θ They have the following forms: constant ´ function function + constant function ´ constant In order to cancel out the square root, we should choose a substitution that will match the argument under the square root with the trig identity of the corresponding form.
(a) There’s not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function 9x2 ´ 16, which has the form (function) ´ (constant). This form matches the trig identity sec2 θ ´ 1 = tan2 θ. We can set x to be whatever we need it to be, but we don’t have the same control over the constant, 16. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 16, we get 16 sec2 θ ´ 16 = 16 tan2 θ What we want is a substitution that gives us 9x2 ´ 16 = 16 sec2 θ ´ 16 So, 9x2 = 16 sec2 θ x = 4 3 sec θ Using this substitution, a 9x2 ´ 16 = a 16 sec2 θ ´ 16 = a 16 tan2 θ = 4| tan θ| So, we eliminated the square root.
(b) There’s not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function 1 ´ 4x2, which has the form (constant) ´ (function). This form matches the trig identity 1 ´ sin2 θ = cos2 θ. What we want is a substitution that gives us 1 ´ 4x2 = 1 ´ sin2 θ So, 4x2 = sin2 θ x = 1 2 sin θ 367 Using this substitution, a 1 ´ 4x2 = a 1 ´ sin2 θ = a cos2 θ = | cos θ| So, we eliminated the square root. We remark that the absolute value signs are not needed in | cos θ|, because, for ´1 2 ď x ď 1 2, we have θ = arcsin(2x) between ´π 2 and π 2 , and cos(θ) ě 0 for those θ’s.
(c) There’s not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the fractional power. (That is, we want to eliminate the square root.) The function under the power is 25 + x2, which has the form (constant) + (function). This form matches the trig identity tan2 θ + 1 = sec2 θ. We can set x to be whatever we need it to be, but we don’t have the same control over the constant, 25. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 25, we get 25 tan2 θ + 25 = 25 sec2 θ What we want is a substitution that gives us 25 + x2 = 25 tan2 θ + 25 So, x2 = 25 tan2 θ x = 5 tan θ Using this substitution, (25 + x2)´5/2 = (25 + 25 tan2 θ)´5/2 = (25 sec2 θ)´5/2 = (5| sec θ|)´5 So, we eliminated the square root. We remark that the absolute value signs are not needed in | sec θ|, because, for ´8 ă x ă 8, we have θ = arctan(x/5) between ´π 2 and π 2 , and sec(θ) ě 0 for those θ’s.
S-2: Just as in Question 1, we want to choose a trigonometric substitution that will allow us to eliminate the square roots. Before we can make that choice, though, we need to complete the square. In subsequent problems, we won’t show the algebra behind completing the square, but for this problem we’ll work it out explicitly. After some practice, you’ll be able to do this step in your head for many cases.
After the squares are completed, the choice of trig substitution follows the logic outlined in the solutions to Question 1, or (equivalently) the template in the text.
(a) The quadratic function under the square root is x2 ´ 4x + 1. To complete the square, we match the non-constant terms to those of a perfect square.
(ax + b)2 = a2x2 + 2abx + b2 x2 ´ 4x + 1 = a2x2 + 2abx + b2 + c for some constant c 368 • Looking at the leading term tells us a = 1.
• Then the second term tells us ´4 = 2ab = 2b, so b = ´2.
• Finally, the constant terms give us 1 = b2 + c = 4 + c, so c = ´3.
ż 1 ?
x2 ´ 4x + 1 dx = ż 1 a (x ´ 2)2 ´ 3 dx = ż 1 b x ´ 2 2 ´ ?
3 2 dx So we use the substitution (x ´ 2) = ?
3 sec u, which eliminates the square root: b (x ´ 2)2 ´ 3 = a 3 sec2 u ´ 3 = a 3 tan2 u = ?
3| tan u| (b) The quadratic function under the square root is ´x2 + 2x + 4 = ´[x2 ´ 2x ´ 4]. To complete the square, we match the non-constant terms to those of a perfect square.
We factored out the negative to make things a little easier–don’t forget to put it back in before choosing a substitution!
(ax + b)2 = a2x2 + 2abx + b2 x2 ´ 2x ´ 4 = a2x2 + 2abx + b2 + c for some constant c • Looking at the leading term tells us a = 1.
• Then the second term tells us ´2 = 2ab = 2b, so b = ´1.
• Finally, the constant terms give us ´4 = b2 + c = 1 + c, so c = ´5.
• Then ´x2 + 2x + 4 = ´[x2 ´ 2x ´ 4] = ´[(x ´ 1)2 ´ 5] = 5 ´ (x ´ 1)2.
ż (x ´ 1)6 (´x2 + 2x + 4)3/2 dx = ż (x ´ 1)6 (5 ´ (x ´ 1)2)3/2 dx = ż (x ´ 1)6 ?
5 2 ´ x ´ 1 23/2 dx So we use the substitution (x ´ 1) = ?
5 sin u, which eliminates the square root (fractional power): (5 ´ (x ´ 1)2)3/2 = 5 ´ 5 sin2 u 3/2 = 5 cos2 u 3/2 = 5 ?
5| cos3 u| (c) The quadratic function under the square root is 4x2 + 6x + 10. To complete the square, we match the non-constant terms to those of a perfect square.
(ax + b)2 = a2x2 + 2abx + b2 4x2 + 6x + 10 = a2x2 + 2abx + b2 + c for some constant c • Looking at the leading term tells us a = 2.
• Then the second term tells us 6 = 2ab = 4b, so b = 3 2.
369 • Finally, the constant terms give us 10 = b2 + c = 9 4 + c, so c = 31 4 .
ż 1 ?
4x2 + 6x + 10 dx = ż 1 b 2x + 3 2 2 + 31 4 dx = ż 1 c 2x + 3 2 2 + ?
31 2 2 dx So we use the substitution 2x + 3 2 = ?
31 2 tan u, which eliminates the square root: d 2x + 3 2 2 + 31 4 = c 31 4 tan2 u + 31 4 = c 31 4 sec2 u = ?
31 2 | sec u| (d) The quadratic function under the square root is x2 ´ x. To complete the square, we match the non-constant terms to those of a perfect square.
(ax + b)2 = a2x2 + 2abx + b2 x2 ´ x = a2x2 + 2abx + b2 + c for some constant c • Looking at the leading term tells us a = 1.
• Then the second term tells us ´1 = 2ab = 2b, so b = ´1 2.
• Finally, the constant terms give us 0 = b2 + c = 1 4 + c, so c = ´1 4.
ż a x2 ´ x dx = ż d x ´ 1 2 2 ´ 1 4 dx = ż d x ´ 1 2 2 ´ 1 2 2 dx So we use the substitution (x ´ 1/2) = 1 2 sec u, which eliminates the square root: d x ´ 1 2 2 ´ 1 4 = c 1 4sec 2u ´ 1 4 = c 1 4 tan2 u = 1 2| tan u| S-3: (a) If sin θ = 1 20 and θ is between 0 and π/2, then we can draw a right triangle with angle θ that has opposite side length 1, and hypotenuse length 20. By the Pythagorean Theorem, the adjacent side has length ?
202 ´ 12 = ?
399. So, cos θ = adj hyp = ?
399 20 .
θ ?
399 1 20 370 We can do a quick “reasonableness” check here: 1 20 is pretty close to 0, so we might expect θ to be pretty close to 0, and so cos θ should be pretty close to 1. Indeed it is: ?
399 20 « ?
400 20 = 20 20 = 1.
Alternately, we can solve this problem using identities.
sin2 θ + cos2 θ = 1 1 20 2 + cos2 θ = 1 cos θ = ˘ c 1 ´ 1 400 = ˘ ?
399 20 Since 0 ď θ ď π 2 , cos θ ě 0, so cos θ = ?
399 20 (b) If tan θ = 7 and θ is between 0 and π/2, then we can draw a right triangle with angle θ that has opposite side length 7 and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length ?
72 + 12 = ?
50 = 5 ?
2. So, csc θ = hyp opp = 5 ?
2 7 .
θ 1 7 5 ?
2 Again, we can do a quick reasonableness check. Since 7 is much larger than 1, the triangle we’re thinking of doesn’t look much like the triangle in our standardized picture above: it’s really quite tall, with a small base. So, the opposite side and hypotenuse are pretty close in length. Indeed, 5 ?
2 7 « 7.071, so this dimension seems reasonable.
(c) If sec θ = ?
x ´ 1 2 and θ is between 0 and π/2, then we can draw a right triangle with angle θ that has hypotenuse length ?
x ´ 1 and adjacent side length 2. By the Pythagorean Theorem, the opposite side has length b?
x ´ 12 ´ 22 = ?
x ´ 1 ´ 4 = ?
x ´ 5. So, tan θ = opp adj = ?
x ´ 5 2 .
θ 2 ?
x ´ 5 ?
x ´ 1 371 We can also solve this using identities. Note that since sec θ exists, θ ‰ π 2 .
tan2 θ + 1 = sec2 θ tan2 θ + 1 = ?
x ´ 1 2 2 = x ´ 1 4 tan θ = ˘ c x ´ 1 4 ´ 1 = ˘ ?
x ´ 5 2 Since 0 ď θ ă π 2 , tan θ ě 0, so tan θ = ?
x ´ 5 2 S-4: (a) Let θ = arccos x 2 . That is, cos(θ) = x 2, and 0 ď θ ď π. Then we can draw the corresponding right triangle with angle θ with adjacent side of signed length x (we note that if θ ą π 2 , then x is negative) and hypotenuse of length 2. By the Pythagorean Theorem, the opposite side of the triangle has length ?
4 ´ x2.
θ x ?
4 ´ x2 2 So, sin arccos x 2 = sin θ = opp hyp = ?
4 ´ x2 2 (b) Let θ = arctan 1 ?
3 . That is, tan(θ) = 1 ?
3, and ´π 2 ď θ ď π 2 .
Solution 1: Then θ = π 6 , so sin θ = 1 2.
Solution 2: Then we can draw the corresponding right triangle with angle θ with opposite side of length 1 and adjacent side of length ?
3. By the Pythagorean Theorem, the hypotenuse of the triangle has length b?
3 2 + 12 = 2.
θ ?
3 1 2 So, 372 sin arctan 1 ?
3 = sin θ = opp hyp = 1 2 (c) Let θ = arcsin (?x). That is, sin(θ) = ?x, and ´π 2 ď θ ď π 2 . Then we can draw the corresponding right triangle with angle θ with opposite side of length ?x and hypotenuse of length 1. By the Pythagorean Theorem, the adjacent side of the triangle has length ?
1 ´ x.
θ ?
1 ´ x ?x 1 So, sec arcsin ?x = sec θ = hyp adj = 1 ?
1 ´ x S-5: Let x = 2 tan θ, so that x2 + 4 = 4 tan2 θ + 4 = 4 sec2 θ and dx = 2 sec2 θ dθ. Then ż 1 (x2 + 4)3/2 dx = ż 1 (4 sec2 θ)3/2 ¨ 2 sec2 θ dθ = ż 2 sec2 θ 8 sec3 θ dθ = 1 4 ż cos θ dθ = 1 4 sin θ + C = 1 4 x ?
x2 + 4 + C θ 2 x ?
x2 + 4 To find sin θ in terms of x, we construct the right triangle above. Since tan θ = x 2 = opp adj , we label the opposite side x and the adjacent side 2. By the Pythagorean Theorem, the hypotenuse has length ?
x2 + 4. Then sin θ = opp hyp = x ?
x2 + 4 .
To see why we could write (sec2 θ)3/2 = sec3 θ, as opposed to (sec2 θ)3/2 = ˇ ˇ sec3 θ ˇ ˇ, in the second line above, see Example 1.9.5 in the CLP-2 text.
As a check, we observe that the derivative of the answer d dx 1 4 x ?
x2 + 4 + C = 1 4 1 ?
x2 + 4 ´ 1 2 ˆ 4 x(2x) x2 + 4 3/2 = x2 4 + 1 ´ x2 4 x2 + 4 3/2 = 1 x2 + 4 3/2 is exactly the integrand.
373 S-6: Solution 1: As in Question 5, substitute x = 2 tan u, dx = 2 sec2 u du. Note that when x = 4 we have 4 = 2 tan u, so that tan u = 2.
ż 4 0 1 (4 + x2)3/2 dx = ż arctan 2 0 1 (4 + 4 tan2 u)3/2 2 sec2 u du = ż arctan 2 0 2 sec2 u (2 sec u)3 du = 1 4 ż arctan 2 0 sec2 u sec3 u du = 1 4 ż arctan 2 0 cos u du u 1 2 ?
5 = 1 4 sin u arctan 2 0 = 1 4 sin(arctan 2) ´ 0 = 1 2 ?
5 To find sin(arctan 2), we use the right triangle above, with angle u = arctan 2. Since tan u = 2 = opp adj , we label the opposite side as 2, and the adjacent side as 1. The Pythagorean Theorem tells us the hypotenuse has length ?
5, so sin u = opp hyp = 2 ?
5.
Solution 2: Using our result from Question 5, ż 4 0 1 (4 + x2)3/2 dx = 1 4 x ?
x2 + 4 4 0 = 1 4 ¨ 4 ?
42 + 4 = 1 2 ?
5 S-7: Make the change of variables x = 5 sin θ, dx = 5 cos θ dθ. Since x = 0 corresponds to θ = 0 and x = 5 2 correponds to sin θ = 1 2 or θ = π 6 , ż 5/2 0 dx ?
25 ´ x2 = ż π/6 0 5 cos θ dθ a 25 ´ 25 sin2 θ = ż π/6 0 dθ = π 6 374 S-8: Substitute x = 5 tan u, so that dx = 5 sec2 u du.
ż 1 ?
x2 + 25 dx = ż 1 a 25 tan2 u + 25 5 sec2 u du = ż 5 sec2 u 5 sec u du = ż sec u du = log ˇ ˇ sec u + tan u ˇ ˇ + C u 5 x ?
x2 + 25 = log ˇ ˇ ˇ c 1 + x2 25 + x 5 ˇ ˇ ˇ + C To find sec u and tan u, we have two options. One is to set up a right triangle with angle u and tan u = x 5. Then we can label the opposite side x and the adjacent side 5, and use Pythagoras to find that the hypotenuse is ?
x2 + 25.
Another option is to look back at our work a little more closely–in fact, we’ve already found what we’re looking for. Since we used the substitution x = 5 tan u, this gives us tan u = x 5. In the denominator of the integrand, we simplified ?
x2 + 25 = 5 sec u, so sec u = 1 5 ?
x2 + 25 = b 1 + x2 25.
To see why we could write ?
x2 + 25 = 5 sec u, as opposed to ?
x2 + 25 = 5| sec u|, see Example 1.9.5 in the CLP-2 text.
S-9: The quadratic formula underneath the square root makes us think of a trig substitution, but in the interest of developing good habits, let’s check for an easier way first. If we let u = 2x2 + 4x, then du = (4x + 4) dx, so 1 4 du = (x + 1) dx. This substitution looks easier than a trig substitution (which would start with completing the square).
ż x + 1 ?
2x2 + 4x dx = 1 4 ż 1 ?u du = 1 2 ?u + C = 1 2 a 2x2 + 4x + C S-10: Substitute x = 4 tan u, dx = 4 sec2 u du.
ż 1 x2?
x2 + 16 dx = ż 1 16 tan2 u a 16 tan2 u + 16 4 sec2 u du = ż sec2 u 16 tan2 u sec u du = 1 16 ż sec u tan2 u du = 1 16 ż cos u sin2 u du To finish off the integral, we’ll substitute v = sin u, dv = cos u du.
ż 1 x2?
x2 + 16 dx = 1 16 ż cos u sin2 u du = 1 16 ż dv v2 = ´ 1 16v + C = ´ 1 16 sin u + C = ´ 1 16 ?
x2 + 16 x + C u 4 x ?
x2 + 16 375 To find sin u, we draw a right triangle with angle u and tan u = x 4. We label the opposite side x and the adjacent side 4, and then from Pythagoras we find that the hypotenuse has length ?
x2 + 16. So, sin u = ?
x2 + 16 x .
As a check, we observe that the derivative of the answer d dx ´ 1 16 ?
x2 + 16 x + C !
= 1 16 ?
x2 + 16 x2 ´ 1 16 x x ?
x2 + 16 = 1 16 (x2 + 16) ´ x2 x2?
x2 + 16 = 1 x2?
x2 + 16 is exactly the integrand.
S-11: Substitute x = 3 sec u with 0 ď u ă π 2 . Then dx = 3 sec u tan u du and ?
x2 ´ 9 = ?
9 sec2 u ´ 9 = ?
9 tan2 u = 3 tan u, so that ż dx x2?
x2 ´ 9 = ż 3 sec u tan u du 9 sec2 u ?
9 tan2 u = 1 9 ż du sec u = 1 9 ż cos u du = 1 9 sin u + C.
u 3 ?
x2 ´ 9 x To evaluate sin u, we make a right triangle with angle u. Since sec u = x 3 = hyp adj , we label the hypotenuse x and the adjacent side 3. Using the Pythagorean Theorem, the opposite side has length ?
x2 ´ 9. So, sin u = ?
x2 ´ 9 x and ż dx x2?
x2 ´ 9 = ?
x2 ´ 9 9x + C.
As a check, we observe that the derivative of the answer d dx ?
x2 ´ 9 9x + C !
= ´ ?
x2 ´ 9 9x2 + x 9x ?
x2 ´ 9 = 1 9 ´(x2 ´ 9) + x2 x2?
x2 ´ 9 = 1 x2?
x2 ´ 9 is exactly the integrand. (We remark that this is the case even for x ď ´3.) S-12: (a) We’ll use the trig identity cos 2θ = 2 cos2 θ ´ 1. It implies that cos2 θ = cos 2θ + 1 2 ù ñ cos4 θ = 1 4 cos2 2θ + 2 cos 2θ + 1 = 1 4 hcos 4θ + 1 2 + 2 cos 2θ + 1 i = cos 4θ 8 + cos 2θ 2 + 3 8 376 So, ż π/4 0 cos4 θ dθ = ż π/4 0 cos 4θ 8 + cos 2θ 2 + 3 8 dθ = sin 4θ 32 + sin 2θ 4 + 3 8θ π/4 0 = 1 4 + 3 8 ¨ π 4 = 8 + 3π 32 as required.
(b) We’ll use the trig substitution x = tan θ, dx = sec2 θ dθ. Note that when θ = ˘π 4 , we have x = ˘1. Also note that dividing the trig identity sin2 θ + cos2 θ = 1 by cos2 θ gives the trig identity tan2 θ + 1 = sec2 θ. So ż 1 ´1 dx (x2 + 1)3 = 2 ż 1 0 dx (x2 + 1)3 (even integrand) = 2 ż π/4 0 sec2 θ dθ (tan2 θ + 1)3 = 2 ż π/4 0 sec2 θ dθ (sec2 θ)3 = 2 ż π/4 0 cos4 θ dθ = 8 + 3π 16 by part (a).
S-13: The integrand is an odd function, and the limits of integration are symmetric, so ż π/12 ´π/12 15x3 (x2 + 1) ?
9 ´ x25 dx = 0.
S-14: Substitute x = 2 sin u, so that dx = 2 cos u du.
ż a 4 ´ x2 dx = ż a 4 ´ 4 sin2 u 2 cos u du = ż a 4 cos2 u 2 cos u du = ż 4 cos2 u du = 2 ż 1 + cos(2u) du = 2u + sin(2u) + C = 2u + 2 sin u cos u + C = 2 arcsin x 2 + x 2 a 4 ´ x2 + C u ?
4 ´ x2 x 2 377 To see why we could write ?
4 cos2 u = 2 cos u, as opposed to ?
4 cos2 u = 2| cos u|, in the third line above, see Example 1.9.2 in the CLP-2 text.
We used the substitution x = 2 sin u, so we know sin u = x 2 and u = arcsin( x 2). We have three options for finding cos u.
First, we can draw a right triangle with angle u. Since sin u = x 2, we label the opposite side x and the hypotenuse 2, then by the Pythagorean Theorem the adjacent side has length ?
4 ´ x2. So, cos u = adj hyp = ?
4 ´ x2 2 .
Second, we can look back carefully at our work. We simplified ?
4 ´ x2 = 2 cos u, so cos u = ?
4 ´ x2 2 .
Third, we could use the identity sin2 u + cos2 u = 1. Then cos u = ˘ a 1 ´ sin2 u = ˘ b 1 ´ x2 4 . Since u = arcsin(x/2), u is in the range of arcsine, which means ´π 2 ď u ď π 2 . Therefore, cos u ě 0, so cos u = b 1 ´ x2 4 = ?
4´x2 2 .
So, ż a 4 ´ x2 dx = 2u + 2 sin u cos u + C = 2 arcsin x 2 + x ¨ ?
4 ´ x2 2 + C S-15: Substitute x = 2 5 sec u with 0 ă u ă π 2 , so that dx = 2 5 sec u tan u du and ?
25x2 ´ 4 = a 4(sec2 u ´ 1) = ?
4 tan2 u = 2 tan u. Then ż ?
25x2 ´ 4 x dx = ż 2 tan u 2 5 sec u ¨ 2 5 sec u tan u du = 2 ż tan2 u du = 2 ż sec2 u ´ 1 du = 2 tan u ´ 2u + C = a 25x2 ´ 4 ´ 2 arcsec 5x 2 + C u 2 ?
25x2 ´ 4 5x To find tan u, we draw a right triangle with angle u. Since sec u = 5x 2 , we label the hypotenuse 5x and the adjacent side 2. Then the Pythagorean Theorem gives us the opposite side as length ?
25x2 ´ 4. Then tan u = opp adj = ?
25x2 ´ 4 2 .
Alternately, we can notice that in our work, we already showed 2 tan u = ?
25x2 ´ 4, so tan u = 1 2 ?
25x2 ´ 4.
378 As a check, we observe that the derivative of the answer d dx a 25x2 ´ 4 ´ 2 arcsec 5x 2 + C = 25x ?
25x2 ´ 4 ´ 2 5 2 ˇ ˇ5x 2 ˇ ˇ b 25x2 4 ´ 1 = 25x ?
25x2 ´ 4 ´ 4 x ?
25x2 ´ 4 since x ą 0 = 25x2 ´ 4 x ?
25x2 ´ 4 = ?
25x2 ´ 4 x is exactly the integrand (provided x ą 2 5).
S-16: The integrand has a quadratic polynomial under a square root, which makes us think of trig substitutions. However, it’s good practice to look for simpler methods before we jump into more complicated ones, and in this case we find something nicer than a trig substitution: the substitution u = x2 ´ 1, du = 2x dx. Then xdx = 1 2du, and x2 = u + 1. When x = ?
10, u = 9, and when x = ?
17, u = 16.
ż ?
17 ?
10 x3 ?
x2 ´ 1 dx = ż ?
17 ?
10 x2 ?
x2 ´ 1 ¨ xdx = 1 2 ż 16 9 u + 1 ?u du = 1 2 ż 16 9 u1/2 + u´1/2 du = 1 2 2 3u3/2 + 2u1/2 16 9 = 1 2 2 3 ¨ 43 + 2 ¨ 4 ´ 2 3 ¨ 33 ´ 2 ¨ 3 = 40 3 S-17: This integrand looks very different from those above. But it is only slightly disguised. If we complete the square ż dx ?
3 ´ 2x ´ x2 = ż dx a 4 ´ (x + 1)2 and make the substitution y = x + 1, dy = dx ż dx ?
3 ´ 2x ´ x2 = ż dx a 4 ´ (x + 1)2 = ż dy a 4 ´ y2 379 we get a typical trig substitution integral. So, we substitute y = 2 sin θ, dy = 2 cos θ dθ to get ż dx ?
3 ´ 2x ´ x2 = ż dy a 4 ´ y2 = ż 2 cos θ dθ a 4 ´ 4 sin2 θ = ż 2 cos θ dθ ?
4 cos2 θ = ż dθ = θ + C = arcsin y 2 + C = arcsin x + 1 2 + C An experienced integrator would probably substitute x + 1 = 2 sin θ directly, without going through y.
S-18: Completing the square, we see 4x2 ´ 12x + 8 = (2x ´ 3)2 ´ 1.
ż 1 (2x ´ 3)3?
4x2 ´ 12x + 8 dx = ż 1 (2x ´ 3)3a (2x ´ 3)2 ´ 1 dx As x ą 2, we have 2x ´ 3 ą 1. We use the substitution 2x ´ 3 = sec θ with 0 ď θ ă π 2 . So 2 dx = sec θ tan θ dθ and a (2x ´ 3)2 ´ 1 = ?
sec2 θ ´ 1 = ?
tan2 θ = tan θ.
= 1 2 ż 1 sec3 θ ?
sec2 θ ´ 1 sec θ tan θ dθ = 1 2 ż 1 sec3 θ tan θ sec θ tan θ dθ = 1 2 ż 1 sec2 θ dθ = 1 2 ż cos2 θ dθ = 1 4 ż (1 + cos(2θ)) dθ = 1 4 θ + 1 2 sin(2θ) + C = 1 4 (θ + sin θ cos θ) + C θ 1 ?
4x2 ´ 12x + 8 2x ´ 3 = 1 4 arccos 1 2x ´ 3 + ?
4x2 ´ 12x + 8 (2x ´ 3)2 !
+ C Since 2x ´ 3 = sec θ, we know cos θ = 1 2x´3 and θ = arccos 1 2x´3 . (Equivalently, θ = arcsec(2x ´ 3).) To find sin θ, we draw a right triangle with adjacent side of length 1, and hypotenuse of length 2x ´ 3. By the Pythagorean Theorem, the opposite side has length ?
4x2 ´ 12x + 8.
380 S-19: We use the substitution x = tan u, dx = sec2 u du. Note tan 0 = 0 and tan π 4 = 1.
ż 1 0 x2 ?
x2 + 1 3dx = ż π/4 0 tan2 u a tan2 u + 1 3 sec2 u du = ż π/4 0 tan2 u ?
sec2 u 3 sec2 u du = ż π/4 0 tan2 u sec u du = ż π/4 0 sec2 u ´ 1 sec u du = ż π/4 0 sec u ´ cos u du = h log |sec u + tan u| ´ sin u iπ/4 0 = log ˇ ˇ ˇ ?
2 + 1 ˇ ˇ ˇ ´ 1 ?
2 ´ (log |1 + 0| ´ 0) = log(1 + ?
2) ´ 1 ?
2 S-20: There’s no square root, but we can still make use of the substitution x = tan θ, dx = sec2 θ dθ.
ż 1 (x2 + 1)2 dx = ż 1 (tan2 θ + 1)2 sec2 θ dθ = ż 1 sec4 θ sec2 θ dθ = ż cos2 θ dθ = 1 2 ż 1 + cos(2θ) dθ = 1 2 θ + 1 2 sin(2θ) + C = 1 2 (θ + sin θ cos θ) + C θ 1 x ?
x2 + 1 = 1 2 arctan x + x x2 + 1 + C Since x = tan θ, we can draw a right triangle with angle θ, opposite side x, and adjacent side 1. Then by the Pythagorean Theorem, its hypotenuse has length ?
x2 + 1, which allows us to find sin θ and cos θ.
S-21: We complete the square to find x2 ´ 2x + 2 = (x ´ 1)2 + 1.
ż x2 ?
x2 ´ 2x + 2 dx = ż x2 a (x ´ 1)2 + 1 dx 381 We use the substitution x ´ 1 = tan θ, which implies dx = sec2 θ dθ and x = tan θ + 1 = ż (tan θ + 1)2 a (tan θ)2 + 1 sec2 θ dθ = ż tan2 θ + 2 tan θ + 1 sec θ sec2 θ dθ = ż (sec2 θ + 2 tan θ) sec θ dθ = ż sec3 θ + 2 tan θ sec θ dθ θ 1 x ´ 1 ?
x2 ´ 2x + 2 = 1 2 sec θ tan θ + 1 2 log | sec θ + tan θ| + 2 sec θ + C = 1 2 a x2 ´ 2x + 2(x ´ 1) + 1 2 log ˇ ˇ ˇ a x2 ´ 2x + 2 + x ´ 1 ˇ ˇ ˇ + 2 a x2 ´ 2x + 2 + C = 3 + x 2 a x2 ´ 2x + 2 + 1 2 log ˇ ˇ ˇ a x2 ´ 2x + 2 + x ´ 1 ˇ ˇ ˇ + C To see why we could write a (tan θ)2 + 25 = sec θ, as opposed to a (tan θ)2 + 25 = | sec θ|, see Example 1.9.5 in the CLP-2 text.
From our substitution, we know tan θ = x ´ 1. To find sec θ, we can notice that in our work we already simplified ?
x2 ´ 2x + 1 = sec θ. Alternately, we can draw a right triangle with angle θ, opposite side x ´ 1, adjacent side 1, and use the Pythagorean Theorem to find the hypotenuse.
S-22: First, we complete the square. The constants aren’t integers, but we can still use the same method as in Question 2. The quadratic function under the square root is 3x2 + 5x.
We match the non-constant terms to those of a perfect square.
(ax + b)2 = a2x2 + 2abx + b2 3x2 + 5x = a2x2 + 2abx + b2 + c for some constant c • Looking at the leading term tells us a = ?
3.
• Then the second term tells us 5 = 2ab = 2 ?
3b, so b = 5 2 ?
3.
• Finally, the constant terms give us 0 = b2 + c = 25 12 + c, so c = ´25 12.
So, 3x2 + 5x = ?
3x + 5 2 ?
3 2 ´ 25 12.
ż 1 ?
3x2 + 5x dx = ż 1 c?
3x + 5 2 ?
3 2 ´ 25 12 dx 382 We use the substitution ?
3x + 5 2 ?
3 = 5 2 ?
3 sec θ, which leads to ?
3dx = 5 2 ?
3 sec θ tan θ dθ, i.e. dx = 5 6 sec θ tan θ dθ.
= ż 1 c 5 2 ?
3 sec θ 2 ´ 25 12 ¨ 5 6 sec θ tan θ dθ = ż 1 b 25 12 sec2 θ ´ 25 12 ¨ 5 6 sec θ tan θ dθ = ż 1 b 25 12 tan2 θ ¨ 5 6 sec θ tan θ dθ = ż 1 5 2 ?
3 tan θ ¨ 5 6 sec θ tan θ dθ = 1 ?
3 ż sec θ dθ = 1 ?
3 log |sec θ + tan θ| + C θ 5 2 ?
9x2 + 15x 6x + 5 = 1 ?
3 log ˇ ˇ ˇ ˇ 6 5x + 1 + 2 5 a 9x2 + 15x ˇ ˇ ˇ ˇ + C Since we used the substitution ?
3x + 5 2 ?
3 = 5 2 ?
3 sec θ, we have sec θ = 6 5x + 1 = 6x+5 5 . To find tan θ in terms of x, we have two options. We can make a right triangle with angle θ, hypotenuse 6x + 5, and adjacent side 5, then use the Pythagorean Theorem to find the opposite side. Or, we can look through our work and see that ?
3x2 + 5 = 5 2 ?
3 tan θ, so tan θ = 2 ?
3 5 ?
3x2 + 5 = 2 5 ?
9x2 + 15.
As a check, we observe that the derivative of the answer d dx 1 ?
3 log ˇ ˇ ˇ ˇ 6 5x + 1 + 2 5 a 9x2 + 15x ˇ ˇ ˇ ˇ + C = 1 ?
3 6 5 + 1 5 18x+15 ?
9x2+15x 6 5x + 1 + 2 5 ?
9x2 + 15x = 1 ?
3 6 + 3 6x+5 ?
9x2+15x (6x + 5) + 2 ?
9x2 + 15x = ?
3 2 + 6x+5 ?
9x2+15x (6x + 5) + 2 ?
9x2 + 15x = ?
3 ?
9x2 + 15x = 1 ?
3x2 + 5x is exactly the integrand.
Remark: in applications, often the numbers involved are messier than they are in textbooks. The ideas of this problem are similar to other problems in this section, but it’s good practice to apply them in a slightly messy context.
383 S-23: We use the substitution x = tan u, dx = sec2 u du.
ż ?
1 + x23 x dx = ż a 1 + tan2 u 3 tan u sec2 u du = ż sec3 u tan u sec2 u du = ż (sec2 u)2 tan u sec u du = ż (tan2 u + 1)2 tan u sec u du = ż tan4 u + 2 tan2 u + 1 tan u sec u du = ż tan3 u sec u du + ż 2 sec u tan u du + ż sec u tan u du For the first integral, we use the substitution w = sec u. The second is the antiderivative of 2 sec u. The third we simplify as sec u tan u = 1 cos u ¨ cos u sin u = csc u .
= ż (sec2 u ´ 1) sec u tan u du + 2 sec u + log | cot u ´ csc u| + C = ż (w2 ´ 1) dw + 2 sec u + log | cot u ´ csc u| + C = 1 3w3 ´ w + 2 sec u + log | cot u ´ csc u| + C = 1 3 sec3 u ´ sec u + 2 sec u + log | cot u ´ csc u| + C u 1 x ?
1 + x2 = 1 3 sec3 u + sec u + log | cot u ´ csc u| + C We began with the substitution x = tan u. Then cot u = 1 x. To find csc u and sec u, we draw a right triangle with angle u, opposite side x, and adjacent side 1. The Pythagorean Theorem gives us the hypotenuse.
= 1 3 a 1 + x23 + a 1 + x2 + log ˇ ˇ ˇ ˇ ˇ 1 x ´ ?
1 + x2 x ˇ ˇ ˇ ˇ ˇ + C = 1 3 a 1 + x2(4 + x2) + log ˇ ˇ ˇ ˇ ˇ 1 ´ ?
1 + x2 x ˇ ˇ ˇ ˇ ˇ + C S-24: The half of the ellipse to the right of the y-axis is given by the equation x = f (y) = 4 c 1 ´ y 2 2 384 The area we want is twice the area between the right-hand side of the curve and the y-axis, from y = ´1 to y = 1. In other words, Area = 2 ż 1 ´1 4 c 1 ´ y 2 2 dy = 8 ż 1 ´1 c 1 ´ y 2 2 dy Since the integrand b 1 ´ y 2 2 is an even function of y, ż 1 ´1 c 1 ´ y 2 2 dy = 2 ż 1 0 c 1 ´ y 2 2 dy ù ñ Area = 16 ż 1 0 c 1 ´ y 2 2 dy We use the substitution y 2 = sin θ, 1 2dy = cos θ dθ. When y = 0, sin θ = 0 so that θ = 0, and when y = 1, sin θ = 1 2 so that θ = π 6 . Hence Area = 16 ż π/6 0 b 1 ´ (sin θ)2 2 cos θ dθ = 32 ż π/6 0 a cos2 θ cos θ dθ = 32 ż π/6 0 cos2 θ dθ = 16 ż π/6 0 1 + cos(2θ) dθ = 16 θ + 1 2 sin(2θ) π/6 0 = 16 π 6 + 1 2 ¨ ?
3 2 = 8π 3 + 4 ?
3 Remark: we also investigated areas of ellipses in Question 16, Section 1.2.
S-25: Note that f (x) is an even function, nonnegative over its entire domain.
(a) To find the area of R, we evaluate Area = ż 1/2 ´1/2 |x| 4 ?
1 ´ x2 dx = 2 ż 1/2 0 x 4 ?
1 ´ x2 dx We use the substitution u = 1 ´ x2, du = ´2x dx.
= ´ ż 3/4 1 1 u1/4 du = ´ 4 3u3/4 3/4 1 = ´4 3 3 4 3/4 ´ 1 !
= 4 3 ´ 4 c 4 3 385 (b) We slice the solid of rotation into circular disks of width dx and radius |x| 4 ?
1 ´ x2.
Volume = ż 1/2 ´1/2 π |x| 4 ?
1 ´ x2 2 dx = 2π ż 1/2 0 x2 ?
1 ´ x2 dx We use the substitution x = sin θ, dx = cos θ dθ, so ?
1 ´ x2 = a 1 ´ sin2 θ = cos θ. Note sin 0 = 0 and sin π 6 = 1 2.
= 2π ż π/6 0 sin2 θ cos θ cos θ dθ = 2π ż π/6 0 sin2 θ dθ = π ż π/6 0 1 ´ cos(2θ) dθ = π θ ´ 1 2 sin(2θ) π/6 0 = π π 6 ´ 1 2 ¨ ?
3 2 = π2 6 ´ ?
3π 4 S-26: If we think of ex as ex/22, the function under the square root suggests the substitution ex/2 = tan θ. Then 1 2ex/2 dx = sec2 θ dθ, so dx = 2 ex/2 sec2 θ dθ = 2 tan θ sec θ dθ.
ż ?
1 + ex dx = ż 2 a 1 + tan2 θ tan θ sec2 θ dθ = 2 ż sec3 θ tan θ dθ = 2 ż sec θ(tan2 θ + 1) tan θ dθ = 2 ż sec θ tan θ + sec θ tan θ dθ = 2 ż sec θ tan θ + csc θ dθ θ 1 ex/2 ?
1 + ex = 2 sec θ + 2 log | cot θ ´ csc θ| + C = 2 ?
1 + ex + 2 log ˇ ˇ ˇ ˇ 1 ex/2 ´ ?
1 + ex ex/2 ˇ ˇ ˇ ˇ + C = 2 ?
1 + ex + 2 log ˇ ˇ ˇ1 ´ ?
1 + ex ˇ ˇ ˇ ´ 2 log(ex/2) + C = 2 ?
1 + ex + 2 log ˇ ˇ ˇ1 ´ ?
1 + ex ˇ ˇ ˇ ´ x + C 386 We used the substitution ex/2 = tan θ, so cot θ = 1 ex/2. To find sec θ and csc θ, we draw a right triangle with opposite side ex/2 and adjacent side 1. They by the Pythagorean Theorem, the hypotenuse has length ?
1 + ex.
Remark: if we use the substitution u = ?
1 + ex, then we can change the integral to ż 2u2 u2 ´ 1 du. We can integrate this using the method of partial fractions, which we’ll learn in the next section. You can explore this option in Question 26, Section 1.10.
S-27: (a) We can save ourselves some trouble by applying logarithm rules before we differentiate.
log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ = log |1 + x| ´ log | a 1 ´ x2| = log |1 + x| ´ 1 2 log |1 ´ x2| = log |1 + x| ´ 1 2 log |(1 + x)(1 ´ x)| = log |1 + x| ´ 1 2 log |1 + x| ´ 1 2 log |1 ´ x| d dx " log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ = d dx " log |1 + x| ´ 1 2 log |1 + x| ´ 1 2 log |1 ´ x| = 1 1 + x ´ 1/2 1 + x + 1/2 1 ´ x = 1/2 1 + x + 1/2 1 ´ x = 1 1 ´ x2 Notice this is the integrand from our work in blue.
(b) False: ż 3 2 1 1 ´ x2 dx is a number, because it is the area under a finite portion of a continuous curve. (We note that the integrand is continuous over the interval [2, 3], although it is not continuous everywhere.) However, log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ x=3 x=2 is not defined, since the denominator takes the square root of a negative number. So, these two expressions are not the same.
(c) The work in the question is not correct. The most salient problem is that when we make the substitution x = sin θ, we restrict the possible values of x to [´1, 1], since this is the range of the sine function. However, the original integral had no such restriction.
How can we be sure we avoid this problem in the future? In the introductory text to Section 1.9 (before Example 1.9.1), the CLP-2 text tells us that we are allowed to write our old variable as a function of a new variable (say x = s(u)) as long as that function 387 is invertible to recover our original variable x. There is one very obvious reason why invertibility is necessary: after we antidifferentiate using our new variable u, we need to get it back in terms of our original variable, so we need to be able to recover x. Moreover, invertibility reconciles potential problems with domains: if an inverse function u = s´1(x) exists, then for any x, there exists a u with s(u) = x. (This was not the case in the work for the question, because we chose x = sin θ, but if x = 2, there is no corresponding θ. Note, however, that x = sin θ is invertible over [´1, 1], so the work is correct if we restrict x to those values.) Remark: in the next section, you will learn to use partial fractions to find ż 1 1 ´ x2 dx = log |1 + x| ´ 1 2 log |1 ´ x|. When ´1 ă x ă 1, this is equivalent to log ˇ ˇ ˇ ˇ 1 + x ?
1 ´ x2 ˇ ˇ ˇ ˇ.
S-28: Remember that for any value X, |X| = " X if X ě 0 ´X if X ď 0 So, |X| ‰ X precisely when X ă 0.
(a) The range of arcsine is ´ π 2 , π 2 . So, since u = arcsin(x/a), u is in the range ´ π 2 , π 2 .
Therefore cos u ě 0. Since a is positive, a cos u ě 0, so a cos u = |a cos u|. That is, a a2 ´ x2 = |a cos u| = a cos u all the time.
(b) The range of arctangent is ´ π 2 , π 2 . So, since u = arctan(x/a), u is in the range ´ π 2 , π 2 . Therefore sec u = 1 cos u ą 0. Since a is positive, a sec u ą 0, so a sec u = |a sec u|.That is, a a2 + x2 = |a sec u| = a sec u all the time.
(c) The range of arccosine is 0, π . So, since u = arcsec(x/a) = arccos(a/x), u is in the range 0, π . (Actually, it’s in the range [0, π 2 ) Y ( π 2 , π], since secant is undefined at π/2.) If |a tan u| ‰ a tan u, then tan u ă 0, which happens when u is in the range π 2 , π). This is the same range over which ´1 ă cos u ă 0, and so ´1 ă a x ă 0. Since a x ă 0, a and x have different signs, so x ă 0. Then since ´1 ă a x, also x ă ´a.
So, a x2 ´ a2 = |a tan u| = ´a tan u ‰ a tan u happens precisely when when x ă ´a.
Solutions to Exercises 1.10 — Jump to TABLE OF CONTENTS 388 S-1: If a quadratic function can be factored as (ax + b)(cx + d) for some constants a, b, c, d, then it has roots ´b a and ´d c. So, if a quadratic function has no roots, it is irreducible: this is the case for the function in graph (d).
If a quadratic function has two different roots, then (ax + b) ‰ α(cx + d) for any constant α. That is, the quadratic function is the product of distinct linear factors. This is the case for the functions graphed in (b) and (c), since these each have two distinct places where they cross the x-axis.
Finally, if a quadratic function has precisely one root, then b a = d c, so: (ax + b)(cx + d) = a(x + b a)(cx + d) = a(x + d c )(cx + d) = a c(cx + d)(cx + d) That is, the quadratic function is the product of a repeated linear factor, and a constant a c (which might simply be a c = 1).
S-2: Our first step is to fully factor the denominator: (x2 ´ 1)2(x2 + 1) = (x ´ 1)2(x + 1)2(x2 + 1) Once a term is linear, it can’t be factored further; for quadratic terms, we should check that they are irreducible. Since x2 + 1 has no real roots (we are familiar with its graph, which is entirely above the x-axis), it is irreducible, so now our denominator is fully factored.
x3 + 3 (x2 ´ 1)2(x2 + 1) = x3 + 3 (x ´ 1)2(x + 1)2(x2 + 1) = A x ´ 1 + B (x ´ 1)2 + C x + 1 + D (x + 1)2 + Ex + F x2 + 1 Notice (x ´ 1) and (x + 1) are (repeated) linear factors, while (x2 + 1) is an irreducible quadratic factor. This accounts for the difference in the numerators of their corresponding terms.
S-3: The partial fraction decomposition has the form 3x3 ´ 2x2 + 11 x2(x ´ 1)(x2 + 3) = A x ´ 1 + various terms When we multiply through by the original denominator, this becomes 3x3 ´ 2x2 + 11 = x2(x2 + 3)A + (x ´ 1)(other terms).
Evaluating both sides at x = 1 yields 3 ¨ 13 ´ 2 ¨ 12 + 11 = 12(12 + 3)A + 0, or A = 3.
S-4: (a) We start by dividing. The leading term of the numerator is x times the leading term of the denominator. The remainder is x + 2.
389 x x2 + 1 x3 + 2x + 2 ´ x3 ´ x x + 2 That is, x3 + 2x + 2 = x(x2 + 1) + (x + 2). So, x3 + 2x + 2 x2 + 1 = x + x + 2 x2 + 1 (b) We start by dividing. The leading term of the numerator is 3x2 times the leading term of the denominator.
3x2 5x2 + 2x + 8 15x4 + 6x3 + 34x2 + 4x + 20 ´ 15x4 ´ 6x3 ´ 24x2 10x2 + 4x + 20 Then 5x2 goes into 10x2 twice, so 3x2 + 2 5x2 + 2x + 8 15x4 + 6x3 + 34x2 + 4x + 20 ´ 15x4 ´ 6x3 ´ 24x2 10x2 + 4x + 20 ´ 10x2 ´ 4x ´ 16 4 Our remainder is 4. That is, 15x4 + 6x3 + 34x2 + 4x + 20 5x2 + 2x + 8 = 3x2 + 2 + 4 5x2 + 2x + 8.
(c) We start by dividing. The leading term of the numerator is x3 times the leading term of the denominator.
x3 2x2 + 5 2x5 + 9x3 + 12x2 + 10x + 30 ´ 2x5 ´ 5x3 4x3 + 12x2 + 10x Then 2x2(2x) gives us 4x3.
x3 + 2x 2x2 + 5 2x5 + 9x3 + 12x2 + 10x + 30 ´ 2x5 ´ 5x3 4x3 + 12x2 + 10x ´ 4x3 ´ 10x 12x2 + 30 390 Finally, 2x2 goes into 12x2 six times.
x3 + 2x + 6 2x2 + 5 2x5 + 9x3 + 12x2 + 10x + 30 ´ 2x5 ´ 5x3 4x3 + 12x2 + 10x ´ 4x3 ´ 10x 12x2 + 30 ´ 12x2 ´ 30 0 Since there is no remainder, 2x5 + 9x3 + 12x2 + 10x + 30 2x2 + 5 = x3 + 2x + 6 Remark: if we wanted to be pedantic about the question statement, we could write our final answer as x3 + 2x + 6 + 0 x, so that we are indeed adding a polynomial to a rational function whose numerator has degree strictly smaller than its denominator.
S-5: (a) The polynomial 5x3 ´ 3x2 ´ 10x + 6 has a repeated pattern: the ratio of the first two coefficients is the same as the ratio of the last two coefficients. We can use this to factor.
5x3 ´ 3x2 ´ 10x + 6 = x2(5x ´ 3) ´ 2(5x ´ 3) = (x2 ´ 2)(5x ´ 3) = (x + ?
2)(x ´ ?
2)(5x ´ 3) (b) The polynomial x4 ´ 3x2 ´ 5 has only even powers of x, so we can (temporarily) replace them with x2 = y to turn our quartic polynomial into a quadratic.
x4 ´ 3x2 ´ 5 = y2 ´ 3y ´ 5 There’s no obvious factoring here, but we can find its roots, if any, using the quadratic equation.
y = 3 ˘ a 32 ´ 4(1)(´5) 2 = 3 ˘ ?
29 2 So, y2 ´ 3y ´ 5 = y ´ 3 + ?
29 2 y ´ 3 ´ ?
29 2 Therefore, x4 ´ 3x2 ´ 5 = x2 ´ 3 + ?
29 2 x2 ´ 3 ´ ?
29 2 391 We’d like to use the difference of two squares to factor these quadratic expressions.
For this to work, the constants must be positive (so their square roots are real). Since ?
29 ą 3, only the first quadratic is factorable. The other is irreducible–it’s always positive, so it had no roots.
x4 ´ 3x2 ´ 5 = x + d 3 + ?
29 2 x ´ d 3 + ?
29 2 x2 + ?
29 ´ 3 2 (c) Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, ´6. So, our candidates for roots are ˘1, ˘2, ˘3, and ˘6. To save time, we don’t need to know exactly the value of our polynomial at these points: only whether or not it is 0. Write f (x) = x4 ´ 4x3 ´ 10x2 ´ 11x ´ 6.
f (´1) = 0 f (´2) ‰ 0 f (´3) ‰ 0 f (´6) ‰ 0 f (1) ‰ 0 f (2) ‰ 0 f (3) ‰ 0 f (6) = 0 Since x = ´1 and x = 6 are roots of our polynomial, it has factors (x + 1) and (x ´ 6).
Note (x + 1)(x ´ 6) = x2 ´ 5x ´ 6. We use long division to figure out what else is lurking in our polynomial.
x2 + x + 1 x2 ´ 5x ´ 6 x4 ´ 4x3 ´ 10x2 ´ 11x ´ 6 ´ x4 + 5x3 + 6x2 x3 ´ 4x2 ´ 11x ´ x3 + 5x2 + 6x x2 ´ 5x ´ 6 ´ x2 + 5x + 6 0 So, x4 ´ 4x3 ´ 10x2 ´ 11x ´ 6 = (x + 1)(x ´ 6)(x2 + x + 1).
We should check whether x2 + x + 1 is reducible or not. If we try to find its roots with the quadratic equation, we get ´1 ˘ ?
´3 2 , which are not real numbers. So, we’re at the end of our factoring.
(d) Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, ´15. So, our candidates for roots are ˘1, ˘3, ˘5, and ˘15. Write f (x) = 2x4 + 12x3 ´ x2 ´ 52x + 15.
f (´1) ‰ 0 f (´3) = 0 f (´5) = 0 f (´5) ‰ 0 f (1) ‰ 0 f (3) ‰ 0 f (5) ‰ 0 f (15) ‰ 0 Since x = ´3 and x = ´5 are roots of our polynomial, it has factors (x + 3) and (x + 5). Note (x + 3)(x + 5) = x2 + 8x + 15. We use long division to move forward.
392 2x2 ´ 4x + 1 x2 + 8x + 15 2x4 + 12x3 ´ x2 ´ 52x + 15 ´ 2x4 ´ 16x3 ´ 30x2 ´ 4x3 ´ 31x2 ´ 52x 4x3 + 32x2 + 60x x2 + 8x + 15 ´ x2 ´ 8x ´ 15 0 So, 2x4 + 12x3 ´ x2 ´ 52x + 15 = (x + 3)(x + 5)(2x2 ´ 4x + 1).
We should check whether 2x2 ´ 4x + 1 is reducible or not. There’s not an obvious way to factor it, but we can use the quadratic equation. This gives us roots 4 ˘ ?
16 ´ 8 4 = 1 ˘ ?
2 2 . So, we have two more linear factors.
Specifically: 2x4 + 12x3 ´ x2 ´ 52x + 15 = (x + 3)(x + 5) x ´ 1 + ?
2 2 x ´ 1 ´ ?
2 2 .
S-6: The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution u = (ax + b). It’s less clear how to find the antiderivative of (2).
S-7: The integrand is a rational function, so it’s a candidate for partial fraction. We quickly rule out any obvious substitution or integration by parts, so we go ahead with the decomposition.
We start by expressing the integrand, i.e. the fraction 1 x+x2 = 1 x(1+x), as a linear combination of the simpler fractions 1 x and 1 x+1 (which we already know how to integrate). We will have 1 x + x2 = 1 x(1 + x) = a x + b x + 1 = a(x + 1) + bx x(1 + x) The fraction on the left hand side is the same as the fraction on the right hand side if and only if the numerator on the left hand side, which is 1 = 0x + 1, is equal to the numerator on the right hand side, which is a(x + 1) + bx = (a + b)x + a. This in turn is the case if and only of a = 1 (i.e. the constant terms are the same in the two numerators) and a + b = 0 (i.e. the coefficients of x are the same in the two numerators). So a = 1 and b = ´1. Now we can easily evaluate the integral ż 2 1 dx x + x2 = ż 2 1 dx x(x + 1) = ż 2 1 1 x ´ 1 x + 1 dx = h log x ´ log(x + 1) i2 1 = log 2 ´ log 3 2 = log 4 3 393 S-8: We’ll first do a partial fraction decomposition. The sneaky way is to temporarily rename x2 to y. Then x4 + x2 = y2 + y and 1 x4 + x2 = 1 y(y + 1) = 1 y ´ 1 y + 1 as we found in Question 7. Now we restore y to x2.
ż 1 x4 + x2 dx = ż 1 x2 ´ 1 x2 + 1 dx = ´1 x ´ arctan x + C S-9: The integrand is of the form N(x)/D(x) with D(x) already factored and N(x) of lower degree. We immediately look for a partial fraction decomposition: 12x + 4 (x ´ 3)(x2 + 1) = A x ´ 3 + Bx + C x2 + 1 .
Multiplying through by the denominator yields 12x + 4 = A(x2 + 1) + (Bx + C)(x ´ 3) (˚) Setting x = 3 we find: 36 + 4 = A(9 + 1) + 0 ù ñ 40 = 10A ù ñ A = 4 Substituting A = 4 in (˚) gives 12x + 4 = 4(x2 + 1) + (Bx + C)(x ´ 3) ù ñ ´4x2 + 12x = (x ´ 3)(Bx + C) ù ñ (´4x)(x ´ 3) = (Bx + C)(x ´ 3) ù ñ B = ´4, C = 0 So we have found that A = 4, B = ´4, and C = 0. Therefore ż 12x + 4 (x ´ 3)(x2 + 1) dx = ż 4 x ´ 3 ´ 4x x2 + 1 dx = 4 log |x ´ 3| ´ 2 log(x2 + 1) + C The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution u = x2 + 1, du = 2x dx.
S-10: The integrand is of the form N(x)/D(x) with D(x) already factored and N(x) of lower degree. With no obvious substitution available, we look for a partial fraction decomposition.
3x2 ´ 4 (x ´ 2)(x2 + 4) = A x ´ 2 + Bx + C x2 + 4 Multiplying through by the denominator gives 3x2 ´ 4 = A(x2 + 4) + (Bx + C)(x ´ 2) (˚) 394 Setting x = 2 we find: 12 ´ 4 = A(4 + 4) + 0 ù ñ 8 = 8A ù ñ A = 1 Substituting A = 1 in (˚) gives 3x2 ´ 4 = (x2 + 4) + (x ´ 2)(Bx + C) ù ñ 2x2 ´ 8 = (x ´ 2)(Bx + C) ù ñ (x ´ 2)(2x + 4) = (x ´ 2)(Bx + C) ù ñ B = 2, C = 4 Thus, we have: 3x2 ´ 4 (x ´ 2)(x2 + 4) = 1 x ´ 2 + 2x + 4 x2 + 4 = 1 x ´ 2 + 2x x2 + 4 + 4 x2 + 4 The first two of these are directly integrable: F(x) = log |x ´ 2| + log |x2 + 4| + ż 4 x2 + 4 dx (The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution u = x2 + 4, du = 2x dx.) For the final integral, we substitute: x = 2y, dx = 2dy, and see that: ż 4 x2 + 4 dx = 2 ż 1 y2 + 1 dy = 2 arctan y + D = 2 arctan(x/2) + D for any constant D. All together we have: F(x) = log |x ´ 2| + log(x2 + 4) + 2 arctan(x/2) + D S-11: This integrand is a rational function, with no obvious substitution. This sure looks like a partial fraction problem. Let’s go through our protocol.
• The degree of the numerator x ´ 13 is one, which is strictly smaller than the dergee of the denominator x2 ´ x ´ 6, which is two. So we don’t need long division to pull out a polynomial.
• Next we factor the denominator.
x2 ´ x ´ 6 = (x ´ 3)(x + 2) • Next we find the partial fraction decomposition of the integrand. It is of the form x ´ 13 (x ´ 3)(x + 2) = A x ´ 3 + B x + 2 To find A and B, using the sneaky method, we cross multiply by the denominator.
x ´ 13 = A(x + 2) + B(x ´ 3) 395 Now we can find A by evaluating at x = 3 3 ´ 13 = A(3 + 2) + B(3 ´ 3) ù ñ A = ´2 and find B by evaluating at x = ´2.
´2 ´ 13 = A(´2 + 2) + B(´2 ´ 3) ù ñ B = 3 (Hmmm. A and B are nice round numbers. Sure looks like a rigged exam or homework question.) Our partial fraction decomposition is x ´ 13 (x ´ 3)(x + 2) = ´2 x ´ 3 + 3 x + 2 As a check, we recombine the right hand side and make sure that it matches the left hand side.
´2 x ´ 3 + 3 x + 2 = ´2(x + 2) + 3(x ´ 3) (x ´ 3)(x + 2) = x ´ 13 (x ´ 3)(x + 2) • Finally, we evaluate the integral.
ż x ´ 13 x2 ´ x ´ 6dx = ż ´2 x ´ 3 + 3 x + 2 dx = ´2 log |x ´ 3| + 3 log |x + 2| + C S-12: Again, this sure looks like a partial fraction problem. So let’s go through our protocol.
• The degree of the numerator 5x + 1 is one, which is strictly smaller than the dergee of the denominator x2 + 5x + 6, which is two. So we do not long divide to pull out a polynomial.
• Next we factor the denominator.
x2 + 5x + 6 = (x + 2)(x + 3) • Next we find the partial fraction decomposition of the integrand. It is of the form 5x + 1 (x + 2)(x + 3) = A x + 2 + B x + 3 To find A and B, using the sneaky method, we cross multiply by the denominator.
5x + 1 = A(x + 3) + B(x + 2) Now we can find A by evaluating at x = ´2 ´10 + 1 = A(´2 + 3) + B(´2 + 2) ù ñ A = ´9 396 and find B by evaluating at x = ´3.
´15 + 1 = A(´3 + 3) + B(´3 + 2) ù ñ B = 14 So our partial fraction decomposition is 5x + 1 (x + 2)(x + 3) = ´9 x + 2 + 14 x + 3 As a check, we recombine the right hand side and make sure that it matches the left hand side ´9 x + 2 + 14 x + 3 = ´9(x + 3) + 14(x + 2) (x + 2)(x + 3) = 5x + 1 (x + 2)(x + 3) • Finally, we evaluate the integral ż 5x + 1 x2 + 5x + 6dx = ż ´9 x + 2 + 14 x + 3 dx = ´9 log |x + 2| + 14 log |x + 3| + C S-13: We have a rational function with no obvious substitution, so let’s use partial fraction decomposition.
• Since the degree of the numerator is the same as the degree of the denominator, we need to pull out a polynomial.
5 x2 ´ 1 5x2 ´ 3x ´ 1 ´ 5x2 + 5 ´ 3x + 4 That is, ż 5x2 ´ 3x ´ 1 x2 ´ 1 dx = ż 5 + ´3x + 4 x2 ´ 1 dx = 5x + ż ´3x + 4 x2 ´ 1 dx • Again, there’s no obvious substitution for the new integrand, so we want to use partial fraction. The denominator factors as (x ´ 1)(x + 1), so our decomposition has this form: ´3x + 4 x2 ´ 1 = ´3x + 4 (x ´ 1)(x + 1) = A x ´ 1 + B x + 1 = (A + B)x + (A ´ B) (x ´ 1)(x + 1) So, (1) A + B = ´3 and (2) A ´ B = 4.
• We solve (2) for A in terms of B, namely A = 4 + B. Plugging this into (1), we see (4 + B) + B = ´3. So, B = ´7 2, and A = 1 2.
397 • Now we can write our integral in a friendlier form and evaluate.
ż 5x2 ´ 3x ´ 1 x2 ´ 1 dx = = 5x + ż ´3x + 4 x2 ´ 1 dx = 5x + ż 1/2 x ´ 1 ´ 7/2 x + 1 dx = 5x + 1 2 log |x ´ 1| ´ 7 2 log |x + 1| + C S-14: The integrand is a rational function with no obvious substitution, so we use partial fraction decomposition.
• The degree of the numerator is the same as the degree of the denominator. Since it’s not smaller, we need to re-write our integrand. We could do this using long division, but this case is simple enough to do more informally.
4x4 + 14x2 + 2 4x4 + x2 = 4x4 + x2 + 13x2 + 2 4x4 + x2 = 4x4 + x2 4x4 + x2 + 13x2 + 2 4x4 + x2 = 1 + 13x2 + 2 4x4 + x2 • The denominator factors as x2(4x2 + 1).
• We want to find the partial fraction decomposition of the fractional part of our simplified integrand.
13x2 + 2 4x4 + x2 = 13x2 + 2 x2(4x2 + 1) = A x + B x2 + Cx + D 4x2 + 1 Multiply through by the original denominator.
13x2 + 2 = Ax(4x2 + 1) + B(4x2 + 1) + (Cx + D)x2 (1) Setting x = 0 gives us: 2 = B We use B = 2 to simplify Equation (1).
13x2 + 2 = Ax(4x2 + 1) + 2(4x2 + 1) + (Cx + D)x2 5x2 = Ax(4x2 + 1) + (Cx + D)x2 5x = A(4x2 + 1) + (Cx + D)x (2) Again, let x = 0.
0 = A Using A = 0, simplify Equation (2).
5x = (Cx + D)x 5 = Cx + D C= 0, D = 5 398 • Now we can write our integral in pieces.
ż 4x4 + 14x2 + 2 4x4 + x2 dx = ż 1 + 13x2 + 2 4x4 + x2 dx = ż 1 + 2 x2 + 5 4x2 + 1 dx = x ´ 2 x + ż 5 (2x)2 + 1 dx Substitute u = 2x, du = 2dx.
= x ´ 2 x + ż 5/2 u2 + 1 du = x ´ 2 x + 5 2 arctan u + C = x ´ 2 x + 5 2 arctan(2x) + C S-15: The integrand is a rational function with no obvious substitution, so we’ll use a partial fraction decomposition.
• Since the numerator has strictly smaller degree than the denominator, we don’t need to start off with a long division.
• We do, however, need to factor the denominator. We can immediately pull out x2; the remaining part is x2 ´ 2x + 1 = (x ´ 1)2.
• Now we can perform our partial fraction decomposition.
x2 + 2x ´ 1 x4 ´ 2x3 + x2 = x2 + 2x ´ 1 x2(x ´ 1)2 = A x + B x2 + C x ´ 1 + D (x ´ 1)2 Multiply both sides by the original denominator.
x2 + 2x ´ 1 = Ax(x ´ 1)2 + B(x ´ 1)2 + Cx2(x ´ 1) + Dx2 (1) To be sneaky, we set x = 0, and find: ´1 = B We also set x = 1, and find: 2 = D We use B and D to simplify Equation (1).
x2 + 2x ´ 1 = Ax(x ´ 1)2´1(x ´ 1)2 + Cx2(x ´ 1) + 2x2 0 = Ax(x ´ 1)2 + Cx2(x ´ 1) = x(x ´ 1)[(A + C)x ´ A] So, 0 = (A + C)x ´ A That is, A = C = 0.
399 • Now we can evaluate our integral.
ż x2 + 2x ´ 1 x4 ´ 2x3 + x2 dx = ż ´1 x2 + 2 (x ´ 1)2 dx = 1 x ´ 2 x ´ 1 + C S-16: Our integrand is a rational function with no obvious substitution, so we’ll use the method of partial fractions.
• The degree of the numerator is less than the degree of the denominator.
• We need to factor the denominator. The first two terms have the same ratio as the last two terms.
2x3 ´ x2 ´ 8x + 4 = x2(2x ´ 1) ´ 4(2x ´ 1) = (x2 ´ 4)(2x ´ 1) = (x ´ 2)(x + 2)(2x ´ 1) • Now we find our partial fraction decomposition.
3x2 ´ 4x ´ 10 2x3 ´ x2 ´ 8x + 4 = 3x2 ´ 4x ´ 10 (x ´ 2)(x + 2)(2x ´ 1) = A x ´ 2 + B x + 2 + C 2x ´ 1 Multiply both sides by the original denominator.
3x2 ´ 4x ´ 10 = A(x + 2)(2x ´ 1) + B(x ´ 2)(2x ´ 1) + C(x ´ 2)(x + 2) Distinct linear factors is the best possible scenario for the sneaky method. First, let’s set x = 2.
3(4) ´ 4(2) ´ 10 = A(4)(3) + B(0) + C(0) A = ´1 2 Now, let x = ´2.
3(4) ´ 4(´2) ´ 10 = A(0) + B(´4)(´5) + C(0) B = 1 2 Finally, let x = 1 2.
3 4 ´ 2 ´ 10 = A(0) + B(0) + C ´3 2 5 2 C = 3 400 • Now we can evaluate our integral in its new form.
ż 3x2 ´ 4x ´ 10 2x3 ´ x2 ´ 8x + 4 dx = ż ´1/2 x ´ 2 + 1/2 x + 2 + 3 2x ´ 1 dx = ´1 2 log |x ´ 2| + 1 2 log |x + 2| + 3 2 log |2x ´ 1| + C = 1 2 log ˇ ˇ ˇ ˇ x + 2 x ´ 2 ˇ ˇ ˇ ˇ + 3 2 log |2x ´ 1| + C S-17: The integrand is a rational function with no obvious substitution, so we use the method of partial fractions.
• The numerator has smaller degree than the denominator.
• We need to factor the denominator. In the absence of any clues, we look for an integer root. The constant term is 5, so the possible integer roots are ˘1 and ˘5.
Name f (x) = 2x3 + 11x2 + 6x + 5.
f (´1) ‰ 0 f (´5) = 0 f (1) ‰ 0 f (5) ‰ 0 So, (x + 5) is a factor of the denominator.
• We use long division to pull out the factor of (x + 5).
2x2 + x + 1 x + 5 2x3 + 11x2 + 6x + 5 ´ 2x3 ´ 10x2 x2 + 6x ´ x2 ´ 5x x + 5 ´ x ´ 5 0 That is, our denominator is (x + 5)(2x2 + x + 1).
• The quadratic function 2x2 + x + 1 is irreducible: we can see this by using the quadratic equation, and finding no real roots. So, we are ready to find our partial fraction decomposition.
10x2 + 24x + 8 2x3 + 11x2 + 6x + 5 = 10x2 + 24x + 8 (x + 5)(2x2 + x + 1) = A x + 5 + Bx + C 2x2 + x + 1 Multiply through by the original denominator.
10x2 + 24x + 8 = A(2x2 + x + 1) + (Bx + C)(x + 5) (1) Set x = ´5.
10(25) ´ 24(5) + 8 = A(2(25) ´ 5 + 1) + (B(´5) + C)(0) A = 3 401 Using our value of A, we simplify Equation (1).
10x2 + 24x + 8 = 3(2x2 + x + 1) + (Bx + C)(x + 5) 4x2 + 21x + 5 = (Bx + C)(x + 5) We factor the left side. We know (x + 5) must be one of its factors.
(4x + 1)(x + 5) = (Bx + C)(x + 5) 4x + 1 = Bx + C So, B = 4 and C = 1.
• Now we can write our integral in smaller pieces.
ż 1 0 10x2 + 24x + 8 2x3 + 11x2 + 6x + 5 dx = ż 1 0 3 x + 5 + 4x + 1 2x2 + x + 1 dx The antiderivative of the left fraction is 3 log |x + 5|. For the right fraction, we use the substitution u = 2x2 + x + 1, du = (4x + 1)dx to antidifferentiate.
= 3 log |x + 5| + log |2x2 + x + 1| 1 0 = 3 log 6 + log 4 ´ 3 log 5 ´ log 1 = log 4 ¨ 63 53 S-18: We follow the example in the text.
ż csc x dx = ż 1 sin x dx = ż sin x sin2 x dx = ż sin x 1 ´ cos2 x dx Let u = cos x, du = ´ sin x dx.
= ż ´1 1 ´ u2 du = ż ´1 (1 + u)(1 ´ u) du We see an opportunity for partial fraction.
´1 (1 + u)(1 ´ u) = A 1 + u + B 1 ´ u Multiply both sides by the original denominator.
´1 = A(1 ´ u) + B(1 + u) Let u = 1.
´1 = 2B ñ B = ´1 2 402 Let u = ´1.
´1 = 2A ñ A = ´1 2 We can now re-write our integral.
ż csc x dx = ż ´1 (1 + u)(1 ´ u) du = ż ´1/2 1 + u + ´1/2 1 ´ u du = ´1 2 log |1 + u| + 1 2 log |1 ´ u| + C = 1 2 log ˇ ˇ ˇ ˇ 1 ´ u 1 + u ˇ ˇ ˇ ˇ + C = 1 2 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ + C Remark: Elsewhere in the text, and in many tables of integrals, the antiderivative of cosecant is given as log | csc x ´ cot x|. We show that this is equivalent to our result.
log | csc x ´ cot x| = 1 2 log ˇ ˇ ˇ(csc x ´ cot x)2ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇcsc2 x ´ 2 csc x cot x + cot2 x ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇ ˇ 1 sin2 x ´ 2 cos x sin2 x + cos2 x sin2 x ˇ ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇ ˇ 1 ´ 2 cos x + cos2 x sin2 x ˇ ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇ ˇ ˇ (1 ´ cos x)2 1 ´ cos2 x ˇ ˇ ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇ ˇ ˇ (1 ´ cos x)2 (1 ´ cos x)(1 + cos x) ˇ ˇ ˇ ˇ ˇ = 1 2 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ S-19: We follow the example in the text.
ż csc3 x dx = ż 1 sin3 x dx = ż sin x sin4 x dx = ż sin x (1 ´ cos2 x)2 dx Let u = cos x, du = ´ sin x dx.
= ż ´1 (1 ´ u2)2 du 403 In Question 18, we saw 1 1´u2 = 1/2 1+u + 1/2 1´u , so ż ´1 (1 ´ u2)2 du = ´ ż 1 1 ´ u2 2 du = ´ ż 1/2 1 + u + 1/2 1 ´ u 2 du = ´1 4 ż 1 (1 + u)2 + 2 1 ´ u2 + 1 (1 ´ u)2 du = ´1 4 ż 1 (1 + u)2 + 1 1 + u + 1 1 ´ u + 1 (1 ´ u)2 du = ´1 4 ´ 1 1 + u + log |1 + u| ´ log |1 ´ u| + 1 1 ´ u + C = ´1 4 2u 1 ´ u2 + log ˇ ˇ ˇ ˇ 1 + u 1 ´ u ˇ ˇ ˇ ˇ + C = ´u 2(1 ´ u2) + 1 4 log ˇ ˇ ˇ ˇ 1 ´ u 1 + u ˇ ˇ ˇ ˇ + C = ´ cos x 2 sin2 x + 1 4 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ + C Remark: In Example 1.8.23 of the CLP-2 text, and in many tables of integrals, the antiderivative of csc3 x is given as ´1 2 cot x csc x + 1 2 log | csc x ´ cot x| + C. This is equivalent to our result. Recall in the remark after the solution to Question 18, we saw 1 2 log ˇ ˇ ˇ 1´cos x 1+cos x ˇ ˇ ˇ = log | csc x ´ cot x|.
´1 2 cot x csc x + 1 2 log | csc x ´ cot x| = ´1 2 cot x csc x + 1 4 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ = ´1 2 cos x sin x 1 sin x + 1 4 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ = ´ cos x 2 sin2 x + 1 4 log ˇ ˇ ˇ ˇ 1 ´ cos x 1 + cos x ˇ ˇ ˇ ˇ S-20: This is a rational function, and there’s no obvious substitution, so we’ll use partial fraction decomposition.
• First, we check that the numerator has strictly smaller degree than the denominator, so we don’t have to use long division.
• Second, we factor the denominator. We can immediately pull out a factor of x2; then we’re left with the quadratic polynomial x2 + 5x + 10. Using the quadratic equation, we check that this has no real roots, so it is irreducible.
• Once we know the factorization of the denominator, we can set up our decomposition.
3x3 + 15x2 + 35x + 10 x4 + 5x3 + 10x2 = 3x3 + 15x2 + 35x + 10 x2(x2 + 5x + 10) = A x + B x2 + Cx + D x2 + 5x + 10 404 We multiply both sides by the original denominator.
3x3 + 15x2 + 35x + 10 = Ax(x2 + 5x + 10) + B(x2 + 5x + 10) + (Cx + D)x2 (1) Following the “Sneaky Method,” we plug in x = 0.
0 + 10 = A(0) + B(10) + (C(0) + D)(0) B = 1 • Knowing B allows us to simplify our Equation (1).
3x3 + 15x2 + 35x + 10 = Ax(x2 + 5x + 10) + 1(x2 + 5x + 10) + (Cx + D)x2 3x3 + 14x2 + 30x = Ax(x2 + 5x + 10) + (Cx + D)x2 We can factor x out of both sides of the equation.
3x2 + 14x + 30 = A(x2 + 5x + 10) + (Cx + D)x (2) • Again, we set x = 0.
0 + 30 = A(10) + (C(0 + D)(0) A = 3 • We simplify Equation (2), using A = 3.
3x2 + 14x + 30 = 3(x2 + 5x + 10) + (Cx + D)x ´x = Cx2 + Dx C = 0, D = ´1 • Now that we have our coefficients, we can re-write our integral in a friendlier form.
ż 2 1 3x3 + 15x2 + 35x + 10 x4 + 5x + 10x2 dx = ż 2 1 3 x + 1 x2 ´ 1 x2 + 5x + 10 dx = 3 log |x| ´ 1 x 2 1 ´ ż 2 1 1 x2 + 5x + 10 dx = 3 log 2 + 1 2 ´ ż 2 1 1 x2 + 5x + 10 dx 405 The remaining integral is the reciprocal of a quadratic polynomial, much like 1 1 + x2, whose antiderivative is arctangent. We complete the square and use the substitution u = 2x+5 ?
15 , du = 2 ?
15 dx.
ż 2 1 1 x2 + 5x + 10 dx = ż 2 1 1 x + 5 2 2 + 15 4 dx = 4 15 ż 2 1 1 2x+5 ?
15 2 + 1 dx = 2 ?
15 ż 9/ ?
15 7/ ?
15 1 u2 + 1 du = 2 ?
15 h arctan u i9/ ?
15 7/ ?
15 = 2 ?
15 arctan 9 ?
15 ´ arctan 7 ?
15 So, all together, ż 2 1 3x3 + 15x2 + 35x + 10 x4 + 5x3 + 10x2 dx = 3 log 2+ 1 2 ´ 2 ?
15 arctan 9 ?
15 ´ arctan 7 ?
15 S-21: Our integrand is already in the nice form that would come out of a partial fractions decomposition. Let’s consider its different pieces.
• First piece: ş 3 x2+2 dx. The fraction looks somewhat like the derivative of arctangent, so we can massage it to find an appropriate substitution.
ż 3 x2 + 2 dx = 3 2 ż 1 x ?
2 2 + 1 dx Use the substitution u = x ?
2, du = 1 ?
2 dx.
= 3 ?
2 ż 1 u2 + 1 du = 3 ?
2 arctan u + C = 3 ?
2 arctan x ?
2 + C • The next piece is ş x´3 (x2+2)2 dx. If the numerator were only x (and no constant), we could use the substitution u = x2 + 2, du = 2x dx. So, to that end, we can break up that fraction into x (x2+2)2 ´ 3 (x2+2)2. For now, we only evaluate the first half.
406 ż x (x2 + 2)2 dx = 1 2 ż 1 u2 du = ´ 1 2u + C = ´ 1 2x2 + 4 + C • That leaves us with the final piece, 3 (x2+2)2, which is the hardest. We saw something similar in Question 20 in Section 1.9: we can use the substitution x = ?
2 tan θ, dx = ?
2 sec2 θ dθ.
ż 3 (x2 + 2)2 dx = ż 3 (2 tan2 θ + 2)2 ?
2 sec2 θ dθ = ż 3 4 sec4 θ ?
2 sec2 θ dθ = 3 2 ?
2 ż cos2 θ dθ = 3 4 ?
2 ż 1 + cos(2θ) dθ = 3 4 ?
2 θ + 1 2 sin(2θ) + C = 3 4 ?
2 (θ + sin θ cos θ) + C θ ?
2 x ?
x2 + 2 = 3 4 ?
2 arctan x ?
2 + x ?
2 x2 + 2 + C From our substitution, tan θ = x ?
2. So, we can draw a right triangle with angle θ, opposite side x, and adjacent side ?
2. Then by the Pythagorean Theorem, the hypotenuse has length ?
x2 + 2, and this gives us sin θ and cos θ.
Now we have our integral.
ż 3 x2 + 2 + x ´ 3 (x2 + 2)2 dx = ż 3 x2 + 2 dx + ż x (x2 + 2)2 dx ´ ż 3 (x2 + 2)2 dx = 3 ?
2 arctan x ?
2 ´ 1 2x2 + 4 ´ 3 4 ?
2 arctan x ?
2 + x ?
2 x2 + 2 + C = 9 4 ?
2 arctan x ?
2 ´ 1 2(x2 + 2) ´ 3x 4(x2 + 2) + C = 9 4 ?
2 arctan x ?
2 ´ 2 + 3x 4(x2 + 2) + C S-22: This is already as simplified as we can make it using partial fraction. Indeed, this is the kind of term that could likely come out of the partial fraction decomposition of a 407 scarier rational function. So, we need to know how to integrate it. Similar to the last piece we integrated in Question 21, we can use the substitution x = tan θ, dx = sec2 θ dθ.
ż 1 (1 + x2)3 dx = ż sec2 θ (1 + tan2 θ)3 dθ = ż sec2 θ (sec2 θ)3 dθ = ż cos4 θ dθ = ż 1 + cos(2θ) 2 2 dθ = 1 4 ż (1 + cos(2θ))2 dθ = 1 4 ż 1 + 2 cos(2θ) + cos2(2θ) dθ = 1 4 ż 1 + 2 cos(2θ) + 1 2(1 + cos(4θ)) dθ = 1 4 ż 3 2 + 2 cos(2θ) + 1 2 cos(4θ)) dθ = 1 4 3 2θ + sin(2θ) + 1 8 sin(4θ) + C = 3 8θ + 1 4 sin(2θ) + 1 32 sin(4θ) + C = 3 8θ + 1 2 sin θ cos θ + 1 16 sin(2θ) cos(2θ) + C = 3 8θ + 1 2 sin θ cos θ + 1 8 sin θ cos θ(cos2 θ ´ sin2 θ) + C = 3 8 arctan x + x 2(1 + x2) + 1 8 x 1 + x2 1 ´ x2 1 + x2 + C θ 1 x ?
1 + x2 = 3 8 arctan x + 3x3 + 5x 8(1 + x2)2 + C To change our variables from θ to x, recall we used the substitution x = tan θ. So, we draw a right triangle with angle θ, opposite side length x, and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length ?
1 + x2. This allows us to find sin θ and cos θ.
S-23: Our integrand is already as simplified as the method of partial fractions can make it. The first term is easy to antidifferentiate. The second term would be easier if it were broken into two pieces: one where the numerator is a constant, and one where the numerator is a multiple of x.
ż 3x + 3x + 1 x2 + 5 + 3x (x2 + 5)2 dx = 3 2x2 + ż 1 x2 + 5 + 3x x2 + 5 + 3x (x2 + 5)2 dx = 3 2x2 + ż 1 x2 + 5dx + ż 3x x2 + 5 + 3x (x2 + 5)2 dx 408 The first integral looks similar to the derivative of arctangent. For the second integral, we use the substitution u = x2 + 5, du = 2x dx.
= 3 2x2 + 1 5 ż 1 x ?
5 2 + 1 dx + ż 3/2 u + 3/2 u2 du For the first integral, use the substitution w = x ?
5, dw = 1 ?
5 dx.
= 3 2x2 + 1 ?
5 ż 1 w2 + 1dw + 3 2 log |u| ´ 3 2u = 3 2x2 + 1 ?
5 arctan w + 3 2 log |x2 + 5| ´ 3 2x2 + 10 + C = 3 2x2 + 1 ?
5 arctan x ?
5 + 3 2 log |x2 + 5| ´ 3 2x2 + 10 + C S-24: If our denominator were all sines, we could use the substitution x = sin θ. To that end, we apply the identity cos2 θ = 1 ´ sin2 θ.
ż cos θ 3 sin θ + cos2 θ ´ 3 dθ = ż cos θ 3 sin θ + 1 ´ sin2 θ ´ 3 dθ = ż cos θ 3 sin θ ´ sin2 θ ´ 2 dθ We use the substitution x = sin θ, dx = cos θ dθ.
= ż 1 3x ´ x2 ´ 2 dx = ż ´1 x2 ´ 3x + 2 dx = ż ´1 (x ´ 1)(x ´ 2) dx Now we can find a partial fraction decomposition.
´1 (x ´ 1)(x ´ 2) = A x ´ 1 + B x ´ 2 ´1 = A(x ´ 2) + B(x ´ 1) Setting x = 1 and x = 2, we see A = 1, B = ´1 Now, we can evaluate our integral.
ż cos θ 3 sin θ + cos2 θ ´ 3 dθ = ż ´1 (x ´ 1)(x ´ 2) dx = ż 1 x ´ 1 ´ 1 x ´ 2 dx = log |x ´ 1| ´ log |x ´ 2| + C = log ˇ ˇ ˇ ˇ x ´ 1 x ´ 2 ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ sin θ ´ 1 sin θ ´ 2 ˇ ˇ ˇ ˇ + C 409 S-25: This looks a lot like a rational function, but with the function et in place of the variable. So, we would like to make the substitution x = et, dx = etdt. Then dt = 1 et dx = 1 x dx.
ż 1 e2t + et + 1 dt = ż 1 x (x2 + x + 1)dx The factor x2 + x + 1 is an irreducible quadratic, so the denominator is completely factored. Now we can use partial fraction decomposition.
1 x (x2 + x + 1) = A x + Bx + C x2 + x + 1 1 = A(x2 + x + 1) + (Bx + C)x 1 = (A + B)x2 + (A + C)x + A The constant terms tell us A = 1; then the coefficient of x tells us C = ´A = ´1. Finally, the coefficient of x2 tells us B = ´A = ´1. Now we can evaluate our integral.
ż 1 e2t + et + 1 dt = ż 1 x (x2 + x + 1)dx = ż 1 x ´ x + 1 x2 + x + 1 dx = ż 1 x ´ x + 1/2 + 1/2 x2 + x + 1 dx (˚) = ż 1 x dx ´ ż x + 1/2 x2 + x + 1 dx ´ ż 1/2 x2 + x + 1 dx = log |x| ´ 1 2 log |x2 + x + 1| ´ ż 1/2 x2 + x + 1 dx In step (˚), we set ourselves up so that we could evaluate the second integral with the substitution u = x2 + x + 1. For the remaining integral, we complete the square, so that the integrand looks something like the derivative of arctangent.
= log |x| ´ 1 2 log |x2 + x + 1| ´ ż 1/2 x + 1 2 2 + 3 4 dx = log |x| ´ 1 2 log |x2 + x + 1| ´ 2 3 ż 1 2x+1 ?
3 2 + 1 dx 410 We use the substitution u = 2x+1 ?
3 , du = 2 ?
3.
= log |x| ´ 1 2 log |x2 + x + 1| ´ 1 ?
3 ż 1 u2 + 1 du = log |x| ´ 1 2 log |x2 + x + 1| ´ 1 ?
3 arctan u + C = log |x| ´ 1 2 log |x2 + x + 1| ´ 1 ?
3 arctan 2x + 1 ?
3 + C = log |et| ´ 1 2 log |e2t + et + 1| ´ 1 ?
3 arctan 2et + 1 ?
3 + C = t ´ 1 2 log |e2t + et + 1| ´ 1 ?
3 arctan 2et + 1 ?
3 + C S-26: Solution 1: We use the substitution u = ?
1 + ex.
Then du = ex 2 ?
1 + ex dx, so dx = 2u u2 ´ 1du.
ż ?
1 + ex dx = ż u ¨ 2u u2 ´ 1du = ż 2u2 u2 ´ 1du = ż 2(u2 ´ 1) + 2 u2 ´ 1 du = ż 2 + 2 u2 ´ 1 du We use a partial fraction decomposition on the fractional part of the integrand.
2 u2 ´ 1 = 2 (u ´ 1)(u + 1) = A u ´ 1 + B u + 1 = (A + B)u + (A ´ B) (u ´ 1)(u + 1) A + B = 0, A ´ B = 2 A = 1, B = ´1 ż ?
1 + ex dx = ż 2 + 2 u2 ´ 1 du = ż 2 + 1 u ´ 1 ´ 1 u + 1 du = 2u + log |u ´ 1| ´ log |u + 1| + C = 2u + log ˇ ˇ ˇ ˇ u ´ 1 u + 1 ˇ ˇ ˇ ˇ + C = 2 ?
1 + ex + log ˇ ˇ ˇ ˇ ?
1 + ex ´ 1 ?
1 + ex + 1 ˇ ˇ ˇ ˇ + C Solution 2: It might not occur to us right away to use the fruitful substitution in Solution 1. More realistically, we might start with the “inside function,” u = 1 + ex.
Then du = ex dx, so dx = 1 u´1du.
ż ?
1 + ex dx = ż ?u u ´ 1du 411 This isn’t quite a rational function, because we have a square root on top. If we could turn it into a rational function, we could use partial fraction. To that end, let w = ?u, dw = 1 2?udu, so du = 2wdw.
= ż w w2 ´ 12wdw = ż 2w2 w2 ´ 1dw = ż 2(w2 ´ 1) + 2 w2 ´ 1 dw = ż 2 + 2 w2 ´ 1dw Now we can use partial fraction decomposition.
2 w2 ´ 1 = 2 (w ´ 1)(w + 1) = A w ´ 1 + B w + 1 = (A + B)w + (A ´ B) (w ´ 1)(w + 1) A + B = 0, A ´ B = 2 A = 1, B = ´1 This allows us to antidifferentiate.
ż ?
1 + ex dx = ż 2 + 2 w2 ´ 1 dw = ż 2 + 1 w ´ 1 ´ 1 w + 1 dw = 2w + log |w ´ 1| ´ log |w + 1| + C = 2w + log ˇ ˇ ˇ ˇ w ´ 1 w + 1 ˇ ˇ ˇ ˇ + C = 2?u + log ˇ ˇ ˇ ˇ ?u ´ 1 ?u + 1 ˇ ˇ ˇ ˇ + C = 2 ?
1 + ex + log ˇ ˇ ˇ ˇ ?
1 + ex ´ 1 ?
1 + ex + 1 ˇ ˇ ˇ ˇ + C Remark: we also evaluated this integral using trigonometric substitution in Section 1.9, Question 26. In that question, we found the antiderivative to be 2 ?
1 + ex + 2 log ˇ ˇ1 ´ ?
1 + exˇ ˇ ´ x + C. These expressions are equivalent: log ˇ ˇ ˇ ˇ ?
1 + ex ´ 1 ?
1 + ex + 1 ˇ ˇ ˇ ˇ = log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ + log ˇ ˇ ˇ ˇ 1 ?
1 + ex + 1 ˇ ˇ ˇ ˇ = log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ + log ˇ ˇ ˇ ˇ 1 ?
1 + ex + 1 1 ´ ?
1 + ex 1 ´ ?
1 + ex ˇ ˇ ˇ ˇ = log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ + log ˇ ˇ ˇ ˇ 1 ´ ?
1 + ex 1 ´ (1 + ex) ˇ ˇ ˇ ˇ = log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ + log ˇ ˇ ˇ ˇ 1 ´ ?
1 + ex ´ex ˇ ˇ ˇ ˇ = log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ + log ˇ ˇ ˇ1 ´ ?
1 + ex ˇ ˇ ˇ ´ log | ´ ex| = 2 log ˇ ˇ ˇ ?
1 + ex ´ 1 ˇ ˇ ˇ ´ x 412 S-27: (a) Let’s graph y = 10 ?
25 ´ x2. We start with the endpoints: (3, 5 2) and (4, 10 3 ). Then we consider the first derivative: d dx " 10 ?
25 ´ x2 = 10x ?
25 ´ x23 Over the interval [3, 4], this is always positive, so our function is increasing over the entire interval. The second derivative, d2 dx2 " 10 ?
25 ´ x2 = d dx # 10x ?
25 ´ x23 + = 10(2x2 + 25) ?
25 ´ x25 , is always positive, so our function is concave up over the entire interval. So, the region R is: 3 4 y = 10 √ 25−x2 x y (b) Let V1 be the solid obtained by revolving R about the x–axis. The portion of V1 with x–coordinate between x and x + dx is obtained by rotating the red vertical strip in the figure on the left below about the x–axis. That portion is a disk of radius 10 ?
25´x2 and thickness dx. The volume of this disk is π 10 ?
25´x2 2 dx. So the total volume of V1 is ż 4 3 π 10 ?
25 ´ x2 2 dx = 100π ż 4 3 1 25 ´ x2 dx = 100π ż 4 3 1 (5 ´ x)(5 + x) dx = 10π ż 4 3 1 5 ´ x + 1 5 + x dx = 10π h ´ log(5 ´ x) + log(5 + x) i4 3 = 10π h ´ log 1 + log 9 + log 2 ´ log 8 i = 10π log 9 4 = 20π log 3 2 3 4 y = 10 √ 25−x2 x y 3 4 5 2 x = q 25 −100 y2 x y (c) We’ll use horizontal washers as in Example 1.6.5 of the CLP-2 text.
• We cut R into thin horizontal strips of width dy as in the figure on the right above.
413 • When we rotate R about the y–axis, each strip sweeps out a thin washer – whose outer radius is rout = 4, and – whose inner radius is rin = b 25 ´ 100 y2 when y ě 10 ?
25´32 = 10 4 = 5 2 (see the red strip in the figure on the right above), and whose inner radius is rin = 3 when y ď 5 2 (see the blue strip in the figure on the right above) and – whose thickness is dy and hence – whose volume is π(r2 out ´ r2 in)dy = π 100 y2 ´ 9 dy when y ě 5 2 and whose volume is π(r2 out ´ r2 in)dy = 7π dy when y ď 5 2 and • As our bottommost strip is at y = 0 and our topmost strip is at y = 10 3 (since at the top x = 4 and y = 10 ?
25´x2 = 10 ?
25´42 = 10 3 ), the volume is ż 10/3 5/2 π 100 y2 ´ 9 dy + ż 5/2 0 7π dy = π h ´ 100 y ´ 9y i10/3 5/2 + 35 2 π = π h ´ 30 + 40 ´ 30 + 45 2 i + 35 2 π = 20π S-28: In order to find the area between the curves, we need to know which one is on top, and which on the bottom. Let’s start by finding where they meet.
4 3 + x2 = 2 x(x + 1) 2x2 + 2x = 3 + x2 x2 + 2x ´ 3 = 0 (x ´ 1)(x + 3) = 0 In the interval [ 1 4, 3], the curves only meet at x = 1. So, to find which is on top and on bottom in the intervals [ 1 4, 1) and (1, 3], it suffices to check some point in each interval.
x 4 3+x2 2 x(x+1) Top: 1/2 16/13 8/3 2 x(x+1) 2 4/7 1/3 4 3+x2 So, 2 x(x+1) is the top function when 1 4 ď x ă 1, and 4 3+x2 is the top function when 1 ă x ď 3. Then the area we want to find is: Area = ż 1 1 4 2 x(x + 1) ´ 4 3 + x2 dx + ż 3 1 4 3 + x2 ´ 2 x(x + 1) dx 414 We’ll need to antidifferentiate both these functions. We can antidifferentiate 2 x(x + 1) using partial fraction decomposition.
2 x(x + 1) = A x + B x + 1 = (A + B)x + A x(x + 1) A = 2, B = ´2 ż 2 x(x + 1) dx = ż 2 x ´ 2 x + 1 dx = 2 log |x| ´ 2 log |x + 1| + C = 2 log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ + C We can antidifferentiate 4 3 + x2 using the substitution u = x ?
3, du = 1 ?
3 dx.
ż 4 3 + x2 dx = ż 4 3 1 + x ?
3 2 dx = ż 4 ?
3 3 (1 + u2) du = 4 ?
3 arctan u + C = 4 ?
3 arctan x ?
3 + C Now, we can find our area.
Area = ż 1 1 4 2 x(x + 1) ´ 4 3 + x2 dx + ż 3 1 4 3 + x2 ´ 2 x(x + 1) dx = 2 log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ ´ 4 ?
3 arctan x ?
3 1 1/4 + 4 ?
3 arctan x ?
3 ´ 2 log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ 3 1 = 2 log 1 2 ´ 4 ?
3 ¨ π 6 ´ 2 log 1 5 + 4 ?
3 arctan 1 4 ?
3 + 4 ?
3 ¨ π 3 ´ 2 log 3 4 ´ 4 ?
3 ¨ π 6 + 2 log 1 2 = 2 log 5 3 + 4 ?
3 arctan 1 4 ?
3 415 S-29: (a) To antidifferentiate 1 t2 ´ 9, we use a partial fraction decomposition.
1 t2 ´ 9 = 1 (t ´ 3)(t + 3) = A t ´ 3 + B t + 3 = (A + B)t + 3(A ´ B) (t ´ 3)(t + 3) A + B = 0, A ´ B = 1 3 A = 1 6, B = ´1 6 F(x) = ż x 1 1 t2 ´ 9 dx = ż x 1 1/6 t ´ 3 ´ 1/6 t + 3 dx = 1 6 log |t ´ 3| ´ 1 6 log |t + 3| x 1 = 1 6 log |x ´ 3| ´ 1 6 log |x + 3| ´ 1 6 log 2 + 1 6 log 4 = 1 6 log ˇ ˇ ˇ ˇ2 ¨ x ´ 3 x + 3 ˇ ˇ ˇ ˇ (b) Rather than differentiate our answer from (a), we use the Fundamental Theorem of Calculus Part 1 to conclude F1(x) = d dx "ż x 1 1 t2 ´ 9dt = 1 x2 ´ 9 Solutions to Exercises 1.11 — Jump to TABLE OF CONTENTS S-1: The absolute error is the difference between the two values: |1.387 ´ 1.5| = 0.113 The relative error is the absolute error divided by the exact value: 0.113 1.387 « 0.08147 The percent error is 100 times the relative error: « 8.147% S-2: Midpoint rule: 416 x y 2 10 Trapezoidal rule: x y 2 10 S-3: (a) Differentiating, we find f 2(x) = ´x2 + 7x ´ 6. Since f 2(x) is quadratic, we have a pretty good idea of what it looks like.
• It factors as f (x) = ´(x ´ 6)(x ´ 1), so its two roots are at x = 6 and x = 1.
• The “flat part” of the parabola is at x = 3.5 (since this is exactly half way between x = 1 and x = 6; alternately, we can check that f 3(3.5) = 0).
• Since the coefficient of x2 is negative, f (x) is increasing from ´8 to 3.5, then decreasing from 3.5 to 8.
Therefore, over the interval [1, 6], the largest positive value of f 2(x) occurs when x = 3.5, and this is f 2(3.5) = ´(3.5 ´ 6)(3.5 ´ 1) = 6.25.
417 x y y = f 2(x) 1 6 6.25 So, we take M = 6.25.
(b) We differentiate further to find f (4)(x) = ´2. This is constant everywhere, so we take L = | ´ 2| = 2.
S-4: Let’s start by differentiating.
f (x) = x sin x + 2 cos x f 1(x) = x cos x + sin x ´ 2 sin x = x cos x ´ sin x f 2(x) = ´x sin x + cos x ´ cos x = ´x sin x For any value of x, | sin x| ď 1. When ´3 ď x ď 2, then |x| ď 3. So, it is true (and not unreasonably sloppy) that f 2(x) ď 3 whenever x is in the interval [´3, 2]. So, we can take M = 3.
Note that |f 2(x)| is actually smaller than 3 whenever x is in the interval [´3, 2], because when x = ´3, sin x ‰ 1. In fact, since 3 is pretty close to π, sin 3 is pretty small. (The actual maximum value of |f 2(x)| when ´3 ď x ď 2 is about 1.8.) However, we find parameters like M for the purpose of computing error bounds. There is often not much to be gained from taking the time to find the actual maximum of a function, so we content ourselves with reasonable upper bounds. Question 31 has a further investigation of “sloppy” bounds like this.
S-5: (a) Let f (x) = cos x. Then f (4)(x) = cos x, so |f (4)(x)| ď 1 when ´π ď x ď π. So, using L = 1, we find the upper bound of the error using Simpson’s rule with n = 4 is: L(b ´ a)5 180n4 = (2π)5 180 ¨ 44 = π5 180 ¨ 8 « 0.2 The error bound comes from Theorem 1.11.12 in the CLP-2 text. We used a calculator to find the approximate decimal value.
418 (b) We use the general form of Simpson’s rule (Equation 1.11.9 in the CLP-2 text) with ∆x = b´a n = 2π 4 = π 2 .
A « ∆x 3 ( f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4)) = π/2 3 f (´π) + 4f (´π 2 ) + 2f (0) + 4f ( π 2 ) + f (π) = π 6 (´1 + 4(0) + 2(1) + 4(0) ´ 1) = 0 (c) To find the actual error in our approximation, we compare the approximation from (b) to the exact value of A. In fact, A = 0: this is a fact you’ve probably seen before by considering the symmetry of cosine, but it’s easy enough to calculate: A = ż π ´π cos x dx = sin π ´ sin(´π) = 0 So, our approximation was exactly the same as our exact value. The absolute error is 0.
Remark: the purpose of this question was to remind you that the error bounds we calculate are not (usually) the same as the actual error. Often our approximations are better than we give them credit for. In normal circumstances, we would be approximating an integral precisely to avoid evaluating it exactly, so we wouldn’t find our exact error. The bound is a quick way of ensuring that our approximation is not too far off.
S-6: Using Theorem 1.11.12 in the CLP-2 text, the error using the trapezoidal rule as described is at most M(b ´ a)3 12 ¨ n2 = M 48 ď 3 48 = 1 16.
So, we’re really being asked to find a function with the maximum possible error using the trapezoidal rule, given its second derivative.
With that in mind, our function should have the largest second derivative possible: let’s set f 2(x) = 3 for every x. Then: f 2(x) = 3 f 1(x) = 3x + C f (x) = 3 2x2 + Cx + D 419 for some constants C and D. Now we can find the exact and approximate values of ż 1 0 f (x) dx.
Exact: ż 1 0 f (x) dx = ż 1 0 3 2x2 + Cx + D dx = 1 2x3 + C 2 x2 + Dx 1 0 = 1 2 + C 2 + D Approximate: ż 1 0 f (x) dx « ∆x 1 2 f (0) + f (1 2) + 1 2 f (1) = 1 2 1 2(D) + 3 8 + C 2 + D + 1 2 3 2 + C + D = 1 2 9 8 + C + 2D = 9 16 + C 2 + D So, the absolute error associated with the trapezoidal approximation is: ˇ ˇ ˇ ˇ 1 2 + C 2 + D ´ 9 16 + C 2 + D ˇ ˇ ˇ ˇ = 1 16 So, for any constants C and D, f (x) = 3 2x2 + Cx + D has the desired error.
Remark: contrast this question with Question 5. In this problem, our absolute error was exactly as bad as the bound predicted, but sometimes it is much better. The thing to remember is that, in general, we don’t know our absolute error. We only guarantee that it’s not any worse than some worst-case-scenario bound.
S-7: Under any reasonable assumptions5, my mother is older than I am.
S-8: (a) Since both expressions are positive, and 1 24 ď 1 12, the inequality is true.
(b) False. The reasoning is the same as in Question 7. The error bound given by Theorem 1.11.12 is always better for the trapezoid rule, but this doesn’t necessarily mean the error is better.
To see how the trapezoid approximation could be better than the corresponding midpoint approximation in some cases, consider the function f (x) sketched below.
5 Anyone caught trying to come up with a scenario in which I am older than my mother will be sent to maximum security grad school.
420 x y a b The trapezoidal approximation of ż b a f (x) dx with n = 1 misses the thin spike, and gives a mild underapproximation. By contrast, the midpoint approximation with n = 1 takes the spike as the height of the entire region, giving a vast overapproximation.
x y a b trapezoidal x y a b midpoint S-9: True. Because f (x) is positive and concave up, the graph of f (x) is always below the top edges of the trapezoids used in the trapezoidal rule.
x y y = f (x) S-10: According to Theorem 1.11.12 in the CLP-2 text, the error associated with the Simpson’s rule approximation is no more than L 180 (b ´ a)5 n4 , where L is a constant such that |f (4)(x)| ď L for all x in [a, b]. If L = 0, then the error is no more than 0 regardless of a, b, or n–that is, the approximation is exact.
Any polynomial f (x) of degree at most 3 has f (4)(x) = 0 for all x. So, any polynomial of degree at most 3 is an acceptable answer. For example, f (x) = 5x3 ´ 27, or f (x) = x2.
421 S-11: • For all three approximations, ∆x = b ´ a n = 30 ´ 0 6 = 5.
• For the trapezoidal rule and Simpson’s rule, the x-values where we evaluate 1 x3 + 1 start at x = a = 0 and move up by ∆x = 5: x0 = 0, x1 = 5, x2 = 10, x3 = 15, x4 = 20, x5 = 25, and x6 = 30.
0 x0 5 x1 10 x2 15 x3 20 x4 25 x5 30 x6 • For the midpoint rule, the x-values where we evaluate 1 x3 + 1 start at x = 2.5 = x0+x1 2 and move up by ∆x = 5: ¯ x1 = 2.5, ¯ x2 = 7.5, ¯ x3 = 12.5, ¯ x4 = 17.5, ¯ x5 = 22.5, and ¯ x6 = 27.5.
0 x0 5 x1 10 x2 15 x3 20 x4 25 x5 30 x6 2.5 ¯ x1 7.5 ¯ x2 12.5 ¯ x3 17.5 ¯ x4 22.5 ¯ x5 27.5 ¯ x6 • Following Equation 1.11.2 in the CLP-2 text, the midpoint rule approximation is: ż 30 0 1 x3 + 1 dx « h f ( ¯ x1) + f ( ¯ x2) + ¨ ¨ ¨ + f ( ¯ xn) i ∆x = 1 (2.5)3+1 + 1 (7.5)3+1 + 1 (12.5)3+1 + 1 (17.5)3+1 + 1 (22.5)3+1 + 1 (27.5)3+1 5 • Following Equation 1.11.6 in the CLP-2 text, the trapezoidal rule approximation is: ż 30 0 1 x3 + 1 dx « h 1 2 f (x0) + f (x1) + f (x2) + ¨ ¨ ¨ + f (xn´1) + 1 2 f (xn) i ∆x = 1/2 03 + 1 + 1 53 + 1 + 1 103 + 1 + 1 153 + 1 + 1 203 + 1 + 1 253 + 1 + 1/2 303 + 1 5 • Following Equation 1.11.9 in the CLP-2 text, the Simpson’s rule approximation is: ż 30 0 1 x3 + 1 dx « h f (x0)+ 4f (x1)+ 2f (x2)+ 4f (x3)+ 2f (x4)+ 4f (x5)+ f (x6) i ∆x 3 = h 1 03 + 1+ 4 53 + 1+ 2 103 + 1+ 4 153 + 1+ 2 203 + 1+ 4 253 + 1+ 1 303 + 1 i5 3 S-12: By Equation 1.11.2 in the CLP-2 text, the midpoint rule approximation to şb a f (x) dx with n = 3 is ż b a f (x) dx « f ( ¯ x1) + f ( ¯ x2) + f ( ¯ x3) ∆x 422 where ∆x = b´a 3 and x0 = a x1 = a + ∆x x2 = a + 2∆x x3 = b ¯ x1 = x0+x1 2 ¯ x2 = x1+x2 2 ¯ x3 = x2+x3 2 For this problem, a = 0, b = π and f (x) = sin x, so that ∆x = π 3 and x0 = 0 x1 = π 3 x2 = 2π 3 x3 = π ¯ x1 = π 6 ¯ x2 = π 2 ¯ x3 = 5π 6 0 x0 π x3 π/3 x1 2π/3 x2 π/6 ¯ x1 π/2 ¯ x2 5π/6 ¯ x3 Therefore, ż π 0 sin x dx « sin π 6 + sin π 2 + sin 5π 6 π 3 = 1 2 + 1 + 1 2 π 3 = 2π 3 S-13: Let f (x) denote the diameter at height x. As in Example 1.6.6 of the CLP-2 text, we slice V into thin horizontal “pancakes”, which in this case are circular.
dx f (x) x • We are told that the pancake at height x is a circular disk of diameter f (x) and so • has cross-sectional area π f (x) 2 2 and thickness dx and hence • has volume π f (x) 2 2dx.
Hence the volume of V is ż 40 0 π h f (x) 2 i2 dx « π 4 10 h 1 2 f (0)2 + f (10)2 + f (20)2 + f (30)2 + 1 2 f (40)2i = π 4 10 h 1 2242 + 162 + 102 + 62 + 1 242i = 688 ˆ 2.5π = 1720π « 5403.5 where we have approximated the integral using the trapezoidal rule with ∆x = 10, and used a calculator to get a decimal approximation.
423 S-14: Let f (x) be the diameter a distance x from the left end of the log. If we slice our log into thin disks, the disks x metres from the left end of the log has • radius f (x) 2 , • width dx, and so • volume π f (x) 2 2 dx = π 4 f (x)2 dx.
x f (x) dx Using Simpson’s Rule with ∆x = 1, the volume of the log is: V = ż 6 0 π 4 f (x)2 dx « π 4 1 3 h f (0)2 + 4f (1)2 + 2f (2)2 + 4f (3)2 + 2f (4)2 + 4f (5)2 + f (6)2i = π 12 h 1.22 + 4(1)2 + 2(0.8)2 + 4(0.8)2 + 2(1)2 + 4(1)2 + 1.22i = π 12(16.72) « 4.377 m3 where we used a calculator to approximate the decimal value.
S-15: At height x metres, let the circumference of the tree be c(x). The corresponding radius is c(x) 2π , so the corresponding cross-sectional area is π c(x) 2π 2 = c(x)2 4π .
424 dx c(x) x The height of a very thin cross-sectional disk is dx, so the volume of a cross-sectional disk is c(x)2 4π dx. Therefore, total volume of the tree is: ż 8 0 c(x)2 4π dx « 1 4π 2 3 h c(0)2 + 4c(2)2 + 2c(4)2 + 4c(6)2 + c(8)2i = 1 6π h 1.22 + 4(1.1)2 + 2(1.3)2 + 4(0.9)2 + 0.22i = 12.94 6π « 0.6865 where we used Simpson’s rule with ∆x = 2 and n = 4 to approximate the value of the integral based on the values of c(x) given in the table.
S-16: For both approximations, ∆x = 10 and n = 6.
(a) The Trapezoidal Rule gives V = ż 60 0 A(h) dh « 10 h 1 2 A(0) + A(10) + A(20) + A(30) + A(40) + A(50) + 1 2 A(60) i = 363,500 (b) Simpson’s Rule gives V = ż 60 0 A(h) dh « 10 3 h A(0) + 4A(10) + 2A(20) + 4A(30) + 2A(40) + 4A(50) + A(60) i = 367,000 425 S-17: Call the curve in the graph y = f (x). It looks like f (2) = 3 f (3) = 8 f (4) = 7 f (5) = 6 f (6) = 4 We’re estimating ş6 2 f (x) dx with n = 4, so ∆x = 6´2 4 = 1.
(a) The trapezoidal rule gives T4 = 3 2 + 8 + 7 + 6 + 4 2 ˆ 1 = 49 2 (b) Simpson’s rule gives S4 = 1 3 3 + 4 ˆ 8 + 2 ˆ 7 + 4 ˆ 6 + 4 ˆ 1 = 77 3 S-18: Let f (x) = sin(x2). Then f 1(x) = 2x cos(x2) and f 2(x) = 2 cos(x2) ´ 4x2 sin(x2).
Since |x2| ď 1 when |x| ď 1, and |sin θ| ď 1 and |cos θ| ď 1 for all θ, we have ˇ ˇ ˇ2 cos(x2) ´ 4x2 sin(x2) ˇ ˇ ˇ ď 2| cos(x2)| + 4x2| sin(x2)| ď 2 ˆ 1 + 4 ˆ 1 ˆ 1 = 2 + 4 = 6 We can therefore choose M = 6, and it follows that the error is at most M[b ´ a]3 24n2 ď 6 ¨ [1 ´ (´1)]3 24 ¨ 10002 = 2 106 = 2 ¨ 10´6 S-19: Setting f (x) = 2x4 and b ´ a = 1 ´ (´2) = 3, we compute f 2(x) = 24x2. The largest value of 24x2 on the interval [´2, 1] occurs at x = ´2, so we can take M = 24 ¨ (´2)2 = 96.
Thus the total error for the midpoint rule with n = 60 points is bounded by M(b ´ a)3 24n2 = 96 ˆ 33 24 ˆ 60 ˆ 60 = 3 100 That is: we are guaranteed our absolute error is certainly no more6 than 3 100, and using the bound stated in the problem we cannot give a better guarantee. (The second part of the previous sentence comes from the fact that we used the smallest possible M: if we had used a larger value of M, we would still have some true statement about the error, for example “the error is no more than 5 100,” but it would not be the best true statement we could make.) 6 This is what the error bound always tells us.
426 S-20: (a) Since a = 0, b = 2 and n = 6, we have ∆x = b´a n = 2´0 6 = 1 3, and so x0 = 0, x1 = 1 3, x2 = 2 3, x3 = 1, x4 = 4 3, x5 = 5 3, and x6 = 2. Since Simpson’s Rule with n = 6 in general is ∆x 3 f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + 4f (x5) + f (x6) , the desired approximation is 1/3 3 (´3)5 + 4 1 3 ´ 3 5 + 2 2 3 ´ 3 5 + 4(´2)5 + 2 4 3 ´ 3 5 + 4 5 3 ´ 3 5 + (´1)5 (b) Here f (x) = (x ´ 3)5, which has derivatives f 1(x) = 5(x ´ 3)4 f 2(x) = 20(x ´ 3)3 f (3)(x) = 60(x ´ 3)2 f (4)(x) = 120(x ´ 3).
For 0 ď x ď 2, (x ´ 3) runs from ´3 to ´1, so the maximum absolute values are found at x = 0, giving M = 20 ¨ |0 ´ 3|3 = 540 and L = 120 ¨ |0 ´ 3| = 360. Consequently, for the Midpoint Rule with n = 100, |EM| ď M(b ´ a)3 24n2 = 540 ˆ 23 24 ˆ 104 = 180 104 ; whereas for Simpson’s Rule with n = 10, |ES| ď 360 ˆ 25 180 ˆ 104 = 64 104.
Since 64 ă 180, Simpson’s Rule results in a smaller error bound.
S-21: In general the error in approximating şb a f (x) dx using Simpson’s rule with n steps is bounded by L(b´a) 180 (∆x)4 where ∆x = b´a n and L ě |f (4)(x)| for all a ď x ď b. In this case, a = 1, b = 5, n = 4 and f (x) = 1 x. We need to find L, so we differentiate.
f 1(x) = ´ 1 x2 f 2(x) = 2 x3 f (3)(x) = ´ 6 x4 f (4)(x) = 24 x5 and ˇ ˇf (4)(x) ˇ ˇ ď 24 for all x ě 1 So we may take L = 24 and ∆x = 5´1 4 = 1, which leads to |Error | ď 24(5 ´ 1) 180 (1)4 = 24 45 = 8 15 427 S-22: In general, the error in approximating şb a f (x) dx using Simpson’s rule with n steps is bounded by L(b ´ a) 180 (∆x)4 where ∆x = b ´ a n and L ě |f (4)(x)| for all a ď x ď b. In this case, a = 0, b = 1, n = 6 and f (x) = e´2x + 3x3. We need to find L, so we differentiate.
f 1(x) = ´2e´2x + 9x2 f 2(x) = 4e´2x + 18x f (3)(x) = ´8e´2x + 18 f (4)(x) = 16e´2x Since e´2x = 1 e2x , we see f (4)(x) is a positive, decreasing function. So, its maximum occurs when x is as small as possible. In the interval [0, 1], that means x = 0.
ˇ ˇf (4)(x) ˇ ˇ ď f (0) = 16 for all x ě 0 So, we take L = 16 and ∆x = 1´0 6 = 1 6.
|Error | ď L(b ´ a) 180 (∆x)4 = 16(1 ´ 0) 180 (1/6)4 = 16 180 ˆ 64 = 1 180 ˆ 34 = 1 14580 S-23: For both approximations, a = 1, b = 2, n = 4, f (x) = 1 x and ∆x = b´a n = 1 4.
Then x0 = 1, x1 = 5 4, x2 = 3 2, x3 = 7 4, and x4 = 2.
1 x0 5/4 x1 3/2 x2 7/4 x3 2 x4 (a) T4 =∆x 1 2 f (x0) + f (x1) + f (x2) + f (x3) + 1 2 f (x4) =∆x 1 2 f (1) + f (5/4) + f (3/2) + f (7/4) + 1 2 f (2) = 1 4 1 2 ˆ 1 + 4 5 + 2 3 + 4 7 + 1 2 ˆ 1 2 (b) S4 = ∆x 3 f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4) = ∆x 3 f (1) + 4f (5/4) + 2f (3/2) + 4f (7/4) + f (2) = 1 12 1 + 4 ˆ 4 5 + 2 ˆ 2 3 + 4 ˆ 4 7 + 1 2 (c) In this case, a = 1, b = 2, n = 4 and f (x) = 1 x. We need to find L, so we differentiate.
f 1(x) = ´ 1 x2 f 2(x) = 2 x3 f (3)(x) = ´ 6 x4 f (4)(x) = 24 x5 428 So, ˇ ˇf (4)(x) ˇ ˇ ď 24 for all x in the interval [1, 2] We take L = 24.
|Error | ď L(b ´ a)5 180 ˆ n4 ď 24(2 ´ 1)5 180 ˆ 44 = 24 180 ˆ 44 = 1 1920 S-24: Set a = 0 and b = 8. Since we have information about s(x) when x is 0, 2, 4, 6, and 8, we set ∆x = b´a n = 2, so n = 4. (Recall with the trapezoid rule and Simpson’s rule, n = 4 intervals actually uses the value of the function at 5 points.) We could perform the trapezoidal approximations with fewer intervals, for example n = 2, but this would involve ignoring some of the points we’re given. Since the question asks for the best estimation we can give, we use n = 4 intervals and no fewer.
(a) T4 = ∆x 1 2s(0) + s(2) + s(4) + s(6) + 1 2s(8) = 2 1.00664 2 + 1.00543 + 1.00435 + 1.00331 + 1.00233 2 = 8.03515 S4 = ∆x 3 s(0) + 4s(2) + 2s(4) + 4s(6) + s(8) = 2 3 1.00664 + 4 ˆ 1.00543 + 2 ˆ 1.00435 + 4 ˆ 1.00331 + 1.00233 « 8.03509 (b) The information ˇ ˇs(k)(x) ˇ ˇ ď k 1000, with k = 2, tells us |s2(x)| ď 2 1000 for all x in the interval [0, 8]. So, we take K2 (also called M in your text) to be 2 1000.
Then the absolute error associated with our trapezoid rule approximation is at most ˇ ˇ ˇ ˇ ż b a f (x) dx ´ Tn ˇ ˇ ˇ ˇ ď K2(b ´ a)3 12n2 ď 2 1000 ¨ 83 12(4)2ď 0.00533 For k = 4, we see |s(4)(x)| ď 4 1000 for all x in the interval [0, 8]. So, we take K4 (also called L in your text) to be 4 1000.
Then the absolute error associated with our Simpson’s rule approximation is at most ˇ ˇ ˇ ˇ ż b a f (x) dx ´ Sn ˇ ˇ ˇ ˇ ď K4(b ´ a)5 180n4 ď 4 1000 ¨ 85 180(4)4ď 0.00284 429 S-25: In this case, a = 1, b = 4. Since ´2 ď f 2(x) ď 0 over the relevant interval, we take M = 2. (Remember M is an upper bound on |f 2(x)|, not f 2(x).) So we need n to obey 2(4 ´ 1)3 12n2 ď 0.001 ð ñ n2 ě 2(3)3 12 1000 = 27000 6 = 9000 2 = 4500 One obvious allowed n is 100. Since ?
4500 « 67.01, and n has to be a whole number, any n ě 68 works.
S-26: Denote by f (x) the width of the pool x feet from the left-hand end. From the sketch, f (0) = 0, f (2) = 10, f (4) = 12, f (6) = 10, f (8) = 8, f (10) = 6, f (12) = 8, f (14) = 10 and f (16) = 0.
A cross-section of the pool x feet from the left end is half of a circular disk with diameter f (x) (so, radius f (x) 2 ) and thickness dx. So, the volume of the part of the pool with x–coordinate running from x to (x + dx) is 1 2π f (x) 2 2 dx = π 8 [ f (x)]2 dx.
The total volume is given by the following integral.
V = π 8 ż 16 0 f (x)2 dx « π 8 ¨ ∆x 3 h f (0)2 + 4f (2)2 + 2f (4)2 + 4f (6)2 + 2f (8)2 + 4f (10)2 + 2f (12)2 + 4f (14)2+ f (16)2i = π 8 ¨ 2 3 h 0 + 4(10)2 + 2(12)2 + 4(10)2 + 2(8)2 + 4(6)2 + 2(8)2 + 4(10)2 + 0 i = 472 3 π « 494 ft3 S-27: (a) The Trapezoidal Rule with n = 4, a = 0, b = 1, and ∆x = 1 4 gives: W = 2π10´6 ż 1 0 rg(r) dr « 2π10´6∆x 1 2x0g(x0) + x1g(x1) + x2g(x2) + x3g(x3) + 1 2x4g(x4) = 2π10´6 1 4 h1 20g(0) + 1 4g 1 4 + 1 2g 1 2 + 3 4g 3 4 + 1 2g(1) i = π10´6 1 2 h8100 4 + 8144 2 + 3 ¨ 8170 4 + 8190 2 i = 32639π 4 ¨ 106 « 0.025635 (b) Using the product rule, the integrand f (r) = 2π10´6rg(r) obeys f 2(r) = 2π10´6 d dr g(r) + rg1(r) = 2π10´6 2g1(r) + rg2(r) and hence, for 0 ď r ď 1, ˇ ˇf 2(r) ˇ ˇ ď 2π10´6 2 ˆ 200 + 1 ˆ 150 = 1.1π10´3 430 So, |Error| ď 1.1π10´3(1 ´ 0)3 12(4)2 ď 1.8 ˆ 10´5 S-28: (a) Let f (x) = 1 x, a = 1, b = 2 and ∆x = b´a 6 = 1 6. Using Simpson’s rule: ż 2 1 1 x dx « ∆x 3 h f (1) + 4f 7 6 + 2f 8 6 + 4f 9 6 + 2f 10 6 + 4f 11 6 + f (2) i = 1 18 h 1 + 24 7 + 12 8 + 24 9 + 12 10 + 24 11 + 1 2 i « 0.6931698 (b) The integrand is f (x) = 1 x. The first four derivatives of f (x) are: f 1(x) = ´ 1 x2, f 2(x) = 2 x3, f (3)(x) = ´ 6 x4, f (4)(x) = 24 x5 On the interval 1 ď x ď 2, the fourth derivative is never bigger in magnitude than L = 24.
|En| ď L(b ´ a)5 180n4 = 24(2 ´ 1)5 180n4 = 4 30n4 So, we want an even number n such that 4 30n4 ď 0.00001 = 1 105 n4 ě 40000 3 n ě 4 c 40000 3 « 10.7 So, any even number greater than or equal to 12 will do.
S-29: (a) From the figure, we see that the magnitude of |f 4(x)| never exceeds 310 for 0 ď x ď 2. So, the absolute error is bounded by 310(2 ´ 0)5 180 ˆ 84 ď 0.01345 (b) We want to choose n such that: 310(2 ´ 0)5 180 ˆ n4 ď 10´4 n4 ě 310 ˆ 25 180 104 n ě 10 4 c 310 ˆ 32 180 « 27.2 For Simpson’s rule, n must be even, so any even integer obeying n ě 28 will guarantee us the requisite accuracy.
431 S-30: Let g(x) = ż x 0 sin( ?
t) dt. By the Fundamental Theorem of Calculus Part 1, g1(x) = sin(?x). By its definition, f (x) = g(x2), so we use the chain rule to differentiate f (x).
f 1(x) = 2xg1(x2) = 2x sin x f 2(x) = 2 sin x + 2x cos x Since | sin x|, | cos x| ď 1, we have |f 2(x)| ď 2 + 2|x| and, for 0 ď t ď 1, |f 2(t)| ď 4. When the trapezoidal rule with n subintervals is applied, the resulting error En obeys En ď 4(1 ´ 0)3 12n2 = 1 3n2 We want an integer n such that 1 3n2 ď 0.000005 n2 ě 4 12 ˆ 0.000005 n ě c 1 3 ˆ 0.000005 « 258.2 Any integer n ě 259 will do.
S-31: (a) When 0 ď x ď 1, then x2 ď 1 and x + 1 ě 1, so |f 2(x)| = x2 |x + 1| ď 1 1 = 1.
(b) To find the maximum value of a function over a closed interval, we test the function’s values at the endpoints of the interval and at its critical points inside the interval. The critical points are where the function’s derivative is zero or does not exist.
The function we’re trying to maximize is |f 2(x)| = x2 |x+1| = x2 x+1 = f 2(x) (since our interval only contains nonnegative numbers). So, the critical points occur when f 3(x) = 0 or does not exist. We find f 3(x) Using the quotient rule.
f 3(x) = (x + 1)(2x) ´ x2 (x + 1)2 = x2 + 2x (x + 1)2 0 = x(x + 2) x + 1 0 = x or x = ´1 or x = ´2 The only critical point in [0, 1] is x = 0. So, the extrema of f 2(x) over [0, 1] will occur at its endpoints. Indeed, since f 3(x) ě 0 for all x in [0, 1], f 2(x) is increasing over this interval, so its maximum occurs at x = 1. That is, |f 2(x)| ď f 2(1) = 1 2 432 (c) The absolute error using the midpoint rule is at most M(b ´ a)3 24n2 . Using M = 1, if we want this to be no more than 10´5, we find an acceptable value of n with the following calculation: M(b ´ a)3 24n2 ď 10´5 1 24n2 ď 10´5 (b ´ a = 1, M = 1) 105 24 ď n2 n ě 65 (d) The absolute error using the midpoint rule is at most M(b ´ a)3 24n2 . Using M = 1 2, if we want this to be no more than 10´5, we find an acceptable value of n with the following calculation: M(b ´ a)3 24n2 ď 10´5 1 48n2 ď 10´5 (b ´ a = 1, M = 1 2) 105 48 ď n2 n ě 46 Remark: how accurate you want to be in these calculations depends a lot on your circumstances. Imagine, for instance, that you were finding M by hand, using this to find n by hand, then programming a computer to evaluate the approximation. For a simple integral like this, the difference between computing time for 65 intervals versus 46 is likely to be miniscule. So, there’s not much to be gained by the extra work in (b).
However, if your original sloppy M gave you something like n = 1000000, you might want to put some time into improving it, to shorten computation time. Moreover, if you were finding the approximation by hand, the difference between adding 46 terms and adding 65 terms would be considerable, and you would probably want to put in the effort up front to find the most accurate M possible.
S-32: Before we can take our Simpson’s rule approximation of ż x 1 1 t dt, we need to know how many intervals to use. That means we need to bound our error, which means we need to bound d4 dt4 !
1 t ) .
d dt "1 t = ´ 1 t2 d2 dt2 "1 t = 2 t3 d3 dt3 "1 t = ´ 6 t4 d4 dt4 "1 t = 24 t5 433 So, over the interval [1, 3], ˇ ˇ ˇ ˇ d4 dt4 "1 t ˇ ˇ ˇ ˇ ď 24.
Now, we can find an appropriate n to ensure our error will be be less than 0.1 for any x in [1, 3]: L(b ´ a)5 180n4 ă 0.1 24(x ´ 1)5 180n4 ă 1 10 n4 ą 24 ¨ (x ´ 1)5 18 Because x ´ 1 ď 2 for every x in [1, 3], if n4 ą 24 ¨ 25 18 , then n4 ą 24 ¨ (x ´ 1)5 18 for every allowed x.
n4 ą 24 ¨ 25 18 = 128 3 n ą 4 c 128 3 « 2.6 Since n must be even, n = 4 is enough intervals to guarantee our error is not too high for any x in [1, 3]. Now we find our Simpson’s rule approximation with n = 4, a = 1, b = x, and ∆x = x ´ 1 4 . The points where we evaluate 1 t are: x0 =1 x1 = 1 + x ´ 1 4 x2 = 1 + 2x ´ 1 4 x3 = 1 + 3x ´ 1 4 x4 = 1 + 4x ´ 1 4 = x + 3 4 = x + 1 2 = 3x + 1 4 = x 1 x0 x+3 4 x1 x+1 2 x2 3x+1 4 x3 x x3 log x = ż x 1 1 t dt « ∆x 3 1 x0 + 4 x1 + 2 x2 + 4 x3 + 1 x4 = x ´ 1 12 1 + 16 x + 3 + 4 x + 1 + 16 3x + 1 + 1 x = f (x) Below is a graph of our approximation f (x) and natural logarithm on the same axes. The natural logarithm function is shown red and dashed, while our approximating function is solid blue. Our approximation appears to be quite accurate for small, positive values of x.
434 x y y = f (x) y = log x 2 4 6 8 10 1 2 3 S-33: First, we want a strategy for approximating arctan 2. Our hints are that involves integrating 1 1 + x2, which is the antiderivative of arctangent, and the number π 4 , which is the same as arctan(1). With that in mind: ż 2 1 1 1 + x2 dx = arctan(2) ´ arctan(1) = arctan(2) ´ π 4 So, arctan(2) = π 4 + ż 2 1 1 1 + x2 dx (˚) We won’t know the value of the integral exactly, but we’ll have an approximation A bounded by some positive error bound ε. Then, ´ε ď ż 2 1 1 1 + x2 dx ´ A !
ď ε A ´ ε ď ż 2 1 1 1 + x2 dx !
ď A + ε So, from (˚), π 4 + A ´ ε ď arctan(2) ď π 4 + A + ε Which approximation should we use? We’re given the fourth derivative of 1 1 + x2, which is the derivative we need for Simpson’s rule. Simpson’s rule is also usually quite efficient, and we’re very interested in not adding up dozens of terms, so we choose Simpson’s rule.
Now that we’ve chosen Simpson’s rule, we should decide how many intervals to use. In order to bound our error, we need to find a bound for the fourth derivative. To that end, define N(x) = 24(5x4 ´ 10x2 + 1). Then N1(x) = 24(20x3 ´ 20x) = 480x(x2 ´ 1), which 435 is positive over the interval [1, 2]. So, N(x) ď N(2) = 24(5 ¨ 24 ´ 10 ¨ 22 + 1) = 984 when 1 ď x ď 2. Furthermore, let D(x) = (x2 + 1)5. If 1 ď x ď 2, then D(x) ě 25. Now we can find a reasonable value of L: |f (4)(x)| = ˇ ˇ ˇ ˇ 24(5x4 ´ 10x2 + 1) (x2 + 1)5 ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ N(x) D(x) ˇ ˇ ˇ ˇ ď 984 25 = 123 4 = 30.75 So, we take L = 30.75.
We want hπ 4 + A ´ ε, π 4 + A + ε i to look something like hπ 4 + 0.321, π 4 + 0.323 i . Note ε is half the length of the first interval. Half the length of the second interval is 0.001 = 1 1000. So, we want a value of ε that is no larger than this. Now we can find our n: L(b ´ a)5 180 ¨ n4 ď 1 1000 30.75 180 ¨ n4 ď 1 1000 n4 ě 30.75 ˆ 1000 180 n ě 4 c 30750 180 « 3.62 So, we choose n = 4), and are guaranteed that the absolute error in our approximation will be no more than 30.75 180 ¨ 44 ă 0.00067.
Since n = 4, then ∆x = b ´ a n = 1 4, so: x0 = 1 x1 = 5 4 x2 = 3 2 x3 = 7 4 x4 = 2 Now we can find our Simpson’s rule approximation A: ż 1 0 1 1 + x2 dx « ∆x 3 f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4) = 1/4 3 f (1) + 4f (5/4) + 2f (3/2) + 4f (7/4) + f (2) = 1 12 1 1 + 1 + 4 25/16 + 1 + 2 9/4 + 1 + 4 49/16 + 1 + 1 4 + 1 = 1 12 1 2 + 4 ¨ 16 25 + 16 + 2 ¨ 4 9 + 4 + 4 ¨ 16 49 + 16 + 1 5 = 1 12 1 2 + 64 41 + 8 13 + 64 65 + 1 5 « 0.321748 = A 436 As we saw before, the error associated with this approximation is at most 30.75 180 ¨ 44 ă 0.00067 = ε. So, A ´ ε ď ż 2 1 1 1 + x2 dx ď A + ε ñ 0.321748 ´ 0.00067 ď ż 2 1 1 1 + x2 dx ď 0.321748 + 0.00067 ñ 0.321078 ď ż 2 1 1 1 + x2 dx ď 0.322418 ñ 0.321 ď ż 2 1 1 1 + x2 dx ď 0.323 ñ π 4 + 0.321 ď ż 2 1 1 1 + x2 dx + π 4 ď π 4 + 0.323 ñ π 4 + 0.321 ď arctan(2) ď π 4 + 0.323 This is precisely what we wanted to show.
Solutions to Exercises 1.12 — Jump to TABLE OF CONTENTS S-1: If b = ˘8, then our integral is improper because one limit is not a real number.
Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, x = 1 and x = ´1. Since one limit of integration is 0, the integral is improper if b ě 1 or if b ď ´1.
Below, we’ve graphed 1 x2´1 to make it clearer why values of b in (´1, 1) are the only values that don’t result in an improper integral when the other limit of integration is a = 0.
x y y = 1 x2´1 ´1 1 437 S-2: Since the integrand is continuous for all real x, the only kind of impropriety available to us is to set b = ˘8.
S-3: For large values of x, |red function| ď (blue function) and 0 ď (blue function). If the blue function’s integral converged, then the red function’s integral would as well (by the comparison test, Theorem 1.12.17 in the CLP-2 text). Since one integral converges and the other diverges, the blue function is g(x) and the red function is f (x).
S-4: False. The inequality goes the “wrong way” for Theorem 1.12.17 in the CLP-2 text: the area under the curve f (x) is finite, but the area under g(x) could be much larger, even infinitely larger.
For example, if f (x) = e´x and g(x) = 1, then 0 ď f (x) ď g(x) and ż 8 1 f (x) dx converges, but ż 8 1 g(x) dx diverges.
S-5: (a) Not enough information to decide. For example, consider h(x) = 0 versus h(x) = ´1. In both cases, h(x) ď f (x). However, ż 8 0 0 dx converges to 0, while ż 8 0 ´1 dx diverges.
Note: if we had also specified 0 ď h(x), then we would be able to conclude that ş8 0 h(x) dx converges by the comparison test.
(b) Not enough information to decide. For example, consider h(x) = f (x) versus h(x) = g(x). In both cases, f (x) ď h(x) ď g(x).
(c) ż 8 0 h(x) dx converges.
• From the given information, |h(x)| ď 2f (x).
• We claim ż 8 0 2f (x) dx converges.
– We can see this by writing ż 8 0 2f (x) dx = 2 ż 8 0 f (x) dx and noting that the second integral converges.
– Alternately, we can use the limiting comparison test, Theorem 1.12.22 in the CLP-2 text. Since f (x) ě 0, ż 8 0 f (x) dx converges, and lim xÑ8 2f (x) f (x) = 2 (the limit exists), we conclude ż 8 0 2f (x) dx converges.
438 • So, comparing h(x) to 2f (x), by the comparison test (Theorem 1.12.17 in the CLP-2 text) ż 8 0 h(x) dx converges.
S-6: The denominator is zero when x = 1, but the numerator is not, so the integrand has a singularity (infinite discontinuity) at x = 1. Let’s replace the limit x = 1 with a variable that creeps toward 1.
ż 1 0 x4 x5 ´ 1 dx = lim tÑ1´ ż t 0 x4 x5 ´ 1 dx To evaluate this integral we use the substitution u = x5, du = 5x4dx. When x = 0 we have u = 0, and when x = t we have u = t5, so ż 1 0 x4 x5 ´ 1 dx = lim tÑ1´ ż t 0 x4 x5 ´ 1 dx = lim tÑ1´ ż u=t5 u=0 1 5(u ´ 1) du = lim tÑ1´ 1 5 log |u ´ 1| t5 0 !
= lim tÑ1´ 1 5 log |t5 ´ 1| = ´8 The limit does not exist, so the integral diverges.
S-7: The denominator of the integrand is zero when x = ´1, but the numerator is not. So, the integrand has a singularity (infinite discontinuity) at x = ´1. This is the only “source of impropriety” in this integral, so we only need to make one break in the domain of integration.
ż 2 ´2 1 (x + 1)4/3 dx = lim tÑ´1´ ż t ´2 1 (x + 1)4/3 dx + lim tÑ´1+ ż 2 t 1 (x + 1)4/3 dx Let’s start by considering the left limit.
lim tÑ´1´ ż t ´2 1 (x + 1)4/3 dx = lim tÑ´1´ ´ 3 (x + 1)1/3 ˇ ˇ ˇ ˇ t ´2 !
= lim tÑ´1´ ´ 3 (t + 1)1/3 + 3 (´1)1/3 = 8 Since this limit diverges, the integral diverges. (A similar argument shows that the second integral diverges. Either one of them diverging is enough to conclude that the original integral diverges.) S-8: First, let’s identify all “sources of impropriety.” The integrand has a singularity when 4x2 ´ x = 0, that is, when x(4x ´ 1) = 0, so at x = 0 and x = 1 4. Neither of these are in our domain of integration, so the only “source of impropriety” is the unbounded domain of integration.
We could antidifferentiate this function (it looks like a nice candidate for a trig substitution), but is seems easier to use a comparison. For large values of x, the term x2 439 will be much larger than x, so we might guess that our integral behaves similarly to ş8 1 1 ?
4x2 dx = ş8 1 1 2x dx.
For all x ě 1, ?
4x2 ´ x ď ?
4x2 = 2x. So, 1 ?
4x2´x ě 1 2x. Note ş8 1 1 2x dx diverges: lim tÑ8 ż t 1 1 2x dx = lim tÑ8 1 2 log x t 1 = lim tÑ8 1 2 log t = 8 So: • 1 2x and 1 ?
4x2´x are defined and continuous for all x ě 1, • 1 2x ě 0 for all x ě 1, • 1 ?
4x2´x ě 1 ?
4x2 = 1 2x for all x ě 1, and • ş8 1 1 2x dx diverges.
By the comparison test, Theorem 1.12.17 in the CLP-2 text, the integral does not converge.
S-9: The integrand is positive everywhere. So, either the integral converges to some finite number, or it is infinite. We want to generate a guess as to which it is.
When x is small, ?x ą x2, so we might guess that our integral behaves like the integral of 1 ?x when x is near to 0. On the other hand, when x is large, ?x ă x2, so we might guess that our integral behaves like the integral of 1 x2 as x goes to infinity. This is the hunch that drives the following work: 0 ď 1 x2 + ?x ď 1 ?x and the integral ż 1 0 dx ?x converges by Example 1.12.9 in the CLP-2 text 0 ď 1 x2 + ?x ď 1 x2 and the integral ż 8 1 dx x2 converges by Example 1.12.8 in the CLP-2 text Note that 1 x2+?x is defined and continuous for all x ą 0, 1 ?x is defined and continuous for x ą 0, and 1 x2 is defined and continuous for x ě 1. So, the integral converges by the comparison test, Theorems 1.12.17 and 1.12.24 in the CLP-2 text.
S-10: There are two “sources of impropriety”: the two (infinite) limits of integration. So, we break our integral into two pieces.
ż 8 ´8 cos x dx = ż 0 ´8 cos x dx + ż 8 0 cos x dx = lim aÑ8 "ż 0 ´a cos x dx # + lim bÑ8 "ż b 0 cos x dx # 440 These are easy enough to antdifferentiate.
= lim aÑ8 [sin 0 ´ sin(´a)] + lim bÑ8 [sin b ´ sin 0] = DNE Since the limits don’t exist, the integral diverges. (It happens that both limits don’t exist; even if only one failed to exist, the integral would still diverge.) S-11: There are two “sources of impropriety”: the ´8 and the +8. So, we break our integral into two pieces.
ż 8 ´8 sin x dx = ż 0 ´8 sin x dx + ż 8 0 sin x dx = lim aÑ8 "ż 0 ´a sin x dx # + lim bÑ8 "ż b 0 sin x dx # = lim aÑ8 [´ cos 0 + cos(´a)] + lim bÑ8 [´ cos b + cos 0] = DNE Since the limits don’t exist, the integral diverges. (It happens that both limits don’t exist; even if only one failed to exist, the integral would diverge.) Remark: it’s very tempting to think that this integral should converge, because as an odd function the area to the right of the x-axis “cancels out” the area to the left when the limits of integration are symmetric. One justification for not using this intuition is given in Example 1.12.11 in the CLP-2 text. Here’s another: In Question 10 we saw that ş8 ´8 cos x dx diverges. Since sin x = cos(x ´ π/2), the area bounded by sine and the area bounded by cosine over an infinite region seem to be the same–only shifted by π/2. So if ş8 ´8 sin x dx = 0, then we ought to also have ş8 ´8 cos x dx = 0, but we saw in Question 10 this is not the case.
x y y = sin x y = cos x S-12: First, we check that the integrand has no singularities. The denominator is always positive when x ě 10, so our only “source of impropriety” is the infinite limit of integration.
We further note that, for large values of x, the integrand resembles x4 x5 = 1 x . So, we have a 441 two-part hunch: that the integral diverges, and that we can show it diverges by comparing it to ż 8 10 1 x dx.
In order to use the comparison test, we’d need to show that x4 ´ 5x3 + 2x ´ 7 x5 + 3x + 8 ě 1 x . If this is true, it will be difficult to prove–and it’s not at all clear that it’s true. So, we will use the limiting comparison test instead, Theorem 1.12.22 in the CLP-2 text, with g(x) = 1 x, f (x) = x4 ´ 5x3 + 2x ´ 7 x5 + 3x + 8 , and a = 10.
• Both f (x) and g(x) are defined and continuous for all x ą 0, so in particular they are defined and continuous for x ě 10.
• g(x) ě 0 for all x ě 10 • ż 8 10 g(x) dx diverges.
• Using l’Hˆ opital’s rule (5 times!), or simply dividing both the numerator and denominator by x5 (the common leading term), tells us: lim xÑ8 f (x) g(x) = lim xÑ8 x4´5x3+2x´7 x5+3x+8 1 x = lim xÑ8 x ¨ x4 ´ 5x3 + 2x ´ 7 x5 + 3x + 8 = lim xÑ8 x5 ´ 5x4 + 2x2 ´ 7x x5 + 3x + 8 = 1 That is, the limit exists and is nonzero.
By the limiting comparison test, we conclude ż 8 10 f (x) dx diverges.
S-13: Our domain of integration is finite, so the only potential “sources of impropriety” are infinite discontinuities in the integrand. To find these, we factor.
ż 10 0 x ´ 1 x2 ´ 11x + 10 dx = ż 10 0 x ´ 1 (x ´ 1)(x ´ 10) dx A removable discontinuity doesn’t affect the integral.
= ż 10 0 1 x ´ 10 dx Use the substitution u = x ´ 10, du = dx. When x = 0, u = ´10, and when x = 10, u = 0.
= ż 0 ´10 1 u du This is a p-integral with p = 1. From Example 1.12.9 and Theorem 1.12.20 in the CLP-2 text, we know it diverges.
442 S-14: You might think that, because the integrand is odd, the integral converges to 0. This is a common mistake– see Example 1.12.11 in the CLP-2 text, or Question 11 in this section. In the absence of such a shortcut, we use our standard procedure: identifying problem spots over the domain of integration, and replacing them with limits.
There are two “sources of impropriety,” namely x Ñ +8 and x Ñ ´8. So, we split the integral in two, and treat the two halves separately. The integrals below can be evaluated with the substitution u = x2 + 1, 1 2du = xdx.
ż +8 ´8 x x2 + 1dx = ż 0 ´8 x x2 + 1dx + ż +8 0 x x2 + 1dx ż 0 ´8 x x2 + 1dx = lim RÑ8 ż 0 ´R x x2 + 1dx = lim RÑ8 1 2 log(x2 + 1) ˇ ˇ ˇ 0 ´R = lim RÑ8 1 2 h log 1 ´ log(R2 + 1) i = lim RÑ8 ´1 2 log(R2 + 1) = ´8 ż +8 0 x x2 + 1dx = lim RÑ8 ż R 0 x x2 + 1dx = lim RÑ8 1 2 log(x2 + 1) ˇ ˇ ˇ R 0 = lim RÑ8 1 2 h log(R2 + 1) ´ log 1 i = lim RÑ8 1 2 log(R2 + 1) = +8 Both halves diverge, so the whole integral diverges.
Once again: after we found that one of the limits diverged, we could have stopped and concluded that the original integrand diverges. Don’t make the mistake of thinking that 8 ´ 8 = 0. That can get you into big trouble. 8 is not a normal number. For example 28 = 8. So if 8 were a normal number we would have both 8 ´ 8 = 0 and 8 ´ 8 = 28 ´ 8 = 8.
S-15: We don’t want to antidifferentiate this integrand, so let’s use a comparison. Note the integrand is positive when x ą 0.
For any x, | sin x| ď 1, so | sin x| x3/2 + x1/2 ď 1 x3/2 + x1/2.
Since x = 0 and x Ñ 8 both cause the integral to be improper, we need to break it into two pieces. Since both terms in the denominator give positive numbers when x is positive, 1 x3/2 + x1/2 ď 1 x3/2 and 1 x3/2 + x1/2 ď 1 x1/2. That gives us two options for comparison.
When x is positive and close to zero, x1/2 ě x3/2, so we guess that we should compare our integrand to 1 x1/2 near the limit x = 0. In contrast, when x is very large, x1/2 ď x3/2, so we guess that we should compare our integrand to 1 x3/2 as x goes to infinity.
443 | sin x| x3/2 + x1/2 ď 1 x1/2 and the integral ż 1 0 dx x1/2 converges by the p-test, Example 1.12.9 in the CLP-2 text | sin x| x3/2 + x1/2 ď 1 x3/2 and the integral ż 8 1 dx x3/2 converges by the p-test, Example 1.12.8 in the CLP-2 text Now we have all the data we need to apply the comparison test, Theorems 1.12.17 and 1.12.24 in the CLP-2 text.
• | sin x| x3/2 + x1/2 , 1 x1/2 , and 1 x3/2 are defined and continuous for x ą 0 • 1 x1/2 and 1 x3/2 are nonnegative for x ě 0 • | sin x| x3/2 + x1/2 ď 1 x1/2 for all x ą 0 and ż 1 0 1 x1/2 dx converges, so ż 1 0 | sin x| x3/2 + x1/2 dx converges.
• | sin x| x3/2 + x1/2 ď 1 x3/2 for all x ě 1 and ż 8 1 1 x3/2 dx converges, so ż 8 1 | sin x| x3/2 + x1/2 dx converges.
Therefore, our integral ż 8 0 | sin x| x3/2 + x1/2 dx converges.
S-16: The integrand is positive everywhere, so either the integral converges to some finite number or it is infinite. There are two potential “sources of impropriety” — a possible singularity at x = 0 and the fact that the domain of integration extends to 8. So we split up the integral.
ż 8 0 x + 1 x1/3(x2 + x + 1) dx = ż 1 0 x + 1 x1/3(x2 + x + 1) dx + ż 8 1 x + 1 x1/3(x2 + x + 1) dx Let’s develop a hunch about whether the integral converges or diverges. When x « 0, x2 and x are both a lot smaller than 1, so we guess we should compare the integrand to 1 x1/3.
x + 1 x1/3(x2 + x + 1) « 1 x1/3(1) = 1 x1/3 Note ş1 0 1 x1/3 dx converges by Example 1.12.9 in the CLP-2 text (it’s a p-type integral), so we guess ş1 0 x+1 x1/3(x2+x+1) dx converges as well.
When x is very large, x2 is much bigger than x, which is much bigger than 1, so we guess we should compare the integrand to 1 x4/3.
x + 1 x1/3(x2 + x + 1) « x x1/3(x2) = 1 x4/3 444 Note ş8 1 1 x4/3 dx converges by Example 1.12.8 in the CLP-2 text (it’s a p-type integral), so we guess ş8 1 x+1 x1/3(x2+x+1) dx converges as well.
Now it’s time to verify our guesses with the limiting comparison test, Theorems 1.12.22 and 1.12.25 in the CLP-2 text. Be careful: our “«” signs are not strong enough to use either the limiting comparison test or the comparison test, they are only enough to suggest a reasonable function to compare to.
• x+1 x1/3(x2+x+1) , 1 x1/3 , and 1 x4/3 are defined and continuous for all x ą 0 • 1 x1/3 and 1 x4/3 are positive for all x ą 0 • ş1 0 1 x1/3 dx and ş8 1 1 x4/3 dx both converge • lim xÑ0 x+1 x1/3(x2+x+1) 1 x1/3 = lim xÑ0 x + 1 x2 + x + 1 = 0 + 1 0 + 0 + 1 = 1; in particular, this limit exists.
• Using the limiting comparison test (Theorem 1.12.25 in the CLP-2 text), ş1 0 x+1 x1/3(x2+x+1) dx converges.
• lim xÑ8 x+1 x1/3(x2+x+1) 1 x4/3 = lim xÑ0 x(x + 1) x2 + x + 1 = 1; in particular, this limit exists.
• Using the limiting comparison test (Theorem 1.12.22 in the CLP-2 text), ş8 1 x+1 x1/3(x2+x+1) dx converges.
We conclude ż 8 0 x + 1 x1/3(x2 + x + 1) dx converges.
S-17: To find the volume of the solid, we cut it into horizontal slices, which are thin circular disks. At height y, the disk has radius x = 1 y and thickness dy, so its volume is π y2dy. The base of the solid is at height y = 1, and its top is at height y = 1 a. So, the volume of the entire solid is: ż 1/a 1 π y2dy = ´π y 1/a 1 = π(1 ´ a) If we imagine sliding a closer and closer to 0, the volume increases, getting closer and closer to π units, but never quite reaching it.
So, the statement is false. For example, if we set M = 4, no matter which a we choose our solid has volume strictly less than M.
Remark: we’ve seen before that ş1 0 1 x dx diverges. If we imagine the solid that would result from choosing a = 0, it would have a scant volume of π cubic units, but a silhouette (side view) of infinite area.
S-18: Our goal is to decide when this integral diverges, and where it converges. We will leave q as a variable, and antidifferentiate. In order to antidifferentiate without knowing 445 q, we’ll need different cases. The integrand is x´5q, so when ´5q ‰ ´1, we use the power rule (that is, ş xn dx = xn+1 n+1 ) to antidifferentiate. Note x(´5q)+1 = x1´5q = 1 x5q´1.
ż t 1 1 x5q dx = $ ’ ’ ’ ’ ’ ’ & ’ ’ ’ ’ ’ ’ % h x1´5q 1´5q it 1 with 1 ´ 5q ą 0 if q ă 1 5 h log x it 1 if q = 1 5 h 1 (1´5q)x5q´1 it 1 with 5q ´ 1 ą 0 if q ą 1 5 = $ ’ & ’ % 1 1´5q(t1´5q ´ 1) with 1 ´ 5q ą 0 if q ă 1 5 log t if q = 1 5 1 5q´1(1 ´ 1 t5q´1) with 5q ´ 1 ą 0 if q ą 1 5.
Therefore, ż 8 1 1 x5q dx = lim tÑ8 ż t 1 1 x5q dx = $ ’ ’ ’ ’ ’ ’ & ’ ’ ’ ’ ’ ’ % 1 1´5q lim tÑ8 t1´5q ´ 1 = 8 if q ă 1 5 lim tÑ8 log t = 8 if q = 1 5 1 5q´1 1 ´ lim tÑ8 1 t5q´1 = 1 5q´1 if q ą 1 5.
The first two cases are divergent, and so the largest such value is q = 1 5. (Alternatively, we might recognize this as a “p-integral” with p = 5q, and recall that the p-integral diverges precisely when p ď 1.) S-19: This integrand is a nice candidate for the substitution u = x2 + 1, 1 2du = xdx.
Remember when we use substitution on a definite integral, we also need to adjust the limits of integration.
ż 8 0 x (x2 + 1)p dx = lim tÑ8 ż t 0 x (x2 + 1)p dx = lim tÑ8 1 2 ż t2+1 1 1 up du = lim tÑ8 1 2 ż t2+1 1 u´p du = $ ’ ’ ’ & ’ ’ ’ % 1 2 lim tÑ8 u1´p 1 ´ p t2+1 1 if p ‰ 1 1 2 lim tÑ8 h log |u| it2+1 1 if p = 1 = $ ’ & ’ % 1 2 lim tÑ8 1 1 ´ p h (t2 + 1)1´p ´ 1 i if p ‰ 1 1 2 lim tÑ8 h log(t2 + 1) i = 8 if p = 1 446 At this point, we can see that the integral diverges when p = 1. When p ‰ 1, we have the limit lim tÑ8 1/2 1 ´ p h (t2 + 1)1´p ´ 1 i = 1/2 1 ´ p lim tÑ8(t2 + 1)1´p ´ 1/2 1 ´ p Since t2 + 1 Ñ 8, this limit converges exactly when the exponent 1 ´ p is negative; that is, it converges when p ą 1, and diverges when p ă 1.
So, the integral in the question converges when p ą 1.
S-20: • First, we notice there is only one “source of impropriety:” the domain of integration is infinite. (The integrand has a singularity at t = 1, but this is not in the domain of integration, so it’s not a problem for us.) • We should try to get some intuition about whether the integral converges or diverges. When t Ñ 8, notice the integrand “looks like” the function 1 t4. We know ş8 1 1 t4 dt converges, because it’s a p-integral with p = 4 ą 1 (see Example 1.12.8 in the CLP-2 text). So, our integral probably converges as well. If we were only asked show it converges, we could use a comparison test, but we’re asked more than that.
• Since we guess the integral converges, we’ll need to evaluate it. The integrand is a rational function, and there’s no obvious substitution, so we use partial fractions.
1 t4 ´ 1 = 1 (t2 + 1)(t2 ´ 1) = 1 (t2 + 1)(t + 1)(t ´ 1) = At + B t2 + 1 + C t + 1 + D t ´ 1 Multiply by the original denominator.
1 = (At + B)(t + 1)(t ´ 1) + C(t2 + 1)(t ´ 1) + D(t2 + 1)(t + 1) (˚) Set t = 1.
1 = 0 + 0 + D(2)(2) ñ D = 1 4 Set t = ´1.
1 = 0 + C(2)(´2) + 0 ñ C = ´1 4 Simplify (˚) using D = 1 4 and C = ´1 4.
1 = (At + B)(t + 1)(t ´ 1)´1 4(t2 + 1)(t ´ 1) + 1 4(t2 + 1)(t + 1) = (At + B)(t + 1)(t ´ 1) + 1 2(t2 + 1) = At3 + B + 1 2 t2 ´ At + 1 2 ´ B 447 By matching up coefficients of corresponding powers of t, we find A = 0 and B = ´1 2.
ż 8 2 1 t4 ´ 1 dt = ż 8 2 ´1/2 t2 + 1 ´ 1/4 t + 1 + 1/4 t ´ 1 dt = lim RÑ8 ż R 2 ´1/2 t2 + 1 ´ 1/4 t + 1 + 1/4 t ´ 1 dt = lim RÑ8 ´1 2 arctan t ´ 1 4 log |t + 1| + 1 4 log |t ´ 1| R 2 = lim RÑ8 ´1 2 arctan t + 1 4 log ˇ ˇ ˇ ˇ t ´ 1 t + 1 ˇ ˇ ˇ ˇ R 2 = lim RÑ8 ´1 2 arctan R + 1 2 arctan 2 + 1 4 log ˇ ˇ ˇ ˇ R ´ 1 R + 1 ˇ ˇ ˇ ˇ ´ 1 4 log ˇ ˇ ˇ ˇ 2 ´ 1 2 + 1 ˇ ˇ ˇ ˇ We can use l’Hˆ opital’s rule to see lim RÑ8 R ´ 1 R + 1 = 1. Also note ´ log(1/3) = log 3.
= ´1 2 π 2 + 1 2 arctan 2 + 1 4 log 1 + 1 4 log 3 = log 3 ´ π 4 + 1 2 arctan 2 S-21: There are three singularities in the integrand: x = 0, x = 1, and x = 2. We’ll need to break up the integral at each of these places.
ż 5 ´5 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx = ż 0 ´5 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx + ż 1 0 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx + ż 2 1 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx + ż 5 2 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx This looks rather unfortunate. Let’s think again. If all of the integrals below converge, then we can write: ż 5 ´5 1 a |x| + 1 a |x ´ 1| + 1 a |x ´ 2| !
dx = ż 5 ´5 1 a |x| dx + ż 5 ´5 1 a |x ´ 1| dx + ż 5 ´5 1 a |x ´ 2| dx That looks a lot better. Also, we have a good reason to guess these integrals converge–they look like p-integrals with p = 1 2. Let’s take a closer look at each one.
ż 5 ´5 1 a |x| dx = ż 0 ´5 1 a |x| dx + ż 5 0 1 a |x| dx = 2 ż 5 0 1 a |x| dx (even function) = 2 ż 5 0 1 ?xdx 448 This is a p-integral, with p = 1 2. By Example 1.12.9 in the CLP-2 text (and Theorem 1.12.20, since the upper limit of integration is not 1), it converges. The other two pieces behave similarly.
ż 5 ´5 1 a |x ´ 1| dx = ż 1 ´5 1 a |x ´ 1| dx + ż 5 1 1 a |x ´ 1| dx Use u = x ´ 1, du = dx = ż 0 ´6 1 a |u| du + ż 4 0 1 a |u| dx = ż 6 0 1 ?udu + ż 4 0 1 ?udx Since our function is even, we use the reasoning of Example 1.2.10 in the CLP-2 text to consider the area under the curve when x ě 0, rather than when x ď 0. Again, these are p-integrals with p = 1 2, so they both converge. Finally: ż 5 ´5 1 a |x ´ 2| dx = ż 2 ´5 1 a |x ´ 2| dx + ż 5 2 1 a |x ´ 2| dx Use u = x ´ 2, du = dx.
= ż 0 ´7 1 a |u| du + ż 3 0 1 a |u| du = ż 7 0 1 ?udu + ż 3 0 1 ?udu Since p = 1 2, so they both converge.
We conclude our original integral, as the sum of convergent integrals, converges.
S-22: We can use integration by parts twice to find the antiderivative of e´x sin x, as in Example 1.7.10 of the CLP-2 text. To keep our work a little simpler, we’ll find the antiderivative first, then take the limit.
Let u = e´x, dv = sin x dx, so du = ´e´x dx and v = ´ cos x.
ż e´x sin x dx = ´e´x cos x ´ ż e´x cos x dx Now let u = e´x, dv = cos x dx, so du = ´e´x dx and v = sin x.
= ´e´x cos x ´ e´x sin x + ż e´x sin x dx = ´e´x cos x ´ e´x sin x ´ ż e´x sin x dx 449 All together, we found ż e´x sin x dx = ´e´x cos x ´ e´x sin x ´ ż e´x sin x dx + C 2 ż e´x sin x dx = ´e´x cos x ´ e´x sin x + C ż e´x sin x dx = ´ 1 2ex (cos x + sin x) + C (Remember, since C is an arbitrary constant, we can rename C 2 to simply C.) Now we can evaluate our improper integral.
ż 8 0 e´x sin x dx = lim bÑ8 ż b 0 e´x sin x dx = lim bÑ8 ´ 1 2ex (cos x + sin x) b 0 = lim bÑ8 1 2 ´ 1 2eb (cos b + sin b) To find the limit, we use the Squeeze Theorem (Theorem 1.4.17 in the CLP-1 text). Since | sin b|, | cos b| ď 1 for any b, we can use the fact that ´2 ď cos b + sin b ď 2 for any b.
´2 2eb ď 1 2eb (cos b + sin b) ď 2 2eb lim bÑ8 ´2 2eb = 0 = 2 2eb So, lim bÑ8 1 2eb (cos b + sin b) = 0 Therefore, 1 2 = lim bÑ8 1 2 ´ 1 2eb (cos b + sin b) That is, ż 8 0 e´x sin x dx = 1 2.
S-23: The integrand is positive everywhere. So either the integral converges to some finite number or it is infinite. There are two potential “sources of impropriety” — a possible singularity at x = 0 and the fact that the domain of integration extends to 8. So, we split up the integral.
ż 8 0 sin4 x x2 dx = ż 1 0 sin4 x x2 dx + ż 8 1 sin4 x x2 dx Let’s consider the first integral. By l’Hˆ opital’s rule (or recall Example 3.7.3 in the CLP-1 text), lim xÑ0 sin x x = lim xÑ0 cos x 1 = cos 0 = 1 450 Consequently, lim xÑ0 sin4 x x2 = lim xÑ0 sin2 x lim xÑ0 sin x x lim xÑ0 sin x x = 0 ˆ 1 ˆ 1 = 0 and the first integral is not even improper.
Now for the second integral. Since | sin x| ď 1, we’ll compare it to ş8 1 1 x2.
• sin4 x x2 and 1 x2 are defined and continuous for every x ě 1 • 0 ď sin4 x x2 ď 14 x2 = 1 x2 for every x ě 1 • ş8 1 1 x2 dx converges by Example 1.12.8 in the CLP-2 text (it’s a p-type integral with p ą 1) By the comparison test, Theorem 1.12.17 in the CLP-2 text, ż 8 1 sin4 x x2 dx converges.
Since ż 1 0 sin4 x x2 dx and ż 8 1 sin4 x x2 dx both converge, we conclude ż 8 0 sin4 x x2 dx converges as well.
S-24: Since the denominator is positive for all x ě 0, the integrand is continuous over [0, 8). So, the only “source of impropriety” is the infinite domain of integration.
Solution 1: Let’s try to use a direct comparison. Note x ex + ?x ě 0 whenever x ě 0. Also note that, for large values of x, ex is much larger than ?x. That leads us to consider the following inequalty: 0 ď x ex + ?x ď x ex If ş8 0 x ex dx converges, we’re in business. Let’s figure it out. The integrand looks like a candidate for integration by parts: take u = x, dv = e´x dx, so du = dx and v = ´e´x.
ż 8 0 x ex dx = lim bÑ8 ż b 0 x ex dx = lim bÑ8 h ´ x ex ib 0 + ż b 0 e´x dx !
= lim bÑ8 ´ b eb + ´e´xb 0 = lim bÑ8 ´ b eb ´ 1 eb + 1 = lim bÑ8 1 ´ b + 1 eb lo omo on numÑ8 denÑ8 = lim bÑ8 1 ´ 1 eb = 1 Using l’Hˆ opital’s rule, we see ş8 0 x ex dx converges. All together: • x ex and x ex+?x are defined and continuous for all x ě 0, • ˇ ˇ ˇ x ex+?x ˇ ˇ ˇ ď x ex , and 451 • ş8 0 x ex dx converges.
So, by Theorem 1.12.17 in the CLP-2 text, our integral ż 8 0 x ex + ?x dx converges.
Solution 2: Let’s try to use a different direct comparison from Solution 1, and avoid integration by parts. We’d like to compare to something like 1 ex , but the inequality goes the wrong way. So, we make a slight modification: we consider 2e´x/2. To that end, we claim x ă 2ex/2 for all x ě 0. We can prove this by noting the following two facts: • 0 ă 2 = 2e0/2, and • d dxtxu = 1 ď ex/2 = d dxt2ex/2u.
So, when x = 0, x ă 2ex/2, and then as x increases, 2ex/2 grows faster than x.
Now we can make the following comparison: 0 ď x ex + ?x ď x ex ă 2ex/2 ex = 2 ex/2 We have a hunch that ş8 0 2 ex/2 dx converges, just like ş8 0 1 ex dx. This is easy enough to prove. We can guess an antiderivative, or use the substitution u = x/2.
ż 8 0 2 ex/2 dx = lim RÑ8 ż R 0 2 ex/2 dx = lim RÑ8 ´ 4 ex/2 R 0 = lim RÑ8 4 e0 ´ 4 eR/2 R 0 = 4 Now we know: • 0 ď x ex+?x ď 2 ex/2, and • ş8 0 2 ex/2 dx converges.
• Furthermore, x ex+?x and 2 ex/2 are defined and continuous for all x ě 0.
By the comparison test (Theorem 1.12.17) in the CLP-2 text, we conclude the integral converges.
Solution 3: Let’s use the limiting comparison test (Theorem 1.12.22 in the CLP-2 text).
We have a hunch that our integral behaves similarly to ş8 0 1 ex dx, which converges (see Example 1.12.18 in the CLP-2 text). Unfortunately, if we choose g(x) = 1 ex (and, of course, f (x) = x ex+?x), then lim xÑ8 f (x) g(x) = lim xÑ8 x ex + ?x ¨ ex = lim xÑ8 x 1 + ?x ex lo omo on Ñ0 = 8 452 That is, the limit does not exist, so the limiting comparison test does not apply. (To find lim xÑ8 ?x ex , you can use l’Hˆ opital’s rule.) This setback encourages us to try a slightly different angle. If g(x) gave larger values, then we could decrease f (x) g(x). So, let’s try g(x) = 1 ex/2 = e´x/2. Now, lim xÑ8 f (x) g(x) = lim xÑ8 x ex + ?x ˜ 1 ex/2 = lim xÑ8 x ex/2 + ?x ex/2 Hmm... this looks hard. Instead of dealing with it directly, let’s use the squeeze theorem, Theorem 1.4.17 in the CLP-1 text.
0 ď x ex/2 + ?x ex/2 ď x ex/2 Using l’Hˆ opital’s rule, lim xÑ8 x ex/2 lo omo on numÑ8 denÑ8 = lim xÑ8 1 1 2ex/2 = 0 = lim xÑ8 0 So, by the squeeze theorem lim xÑ0 x ex+?x 1 ex/2 = 0. Since this limit exists, 1 ex/2 is a reasonable function to use in the limiting comparison test (provided its integral converges). So, we need to show that ş8 0 1 ex/2 dx converges. This can be done by simply evaluating it: ż 8 0 1 ex/2 dx = lim bÑ8 ż b 0 e´x/2 dx = lim bÑ8 ´1 2 e´x/2b 0 = lim bÑ8 ´1 2 1 eb/2 ´ 1 = 1 2 So, all together: • The functions x ex+?x and 1 ex/2 are defined and continuous for all x ě 0, and 1 ex/2 ě 0 for all x ě 0.
• ş8 0 1 ex/2 dx converges.
• The limit lim xÑ8 x ex +?x 1 ex/2 exists (it’s equal to 0).
• So, the limiting comparison test (Theorem 1.12.17 in the CLP-2 text) tells us that ş8 0 x ex+?x dx converges as well.
S-25: There are two sources of error: the upper bound is t, rather than infinity, and we’re using an approximation with some finite number of intervals, n. Our plan is to first find a value of t that introduces an error of no more than 1 210´4. That is, we’ll find a value of t such that ş8 t e´x 1+x dx ď 1 210´4. After that, we’ll find a value of n that approximates 453 şt 0 e´x x+1 dx to within 1 210´4. Then, all together, our error will be at most 1 210´4 + 1 210´4 = 10´4, as desired. (Note we could have broken up the error in another way—it didn’t have to be 1 210´4 and 1 210´4. This will give us one of many possible answers.) Let’s find a t such that ş8 t e´x 1+x dx ď 1 210´4. For all x ě 0, 0 ă e´x 1+x ď e´x, so ż 8 t e´x 1 + x dx ď ż 8 t e´x dx = e´t (˚) ď 1 210´4 where (˚) is true if t ě ´ log 1 210´4 « 9.90 Choose, for example, t = 10.
Now it’s time to decide how many intervals we’re going to use to approximate ż t 0 e´x x + 1 dx. Again, we want our error to be less than 1 210´4. To bound our error, we need to know the second derivative of e´x x+1.
f (x) = e´x 1 + x ù ñ f 1(x) = ´ e´x 1 + x ´ e´x (1 + x)2 ù ñ f 2(x) = e´x 1 + x + 2 e´x (1 + x)2 + 2 e´x (1 + x)3 Since f 2(x) is positive, and decreases as x increases, |f 2(x)| ď f 2(0) = 5 ù ñ |En| ď 5(10 ´ 0)3 24n2 = 5000 24n2 = 625 3n2 and |En| ď 1 210´4 if 625 3n2 ď 1 210´4 ð ñ n2 ě 1250 ˆ 104 3 ð ñ n ě c 1.25 ˆ 107 3 « 2041.2 So t = 10 and n = 2042 will do the job. There are many other correct answers.
S-26: (a) Since f (x) is odd, using the reasoning of Example 1.2.11 in the CLP-2 text, ż ´1 ´8 f (x) dx = lim tÑ8 ż ´1 ´t f (x) dx = lim tÑ8 ´ ż t 1 f (x) dx = ´ lim tÑ8 ż t 1 f (x) dx Since ż 8 1 f (x) dx converges, the last limit above converges. Therefore, ż ´1 ´8 f (x) dx converges.
454 (b) Since f (x) is even, using the reasoning of Example 1.2.10 in the CLP-2 text, ż ´1 ´8 f (x) dx = lim tÑ8 ż ´1 ´t f (x) dx = lim tÑ8 ż t 1 f (x) dx = lim tÑ8 ż t 1 f (x) dx Since ż 8 1 f (x) dx converges, the last limit above converges. Since f (x) is continuous everywhere, by Theorem 1.12.20 in the CLP-2 text, ż 8 ´1 f (x) dx converges (note the adjusted lower limit). Then, since ż 8 ´8 f (x) dx = ż ´1 ´8 f (x) dx + ż 8 ´1 f (x) dx and both terms converge, our original integral converges as well.
S-27: Define F(x) = şx 0 1 etdt.
F(x) = ż x 0 1 et dt = ´ 1 et x 0 = 1 e0 ´ 1 ex ă 1 e0 = 1 So, the statement is false: there is no x such that F(x) = 1. For every real x, F(x) ă 1 e0 = 1.
We note here that lim xÑ8 ż x 0 1 et dt = 1. So, as x grows larger, the gap between F(x) and 1 grows infintesimally small. But there is no real value of x where F(x) is exactly equal to 1.
Solutions to Exercises 1.13 — Jump to TABLE OF CONTENTS S-1: (A) Note ş f 1(x) f (x) dx = ş u du if we substitute u = f (x). This is the kind of integrand described in (I). It’s quite possible that a u = f (x) substitution would work on the others, as well, but (I) is the most reliable kind of integrand for a u = f (x) substitution.
(B) A trigonometric substitution usually allows us to cancel out a square root containing a quadratic function, as in (IV).
(C) We can often antidifferentiate the product of a polynomial with an exponential function using integration by parts: see Examples 1.7.1, 1.7.6 in the CLP-2 text. If we let u be the polynomial function and dv be the exponential, as long as we can antidifferentiate dv, we can repeatedly apply integration by parts until the polynomial function goes away. So, we go with (II) (D) We apply partial fractions to rational functions, (III).
Note: without knowing more about the functions, there’s no guarantee that the methods we chose will be the best methods, or even that they will work (with the exception of (I)).
With practice, you gain intuition about likely methods for different integrals. Luckily for you, there’s lots of practice below.
455 S-2: The integrand is a product of powers of sine and cosine. Since cosine has an odd power, we want to substitute u = sin x, du = cos xdx. Therefore, we should: • reserve one cosine for the derivative of sine in our substitution, and • change the rest of the cosines to sines using the identity sin2 x + cos2 x = 1.
ż π/2 0 sin4 x cos5 xdx = ż π/2 0 sin4 x(cos2 x)2 cos xdx = ż π/2 0 sin4 x(1 ´ sin2 x)2 cos xdx looomooon du = ż sin(π/2) sin(0) u4(1 ´ u2)2du = ż 1 0 u4(1 ´ 2u2 + u4)du = ż 1 0 (u4 ´ 2u6 + u8)du = 1 5u5 ´ 2 7u7 + 1 9u9 u=1 u=0 = 1 5 ´ 2 7 + 1 9 ´ 0 = 8 315 S-3: We notice that there is a quadratic equation under the square root. If that equation were a perfect square, we could get rid of the square root: so we’ll mould it into a perfect square using a trig substitution.
Our candidates will use one of the following identities: 1 ´ sin2 θ = cos2 θ tan2 θ + 1 = sec2 θ sec2 θ ´ 1 = tan2 θ We’ll be substituting x =(something), so we notice that 3 ´ 5x2 has the general form of (constant)´(function), as does 1 ´ sin2 θ. In order to get the constant right, we multiply through by three: 3 ´ 3 sin2 θ = 3 cos2 θ Our goal is to get 3 ´ 5x2 = 3 ´ 3 sin2 θ; so we solve this equation for x and decide on the substitution x = c 3 5 sin θ, dx = c 3 5 cos θdθ 456 Now we evaluate our integral.
ż a 3 ´ 5x2dx = ż g f f e3 ´ 5 c 3 5 sin θ !2c 3 5 cos θdθ = ż a 3 ´ 3 sin2 θ a 3/5 cos θdθ = ż a 3 cos2 θ a 3/5 cos θdθ = ż ?
3 cos θ a 3/5 cos θdθ = 3 ?
5 ż cos2 θdθ = 3 ?
5 ż 1 + cos 2θ 2 dθ = 3 2 ?
5 ż (1 + cos 2θ)dθ = 3 2 ?
5 θ + 1 2 sin(2θ) + C = 3 2 ?
5 [θ + sin θ cos θ] + C From our substitution x = ?
3/5 sin θ, we glean sin θ = x ?
5/3, and θ = arcsin x ?
5/3 .
To figure out cos θ, we draw a right triangle. Let θ be one angle, and since sin θ = x ?
5 ?
3 , we let the hypotenuse be ?
3 and the side opposite θ be x ?
5. By Pythagoras, the missing side (adjacent to θ) has length ?
3 ´ 5x2.
θ ?
3 ´ 5x2 x ?
5 ?
3 Therefore, cos θ = adj hyp = ?
3 ´ 5x2 ?
3 . So our integral evaluates to: 3 2 ?
5 [θ + sin θ cos θ] + C = 3 2 ?
5 " arcsin(x a 5/3) + x a 5/3 ¨ ?
3 ´ 5x2 ?
3 # + C = 3 2 ?
5 arcsin(x a 5/3) + x 2 ¨ a 3 ´ 5x2 + C S-4: First, we note the integral is improper. So, we’ll need to replace the top bound with a variable, and take a limit. Second, we’re going to have to antidifferentiate. The integrand 457 is the product of an exponential function, e´x, with a polynomial function, x ´ 1, so we use integration by parts with u = x ´ 1, dv = e´xdu, du = dx, and v = ´e´x.
ż x ´ 1 ex dx = ´(x ´ 1)e´x + ż e´x dx = ´(x ´ 1)e´x ´ e´x + C = ´xe´x + C So, ż 8 0 x ´ 1 ex dx = lim bÑ8 ż b 0 x ´ 1 ex dx = lim bÑ8 h ´ x ex ib 0 = lim bÑ8 ´ b eb lo omo on numÑ8 denÑ8 (˚) = lim bÑ8 ´ 1 eb = 0 (In the equality marked (˚), we used l’Hˆ opital’s rule.) So, ż 8 0 x ´ 1 ex dx = 0.
Remark: this shows that, interestingly, ż 8 0 x ex dx = ż 8 0 1 ex dx.
S-5: Solution 1: Notice the denominator factors as (x + 1)(3x + 1). Since the integrand is a rational function (the quotient of two polynomials), we can use partial fraction decomposition.
´2 3x2 + 4x + 1 = ´2 (x + 1)(3x + 1) = A x + 1 + B 3x + 1 = A(3x + 1) + B(x + 1) (x + 1)(3x + 1) = (3A + B)x + (A + B) (x + 1)(3x + 1) So: ´2 = (3A + B)x + (A + B) 0 = 3A + B and ´ 2 = A + B B = ´3A and hence ´ 2 = A + (´3A) A= 1 so then B = ´3 458 So now: ´2 3x2 + 4x + 1 = 1 x + 1 ´ 3 3x + 1 ż ´2 3x2 + 4x + 1dx = ż 1 x + 1 ´ 3 3x + 1 dx = log |x + 1| ´ log |3x + 1| + C = log ˇ ˇ ˇ ˇ x + 1 3x + 1 ˇ ˇ ˇ ˇ + C Solution 2: The previous solution is probably the nicest. However, for the foolhardy or the brave, this integral can also be evaluated using trigonometric substitution.
We start by completing the square on the denominator.
3x2 + 4x + 1 = 3 x2 + 4 3x + 1 3 = 3 x2 + 2 ¨ 2 3x + 4 9 ´ 4 9 + 1 3 = 3 x + 2 3 2 ´ 4 9 + 3 9 !
= 3 x + 2 3 2 ´ 1 9 !
= 3 x + 2 3 2 ´ 1 3 This has the form of a function minus a constant, which matches the trigonometric identity sec2 θ ´ 1 = tan2 θ. Multiplying through by 1 3, we see we can use the identity 1 3 sec2 θ ´ 1 3 = 1 3 tan2 θ. So, to get the substitution right, we want to choose a substitution that makes the following true: 3 x + 2 3 2 ´ 1 3 = 1 3 sec2 θ ´ 1 3 3 x + 2 3 2 = 1 3 sec2 θ 9 x + 2 3 2 = sec2 θ 3x + 2 = sec θ And, accordingly: 3dx = sec θ tan θdθ 459 Now, let’s simplify a little and use this substitution on our integral: ż ´2 3x2 + 4x + 1dx = ż ´2 3 x + 2 3 2 ´ 1 3 dx = ż ´2 9 x + 2 3 2 ´ 1 3dx = ż ´2 (3x + 2) 2 ´ 13dx = ż ´2 (sec θ) 2 ´ 1sec θ tan θdθ = ż ´2 tan2 θ sec θ tan θdθ = ż ´2sec θ tan θdθ = ż ´2 1 cos θ ¨ cos θ sin θ dθ = ż ´2 1 sin θdθ = ż ´2 csc θdθ Using the result of Example 1.8.21 in the CLP-2 text, or a table of integrals: = 2 log |csc θ + cot θ| + C Our final task is to translate this back from θ to x. Recall we used the substitution 3x + 2 = sec θ. Using this information, and sec θ = hypotenuse adjacent , we can fill in two sides of a right triangle with angle θ. The Pythagorean theorem tells us the third side (opposite to θ) has measure a (3x + 2)2 ´ 1 = ?
9x2 ´ 12x + 3.
460 θ 1 ?
9x2 + 12x + 3 3x + 2 2 log |csc θ + cot θ| + C = 2 log ˇ ˇ ˇ ˇ 3x + 2 ?
9x2 + 12x + 3 + 1 ?
9x2 + 12x + 3 ˇ ˇ ˇ ˇ + C = 2 log ˇ ˇ ˇ ˇ 3x + 3 ?
9x2 + 12x + 3 ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ ˇ (3x + 3)2 ?
9x2 + 12x + 3 2 ˇ ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ (3x + 3)2 9x2 + 12x + 3 ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ 9(x + 1)2 3(3x + 1)(x + 1) ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ 3(x + 1)2 (3x + 1)(x + 1) ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ 3(x + 1) 3x + 1 ˇ ˇ ˇ ˇ + C = log ˇ ˇ ˇ ˇ x + 1 3x + 1 ˇ ˇ ˇ ˇ + log 3 + C Since C is an arbitrary constant, we can write our final answer as log ˇ ˇ ˇ ˇ x + 1 3x + 1 ˇ ˇ ˇ ˇ + C S-6: We see that we have two functions multiplied, but they don’t simplify nicely with each other. However, if we differentiate logarithm, and integrate x2, we’ll get a polynomial.
So, let’s use integration by parts.
u = log x dv = x2dx du = (1/x)dx v = x3/3 461 First, let’s antidifferentiate. We’ll deal with the limits of integration later.
ż x2 log xdx = (log x) loomoon u (x3/3) loomoon v ) ´ ż (x3/3) loomoon v (1/x)dx looomooon du = 1 3x3 log x ´ 1 3 ż x2dx = 1 3x3 log x ´ 1 3 ¨ 1 3x3 + C = 1 3x3 log x ´ 1 9x3 + C We use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.
ż 2 1 x3 log xdx = 1 3x3 log x ´ 1 9x3 2 1 = 1 323 log 2 ´ 1 923 ´ 1 313 log 1 ´ 1 913 = 8 log 2 3 ´ 8 9 ´ 0 + 1 9 = 8 3 log 2 ´ 7 9 S-7: The derivative of the denominator shows up in the numerator, only differing by a constant, so we perform a substitution. Specifically, substitute u = x2 ´ 3, du = 2x dx.
This gives ż x x2 ´ 3dx = ż du/2 u = 1 2 log |u| + C = 1 2 log ˇ ˇx2 ´ 3 ˇ ˇ + C S-8: (a) Although a quadratic under a square root often suggests trigonometric substitution, in this case we have an easier substitution. Specifically, let y = 9 + x2. Then dy = 2xdx, xdx = dy 2 , y(0) = 9, and y(4) = 25.
ż 4 0 x ?
9 + x2 dx = ż 25 9 1 ?y dy 2 = 1 2 ¨ ?y 1/2 ˇ ˇ ˇ 25 9 = 5 ´ 3 = 2 (b) The power of cosine is odd, so we can reserve one cosine for the differential and change the rest to sines. Substituting y = sin x, dy = cos x, dx, y(0) = 0, y(π/2) = 1, cos2 x = 1 ´ y2: ż π/2 0 cos3 x sin2 x dx = ż π/2 0 cos2 x sin2 x cos x dx = ż 1 0 (1 ´ y2)y2 dy = ż 1 0 (y2 ´ y4) dy = y3 3 ´ y5 5 1 0 = 1 3 ´ 1 5 = 2 15 462 (c) The integrand is the product of two different kinds of functions, with no obvious substitution or simplification. If we differentiate log x, it will match better with the polynomial nature of the rest of the integrand. So, integrate by parts with u(x) = log x and dv = x3 dx, then du = 1 x dx and v = x4/4.
ż e 1 x3 log x dx = x4 4 log x ˇ ˇ ˇ ˇ e 1 ´ ż e 1 x4 4 ¨ 1 x dx = e4 4 ´ ż e 1 x3 4 dx = e4 4 ´ x4 16 ˇ ˇ ˇ ˇ e 1 = 3e4 16 + 1 16 S-9: (a) Integrate by parts with u = x and dv = sin x dx so that du = dx and v = ´ cos x.
ż x sin x dx = ´x cos x ´ ż (´ cos x) dx = ´x cos x + sin x + C So, ż π/2 0 x sin x dx = h ´ x cos x + sin x iπ/2 0 = 1 (b) The power of cosine is odd, so we can reserve one cosine for du and change the rest into sines. Make the substitution u = sin x, du = cos x dx.
ż π/2 0 cos5 x dx = ż π/2 0 1 ´ sin2 x 2 cos x dx = ż 1 0 1 ´ u22 du = ż 1 0 1 ´ 2u2 + u4 du = u ´ 2 3u3 + 1 5u5 1 0 = 1 ´ 2 3 + 1 5 = 8 15 S-10: (a) This is a classic integration-by-parts example. If we integrate ex, it doesn’t change, and if we differentiate x it becomes a constant. So, let u = x and dv = ex dx, so that du = dx and v = ex.
ż 2 0 xex dx = h xexi2 0 ´ ż 2 0 ex dx = 2e2 ´ h exi2 0 = e2 + 1 (b) We have a quadratic function underneath a square root. In the absence of an easier substitution, we can get rid of the square root with a trigonometric substitution.
Substitute x = tan y, dx = sec2 y dy. When x = 0, tan y = 0 so y = 0. When x = 1, tan y = 1 so y = π 4 . Also ?
1 + x2 = b 1 + tan2 y = a sec2 y = sec y, since sec y ě 0 for all 0 ď y ď π 4 .
ż 1 0 1 ?
1 + x2 dx = ż π/4 0 sec2 y dy sec y = ż π/4 0 sec y dy = h log | sec y + tan y| iπ/4 0 = log ˇ ˇ ˇsec π 4 + tan π 4 ˇ ˇ ˇ ´ log |sec 0 + tan 0| = log ˇ ˇ ˇ ?
2 + 1 ˇ ˇ ˇ ´ log |1 + 0| = log( ?
2 + 1) So, ż 1 0 1 ?
1 + x2 dx = log( ?
2 + 1) 463 (c) The integral is a rational function. In the absence of an obvious substitution, we use partial fractions.
4x (x2 ´ 1)(x2 + 1) = 4x (x ´ 1)(x + 1)(x2 + 1) = a x ´ 1 + b x + 1 + cx + d x2 + 1 Multiplying by the denominator, 4x = a(x + 1)(x2 + 1) + b(x ´ 1)(x2 + 1) + (cx + d)(x ´ 1)(x + 1) (˚) Setting x = 1 gives 4a = 4, so a = 1. Setting x = ´1 gives ´4b = ´4, so b = 1.
Substituting in a = b = 1 in (˚) gives: 4x = (x + 1)(x2 + 1) + (x ´ 1)(x2 + 1) + (cx + d)(x ´ 1)(x + 1) 4x = 2x(x2 + 1) + (cx + d)(x ´ 1)(x + 1) 4x ´ 2x(x2 + 1) = (cx + d)(x ´ 1)(x + 1) ´2x(x2 ´ 1) = (cx + d)(x2 ´ 1) ´2x = cx + d c = ´2, d = 0 So, ż 5 3 4x (x2 ´ 1)(x2 + 1) dx = ż 5 3 1 x ´ 1 + 1 x + 1 ´ 2x x2 + 1 dx = h log |x ´ 1| + log |x + 1| ´ log(x2 + 1) i5 3 = log 4 + log 6 ´ log 26 ´ log 2 ´ log 4 + log 10 = log 6 ˆ 10 26 ˆ 2 = log 15 13 « 0.1431 S-11: (a) ş3 0 ?
9 ´ x2 dx is the area of the portion of the disk x2 + y2 ď 9 that lies in the first quadrant. It is 1 4π33 = 9 4π . Alternatively, you could also evaluate this integral using the substitution x = 3 sin y, dx = 3 cos y dy.
ż 3 0 a 9 ´ x2 dx = ż π/2 0 b 9 ´ 9 sin2 y (3 cos y) dy = 9 ż π/2 0 cos2 y dy = 9 2 ż π/2 0 [1 + cos(2y)] dy = 9 2 h y + sin(2y) 2 iπ/2 0 = 9 4π 464 x y y = ?
9 ´ x2 3 (b) It’s not immediately obvious what to do with this one, but remember we found ş log xdx using integration by parts with u = log x and dv = dx. Let’s hope a similar trick works here. Integrate by parts, using u = log(1 + x2) and dv = dx, so that du = 2x 1+x2 dx, v = x.
ż 1 0 log(1 + x2) dx = h x log(1 + x2) i1 0 ´ ż 1 0 x 2x 1 + x2 dx = log 2 ´ 2 ż 1 0 x2 1 + x2 dx = log 2 ´ 2 ż 1 0 1 ´ 1 1 + x2 dx = log 2 ´ 2 x ´ arctan x 1 0 = log 2 ´ 2 + π 2 « 0.264 (c) The integrand is a rational function with no obvious substitution, so we use partial fractions.
x (x ´ 1)2(x ´ 2) = a (x ´ 1)2 + b x ´ 1 + c x ´ 2 = a(x ´ 2) + b(x ´ 1)(x ´ 2) + c(x ´ 1)2 (x ´ 1)2(x ´ 2) Multiply by the denominator.
x = a(x ´ 2) + b(x ´ 1)(x ´ 2) + c(x ´ 1)2 Setting x = 1 gives a = ´1. Setting x = 2 gives c = 2. Substituting in a = ´1 and c = 2 gives b(x ´ 1)(x ´ 2) = x + (x ´ 2) ´ 2(x ´ 1)2 = ´2x2 + 6x ´ 4 = ´2(x ´ 1)(x ´ 2) ù ñ b = ´2 465 Hence ż 8 3 x (x ´ 1)2(x ´ 2) dx = lim MÑ8 ż M 3 ´ 1 (x ´ 1)2 ´ 2 x ´ 1 + 2 x ´ 2 dx = lim MÑ8 1 x ´ 1 ´ 2 log |x ´ 1| + 2 log |x ´ 2| M 3 = lim MÑ8 1 x ´ 1 + 2 log ˇ ˇ ˇ ˇ x ´ 2 x ´ 1 ˇ ˇ ˇ ˇ M 3 = lim MÑ8 1 M ´ 1 + 2 log ˇ ˇ ˇ ˇ M ´ 2 M ´ 1 ˇ ˇ ˇ ˇ ´ 1 3 ´ 1 + 2 log ˇ ˇ ˇ ˇ 3 ´ 2 3 ´ 1 ˇ ˇ ˇ ˇ = 2 log 2 ´ 1 2 « 0.886 since lim MÑ8 log M ´ 2 M ´ 1 = lim MÑ8 log 1 ´ 2/M 1 ´ 1/M = log 1 = 0 and log 1 2 = ´ log 2 S-12: This looks quite a lot like a rational function, but with variable sin θ instead of x. So, we use the substitution x = sin θ, dx = cos θ dθ.
ż sin4 θ ´ 5 sin3 θ + 4 sin2 θ + 10 sin θ sin2 θ ´ 5 sin θ + 6 cos θ dθ = ż x4 ´ 5x3 + 4x2 + 10x x2 ´ 5x + 6 dx Since the numerator does not have smaller degree than the denominator, we need to do some long division before we can set up our partial fractions decomposition.
x2 ´ 2 x2 ´ 5x + 6 x4 ´ 5x3 + 4x2 + 10x ´ x4 + 5x3 ´ 6x2 ´ 2x2 + 10x 2x2 ´ 10x + 12 12 That is, x4 ´ 5x3 + 4x2 + 10x x2 ´ 5x + 6 = x2 ´ 2 + 12 x2 ´ 5x + 6 = x2 ´ 2 + 12 (x ´ 2)(x ´ 3) We use partial fractions decomposition on the rightmost term.
12 (x ´ 2)(x ´ 3) = A x ´ 2 + B x ´ 3 12 = A(x ´ 3) + B(x ´ 2) 466 Setting x = 3 and x = 2 gives us B = 12, A = ´12 Now we can evaluate our integral.
ż sin4 θ ´ 5 sin3 θ + 4 sin2 θ + 10 sin θ sin2 θ ´ 5 sin θ + 6 cos θdθ = ż x4 ´ 5x3 + 4x2 + 10x x2 ´ 5x + 6 dx = ż x2 ´ 2 + 12 (x ´ 2)(x ´ 3) dx = ż x2 ´ 2´ 12 x ´ 2 + 12 x ´ 3 dx = 1 3x3 ´ 2x ´ 12 log |x ´ 2| + 12 log |x ´ 3| + C = 1 3x3 ´ 2x + 12 log ˇ ˇ ˇ ˇ x ´ 3 x ´ 2 ˇ ˇ ˇ ˇ + C = 1 3 sin3 θ ´ 2 sin θ + 12 log ˇ ˇ ˇ ˇ sin θ ´ 3 sin θ ´ 2 ˇ ˇ ˇ ˇ + C S-13: (a) It doesn’t matter to us right now that the arguments of sine and cosine are 2x rather than x. This is still the integral of powers of products of sines and cosines. Since cosine has an odd power, we make the substitution u = sin(2x), du = 2 cos(2x) dx.
ż π 4 0 sin2(2x) cos3(2x) dx = ż π 4 0 sin2(2x) 1 ´ sin2(2x) cos(2x) dx = 1 2 ż 1 0 u2 1 ´ u2 du = 1 2 ż 1 0 u2 ´ u4 du = 1 2 h1 3u3 ´ 1 5u5i1 0 = 1 15 (b) Make the substitution x = 3 tan t, dx = 3 sec2 t dt and use the trig identity 9 + 9 tan2 t = 9 sec2 t.
ż 9 + x2´ 3 2 dx = ż 9 + 9 tan2 t ´ 3 2 3 sec2 t dt = ż 3 sec t ´3 3 sec2 t dt = 1 9 ż cos t dt = 1 9 sin t + C = 1 9 x ?
x2 + 9 + C To convert back to x, in the last step, we used the triangle below, which is rigged to have tan t = x 3.
t 3 x ?
x2 + 9 467 (c) Seeing a rational function with no obvious substitutions, we use partial fractions.
1 (x ´ 1)(x2 + 1) = a x ´ 1 + bx + c x2 + 1 = a(x2 + 1) + (bx + c)(x ´ 1) (x ´ 1)(x2 + 1) Multiply by the original denominator.
1 = a(x2 + 1) + (bx + c)(x ´ 1) (˚) Setting x = 1 gives 2a = 1 or a = 1 2. Substituting in a = 1 2 in (˚) gives 1 2(x2 + 1) + (bx + c)(x ´ 1) = 1 ð ñ (bx + c)(x ´ 1) = 1 2(1 ´ x2) = ´1 2(x ´ 1)(x + 1) ð ñ (bx + c) = ´1 2(x + 1) ð ñ b = c = ´1 2 So, ż dx (x ´ 1)(x2 + 1) = ż h 1/2 x ´ 1 ´ 1 2(x + 1) x2 + 1 i dx To antidifferentiate the second piece, we split it into two integrals: one that can be handled with the substitution u = x2 + 1, and another that looks like the derivative of arctangent.
= ż 1/2 x ´ 1 ´ x/2 x2 + 1 ´ 1/2 x2 + 1 dx = ż 1/2 x ´ 1 ´ 1 4 ¨ 2x x2 + 1 ´ 1/2 x2 + 1 dx = 1 2 log |x ´ 1| ´ 1 4 log(x2 + 1) ´ 1 2 arctan x + C (d) We know the derivative of arctangent, and it would integrate nicely if multiplied to the antiderivative of x. So, we integrate by parts with u = arctan x and dv = x dx so that du = 1 1+x2dx and v = 1 2x2. Then ż x arctan x dx = 1 2x2 arctan x ´ 1 2 ż x2 1 + x2 dx = 1 2x2 arctan x ´ 1 2 ż 1 + x2 ´ 1 1 + x2 dx = 1 2x2 arctan x ´ 1 2 ż 1 ´ 1 1 + x2 dx = 1 2 x2 arctan x ´ x + arctan x + C 468 S-14: (a) We substitute y = sin(2x), dy = 2 cos(2x) dx. Note sin(2 ¨ 0) = 0 and sin(2 ¨ π 4 ) = 1.
ż π/4 0 sin5(2x) cos(2x) dx = ż 1 0 y5 dy 2 = 1 12 y6 1 0 = 1 12 (b) We can get rid of the square root with a trig substitution. Substituting x = 2 sin y, dx = 2 cos y dy, ż a 4 ´ x2 dx = ż b 4 ´ 4 sin2 y 2 cos y dy = 4 ż cos2 y dy = 2 ż 1 + cos(2y) dy = 2y + sin(2y) + C = 2y + 2 sin y cos y + C = 2 sin´1 x 2 + x c 1 ´ x2 4 + C since sin y = x 2 and cos y = b 1 ´ sin2 y = b 1 ´ x2 4 . Alternately, we can draw a triangle with sin y = x 2, and use the Pythagorean theorem to find the adjacent side.
y ?
4 ´ x2 x 2 (c) Seeing a rational function with no obvious substitution, we use the method of partial fractions. The denominator is already completely factored.
x + 1 x2(x ´ 1) = A x + B x2 + C x ´ 1 x + 1 = Ax(x ´ 1) + B(x ´ 1) + Cx2 Setting x = 1 gives us C = 2. Setting x = 0 gives us B = ´1. Furthermore, the coefficient of x2 on the left hand side (after collecting like terms), namely A + C, must be the same as the coefficient of x2 on the right hand side, namely 0. So A + C = 0 and A = ´2.
Checking, ´2x(x ´ 1) ´ (x ´ 1) + 2x2 = ´2x2 + 2x ´ x + 1 + 2x2 = x + 1 as desired. Thus, ż x + 1 x2(x ´ 1) dx = ż h ´2 x´ 1 x2 + 2 x ´ 1 i dx = ´2 log |x| + 1 x + 2 log |x ´ 1| + C S-15: (a) Define I1 = ż 8 0 e´x sin(2x) dx I2 = ż 8 0 e´x cos(2x) dx 469 We integrate by parts, with u = sin(2x) or cos(2x) and dv = e´x dx. That is, v = ´e´x.
I1 = ż 8 0 e´x sin(2x) dx = lim RÑ8 ż R 0 e´x sin(2x) dx = lim RÑ8 h ´ e´x sin(2x) iR 0 + 2 ż R 0 e´x cos(2x) dx = 2I2 I2 = ż 8 0 e´x cos(2x) dx = lim RÑ8 ż R 0 e´x cos(2x) dx = lim RÑ8 h ´ e´x cos(2x) iR 0 ´ 2 ż R 0 e´x sin(2x) dx = 1 ´ 2I1 Substituting I2 = 1 2 I1 into I2 = 1 ´ 2I1 gives 5 2 I1 = 1, or ż 8 0 e´x sin(2x) dx = 2 5.
(b) We can cancel out the square root if we use a trig substitution. Substitute x = ?
2 tan y, dx = ?
2 sec2 y dy.
ż ?
2 0 1 (2 + x2)3/2 dx = ?
2 ż π/4 0 sec2 y (2 + 2 tan2 y)3/2 dy = 1 2 ż π/4 0 cos y dy = 1 2 sin y π/4 0 = 1 2 ?
2 (c) Solution 1: Integrate by parts, using u = log(1 + x2) and dv = x dx, so that du = 2x 1+x2, v = x2 2 .
ż 1 0 x log(1 + x2) dx = h1 2x2 log(1 + x2) i1 0 ´ ż 1 0 x3 1 + x2 dy = 1 2 log 2 ´ ż 1 0 h x ´ x 1 + x2 i dx = 1 2 log 2 ´ hx2 2 ´ 1 2 log(1 + x2) i1 0 = log 2 ´ 1 2 « 0.193 Solution 2: First substitute y = 1 + x2, dy = 2x dx.
ż 1 0 x log(1 + x2) dx = 1 2 ż 2 1 log y dy Then integrate by parts, using u = log y and dv = dy, so that du = 1 y, v = y.
ż 1 0 x log(1 + x2) dx = 1 2 ż 2 1 log y dy = h1 2y log y i2 1 ´ 1 2 ż 2 1 y1 y dy = log 2 ´ 1 2 « 0.193 (d) Seeing a rational function with no obvious substitution, we use partial fractions.
1 (x ´ 1)2(x ´ 2) = a (x ´ 1)2 + b x ´ 1 + c x ´ 2 1 = a(x ´ 2) + b(x ´ 1)(x ´ 2) + c(x ´ 1)2 (˚) 470 Setting x = 1 gives a = ´1. Setting x = 2 gives c = 1. Substituting in a = ´1 and c = 1 to (˚) gives b(x ´ 1)(x ´ 2) = 1 + (x ´ 2) ´ (x ´ 1)2 = ´x2 + 3x ´ 2 = ´(x ´ 1)(x ´ 2) ù ñ b = ´1 Hence: ż 8 3 1 (x ´ 1)2(x ´ 2) dx = lim MÑ8 ż M 3 ´ 1 (x ´ 1)2 ´ 1 x ´ 1 + 1 x ´ 2 dx = lim MÑ8 h 1 x ´ 1 ´ log(x ´ 1) + log(x ´ 2) iM 3 = lim MÑ8 h 1 M ´ 1 + log M ´ 2 M ´ 1 i ´ h 1 3 ´ 1 + log 3 ´ 2 3 ´ 1 i = log 2 ´ 1 2 « 0.193 since lim MÑ8 log M ´ 2 M ´ 1 = lim MÑ8 log 1 ´ 2/M 1 ´ 1/M = log 1 = 0 S-16: (a) Integrate by parts with u = log x and dv = x dx, so that du = dx x and v = 1 2x2.
ż x log x dx = 1 2x2 log x ´ 1 2 ż x2 ¨ 1 x dx = 1 2x2 log x ´ 1 4x2 + C (b) The denominator is an irreducible quadratic, so partial fractions can’t get us any further. To integrate a function whose denominator is quadratic, we split the numerator up so that one piece can be evaluated with a u-substitution, and the other piece looks like arctangent.
ż (x ´ 1)dx x2 + 4x + 5 = ż x + 2 ´ 3 x2 + 4x + 5dx = 1 2 ż 2x + 4 x2 + 4x + 5dx ´ ż 3 x2 + 4x + 5dx = 1 2 ż 2x + 4 x2 + 4x + 5dx ´ 3 ż 1 (x + 2)2 + 1dx = 1 2 log[x2 + 4x + 5] ´ 3 arctan(x + 2) + C For the last step, you can guess the antiderivative, or use the substitutions u1 = x2 + 4x + 5 and u2 = x + 2, respectively, for the two integrals.
471 (c) We use partial fractions.
1 x2 ´ 4x + 3 = 1 (x ´ 3)(x ´ 1) = a x ´ 3 + b x ´ 1 1 = a(x ´ 1) + b(x ´ 3) Setting x = 3 gives a = 1 2. Setting x = 1 gives b = ´1 2. So, ż dx x2 ´ 4x + 3 = ż 1/2 x ´ 3´ 1/2 x ´ 1 dx = 1 2 log |x ´ 3| ´ 1 2 log |x ´ 1| + C (d) Substitute y = x3, dy = 3x2 dx.
ż x2 dx 1 + x6 = 1 3 ż dy 1 + y2 = 1 3 arctan y + C = 1 3 arctan x3 + C S-17: (a) Integrate by parts with u = arctan x, dv = dx, du = dx 1+x2 and v = x. This gives ż 1 0 arctan x dx = x arctan x 1 0 ´ ż 1 0 x 1 + x2dx = arctan 1 ´ h1 2 log(1 + x2) i1 0 = π 4 ´ 1 2 log 2 (b) Note that the derivative of the denominator is 2x ´ 2, which differs from the numerator only by 1.
ż 2x ´ 1 x2 ´ 2x + 5 dx = ż 2x ´ 2 x2 ´ 2x + 5 dx + ż 1 x2 ´ 2x + 5 dx = ż 2x ´ 2 x2 ´ 2x + 5 dx + ż 1 (x ´ 1)2 + 4 dx = log |x2 ´ 2x + 5| + 1 2 arctan x ´ 1 2 + C In the last step, you can guess the antiderivative, or use the substitutions u1 = x2 ´ 2x + 5 and u2 = (x ´ 1)/2, respectively.
S-18: (a) Substituting u = x3 + 1, du = 3x2 dx ż x2 (x3 + 1)101dx = ż 1 u101 ¨ du 3 = u´100 ´100 ¨ 1 3 + C = ´ 1 300(x3 + 1)100 + C (b) Substituting u = sin x, du = cos x dx, cos2 x = 1 ´ sin2 x = 1 ´ u2, ż cos3x sin4x dx = ż cos2x sin4x cos x dx = ż (1 ´ u2)u4 du = ż (u4 ´ u6) du = u5 5 ´ u7 7 + C = sin5x 5 ´ sin7x 7 + C 472 S-19: First, we note that the integral is improper, because sin π = 0. So, we’ll have to use a limit.
Second, we need to antidifferentiate. The substitution u = sin x, du = cos x dx fits just right.
ż π π/2 cos x ?
sin x dx = lim bÑπ´ ż b π/2 cos x ?
sin x dx = lim bÑπ´ ż sin b 1 1 ?u du = lim bÑπ´ h 2?u isin b 1 = 2 ?
0 ´ 2 ?
1 = ´2 S-20: (a) If the integrand had x’s instead of ex’s it would be a rational function, ripe for the application of partial fractions. So let’s start by making the substitution u = ex, du = ex dx: ż ex (ex + 1)(ex ´ 3)dx = ż du (u + 1)(u ´ 3) Now, we follow the partial fractions protocol, starting with expressing 1 (u + 1)(u ´ 3) = A u + 1 + B u ´ 3 To find A and B, the sneaky way, we cross multiply by the denominator 1 = A(u ´ 3) + B(u + 1) and find A and B by evaluating at u = ´1 and u = 3, respectively.
1 = A(´1 ´ 3) + B(´1 + 1) ð ñ A = ´1 4 1 = A(3 ´ 3) + B(3 + 1) ð ñ B = 1 4 Finally, we can do the integral: ż ex (ex + 1)(ex ´ 3)dx = ż du (u + 1)(u ´ 3) = ż ´1/4 u + 1 + 1/4 u ´ 3 du = ´1 4 log |u + 1| + 1 4 log |u ´ 3| + C = ´1 4 log |ex + 1| + 1 4 log |ex ´ 3| + C (b) The argument of the square root is 12 + 4x ´ x2 = 12 ´ (x ´ 2)2 + 4 = 16 ´ (x ´ 2)2 473 Hmmm. The numerator is x2 ´ 4x + 4 = (x ´ 2)2. So let’s make the integral look somewhat simpler by substituting u = x ´ 2, du = dx. When x = 2 we have u = 0, and when x = 4 we have u = 2, so: ż x=4 x=2 x2 ´ 4x + 4 ?
12 + 4x ´ x2dx = ż u=2 u=0 u2 ?
16 ´ u2du This is perfect for the trig substitution u = 4 sin θ, du = 4 cos(θ) dθ. When u = 0 we have 4 sin θ = 0 and hence θ = 0. When u = 2 we have 4 sin θ = 2 and hence θ = π 6 . So ż u=2 u=0 u2 ?
16 ´ u2du = ż θ=π/6 θ=0 16 sin2 θ a 16 ´ 16 sin2 θ 4 cos θdθ = 16 ż π/6 0 sin2 θ dθ = 8 ż π/6 0 1 ´ cos(2θ) dθ = 8 θ ´ 1 2 sin(2θ) π/6 0 = 8 π 6 ´ 1 2 ¨ ?
3 2 = 4π 3 ´ 2 ?
3 S-21: (a) Substituting y = cos x, dy = ´ sin x dx, sin2 x = 1 ´ cos2 x = 1 ´ y2 ż sin3 x cos3 x dx = ż sin2 x cos3 x sin x dx = ż 1 ´ y2 y3 (´dy) = ´ ż y´3 ´ y´1 dy = ´y´2 ´2 + log |y| + C = 1 2 sec2 x + log | cos x| + C (b) The integrand is an even function, and the limits of integration are symmetric. So, we can slightly simplify the integral by replacing the lower limit with 0, and doubling the integral.
We’d rather not use partial fractions here, because it would be pretty complicated.
Instead, notice that the numerator is only off by a constant from the derivative of x5.
Substituting x5 = 4y, 5x4 dx = 4 dy, and using that x = 2 ù ñ 25 = 4y ù ñ y = 8, ż 2 ´2 x4 x10 + 16 dx = 2 ż 2 0 x4 x10 + 16 dx = 2 ¨ 4 5 ż 8 0 1 16y2 + 16 dy = 1 10 ż 8 0 1 y2 + 1 dy = 1 10 arctan 8 « 0.1446 S-22: 474 Solution 1: Let’s use the substitution u = x ´ 1, du = dx.
ż x ?
x ´ 1 dx = ż (u + 1)?u du = ż u3/2 + u1/2 du = 2 5u5/2 + 2 3u3/2 + C = 2 5(x ´ 1)5/2 + 2 3(x ´ 1)3/2 + C Solution 2: We have an integrand with x multiplied by something integrable. So, if we use integration by parts with u = x and dv = ?
x ´ 1 dx, then du = dx (that is, the x goes away) and v = 2 3(x ´ 1)3/2.
ż x ?
x ´ 1 dx = 2 3x ?
x ´ 1 3 ´ 2 3 ż (x ´ 1)3/2dx = 2 3x ?
x ´ 1 3 ´ 2 3 2 5(x ´ 1)5/2 + C = 2 3 ?
x ´ 1 x(x ´ 1) ´ 2 5(x ´ 1)2 + C = 2 15 ?
x ´ 1 ¨ (3x2 ´ x ´ 2) + C = 2 15 ?
x ´ 1 ¨ (3(x2 ´ 2x + 1) + 5x ´ 5) + C = 2 15 ?
x ´ 1 ¨ (3(x ´ 1)2 + 5(x ´ 1)) + C = 2 15 ¨ 3 ?
x ´ 1 5 + 2 15 ¨ 5 ?
x ´ 1 3 + C = 2 5 ?
x ´ 1 5 + 2 3 ?
x ´ 1 3 + C We have just seen two solutions. There are other solutions too. For example, one could use the substitution u = ?
x ´ 1. Or, as another example, one could write x ?
x ´ 1 = (x ´ 1) ?
x ´ 1 + ?
x ´ 1 = (x ´ 1)3/2 + (x ´ 1)1/2 and just guess a function whose derivative is (x ´ 1)3/2 and a function whose derivative is (x ´ 1)1/2.
S-23: We are to integrate ż ?
x2 ´ 2 x2 dx We notice that there is a quadratic function under the square root. If that function were a perfect square, we could get rid of the square root: so we’ll mould it into a perfect square using a trig substitution.
475 Our candidates are the following identities: 1 ´ sin2 θ = cos2 θ tan2 θ + 1 = sec2 θ sec2 θ ´ 1 = tan2 θ We’ll be substituting x =(something), so we notice that x2 ´ 2 has the general form of (function)´(constant), as does sec2 θ ´ 1. In order to get the constant right, we multiply through by two: 2 sec2 θ ´ 2 = 2 tan2 θ or: ( ?
2 sec θ)2 ´ 2 = 2 tan2 θ So we decide to use the substitution x = ?
2 sec θ 0 ď θ ă π/2 (Recall that x ě ?
2.) dx = ?
2 sec θ tan θ dθ a x2 ´ 2 = a 2 sec2 θ ´ 2 = a 2 tan2 θ = ?
2 | tan θ| = ?
2 tan θ since 0 ď θ ă π/2 Now that we’ve chosen the substitution, we evaluate the integral.
ż ?
x2 ´ 2 x2 dx = ż ?
2 tan θ 2 sec2 θ ?
2 sec θ tan θ dθ = ż tan2 θ sec θ dθ = ż sec2 θ ´ 1 sec θ dθ = ż sec θ ´ cos θ dθ = log | sec θ + tan θ| ´ sin θ + C Now we need everything back in terms of x. We need a triangle. Since x = ?
2 sec θ, that means that if we label an angle θ, its secant (hypotenuse over adjacent side) is x ?
2. By Pythagoras, the opposite side is ?
x2 ´ 2.
θ ?
2 ?
x2 ´ 2 x So tan θ = opp adj = ?
x2 ´ 2 ?
2 , and sin θ = opp hyp = ?
x2 ´ 2 x . Then the value of the integral is: log | sec θ + tan θ| ´ sin θ + C = log ˇ ˇ ˇ ˇ ˇ x ?
2 + ?
x2 ´ 2 ?
2 ˇ ˇ ˇ ˇ ˇ ´ ?
x2 ´ 2 x + C = log ˇ ˇ ˇx + a x2 ´ 2 ˇ ˇ ˇ ´ log ?
2 ´ ?
x2 ´ 2 x + C = log ˇ ˇ ˇx + a x2 ´ 2 ˇ ˇ ˇ ´ ?
x2 ´ 2 x + C 476 Note that the simplification in the last step is possible because C is an arbitrary constant.
So, C ´ log ?
2 is just another arbitrary constant and can be renamed to C.
S-24: This is the product of secants and tangents, as in Section 1.8.2 of the CLP-2 text. If u = tan x, then du = sec2 xdx. We can get the remaining two secants to turn into tangents with the identity sec2 x = 1 + tan2 x, so we’ll use this substitution.
ż π/4 0 sec4 x tan5 xdx = ż π/4 0 sec2 x tan5 x sec2 xdx = ż π/4 0 (1 + tan2 x) tan5 x sec2 xdx looomooon du = ż tan(π/4) tan(0) (1 + u2)u5du = ż 1 0 (u5 + u7)du = 1 6u6 + 1 8u8 1 0 = 1 6 + 1 8 ´ 0 = 7 24 S-25: We can use partial fraction decomposition to break this into chunks that we can deal with. The denominator has a repeated linear factor, so it can be decomposed as the sum of constants divided by powers of that factor.
3x2 + 4x + 6 (x + 1)3 = A x + 1 + B (x + 1)2 + C (x + 1)3 = A(x + 1)2 + B(x + 1) + C (x + 1)3 ñ 3x2 + 4x + 6 = A(x + 1)2 + B(x + 1) + C = Ax2 + (2A + B)x + (A + B + C) So, by matching coefficients: A = 3, 2A + B = 4, and A + B + C = 6 A = 3, B = ´2, C = 5 Therefore: 3x2 + 4x + 6 (x + 1)3 = 3 x + 1 + ´2 (x + 1)2 + 5 (x + 1)3 477 Now, the integration is easy, with a substitution of u = x + 1 and du = dx: ż 3x2 + 4x + 6 (x + 1)3 dx = ż 3 x + 1 + ´2 (x + 1)2 + 5 (x + 1)3 dx = ż 3u´1 ´ 2u´2 + 5u´3 du = 3 log |u| + 2u´1 ´ 5 2u´2 + C = 3 log |x + 1| + 2 x + 1 ´ 5 2(x + 1)2 + C S-26: If the denominator were x2 + 1, the antiderivative would be arctangent. So, by completing the square, let’s aim for the fraction to look like 1 u2 + 1, for some u. This is a good strategy for integrating an irreducible quadratic under a constant.
First: complete the square ż 1 x2 + x + 1dx = ż 1 x2 + x + 1 4 + 3 4 dx = ż 1 x + 1 2 2 + 3 4 dx Second: get the denominator in the form u2 + 1. To do this, we need to fix the constant = ż 1 x + 1 2 2 + 3 4 4 3 4 3 !
dx = 4 3 ż 1 4 3 ¨ x + 1 2 2 + 1 dx Now a quick wiggle to make that first part of the denominator into something squared again: = 4 3 ż 1 2 ?
3x + 1 ?
3 2 + 1 dx Now we see that u = 2 ?
3x + 1 ?
3, du = 2 ?
3dx will do the job = 4 3 ż 1 u2 + 1 ¨ ?
3 2 du = 2 ?
3 ż 1 u2 + 1du = 2 ?
3 arctan u + C = 2 ?
3 arctan 2 ?
3x + 1 ?
3 + C 478 S-27: Since tan x = sin x cos x, ż sin x cos x tan x dx = ż sin2 x dx = ż 1 2 1 ´ cos(2x) dx = 1 2 x ´ 1 2 sin(2x) + C = 1 2 (x ´ sin x cos x) + C S-28: We have the integral of a rational function with no obvious substitution, so we use partial fractions. That means we need to factor the denominator. We see that x = ´1 is a root of the denominator, so x + 1 is a factor. You might be able to figure out the rest of the factorization by inspection, or from having seen this common expression before; alternately, we can use long division.
x2 ´ x + 1 x + 1 x3 + 1 ´ x3 ´ x2 ´ x2 x2 + x x + 1 ´ x ´ 1 0 Note x2 ´ x + 1 is an irreducible quadratic.
1 x3 + 1 = 1 (x + 1)(x2 ´ x + 1) = A x + 1 + Bx + C x2 ´ x + 1 1 = A(x2 ´ x + 1) + (Bx + C)(x + 1) (˚) When x = ´1, we see 1 = 3A, so 1 3 = A. We plug this into (˚).
1 = 1 3(x2 ´ x + 1) + (Bx + C)(x + 1) ´1 3x2 + 1 3x + 2 3 = Bx2 + (B + C)x + C Matching up coefficients of corresponding power of x, we see B = ´1 3 and C = 2 3.
ż 1 x3 + 1 dx = ż 1/3 x + 1´ 1 3x ´ 2 3 x2 ´ x + 1 !
dx 479 To integrate the second fraction, we break it up into two pieces: one we can integrate using the substitution u = x2 ´ x + 1, the other will look like the derivative of arctangent.
= 1 3 log |x + 1| ´ ż 1 3x ´ 1 6 ´ 1 2 x2 ´ x + 1 dx = 1 3 log |x + 1| ´ 1 6 ż 2x ´ 1 x2 ´ x + 1 dx + 1 2 ż 1 (x ´ 1 2)2 + 3 4 dx = 1 3 log |x + 1| ´ 1 6 log |x2 ´ x + 1| + 1 2 ż 1 3 4 2x´1 ?
3 2 + 1 dx = 1 3 log |x + 1| ´ 1 6 log |x2 ´ x + 1| + 2 3 ż 1 2x´1 ?
3 2 + 1 dx Let u = 2x´1 ?
3 , du = 2 ?
3 dx.
= 1 3 log |x + 1| ´ 1 6 log |x2 ´ x + 1| + 1 ?
3 ż 1 u2 + 1 dx = 1 3 log |x + 1| ´ 1 6 log |x2 ´ x + 1| + 1 ?
3 arctan 2x ´ 1 ?
3 + C S-29: By process of elimination, we decide to use integration by parts. We won’t get anything better by antidifferentiating arcsine, so let’s plan on differentiating it: u = arcsin x dv = (3x)2dx du = 1 ?
1 ´ x2dx v = 3x3 ż (3x)2 arcsin xdx = arcsin x looomooon u ¨ 3x3 lo omo on v ´ ż 3x3 lo omo on v ¨ 1 ?
1 ´ x2dx looooomooooon du = 3x3 arcsin x ´ ż 3x3 ?
1 ´ x2dx So: we’ve gotten rid of the ugly pairing of arcsine with a polynomial, but now we’re in another pickle. From here, two options present themselves. We could use the substitution u = 1 ´ x2, or we could use a trig substitution.
480 Option 1: Let u = 1 ´ x2. Then ´1 2du = dx, and x2 = 1 ´ u.
ż (3x)2 arcsin xdx = 3x3 arcsin x ´ ż 3x3 ?
1 ´ x2dx = 3x3 arcsin x ´ 3 ż x2 ?
1 ´ x2 ¨ x dx = 3x3 arcsin x + 3 2 ż 1 ´ u ?u du = 3x3 arcsin x + 3 2 ż u´1/2 ´ u1/2 du = 3x3 arcsin x + 3 2 2u1/2 ´ 2 3u3/2 + C = 3x3 arcsin x + 3 a 1 ´ x2 ´ a 1 ´ x23 + C Option 2: If we let x = sin θ, then ?
1 ´ x2 = ?
cos2 θ = cos θ. So let’s use the substitution x = sin θ, dx = cos θdθ.
ż (3x)2 arcsin xdx = 3x3 arcsin x ´ ż 3x3 ?
1 ´ x2dx = 3x3 arcsin x ´ ż 3 sin3 θ a 1 ´ sin2 θ cos θdθ = 3x3 arcsin x ´ ż 3 sin3 θdθ And now: a substitution from Section 1.8.1 of the CLP-2 text, u = cos x and du = ´ sin xdx 3x3 arcsin x ´ ż 3 sin3 θdθ = 3x3 arcsin x ´ 3 ż sin2 θ sin θdθ = 3x3 arcsin x ´ 3 ż (1 ´ cos2 θ) sin θdθ = 3x3 arcsin x + 3 ż (1 ´ u2)du = 3x3 arcsin x + 3 u ´ 1 3u3 + C = 3x3 arcsin x + 3u ´ u3 + C θ ?
1 ´ x2 x 1 = 3x3 arcsin x + 3 cos θ ´ cos3 θ + C Recall x = sin θ; so we draw a triangle with angle θ, opposite side x, hypotenuse 1.
Then by Pythagoras, adjacent side is ?
1 ´ x2, so cos θ = ?
1 ´ x2.
= 3x3 arcsin x + 3 a 1 ´ x2 ´ (1 ´ x2)3/2 + C 481 S-30: We would like to not have that square root there. Luckily, there’s a way of turning cosine into cosine squared: the identity cos(2x) = 2 cos2 x ´ 1. If we take 2x = t, then cos t = 2 cos2(t/2) ´ 1.
ż π/2 0 ?
cos t + 1 dt = ż π/2 0 b 2 cos2(t/2) dt = ?
2 ż π/2 0 | cos(t/2)| dt Over the interval [0, π 2 ], cos(t/2) ą 0, so we can drop the absolute values.
= ?
2 ż π/2 0 cos(t/2) dt = ?
2 2 sin t 2 π/2 0 = 2 ?
2 sin π 4 = 2 S-31: Solution 1: Using logarithm rules, log ?x = log x1/2 = 1 2 log x, so we can simplify: ż e 1 log ?x x dx = ż e 1 log x 2x dx We use the substitution u = log x, du = 1 xdx: ż e 1 log x 2x dx = 1 2 ż e 1 log(x) lo omo on u ¨ 1 x dx lo omo on du = 1 2 ż log(e) log(1) u du = 1 2 ż 1 0 u du = 1 2 1 2u2 1 0 = 1 2 1 2 ´ 0 = 1 4 Solution 2: We use the substitution u = log ?x. Then du dx = 1 ?x ¨ 1 2?x = 1 2x, hence 2du = 1 xdx. This fits our integral nicely!
ż e 1 log ?x x dx = ż log ?e log ?
1 u ¨ 2du = h u2i1/2 0 = 1 2 2 ´ 02 = 1 4 482 S-32: ż 0.2 0.1 tan x log(cos x)dx It might not be immediately obvious how to proceed on this one, so this is another example of an integral where you should not be discouraged by finding methods that don’t work. One thing that’s worked for us in the past is to use a u-substitution with the denominator. With that in mind, let’s find the derivative of the denominator.
d dx tlog(cos x)u = 1 cos x ¨ (´ sin x) = ´ sin x cos x = ´ tan x So, if we let u = log(cos x), we see ´du = tan xdx, which will work for a substitution.
ż 0.2 0.1 tan x log(cos x)dx = ż log(cos(0.2)) log(cos(0.1)) ´du u = h ´ log |u| ilog(cos(0.2)) log(cos(0.1)) = ´ log | log(cos 0.2)| + log | log(cos 0.1)| = log ˇ ˇ ˇ ˇ log(cos(0.1)) log(cos(0.2)) ˇ ˇ ˇ ˇ = log log(cos(0.1)) log(cos(0.2)) Things to notice: the integrand is only defined when log(cos x) exists AND is nonzero.
So, for instance, it is not defined when x = 0, because then log cos x = log 1 = 0, and we can’t divide by zero.
In the final simplification, since 0.1 and 0.2 are between 0 and π/2, the cosine term is positive but less than one, so log(cos 0.1) and log(cos 0.2) are both negative; then their quotient is positive, so we can drop the absolute value signs.
Using the base change formula, we can also write the final answer as log logcos(0.2) cos(0.1) .
S-33: (a) Without any other ideas, we see we have a compound function–a function of a function. We often find it useful to substitute for the “inside” function. So, we substitute u = log x, du = 1 x dx. Then dx = x du = eu du.
ż sin(log x) dx = ż sin(u) eu du We have already seen, in Example 1.7.11 of the CLP-2 text, that ż sin(u) eu du = 1 2eu sin u ´ cos u + C 483 So, ż sin(log x) dx = 1 2x sin(log x) ´ cos(log x) + C (b) The integrand is of the form N(x)/D(x) with N(x) of lower degree than D(x). So we factor D(x) = (x ´ 2)(x ´ 3) and look for a partial fractions decomposition: 1 (x ´ 2)(x ´ 3) = A x ´ 2 + B x ´ 3.
Multiplying through by the denominator yields 1 = A(x ´ 3) + B(x ´ 2) Setting x = 2 we find: 1 = A(2 ´ 3) + 0 ù ñ A = ´1 Setting x = 3 we find: 1 = 0 + B(3 ´ 2) ù ñ B = 1 So we have found that A = ´1 and B = 1. Therefore ż 1 (x ´ 2)(x ´ 3) dx = ż 1 x ´ 3 ´ 1 x ´ 2 dx = log |x ´ 3| ´ log |x ´ 2| + C and the definite integral ż 1 0 1 (x ´ 2)(x ´ 3) dx = h log |x ´ 3| ´ log |x ´ 2| i1 0 = log 2 ´ log 1 ´ log 3 ´ log 2 = 2 log 2 ´ log 3 = log 4 3 S-34: (a) If we expand the integrand, one part of it is quite familiar–a portion of a circle.
So, we split the specified integral in two.
ż 3 0 (x + 1) a 9 ´ x2 dx = ż 3 0 a 9 ´ x2 dx + ż 3 0 x a 9 ´ x2 dx The first piece represents the area above the x–axis and below the curve y = ?
9 ´ x2, i.e.
x2 + y2 = 9, with 0 ď x ď 3. That’s the area of one quadrant of a disk of radius 3. So ż 3 0 a 9 ´ x2 dx = 1 4(π ¨ 32) = 9 4π 484 For the second part, we substitute u = 9 ´ x2, du = ´2x dx. Note u(0) = 9 and u(3) = 0.
So, ż 3 0 x a 9 ´ x2 dx = ż 0 9 ?u du ´2 = ´1 2 " u3/2 3/2 #0 9 = ´1 2 ´ 27 3/2 = 9 All together, ż 3 0 (x + 1) a 9 ´ x2 dx = 9 4π + 9 (b) The integrand is of the form N(x)/D(x) with D(x) already factored and N(x) of lower degree. We immediately look for a partial fractions decomposition: 4x + 8 (x ´ 2)(x2 + 4) = A x ´ 2 + Bx + C x2 + 4 .
Multiplying through by the denominator yields 4x + 8 = A(x2 + 4) + (Bx + C)(x ´ 2) (˚) Setting x = 2 we find: 8 + 8 = A(4 + 4) + 0 ù ñ 16 = 8A ù ñ A = 2 Substituting A = 2 in (˚) gives 4x + 8 = A(x2 + 4) + (Bx + C)(x ´ 2) ù ñ ´2x2 + 4x = (x ´ 2)(Bx + C) ù ñ (´2x)(x ´ 2) = (Bx + C)(x ´ 2) ù ñ B = ´2, C = 0 So we have found that A = 2, B = ´2, and C = 0. Therefore ż 4x + 8 (x ´ 2)(x2 + 4) dx = ż 2 x ´ 2´ 2x x2 + 4 dx = 2 log |x ´ 2| ´ log(x2 + 4) + C Here the second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution u = x2 + 4, du = 2x dx.
(c) The given integral is improper, but only because of its infinite limits of integration.
(The integrand is continuous for all real numbers.) So, we’ll have to take two limits.
Before we do that, though, let’s find the antiderivative. We would like to use the substitution u = ex, du = ex dx. That is, 1 u du = dx.
ż 1 ex + e´x dx = ż 1 u(u + 1 u) du = ż 1 u2 + 1 = arctan u + C = arctan(ex) + C 485 Now we can deal with the limits of integration.
ż 8 ´8 1 ex + e´x dx = ż 0 ´8 1 ex + e´x dx + ż 8 0 1 ex + e´x dx = lim aÑ´8 "ż 0 a 1 ex + e´x dx # + lim bÑ8 "ż b 0 1 ex + e´x dx # = lim aÑ´8 arctan(ex) 0 a + lim bÑ8 arctan(ex) b 0 = lim aÑ´8 arctan(e0) ´ arctan(ea) + lim bÑ8 arctan(eb) ´ arctan(e0) = lim aÑ´8 ´ arctan(ea) + lim bÑ8 arctan(eb) = ´ arctan(0) + π 2 = π 2 S-35: It’s not immediately clear where to start, but a common method we’ve seen is to use the denominator in a u-substitution, especially when square roots are involved.
Let u = ?
1 ´ x, du = ´ 1 2 ?
1´xdx. Then u2 = 1 ´ x, so x = 1 ´ u2.
ż c x 1 ´ x dx = 2 ż ?x 2 ?
1 ´ x dx = ´2 ż a 1 ´ u2 du Now we’re back in familiar territory. Let u = sin θ, du = cos θ dθ.
= ´2 ż a 1 ´ sin2 θ cos θ dθ = ´2 ż cos2 θ dθ = ´ ż 1 + cos(2θ) dθ = ´θ ´ 1 2 sin(2θ) + C = ´θ ´ sin θ cos θ + C = ´ arcsin u ´ u a 1 ´ u2 + C (˚) θ ?
1 ´ u2 u 1 = ´ arcsin( ?
1 ´ x) ´ ?
1 ´ x?x + C In (˚), to convert from θ to u, our substitution u = sin θ tells us θ = arcsin u. To find cos θ, we can either trace our work backwards to see that we already simplified ?
1 ´ u2 into cos θ, or we can draw a right triangle with angle θ and sin θ = u, then use the Pythagorean theorem to find the length of the adjacent side of the triangle and cos θ.
S-36: Let’s use the substitution u = ex. There are a few reasons to think this is a good choice. It’s an “inside function,” in that if we let f (x) = ex, then f (ex) = eex, which is a 486 piece of our integrand. Also its derivative, ex, is multiplied by the rest of the integrand, since e2x = ex ¨ ex .
Let u = ex, du = exdx. When x = 0, u = 1, and when x = 1, u = e.
ż 1 0 e2xeexdx = ż 1 0 exeexexdx = ż e 1 ueu du This is more familiar. We use integration by parts with dv = eu du, v = eu. Conveniently, the “u” we brought in with the substitution is what we want to use for the “u” in integration by parts, so we don’t have to change the names of our variables.
= ueue 1 ´ ż e 1 eu du = e ¨ ee ´ e ´ ee + e = ee(e ´ 1) S-37: The substitution u = x + 1 looks promising at first, but doesn’t result in something easily integrable. We can’t use partial fractions because our integration isn’t rational.
This doesn’t look like something from the trig-substitution family. So, let’s think about integration by parts. There’s a lot of different ways we could break up the integrand into two parts. For example, we could view it as x (x+1)2 ex , or we could view it as x x+1 ex x+1 . After some trial and error, we settle on u = xex and dv = (x + 1)´2 dx.
Then du = ex(x + 1) and v = ´1 x+1.
ż xex (x + 1)2 dx = ´ xex x + 1 + ż ex(x + 1) x + 1 dx = ´ xex x + 1 + ż exdx = ´ xex x + 1 + ex + C = ex x + 1 + C S-38: It would be nice to use integration by parts with u = x, because then we would integrate ş v du, and du = dx. That is, the x would go away, and we’d be left with a pure trig integral. If we use u = x, then dv = sin x cos2 x. We need to find v: v = ż sin x cos2 x dx = ż tan x sec x dx = sec x Now we use integration by parts.
ż x sin x cos2 x dx = x sec x ´ ż sec x dx = x sec x ´ log | sec x + tan x| + C 487 S-39: If the unknown exponent gives you the jitters, think about what this looks like in easier cases. If n is a whole number, the integrand is a polynomial. Not so scary, right?
However, it’s a little complicated to expand. (You can do it using the very handy binomial theorem.) Let’s think of an easier way.
If we had simply the variable x raised to the power n, rather than the binomial x + a, that might be nicer. So, let’s use the substitution u = x + a, du = dx. Note x = u ´ a.
ż x(x + a)n dx = ż (u ´ a)un dx = ż un+1 ´ aun du Now, if n ‰ ´1 and n ‰ ´2, we can just use the power rule: = u(n+2) n + 2 ´ a un+1 n + 1 + C = (x + a)(n+2) n + 2 ´ a(x + a)n+1 n + 1 + C If n = ´1, then ż x(x + a)n dx = ż un+1 ´ aun du = ż 1 ´ a u du = u ´ a log |u| + C = (x + a) ´ a log |x + a| + C If n = ´2, then ż x(x + a)n dx = ż un+1 ´ aun du = ż 1 u ´ au´2 du = log |u| + a u + C = log |x + a| + a x + a + C All together, ż x(x + a)n dx = $ ’ & ’ % (x+a)(n+2) n+2 ´ a (x+a)n+1 n+1 + C if n ‰ ´1, ´2 (x + a) ´ a log |a + x| + C if n = ´1 log |x + a| + a x+a + C if n = ´2 S-40: We’ve seen how to antidifferentiate arctan x: integration by parts. Let’s hope the same thing will work here.
Step 1: integration by parts.
Let u = arctan(x2) and dv = dx. Then du = 2x x4+1 du and v = x.
ż arctan(x2) dx = x arctan(x2) ´ ż 2x2 x4 + 1 dx Now we have a rational function. There’s no obvious substitution, but we can use partial fractions. The degree of the numerator is strictly less than the degree of the denominator, 488 so we don’t need to long divide first. We do, however, need to factor the denominator.
It’s a common function, so you might already know the factorization, or you might be able to guess it. Below, we show another way to find the factorization, similar to the method of partial fractions.
Step 2: factor x4 + 1.
For any real x, note x4 + 1 ą 0. Since it has no roots, it has no linear factors. That means it factors as the product of two irreducible quadratics. That is, x4 + 1 = (ax2 + bx + c)(dx2 + ex + f ) Since the coefficient of x4 on the left-hand is 1, we may assume a = d = 1.
x4 + 1 = (x2 + bx + c)(x2 + ex + f ) Since the constant term is 1, c f = 1. That is, f = 1 c.
x4 + 1 = (x2 + bx + c)(x2 + ex + 1/c) = x4 + (b + e) loomoon (1) x3 + 1 c + be + c looooooomooooooon (3) x2 + b c + ec loooomoooon (2) x + 1 (1) The coefficient of x3 tells us e = ´b.
(2) Then the coefficient of x tells us 0 = b c + ec = b c ´ bc. So, c = 1 c, hence c = ˘1.
(3) Finally, the coefficient of x2 tells us 0 = 1 c + be + c = 1 c ´ b2 + c. Since ´b2 is negative (or zero), 1 c + c is positive, so c = 1. That is, 0 = 1 ´ b2 + 1. So, b = ?
2.
All together, x4 + 1 = (x2 + ?
2x + 1)(x2 ´ ?
2x + 1) Step 3: partial fraction decomposition.
Now that we have the denominator factored into irreducible quadratics, we can find the partial fraction decomposition of the integrand.
2x2 x4 + 1 = Ax + B x2 + ?
2x + 1 + Cx + D x2 ´ ?
2x + 1 2x2 = (Ax + B)(x2 ´ ?
2x + 1) + (Cx + D)(x2 + ?
2x + 1) = (A + C)x3 + (B + D ´ ?
2A + ?
2C)x2 + (A + C ´ ?
2B + ?
2D)x + (B + D) From the coefficient of x3, we see C = ´A.
2x2 = (B + D ´ 2 ?
2A)x2 + (´ ?
2B + ?
2D)x + (B + D) From the constant term, we see D = ´B.
2x2 = (´2 ?
2A)x2 + (´2 ?
2B)x 489 From the coefficient of x2, we see ´2 ?
2A = 2, so A = ´1/ ?
2. Since C = ´A, then C = 1/ ?
2.
From the coefficient of x, we see B = 0. Since D = ´B, also D = 0.
Step 4: integration.
ż 2x2 x4 + 1 dx = ż (´1/ ?
2)x x2 + ?
2x + 1 + (1/ ?
2)x x2 ´ ?
2x + 1 dx = 1 ?
2 ż ´x x2 + ?
2x + 1 + x x2 ´ ?
2x + 1 dx To integrate, we want to break the fractions into two pieces each: one we can integrate with a substitution u = x2 ˘ ?
2x + 1 , du = 2x ˘ ?
2 dx (shown in blue), and one that looks like the derivative of arctangent (shown in red).
= 1 ?
2 ż ´x ´ ?
2 2 + ?
2 2 x2 + ?
2x + 1 + x ´ ?
2 2 + ?
2 2 x2 ´ ?
2x + 1 !
dx = 1 ?
2 ż ´1 2(2x + ?
2) x2 + ?
2x + 1 + ?
2 2 x2 + ?
2x + 1 + 1 2(2x ´ ?
2) x2 ´ ?
2x + 1 + ?
2 2 x2 ´ ?
2x + 1 !
dx = 1 ?
2 ´1 2 log ˇ ˇ ˇx2 + ?
2x + 1 ˇ ˇ ˇ + ż ?
2 2 x2 + ?
2x + 1 dx + 1 2 log |x2 ´ ?
2x + 1| + ż ?
2 2 x2 ´ ?
2x + 1dx !
We use logarithm rules to compress our work. In order to evaluate the remaining integrals, we complete the squares of the denominators.
= 1 ?
2 1 2 log ˇ ˇ ˇ x2´ ?
2x+1 x2+ ?
2x+1 ˇ ˇ ˇ + ż ?
2 2 x + 1 ?
2 2 + 1 2 dx + ż ?
2 2 x ´ 1 ?
2 2 + 1 2 dx !
= 1 ?
2 1 2 log ˇ ˇ ˇ x2´ ?
2x+1 x2+ ?
2x+1 ˇ ˇ ˇ + ż ?
2 ?
2x + 1 2 + 1 dx + ż ?
2 ?
2x ´ 1 2 + 1 dx !
Now, we can either guess the antiderivatives of the remaining integrals, or use the substitutions u = ( ?
2x ˘ 1).
= 1 ?
2 1 2 log ˇ ˇ ˇ x2´ ?
2x+1 x2+ ?
2x+1 ˇ ˇ ˇ + arctan ?
2x + 1 + arctan ?
2x ´ 1 !
+ C Step 5: finishing touches.
Finally, we can put our work together. (Remember way back in Step 1, we used 490 integration by parts.) ż arctan(x2) dx = x arctan(x2) ´ ż 2x2 x4 ´ 1 dx = x arctan(x2) ´ 1 ?
2 1 2 log ˇ ˇ ˇ x2´ ?
2x+1 x2+ ?
2x+1 ˇ ˇ ˇ + arctan ?
2x + 1 + arctan ?
2x ´ 1 !
+ C Remark: although this integral calculation was longer than average, it didn’t use any new ideas (except for the factoring of x4 + 1 mentioned in the hint). It’s good exercise to apply familiar techniques in challenging situations, to deepen your mastery.
Solutions to Exercises 2.1 — Jump to TABLE OF CONTENTS S-1: Force is mass ˆ acceleration (with acceleration equal to g in this problem), and both in this scenario are constant, so we don’t need an integral–only a product–to calculate the force acting on the block.
To find the force in newtons, recall one newton is one kg¨m sec2 , so we need the mass of our block in kg. Specifically, our block has mass 3 1000 kg. So, the force involved is F = 3 1000 kg ˆ 9.8 m sec2 = 0.0294 kg ¨ m sec2 = 0.0294 N To find the work in joules, recall one joule is one newton-metre: that is, one newton of force acting over one metre. So, we need our distance in metres.
W = 0.0294 N ˆ 1 10 m = 0.00294 N ¨ m = 0.00294 J S-2: The force of the rock is one newton, or one kilogram-metre per second squared, so 1 kg ¨ m sec2 = (x kg) 9.8 m sec2 Therefore, the mass of the rock is 1 9.8 kg, or about 102 grams.
Now, since one joule is one newton-metre, the amount of work required to counteract 1 N of gravitational force for one metre is precisely one joule.
Remark: having an idea of how much work a joule is, and how much force a newton is, is a good tool for checking the reasonableness of your work. For example, after this question, if you calculate that a marble weighs 100 N, you can be pretty sure there’s an error in your calculation.
S-3: 491 (a) We defined ∆x = b´a n : that is, the length of one interval, when we chop [a, b] into n of them. If b and a are measured in metres, then ∆x is measured in metres as well. So, the units of ∆x are metres.
Put another way, since a and b both describe a quantity in metres, b ´ a describes a quantity in metres as well. (When we add or subtract quantities of the same units, their sum or difference is given in the same units.) Since n is a unitless quantity (simply a number: not “n kg” or “n m”), b´a n still describes a quantity in metres. (If I have 6 metres of cloth, and I cut it into 3 pieces, each piece has 6 3 = 2 metres–not 2 kilograms, or 2 metres per second.) (b) Since F(x) is measured in kilogram-metres per second squared (newtons), the units of F(xi) are kilogram-metres per second squared (newtons).
(c) W is calculated by adding up summands of the form F(xi)∆x. The units of F(xi)∆x are the products of the units of F(xi) with the units of ∆x. That is, the units of F(xi)∆x are kg¨m sec2 (m) = kg¨m2 sec2 = J. The sum of terms given in joules is itself given in joules, so the units of W are joules.
S-4: As we saw in Question 3, the units of şb a f (x) dx are simply the units of the integrand, f (x), multiplied by the units of the variable of integration, x. In this case, that yields smoot¨barn megaFonzie (that is, smoot-barns per megaFonzie).
S-5: Hooke’s law says that the force required to stretch a spring x units past its natural length is proportional to x; that is, there is some constant k associated with the individual spring such that the force required to stretch it x m past its natural length is kx.
Solution 1: Since the force required to stretch the spring is proportional to the amount stretched, and the force acting on the spring is proportional to the mass hanging from it, we conclude the amount the spring stretches is proportional to the mass hung from it. So, if 1 kg stretches it 1 cm, then 10 kg will stretch it 10 cm. We should mark the wall 10 cm below the bottom of the spring as it hangs unloaded.
Solution 2: We can find k from the test with the bag of water. The force exerted by the bag of water was (1 kg)(9.8 m/sec2) = 9.8 N= k(1 cm). So, k = 9.8 kg¨m sec2 0.01 m = 980 kg sec2 If we hang 10 kg from the spring, gravity exerts a force of (10 kg)(9.8 m/sec2) = 98 kg¨m sec2 . This will be matched by the spring with a force of 492 kx newtons, where k is the spring constant and x is the amount stretched.
kx = 98 kg ¨ m sec2 980 kg sec2 (x m) = 98 kg ¨ m sec2 x = 1 10 m = 10 cm So, we should put the mark at 10 cm below the natural length of the spring.
S-6: Definition 2.1.1 in the CLP-2 text tells us the work done by the force is W(b) = şb 1 F(x) dx, where F(x) is the force on the object at position x. So, by the Fundamental Theorem of Calculus Part 1, d dbtW(b)u = d db #ż b 1 F(x) dx + = F(b) d db !
´b3 + 6b2 ´ 9b + 4 ) = F(b) ´3b2 + 12b ´ 9 = F(b) ´3(b ´ 1)(b ´ 3) = F(b) So, F(x) is the quadratic polynomial ´3(x ´ 1)(x ´ 3).
x y y = F(x) 1 2 3 The largest absolute value of F(x) over [1, 3] occurs at x = 2. At this point, we have our strongest force.
S-7: By Definition 2.1.1 in the CLP-2 text, the work done in moving the object from x = 1 meters to x = 16 meters by the force F(x) is W = ż 16 1 F(x) dx = ż 16 1 a ?x dx = h 2a?x ix=16 x=1 = 6a To have W = 18, we need a = 3.
As a side remark, F(x) = a ?x should have units Newtons. Since x, a distance, is measured in meters, a has to have the bizarre units newton-?
meters.
493 S-8: (a) Since c ℓ´x is measured in newtons, and ℓand x (and therefore ℓ´ x) are measured in metres, the units of c are newton-metres, i.e. joules.
(b) Following Definition 2.1.1 in the CLP-2 text, the work done compressing the air is W = ż 1.5 1 F(x) dx where F(x) is the amount of force applied when the plunger is x metres past its natural position. The amount of force applied is equal in magnitude to the amount of force supplied by the tube: c ℓ´x N. Note ℓand c are constants. We can guess the antiderivative, or use the substitution u = ℓ´ x, du = ´dx.
W = ż 1.5 1 c ℓ´ x dx = ´ c log |ℓ´ x| 1.5 1 = ´c log |ℓ´ 1.5| ´ log |ℓ´ 1| = ´c log ℓ´ 1.5 ℓ´ 1 = c log ℓ´ 1 ℓ´ 1.5 J Note that, because ℓą 1.5, the argument of logarithm is positive, so we don’t need the absolute value signs. Furthermore, ℓ´ 1 ą ℓ´ 1.5, so ℓ´1 ℓ´1.5 ą 1, hence log ℓ´1 ℓ´1.5 ą 0.
S-9: By Hooke’s Law, the force exerted by the spring at displacement x m from its natural length is F = kx, where k is the spring constant. Measuring distance in meters and force in newtons (since one joule is one newton-metre), the total work is ż 0.1 m 0 kx dx = 1 2kx2 0.1 m 0 = 1 2 ¨ 50 lo omo on N/m ¨ (0.1)2 lo omo on m2 = 1 4J.
Note the units of the integrand (kx) are newtons, and the units of the variable of integration, x, are metres. So, the evaluated integral has units newton-metres, or joules.
S-10: First note that newtons and joules are SI units with one joule equal to one newton-metre, so we should measure distances in meters rather than centimeters. Next recall that a spring with spring constant k exerts a force F(x) = kx when the spring is stretched x m beyond its natural length. So in this case (0.05 m)(k) = 10 N, or k = 200 N/m. The work done is: ż 0.5 m 0 F(x) dx = ż 0.5 0 200x dx = h 100x2i0.5 0 = 25 J 494 Note the units of the integrand (F(x) = kx = 200x) are newtons (k is given in N/m, and x is given in m). The units of the variable of integration, x are metres. So, the evaluated integral has units newton-metres, or joules.
S-11: Note that the cable has mass density 8 5 kg/m. When the bucket is at height y, the cable that remains to be lifted has length (5 ´ y) m and mass 8 5(5 ´ y) = 8 1 ´ y 5 kg. So, at height y, the cable is subject to a downward gravitational force of 8 1 ´ y 5 ¨ 9.8 N; to raise the cable we need to apply a compensating upward force of 8 1 ´ y 5 ¨ 9.8 N. So, the work required is ż 5 0 8 1 ´ y 5 ¨ 9.8 dy = 8 y ´ y2 10 ¨ 9.8 5 0 = 8 ¨ 2.5 ¨ 9.8 N ¨ m = 196 J.
Alternatively, the cable has linear density 8 kg/5 m = 1.6 kg/m, and so the work required to lift a small piece of the cable (of length ∆y) from height y m to height 5 m is 1.6∆y lo omo on mass ¨ 9.8 lo omo on gravity looooooomooooooon force ¨ (5 ´ y) loomoon distance .
The total work required is therefore ż 5 0 1.6 ¨ 9.8(5 ´ y) dy = 1.6 ¨ 9.8 5y ´ 1 2y2 5 0 = 1.6 ¨ 9.8 ¨ 25 ´ 25 2 = 196 J as before.
S-12: Imagine pumping out a thin, horizontal layer of water that is at height y–that is, y metres above the bottom of the tank. Let the width of the layer be dy.
dy • The volume of water in the layer is 3dy m3 (since the cross-section has area 3 m3).
• One cubic metre is equal to 1003 cubic centimetres. So, the mass of water in one cubic metre is 1003 1000 = 1 000 kg.
• Therefore, the mass of water in our layer is (3 000 dy) kg.
495 • The force of gravity acting on it is (´9.8 ˆ 3 000 dy) N, so we need to pump with a compensating force of (9.8 ˆ 3 000 dy) N.
• The water needs to be pumped a distance of 1 ´ y metres.
• So, the work required to pump out the thin layer of water at height y is (9.8 ˆ 3 000 ˆ (1 ´ y) dy) J.
So, all together, the work to pump out the entire tank is ż 1 0 9.8 ˆ 3 000 ˆ (1 ´ y) dy = 9.8 ˆ 3000 ˆ y ´ 1 2y2 1 0 = 14 700 J S-13: We can model the sculpture as a collection of thin horizontal plates of width dz.
Remember work is force times distance; a horizontal plate at height z moved z + 2 metres from the basement to its final position. So, we need to know the force acting on the plate, which is the product of the mass of the plate with the acceleration due to gravity. Since we are given the density of iron, if we find the volume of the plate, then we can calculate its mass.
dz The plate at height z • has side length 3 ´ z m and hence • has area (3 ´ z)2 m2 and hence • has volume (3 ´ z)2 dz m3 and hence • has mass 8000(3 ´ z)2 dz kg and hence • is subject to a gravitational force of 9.8 ˆ 8000(3 ´ z)2 dz N and hence • requires work 9.8 ˆ 8000(2 + z)(3 ´ z)2 dz J to raise it from 2 m below ground level to z m above ground level.
So the total work is ż 3 0 9.8 ˆ 8000(2 + z)(3 ´ z)2 dz joules S-14: From the information given about the hanging kilogram, we can find the spring constant k. One kilogram generates a force of 9.8 N under gravity. (We find this by the 496 calculation (1 kg) ˆ (9.8 m/sec2) = 9.8 N.) This force is matched by the force of the spring, which by Hooke’s law is equal to k( 1 20 m). So, k = 9.8 N 1 20 m = 196 N m Again by Hooke’s law, the force required to stretch the spring x metres past its natural length is 196x N (when x is measured in metres).
So, the work required to stretch the spring from 5 cm past its natural length to 7 cm past its natural length is W = ż 0.07 0.05 196x dx = 98x20.07 0.05 = 0.2352 J S-15: Let M be the mass of the rope. Then its density is M 4 kg/m. Following the method of Example 2.1.6 in the CLP-2 text, we let y be the height of the firewood above the ground, so the wood is raised from y = 0 to y = 4. When the wood is at height y, • the rope that remains to be lifted has length 4 ´ y, and so it has mass M 4 (4 ´ y) kg, • and the firewood still has mass 10 kg.
• The remaining rope and the wood are subject to a downward gravitational force of magnitude M 4 (4 ´ y) + 10 loooooooooomoooooooooon mass ˆ 9.8 N.
• So, to raise the firewood from height y to height (y + dy), we need to apply a compensating upward force of M 4 (4 ´ y) + 10 ˆ 9.8 through distance dy. This takes work M 4 (4 ´ y) + 10 ˆ 9.8 dy J.
All together, the work involved in hauling up the wood is ż 4 0 M 4 (4 ´ y) + 10 ˆ 9.8 dy = 9.8 ż 4 0 (M + 10) ´ M 4 y dy = 9.8(2M + 40) J Since the work was 400 joules, solving 400 = 9.8(2M + 40) for M tells us the mass of the rope is 200 9.8 ´ 20 = 20 49 kg, or about 408 g.
Alternately, the work involved in lifting up the wood is 10 ˆ 9.8 ˆ 4 = 392 J, so the work in lifting up the rope is 8 J. A small section of rope of length dy, that starts at height y above the ground, has mass M 4 dy kg and is lifted (4 ´ y) metres, so the work involved in lifting this section of rope is 9.8 ˆ (4 ´ y) ˆ M 4 dy. Then the amount of work to lift the whole rope (but not the wood) is 8 J = ż 4 0 9.8 ˆ (4 ´ y) ˆ M 4 dy = 9.8 ˆ M 4 ż 4 0 (4 ´ y) dy = 9.8 ˆ M 4 ˆ 8 which again results in M = 4 9.8 = 20 49 kg.
497 S-16: For Questions 11 and 15 in this section, we gave two methods for finding the work involved in pulling up a cable: one where we consider pulling up the entire remaining cable a tiny distance of dy, and one where we consider pulling a tiny slice of cable of length dy the entire distance up.
There is another variation we can consider with the weight: we can either calculate the work done on the weight and the work done on the rope separately, or we can calculate them together. If we calculate them together, then there are two cases to consider: the work done pulling up the first 5 metres of rope involves the weight, while the last 5 metres does not. These two choices (how to model the rope, and how to deal with the weight) actually lead to four solutions, but to avoid unnecessary repetition only two are presented below.
Solution 1: In this solution, we consider the work on the rope separately from work on the weight, and we imagine lifting a tiny piece of rope the entire distance to the window.
The weight has a mass of 5 kg, and is lifted a distance of 5 m to the window. The force of gravity acting on the weight is (5 kg)(9.8 m/sec2) = 49 N, so the work to lift it 5 metres is (49 N)(5 m) = 245 J.
y dy The density of the rope is 1 10 kg/m. A tiny piece of rope of length dy, hanging y metres from the window, has mass ( 1 10 dy) kg, and needs to be lifted y metres. So, the force of gravity acting on the piece of rope is ( 1 10 dy kg) 9.8 m/sec2 = 0.98 dy N, and the work to pull it up to the window is (0.98y dy) J. So, the total work to pull up the rope is ż 10 0 0.98y dy = 0.98 y2 2 10 0 = 49 J All together, the work to pull up the rope with the weight is 245 + 49 = 294 J.
Solution 2: In this solution, we consider the work on the rope together with the weight, and we imagine lifting the remaining rope a tiny distance to the window.
Suppose y metres of the rope have been pulled in, and 0 ď y ď 5 (shown on the left, below). Then the remaining rope has length 10 ´ y, and contains the weight, so the 498 mass remaining to be pulled up is 1 10(10 ´ y) looooomooooon rope + 5 lo omo on weight = 6 ´ y 10 kg. Then the force of gravity acting on the dangling rope and weight is (9.8 m/sec2)((6 ´ y 10) kg) = (58.8 ´ 0.98y) N. The work needed to lift this rope dy metres is (58.8 ´ 0.98y) dy J.
10 ´ y 10 ´ y Now, suppose y metres of the rope have been pulled in, and 5 ă y ď 10 (shown above, right). Then the remaining rope has length 10 ´ y, but does not contain the weight, so the mass remaining to be pulled up is 1 10(10 ´ y) = 1 ´ y 10 kg. Then the force of gravity acting on the dangling rope is (9.8 m/sec2)((1 ´ y 10) kg) = (9.8 ´ 0.98y) N. The work needed to lift this rope dy metres is (9.8 ´ 0.98y) dy J.
All together, the work needed to lift the rope is W = ż 10 0 F(y) dy = ż 5 0 (58.8 ´ 0.98y) dy + ż 10 5 (9.8 ´ 0.98y) dy = h 58.8y ´ 0.49y2i5 0 + h 9.8y ´ 0.49y2i10 5 = 294 J S-17: (a) The frictional force is µ ˆ m ˆ g = 0.4 (10 kg) 9.8 m sec2 = 39.2 kg¨m sec2 = 39.2 N. Since this constant force acts over a distance of 3 metres, the work is 3 ˆ 39.2 = 117.6 J.
In the case of a constant force, we don’t need to use an integral, but we could if we wanted: W = ż 3 0 39.2 dx = 39.2x 3 0 = 39.2 ˆ 3 = 117.6 J.
(b) Since the box is moving at a speed of 1 m/sec, at time t we can say the box is at position t, 0 ď t ď 3. At position t, the mass of the box is (10 ´ ?
t) kg, so the frictional force is 0.4 ˆ m ˆ g = 0.4 10 ´ ?
t kg 9.8 m sec2 = 3.92(10 ´ ?
t) N. Now 499 that we know the force, to find the work we simply integrate, following Definition 2.1.1 in the CLP-2 text: W = ż 3 0 3.92(10 ´ ?
t) dt = 3.92 10t ´ 2 3t3/2 3 0 = 3.92 30 ´ 2 3 ?
3 3 = 3.92 h 30 ´ 2 ?
3 i « 104 J S-18: Definition 2.1.1 in the CLP-2 text is justified by showing that the work done by a force acting on a particle is equal to the change in the kinetic energy of that particle. We can use Hooke’s law to calculate the work done stretching the spring. That work will be equal to the change in kinetic energy of the ball.
The ball initially has kinetic energy 1 2(1 kg)(v0 m/sec)2 = v2 0 2 kg¨m2 sec2 = v2 0 2 J. At the time the spring is stretched its farthest, the ball’s velocity is 0 m/sec, so its kinetic energy is 1 2(1 kg)(0 m/sec)2 = 0 J. So, the change in kinetic energy of the ball is v2 0 2 J.
Now let’s find the work done by the spring. Its spring constant is k = 5 N/m, so, the force on the spring when it is stretched x metres past its natural length is 5x N. The spring is stretched from its natural length to 10 cm, which is 0.1 m. Then the work done by the spring is W = ż 0.1 0 kx dx = ż 0.1 0 5x dx = 5 2x2 0.1 0 = 1 40 J Now we can find v0.
v2 0 2 = 1 40 ù ñ v0 = 1 ?
20 m/sec « 22.36 cm/sec S-19: The setup to answer this question is similar to Question 18 in this section: the work done by a spring on the occupied vehicle will be equal to the change in kinetic energy of that occupied vehicle. So, we need to find the work done by the spring, and the kinetic energy lost by the falling car. In order to find the work done by the spring, we need to find the spring constant.
Spring constant: The car’s mass of 2000 kg compresses the struts 2 cm past their natural length. The force of the car under gravity is (2000 kg) ˆ (9.8 m/sec2) = 19 600 N.
This force is exactly the same as that exerted by the spring, k(0.02 m). So, k = 980 000 N/m.
Work done by spring: The spring can safely compress 20 cm. So, the amount of work done by the spring compressing that far gives us the maximum amount of work the spring can safely do. While the car is falling, the spring is at its natural length, so the work done to compress it to 20 cm (0.2 m) shorter is: ż 0.2 0 kx dx = ż 0.2 0 980 000x dx = 490 000x20.2 0 = 19 600 J 500 Change in kinetic energy: When the car first hits the pavement, it’s falling at 4 m/sec, so it has kinetic energy 1 2(2100 kg)(4 m/sec)2 = 16 800 J. When the car compresses the springs as far as they go and it starts to rebound, it has kinetic energy 0, since its instantaneous velocity is zero. So, the change in kinetic energy is 16 800 J.
Since the change in kinetic energy is 16 800 J, and the struts can safely do a work of (up to) 19 600 J, the jump is within the (meagre) safety limits set by the question.
S-20: Let’s consider sucking up a flat, horizontal layer of water. If the water is y metres above bottom of the cone, then it needs to be raised 0.15 ´ y metres. So, if its mass is m kg, then the force of gravity acting on it is 9.8m N and the work involved in slurping it to the top of the cone is 9.8m(0.15 ´ y) J. So, what we need to find is the mass of a layer of water y metres from the bottom of the cone.
dy r y 0.15 0.05 y r A horizontal cross-section of the cone is a circle. To find its radius, we use similar triangles: r y = 0.05 0.15, so r = 1 3y. Therefore, the area of the cross-section of the cone y metres above its bottom is π 1 3y 2 = π 9 y2 m2. If this layer has height dy, then its volume is π 9 y2 dy m3, and its mass is 1000π 9 y2 dy kg.
Now, we know that the work to suck up the layer of water y metres from the bottom of the cup is 9.8(0.15 ´ y) 1000π 9 y2 dy J. So, the work involved in drinking all the water is: W = ż 0.15 0 9.8(0.15 ´ y) 1000π 9 y2 dy = 9800π 9 ż 0.15 0 0.15y2 ´ y3 dy = 9800π 9 0.15 3 y3 ´ 1 4y4 0.15 0 = 9800π 9 0.05(0.15)3 ´ 1 4(0.15)4 « 0.144 J Even drinking water takes work. Life is hard.
501 S-21: Imagine slicing the water into horizontal pancakes of thickness dx as in the sketch below.
1m 3m x Denote by x the distance of a pancake below the surface of the water. (So, x runs from 0 to 3.) Each pancake: • has radius ?
32 ´ x2 m (by Pythagoras) and hence • has cross–sectional area π(9 ´ x2) m2 and hence • has volume π(9 ´ x2) dx m3 and hence • has mass 1000π(9 ´ x2) dx kg and hence • is subject to a gravitational force of 9.8 ˆ 1000π(9 ´ x2) dx N and hence • requires work 9800π(9 ´ x2)(x + 4) dx J to raise it to the spout. (It has to be raised x m to bring it to the height of the centre of the sphere, then 3 m more to bring it to the top of the sphere, and finally 1 m more to bring it to the spout.) The total work is: ż 3 0 9800π(9 ´ x2)(x + 4) dx = ż 3 0 9800π 36 + 9x ´ 4x2 ´ x3 dx = 9800π h 36x + 9 2x2 ´ 4 3x3 ´ 1 4x4i3 0 = 9800 ¨ 369 4 π = 904,050π joules S-22: Solution 1: Let’s consider the work involved in lifting up a small section of cable, with length dy, distance y from the bottom end of the cable.
502 5 0 5 ´ y y dy The distance this section must travel is (5 ´ y) metres, so if its mass is M(y), then the work involved is W = ż 5 0 9.8 ˆ (5 ´ y) ˆ M(y) So, we need to find M(y). The length of the section of cable is dy, and its distance from the end of the cable is y, so the mass of the section is (10 ´ y) dy. Therefore, W = ż 5 0 9.8 ˆ (5 ´ y) ˆ M(y) = ż 5 0 9.8 ˆ (5 ´ y) ˆ (10 ´ y) dy = 9.8 ż 5 0 50 ´ 15y + y2 dy = 9.8 50y ´ 15 2 y2 + 1 3y3 5 0 = 6125 6 = 10205 6 J Solution 2: Alternately, we can continue to use the basic method of Example 2.1.6 in the CLP-2 text, noticing that the density of the cable is no longer constant.
Let’s consider pulling the cable up a tiny distance of dy metres, after we have already lifted it y metres (so (5 ´ y) metres of the cable is still in the hole).
503 5 0 y 5 ´ y If R(y) is the mass of the remaining cable (in kg), then the force of gravity is ´9.8 ˆ R(y), so the work done is 9.8 ˆ R(y) ˆ dy. Once we find R(y), we can calculate the total work done: W = ż 5 0 9.8 ˆ R(y) ˆ dy (˚) As given in the question statement, the density of the cable is (10 ´ x) kg/m, where x is the distance from the bottom end of the cable. Consider a tiny section of cable x metres from the bottom end, of length dx.
0 x dx The mass of this tiny section is 10 ´ x kg m ˆ (dx m) = (10 ´ x) dx kg. The section of cable dangling is the last (5 ´ y) metres of cable. So, the combined mass of the section of cable dangling, after we’ve already pulled up y metres of it, is R(y) = ż 5´y 0 (10 ´ x) dx = 10x ´ 1 2x2 5´y 0 = 75 2 ´ 5y ´ 1 2y2 Now we can calculate the total work involved in pulling up the entire cable, using 504 equation (˚).
W = ż 5 0 9.8 ˆ R(y) ˆ dy = ż 5 0 9.8 ˆ 75 2 ´ 5y ´ 1 2y2 ˆ dy = 9.8 75 2 y ´ 5 2y2 ´ 1 6y3 5 0 = 6125 6 = 10205 6 J S-23: (a) The force of the depends on depth, which varies. So, consider a thin rectangle of the plunger at depth y, with height dy and width 1 m (the width of the entire plunger).
Let the area of this rectangle be dA.
depth 0 y dy The area of this rectangle is 1dy m2, so the force of the water acting on it is F = P ¨ dA = 9800 N m3 looooomooooon c (y m) lo omo on d dy m2 loooomoooon dA = 9800y dy N.
The depth at the top of the plunger is y = 0. To find the depth at the bottom of the plunger, note that the water has a volume of 3 m3, and is in a rectangular container with base 1 m by 3 m. So, its height is 1 m.
The force over the entire plunger, from depth y = 0 to y = 1, is ż 1 0 9800y dy = 4900y21 0 = 4900 N (b) Let’s follow our work from part (a), but with the width of the length of the base as x m.
Still, a thin rectangle of plunger has width 1 m and height dy m, so it has area dy m2.
At depth y, it has a force from the water of 9800y dy N. This hasn’t changed from (a).
Now, let’s consider the depth of the water. The volume of water is 3 m3, and it is in a rectangular container with base 1 m by x m. So, its depth is 3/x m. Therefore, the 505 force on the entire plunger must be calculated from y = 0 to y = 3/x.
F(x) = ż 3/x 0 9800y dy = 4900y23/x 0 = 9 x24900 = 44100 x2 N Let’s check that this answer makes sense: F(3) = 4900 N, which matches our answer from (a).
(c) If the force of water acting on the plunger, when the length of the base is x metres, is given by F(x), then we push the plunger with a force of ´F(x). Then the work we’re looking for is W = ż 1 3 ´F(x) dx = ż 3 1 F(x) dx F(x) is exactly what we found in (b): F(x) = 44100 x2 N.
W = ż 3 1 F(x) dx = ż 3 1 44100 x2 dx = 44100 ´1 x 3 1 = 44100 ¨ 2 3 = 29 400 J S-24: Let’s start by converting from time spent pulling to amount pulled. When y metres of rope have been pulled up, 2y seconds have passed, so 1 5y litres of water have leaked out of the bucket, leaving 5 ´ 1 5y litres. (This only makes sense when 1 5y ď 5, but we only consider values of y from 0 to 5, so it’s not a problem. That is, we’re never hauling up an empty bucket that can’t leak any more.) When we’ve pulled up y metres of rope, the mass in the bucket is (5 ´ 1 5y) kg, so the force of gravity acting on it is 9.8(5 ´ 1 5y) N. Since we pull up 5 metres of rope, the work done is: W = ż 5 0 9.8 5 ´ 1 5y dy = 9.8 5y ´ 1 10y2 5 0 = 9.8 [25 ´ 2.5] = 220.5 J S-25: According to the formula for gravity between two objects, the earth and moon will gravitationally attract one another no matter how far apart they are, so what we’re looking for is the work to separate them infinitely far. That is, we want to calculate ż 8 a F(r) dr, where a = 400 000 000 m.
If we take m1 and m2 to be the mass of the earth and moon as given in the question statement, then: Gm1m2 = 6.7 ˆ 10´11 m3 kg ¨ sec2 6 ˆ 1024 kg 7 ˆ 1022 kg « 2.8 ˆ 1037 kg ¨ m3 sec2 506 With that out of the way, let’s calculate our work.
W = ż 8 a F(r) dr = lim bÑ8 ż b a Gm1m2 r2 dr = (Gm1m2) lim bÑ8 ´1 r b a = (Gm1m2) lim bÑ8 1 a ´ 1 b = Gm1m2 a « 2.8 ˆ 1037 kg¨m3 sec2 4 ˆ 108 m = 7 ˆ 1028 J Remark: since the force of gravity between the earth and the moon gets weaker as they are farther apart, it takes less and less work to move them each kilometre. If we move them a finite distance apart, the work involved will always be less than 7 ˆ 1028 joules, no matter how huge that finite distance is. If we move them a very, very long (but finite) distance apart, the work we did will be quite close to (but still less than) 7 ˆ 1028 joules.
S-26: A ball of mass m experiences a gravitational force of mg, so lifting it a height of ℓ/2 involves a work of 1 2mgℓ.
The cable has density m/ℓ. A tiny section of cable with length dy has mass m ℓdy, and so gravity acts on it with a force of mg ℓdy. If the tiny section of cable is y units from the top of the cable, it needs to be pulled up y units, so the work on that section is mg ℓy dy.
Therefore, the work to pull up the entire cable is ż ℓ 0 mg ℓy dy = hmg 2ℓy2iℓ 0 = mg 2ℓℓ2 = 1 2mgℓ So, the work to pull up a cable with uniform density is the same as the work to pull up a ball with the same mass from the middle height of the cable.
Remark: this is a nice fact to use when you’re checking your computations for “pulling up cable” problems, but keep in mind it depends on the cable being of uniform density.
S-27: Like our other tank-pumping problems (e.g. Questions 12, 20, and 21 in this section, and Example 2.1.4 in the CLP-2 text), we can find the work done by considering thin layers of liquid. If the layer of liquid h metres above the bottom of the tank with thickness dh has mass M(h), then the force of gravity acting on it is ´9.8M(h) N and the work required to pump it to the top of the tank (1 ´ h metres away) is 9.8(1 ´ h)M(h) J.
So, the work to empty the entire tank is W = ż 1 0 9.8(1 ´ h)M(h) (˚) 507 Our remaining task is to find M(h). There are two things that vary with height: the density of the liquid, and the area of the cross-section of the tank.
At height h metres, the cross-section of the tank is shaped like the finite region bounded by the curves y = x2 and y = 2 ´ h ´ 3x2. To find this area, we need an integral (see Section 1.5 for a refresher), and to find the limits of integration, we need to know where the two curves meet. By solving x2 = 2 ´ h ´ 3x2, we find that they meet at x = ˘1 2 ?
2 ´ h. (Recall h is between 0 and 1, so ?
2 ´ h is a real number, i.e. the curves do indeed meet.) Furthermore, when ´1 2 ?
2 ´ h ď x ď 1 2 ?
2 ´ h, then x2 ď 2 ´ h ´ 3x2, so y = x2 is the bottom function and 2 ´ h ´ 3x2 is the top function.
x y ´1 2 ?
2 ´ h y = x2 y = 2 ´ h ´ 3x2 So, (taking advantage of the fact that our region has even symmetry) the area of the cross-section of the tank at height h is A(h) = ż 1 2 ?
2´h ´ 1 2 ?
2´h 2 ´ h ´ 3x2 ´ x2 dx = 2 ż 1 2 ?
2´h 0 2 ´ h ´ 4x2 dx = 2 (2 ´ h)x ´ 4 3x3 1 2 ?
2´h 0 = 2 (2 ´ h) ¨ 1 2 ?
2 ´ h ´ 4 3 ¨ 1 8 ?
2 ´ h 3 = (2 ´ h)3/2 1 ´ 1 3 = 2 3(2 ´ h)3/2 Now, we can calculate the volume of a slice at height h of thickness dh.
V(h) = 2 3(2 ´ h)3/2 dh The density of the liquid at height h is 1000 ?
2 ´ h kg/m3, so M(h) = 1000 ?
2 ´ h ˆ 2 3(2 ´ h)3/2 dh = 2000 3 (2 ´ h)2dh 508 Now we use (˚) to find the work done pumping out the tank.
W = ż 1 0 9.8(1 ´ h)M(h) = ż 1 0 9.8(1 ´ h) ¨ 2000 3 (2 ´ h)2dh = 19600 3 ż 1 0 4 ´ 8h + 5h2 ´ h3 dh = 19600 3 4h ´ 4h2 + 5 3h3 ´ 1 4h4 1 0 = 19600 3 4 ´ 4 + 5 3 ´ 1 4 = 19600 3 ˆ 17 12 = 83300 9 = 92555 9 J S-28: Since the only work done is against the force of gravity, we only need to know how high the sand was lifted, not how it got there. So, we don’t really need to worry about its semicircular path: we can imagine that every grain of sand was lifted from its old position to its new position.
Consider a thin, horizontal layer of sand in the hourglass, y metres below the vertical centre of the hourglass.
• Its final position is y metres above the centre of the hourglass. That is, it was lifted 2y metres against the force of gravity.
• The layer is shaped like a circle with radius y2 + 0.01 and height dy, so its volume is π y2 + 0.01 2 dy cubic metres.
• To find the mass of the layer, we need to know the density of the sand. Let the volume of sand in the hourglass be V. We are given its mass M. Then π y2 + 0.01 2 dy cubic metres has a mass of M V π y2 + 0.01 2 dy kilograms.
• So, the force of gravity acting on the layer is 9.8 M V π y2 + 0.01 2 dy N, acting vertically downwards.
• To lift the layer to its final position, we apply a compensating force over a distance of 2y metres, for a total work of 9.8 M V π y2 + 0.01 2 2y dy J.
• Since the hourglass has height 0.2 m, and exactly half of it is filled with sand, the top layer of sand is exactly at the vertical centre of the hourglass, and the bottom layer of sand is 0.1 metres below.
Using V is the volume of sand in the hourglass, and M is its mass (we’re given M = 1 7 kg) then the total work flipping all the sand is: 509 W = ż 0.1 0 9.8 M V π y2 + 0.01 2 2y dy = 19.6Mπ V ż 0.1 0 y5 + 0.02y3 + 0.0001y dy = 19.6Mπ V ż 1/10 0 y5 + 2 102 y3 + 1 104 y dy = 19.6Mπ V 1 6y6 + 1 2 ¨ 102 y4 + 1 2 ¨ 104 y2 1/10 0 = 19.6Mπ V 1 6 ¨ 106 + 1 2 ¨ 106 + 1 2 ¨ 106 ´ 0 = 19.6Mπ V 7 6 ¨ 106 (˚) It remains to find V: the volume of sand in the hourglass. We know the sand is in the shape of a solid of rotation. Recall from Section 1.6 that we can find the volume of such shapes by slicing them into thin disks.
0 x 0.1 x2 + 0.01 dx In the picture above, we’ve used an axis that matches the way we’ve been describing our solid: 0 is the vertical centre of the hourglass, which is where the top of the sand is, and the bottom of the sand is 0.1 metres from 0.
To find the volume of this solid, we slice it into horizontal disks. The disk that is x metres from the centre of the hourglass has radius x2 + 0.01 and thickness dx, so it has volume π(x2 + 0.01)2 dx. The volume of the entire solid, i.e. the volume of the sand, is: V = ż 0.1 0 π(x2 + 0.01)2 dx = π ż 0.1 0 (x4 + 0.02x2 + 0.0001) dx = π 1 5x5 + 2 3 ˆ 102 x3 + 1 104 x 0.1 0 = π 1 5 ˆ 105 + 2 3 ˆ 105 + 1 105 ´ 0 = 28π 15 ˆ 105 = 28π 1.5 ˆ 106 So, the volume of sand in the hourglass is V = 28π 1.5ˆ106 cubic metres.
510 Using (˚), we can find the total work done quickly flipping the hourglass.
W = 19.6Mπ V 7 6 ¨ 106 = 19.6 ˆ 1 7π 28π 1.5ˆ106 7 6 ¨ 106 = 19.6 ˆ 1.5 28 ˆ 6 = 29.4 168 = 7 40 = 0.175 J S-29: Using Definition 2.1.1 in the CLP-2 text, the work involved is W = ż 1/2 0 a 1 ´ x4 dx J.
However, the function F(x) = ?
1 ´ x4 happens to not have an antiderivative that can be expressed as an elementary function. That means we can’t use the Fundamental Theorem of Calculus Part 2 to evaluate this integral (at least, not without knowing a bit more about functions than is prerequisite for this course). Instead, we can use numerical methods, like the midpoint rule or Simpson’s rule, to approximate its value.
It’s not immediately clear which rule (Simpson’s, midpoint, or trapezoidal) will lead us down the easiest path. For Simpson’s rule, we need to know the fourth derivative of F(x), which is not a simple task. But, we often need fewer intervals for Simpson’s rule than for the midpoint or trapezoid rules. In this case, we’ll show below that n = 2 intervals suffice to guarantee a low enough error using the midpoint rule, so Simpson’s rule won’t let us get away with fewer intervals. Below, we find the approximation using the midpoint rule–but there are other ways as well.
In order to decide how many intervals we should use with the midpoint rule, we need to know the second derivative of F(x).
F(x) = (1 ´ x4)1/2 F1(x) = 1 2 1 ´ x4´1/2 ¨ (´4x3) = ´2x3 1 ´ x4´1/2 F2(x) = (´2x3) ´1 2 1 ´ x4´3/2 (´4x3) + (´6x2) 1 ´ x4´1/2 = ´4x6 1 ´ x4´3/2 + (´6x2) 1 ´ x4´1/2 = 2x2 1 ´ x4´3/2 ´2x4 ´ 3 1 ´ x4 = 2x2 1 ´ x4´3/2 x4 ´ 3 = 2x2 x4 ´ 3 (1 ´ x4)3/2 For values of x between 0 and 1 2, 511 • the denominator 1 ´ x43/2 is always at least as big as 1 ´ 1 2 43/2 , • the factor x2 in the numerator is never bigger than 1 2 2, and • the factor x4 ´ 3 in the numerator has magnitude at most 3, so that |F2(x)| = 2x2 ˇ ˇx4 ´ 3 ˇ ˇ (1 ´ x4)3/2 ď 2(1 2)2(3) 1 ´ (1 2)4 3/2 = 3 2 153/2 64 = 32 5 ?
15 ă 2 Using Theorem 1.11.12 in the CLP-2 text with M = 2, a = 0, and b = 1/2, the error in a midpoint approximation with n intervals is at most M 24 ¨ (b ´ a)3 n2 = 2 24 ¨ 1/8 n2 = 1 96n2 If n ě 2, then our error is certainly less than 1 100.
If we use the midpoint rule with n = 2, then ¯ x1 = 1 8 and ¯ x2 = 3 8.
x 0 1 2 1 4 3 8 1 8 The midpoint rule approximation of ş1/2 0 ?
1 ´ x4 dx with n = 2 and ∆x = 1 4 is: ż 1/2 0 a 1 ´ x4 dx « ∆x b 1 ´ ¯ x4 1 + b 1 ´ ¯ x4 2 = 1 4 d 1 ´ 1 8 4 + d 1 ´ 3 8 4 Since we used n = 2, by our previous work the error in this approximation is less than 1 96ˆ22 = 1 384, which is certainly less than 0.01, as required.
Solutions to Exercises 2.2 — Jump to TABLE OF CONTENTS S-1: Since the average of f (x) on the interval [0, 5] is A, using Definition 2.2.2 in the CLP-2 text, A = 1 5 ż 5 0 f (x) dx 5A = ż 5 0 f (x) dx So, a rectangle with width 5 and height A has area ş5 0 f (x) dx.
That is: if we replace f (x) with the constant function g(x) = A, then on the interval [0, 5], the area under the curve is unchanged.
512 x y 5 A (There are many rectangles with area 5A; we drew the one we consider to be the most straightforward in this context.) S-2: Average velocity, as discussed in Example 2.2.5 of the CLP-2 text, is change in position divided by change in time. So, the change in position (i.e. distance travelled) is (100 km/h)(5 h) = 500 km.
S-3: The work done is W = ż b a F(x) dx so the average value of F(x) is 1 b ´ a ż b a F(x) dx = 1 b ´ a(W).
We can quickly check our units: since W is in joules (that is, newton-metres), and b ´ a is in metres, so W b´a is in newtons.
S-4: (a) The entire interval has length b ´ a, and we’re cutting it into n pieces, so the length of one piece (and hence the distance between two consecutive samples) is b´a n .
(b) The first sample, as given in the question statement, is taken at x = a. The second sample, then, is at x = a + b´a n , this third is at x = 1 + 2b´a n , and the fourth is at a + 3b´a n .
(c) The y-value of the fourth sample is simply f a + 3b´a n . Note this is the number we use in our average, not the x-value.
(d) Our samples are f (a), f a + b´a n , f a + 2b´a n , f a + 3b´a n , etc. Since there are n 513 of them, we divide their sum by n. So, the average is: f (a) + f a + b´a n + f a + 2b´a n + ¨ ¨ ¨ + f a + (n ´ 1) b´a n n = 1 n f (a) + f a + b ´ a n f a + 2b ´ a n + ¨ ¨ ¨ + f a + (n ´ 1)b ´ a n = 1 n n ÿ i=1 f a + (i ´ 1)b ´ a n Remark: if we multiply and divide by b ´ a, we see this expression is equivalent to a left Riemann sum, divided by the length of our interval.
= 1 b ´ a n ÿ i=1 f a + (i ´ 1)b ´ a n b ´ a n = 1 b ´ a n ÿ i=1 f (a + (i ´ 1)∆x) ∆x As n gets larger and larger, using the definition of a definite integral, this expression gets closer and closer to 1 b´a şb a f (x) dx. This is one way of justifying our definition of an average of a function on an interval.
S-5: (a) Yes, the average of f (x) is less than or equal to the average of g(x) on [0, 10]. The reason is that, if f (x) ď g(x) for all x in [0, 10], then: 1 10 ż 10 0 f (x) dx ď 1 10 ż 10 0 g(x) dx.
(b) There is not enough information to tell. It’s certainly possible: for instance, take f (x) = 0 and g(x) = 1 for all x in [0, 10]. Then f (x) ď g(x) and the average of f (x) is 0, which is less than 1, the average of g(x).
However, consider f (x) = # 100 if 0 ď x ď 0.01 0 else and g(x) = 0. Then f (x) ď g(x) for all x in [0.01, 10], but the average of f (x) is 0.1, while the average of g(x) is 0.
S-6: Recall the definition of an odd function: f (´x) = ´f (x). Since the domain of integration is symmetric, the signed area on one side of the y-axis “cancels out” the signed area on the other–this is Theorem 1.2.12 in the CLP-2 text.
1 20 ż 10 ´10 f (x) dx = 1 20(0) = 0 514 S-7: By definition, the average value is 1 π ż π/2 ´π/2 sin(5x) + 1 dx We now observe that sin(5x) is an odd function, and hence its integral over the symmetric interval [´π 2 , π 2 ] equals zero. So the average value of f (x) on this interval is 1: 1 π ż π/2 ´π/2 sin(5x) + 1 dx = 1 π ż π/2 ´π/2 sin(5x) dx + 1 π ż π/2 ´π/2 1 dx = 1 π ż π/2 ´π/2 1 dx = 1 Alternatively, using the fundamental theorem of calculus, the average equals: 1 π ´ cos(5x) 5 + x π/2 ´π/2 = 1 π "´ cos(5π/2) 5 + π 2 ´ ´ cos(´5π/2) 5 + ´π 2 = π π = 1 S-8: By definition, the average is 1 e ´ 1 ż e 1 x2 log x dx To antidifferentiate, we use integration by parts with u = log x and dv = x2 dx, hence du = 1 x dx and v = 1 3x3.
1 e ´ 1 ż e 1 x2 log x dx = 1 e ´ 1 1 3x3 log x e 1 ´ ż e 1 1 3x2 dx = 1 e ´ 1 x3 3 log x ´ x3 9 x=e x=1 = 1 e ´ 1 e3 3 ´ e3 9 + 1 9 = 1 e ´ 1 2 9e3 + 1 9 S-9: By definition, the average value in question equals 1 π/2 ´ 0 ż π/2 0 (3 cos3 x + 2 cos2 x) dx = 2 π ż π/2 0 3 cos3 x dx + ż π/2 0 2 cos2 x dx For the first integral we use the substitution u = sin x, du = cos x dx, cos2 x = 1 ´ sin2 x = 1 ´ u2. Note that the endpoints x = 0 and x = π 2 become u = 0 and 515 u = 1, respectively.
ż π/2 0 3 cos3 x dx = ż π/2 0 3 cos2 x cos x dx = ż 1 0 3(1 ´ u2) du = (3u ´ u3) ˇ ˇ ˇ 1 0 = 2.
For the second integral we use the trigonometric identity cos2 x dx = 1+cos(2x) 2 .
2 ż π/2 0 cos2 x dx = ż π/2 0 1 + cos(2x) dx = x + 1 2 sin(2x) π/2 0 = π 2 Therefore, the average value in question is 2 π ż π/2 0 3 cos3 x dx + ż π/2 0 2 cos2 x dx = 2 π 2 + π 2 = 4 π + 1.
S-10: By definition, the average value in question equals Ave = 1 π/k ´ 0 ż π/k 0 sin(kx) dx To evaluate the integral, we use the substitution u = kx, du = k dx. Note that the endpoints x = 0 and x = π/k become u = 0 and u = π, respectively. So Ave = k π ż π 0 sin(u) du k = 1 π h ´ cos(u) iπ 0 = 2 π Remark: the average does not depend on k. To see why this is, note that sin(kx) runs between ´1 and 1 as x changes. When x = 0, kx = 0, and when x = π/k, kx = π. So, our function sin(kx) runs exactly from sin 0 = 0 to sin(π/2) = 1, then back down to sin π = 0.
x y π k π 2k 1 y = sin(kx) 516 S-11: By definition, the average temperature is 1 3 ż 3 0 T(x) dx = 1 3 ż 3 0 80 16 ´ x2 dx We don’t see an obvious substitution, but integrand is a rational function. The degree of the numerator is strictly less than the degree of the denominator, so we factor the denominator and use a partial fraction decomposition.
80 16 ´ x2 = 80 (4 ´ x)(4 + x) = A 4 ´ x + B 4 + x 80 = A(4 + x) + B(4 ´ x) Setting x = 4, we see 80 = 8A, so A = 10. Setting x = ´4, we see 80 = 8B, so B = 10.
1 3 ż 3 0 80 16 ´ x2 dx = 1 3 ż 3 0 80 (4 ´ x)(4 + x) dx = 1 3 ż 3 0 h 10 4 ´ x + 10 4 + x i dx = 1 3 ż 3 0 h ´ 10 x ´ 4 + 10 4 + x i dx = 10 3 h ´ log |x ´ 4| + log |x + 4| i3 0 = 10 3 log ˇ ˇ ˇx + 4 x ´ 4 ˇ ˇ ˇ ˇ ˇ ˇ ˇ 3 0 = 10 3 [log 7 ´ log 1] = 10 3 log 7 degrees Celsius S-12: By definition, the average value is 1 e ´ 1 ż e 1 log x x dx To integrate, we use the substitution u = log x, du = 1 x dx. Then the limits of integration become 0 and 1, respectively.
1 e ´ 1 ż e 1 log x x dx = 1 e ´ 1 ż 1 0 u du = 1 e ´ 1 u2 2 1 0 = 1 2(e ´ 1) S-13: By definition, the average value is: 1 2π ż 2π 0 cos2 x dx = 1 2π ¨ 1 2 ż 2π 0 cos(2x) + 1 dx = 1 4π hsin(2x) 2 + x i2π 0 = 1 4π ¨ 2π = 1 2 517 S-14: Before we start answering questions, let’s look at our function a little more carefully. The term 50 cos t 12π has a period of 24 hours, while the term 200 cos t 4380π has a period of one year. So, the former term describes a standard daily variation, while the latter gives a seasonal variation over the year.
(a) Using the definition of an average, the average concentration over one year (t = 0 to 8760) is: 1 8760 ż 8760 0 400 + 50 cos t 12π + 200 cos t 4380π dt = 1 8760 ż 8760 0 400 dt + 50 8760 ż 8760 0 cos t 12π dt + 200 8760 ż 8760 0 cos t 4380π dt = 400 + 5 876 12 π sin t 12π 8760 0 + 5 219 4380 π sin t 4380π 8760 0 Since 8760 12 = 730, which is even, sin 8760 12 π = sin(0) = 0. Also, sin 8760 4380π = sin(2π) = 0.
= 400 + 5 876(0) + 5 219(0) = 400 ppm Remark: for the portions of the integral in red and blue, we also could have noticed that the integrand goes through a whole (integer) number of periods. For every period, the net signed area between the curve and the x-axis is zero, so we could have seen from the very beginning these terms would contribute 0 to the final average.
(b) Using the definition of an average, the average concentration over the first day (t = 0 to t = 24) is: 1 24 ż 24 0 400 + 50 cos t 12π + 200 cos t 4380π dt = 1 24 ż 24 0 400 dt + 50 24 ż 24 0 cos t 12π dt + 200 24 ż 24 0 cos t 4380π dt Note t = 0 to t = 24 is one complete period for the integrand in red, so the red integral will evaluate to zero. However, t = 0 to t = 24 is less than one cycle for the integrand in blue, so we expect this will contribute some non-zero quantity to the average.
= 400 + 0 + 200 24 4380 π sin t 4380π 24 0 = 400 + 25 3 ¨ 4380 π sin 24 4380π = 400 + 25 3 ¨ 4380 π sin 2 365π « 400 + 199.99 = 599.99 ppm 518 Remark: C(0) = 400 + 50 + 200. The red term comes from the daily variation, and over the first day this will have an average of 0. The blue term comes from the seasonal variation, which changes dramatically over the course of an entire year but won’t change very much over the course of a single day. So, it is reasonable that the average concentration over the first day should be close to (but not exactly) 400 + 200 ppm.
(c) The average of N(t) over [0, 8760] is: 1 8760 ż 8760 0 350 + 200 cos t 4380π dt = 350 + 200 8760 4380 π sin t 4380π 8760 0 = 350 + 200 8760 4380 π sin 8760 4380π = 350 + 100 π sin (2π) = 350 Since the average of C(t) was 400, this gives us an absolute error of |400 ´ 350| = 50 ppm, for a relative error of 50 400 = 0.125, or 12.5%.
That is: sampling at the same time every day, rather than throughout the day, lead to an error of 12.5% in the yearly average concentration of carbon dioxide.
S-15: (a) The cross-section of S at x is a circle with radius x2, so area πx4. The average of these values, 0 ď x ď 2, is A = 1 2 ´ 0 ż 2 0 πx4 dx = 1 2 hπ 5 x5i2 0 = 16π 5 (b) To find the volume of S, imagine cutting it into thin circular disks of radius x2 and thickness dx. The volume of one such disk is πx4 dx, so the volume of S is ż 2 0 πx4 dx = hπ 5 x5i2 0 = 32π 5 (c) The volume of a cylinder is the product of its base area with its length. A cylinder with circular cross-sections of area 16π 5 and length 2 has volume 32π 5 .
Remark: this is the same as the volume of S, so the average cross-sectional area of S tells us the cross-sectional area of a cylinder with the same length and volume as S.
Compare this to Question 1, where we saw the average value of a function gave the height of a rectangle with the same area as the function over the given interval.
S-16: 519 (a) We can see without calculation that the average will be zero, since f (x) = x is an odd function and [´3, 3] is a symmetric interval. Alternately, we can use the definition of an average to calculate 1 6 ż 3 ´3 x dx = 1 12x2 3 ´3 = 1 12(9 ´ 9) = 0 (b) Using the definition provided for root mean square: RMS = d 1 6 ż 3 ´3 x2 dx = d 1 18x3 3 ´3 = c 27 18 ´ ´27 18 = ?
3 S-17: Using the definition provided, RMS = d 2 π ż π/4 ´π/4 tan2 x dx = d 2 π ż π/4 ´π/4 (sec2 x ´ 1) dx = d 2 π h tan x ´ x iπ/4 ´π/4 = c 2 π h 1 ´ π 4 ´ ´1 + π 4 i = c 2 π 2 ´ π 2 = c 4 π ´ 1 « 0.52 S-18: (a) Using Hooke’s law, when the spring is stretched (or compressed) f (t) metres past its natural length, the force exerted is k f (t), where k is the spring constant. In this case, the force is F(x) = (3 N/cm)( f (t) cm) = 3 sin (tπ) N (b) Our interval encompasses three full periods of sine, so the average will be zero.
Alternately, we can compute, using the definition of an average: Avg = 1 6 ż 6 0 3 sin(tπ) dt = 1 6 ´ 3 π cos(tπ) 6 0 = 1 2π [cos 0 ´ cos(6π)] = 0 This it doesn’t tell us very much about the “normal” amount of force from the spring during our time period. It only tells us that force in one direction at is “cancelled out” by force in the opposite direction at another time.
520 (c) Using the definition given for root mean square, RMS = d 1 6 ż 6 0 (3 sin(tπ))2 dt = d 3 2 ż 6 0 sin2(tπ) dt = d 3 4 ż 6 0 (1 ´ cos(2tπ)) dt = d 3 4 t ´ 1 2π sin(2tπ) 6 0 = d 3 4 6 ´ 1 2π sin(12π) ´ 0 = c 3 4(6) = 3 ?
2 « 2.12 S-19: (a) Let v(t) be the speed of the car at time t. Then, by the trapezoidal rule with a = 0, b = 2, ∆t = 1/3, the distance traveled is ż 2 0 v(t) dt « ∆t h 1 2v(0) + v(1/3) + v(2/3) + v(3/3) + v(4/3) + v(5/3) + 1 2v(2) i = 1 3 h 1 250 + 70 + 80 + 55 + 60 + 80 + 1 240 i = 130 km (b) The average speed is dist time « 130 km 2 hr = 65 km/hr.
S-20: (a) Using the definition of an average, A = 1 1 ´ 0 ż 1 0 et dt = e ´ 1 (b) Since s(t) ´ A = et ´ e + 1, its average on [0, 1] is 1 1 ´ 0 ż 1 0 et ´ e + 1 dt = et ´ et + t 1 0 = (e ´ e + 1) ´ (1) = 0 Remark: what’s happening here is that the average difference between s(t) and A is zero, because the values of s(t) that are larger than A (and give a positive value of s(t) ´ A) exactly cancel out the values of s(t) that are smaller than A (and give a negative value of s(t) ´ A). However, knowing how far the average value is from our calculated average is a reasonable thing to measure. That’s where (c) comes in.
(c) Using the definition of an average, the quantity we want is: 1 1 ´ 0 ż 1 0 ˇ ˇet ´ e + 1 ˇ ˇ dt 521 To deal with the absolute value, we consider the integral over two intervals: one where et ´ e + 1 is positive, and one where it’s negative. To decide where to break the limits of integration, notice et ´ e + 1 ą 0 exactly when et ą e ´ 1, so t ą log(e ´ 1).
1 1 ´ 0 ż 1 0 ˇ ˇet ´ e + 1 ˇ ˇ dt = ż log(e´1) 0 | et ´ e + 1 loooomoooon negative | dt + ż 1 log(e´1) | et ´ e + 1 loooomoooon positive | dt = ż log(e´1) 0 ´ et + e ´ 1 dt + ż 1 log(e´1) et ´ e + 1 dt = h ´ et + (e ´ 1)t ilog(e´1) 0 + h et ´ (e ´ 1)t i1 log(e´1) = [´(e ´ 1) + (e ´ 1) log(e ´ 1) + 1] + [e ´ (e ´ 1) ´ (e ´ 1) + (e ´ 1) log(e ´ 1)] = 4 ´ 2e + 2(e ´ 1) log(e ´ 1) « 0.42 Remark: what we just measured is how far s(t) is, on average, from A. We had to neglect whether s(t) was above or below A, because (as we saw in (b)) the values above A “cancel out” the values below A. That’s where the absolute value came in.
Knowing how well most of your function’s values match the average is an important measure, but dealing with absolute values can be a little clumsy. Therefore, the variance of a function squares the differences, rather than taking their absolute value.
(In our example, that means looking at (s(t) ´ A)2, rather than |s(t) ´ A|.) To compensate for the change in magnitude involve in squaring, the standard deviation is the square root of the variance. These are two very commonly used measures of how similar a function is to its average. Compare standard deviation to root-square-mean voltage from Example 2.2.6 in the CLP-2 text and Questions 16 to 18.
S-21: (a) Neither: the average of both these functions is zero. We saw this with a particular function in Question 20 (b), but it’s actually true in general. It’s a quick calculation to prove.
The average of f (x) ´ A is: 1 4 ´ 0 ż 4 0 f (x) ´ A dx = 1 4 ż 4 0 f (x) dx loooooomoooooon A ´A = A ´ A = 0 Similarly, the average of g(x) ´ A is: 1 4 ´ 0 ż 4 0 g(x) ´ A dx = 1 4 ż 4 0 g(x) dx loooooomoooooon A ´A = A ´ A = 0 522 (b) The function |f (x) ´ A| tells us how far f (x) is from A, without worrying whether f (x) is larger or smaller. Looking at our graph, for most values of x in [0, 4], f (x) is quite far away from A, so |f (x) ´ A| is usually a large, positive quantity.
By contrast, |g(x) ´ A| is a small positive quantity for most values of x. The function g(x) is quite close to A for all values of x in [0, 4].
So, since |g(x) ´ A| generally has much smaller values than |f (x) ´ A|, the average of |f (x) ´ A| on [0, 4] will be larger than the average of |g(x) ´ A| on [0, 4].
As discussed in Question 20(c), the average of |f (x) ´ A| is a measure of how closely f (x) resembles its average. We see from the graph that f (x) doesn’t resemble the constant function y = A much at all, while g(x) seems much more similar to the constant function y = A.
This kind of measure–how similar a function is to its average–is also the idea behind the root square mean.
S-22: When we rotate f (x) about the x-axis, we form a solid whose radius at x is |f (x)|.
So, its circular cross-sections have area π|f (x)|2 = π f 2(x). If we slice this solid into circular disks of thickness dx, then the disks have volume π f 2(x) dx. Therefore, the volume of the entire solid is ż b a π f 2(x) dx. All we need to do now is get this into a form where we can replace the integral with the root mean square, R.
V = ż b a π f 2(x) dx = π b ´ a b ´ a ż b a f 2(x) dx = π(b ´ a) d 1 b ´ a ż b a f 2(x) dx 2 = π(b ´ a)R2 Remark: the volume of a cylinder with length b ´ a and radius r is π(b ´ a)r2. So, the root mean square of f (x) gave us the radius of a cylinder with the same volume as the solid formed by rotating f (x). Recall the average of f (x) gave us the height of a rectangle with the same area as f (x). Compare this to the geometric interpretations of averages in Questions 1 and 15.
S-23: The question tells you 1 1´0 ş1 0 f (x) dx = f (0)+ f (1) 2 . Note f (0) = c, and 523 f (1) = a + b + c.
1 1 ´ 0 ż 1 0 f (x) dx = f (0) + f (1) 2 = c + (a + b + c) 2 ż 1 0 ax2 + bx + c dx = a + b + 2c 2 a 3x3 + b 2x2 + cx 1 0 = a 2 + b 2 + c a 3 + b 2 + c = a 2 + b 2 + c a 3 = a a = 0 That is, f (x) is linear.
S-24: The information given in the question is: as2 + bs + c + at2 + bt + c 2 = 1 t ´ s ż t s ax2 + bx + c dx = 1 t ´ s a 3x3 + b 2x2 + cx t s = 1 t ´ s a 3(t3 ´ s3) + b 2(t2 ´ s2) + c(t ´ s) = a 3(t2 + st + s2) + b 2(t + s) + c a 2(s2 + t2) + b 2(s + t) + c = a 3(t2 + st + s2) + b 2(t + s) + c a 2(s2 + t2) = a 3(t2 + st + s2) a s2 + t2 2 ´ t2 + st + s2 3 = 0 a s2 ´ 2st + t2 6 = 0 a(s ´ t)2 = 0 a = 0 OR s = t So, unless s = t (and we’re taking the very boring average of a single point!) then a = 0.
That is: f (x) is linear whenever s ‰ t.
S-25: The function g(x) = f (a + b ´ x), on the interval [a, b], is a mirror of the function f (x), with g(a) = f (b) and g(b) = f (a). So, şb a f (a + b ´ x) dx = şb a f (x) dx, and hence 1 b´a şb a f (a + b ´ x) dx = 1 b´a şb a f (x) dx, so the average value of f (a + b ´ x) on [a, b] is A.
524 Alternately, we can evaluate 1 b´a şb a f (a + b ´ x) dx directly, using the substitution u = a + b ´ x, dx = ´dx: 1 b ´ a ż b a f (a + b ´ x) dx = ´1 b ´ a ż u(b) u(a) f (u) du = ´1 b ´ a ż a b f (u) du = 1 b ´ a ż b a f (u) du = A S-26: (a) The function A(x) only gives us information about an integral when one limit of integration is zero. We can get around this by using properties of definite integrals from Section 1.2 to break our integral into two integrals, each of which has 0 as one limit of integration. So, we find the average of f (t) on [a, b] as follows: 1 b ´ a ż b a f (t) dt = 1 b ´ a ż 0 a f (t) dt + ż b 0 f (t) dt !
= 1 b ´ a ´ ż a 0 f (t) dt + ż b 0 f (t) dt !
= 1 b ´ a ´a ¨ 1 a ż a 0 f (t) dt loooooomoooooon A(a) +b ¨ 1 b ż b 0 f (t) dt loooooomoooooon A(b) = 1 b ´ a (´aA(a) + bA(b)) = bA(b) ´ aA(a) b ´ a (b) From the definition of A(x), we know A(x) = 1 x ż x 0 f (t) dt That is, xA(x) = ż x 0 f (t) dt To find f (x), we differentiate both sides. For the left side, we use the product rule; for the right side, we use the Fundamental Theorem of Calculus part 1.
A(x) + xA1(x) = f (x) So, f (t) = A(t) + tA1(t).
525 S-27: (a) One of many possible answers: f (x) = # ´1 if x ď 0 1 if x ą 0.
(b) No such function exists.
Note 1: Suppose f (x) ą 0 for all x in [´1, 1]. Then 1 2 ş1 ´1 f (x) dx ą 1 2 ş1 ´1 0 dx = 0.
That is, the average value of f (x) on the interval [´1, 1] is not zero–it’s something greater than zero.
Note 2: Suppose f (x) ă 0 for all x in [´1, 1]. Then 1 2 ş1 ´1 f (x) dx ă 1 2 ş1 ´1 0 dx = 0.
That is, the average value of f (x) on the interval [´1, 1] is not zero–it’s something less than zero.
So, if the average value of f (x) is zero, then f (x) ě 0 for some x in [´1, 1], and f (y) ď 0 for some y P [´1, 1]. Since f is a continuous function, and 0 is between f (x) and f (y), by the intermediate value theorem (Theorem 1.6.12 in the CLP-1 text) there is some value c between x and y such that f (c) = 0. Since x and y are both in [´1, 1], then c is as well. Therefore, no function exists as described in the question.
S-28: This seems like it might be true: if f is getting closer and closer to zero, as x grows towards infinity, then over time the later values will become a larger and larger portion of the total interval we’re looking at, and so the average should look more and more like f (x) when x is large–that is, like 0. That’s some intuition to start us out, but it isn’t a rigorous argument. To be sure we haven’t overlooked something, let’s use the definition of an average to express A(x).
A(x) = 1 x ż x 0 f (t) dt lim xÑ8 A(x) = lim xÑ8 1 x ż x 0 f (t) dt If ş8 0 f (t) dt converges, then this limit is 0, and the statement is true. So, suppose it does not converge. Since f (x) is positive, that means lim xÑ8 şx 0 f (t) d(t) = 8, so we can use l’Hˆ opital’s rule. To differentiate the numerator, we use the Fundamental Theorem of Calculus part 1.
lim xÑ8 A(x) = lim xÑ8 şx 0 f (t) dt x = lim xÑ8 f (t) 1 = 0 So, the statement is true whether şx 0 f (t) dt converges or not.
S-29: Note f (t) is a continuous function that takes only positive values, and lim tÑ8 f (t) = 0.
By the result of Question 28, lim xÑ8 A(x) = 0.
526 Solutions to Exercises 2.3 — Jump to TABLE OF CONTENTS S-1: Note ´x2 + 2x + 1 = 2 ´ (x ´ 1)2. So, both parabolas are symmetric about the line x = 1, and the x-coordinate of the centroid is x = 1.
x y y = (x ´ 1)2 y = ´x2 + 2x + 1 x = 1 The parabolas meet when: (x ´ 1)2 = 2 ´ (x ´ 1)2 2(x ´ 1)2 = 2 |x ´ 1| = 1 x = 0, x = 2 At both these points, y = 1, so we see the figure is symmetric about the line y = 1. Then the y-coordinate of the centroid is 1.
x y y = (x ´ 1)2 y = ´x2 + 2x + 1 y = 1 Therefore, the centroid is at (1, 1).
S-2: The circle and the cut-out rectangle are symmetric about the x-axis, and about the y-axis, so the centroid is the origin.
Remark: the centroid of a region doesn’t have to be a point in the region!
S-3: In general, this is false: weights farther out from the centre “count more” when we calculate the centre of mass. For instance, a rod with a 1-kg weight at x = ´10 and a 10-kg weight at x = 1 will balance at x = 0. There’s far more mass to one side of x = 0 than the other.
527 S-4: Following Equation 2.3.1 in the CLP-2 text, the centre of mass of the rod is at: ¯ x = ř (mass)ˆ(position) ř (mass) = 1 ˆ 1 + 2 ˆ 3 + 2 ˆ 4 + 1 ˆ 6 1 + 2 + 2 + 1 = 21 6 = 7 2 That is, the centre of mass is 3.5 metres from the left end.
S-5: (a) If we were to set this figure on a pencil lined up along the vertical line x = a, it seems pretty clear that it would fall to the left. So, the centre of mass is to the left of the line x = a. The same is true in (b): the added density on the left makes it only more lopsided. However, in (c), the right side is denser than the left, which could counterbalance the left. Without knowing more about the dimensions and the density, we can’t say where the centre of mass is in relation to the line x = a.
(d) Consider a section of the figure, consisting of all points (x, y) in the figure with b ď x ď c, and its “mirror” section on the other side of the line x = a. These two sections, which are drawn in red in the sketches below, will have the same area, at the same distance from x = a. Since we only care about the x-coordinate of the centre of mass, it doesn’t matter that the two halves are at different y-coordinates. The centre of mass falls along the line x = a.
A A B B a x y A A B B a x y (e) There is the same amount of area to the left and right of the line x = a, as in part (d).
However, the area to the right is “stretched out” more, so that it occupies space farther away from the line x = a. So, the centre of mass will be to the right of the line x = a.
S-6: • The volume of water in Tank A is 4 3π(1)3 = 4 3π cubic metres.
• The mass of water is 4000 3 π kg.
• By symmetry, the centre of mass of the water when it fills Tank A is exactly in the centre of the sphere, at height ¯ y1 = 4 metres above the ground (one metre above the bottom of Tank A, which is three metres above the ground).
• When the water is entirely in Tank B, its height is 2 3π metres. (The base of Tank B has area 2 m2, and the volume of water is 4 3π m3.) By symmetry, the centre of mass is exactly halfway up, at height ¯ y2 = 1 3π metres.
• So, the point mass in our model is moved from ¯ y1 = 4 to ¯ y2 = 1 3π, a distance of 4 ´ 1 3π metres, by gravity.
528 • The work involved is: 4000 3 π kg ˆ 4 ´ 1 3π m ˆ 9.8 m sec2 = 39200π 9 (12 ´ π) « 121, 212 J S-7: (a) A thin slice of S at position x has height 1 x, so if its width is dx, its area is 1 x dx.
(b) A small piece of R at position x has density 1 x, so if its length is dx, its mass is 1 x dx.
(c) Adding up all our tiny slices from (a) gives us the total area of S: ż 3 1 1 x dx = log 3 (d) Adding up all our tiny pieces from (b) gives us the total mass of R: ż 3 1 1 x dx = log 3 (e) Using Equation 2.3.3 in the CLP-2 text, the x-coordinate of the centroid of S is ş3 1 x ¨ 1 x dx ş3 1 1 x dx = ş3 1 1 dx log 3 = 2 log 3 (f) Using Equation 2.3.2 in the CLP-2 text, the centre of mass of R is ş3 1 x ¨ 1 x dx ş3 1 1 x dx = ş3 1 1 dx log 3 = 2 log 3 Remark: following the derivation of Equation 2.3.3 in the CLP-2 text, if we wanted to find the x-coordinate of the centroid of S, we would set up a rod that had exactly the characteristics of R. That’s why all the answers were repeated.
S-8: (a) If we chop R into n pieces, each piece has length b´a n . Then our ith cut is at position a + i b´a n , so our ith piece runs from a + (i ´ 1) b´a n to a + i b´a n . The approximation of the mass of this piece comes from the density at its midpoint, mi = h a + (i ´ 1) b´a n i + h a + i b´a n i 2 = a + (i ´ 1 2) b ´ a n 529 a b b´a n m1 m2 mn So, the ith piece has length b´a n , with approximate density ρ (mi) = ρ a + (i ´ 1 2) b´a n . We approximate that the ith piece has mass b´a n ¨ ρ (mi) and position mi. Using Equation 2.3.1 in the CLP-2 text, the centre of mass of R is approximately at position: ¯ xn = n ř i=1 (mass of ith piece) ˆ (position of ith piece) n ř i=1 (mass of ith piece) = n ř i=1 h b´a n ρ(mi) ˆ mi i n ř i=1 b´a n ρ(mi) = n ř i=1 h b´a n ρ a + i ´ 1 2 ( b´a n ) ˆ a + (i ´ 1 2) b´a n i n ř i=1 b´a n ρ a + (i ´ 1 2) b´a n (b) Remember the definition of a midpoint Riemann sum: ż b a f (x) dx « n ÿ i=1 b ´ a n ¨ f a + (i ´ 1 2) b ´ a n The numerator of our approximation in part (a) is, therefore, a midpoint Riemann sum of şb a ρ(x) ˆ x dx, and the denominator is a midpoint Riemann sum of şb a ρ(x) dx.
Using the definition of a definite integral (Definition 1.1.9 in the CLP-2 text), we see the limit of the approximation in (a) as x goes to infinity is ¯ x = şb a xρ(x) dx şb a ρ(x) dx This gives us the exact centre of mass of our rod.
Remark: this is Equation 2.3.2 in the CLP-2 text.
S-9: 530 (a) On the left-most corner of S, T(x) = B(x), so the height of S is zero; that is, the area of a very small vertical strip is very close to zero, so the density of R is close to 0. As we move closer to the position labeled a1, the height of the strips increases, so the areas of the strips increases, so the density of R increases. Then, between the points labeled a1 and b1, the height of S remains constant, since T(x) and B(x) are parallel here, so the areas of the strips of S remain constant, and the density of R remains constant. Then, between b1 and b, the height of S decreases, so the area of the strips decrease, so the density of R decreases.
x y T(x) B(x) a b a1 b1 R (b) At position x, the height of S is T(x) ´ B(x), so a rectangle with width dx and this height would have area (T(x) ´ B(x)) dx.
(c) According to our model, the tiny section of R at position x with width dx has mass (T(x) ´ B(x)) dx (that is, the area of S over this same tiny interval), so its density is ρ(x) = mass length = (T(x)´B(x)) dx dx = T(x) ´ B(x).
(d) Imagine S were a solid, of constant density. The mass of a portion of S is proportional to the area of that portion. To find the x-coordinate where the solid would balance, we imagine compressing together the vertical dimension of S until it’s a rod. That is, we would take a very thin vertical strip of S, and turn it into a small segment of a rod, with the same mass. Then the centre of mass of that rod would be exactly the x-coordinate of the centre of mass of the solid–that is, the x-coordinate of the centroid of S.
The compressed rod we form in this way is exactly R (perhaps multiplied by a constant, to account for the density of S, but this doesn’t affect where R balances). So, the x-coordinate of the centroid has the same position as the centre of mass of R.
Our result from Question 8(b) tells us the centre of mass of R is şb a xρ(x) dx şb a ρ(x) dx In (c), we found ρ(x) = T(x) ´ B(x). So, for the solid S bounded by T(x) and B(x) on the interval [a, b], ¯ x = şb a x(T(x) ´ B(x)) dx şb a(T(x) ´ B(x)) dx 531 Remark: the denominator is the area of S. This formula is the same as the formula found in Equation 2.3.3 of the CLP-2 text.
S-10: (a) To begin with, we’ll sketch some strips, and put a dot at the centre of mass of each one (its vertical centre).
x y T(x) B(x) a b a1 b1 In our model, each of these strips corresponds to a weight on R, positioned at its centre of mass (the height of the dot), and with a mass equal to the strip’s area. For the portion of S with a1 ď x ď b1, each centre of mass is at a slightly different height, but the areas of the slices are the same. So, the corresponding weights along R are at different heights, but all have the same mass, as shown below. (Note the rod R below only contains the weights from the middle of S–we’ll add the rest later.) For clarity, the diagrams below are zoomed in.
x y T(x) B(x) a b a1 b1 R By contrast to the slices in the interval [a1, b1], the slices of S along [a, a1] all have the same centre of mass, but different areas. So, there is one position along R that has a number of weights all stacked on top of one another, of varying masses.
The same situation applies to the slices of S along [b, b1]. So, all together, our rod looks something like this: 532 x y T(x) B(x) a b a1 b1 R Remark: if we had sketched the density of R, it would have looked something like this: x y T(x) B(x) a b a1 b1 R because from our sketch, we see that the density of R: • is 0 at either end, • is suddenly very high where the blue weights are, and • is constant and lower between the blue weights.
(b) At position x, the height of S is T(x) ´ B(x), and the width of the strip is dx, so the area of the strip is (T(x) ´ B(x)) dx.
Since the density of S is uniform, the centre of mass of the strip is halfway up: at T(x) + B(x) 2 .
(c) If we cut S into n strips, then the strip at position xi has area (T(xi) ´ B(xi))∆x, where ∆x = b´a n , and its centre of mass is at height T(xi) + B(xi) 2 . So, our 533 approximation of the centre of mass of the rod is: ¯ yn = n ř i=1 (Mi ˆ yi) n ř i=1 Mi = n ř i=1 ((T(xi) ´ B(xi))∆x) ˆ T(xi) + B(xi) 2 n ř i=1 (T(xi) ´ B(xi))∆x = n ř i=1 (T(xi)2 ´ B(xi)2)∆x 2 n ř i=1 (T(xi) ´ B(xi))∆x We use the definition of a definite integral (Definition 1.1.9 in the CLP-2 text) to re-write the limit of the above function.
¯ y = lim nÑ8 n ř i=1 (T(xi)2 ´ B(xi)2)∆x 2 n ř i=1 (T(xi) ´ B(xi))∆x = şb a T(x)2 ´ B(x)2 dx 2 şb a T(x) ´ B(x) dx Remark: the denominator is twice the area of S. This equation for the y-coordinate of the centroid is the same as the one given in Equation 2.3.3 in the CLP-2 text.
S-11: We use vertical strips, as in the sketch below. (To use horizontal strips we would have to split the domain of integration in two: ´3 ď y ď 0 and 0 ď y ď 3.) (−1, −3) (−1, 3) (0, 0) y = −3x y = 3x The equations of the top and bottom of the triangle are y = T(x) = ´3x and y = B(x) = 3x.
The area of the triangle is A = 1 2(6)(1) = 3. Now, we can apply the vertical-slice versions of Equation 2.3.3 in the CLP-2 text.
¯ x = 1 A ż 0 ´1 x T(x) ´ B(x) dx = 1 3 ż 0 ´1 x (´3x) ´ (3x) dx = ´1 3 ż 0 ´1 6x2 dx 534 S-12: Applying Equation 2.3.2 in the CLP-2 text, ¯ x = ş7 0 x ¨ x dx ş7 0 x dx = h 1 3x3i7 0 h 1 2x2 i7 0 = 1 3(73) 1 2(72) = 14 3 S-13: Applying Equation 2.3.2 in the CLP-2 text, ¯ x = ş10 ´3 x ¨ 1 1+x2 dx ş10 ´3 1 1+x2 dx For the numerator, we use the substitution u = 1 + x2, du = 2x dx.
= 1 2 ş101 10 1 u du h arctan x i10 ´3 = 1 2 h log u i101 10 arctan 10 ´ arctan(´3) = h log 101 ´ log 10 i 2(arctan 10 + arctan(3)) = log 10.1 2(arctan 10 + arctan(3)) « 0.43 Since arctangent is an odd function, arctan(´3) = ´ arctan(3); using logarithm rules, log 101 ´ log 10 = log 101 10 = log 10.1.
S-14: If we use horizontal strips, then we need to break the region into two pieces: y ě ´1 = ´e0, and y ď ´1. However, if we use vertical strips, the equation of the top of the region is y = T(x) = 1, and the equation of the bottom of the region is y = B(x) = ´ex, for all x from a = 0 to b = 1. So, we use vertical strips.
x y 1 y = ´ex Using Equation 2.3.3 in the CLP-2 text, the y-coordinate of the centre of mass is ¯ y = 1 2A ż 1 0 T(x)2 ´ B(x)2 dx = 1 2e ż 1 0 1 ´ e2x dx = 1 2e x ´ 1 2e2x 1 0 = 1 2e h 1 ´ e2 2 ´ 0 + 1 2 i = 3 4e ´ e 4 535 S-15: (a) The lines y = 0, x = 0, and x = 2 are easy enough to sketch. Let’s get some basic information about y = T(x) = 1 ?
16´x2 on the interval [0, 2].
• For all x in its domain, T(x) ě 0. In particular, it’s always the top of our region (so T(x) is a reasonable name for it), while the bottom is B(x) = 0.
• T(0) = 1 4, and T(2) = 1 2 ?
3 • T1(x) = x (16´x2)3/2, which is positive on [0, 2], so T(x) is increasing.
Remark: to see that T(x) is increasing, we can also just break it into pieces: – When x ě 0, x2 is increasing, so – 16 ´ x2 is decreasing, so – ?
16 ´ x2 is decreasing, so – 1 ?
16´x2 = T(x) is increasing.
• T2(x) = 2x2+16 (16´x2)5/2, which is positive, so T(x) is concave up.
x = 2 y = 1 √ 16−x2 x y Remark: If we only wanted to solve (b), it would still be nice to have a sketch of the region, but it wouldn’t need to be so detailed. Knowing that T(x) is always greater than 0 would be enough to tell us we could use vertical slices with T(x) as the top and y = 0 as the bottom.
If we wanted to use horizontal slices (we don’t... but we could!) we would additionally want to know that T(x) is increasing over [0, 2], T(0) = 1 4, and T(2) = 1 2 ?
3. This would tell us that: • the right endpoint of a horizontal strip is always x = 2, • the left endpoint is determined by T(x) from y = 1 4 to y = 1 2 ?
3, and • the left endpoint is x = 0 for 0 ď y ď 1 4.
(b) 536 x = 2 y = 1 √ 16−x2 x y The part of the region with x coordinate between x and x + dx is a strip of width dx running from y = 0 to y = 1 ?
16´x2. It is illustrated in red in the figure above. So, the area of the region is A = ż 2 0 1 ?
16 ´ x2 dx = ż arcsin(1/2) 0 1 4 cos t4 cos t dt = arcsin 1 2 = π 6 where we made the substitution x = 4 sin t, dx = 4 cos tdt, ?
16 ´ x2 = 4 cos t.
Using Equation 2.3.3 in the CLP-2 text, ¯ y = ş2 0 T(x)2 ´ B(x)2 dx 2A = ż 2 0 " 1 ?
16 ´ x2 2 ´ 02 # dx 2A = 1 2A ż 2 0 1 16 ´ x2 dx = 1 2A ż 2 0 1 (4 ´ x)(4 + x) dx Using the method of partial fractions, we see 1 16 ´ x2 = 1/8 4 + x + 1/8 4 ´ x.
= 1 2A ż 2 0 h 1/8 4 + x + 1/8 4 ´ x i dx = 1 16A ż 2 0 h 1 x + 4 ´ 1 x ´ 4 i dx = 1 16A h log |x + 4| ´ log |x ´ 4| i2 0 = 6 16π log 6 ´ log 2 ´ log 4 + log 4 = 3 log 3 8π S-16: x y π 4 y = sin x y = cos x 537 The top of the region is y = T(x) = cos(x) and the bottom of the region is y = B(x) = sin(x). So, the area of the region is A = ż π/4 0 T(x) ´ B(x) dx = ż π/4 0 cos(x) ´ sin(x) dx = h sin(x) + cos(x) iπ/4 0 = 1 ?
2 + 1 ?
2 ´ [0 + 1] = ?
2 ´ 1 If we use horizontal slices, we’ll need to break up the object into two regions, so let’s use vertical slices. Using Equation 2.3.3 in the CLP-2 text, the region has centroid ( ¯ x, ¯ y) with: ¯ x = 1 A ż π/4 0 x T(x) ´ B(x) dx = 1 A ż π/4 0 x cos(x) ´ sin(x) dx We use integration by parts with u = x, dv = (cos x ´ sin x)dx; du = dx, v = sin x + cos x.
= 1 A h x(sin x + cos x) iπ/4 0 ´ ż π/4 0 (sin x + cos x) dx !
= 1 A h x sin(x) + x cos(x) + cos x ´ sin x iπ/4 0 = 1 A π 4 ¨ 1 ?
2 + π 4 ¨ 1 ?
2 + 1 ?
2 ´ 1 ?
2 ´ 1 = π 4 ?
2 ´ 1 A = π 4 ?
2 ´ 1 ?
2 ´ 1 Again using Equation 2.3.3 in the CLP-2 text, ¯ y = 1 2A ż π/4 0 T(x)2 ´ B(x)2 dx = 1 2A ż π/4 0 cos2(x) ´ sin2(x) dx = 1 2A ż π/4 0 cos(2x) dx = 1 2A h1 2 sin(2x) iπ/4 0 = 1 4( ?
2 ´ 1) S-17: (a) Since k is positive, k ?
1+x2 ą 0 for every x. Then the top of our region is defined by T(x) = k ?
1+x2, and the bottom is defined by B(x) = 0.
If we make vertical slices, we don’t have to turn our region into two parts, so let’s use vertical slices. The question asks for our final answer in terms of the area A of the region, so we don’t need to find A explicitly. Using Equation 2.3.3 in the CLP-2 text, the x–coordinate of the centroid is ¯ x = 1 A ż 1 0 x(T(x) ´ B(x)) dx = 1 A ż 1 0 x k ?
1 + x2 dx 538 Although we have a quadratic function underneath a square root, we find an easier method than a trig substitution: the substitution u = 1 + x2, du = 2x dx. This changes the limits of integration to 1 + 02 = 1 and 1 + 12 = 2, respectively.
= 1 A ż 2 1 k ?u du 2 = k 2A ?u 1/2 2 1 = k A ?
2 ´ 1 Again using Equation 2.3.3 in the CLP-2 text, the y–coordinate of the centroid is ¯ y = 1 2A ż 1 0 (T(x)2 ´ B(x)2) dx = 1 2A ż 1 0 k2 1 + x2 dx = k2 2A ż 1 0 1 1 + x2 dx = k2 2A h arctan 1 ´ arctan 0 i = k2 2A ¨ π 4 = k2π 8A (b) We have ¯ x = ¯ y if and only if k A ?
2 ´ 1 = k2π 8A Since k and A are a positive constants (hence neither is equal to 0), we can divide both sides by k and multiply both sides by A: ?
2 ´ 1 = kπ 8 k = 8 π ?
2 ´ 1 S-18: (a) The curve y = x2 ´ 3x is a parabola, pointing up, with x-intercepts at x = 0 and x = 3.
The curve y = x ´ x2 is a parabola, pointing down, with x-intercepts at x = 0 and x = 1.
To find where the two curves meet, we set them equal to each other: x2 ´ 3x = x ´ x2 2x2 ´ 4x = 0 2x(x ´ 2) = 0 x = 0 and x = 2 This is enough information to sketch the figure, on the left below.
y=x−x2 y=x2−3x x y (2, −2) y=x−x2 y=x2−3x x y (2, −2) 539 (b) As we found in (a), the curves cross when x = 0, x = 2. The corresponding values of y are y = 0 and y = 2 ´ 22 = ´2. Note the top curve is T(x) = x ´ x2, and the bottom curve is B(x) = x2 ´ 3x. Using vertical strips, as in the figure on the right above, the area of R is ż 2 0 (x ´ x2) ´ (x2 ´ 3x) dx = ż 2 0 4x ´ 2x2 dx = 2x2 ´ 2 3x3 2 0 = 8 ´ 16 3 = 8 3 (c) Using Equation 2.3.3 in the CLP-2 text, the x–coordinate of the centroid of R (i.e. the weighted average of x over R) is ¯ x = 3 8 ż 2 0 x (x´x2) ´ (x2´3x) dx = 3 8 ż 2 0 4x2 ´ 2x3 dx = 3 8 h4 3x3 ´ 1 2x4i2 0 = 3 8 h32 3 ´ 8 i = 1 S-19: Using Equation 2.3.3 in the CLP-2 text, the x–coordinate of the centroid is ¯ x = ş1 0 x 1 1+x2 dx ş1 0 1 1+x2 dx We can guess the antiderivative in the numerator, or use the substitution u = 1 + x2, du = 2x dx.
= 1 2 log(1 + x2) ˇ ˇ1 0 arctan x ˇ ˇ1 0 = 1 2 log 2 π/4 = 2 π log 2 « 0.44127 S-20: By symmetry, the centroid lies on the y–axis, so ¯ x = 0.
The area of the figure is the area of a half-circle of radius 3, and a rectangle of width 6 and height 2. So, A = 1 2π(9) + 6 ˆ 2 = 9 2π + 12.
We’ll use vertical strips as in the sketch below.
y x −2 −1 1 2 3 3 2 1 −1 −2 −3 y = √ 32 −x2 y = −2 540 The top function of our figure is T(x) = ?
9 ´ x2, and the bottom function of our figure is B(x) = ´2. Using Equation 2.3.3 in the CLP-2 text, the y–coordinate of the centroid is: ¯ y = 1 2A ż b a T(x)2 ´ B(x)2 dx = 1 2A ż 3 ´3 a 9 ´ x22 ´ (´2)2 dx = 1 2A ż 3 ´3 5 ´ x2 dx = 1 2A 5x ´ 1 3x3 3 ´3 = 1 2A [15 ´ 9 + 15 ´ 9] = 6 A = 6 9 2π + 12 = 12 9π + 24 S-21: (a) Notice that when x = 0, y = 3 and as x2 increases, y decreases until y hits zero at x2 = 9 4, i.e. at x = ˘3 2. For x2 ą 9 4, y is not even defined. So, on D, x runs from ´3 2 to + 3 2 and, for each x, y runs from 0 to ?
9 ´ 4x2. Here is a sketch of D.
y = √ 9 −4x2 (−3/2, 0) (3/2, 0) x y As an aside, we can rewrite y = ?
9 ´ 4x2 as 4x2 + y2 = 9, y ě 0, which is the top half of the ellipse which passes through (˘a, 0) and (0, ˘b) with a = 3 2 and b = 3. The area of the full ellipse is πab = 9 2π. The area of D is half of that, which is 9 4π. But we are told to use an integral, so we will do so.
The area is Area = ż 3/2 ´3/2 a 9 ´ 4x2 dx We can evaluate this integral by substituting x = 3 2 sin θ, dx = 3 2 cos θ dθ and using x = ˘3 2 ð ñ sin θ = ˘1 541 So ´π 2 ď θ ď π 2 and Area = ż π/2 ´π/2 b 9 ´ 4 3 2 sin θ 2 3 2 cos θ dθ = ż π/2 ´π/2 a 9 ´ 9 sin2 θ 3 2 cos θ dθ = 9 2 ż π/2 ´π/2 cos2 θ dθ = 9 2 ż π/2 ´π/2 cos(2θ) + 1 2 dθ = 9 4 hsin(2θ) 2 + θ iπ/2 ´π/2 = 9 4π (b) The region D is symmetric about the y axis. So the centre of mass lies on the y axis.
That is, ¯ x = 0. Since D has area A = 9 4π, top equation y = T(x) = ?
9 ´ 4x2 and bottom equation y = B(x) = 0, with x running from a = ´3 2 to b = 3 2, Equation 2.3.3 in the CLP-2 text gives us ¯ y: ¯ y = 1 2A ż b a T(x)2 ´ B(x)2 dx = 2 9π ż 3/2 ´3/2 9 ´ 4x2 dx = 4 9π ż 3/2 0 9 ´ 4x2 dx = 4 9π h 9x ´ 4 3x3i3/2 0 = 4 9π h 9 ¨ 3 2 ´ 4 3 ¨ 33 23 i = 4 9π h 9 ¨ 3 2 ´ 9 ¨ 1 2 i = 4 π S-22: Let’s start by sketching the region at hand. We know the general shape of arcsine (it’s like half a period of sine, if you swapped the x and y axes); we can sketch the curve y = arcsin(2 ´ x) by mirroring y = arcsin x about the line x = 1.
x y ´1 1 2 3 ´π 2 π 2 y = arcsin x y = arcsin(2 ´ x) If we use vertical strips, then we need two separate regions, because T(x) = arcsin x when x ď 1, and T(x) = arcsin(2 ´ x) when x ą 1. Also, we’d have to antidifferentiate functions that have arcsine in them. Let’s think about horizontal strips. If y = arcsin x, then x = sin y, and if y = arcsin(2 ´ x) then x = 2 ´ sin y. For all y from ´π 2 to π 2 , the left endpoint of a strip is given by L(y) = sin y, and the right endpoint is given by R(y) = 2 ´ sin y.
First, let’s use our horizontal slices7 to find the area of our region, A.
A = ż π/2 ´π/2 (2 ´ sin y) ´ (sin y) dy = ż π/2 ´π/2 2 ´ 2 sin y dy = h 2y + 2 cos y iπ/2 ´π/2 = (π + 0) ´ (´π + 0) = 2π 7 There’s also a sneaky way to find the area of A: look for a way to snip and rearrange bits of the figure to turn it into a rectangle!
542 From symmetry, it is clear that ¯ x = 1. We find ¯ y using Equation 2.3.3 in the CLP-2 text.
¯ y = şπ/2 ´π/2 y [R(y) ´ L(y)] dy A = şπ/2 ´π/2 y [(2 ´ sin y) ´ (sin y)] dy 2π = 1 2π ż π/2 ´π/2 y(2 ´ 2 sin y) dy = 1 π ż π/2 ´π/2 y dy ´ 1 π ż π/2 ´π/2 (y sin y) dy Since y is an odd function, and the domain of integration is symmetric, the first integral evaluates to 0. Since y sin y is an even function (recall the product of two odd functions is an even function), we can simplify our limits of integration.
= ´ 2 π ż π/2 0 y sin y dy We use integration by parts with u = y, dv = sin y dy; du = dy, v = ´ cos y.
= ´ 2 π ´ y cos y π/2 0 + ż π/2 0 cos y dy !
= ´ 2 π ´ y cos y + sin y π/2 0 = ´ 2 π [(0 + 1) ´ 0] = ´ 2 π S-23: We’ll start by sketching the region.
543 x y 2 1 1 3 y = 3(x ´ 1) y = ex y = ex If we use horizontal slices, we need to divide our figure into three regions, as in the figure below, because the left and right functions change at the dashed lines.
x y 2 1 1 3 y=3(x´1) x=1+ y 3 y=ex x=log y L(x) = 0 L(y) = log y R(y) = 1 + y 3 R(y) = 2 If we use vertical slices, we only need two regions (shown below) to account for the different top and bottom functions. This seems easier than three regions, so we use vertical slices.
544 x y 2 1 1 3 y = 3(x ´ 1) y = ex y = ex T(x) = ex B(x) = 3(x ´ 1) B(x) = 0 When 0 ď x ď 2, T(x) = ex. When 0 ď x ď 1, B(x) = 0, and when 1 ď x ď 2, B(x) = 3(x ´ 1).
The area of the figure is: A = ż 2 0 (T(x) ´ B(x)) dx = ż 1 0 (ex ´ 0) dx + ż 2 1 (ex ´ 3(x ´ 1)) dx = ż 2 0 ex dx ´ ż 2 1 3(x ´ 1) dx = h exi2 0 ´ 3 2(x ´ 1)2 2 1 = e2 ´ 1 ´ 3 2 = e2 ´ 5 2 Using Equation 2.3.3 in the CLP-2 text: ¯ x = ş2 0 x(T(x) ´ B(x)) dx A = 1 e2 ´ 5/2 "ż 1 0 x (ex ´ 0) dx + ż 2 1 x (ex ´ 3(x ´ 1)) dx # = 1 e2 ´ 5/2 "ż 2 0 xex dx ´ ż 2 1 3x(x ´ 1) dx # 545 For the left integral, we use integration by parts with u = x, dv = ex dx; du = dx, v = ex.
= 1 e2 ´ 5/2 " [xex]2 0 ´ ż 2 0 ex dx ´ 3 ż 2 1 (x2 ´ x) dx # = 1 e2 ´ 5/2 [xex ´ ex]2 0 ´ 3 1 3x3 ´ 1 2x2 2 1 !
= 1 e2 ´ 5/2 (2e2 ´ e2) ´ (´1) ´ 3 8 3 ´ 2 ´ 1 3 + 1 2 = e2 ´ 3/2 e2 ´ 5/2 « 1.2 Using Equation 2.3.3 in the CLP-2 text again: ¯ y = ş2 0 T(x)2 ´ B(x)2 dx 2A = 1 2(e2 ´ 5/2) "ż 1 0 e2x ´ 0 dx + ż 2 1 e2x ´ 9(x ´ 1)2 dx # = 1 2(e2 ´ 5/2) "ż 2 0 e2x dx ´ ż 2 1 9(x ´ 1)2 dx # = 1 2(e2 ´ 5/2) 1 2e2x 2 0 ´ h 3(x ´ 1)3i2 1 !
= 1 2e2 ´ 5 1 2e4 ´ 1 2 ´ 3 = e4 ´ 7 4e2 ´ 10 « 2.4 S-24: The area of the region is A = ż 8 1 8 x3 dx = lim tÑ8 ż t 1 8 x3 dx = lim tÑ8 ´ 4 x2 t 1 = lim tÑ8 ´ 4 t2 + 4 12 = 0 + 4 We’ll now compute ¯ y twice, once with vertical strips, as in the figure in the left below, and once with horizontal strips as in the figure on the right below.
(1, 8) y = 8 x3 x y 1 (1, 8) x = 2 y1/3 x y 1 546 Vertical strips: The equation of the top of the region is y = T(x) = 8 x3 and the equation of the bottom of the region is y = B(x) = 0. Using vertical strips, as in the figure on the left above, the y-coordinate of the centre of mass is ¯ y = 1 2A ż 8 1 T(x)2 ´ B(x)2 dx = 1 8 ż 8 1 8 x3 2 dx = lim tÑ8 ż t 1 8 x6 dx = lim tÑ8 ´ 8 5x5 t 1 = lim tÑ8 ´ 8 5t5 + 8 5 ˆ 15 = 8 5 Vertical strips: Since y = 8 x3 is equivalent to x = 3 d 8 y, the equation of the right-hand side of the region is x = R(y) = 2 y1/3 and the equation of the left hand side of the region is x = L(y) = 1. The point at the top of the region is (1, 8). Thus y runs from 0 to 8. So, using horizontal strips, as in the figure on the right above, the y-coordinate of the centre of mass is ¯ y = 1 A ż 8 0 y R(y) ´ L(y) dy = 1 4 ż 8 0 y 2y´1/3 ´ 1 dy = 1 4 ż 8 0 2y2/3 ´ y dy = 1 4 6 5y5/3 ´ y2 2 8 0 = 1 4 6 ˆ 32 5 ´ 8 ˆ 8 2 = 8 6 5 ´ 1 = 8 5 S-25: (a) The two curves cross at points (x, y) that satisfy both y = x2 and y = 6 ´ x, and hence x2 = 6 ´ x ð ñ x2 + x ´ 6 = 0 ð ñ (x + 3)(x ´ 2) = 0 So we see that the two curves intersect at x = 2 (as well as x = ´3, which is to the left of the y-axis and therefore irrelevant). Here is a sketch of A.
547 (2, 4) y = x2 y = 6 −x x y A The top of A has equation y = T(x) = 6 ´ x, the bottom has equation y = B(x) = x2 and x runs from 0 to 2. So, using vertical strips, ¯ x = 1 A ż 2 0 x T(x) ´ B(x) dx = 1 22/3 ż 2 0 x (6 ´ x) ´ x2 dx = 3 22 ż 2 0 (6x ´ x2 ´ x3) dx = 3 22 3x2 ´ x3 3 ´ x4 4 2 0 = 3 22 h 12 ´ 8 3 ´ 4 i = 3 22 16 3 = 8 11 and ¯ y = 1 2A ż 2 0 T(x)2 ´ B(x)2 dx = 1 2 ¨ 1 22/3 ż 2 0 (6 ´ x)2 ´ x4 dx = 3 44 ´(6 ´ x)3 3 ´ x5 5 2 0 = 3 44 ´ 64 ´ 216 3 ´ 32 5 = 3 44 ¨ 664 15 = 166 55 The integral was evaluated by guessing an antiderivative for the integrand. It could also be evaluated as 3 44 ż 2 0 36 ´ 12x + x2 ´ x4 dx = 3 44 36x ´ 6x2 + x3 3 ´ x5 5 2 0 = 3 44 72 ´ 24 + 8 3 ´ 32 5 = 3 44 664 15 = 166 55 (b) The question specifies the use of horizontal slices (as in Example 1.6.5 of the CLP-2 text). The radius of the slice at height y is the x-value of the right-hand boundary of the region at that point. So, we start by converting both equations y = 6 ´ x and y = x2 into equations of the form x = f (y). To do so we solve for x in both equations, yielding x = ?y and x = 6 ´ y.
548 (2, 4) (0, 6) x = √y x = 6 −y x y • We use thin horizontal strips of width dy as in the figure above.
• When we rotate about the y–axis, each strip sweeps out a thin disk – whose radius is r = 6 ´ y when 4 ď y ď 6 (see the blue strip in the figure above), and whose radius is r = ?y when 0 ď y ď 4 (see the red strip in the figure above) and – whose thickness is dy and hence – whose volume is πr2 dy = π(6 ´ y)2 dy when 4 ď y ď 6 and whose volume is πr2 dy = πy dy when 0 ď y ď 4.
• As our bottommost strip is at y = 0 and our topmost strip is at y = 6, the total volume is π ż 4 0 y dy + π ż 6 4 (6 ´ y)2 dy S-26: (a) Here is a sketch of the specified region, which we shall call R.
(0, −1) (1, −1) y = ex x y R y = −1 The top of R has equation y = T(x) = ex, the bottom has equation y = B(x) = ´1 and x runs from 0 to 1. So, using vertical strips, we see that R has area A = ż 1 0 T(x) ´ B(x) dx = ż 1 0 ex ´ (´1) dx = ż 1 0 ex + 1 dx = ex + x 1 0 = e 549 and ¯ y = 1 2A ż 1 0 T(x)2 ´ B(x)2 dx = 1 2e ż 1 0 e2x ´ 1 dx = 1 2e e2x 2 ´ x 1 0 = 1 2e e2 2 ´ 1 ´ 1 2 = e 4 ´ 3 4e (b) To compute the volume when R is rotated about the line y = ´1 • we use thin vertical strips of width dx as in the figure above.
• When we rotate about the line y = ´1, each strip sweeps out a thin disk – whose radius is r = T(x) ´ B(x) = ex + 1 and – whose thickness is dx and hence – whose volume is πr2 dx = π(ex + 1)2 dx.
• As our leftmost strip is at x = 0 and our rightmost strip is at x = 1, the total volume is π ż 1 0 (ex + 1)2 dx = π ż 1 0 (e2x + 2ex + 1) dx = π e2x 2 + 2ex + x 1 0 = π e2 2 + 2e + 1 ´ 1 2 + 2 + 0 = π e2 2 + 2e ´ 3 2 S-27: By symmetry, ¯ y = 1.5. We can’t immediately use Equation 2.3.3 in the CLP-2 text to find ¯ x, because the density is not constant. Instead, we’ll go through the derivation of Equation 2.3.3, to figure out what to do with a non-constant density. (This is a good time to review Questions 9 and 10 in this section.) Our model is that we’re making a rod R that reaches from x = 0 to x = 4, and the mass of the section of the rod along [a, b] is equal to the mass of the strip of our rectangle along [a, b]. If we have a formula ρ(x) for the density of R, we can find the centre of mass of R, which is also the x-coordinate of the centre of mass of the rectangle.
A thin vertical strip of the rectangle with length dx at position x has area 3dx m2 and density x2 kg/m2, so it has mass 3x2 dx kg. Therefore, a short section of R at position x with length dx ought to have mass 3x2 dx kg as well. Then its density at x is ρ(x) = 3x2 dx kg dx m = 3x2 kg/m.
Now, we can use Equation 2.3.2 in the CLP-2 text to find the centre of mass of the rod, 550 which is also the x-coordinate of the centre of mass of our rectangle: ¯ x = ş4 0 xρ(x) dx ş4 0 ρ(x) dx = ş4 0 3x3 dx ş4 0 3x2 dx = 3 4x44 0 x34 0 = 3 ¨ 43 43 = 3 The centre of mass of our rectangle is (3, 1.5).
S-28: By symmetry, the x-coordinate of the centre of mass will be ¯ x = 0; that is, exactly in the middle, horizontally. To find the y-coordinate of the centre of mass, we need to consider the origin of Equation 2.3.3 in the CLP-2 text.
We can make vertical strips or horizontal strips. A vertical strip of the circle has a density that varies from the bottom of the strip to the top, but a horizontal strip has a constant density (assuming the strip is very thin). So it seems that horizontal strips in this case will be the easier route.
Following the derivation of Equation 2.3.3 in the CLP-2 text, we model our circle as a vertical rod R, filling the y-interval [0, 6]. A portion of the rod with a ď y ď b should have the same mass as the portion of the circle with a ď y ď b. To achieve this, we slice the circle into thin horizontal strips of thickness dy, calculate their mass, then use that to find ρ(y), the density of R.
First, let’s find a formula for the mass of a thin horizontal strip of the circle at position y with height dy.
dy y x y x The circle with radius 3 centred at (0, 3) has equation x2 + (y ´ 3)2 = 9. So, the right half of the circle has equation x = a 9 ´ (y ´ 3)2, and the left half of the circle has equation x = ´ a 9 ´ (y ´ 3)2. So, the width of a strip at height y is 2 a 9 ´ (y ´ 3)2 m. Its height is dy m, so its area is 2 a 9 ´ (y ´ 3)2 dy m2. Its density is 2 + y kg m2 , so its mass is 2(2 + y) a 9 ´ (y ´ 3)2 dy kg.
Now we can find ρ(y), the density of R at position y. The mass of the section of R at position y with length dy is 2(2 + y) a 9 ´ (y ´ 3)2 dy kg (the mass of the strip in the paragraph above), so its density is 2(2+y)?
9´(y´3)2 dy kg dy m = 2(2 + y) a 9 ´ (y ´ 3)2 kg m = ρ(y).
Now, Equation 2.3.2 in the CLP-2 text will tell us the centre of mass of R, which is also the y-coordinate of the centre of mass of the circle.
551 ¯ y = şb a yρ(y) dy şb a ρ(y) dy = ş6 0 y ˆ 2(2 + y) a 9 ´ (y ´ 3)2 dy ş6 0 2(2 + y) a 9 ´ (y ´ 3)2 dy = ş6 0 y(2 + y) a 9 ´ (y ´ 3)2 dy ş6 0(2 + y) a 9 ´ (y ´ 3)2 dy To make things look a little cleaner, we use the substitution u = y ´ 3, du = dy. Then the limits of integration become ´3 and 3, respectively, and y = u + 3. (Geometrically, we’re re-centring the circle at the origin, instead of at the point (0,3).) = ş3 ´3(u + 3)(2 + u + 3) ?
9 ´ u2 du ş3 ´3 2 + u + 3 ?
9 ´ u2 du = ş3 ´3 u2 + 8u + 15 ?
9 ´ u2 du ş3 ´3 u + 5 ?
9 ´ u2 du = N D (˚) Let’s start by finding D, the integral of the denominator. If we break it into two pieces, we can use symmetry and geometry to evaluate it.
D = ż 3 ´3 u a 9 ´ u2 du + 5 ż 3 ´3 a 9 ´ u2 du The left integrand is odd, so its integral over a symmetric interval is 0. (You can also evaluate this using the substitution w = 9 ´ u2, dw = ´2u du.) The right integral represents the area underneath half a circle of radius 3, centred at the origin.
D = 0 + 5 ¨ 1 2π ¨ 32 = 45 2 π Now, let’s evaluate our numerator integral from (˚), N = ş3 ´3 u2 + 8u + 15 ?
9 ´ u2 du. If we break it into three pieces, we can simplify the integration somewhat.
N = ż 3 ´3 u2a 9 ´ u2 du + 8 ż 3 ´3 u a 9 ´ u2 du + 15 ż 3 ´3 a 9 ´ u2 du The first integrand is even, with a symmetric interval of integration, so we can simplify its limits of integration a little bit. The middle integrand is odd, so its integral over the symmetric interval [´3, 3] is zero. The last integral is the area of half a circle of radius 3.
N = 2 ż 3 0 u2a 9 ´ u2 du + 0 + 15 ¨ π ¨ 32 = 135 2 π + 2 ż 3 0 u2a 9 ´ u2 du 552 The remaining integral has a quadratic function underneath a square root with no obvious substitution, so we use a trigonometric substitution. Let u = 3 sin θ, du = 3 cos θ dθ. Note 3 sin(0) = 0 and 3 sin(π/2) = 3, so the limits of integration become 0 and π 2 .
N = 135 2 π + 2 ż π/2 0 3 sin θ 2b 9 ´ 3 sin θ 2 ¨ 3 cos θ dθ = 135 2 π + 2 ż π/2 0 9 sin2 θ ¨ a 9 ´ 9 sin2 θ ¨ 3 cos θ dθ = 135 2 π + 54 ż π/2 0 sin2 θ ¨ a 9 cos2 θ ¨ cos θ dθ = 135 2 π + 54 ż π/2 0 sin2 θ ¨ 3 cos θ ¨ cos θ dθ = 135 2 π + 162 ż π/2 0 sin2 θ ¨ cos2 θ dθ Using the identity sin(2θ) = 2 sin θ cos θ, we see sin2 θ cos2 θ = sin θ cos θ 2 = 1 4 sin2(2θ) N = 135 2 π + 162 ż π/2 0 1 4 sin2(2θ) dθ Now, we use the identity sin2 x = 1 2(1 ´ cos(2x)), with x = 2θ.
N = 135 2 π + 162 ż π/2 0 1 8 1 ´ cos(4θ) dθ = 135 2 π + 81 4 ż π/2 0 1 ´ cos(4θ) dθ = 135 2 π + 81 4 θ ´ 1 4 sin(4θ) π/2 0 = 135 2 π + 81 4 π 2 = 621 8 π Now, using equation (˚), we find ¯ y: ¯ y = N D = 621 8 π 45 2 π = 69 20 = 3.45 Let’s quickly check that this makes sense: if the circle has uniform density, its centre of mass would lie at (0, 3). Since it’s denser at the top, the centre of mass should be higher, and indeed 3.45 is higher than 3 (without being so high it’s above the entire circle).
S-29: 553 (a) To find the centre of mass of the rod R, we need to know its density at height y, ρ(y).
Since the mass of a section of R is the same as the volume of a section of the cone, let’s find the volume of a thin horizontal slice of the cone at height y, with thickness dy. To find its radius s, we use similar triangles. The diagram below represents a vertical cross-section of the cone.
x y h y r s Since r h = s h´y, the radius of our slice at height y is s = r h(h ´ y). Then the volume of the slice is πs2dy = π r h(h ´ y) 2 dy. Correspondingly, the mass of the piece of the rod at position y with length dy is π r h(h ´ y) 2 dy, so its density is ρ(y) = π r h(h ´ y) 2 dy dy = π r h(h ´ y) 2 .
Now, we can find the centre of mass of R: ¯ y = şh 0 yρ(y) dy şh 0 ρ(y) dy = şh 0 yπ r h(h ´ y) 2 dy şh 0 π r h(h ´ y) 2 dy = r2 h2 π şh 0 y (h ´ y)2 dy r2 h2 π şh 0 (h ´ y)2 dy = şh 0 h2y ´ 2hy2 + y3 dy şh 0 (h2 ´ 2hy + y2) dy = h h2 2 y2 ´ 2h 3 y3 + 1 4y4ih 0 h h2y ´ hy2 + 1 3y3 ih 0 = h4 2 ´ 2h4 3 + h4 4 h3 ´ h3 + 1 3h3 = h4 h3 ¨ 1 2 ´ 2 3 + 1 4 1 3 = h 4 554 So, the centre of mass of the cone occurs h 4 metres above its base.
Remark: it is quite interesting that the centre of mass does not depend on the radius of the cone!
(b) To find the centre of mass of a truncated cone, we simply consider a truncated rod. If the top h ´ k metres are missing, then the height of the cone (and also the rod) is k.
Then the centre of mass has height: ¯ y = şk 0 yρ(y) dy şh 0 ρ(y) dy = şk 0 yπ r h(h ´ y) 2 dy şh 0 π r h(h ´ y) 2 dy = r2 h2 π şk 0 y (h ´ y)2 dy r2 h2 π şk 0 (h ´ y)2 dy = şk 0 h2y ´ 2hy2 + y3 dy şk 0 (h2 ´ 2hy + y2) dy = h h2 2 y2 ´ 2h 3 y3 + 1 4y4ik 0 h h2y ´ hy2 + 1 3y3 ik 0 = 1 2h2k2 ´ 2 3hk3 + 1 4k4 h2k ´ hk2 + 1 3k3 = 1 2h2k ´ 2 3hk2 + 1 4k3 h2 ´ hk + 1 3k2 S-30: To use the result of Question 29, we need to know the dimensions of the cone that was truncated to make the hourglass. The bottom (or top) half of our hourglass has base radius 5 cm, height 9 cm, and top radius 0.5 cm. Imagine extending it to a full cone. Let t be the distance from the top of the half hourglass to the tip of the full cone.
5 9 0.5 t 555 Using similar triangles, t 0.5 = t + 9 5 so 5t = 1 2(t + 9) 4.5t = 4.5 t = 1 Then the height of the full cone (that we imagined truncating to make half of the hourglass) is h = 10 cm.
Before the hourglass is turned over, the sand forms a truncated cone of height 6 cm. So, it’s the bottom k = 6 cm of a cone of height h = 10 cm. Using the result of Question 29, its centre of mass is at height: 1 2h2k ´ 2 3hk2 + 1 4k3 h2 ´ hk + 1 3k2 = 1 2102 ¨ 6 ´ 2 310 ¨ 62 + 1 463 102 ´ 10 ¨ 6 + 1 362 = 57 26 « 2.2 Next, let’s find the centre of mass of the sand after it’s been rotated. We have to be a little careful with our vocabulary here: usually we imagine a cone sitting on its base, with its tip pointing up. The upturned sand is in the opposite configuration. When we say the “base” of the cone, we mean the larger horizontal face–the top of the sand as it sits in the hourglass.
The formula we have from Question 29 gives us our centre of mass as a distance from the base of the truncated cone (that is, the distance from the top of the upturned sand). If k is the height the sand actually occupies, then we were told we may assume k = 8.8 cm. It’s missing its “tip” of height 1 cm, so h, the height of the “untruncated” cone, is 9.8 cm.
Using our model from Question 29, we don’t care about the empty, uppermost piece of the hourglass. The shape of the sand is of a cone of height 9.8 cm (not 10 cm), with a tip of height 1 cm chopped off.
0.2 cm of empty hourglass: ignored in our calculation of sand’s centre of mass cut-off tip: height 1 cm sand: height k = 8.8 cm h = 9.8 1 2h2k ´ 2 3hk2 + 1 4k3 h2 ´ hk + 1 3k2 = 1 29.82 ¨ 8.8 ´ 2 39.8 ¨ 8.82 + 1 48.83 9.82 ´ 9.8 ¨ 8.8 + 1 38.82 « 2.443 556 That is, the centre of mass of the upturned sand is about 2.443 centimetres below its top, which is at height 8.8 + 10 = 18.8 cm above the very bottom of the hourglass. So, the centre of mass of the upturned sand is at height y = 18.8 ´ 2.443 = 16.357 cm.
Now, we have our model: the sand, viewed as a point mass, is moved from y = 57 26 to y = 16.357 cm. That is, it moved about 14.165 cm, or about 0.14165 m. It has a mass of 0.6 kg, so the force required to lift it against gravity is (0.6 kg) ˆ 9.8 m sec2 ˆ (0.14165 m) « 0.833 newtons S-31: The techniques of Section 2.1 get pretty complicated here, so we will use the techniques we developed in Questions 6, 29 and 30 in this section. That is, (1) find the centre of mass of the water in its starting and ending positions, and then (2) compute the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second. For the centre of mass, all we need to know is the height–for one thing, we could find the other coordinates by symmetry, but we don’t need them.
The height moved by the water is all that matters if we’re calculating the work done opposing gravity.
Let’s start by calculating the volume of the water. The volume of a sphere of radius 1 is 4 3π ¨ 13, so the volume of water is 2 3π m3.
Then the mass of the water is 2000 3 π kg.
Next, we calculate the centre of mass of Tank A, and the work done to pump the water out of Tank A to a height of 3 metres. Symmetry alone won’t tell us the height of the centre of mass. We’ll show you two ways to go about this.
Option 1: As in Question 29, we’ll model the tank of water as a vertical rod, along the y–axis spanning the interval [0, 1], such that the mass of a piece of the rod along [a, b] is the same as the mass of the water from height y = a to height y = b. Then, the centre of mass of the rod will be the same as the centre of mass of the water.
Consider a horizontal slice of water at height y, with thickness dy. If the radius of this slice is r(y), then the volume of the slice is πr(y)2 dy m3, so its mass is 1000πr(y)2 dy kg. Then the mass of the slice of the rod at position y with length dy is 1000πr(y)2 dy kg, so its density ρ(y) is ρ(y) = 1000πr(y)2 dy kg dy m = 1000πr(y)2 kg m .
So, let’s find r(y), the radius of the slice of water at height y.
557 x y r y 1 Using the Pythagorean Theorem, r = a 1 ´ y2. Therefore, ρ(y) = 1000π(1 ´ y2) We use Equation 2.3.2 in the CLP-2 text to calculate the centre of mass of the rod, which is the height of the centre of mass of Tank A: ¯ yA = ş1 0 yρ(y) dy ş1 0 ρ(y) dy = ş1 0 1000πy(1 ´ y2) dy ş1 0 1000π(1 ´ y2) dy = ş1 0(y ´ y3) dy ş1 0(1 ´ y2) dy = h 1 2y2 ´ 1 4y4i1 0 h y ´ 1 3y3 i1 0 = 1 2 ´ 1 4 1 ´ 1 3 = 3 8 m From here, we can find the work done moving pumping the water to a height of 3 metres. We’ve moved the centre of mass from ¯ yA = 3 8 metres to 3 metres.
W = 2000 3 π kg ˆ 3 ´ 3 8 m ˆ 9.8 m sec2 = 17, 150π J Option 2: We can use the techniques of Section 2.1 in the CLP-2 text to calculate the amount of work it takes to pump the water from tank A to a height of 3 metres.
That solves part (a), and we can use the amount of work to figure out the centre of gravity of the water in Tank A to help us solve part (b).
At height y, a horizontal layer of water in Tank A forms a disk with thickness dy and radius a 1 ´ y2. (The radius comes from the Pythagorean Theorem–see the diagram below.) x y y 1 558 The volume of the layer at height y is π a 1 ´ y2 2 dy = π(1 ´ y2) dy m3, so its mass is 1000π(1 ´ y2) dy kg.
The layer at height y needs to be pumped a distance of 3 ´ y metres. So, the work involved pumping the layer at height y is: dW = 1000π(1 ´ y2) dy kg ˆ (3 ´ y m) ˆ 9.8 m/sec2 = 9800π(y3 ´ 3y2 ´ y + 3) dy J Then the work involved pumping out the entire tank to a height of 3 metres is: W = ż 1 0 9800π(y3 ´ 3y2 ´ y + 3) dy = 9800π 1 4y4 ´ y3 ´ 1 2y2 + 3y 1 0 = 17, 150π J This gives us an answer to part (a). To find the centre of mass of the water in Tank A, note that the work done is equivalent to moving a point mass from the centre of mass of the tank to a height of 3 metres. We know the water in Tank A has mass 2000 3 π kg. So, if ¯ yA is the centre of mass of the water in Tank A: W = 2000 3 π kg ˆ (3 ´ ¯ yA m) ˆ 9.8 m/sec2 17, 150π = 2000 3 π (3 ´ ¯ yA) (9.8) 21 8 = 3 ´ ¯ yA ¯ yA = 3 8 m Next let’s calculate the centre of mass of the water in Tank B. Since the volume of the water in Tank B is 2 3π m3, and the base of Tank B has area 1 m2, the height of the water in Tank B is 2 3π m. Since the water is of uniform density, and Tank B has uniform horizontal cross-sections, by symmetry the centre of mass of the water in Tank B is at ¯ yB = 1 3π m.
Now, we can calculate the work done by moving the water directly from Tank A to its final position in Tank B. The work done moving a point mass of 2000 3 π kg a distance of ¯ yB ´ ¯ yA = 1 3π ´ 3 8 m against the gravity, g = 9.8 m/sec2, is: W = 2000 3 π kg ˆ 1 3π ´ 3 8 m ˆ 9.8m/sec2 = 2450 9 π (8π ´ 9) « 13, 797 J 559 Finally, the “wasted” work is: ∆W = 17, 150π ´ 2450 9 π (8π ´ 9) = 2450π 7 ´ 8π ´ 9 9 = 2450π 8 ´ 8π 9 = 19, 600π 1 ´ π 9 As a percentage of 17,150π, this is: waste = 19, 600π 1 ´ π 9 17, 150π !
ˆ 100 = 8 7 1 ´ π 9 ˆ 100 « 74% S-32: Using Equation 2.3.3 in the CLP-2 text with T(x) = 2x sin(x2) and B(x) = 0, ¯ x = ż ?
π/2 0 2x2 sin(x2) dx ż ?
π/2 0 2x sin(x2) dx We can evaluate the bottom integral exactly with the substitution u = x2, du = 2xdx.
When x = 0, u = 0, and when x = ?
π/2, u = π/2.
ż ?
π/2 0 2x sin(x2) dx = ż π/2 0 sin u du = h ´ cos u iπ/2 0 = 1 So, ¯ x = ż ?
π/2 0 2x2 sin(x2) dx Evaluating the integral ş x2 sin(x2) dx is not so simple8, so we use a numerical approximation. Since we’re given an upper bound on the fourth derivative, we decide to use Simpson’s rule. The error involved using Simpson’s rule with n intervals is at most L(b´a)5 180n4 . For our approximation, a = 0 and b = ?
π/2. According to the information 8 Indeed, the antiderivative of 2x2 sin(x2) is not expressible as an elementary function.
560 given in the problem statement, ˇ ˇ ˇ d4 dx4 ␣ 2x2 sin(x2) (ˇ ˇ ˇ ď 415 over the interval h 0, b π 2 i , so we set L = 415.
We want our final error to be no more than 1 100, so we want to find an even n such that: 415 b π 2 ´ 0 5 180n4 ď 1 100 n4 ě 415 ¨ 100 π 2 5/2 180 = 2075π5/2 36 ?
2 n ě 4 d 2075π5/2 36 ?
2 « 5.17 So, n = 6 intervals suffices. Then ∆x = b´a 6 = 1 6 b π 2 and our grid points are x0 = 0, x1 = 1 6 b π 2 , x2 = 1 3 b π 2 , x3 = 1 2 b π 2 , x4 = 2 3 b π 2 , x5 = 5 6 b π 2 , and , x6 = b π 2 .
0 1 6 b π 2 1 3 b π 2 1 2 b π 2 2 3 b π 2 5 6 b π 2 b π 2 Following Equation 1.11.9 in the CLP-2 text, the Simpson’s rule approximation of ż ?
π/2 0 2x2 sin(x2) dx is: ∆x 3 h f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + 4f (x5) + f (x6) i =1 6 c π 2 ¨ 1 3 0 + 4 ˆ 2π 72 sin π 72 + 2 ˆ 2π 18 sin π 18 + 4 ˆ 2π 8 sin π 8 + 2 ˆ 8π 18 sin 4π 18 + 4 ˆ 50π 72 sin 25π 72 + 2π 2 sin π 2 = 1 18 c π 2 π 9 sin π 72 + 2π 9 sin π 18 + π sin π 8 + 8π 9 sin 2π 9 + 25π 9 sin 25π 72 + π = π 18 c π 2 1 9 sin π 72 + 2 9 sin π 18 + sin π 8 + 8 9 sin 2π 9 + 25 9 sin 25π 72 + 1 = π 162 c π 2 sin π 72 + 2 sin π 18 + 9 sin π 8 + 8 sin 2π 9 + 25 sin 25π 72 + 9 « 0.976 The absolute error in our answer is at most: L(b ´ a)5 180n4 = 415 ˆ b π 2 5 180 ˆ 64 = 82?π5 186624 ?
2 « 0.005 561 Remark: combining the error with our approximation, we see the actual value of ¯ x is in the interval [0.976 ´ 0.005, 0.976 + 0.005] = [0.971, .981] A computer algebra system approximates ¯ x as 0.977451.
Solutions to Exercises 2.4 — Jump to TABLE OF CONTENTS S-1: (a) If y = 5(ex ´ 3x2 ´ 6x ´ 6), then dy dx = 5(ex ´ 6x ´ 6). Let’s see whether this is equal to y + 15x2: y + 15x2 = 5(ex ´ 3x2 ´ 6x ´ 6) + 15x2 = 5(ex ´ 3x2 ´ 6x ´ 6 + 3x2) = 5(ex ´ 6x ´ 6) = dy dx So, y = 5(ex ´ 3x2 ´ 6x ´ 6) is indeed a solution to the differential equation dy dx = y + 15x2.
(b) If y = ´2 x2 + 1, then dy dx = 4x (x + 1)2. Let’s see whether this is equal to xy2: xy2 = x ´2 x2 + 1 2 = 4x (x2 + 1)2 = dy dx So, y = ´2 x2 + 1 is indeed a solution to the differential equation dy dx = xy2.
(c) If y = x3/2 + x, then dy dx = 3 2 ?x + 1.
dy dx 2 + dy dx = 3 2 ?x + 1 2 + 3 2 ?x + 1 = 9 4x + 9 2 ?x + 2 ‰ dy dx So, y = x3/2 + x is not a solution to the differential equation dy dx 2 + dy dx = y.
562 S-2: (a) 3y dy dx = x sin y can be written as dy dx = x sin y 3y , which fits the form of a separable equation with f (x) = x, g(y) = sin y 3y .
(b) dy dx = ex+y = exey which fits the form of a separable equation using f (x) = ex, g(y) = ey.
(c) dy dx + 1 = x can be written as dy dx = (x ´ 1), which fits the form of a separable equation using f (x) = x ´ 1, g(y) = 1. (We can solve it by simply antidifferentiating.) (d) Notice the left side of the equation dy dx 2 ´ 2x dy dx + x2 = 0 is a perfect square. So, this equation is equivalent to dy dx ´ x 2 = 0, that is, dy dx = x. This has the form of a separable equation with f (x) = x, g(y) = 1.
S-3: The mnemonic allows us to skip from the separable differential equation we want to solve (very first line) to the equation ż 1 g(y) dy = ż f (x) dx So, the mnemonic is just a shortcut for the substitution we performed to get this point.
We also generally skip the explanation about C1 and C2 being replaced with C.
S-4: To say y = f (x) + C is a solution to the differential equation means: d dxtf (x) + Cu = x( f (x) + C) Since y = f (x) is a solution, we know d dxtf (x)u = x f (x). Also, d dxtf (x) + Cu = d dxtf (x)u.
So, d dxtf (x) + Cu = x f (x).
x f (x) = x( f (x) + C) 0 = xC Our equation should hold for all x in our domain, and for the derivative to y with respect to x to make sense, our domain should not be a single point. So, there is some x in our domain such that x ‰ 0. Therefore, the C must be zero. So, f (x) + C is not a solution to the differential equation for any constant C.
When we’re finding a general antiderivative, we add “+C” at the end. When we’re finding a general solution to a differential equation, the “+C” gets added when we antidifferentiate–we don’t add another one at the end of our work.
S-5: (a) Since |y| ě 0 no matter what y is, we see Cx ě 0 for all x in the domain of f (x). Since C is positive, that means the domain of f (x) only includes nonnegative numbers. So, the largest possible domain of f (x) is [0, 8).
563 (b) None exists.
The graph of Cx is given below for some positive constant C, also with the graph of ´Cx. If y = f (x) were sometimes the top function, and other times the bottom function, then there would be a jump discontinuity where it switched. Then the derivative of f (x) would not exist, violating the second property.
x y y = Cx y = ´Cx A tiny technical note is that it’s possible that f (x) = Cx when x = 0 and f (x) = ´Cx when x ą 0 (or vice-versa). This would not introduce a jump discontinuity, but it also does not satisfy that f (x) ą 0 for some values of x.
Remark: in several instances below, solving a differential equation will lead us to conclude something like |y| = g(x). In these cases, we choose either y = g(x), or y = ´g(x), but not y = ˘g(x) (which is not a function) or that y is sometimes g(x), and other times ´g(x). The reasoning above somewhat explains this choice: if y were sometimes positive and sometimes negative, then dy dx would not exist at the values of x where the sign of y switches, unless that switch occurrs at a root of g(x). Since that’s a pretty specific occurrence, we usually feel safe ignoring it to avoid getting bogged down in technical details.
S-6: Let Q(t) be the quantity of morphine in a patient’s bloodstream at time t, where t is measured in minutes.
Using the definition of a derivative, dQ dt = lim hÑ0 Q(t + h) ´ Q(t) h « Q(t + 1) ´ Q(t) 1 So, dQ dt is roughly the change in the amount of morphine in one minute, from t to t + 1.
The sentence tells us that the change in the amount of morphine in one minute is about ´0.003Q, where Q is the quantity in the bloodstream. That is: dQ dt = ´0.003Q(t) 564 S-7: If p(t) is the proportion of times speakers use the new form, measured between 0 and 1, then 1 ´ p(t) is the proportion of times speakers use the old form.
The law, then, states that dp dt is proportional to p(t) ˆ 1 ´ p(t) . When we say two quantities are proportional, we mean that one is a constant multiple of the other. So, the law says dp dt = αp(t) 1 ´ p(t) for some constant α.
Remark: it follows from this model that, when a new form is either very rare or entirely ubiquitous, the rate of change of its adoption is small. This makes sense: if the new form is used all the time (p(t) « 1), there’s nobody left to convert; if the new form is almost never used (p(t) « 0) then people don’t know about it, so they won’t pick it up.
S-8: (a) When y = 0, y1 = 0 2 ´ 1 = ´1.
(b) When y = 2, y1 = 2 2 ´ 1 = 0.
(c) When y = 3, y1 = 3 2 ´ 1 = 0.5.
(d) The small red lines have varying slopes. The red lines on points with y-coordinate 2 have slopes of 0; this matches y1 when y = 0, as we saw above. The red lines on points with y-coordinate 0 have slopes of approximately ´1; again, this matches what we found for y1 when y = 0.
The red lines correspond to a tiny section of y(x), if y(x) passes through that point.
So, we can sketch a possible curve y(x) satisfying the equation by starting somewhere, then following the slopes.
For example, suppose we start at the origin.
x y 1 1 565 Then our function is decreasing at that point, which leads us to a coordinate where (as we see from the red marks) the function is decreasing slightly faster.
x y 1 1 Following the red marks leads us down even further, so our function y(x) might look something like this: x y 1 1 However, we didn’t have to start at the origin. Suppose y(0) = 3. Then at x = 0, y is increasing, with slope 1 2.
566 x y 1 1 Our red marks run out that high up, but we now y1 = 1 2y ´ 1, so y1 increases as y increases. That means our function keeps getting steeper and steeper, possibly something like this: x y 1 1 If y(0) = 2, we see another possible curve is the constant function y(x) = 2.
Remark: from Theorem 2.4.4 in the CLP-2 text, we see the solutions to the equation y1 = 1 2y ´ 1 = 1 2(y ´ 2) are of the form y(x) = Cex/2 + 2 for some constant C. Check that the curves you’re sketching look exponential.
S-9: (a) If y(1) = 0, then y1(1) = 0 ´ 1 2 = ´1 2.
(b) If y(1) = 2, then y1(1) = 2 ´ 1 2 = 3 2.
(c) If y(1) = ´2, then y1(1) = ´2 ´ 1 2 = ´5 2.
567 (d) There are 7 ˆ 7 = 49 points on the grid; we don’t want to make 49 separate calculations. Let’s find some shortcuts.
• If y1(x) = 0, then y = x 2, which applies to the points (0, 0), (2, 1), (4, 2) and (6, 3).
These are the orange dots in the sketch below.
• If y1(x) = 1, then y = 1 + x 2, which applies to the points (0, 1), (2, 2), and (4, 3).
(Note these are exactly 1 unit above the points with y1 = 0.) These are the red dots in the sketch below.
• If y1(x) = ´1, then y = ´1 + x 2, which applies to the points (0, ´1), (2, ´2), and (4, ´3). (Note these are exactly 1 unit below the points with y1 = 0.) These are the yellow dots in the sketch below.
• If x increases and y stays the same, y decreases.
• If y increases and x stays the same, y increases.
• If we draw a straight line of slope 1 2 on our sketch, for every point on that line, our mark has the same slope: for instance, the points where we draw a mark with slope 0 are (0, 0), (2, 1), and (4, 2), and these all lie on the line f (x) = x 2.
This is enough to give us a pretty good sketch. The points whose slopes we found explicitly have dots; the rest can be sketched as either steeper or less steep than what’s near them.
x y 1 1 (e) To sketch a possible graph of y(x), we choose a point (x, y(x)), then follow the red lines.
For example, if we suppose that y(4) = 2, then near (4, 2), the lines tell us y(x) is fairly flat; and it is increasing to the left of x = 4, and decreasing to the right.
568 x y 1 1 Following the red lines a little farther in each direction brings us somewhere like this: x y 1 1 Extending yet further, we might sketch something like the following: 569 x y 1 1 By choosing another point (x, y(x)) to be on the curve, we might find other potential curves. Some examples are shown below.
Passing through (0, 0): x y 1 1 Passing through (0, 1): x y 1 1 570 Passing through (6, 3): x y 1 1 Passing through (6, ´3): x y 1 1 Remark: the differential equation y1 = y ´ x 2 is not separable, so we haven’t talked about how to solve it. The solutions have the form y(x) = Cex + x+1 2 . You can verify that these functions satisfy y1 = y ´ x 2.
S-10: Rearranging, we have: ey dy = 2x dx.
Integrating both sides: ż ey dy = ż 2x dx ey = x2 + C Since y = log 2 when x = 0, we have elog 2 = 02 + C 2 = C, and therefore ey = x2 + 2 y = log(x2 + 2) 571 S-11: Using separation of variables: dy dx = xy x2 + 1 dy y = x x2 + 1dx ż dy y = ż x x2 + 1dx log |y| = 1 2 log(1 + x2) + C To satisfy y(0) = 3, we need log 3 = 1 2 log(1 + 0) + C, so C = log 3. Thus: log |y| = 1 2 log(1 + x2) + log 3 = log a 1 + x2 + log 3 = log 3 a 1 + x2 So, |y| = 3 a 1 + x2 We are told to find a function y(x). So far, we have two possible functions from the work above: maybe y = 3 ?
1 + x2, and maybe y = ´3 ?
1 + x2. It’s important to note that y = ˘3 ?
1 + x2 is not a function: for an equation to represent a function, for every input in the domain, there must only be one output. That is, functions pass the vertical line test. (Definition 0.4.1 in the CLP-1 text gives a formal definition of a function.) So, we need to decide whether our function is y = 3 ?
1 + x2 or y = ´3 ?
1 + x2. Since y(0) = 3, we conclude y(x) = 3 a 1 + x2 S-12: The given differential equation is separable and we solve it accordingly.
y1 = e y 3 cos t e´y/3dy = cos t dt ż e´y/3dy = ż cos t dt ´3e´y/3 = sin t + C 1 ey/3 = sin t + C ´3 ey/3 = ´3 C + sin t y 3 = log ´3 C + sin t y(t) = 3 log ´3 C + sin t 572 for any constant C.
Since the domain of logarithm is (0, 8), the solution only exists when C + sin t ă 0.
S-13: The given differential equation is separable and we solve it accordingly.
dy dx = xex2´log(y2) = xex2 y2 y2dy = xex2 dx ż y2dy = ż xex2 dx We can guess the antiderivative of xex2, or use the substitution u = x2, du = 2xdx.
y3 3 = 1 2ex2 + C1 y3 = 3 2ex2 + 3C1 Since C1 can be any constant in (´8, 8), then also 3C1 can be any constant in (´8, 8), so we replace 3C1 with the arbitrary constant C.
y3 = 3 2ex2 + C y = 3 c 3 2ex2 + C for any constant C.
S-14: The given differential equation is separable and we solve it accordingly.
dy dx = xey dy ey = x dx ż dy ey = ż x dx ´e´y = 1 2x2 + C e´y = ´1 2x2 ´ C Since C can be any constant in (´8, 8), then also ´C can be any constant in (´8, 8), so we write C instead of ´C.
e´y = C ´ 1 2x2 ´y = log C ´ x2 2 y = ´ log C ´ x2 2 573 for any constant C.
The solution only exists for C ´ x2 2 ą 0. For this to happen, we need C ą 0, and then the domain of the function is those values x for which |x| ă ?
2C.
S-15: The given differential equation is separable and we solve it accordingly.
Cross–multiplying, we rewrite the equation as y2dy dx = ex ´ 2x y2 dy = (ex ´ 2x) dx.
Integrating both sides, we find ż y2 dy = ż (ex ´ 2x) dx 1 3y3 = ex ´ x2 + C Setting x = 0 and y = 3, we find 1 333 = e0 ´ 02 + C and hence C = 8.
1 3y3 = ex ´ x2 + 8 y = (3ex ´ 3x2 + 24)1/3 S-16: This is a separable differential equation that we solve in the usual way.
dy dx = ´xy3 ´dy y3 = x dx ż ´dy y3 = ż x dx ´y´2 ´2 = x2 2 + C y´2 = x2 + 2C.
(˚) To have y = ´1 4 when x = 0, we must choose C to obey ´ 1 4 ´2 = 0 + 2C 16 = 2C So, from (˚), y´2 = x2 + 2C = x2 + 16 y2 = 1 x2 + 16 574 Now, we have two potential candidates for y(x): y = 1 ?
x2 + 16 OR y = ´ 1 ?
x2 + 16 We know y = ´1 4 when x = 0. The only function above that fits this is y = ´ 1 ?
x2 + 16 So, f (x) = ´ 1 ?
x2 + 16 .
S-17: This is a separable differential equation that we solve in the usual way.
Cross-multiplying and integrating, y dy = (15x2 + 4x + 3) dx ż y dy = ż (15x2 + 4x + 3) dx y2 2 = 5x3 + 2x2 + 3x + C.
Plugging in x = 1 and y = 4 gives 42 2 = 5 + 2 + 3 + C, and so C = ´2. Therefore y2 2 = 5x3 + 2x2 + 3x ´ 2 y2 = 10x3 + 4x2 + 6x ´ 4 This leaves us with two possible functions for y: y = a 10x3 + 4x2 + 6x ´ 4 OR y = ´ a 10x3 + 4x2 + 6x ´ 4 When x = 1, y = 4. This only fits the first equation, so y = a 10x3 + 4x2 + 6x ´ 4 S-18: The given differential equation is separable and we solve it accordingly.
dy dx = x3y dy y = x3 dx ż dy y = ż x3 dx log |y| = x4 4 + C |y| = ex4/4+C = ex4/4eC 575 We are told that y = 1 when x = 0. That is, 1 = e0eC, so eC = 1. That is, C = 0.
|y| = ex4/4 This leaves us with two potential functions: y = ex4/4 OR y = ´ex4/4 The first is always positive, and the second is always negative. Since y = 1 (a positive number) when x = 0, we see y = ex4/4 S-19: This is a separable differential equation, even if it doesn’t quite look like it. First move the y from the left hand side to the right hand side.
xdy dx + y = y2 xdy dx = y2 ´ y = y(y ´ 1) dy y(y ´ 1) = dx x Using the method of partial fractions, we see 1 y(y´1) = 1 y´1 ´ 1 y.
1 y ´ 1 ´ 1 y dy = dx x ż 1 y ´ 1 ´ 1 y dy = ż dx x log |y ´ 1| ´ log |y| = log |x| + C log |y ´ 1| |y| = log |x| + C (˚) To determine C we set x = 1 and y = ´1.
log | ´ 2| | ´ 1| = log |1| + C log 2 = C Returning to (˚), log |y ´ 1| |y| = log |x| + log 2 log ˇ ˇ ˇ ˇ y ´ 1 y ˇ ˇ ˇ ˇ = log |2x| ˇ ˇ ˇ ˇ y ´ 1 y ˇ ˇ ˇ ˇ = |2x| 576 As y(1) = ´1 is an initial condition, we have that x ě 1 and |2x| = 2x. For x = 1, we have y = ´1. So at least for x near 1, we have y near ´1, so that y´1 y is positive and we may drop the absolute value signs. There remains the possibility that y(x)´1 y(x) changes sign for some larger x ą 1. For now, we will simply ignore that possibility. At the end, we will explicitly check that the y(x) we come up with really does satisfy the differential equation x dy dx + y = y2 and the initial condition y(1) = ´1.
y ´ 1 y = 2x y ´ 1 = 2xy y ´ 2xy = 1 y(1 ´ 2x) = 1 y = 1 1 ´ 2x As a check, we compute: xdy dx + y = x d dx " 1 1 ´ 2x + y = x 2 (1 ´ 2x)2 + 1 1 ´ 2x = 2x + (1 ´ 2x) (1 ´ 2x)2 = 1 (1 ´ 2x)2 = y2 So, our differential equation is satisfied. Furthermore: y(1) = 1 1 ´ 2 ˆ 1 = ´1 as desired. This confirms that our solution is correct.
S-20: The unknown function f (x) satisfies an equation that involves the derivative of f.
That means we’re in differential equation territory. Specifically, we are told that y = f (x) obeys the separable differential equation dy dx = xy.
dy dx = xy dy y = x dx ż dy y = ż x dx log |y| = x2 2 + C 577 To determine C we set x = 0 and y = e.
log e = 02 2 + C 1 = C So, the solution is log |y| = x2 2 + 1 We are told that y = f (x) ą 0, so may drop the absolute value signs.
log y = x2 2 + 1 y = e1+ 1 2 x2 = e ¨ ex2/2 S-21: This is a separable differential equation.
dy dx = 1 (x2 + x)y y dy = dx x(x + 1) Using partial fractions decomposition, we find 1 x(x+1) = 1 x ´ 1 x+1.
y dy = 1 x ´ 1 x + 1 dx ż y dy = ż 1 x ´ 1 x + 1 dx y2 2 = log |x| ´ log |x + 1| + C = log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ + C To satisfy the initial condition y(1) = 2 we must choose C to obey 22 2 = log ˇ ˇ ˇ ˇ 1 1 + 1 ˇ ˇ ˇ ˇ + C 2 = log 1 2 + C C = 2 ´ log 1 2 So, y2 2 = log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ + 2 ´ log 1 2 y2 = 2 log ˇ ˇ ˇ ˇ x x + 1 ˇ ˇ ˇ ˇ + 4 ´ 2 log 1 2 578 Note that the question specifies that y(1) = 2 is an initial condition. So we always have x ě 1. Then x x+1 is positive, and we can drop the absolute values.
y2 = 2 log x x + 1 + 4 ´ 2 log 1 2 This leaves two options for y(x): the positive or negative square root of the right hand side above. Since y(1) = 2, which is positive, we must choose the positive square root.
y(x) = c 2 log x x + 1 ´ log 1 2 + 2 = c 4 + 2 log 2x x + 1 You might worry that y(x) could pass through zero, changing sign, at some x ą 1. But the differential equation says that dy dx = 1 (x2+x)y is positive whenever y ą 0 and x ě 1. So y(x) is an increasing function whenever y ą 0 and x ě 1. As y(1) = 2, we have y(x) ě 2 for all x ě 1.
S-22: This is a separable differential equation.
1 + a y2 ´ 4 tan x dy dx = sec x y y 1 + b y2 ´ 4 dy = sec x tan x dx ż y 1 + b y2 ´ 4 dy = ż sec x tan x dx For the integral on the left, we use the substitution u = y2 ´ 4, 1 2du = y dy.
1 2 ż 1 + ?u du = sec x + C 1 2 u + 2 3u3/2 = sec x + C 1 2 y2 ´ 4 + 2 3(y2 ´ 4)3/2 = sec x + C y2 + 2 3(y2 ´ 4)3/2 = 2 sec x + 2C + 4 To find C we set x = 0 and y = 2.
4 + 2 3 ?
4 ´ 4 3 = 2 sec(0) + 2C + 4 4 = 2 + 2C + 4 2 = 2C + 4 So, y2 + 2 3(y2 ´ 4)3/2 = 2 sec x + 2 579 S-23: The given differential equation is separable and we solve it accordingly.
dP dt = ´k ?
P dP ?
P = ´k dt ż dP ?
P = ż ´k dt 2 ?
P = ´kt + C At t = 0, P = 90, 000 so 2 a 90, 000 = ´k ˆ 0 + C C = 2 ˆ 300 = 600 Therefore, 2 ?
P = ´kt + 600 (˚) Now, we find k. Let t be measured in weeks. Then when t = 6, P = 40, 000.
2 a 40, 000 = ´6k + 600 2 ¨ 200 = ´6k + 600 k = 200 6 = 100 3 Substituting our value of k into (˚): 2 ?
P = ´100 3 t + 600 To find when the population will be 10,000, we set P = 10, 000 and solve for t.
2 a 10, 000 = ´100 3 t + 600 2 ¨ 100 = ´100 3 t + 600 100 3 t = 400 t = 12 Since we measured t in weeks when we found k, we see that in 12 weeks the population will decrease to 10,000 individuals.
S-24: The given differential equation is separable and we solve it accordingly.
mdv dt = ´(mg + kv2) m mg + kv2 dv = ´dt ż m mg + kv2 dv = ż ´dt 580 The left integral looks something like the antiderivative of arctangent. Let’s factor out that mg from the denominator.
1 mg ż m 1 + k mgv2 dv = ´t + C 1 g ż 1 1 + b k mgv 2 dv = ´t + C Now it looks even more like the derivative of arctangent. We can guess the antiderivative from here, or use the substitution u = b k mgv, du = b k mg dv.
1 g c mg k arctan d k mgv !
= ´t + C c m gk arctan d k mgv !
= ´t + C (˚) At t = 0, v = v0, so: c m gk arctan d k mgv0 !
= C Plug C into (˚).
c m gk arctan d k mgv !
= c m gk arctan d k mgv0 !
´ t At its highest point, the object has velocity v = 0. This happens when t obeys: c m gk arctan d k mg0 !
= c m gk arctan d k mgv0 !
´ t 0 = c m gk arctan d k mgv0 !
´ t t = c m gk arctan d k mgv0 !
S-25: (a) The given differential equation is separable and we solve it accordingly.
dv dt = ´k v2 ´dv v2 = k dt ż ´dv v2 = ż k dt 1 v = kt + C 581 At t = 0, v = 40 so 1 40 = k ˆ 0 + C C = 1 40 Therefore, v(t) = 1 kt + C = 1 kt + 1/40 = 40 40kt + 1 (˚) The constant of proportionality k is determined by v(10) = 20 20 = 40 40k ˆ 10 + 1 1 2 = 1 400k + 1 400k + 1 = 2 k = 1 400 (b) Subbing in the value of k to (˚), v(t) = 40 40kt + 1 = 40 t/10 + 1 We want to know the value of t that gives v(t) = 5.
5 = 40 t/10 + 1 t 10 + 1 = 8 t = 70 sec S-26: (a) The given differential equation is separable and we solve it accordingly.
dx dt = k(3 ´ x)(2 ´ x) dx (x ´ 2)(x ´ 3) = kdt 582 Using the method of partial fractions, we find 1 (x´2)(x´3) = 1 x´3 ´ 1 x´2.
ż h 1 x ´ 3 ´ 1 x ´ 2 i dx = ż kdt log |x ´ 3| ´ log |x ´ 2| = kt + C log ˇ ˇ ˇ ˇ x ´ 3 x ´ 2 ˇ ˇ ˇ ˇ = kt + C ˇ ˇ ˇ ˇ x ´ 3 x ´ 2 ˇ ˇ ˇ ˇ = ekt+C = ekteC x ´ 3 x ´ 2 = Dekt where D = ˘eC. When t = 0, x = 1, forcing 1 ´ 3 1 ´ 2 = De0 D = 2 Hence x ´ 3 x ´ 2 = 2ekt x ´ 3 = 2ekt(x ´ 2) x ´ 2ektx = 3 ´ 4ekt x(t) = 3 ´ 4ekt 1 ´ 2ekt (b) To evaluate the limit, we could use l’Hˆ opital’s rule, but we could also just multiply the numerator and denominator by e´kt. Note lim tÑ8 e´tk = 0.
lim tÑ8 x(t) = lim tÑ8 3 ´ 4ekt 1 ´ 2ekt looomooon numÑ´8 denÑ´8 = lim tÑ8 3 ´ 4ekt 1 ´ 2ekt ¨ e´kt e´kt = lim tÑ8 3e´kt ´ 4 e´kt ´ 2 = 0 ´ 4 0 ´ 2 = 2 S-27: (a) The given differential equation is separable and we solve it accordingly.
dP dt = 4P ´ P2 dP 4P ´ P2 = dt dP P(4 ´ P) = dt 583 Using the method of partial fractions, we see 1 P(4´P) = 1/4 P + 1/4 4´P.
1 4 h 1 P + 1 4 ´ P i dP = dt ż 1 4 h 1 P + 1 4 ´ P i dP = ż dt 1 4 log |P| ´ log |4 ´ P| i = t + C When t = 0, P = 2, so 1 4 log |2| ´ log |2| = C ù ñ C = 0. So, 1 4 log ˇ ˇ ˇ P 4 ´ P ˇ ˇ ˇ = t At time t = 0, P 4´P = 1 ą 0. The ratio may not change sign at any finite time, because this could only happen if at some finite time P took either the value 0 or the value 4. But at this time t = 1 4 log ˇ ˇ P 4´P ˇ ˇ would have to be infinite. So P 4´P ą 0 for all time and: 1 4 log P 4 ´ P = t log P 4 ´ P = 4t P 4 ´ P = e4t P = (4 ´ P)e4t P + Pe4t = 4e4t P = 4e4t 1 + e4t = 4 1 + e´4t (b) At t = 1 2, P = 4 1+e´2 « 3.523.
lim tÑ8 P(t) = lim tÑ8 4 1 + e´4t = 4 1 + 0 = 4 S-28: (a) The rate of change of speed at time t is ´kv(t)2 for some constant of proportionality k (to be determined–but we assume it is positive, since the speed is decreasing). So v(t) obeys the differential equation dv dt = ´kv2 .
(b) The equation dv dt = ´kv2 is a separable differential equation, which we can solve in 584 the usual way.
dv dt = ´kv2 dv ´v2 = kdt ż ´dv v2 = ż kdt 1 v = kt + C At time t = 0, v = 400, so C = 1 400. Then: 1 v = kt + 1 400 (˚) At time t = 1, v = 200, so 1 200 = k + 1 400 k = 1 400 Therefore, from (˚), 1 v = t 400 + 1 400 = t + 1 400 v = 400 t + 1 (c) To find when the speed is 50, we set v = 50 in the equation from (b) and solve for t.
50 = 400 t + 1 50(t + 1) = 400 t + 1 = 8 t = 7 S-29: (a) The given differential equation is separable and we solve it accordingly.
dB dt = (0.06 + 0.02 sin t)B dB B = (0.06 + 0.02 sin t) dt ż dB B = ż (0.06 + 0.02 sin t) dt log |B(t)| = 0.06t ´ 0.02 cos t + C1 585 Since B(t) is our bank account balance and we’re not withdrawing money, B(t) is positive, so we can drop the absolute value signs.
log B(t) = 0.06t ´ 0.02 cos t + C1 B(t) = e0.06t´0.02 cos teC1 B(t) = Ce0.06t´0.02 cos t for arbitrary constants C1 and C = eC1 ě 0.
Remark: the function B(t) = 0 obeys the differential equation so that C = 0 is allowed, even though it is not of the form C = eC1. This seeming discrepancy arose because, in our very first step of part (a), we divided both sides of the differential equation by B, which is only allowable if B ‰ 0. So, in this step, we implicitly assumed B was nonzero.
(b) We are told that B(0) = 1000. This allows us to find C.
1000 = B(0) = Ce0´0.02 cos 0 = Ce´0.02 C = 1000e0.02 So, when t = 2, B(2) = 1000e0.02 looomooon C e0.06ˆ2´0.02 cos 2 = $1159.89 rounded to the nearest cent.
Note that cos 2 is the cosine of 2 radians, cos 2 « ´0.416.
S-30: (a) The given differential equation is separable and we could solve it accordingly. In fact we have already done so. If we rewrite the equation in the form dB dt = a B ´ m a it is of the form covered by Theorem 2.4.4 in the CLP-2 text. So that theorem tells us that the solution is B(t) = B(0) ´ m a eat + m a In this problem we are told that a = 0.02 = 1 50, so B(t) = tB(0) ´ 50mu et/50 + 50m = t30000 ´ 50mu et/50 + 50m (b) The solution of part (a) is independent of time if and only if 30000 ´ 50m = 0. So we need m = 30000 50 = $600 586 S-31: What we’re given is an equation relating y to the integral of a function of y. What we know how to solve is an equation relating the derivative of y to a function of y. We can create this by differentiating the given integral equation. By the Fundamental Theorem of Calculus, part 1: y1(x) = d dx "ż x 0 y(t)2 ´ 3y(t) + 2 sin t dt = y(x)2 ´ 3y(x) + 2 sin x So y(x) satisfies the differential equation y1 = y2 ´ 3y + 2 sin x = (y ´ 2)(y ´ 1) sin x and the initial equation y(0) = 3 (just substitute x = 0 into (˚)). For y ‰ 1, 2: dy dx = (y ´ 2)(y ´ 1) sin x dy (y ´ 2)(y ´ 1) = sin x dx ż dy (y ´ 2)(y ´ 1) = ż sin x dx Using the method of partial fractions, we see 1 (y´2)(y´1) = 1 y´2 ´ 1 y´1.
ż h 1 y ´ 2 ´ 1 y ´ 1 i dy = ż sin x dx log |y ´ 2| ´ log |y ´ 1| = ´ cos x + c log ˇ ˇ ˇ ˇ y ´ 2 y ´ 1 ˇ ˇ ˇ ˇ = ´ cos x + c ˇ ˇ ˇ ˇ y ´ 2 y ´ 1 ˇ ˇ ˇ ˇ = ec´cos x The condition y(0) = 3 forces ˇ ˇ3´2 3´1 ˇ ˇ = ec´1 or ec = 1 2e, hence ˇ ˇ ˇ ˇ y ´ 2 y ´ 1 ˇ ˇ ˇ ˇ = 1 2e1´cos x Observe that, when x = 0, y´2 y´1 = 1 2 ą 0. Furthermore 1 2e1´cos x, and hence ˇ ˇy´2 y´1 ˇ ˇ, can never take the value zero. As y(x) varies continuously with x, y(x) must remain larger than 2.
Consquently, y´2 y´1 remains positive and we may drop the absolute value signs. Hence y ´ 2 y ´ 1 = 1 2e1´cos x Solving for y, y ´ 2 y ´ 1 = 1 2e1´cos x 2(y ´ 2) = e1´cos x(y ´ 1) 2y ´ 4 = ye1´cos x ´ e1´cos x y 2 ´ e1´cos x = 4 ´ e1´cos x y = 4 ´ e1´cos x 2 ´ e1´cos x 587 To avoid division by zero in the last step, we need e1´cos x ‰ 2 1 ´ cos x ‰ log 2 cos x ‰ 1 ´ log 2 Let L = 1 ´ log 2, for brevity, and note that L ą 0. (This can be seen by observing 2 ă e, so, log 2 ă log e = 1, hence 1 ´ log 2 ą 0.) x Y L arccos(L) ´ arccos(L) Y = cos x We know x = 0 is in the domain of our function, but the points x = ˘ arccos(L) = ˘ arccos(1 ´ log 2) are not.
x 0 in domain arccos(1 ´ log 2) not in domain ´ arccos(1 ´ log 2) not in domain Therefore, the largest interval for which our answer makes sense is ´ arccos(1 ´ log 2)) ą x ą arccos(1 ´ log 2) or approximately ´1.259 ă x ă 1.259.
S-32: Suppose that in a very short time interval dt, the height of water in the tank changes by dh (which is negative). Then in this time interval the amount of the water in the tank decreases by dV = ´π(3)2dh. This must be the same as the amount of water that flows through the hole in this time interval. The water flowing through the hole makes a cylinder of radius 1 cm (that is, 0.01 m) with length v(t)dt, the distance the water moves out of the hole in dt seconds. So, the amount of water leaving the hole over the time interval dt is π(0.01)2v(t) dt = π(0.01)2a 2gh(t) dt.
588 dh dV This gives us a separable differential equation. Recall g is a constant.
´π(3)2dh = π(0.01)2b 2gh(t) dt dh ?
h = ´ 0.01 3 2a 2g dt ż dh ?
h = ż ´ 0.01 3 2a 2g dt 2 ?
h = ´ 0.01 3 2a 2g t + C At time 0, the height is 6, so C = 2 ?
6 and 2 ?
h = ´ 0.01 3 2a 2g t + 2 ?
6 We want to know when the height of the water in the tank is 0.
0 = ´ 0.01 3 2a 2g t + 2 ?
6 0.01 3 2a 2g t = 2 ?
6 t = 2 ?
6 0.01 3 2a 2g = 2 3 0.01 2d 3 g = 180, 000 d 3 g « 99, 591 sec « 27.66 hr S-33: Suppose that at time t, the mercury in the tank has height h, which is between 0 and 12 feet. At that time, the top surface of the mercury forms a circular disk of radius 589 6 6 h −6 a 62 ´ (h ´ 6)2. (We found this by applying the Pythagorean Theorem to the triangle in the diagram above. In the diagram, h is shown as being larger than 6, but the same equation holds for all h in [0, 12].) Now suppose that in a very short time interval dt, the height of mercury in the tank changes by dh (which is negative). Then in this time interval the amount of the mercury in the tank decreases by ´π a 62 ´ (h ´ 6)2 2dh.
(That’s the volume of the red disk in the figure above.) This must be the same as the amount of mercury that flows through the hole in this time interval. The mercury comes out of the hole as a cylinder. Its radius is the radius of the hole, 1 12 foot, and its length is the distance the mercury travels in dt seconds, v(t)dt feet. So, the volume of escaped mercury is π 1 12 2v dt = π 1 12 2a 2gh dt. This gives us a separable differential equation.
´π b 62 ´ (h ´ 6)2 2 dh = π 1 12 2a 2gh dt ´ 36 ´ (h2 ´ 12h + 36) dh = 1 12 2a 2gh dt h2 ´ 12h dh = 1 144 a 2g ?
h dt h3/2 ´ 12h1/2 dh = 1 144 a 2g dt ż h3/2 ´ 12h1/2 dh = ż 1 144 a 2g dt h5/2 5/2 ´ 12h3/2 3/2 = 1 144 a 2g t + C At time 0, the height is 12, so C = 125/2 5/2 ´ 12123/2 3/2 = 125/2 2 5 ´ 2 3 = ´ 4 15125/2, which yields h5/2 5/2 ´ 12h3/2 3/2 = 1 144 a 2g t ´ 4 15125/2 We want to find the time t when the height is h = 0.
0 = 1 144 a 2g t ´ 4 15125/2 1 144 a 2g t = 4 15125/2 t = 4 ˆ 144 15 d 125 2g = 38.4 d 124416 g « 2, 394 sec « 0.665 hr 590 S-34: (a) Setting x = 0 gives f (0) = 3 + ż 0 0 f (t) ´ 1 f (t) ´ 2 dt = 3 (b) By the Fundamental Theorem of Calculus part 1, f 1(x) = d dx ż x 0 f (t) ´ 1 f (t) ´ 2 dt = f (x) ´ 1 f (x) ´ 2 Thus y = f (x) obeys the differential equation y1 = (y ´ 1)(y ´ 2).
(c) If y ‰ 1, 2, dy dx = (y ´ 1)(y ´ 2) dy (y ´ 1)(y ´ 2) = dx ż dy (y ´ 1)(y ´ 2) = ż dx Using the method of partial fractions, ż 1 y ´ 2 ´ 1 y ´ 1 dy = ż dx log |y ´ 2| ´ log |y ´ 1| = x + C log ˇ ˇ ˇ ˇ y ´ 2 y ´ 1 ˇ ˇ ˇ ˇ = x + C Observe that dy dx = (y ´ 1)(y ´ 2) ą 0 for all y ě 2. That is, f (x) is increasing at all x for which f (x) ą 2. As f (0) = 3, f (x) increases for all x ě 0, and f (x) ě 3 for all x ě 0. So we may drop the absolute value signs.
log f (x) ´ 2 f (x) ´ 1 = x + C f (x) ´ 2 f (x) ´ 1 = eCex At x = 0, f (x)´2 f (x)´1 = 1 2 so eC = 1 2.
f (x) ´ 2 f (x) ´ 1 = 1 2ex 2f (x) ´ 4 = [ f (x) ´ 1]ex [2 ´ ex] f (x) = 4 ´ ex f (x) = 4 ´ ex 2 ´ ex 591 S-35: Suppose that at time t (measured in hours starting at, say, noon), the water in the tank has height y, which is between 0 and 2 metres. At that time, the top surface of the water forms a circular disk of radius r = yp and area A(y) = πy2p. Thus, by Torricelli’s law, πy2p dy dt = ´c?y ´π c ¨ y2p´ 1 2dy = dt ż ´π c ¨ y2p´ 1 2dy = ż dt ´π c ¨ y2p+ 1 2 2p + 1 2 + d = t for some constant d. At time t = 0, the height is y = 2, so d = π c ¨ 22p+ 1 2 2p + 1 2 .
t = π c 22p+ 1 2 2p + 1 2 ´ y2p+ 1 2 2p + 1 2 = π c(2p + 1 2) 22p+ 1 2 ´ y2p+ 1 2 The time at which the height is 1 is obtained by subbing y = 1 into this formula. The time at which the height is 0 is obtained by subbing y = 0 into this formula. Thus the condition that the top half (y = 2 to y = 1) takes exactly the same amount of time to drain as the bottom half (y = 1 to y = 0) is: t(2) ´ t(1) = t(1) ´ t(0) 0 ´ t(1) = t(1) ´ t(0) t(0) = 2t(1) π c(2p + 1 2) 22p+ 1 2 ´ 02p+ 1 2 = 2 π c(2p + 1 2) 22p+ 1 2 ´ 12p+ 1 2 22p+ 1 2 = 2 22p+ 1 2 ´ 1 22p+ 1 2 = 2 ¨ 22p+ 1 2 ´ 2 2 = 22p+ 1 2 1 = 2p + 1 2 p = 1 4 S-36: (a) If we let f (t) = 0 for all t, then its average over any interval is 0, as is its root mean square.
592 (b) Let’s start by simplifying the given equation.
1 x ´ a ż x a f (t) dt = d 1 x ´ a ż x a f 2(t) dt 1 ?x ´ a ż x a f (t) dt = dż x a f 2(t) dt (3.1) d dx " 1 ?x ´ a ż x a f (t) dt = d dx #dż x a f 2(t) dt + (3.2) For the derivative on the left, we use the product rule and the Fundamental Theorem of Calculus, part 1.
d dx " 1 ?x ´ a ż x a f (t) dt = d dx " 1 ?x ´ a ż x a f (t) dt + 1 ?x ´ a ¨ d dx "ż x a f (t) dt = ´ 1 2?x ´ a3 ż x a f (t) dt + f (x) ?x ´ a = 1 ?x ´ a f (x) ´ 1 2(x ´ a) ż x a f (t) dt For the derivative on the right in Equation (3.2), we use the chain rule and the Fundamental Theorem of Calculus, part 1.
d dx #dż x a f 2(t) dt + = 1 2 ż x a f 2(t) dt ´ 1 2 ¨ d dx "ż x a f 2(t) dt = f 2(x) 2 bşx a f 2(t) dt So, Equation (3.2) yields the following: 1 ?x ´ a f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) 2 bşx a f 2(t) dt (3.3) (c) From Equation (3.1), bşx a f 2(t) dt = 1 ?x´a şx a f (t) dt.
1 ?x ´ a f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) 2 1 ?x´a şx a f (t) dt 2 x ´ a ż x a f (t) dt f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) (d) Now what we have is a differential equation, although it might not look like it. Let Y = şx a f (t) dt. Then dY dx = f (x).
2 x ´ aY dY dx ´ 1 2(x ´ a)Y = dY dx 2 (3.4) 593 We’re used to solving differential equations of the form dY dx =(something). So, let’s manipulate our equation until it has this form.
dY dx 2 ´ 2Y x ´ a dY dx + Y x ´ a 2 = 0 This is a quadratic equation, with variable dY dx . Its solutions are: dY dx = 2Y x´a ˘ c 2Y x´a 2 ´ 4 ¨ Y x´a 2 2 = 2Y x´a ˘ 0 2 = Y x ´ a This gives us the separable differential equation dY dx = Y x ´ a dY Y = dx x ´ a (3.5) ż dY Y = ż dx x ´ a log |Y| = log |x ´ a| + C |Y| = elog(x´a)+C = (x ´ a)eC Y = D(x ´ a) where D is some constant, eC or ´eC. Note this covers all real constants except D = 0.
If D = 0, then Y(x) = 0 for all x. This function also satisfies Equation (3.4), so indeed, Y(x) = D(x ´ a) (3.6) for any constant D is the family of equations satisfying our differential equation.
Remark: the reason we “lost” the solution Y(x) = 0 is that in Equation (3.5), we divided by Y, thus tacitly assuming it was not identically 0.
(e) Remember Y = şx a f (t) dt. So, Equation (3.6) tells us: ż x a f (t) dt = D(x ´ a) d dx "ż x a f (t) dt = d dxtD(x ´ a)u f (x) = D We should check that this function works.
favg = 1 x ´ a ż x a D dt = 1 x ´ a h Dt it=x t=a = Dx ´ Da x ´ a = D fRMS = d 1 x ´ a ż x a D2 dt = c 1 x ´ a h D2x it=x t=a = d D2x ´ D2a x ´ a = ?
D2 = |D| 594 So, f (x) = D works only if D is nonnegative.
That is: the only functions whose average matches their root square mean over every interval are constant, nonnegative functions.
Remark: it was step (c) where we introduced the erroneous answer f (x) = D, D ă 0 to our solution. In Equation (3.3), f (x) = D is not a solution if D ă 0: 1 ?x ´ a f (x) ´ 1 2(x ´ a) ż x a f (t) dt = f 2(x) 2 bşx a f 2(t) dt 1 ?x ´ a D ´ 1 2(x ´ a) ż x a D dt = D2 2 bşx a D2 dt 1 ?x ´ a D ´ 1 2(x ´ a)D(x ´ a) = D2 2 a D2(x ´ a) 1 ?x ´ a 1 2D = D2 2|D|?x ´ a D = D2 |D| = |D| In (c), we replace bşx a f 2(t) dt, which cannot be negative, with 1 ?x´a şx a f (t) dt, which could be negative if f (t) = D ă 0. Indeed, if f (t) = D, then bşx a f 2(t) dt = |D|?x ´ a, while 1 ?x´a şx a f (t) dt = D?x ´ a. It is at this point that negative functions creep into our solution.
S-37: We start by antidifferentiating both sides with respect to x.
ż d2y dx2 dx = ż 2 y3 ¨ dy dx dx The right integral is in exactly the form we would use for a change of variables (substitution) to y.
dy dx = ż 2 y3 dy = ´ 1 y2 + C When y = 1, dy dx = 3.
3 = ´1 1 + C C = 4 So, dy dx = ´ 1 y2 + 4 595 This is a separable differential equation.
dy dx = 4y2 ´ 1 y2 y2 4y2 ´ 1 dy = dx ż y2 4y2 ´ 1 dy = ż dx (˚) We can evaluate the left integral with partial fractions, but because the numerator has the same degree as the denominator, we have to simplify first. We do this by inspection, but you can also use long division.
y2 4y2 ´ 1 = 1 4(4y2 ´ 1) + 1 4 4y2 ´ 1 = 1 4 1 + 1 4y2 ´ 1 = 1 4 1 + 1 (2y ´ 1)(2y + 1) = 1 4 1 + 1/2 2y ´ 1 ´ 1/2 2y + 1 Now, we return to (˚).
ż dx = ż y2 4y2 ´ 1 dy = ż 1 4 1 + 1/2 2y ´ 1 ´ 1/2 2y + 1 dy = 1 4 y + 1 4 log |2y ´ 1| ´ 1 4 log |2y + 1| = 1 4 y + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ x + C = 1 4 y + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ When x = ´ 1 16 log 3, y = 1.
´ 1 16 log 3 + C = 1 4 1 + 1 4 log ˇ ˇ ˇ ˇ 2 ´ 1 2 + 1 ˇ ˇ ˇ ˇ = 1 4 + 1 16 log 1 3 C = 1 4 So, x + 1 4 = 1 4 y + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ x = 1 4 y ´ 1 + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ 596 We can check our answer by differentiating with respect to x.
x = 1 4 y ´ 1 + 1 4 log ˇ ˇ ˇ ˇ 2y ´ 1 2y + 1 ˇ ˇ ˇ ˇ 4x = y ´ 1 + 1 4 log |2y ´ 1| ´ 1 4 log |2y + 1| d dxt4xu = d dx " y ´ 1 + 1 4 log |2y ´ 1| ´ 1 4 log |2y + 1| 4 = dy dx + 1 4 ¨ 2dy dx 2y ´ 1 ´ 1 4 ¨ 2dy dx 2y + 1 4 = dy dx 1 + 1/2 2y ´ 1 ´ 1/2 2y + 1 = dy dx 4y2 4y2 ´ 1 dy dx = 4y2 ´ 1 y2 = 4 ´ 1 y2 (˚˚) Differentiating with respect to x again, using the chain rule, d2y dx2 = 2 y3 ¨ dy dx This is exactly the differential equation we were meant to solve.
Solutions to Exercises 3.1 — Jump to TABLE OF CONTENTS S-1: (a) The values of the sequence seem to be getting closer and closer to -2, so we guess the limit of this sequence is -2.
(b) Overall, the values of the sequence seem to be getting extremely close to 0, so we approximate the limit of this sequence as 0. It doesn’t matter that the sequence changes signs, or that the numbers are sometimes farther from 0, sometimes closer.
(c) This limit does not exist. The sequence is sometimes 0, sometimes -2, and not consistently staying extremely near to either one.
S-2: True. We consider the end behaviour of the sequences, which does not depend on any finite number of terms at their beginning.
S-3: (a) We follow the arithmetic of limits, Theorem 3.1.8 in the CLP-2 text: A ´ B C (b) Since lim nÑ8 cn is some real number, and n grows without bound, lim nÑ8 cn n = 0.
(c) We note lim nÑ8 a2n+5 = lim nÑ8 an, so a2n+5 bn = A B .
S-4: There are many possible answers. One is: an = # 3000 ´ n if n ď 1000 ´2 + 1 n if n ą 1000 597 where we have a series that looks different before and after its thousandth term. Note every term is smaller than the term preceding it.
Another sequence with the desired properties is: an = 1, 002, 001 n ´ 2 When n ď 1000, an ě 1,002,001 1000 ´ 2 ą 1,002,000 1,000 ´ 2 = 1000. That is, an ą 1000 when n ď 1000.
As n gets larger, an gets smaller, so an+1 ă an for all n. Finally, lim nÑ8 an = 0 ´ 2 = ´2.
S-5: One possible answer is an = (´1)n = t´1, 1, ´1, 1, ´1, 1, ´1, . . .u.
Another is an = n(´1)n = t´1, 2, ´3, 4, ´5, 6, ´7, . . .u.
S-6: If the terms of a sequence are alternating sign, but the limit of the sequence exists, the limit must be zero. (If it were a positive number, the negative terms would not get very close to it; if it were a negative number, the positive terms would not get very close to it.) This gives us the idea to modify an answer from Question 5. One possible sequence: an = (´1)n n = " ´1, 1 2, ´1 3, 1 4, ´1 5, 1 6, . . .
S-7: (a) Since ´1 ď sin n ď 1 for all n, one potential set of upper and lower bound is ´1 n ď sin n n ď 1 n Note lim nÑ8 ´1 n = lim nÑ8 1 n, so these are valid comparison sequences for the squeeze theorem.
(b) Since ´1 ď sin n ď 1 and ´5 ď ´5 cos n ď 5 for all n, we see 7´1´5 ď 7 + sin n ´ 5 cos n ď 7 + 1 + 5 1 ď 7 + sin n ´ 5 cos n ď 13 This gives us the idea to try the bounds n2 13en ď n2 en(7 + sin n ´ 5 cos n) ď n2 en We check that lim nÑ8 n2 13en = lim nÑ8 n2 en (they’re both 0–you can verify using l’Hˆ opital’s rule), so these are indeed reasonable bounds to choose to use with the squeeze theorem. Alternatively, since 0 ď n2 13en, we can also use 0 ď n2 en(7 + sin n ´ 5 cos n) ď n2 en 598 (c) Since (´n)´n = 1 (´n)n = (´1)n nn , we see ´1 nn ď (´n)´n ď 1 nn Since both lim nÑ8 ´1 nn and lim nÑ8 1 nn are 0, these are reasonable bounds to use with the squeeze theorem.
S-8: (a) • Note an = bn since n = |n| for all n ě 1. Then an = bn = 1 + 1 n = n + 1 n . So, whenever n is a whole number, an and bn are the same as h(n) and i(n). (Be careful here: h(x) ‰ i(x) when x is not a whole number.) • cn = e´n = 1 en = j(n) • For any integer n, cos(πn) = (´1)n. So, dn = f (n).
• Similarly, en = g(n).
(b) According to Theorem 3.1.6 in the CLP-2 text, if any of the functions on the right have limits that exist as x Ñ 8, then these limits match the limits of their corresponding sequences. So, we only have to be suspicious of f (x) and i(x), since these do not converge.
The limit lim xÑ8 f (x) does not exist, and f (n) = dn; the limit lim nÑ8 dn also does not exist.
(We generally don’t write equality for two things that don’t exist: equality refers to numerical value, and these have none.9) The limit lim xÑ8 i(x) does not exist, because i(x) = 0 when x is not a whole number, while i(x) approaches 1 when x is a whole number. However, lim lim nÑ8 an = lim nÑ8 bn = 1.
So, using our answers from part (a), we match the following: • lim nÑ8 an = lim nÑ8 bn = lim xÑ8 h(x) = 1 • lim nÑ8 cn = lim nÑ8 en = lim xÑ8 g(x) = lim xÑ8 j(x) = 0 • lim nÑ8 dn, lim xÑ8 f (x) and lim xÑ8 i(x) do not exist.
S-9: (a) We want to find odd multiples of π that are close to integers.
Solution 1: One way to do that is to remember that π is somewhat close to 22 7 . Then when we multiply π by a multiple of 7, we should get something close to an 9 The idea “two things that both don’t exist are equal” is also rejected because it can lead to contradictions.
For example, in the real numbers ?
´1 and ?
´2 don’t exist; if we write ?
´1 = ?
´2, then squaring both sides yields the inanity ´1 = ´2.
599 integer. In particular, 7π, 21π, and 35π should be reasonably close to 7 22 7 = 22, 21 22 7 = 66, and 35 22 7 = 110, respectively. We check whether they are close enough: 7π « 21.99 21π « 65.97 35π « 109.96 So indeed, 22, 66, and 110 are all within 0.1 of some odd multiple of π.
Since the cosine of an odd multiple of π is ´1, we expect all of the sequence values to be close to ´1. Using a calculator: a22 = cos(22) « ´0.99996, a66 = cos(66) « ´0.99965, a110 = cos(110) « ´0.99902 Solution 2: Alternately, we could have just listed odd multiple of π until we found three that are close to integers.
2k + 1 (2k + 1)π π π 1 3.14 3 9.42 5 15.71 7 21.99 9 28.27 11 34.56 13 40.84 15 47.12 17 53.41 19 59.69 21 65.97 23 72.26 25 78.54 27 84.82 29 91.11 31 97.39 33 103.67 35 109.96 Some earlier odd multiples of π (like 15π and 29π) get fairly close to integers, but not within 0.1.
(b) If x = 2k + 1 2 π for some integer k (that is, x is an odd multiple of π/2), then cos x = 0.
So, we can either list out the first few terms of an until we find three that are very close to 0, or we can use our approximation π « 22 7 to choose values of n that are close to 2k + 1 2 π.
600 Solution 1: 2k + 1 2 π « (2k + 1) ˆ 22 2 ˆ 7 = 11 ˆ 2k + 1 7 So, we expect our values to be close to integers when 2k + 1 is a multiple of 7. For example, 2k + 1 = 7, 2k + 1 = 21, and 2k + 1 = 35.
We check: x n an 7 ˆ π 2 « 10.99557 11 a11 « 0.0044 21 ˆ π 2 « 32.98672 33 a33 « ´0.0133 35 ˆ π 2 « 54.97787 55 a55 « 0.0221 These seem like values of an that are all pretty close to 0.
Solution 2: We could have listed the first several values of an, and looked for some that are close to 0.
n an 1 0.54 2 ´0.42 3 ´0.99 4 ´0.65 5 0.28 6 0.96 7 0.75 8 ´0.15 9 ´0.91 10 ´0.84 Oof. Nothing very close yet. Maybe a better way is to list values of 2k+1 2 π, and see which ones are close to integers.
601 2k + 1 2k+1 2 π π π 1 1.57 3 4.71 5 7.85 7 10.996 9 14.14 11 17.28 13 20.42 15 23.56 17 26.70 19 29.85 21 32.99 23 36.13 25 39.27 27 42.41 29 45.55 31 48.69 33 51.84 35 54.98 We find roughly the same candidates we did in Solution 1, depending on what we’re ready to accept as “close”.
(c) One can use the same strategies as we did for parts (a) and (b). But since we already know some n’s with cos(n) close to ´1, it’s easier to use the trig identity cos(2m) = 2 cos2(m) ´ 1 This identity shows that if cos(m) is close to ´1, then cos(2m) is close to +1. So let’s try n = 2m = 2 ˆ 22 = 44, 2 ˆ 66 = 132 and 2 ˆ 110 = 220.
n an 44 0.9998 132 0.9986 220 0.9961 They do the trick.
Remark: it is possible to turn the ideas of this question into a rigorous proof that lim nÑ8 cos n is undefined.
• Let, for each integer k ě 1, nk be the integer that is closest to 2kπ. Then 2kπ ´ 1 2 ď nk ď 2kπ + 1 2 so that cos(nk) ě cos 1 2 ě 0.8. Consequently, if lim nÑ8 cos n = c exists, we must have c ě 0.8.
• Let, for each integer k ě 1, n1 k be the integer that is closest to (2k + 1)π. Then (2k + 1)π ´ 1 2 ď n1 k ď (2k + 1)π + 1 2 so that cos(n1 k) ď ´ cos 1 2 ď ´0.8. Consequently, if lim nÑ8 cos n = c exists, we must have c ď ´0.8.
602 • It is impossible to have both c ě 0.8 and c ď ´0.8, so lim nÑ8 cos n does not exist.
Remark: This question also hints at a property of the set of all numbers cos(n), with n running over the integers. Mathematicians say that this set is “dense in the interval [´1, 1]”. This means that if you pick any number ´1 ď r ď 1, there is an integer n such that cos(n), while not necessarily being exactly r, is as close to r as you like. If you feed “cos(n) dense” into your favourite search engine, you can find out more.
S-10: When determining the end behaviour of rational functions, recall from last semester that we can either cancel out the highest power of n from the numerator and denominator, or skip this step and compare the highest powers of the numerator and denominator.
(a) Since the numerator has a higher degree than the denominator, this sequence will diverge to positive or negative infinity; since its terms are positive for large n, its limit is (positive) infinity. (You can imagine that the numerator is growing much, much faster than the denominator, leading the terms to have a very, very large absolute value.) Calculating the longer way: an = 3n2 ´ 2n + 5 4n + 3 1 n 1 n !
= 3n ´ 2 + 5 n 4 + 3 n lim nÑ8 an = lim nÑ8 3n ´ 2 + 5 n 4 + 3 n = lim nÑ8 3n ´ 2 + 0 4 + 0 = 8 (b) Since the numerator has the same degree as the denominator, as n goes to infinity, this sequence will converge to the ratio of their leading coefficients: 3 4. (You can imagine that the numerator is growing at roughly the same rate as the denominator, so the terms settle into an almost-constant ratio.) Calculating the longer way: bn = 3n2 ´ 2n + 5 4n2 + 3 1 n2 1 n2 !
= 3 ´ 2 n + 5 n2 4 + 3 n2 lim nÑ8 bn = lim nÑ8 3 ´ 2 n + 5 n2 4 + 3 n2 = 3 ´ 0 + 0 4 + 0 = 3 4 (c) Since the numerator has a lower degree than the denominator, this sequence will converge to 0 as n goes to infinity. (You can imagine that the denominator is growing much, much faster than the numerator, leading the terms to be very, very small.) 603 Calculating the longer way: cn = 3n2 ´ 2n + 5 4n3 + 3 1 n3 1 n3 !
= 3 n ´ 2 n2 + 5 n3 4 + 3 n3 lim nÑ8 cn = lim nÑ8 3 n ´ 2 n2 + 5 n3 4 + 3 n3 = 0 ´ 0 + 0 4 + 0 = 0 S-11: At first glance, we see both the numerator and denominator grow huge as n increases, so we’ll need to think a little further to find the limit.
We don’t have a rational function, but we can still divide the top and bottom by ne to get a clearer picture.
an = 4n3 ´ 21 ne + 1 n 1 ne 1 ne !
= 4n3´e ´ 21 ne 1 + 1 ne+1 Since e ă 3, we see 3 ´ e is positive, so lim nÑ8 n3´e = 8.
lim nÑ8 an = lim nÑ8 4n3´e ´ 21 ne 1 + 1 ne+1 = lim nÑ8 4n3´e ´ 0 1 + 0 = 8 S-12: This isn’t a rational sequence, but factoring out ?n from the top and bottom will still clear things up.
bn = 4 ?n + 1 ?
9n + 3 1 ?n 1 ?n !
= 1 4 ?n + 1 ?n b 9 + 3 n lim nÑ8 bn = lim nÑ8 1 4 ?n + 1 ?n b 9 + 3 n = 0 + 0 ?
9 + 0 = 0 S-13: First, let’s start with a tempting fallacy.
The denominator grows without bound, so lim nÑ8 cos(n + n2) n = 0.
It’s certainly true that if the limit of the numerator is a real number, and the denominator grows without bound, then the limit of the sequence is zero. However, in our case, the limit of the numerator does not exist. To apply the limit arithmetic rules from the CLP-2 text (Theorem 3.1.8), our limits must actually exist.
A better reasoning looks something like this: 604 The denominator grows without bound, and the numerator never gets very large, so lim nÑ8 cos(n + n2) n = 0.
To quantify this reasoning more precisely, we use the squeeze theorem, Theorem 3.1.10 in the CLP-2 text. There are two parts to the squeeze theorem: finding two bounding functions, and making sure these functions have the same limit.
• Since ´1 ď cos(n + n2) ď 1 for all n, we choose functions an = ´1 n and bn = 1 n. Then an ď cn ď bn for all n.
• Both lim nÑ8 an = 0 and lim nÑ8 bn = 0.
So, by the squeeze theorem, lim nÑ8 cos(n + n2) n = 0.
S-14: The denominator of this sequence grows without bound. The numerator is unpredictable: imagine that n is large. When sin n is close to ´1, nsin n puts a power of n “in the denominator,” so we can have nsin n very close to 0. When sin n is close to 1, nsin n is close to n, which is large.
To control for these variations, we’ll use the squeeze theorem.
• Since ´1 ď sin n ď 1 for all n, let bn = n´1 n2 = 1 n3 and cn = n n2 = 1 n. Then bn ď an ď cn.
• Both lim nÑ8 bn = 0 and lim nÑ8 cn = 0.
So, by the squeeze theorem, lim nÑ8 nsin n n2 = 0 as well.
Remark: we also could have used bn = 0 for our lower bound, since an ě 0 for all n.
S-15: dn = e´1/n = 1 e1/n lim nÑ8 dn = lim nÑ8 1 e1/n = 1 e0 = 1 1 = 1 S-16: Solution 1: Let’s use the squeeze theorem. Since sin(n2) and sin n are both between ´1 and 1 for all n, we note: 1 + 3(´1) ´ 2(1) ď 1 + 3 sin(n2) ´ 2 sin n ď 1 + 3(1) ´ 2(´1) ´4 ď 1 + 3 sin(n2) ´ 2 sin n ď 6 This allows us to choose suitable bounding functions for the squeeze theorem.
• Let bn = ´4 n and cn = 6 n. From the work above, we see bn ď an ď cn for all n.
605 • Both lim nÑ8 bn = 0 and lim nÑ8 cn = 0.
So, by the squeeze theorem, lim nÑ8 1 + 3 sin(n2) ´ 2 sin n n = 0.
Solution 2: We simplify slightly to begin.
an = 1 + 3 sin(n2) ´ 2 sin n n = 1 n + 3 ¨ sin(n2) n ´ 2 ¨ sin n n We apply the squeeze theorem to the pieces sin(n2) n and sin n n .
• Let bn = ´1 n and cn = 1 n. Then bn ď sin(n2) n ď cn, and bn ď sin n n ď cn.
• Both lim nÑ8 bn = 0 and lim nÑ8 cn = 0.
So, by the squeeze theorem, lim nÑ8 sin(n2) n = 0 and lim nÑ8 sin n n = 0.
Now, using the arithmetic of limits from Theorem 3.1.8 in the CLP-2 text, lim nÑ8 an = lim nÑ8 1 n + 3 ¨ sin(n2) n ´ 2 ¨ sin n n = 0 + 3 ¨ 0 ´ 2 ¨ 0 = 0 S-17: First, we note that both numerator and denominator grow without bound. So, we have to decide whether one outstrips the other, or whether they reach a stable ratio.
Solution 1: Let’s try dividing the numerator and denominator by 2n (the dominant term in the denominator; this is the same idea behind factoring out the leading term in rational expressions).
bn = en 2n + n2 1 2n 1 2n !
= e 2 n 1 + n2 2n Since e ą 2, we see e 2 ą 1, and so lim nÑ8 e 2 n = 8. Since exponential functions grow much, much faster than polynomial functions, we also see lim nÑ8 n2 2n = 0. So, lim nÑ8 bn = lim nÑ8 e 2 n 1 + n2 2n = lim nÑ8 e 2 n 1 + 0 = 8 Solution 2: Since the numerator and denominator both increase without bound, we 606 apply l’Hˆ opital’s rule. Recall d dxt2xu = 2x log 2.
lim nÑ8 bn = lim nÑ8 en 2n + n2 looomooon numÑ8 denÑ8 = lim nÑ8 en 2n log 2 + 2n looooooomooooooon numÑ8 denÑ8 = lim nÑ8 en 2n(log 2)2 + 2 looooooomooooooon numÑ8 denÑ8 = lim nÑ8 en 2n(log 2)3 = 1 (log 2)3 lim nÑ8 e 2 n = 8 Since e ą 2, we see e 2 ą 1, and so lim nÑ8 e 2 n = 8.
S-18: First, we simplify. Remember n! = n(n ´ 1)(n ´ 2) ¨ ¨ ¨ (2)(1) for any whole number n, so (k + 1)! = (k + 1)k! .
ak = k! sin3 k (k + 1)! = k! sin3 k (k + 1)k! = sin3 k k + 1 Now, we can use the squeeze theorem.
• ´1 ď sin k ď 1 for all k, so ´1 ď sin3 k ď 1. Let bk = ´1 k+1 and ck = 1 k+1. Then bk ď ak ď ck.
• Both lim kÑ8 bk = 0 and lim kÑ8 ck = 0.
So, by the squeeze theorem, also lim kÑ8 ak = 0.
S-19: Note lim nÑ8(´1)n doesn’t exist, but ´1 ď (´1)n ď 1 for all n. Let’s use the squeeze theorem.
• Let an = ´ sin 1 n and bn = sin 1 n . Then an ď (´1)n sin 1 n ď bn.
• Both lim nÑ8 ´ sin 1 n = 0 and lim nÑ8 sin 1 n = 0, since lim nÑ8 1 n = 0 and sin 0 = 0.
By the squeeze theorem, the sequence !
(´1)n sin 1 n ) converges to 0.
607 S-20: First, we note that lim nÑ8 6n2 + 5n n2 + 1 = 6. We see this either by comparing the leading terms in the numerator and denominator, or by factoring out n2 from the top and the bottom.
Second, since lim nÑ8 1 n2 = 0, we see lim nÑ8 cos 1 n2 = cos 0 = 1.
Using arithmetic of limits, Theorem 3.1.8 in the CLP-2 text, we conclude lim nÑ8 6n2 + 5n n2 + 1 + 3 cos(1/n2) = 6 + 3(1) = 9.
S-21: Let’s take stock: sin(1/n) Ñ sin(0) = 0 as n Ñ 8, so log (sin(1/n)) Ñ ´8.
However, log(2n) Ñ 8. So, we have some tension here: the two pieces behave in ways that pull the terms of the sequence in different directions. (Recall we cannot conclude anything like “´8 + 8 = 0.”) We try using logarithm rules to get a clearer picture.
log sin 1 n + log(2n) = log 2n sin 1 n Still, we have indeterminate behaviour: 2n sin(1/n) is the product of 2n, which grows without bound, and sin(1/n), which approaches zero. In the past, we learned that we can handle the indeterminate form 0 ¨ 8 with l’Hˆ opital’s rule (after a little algebra), but there’s a slicker way. Note 1/n Ñ 0 as n Ñ 8. If we write 1 n = x, then this piece of our limit resembles something familiar.
2n sin 1 n = 2 sin x x If n Ñ 8, then x = 1 n Ñ 0.
lim nÑ8 2n sin 1 n = 2 lim xÑ0 sin x x That limit is familiar: = 2(1) = 2 Then: lim nÑ8 log 2n sin 1 n = log 2 Note: if you have forgotten that lim xÑ0 sin x x = 1, you can also evaluate this limit using l’Hˆ opital’s rule: lim xÑ0 sin x x loooomoooon numÑ0 denÑ0 = lim xÑ0 cos x 1 = cos 0 = 1 608 S-22: First, although this sequence is not defined for some small values of n, it is defined as long as n ě 5, so it’s not a problem to take the limit as n Ñ 8. Second, we notice that our limit has the indeterminate form 8 ´ 8. Since this form is indeterminate, more work is needed to find our limit, if it exists.
A standard trick we saw last semester with functions of this form was to multiply and divide by the conjugate of the expression, ?
n2 + 5n + ?
n2 ´ 5n. Then the denominator will be the sum of two similar things, rather than their difference. See the work below to find out why that is helpful.
a n2 + 5n ´ a n2 ´ 5n = a n2 + 5n ´ a n2 ´ 5n ?
n2 + 5n + ?
n2 ´ 5n ?
n2 + 5n + ?
n2 ´ 5n !
= (n2 + 5n) ´ (n2 ´ 5n) ?
n2 + 5n + ?
n2 ´ 5n = 10n ?
n2 + 5n + ?
n2 ´ 5n Now, we’ll cancel out n from the top and the bottom. Note n = ?
n2.
= 10n ?
n2 + 5n + ?
n2 ´ 5n 1 n 1 n !
= 10n ?
n2 + 5n + ?
n2 ´ 5n 1 n 1 ?
n2 !
= 10 b 1 + 5 n + b 1 ´ 5 n Now, the limit is clear.
lim nÑ8 10 b 1 + 5 n + b 1 ´ 5 n = 10 ?
1 + 0 + ?
1 + 0 = 10 1 + 1 = 5 S-23: First, although this sequence is not defined for some small values of n, it is defined as long as n ě ?
2.5, so it’s not a problem to take the limit as n Ñ 8. Second, we notice that our limit has the indeterminate form 8 ´ 8. Since this form is indeterminate, more work is needed to find our limit, if it exists.
In Question 22, we saw a similar limit, and made use of the conjugate. However, in this 609 case, there’s an easier path: let’s factor out n from each term.
a n2 + 5n ´ a 2n2 ´ 5 = d n2 1 + 5 n ´ d n2 2 ´ 5 n2 = n c 1 + 5 n ´ n c 2 ´ 5 n2 = n c 1 + 5 n ´ c 2 ´ 5 n2 !
Now, the limit is clear.
lim nÑ8 ha n2 + 5n ´ a 2n2 ´ 5 i = lim nÑ8 " n c 1 + 5 n ´ c 2 ´ 5 n2 !# = lim nÑ8 h n ?
1 + 0 ´ ?
2 ´ 0 i = lim nÑ8 [n (´1)] = ´8 Remark: check Question 22 to see whether a similar trick would work there. Why or why not?
S-24: First, we note that we have in indeterminate form: as n grows, 2 + 1 n Ñ 2, so n 2 + 1 n 100 ´ 2100 has the form 8 ¨ 0. To overcome this difficulty, we could use some algebra and l’Hˆ opital’s rule, but there’s a slicker way. If we let h = 1 n, then h Ñ 0 as n Ñ 8, and our limit looks like: lim nÑ8 n " 2 + 1 n 100 ´ 2100 # = lim hÑ0 (2 + h)100 ´ 2100 h This reminds us of the definition of a derivative.
d dx !
x100) = lim hÑ0 (x + h)100 ´ x100 h So, if we set f (x) = x100, our limit is simply f 1(2). That is, 100x99 x=2 = 100 ¨ 299.
S-25: Using the definition of a derivative, f 1(a) = lim hÑ0 f (a + h) ´ f (a) h We want n Ñ 8, so we set h = 1 n.
= lim 1 nÑ0 f a + 1 n ´ f (a) 1 n = lim nÑ8 n f a + 1 n ´ f (a) 610 We also could have chosen h = ´ 1 n, which leads to the following: lim hÑ0 f (a + h) ´ f (a) h = lim ´ 1 nÑ0 f a ´ 1 n ´ f (a) ´1/n = lim nÑ8 ´n f a ´ 1 n ´ f (a) = lim nÑ8 n f (a) ´ f a ´ 1 n S-26: (a) To find the area An, note that the figure with n sides can be divided up into n isosceles triangles, each with two sides of length 1 and angle between them of 2π n : 1 1 2π n Each of these triangles has area 1 2 sin 2π n : 2π n sin 2π n 1 1 All together, the area of the n-sided figure is An = n 2 sin 2π n .
(b) We will discuss two ways to find lim nÑ8 An, which has the indeterminate form 8 ˆ 0.
First, note that as n Ñ 8, our figures look more and more like a circle of radius 1. So, we see An is approaching the area of a circle of radius 1. That is, lim nÑ8 An = π.
Alternately, we can make use of the limit lim xÑ0 sin x x = 1. Let x = 2π n . Note if n Ñ 8, then x Ñ 0.
lim nÑ8 An = lim nÑ8 n 2 sin 2π n = lim nÑ8 π 2π n sin 2π n = lim xÑ0 πsin x x = π ˆ 1 = π 611 S-27: (a) f2(x) = # 1 2 ď x ă 3 0 else x y 1 2 3 (b) f3(x) = # 1 3 ď x ă 4 0 else x y 1 4 3 (c) For any n, fn(x) = 1 for an interval of length 1, and fn(x) = 0 for all other x. So, the area under the curve is a square of side length one.
x y 1 Then An = ş8 0 fn(x) dx = 1 for all n. That is, the sequence tAnu is simply t1, 1, . . . , 1u, a sequence of all 1s.
(d) Given the description above, lim nÑ8 An = 1.
(e) For any fixed x, recall tfn(x)u = t0, . . . , 0, 1, 0, . . . 0, 0, 0, 0, 0, . . .u. In particular, there are infinitely many zeroes at its end. So, lim nÑ8 fn(x) = 0. Then g(x) = 0 for every x.
(f) Given the description above, ż 8 0 g(x) dx = ż 8 0 0 dx = 0.
Remark: what we’ve shown here is that, for this particular fn(x), lim nÑ8 ż 8 0 fn(x) dx ‰ ż 8 0 lim nÑ8 fn(x) dx That is, we can’t necessarily swap a limit with an integral (which is, in this case, another 612 limit, since the integral is improper). The interested reader can look up “uniform convergence” to learn about the conditions under which these can be swapped.
S-28: If we naively try to find the limit, we run up against the indeterminate form 18.
We’d like to use l’Hˆ opital’s rule, but we don’t have the form 8 8 or 0 0–we’ll need to use a logarithm. Additionally, l’Hˆ opital’s rule applies to differentiable functions defined for real numbers–so we’ll consider a function, rather than the sequence.
Note the terms of the sequence are all positive.
Solution 1: Define x = 1 n, and f (x) = 1 + 3x + 5x21/x. Then bn = f 1 n = f (x), and lim nÑ8 f 1 n = lim xÑ0+ f (x).
If this limit exists, it is equal to lim nÑ8 bn.
lim xÑ0+ f (x) = lim xÑ0+ 1 + 3x + 5x21/x lim xÑ0+ log[ f (x)] = lim xÑ0+ log 1 + 3x + 5x21/x = lim xÑ0+ log 1 + 3x + 5x2 x loooooooooooooomoooooooooooooon numÑ0 denÑ0 = lim xÑ0+ 3+10x 1+3x+5x2 1 = 3 lim xÑ0+ f (x) = e3 Since the limit exists, lim nÑ8 bn = e3.
Solution 2: If we didn’t see the nice simplifying trick of letting x = 1 n, we can still solve the problem using g(x) = 1 + 3 x + 5 x2 x : g(x) = 1 + 3 x + 5 x2 x log[g(x)] = x log 1 + 3 x + 5 x2 = log h 1 + 3 x + 5 x2 i 1/x looooooooomooooooooon numÑ0 denÑ0 lim xÑ8 log[g(x)] = lim xÑ8 ´ 3 x2 ´ 10 x3 1+ 3 x + 5 x2 ´1 x2 = lim xÑ8 x2 3 x2 + 10 x3 1 + 3 x + 5 x2 = lim xÑ8 3 + 10 x 1 + 3 x + 5 x2 = 3 + 0 1 + 0 + 0 = 3 lim xÑ8 f (x) = e3 Since the limit exists, lim nÑ8 bn = e3 613 S-29: (a) When a1 = 4, we see a2 = 4 + 8 3 = 4, and so on. That is, an = 4 for every n. So, lim nÑ8 an = 4.
(b) Cross-multiplying, we see 3x = x + 8, hence x = 4.
(c) In order for our sequence to converge to 4, the terms should be getting infinitely close to 4. So, we find the relationship between an+1 ´ 4 and an ´ 4.
an+1 = an + 8 3 an+1 ´ 4 = an + 8 3 ´ 4 = an ´ 4 3 So, the distance between our sequence terms and the number 4 is decreasing by a factor of 3 each term. This implies that the terms get infinitely close to 4 as n grows.
That is, lim nÑ8 an = 4.
S-30: (a) Since w1 has the highest frequency, w2 has the next-highest frequency, and so on, we know f1 is larger than the other members of its sequence, f2 is the next largest, etc.
So, tfnu is a decreasing sequence.
(b) The most-used word in a language is w1, while the n-th most used word in a language is wn. So, we re-state the law as: f1 = n fn Then we can rewrite this fomula a little more naturally as fn = 1 n f1.
(c) Then f3 = 1 3 f1. In this case, we expect the third-most used word to account for 1 3(6%) = 2% of all words.
(d) From (b), we know f10 = 1 10 f1. Note f1 = 6f6 = 6(0.3%). Then: f10 = 1 10 f1 = 1 106f6 = 1 10(6)(0.3%) = 1.8 10 % = 0.18% So, f10 should be 0.18% of all words.
(e) The use of the word “frequency” in the statement of Zipf’s law implies fn = uses of wn total number of words. The question asks for the total uses of wn. If we call this quantity tn, and the total number of all words is T, then Zipf’s law tells us tn T = 1 n t1 T , hence tn = 1 nt1.
With this notation, the problem states t1 = 22, 038, 615, w1 = the, w2 = be, and w3 = and.
614 Following Zipf’s law, tn = 1 nt1. So, we expect t2 = t1 2 = 11, 019, 307.5; since this isn’t an integer, let’s say we expect t2 « 11, 019, 308. Similarly, we expect t3 = t1 3 = 7, 346, 205.
Remark: The 450-million-word source material that used “the” 22,038,615 times also contained 12,545,825 instances of “be,” and 10,741,073 instances of “and.” While Zipf’s Law might be a nice model for our data overall, in these few instances it does not appear to be extremely accurate.
Solutions to Exercises 3.2 — Jump to TABLE OF CONTENTS S-1: The Nth term of the sequence of partial sums, SN, is the sum of the first N terms of the series 8 ÿ n=1 1 n.
N SN 1 1 2 1 + 1 2 3 1 + 1 2 + 1 3 4 1 + 1 2 + 1 3 + 1 4 5 1 + 1 2 + 1 3 + 1 4 + 1 5 S-2: If there were a total of 17 cookies before Student 11 came, and 20 cookies after, then Student 11 brought 3 cookies.
C10 C11 1 2 3 4 5 6 7 8 9 10 11 S-3: (a) We find tanu from tSNu using the same logic as Question 2. SN is the sum of the first N terms of tanu, and SN´1 is the sum of all the same terms except aN. So, aN = SN ´ SN´1 when N ě 2. Written another way: SN = a1 + a2 + a3 + ¨ ¨ ¨ + aN´2 + aN´1 + aN SN´1 = a1 + a2 + a3 + ¨ ¨ ¨ + aN´2 + aN´1 615 So, SN ´ SN´1 = h a1 + a2 + a3 + ¨ ¨ ¨ + aN´2 + aN´1 + aN i ´ h a1 + a2 + a3 + ¨ ¨ ¨ + aN´2 + aN´1 i = aN So, we calculate aN = SN ´ SN´1 = N N + 1 ´ N ´ 1 N ´ 1 + 1 = N2 N(N + 1) ´ N2 ´ 1 N(N + 1) = 1 N(N + 1) Therefore, an = 1 n(n + 1) Remark: the formula given for SN has S0 = 0, which makes sense: the sum of no terms at all should be 0. However, it is common for a sequence of partial sums to start at N = 1. (This fits our definition of a partial sum–we don’t really define the “sum of no terms.”) In this case, a1 must be calculated separately from the other terms of tanu. To find a1, we simply set a1 = S1, which (to reiterate) might not be the same as S1 ´ S0.
(b) lim nÑ8 an = lim nÑ8 1 n(n + 1) = 0.
That is, the terms we’re adding up are getting very, very small as we go along.
(c) By Definition 3.2.3 in the CLP-2 text, 8 ÿ n=1 an = lim NÑ8 SN = lim NÑ8 N N + 1 = 1 That is, as we add more and more terms of our series, our cumulative sum gets very, very close to 1.
S-4: As in Question 3, aN = SN ´ SN´1 = (´1)N + 1 N ´ (´1)N´1 + 1 N ´ 1 = (´1)N ´ (´1)N´1 + 1 N ´ 1 N ´ 1 = (´1)N + (´1)N + N ´ 1 N(N ´ 1) ´ N N(N ´ 1) = 2(´1)N ´ 1 N(N ´ 1) 616 Note, however, that aN is only the same as SN ´ SN´1 when N ě 2: otherwise, we’re trying to calculate S1 ´ S0, but S0 is not defined. So, we find a1 separately: a1 = S1 = (´1)1 + 1 1 = 0 All together: an = # 0 if n = 1 2(´1)n ´ 1 n(n´1) else S-5: If f 1(N) ă 0, that means f (N) is decreasing. So, adding more terms makes for a smaller sum. That means the terms we’re adding are negative. That is, an ă 0 for all n ě 2.
S-6: (a) To generate the pattern, we repeat the following steps: • divide the top triangle into four triangles of equal area, • colour the bottom two of them black, and • leave the middle one white.
Every time we repeat this sequence, we divide up a triangle with an area one-quarter the size of our previous triangle, and take two of the four resulting pieces. So, our area should end up as a geometric sum with common ratio r = 1 4, and coefficient a = 2. This is shown more explicitly below.
Since the entire triangle (outlined in red) has area 1, the four smaller triangles below each have area 1 4. The two black triangles will be added to our total black area; the blue triangle will be subdivided.
1 4 1 4 The blue triangle had area 1 4, so each of the small black triangles below has area 1 4 1 4 = 1 42.
617 1 4 1 4 1 42 1 42 Each time we make another subdivision, we add two black triangles, each with 1 4 the area of the previous black triangles. So, our total black area is: 2 1 4 + 2 1 42 + 2 1 43 + 2 1 44 + ¨ ¨ ¨ = 8 ÿ n=1 2 4n (b) To evalutate the series, we imagine gathering up all our little triangles and sorting them into three identical piles: the bottom three triangles go in three different piles, the three triangles directly above them go in three different piles, etc. (In the picture below, different colours correspond to different piles.) Since the piles all have equal area, each pile has a total area of 1 3. The black area shaded in the problem corresponds to two piles (red and blue above), so 8 ÿ n=1 2 4n = 2 3 S-7: (a) The pattern can be described as follows: divide the innermost square into 9 equal parts (a 3 ˆ 3 grid), choose one square to be black, and another square to subdivide.
618 The area of the red (outermost) square is 1, so the area of the largest black square is 1 9.
The area of the central, blue square below is also 1 9.
1 9 1 9 When we subdivide the blue square, the subdivisions each have one-ninth its area, or 1 92.
1 9 1 92 1 92 We continue taking squares that are one-ninth the area of the previous square. So, our total black area is 1 9 + 1 92 + 1 93 + ¨ ¨ ¨ = 8 ÿ n=1 1 9n (b) If we cut up this square along the marks, we can easily share it equally among 8 friends: there are eight squares of area 1 9 along the outer ring, eight squares of area 1 92 along the next ring in, and so on.
619 Since the eight friends all get the same total area, the area each friend gets is 1 8. The area shaded in black in the question corresponds to the pile given to one friend. So, 8 ÿ n=1 1 9n = 1 8 S-8: If we start with a shape of area 1, and iteratively divide it into thirds, taking one of the three newly created pieces each time, then the area we take will be equal to the desired series, 8 ÿ n=1 1 3n .
One way to do this is to start with a rectangle, make three vertical strips, then keep the left strip and subdivide the middle strip.
We see that the total area we take approaches one-half the total area of the figure, so 8 ÿ n=1 1 3n = 1 2.
Alternately, instead of always taking vertical strips, we could alternate vertical and horizontal slices.
620 In this setup, we notice that our strips come in pairs: two large vertical strips, two smaller horizontal strips, two smaller vertical strips, etc. We shaed exactly one of each, so the shaded area is one-half the total area: 8 ÿ n=1 1 3n = 1 2.
Other solutions are possible, as well.
S-9: Equation 3.2.1 in the CLP-2 text tells us N ÿ n=0 arn = a1 ´ rN+1 1 ´ r , for r ‰ 1. Our geometric sum has a = 1, r = 1 5, and N = 100. So: 100 ÿ n=0 1 5n = 1 ´ 1 5101 1 ´ 1 5 = 5101 ´ 1 4 ¨ 5100 S-10: After twenty students have brought their cookies, the pile numbers 53 cookies. 17 of these cookies were brought by students one through ten. So, the remainder (53 ´ 17 = 36) is the number of cookies brought by students 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20, together.
621 C10 C20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 S-11: Solution 1: Using the ideas of Question 10, we see: 100 ÿ n=50 1 5n = 100 ÿ n=0 1 5n ´ 49 ÿ n=0 1 5n That is, we want start with the sum of all the terms up to 1 5100, and then subtract off the ones we actually don’t want, which is everything up to 1 549. Now, both series are in a form appropriate for Equation 3.2.1 in the CLP-2 text.
100 ÿ n=0 1 5n ´ 49 ÿ n=0 1 5n = 1 ´ 1 5101 1 ´ 1 5 ´ 1 ´ 1 550 1 ´ 1 5 = 5101 ´ 1 4 ¨ 5100 ´ 550 ´ 1 4 ¨ 549 551 551 = 5101 ´ 1 4 ¨ 5100 ´ 5101 ´ 551 4 ¨ 5100 = 551 ´ 1 4 ¨ 5100 Solution 2: If we write out the first few terms of our series, we see we can factor out a constant to change the starting index.
100 ÿ n=50 1 5n = 1 550 + 1 551 + 1 552 + 1 553 + ¨ ¨ ¨ + 1 5100 = 1 550 1 50 + 1 51 + 1 52 + 1 53 + ¨ ¨ ¨ + 1 550 = 50 ÿ n=0 1 550 ¨ 1 5n 622 Now, our sum is in the form of Equation 3.2.1 in the CLP-2 text with a = 1 550, r = 1 5, and N = 50.
50 ÿ n=0 1 550 ¨ 1 5n = 1 550 ¨ 1 ´ 1 551 1 ´ 1 5 = 1 ´ 1 551 4 ¨ 549 551 551 = 551 ´ 1 4 ¨ 5100 S-12: (a) The table below is a record of our account, with black entries representing the money your friend gives you, and red entries representing the money you give them (which is why the red entries are negative).
d ´ 1 d+1 1 d total after day d 1 ´1 2 1 1 2 2 ´1 3 1 2 2 3 3 ´1 4 1 3 3 4 4 ´1 5 1 4 4 5 5 ´1 6 1 5 5 6 6 ´1 7 1 6 6 7 After the exchange of day n, the amount you’re left with is $ 1 ´ 1 n+1 . We see this by the cancellation in the table: the $1 2 you gave your friend on day 1 was returned on day 2; the $1 3 you gave your friend on day 2 was returned on day 3, etc.
So, after a long time, you’ll have gained close to (but always slightly less than) one dollar.
(b) The series 8 ÿ d=1 1 d ´ 1 (d + 1) describes the scenario in (a), so by our reasoning there, 8 ÿ d=1 1 d ´ 1 (d + 1) = lim nÑ8 1 ´ 1 n + 1 = 1 (c) Again, let’s set up an account book.
623 d d + 1 ´(d + 2) total 1 2 ´3 ´1 2 3 ´4 ´2 3 4 ´5 ´3 4 5 ´6 ´4 5 6 ´7 ´5 6 7 ´8 ´6 By day d, you’ve lost $ d to your so-called friend. As time goes on, you lose more and more.
(d) The series 8 ÿ d=1 ((d + 1) ´ (d + 2)) exactly describes the scenario in part (c), so it diverges to ´8. You can also see this by writing 8 ÿ d=1 ((d + 1) ´ (d + 2)) = 8 ÿ d=1 (´1) = ´1 ´ 1 ´ 1 ´ 1 ´ 1 ´ ¨ ¨ ¨ .
Be careful to avoid a common mistake with telescoping series: if we look back at our account book, we see that every negative term will cancel with a positive term, with the initial +2 as the only term that never cancels. Your friend takes $3, which they return the next day; then they take $4, which they return the next day; then they take $5, which they return the next day, and so on. It’s extremely tempting to say that the series adds up to $2, since every other term cancels out eventually. This is where we lean on Definition 3.2.3 in the CLP-2 text: we evaluate the partial sums, which always leave your friend’s last withdrawal unreturned. This definition makes sense: saying “I gained two bucks from this exchange” doesn’t really capture the reality of your increasing debt.
S-13: Using arithmetic of series, Theorem 3.2.8 in the CLP-2 text, we see 8 ÿ n=1 (an + bn + cn+1) = A + B + 8 ÿ n=1 cn+1 The question remaining is what do to with the last series. If we write out the terms, we see the difference between 8 ÿ n=1 cn and 8 ÿ n=1 cn+1 is simply that the latter is missing c1: 8 ÿ n=1 cn+1 = c2 + c3 + c4 + c5 + ¨ ¨ ¨ = ´c1 + c1 + c2 + c3 + c4 + c5 + ¨ ¨ ¨ = ´c1 + 8 ÿ n=1 cn 624 So, 8 ÿ n=1 (an + bn + cn+1) = A + B + C ´ c1 S-14: Theorem 3.2.8 in the CLP-2 text, arithmetic of series, doesn’t mention division, because in general it doesn’t work the way the question suggests. For example, let tanu = tbnu = 1 2n. Then: • ř8 n=0 an = ř8 n=0 bn = 1 1´ 1 2 = 2, while • ř8 n=0 an bn = ř8 n=0 1 = 8.
For the statement in the question, we can take tanu = tbnu = 1 2n, A = B = 2, tcnu = t0, 0, 0, . . .u, and C = 0. We see the statement is false in this case.
So, in general, the statement given is false.
S-15: We recognize that this is a geometric series: 1 + 1 3 + 1 9 + 1 27 + 1 81 + 1 243 + ¨ ¨ ¨ = 1 30 + 1 31 + 1 32 + 1 33 + 1 34 + 1 35 + ¨ = 8 ÿ n=0 1 3n Using Equation 3.2.2 in the CLP-2 text with r = 1 3 and a = 1, = 1 1 ´ 1 3 = 3 2.
S-16: This is a geometric series, with ratio r = 1 8. However, it doesn’t start at k = 0, which is what we’re used to.
Solution 1: We write out the first few terms of the series to figure out a convenient constant to factor out.
8 ÿ k=7 1 8k = 1 87 + 1 88 + 1 89 + ¨ ¨ ¨ = 1 87 1 80 + 1 81 + 1 82 + ¨ ¨ ¨ = 8 ÿ k=0 1 87 ¨ 1 8n 625 We now evaluate the series using Equation 3.2.2 in the CLP-2 text with r = 1 8, a = 1 87.
= 1 87 ¨ 1 1 ´ 1 8 = 1 7 ˆ 86 Solution 2: Using the idea of Question 10, we express the series we’re interested in as the difference of two series that we can easily evaluate.
8 ÿ k=7 1 8k = 8 ÿ k=0 1 8k ´ 6 ÿ k=0 1 8k Using Equations 3.2.2 and 3.2.1 in the CLP-2 text, = 1 1 ´ 1 8 ´ 1 ´ 1 87 1 ´ 1 8 = 1 7 ˆ 86 S-17: We recognize this as a telescoping series.
k 6 k2 ´ 6 (k+1)2 sk 1 6 ´6 4 6 ´ 6 4 2 6 4 ´6 9 6 ´ 6 9 3 6 9 ´ 6 16 6 ´ 6 16 4 6 16 ´ 6 25 6 ´ 6 25 5 6 25 ´ 6 36 6 ´ 6 36 6 6 36 ´ 6 47 6 ´ 6 47 .
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When we compute the nth partial sum, i.e. the sum of of the first n terms, successive terms cancel and only the first half of the first term, 6 k2 ´ 6 (k+1)2 ˇ ˇ ˇ k=1, and the second half of the nth term, 6 k2 ´ 6 (k+1)2 ˇ ˇ ˇ k=n, survive. That is: sn = n ÿ k=1 6 k2 ´ 6 (k + 1)2 = 6 12 ´ 6 (n + 1)2 Therefore, we can see directly that the sequence of partial sums tsnu is convergent: lim nÑ8 sn = lim nÑ8 6 ´ 6 (n + 1)2 = 6 626 By Definition 3.2.3 in the CLP-2 text the series is also convergent, with limit 6.
S-18: We recognize that this is a telescoping series, and set up a table to find the sequence of partial sums.
n cos ß n ´ cos ß n+1 sn 3 cos π 3 ´ cos π 4 1 2 ´ cos π 4 4 cos π 4 ´ cos π 5 1 2 ´ cos π 5 5 cos π 5 ´ cos π 6 1 2 ´ cos π 6 6 cos π 6 ´ cos π 7 1 2 ´ cos π 7 7 cos π 7 ´ cos π 8 1 2 ´ cos π 8 8 cos π 8 ´ cos π 9 1 2 ´ cos π 9 .
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In the partial sum sN = N ÿ n=3 cos π n ´ cos π n + 1 every term cancels except the first part of the first term (cos π 3 = 1 2) and the second part of the last term (´ cos( π N+1)). So sN = cos π 3 ´ cos π N + 1 = 1 2 ´ cos π N + 1 .
As N Ñ 8, the argument π N+1 converges to 0, and cos x is continuous at x = 0. By Definition 3.2.3 in the CLP-2 text, the value of the series is lim NÑ8 sN = lim NÑ8 1 2 ´ cos π N + 1 = 1 2 ´ cos(0) = ´1 2 S-19: (a) As in Question 2, since sn´1 = a1 + a2 + ¨ ¨ ¨ + an´1 sn = a1 + a2 + ¨ ¨ ¨ + an´1 + an 627 we can find an by subtracting: an = sn ´ sn´1 = 1 + 3n 5 + 4n ´ 1 + 3(n ´ 1) 5 + 4(n ´ 1) = 3n + 1 4n + 5 ´ 3n ´ 2 4n + 1 = (3n + 1)(4n + 1) ´ (3n ´ 2)(4n + 5) (4n + 1)(4n + 5) = 11 16n2 + 24n + 5 (b) Using Definition 3.2.3 in the CLP-2 text, 8 ÿ n=1 an = lim nÑ8 sn = lim nÑ8 1 + 3n 5 + 4n = lim nÑ8 1/n + 3 5/n + 4 = 0 + 3 0 + 4 = 3 4 The series converges to 3 4.
S-20: What we have is a geometric series, but we need to get it into the proper form before we can evaluate it.
8 ÿ n=2 3 ¨ 4n+1 8 ¨ 5n = 8 ÿ n=2 3 ¨ 4 ¨ 4n 8 ¨ 5n = 3 2 8 ÿ n=2 4 5 n Solution 1: If we factor our 4 5 2 , we can change our index to something more convenient.
3 2 8 ÿ n=2 4 5 n = 3 2 8 ÿ n=2 4 5 2 4 5 n´2 = 3 2 8 ÿ n=0 4 5 2 4 5 n We use Equation 3.2.2 in the CLP-2 text with r = 4 5.
= 3 2 4 5 2 ¨ 1 1 ´ 4 5 = 24 5 Solution 2: Using the idea of Question 10, we view our series as a more convenient series, minus a few initial terms.
3 2 8 ÿ n=2 4 5 n = 3 2 " 8 ÿ n=0 4 5 n# ´ 4 5 1 ´ 4 5 0!
= 3 2 8 ÿ n=0 4 5 n ´ 9 5 !
628 We use Equation 3.2.2 in the CLP-2 text with r = 4 5.
= 3 2 1 1 ´ 4 5 ´ 9 5 !
= 24 5 S-21: The number is: 0.2 + 3 100 + 3 1000 + 3 10000 + ¨ ¨ ¨ = 1 5 + 3 102 + 3 103 + 3 104 + ¨ ¨ ¨ = 1 5 + 3 102 1 100 + 1 101 + 1 102 + ¨ ¨ ¨ = 1 5 + 3 102 8 ÿ n=0 1 10n We use Equation 3.2.2 in the CLP-2 text with r = 1 10.
= 1 5 + 3 102 ¨ 1 1 ´ 1 10 = 1 5 + 1 30 = 7 30 S-22: The number is: 2 + 65 100 + 65 10000 + 65 1000000 + ¨ ¨ ¨ = 2 + 65 100 + 65 1002 + 65 1003 + ¨ ¨ ¨ = 2 + 65 100 8 ÿ n=0 1 100n We use Equation 3.2.2 in the CLP-2 text with r = 1 100.
= 2 + 65 100 ¨ 1 1 ´ 1 100 = 2 + 65 99 = 263 99 S-23: The number is: 0.321 = 0.321321321 . . .
= 321 1000 + 321 106 + 321 109 + ¨ ¨ ¨ = 321 1000 1 103 0 + 1 103 1 + 1 103 2 + ¨ ¨ ¨ !
= 321 1000 8 ÿ n=0 1 103 n 629 We use Equation 3.2.2 in the CLP-2 text with r = 1 103.
= 321 1000 ¨ 1 1 ´ 1 103 = 321 999 = 107 333 S-24: We split the sum into two parts.
8 ÿ n=2 2n+1 3n + 1 2n ´ 1 ´ 1 2n + 1 = 8 ÿ n=2 2n+1 3n + 8 ÿ n=2 1 2n ´ 1 ´ 1 2n + 1 The first part is a geometric series.
8 ÿ n=2 2n+1 3n = 8 ÿ n=0 2n+3 3n+2 = 8 ÿ n=0 23 32 ¨ 2 3 n We use Equation 3.2.2 in the CLP-2 text with r = 2 3 and a = 8 9.
= 8 9 ¨ 1 1 ´ 2 3 = 8 3 The second part is a telescoping series. Let’s make a table to see how it cancels.
n 1 2n´1 ´ 1 2n+1 sn 2 1 3 ´1 5 1 3 ´ 1 5 3 1 5 ´1 7 1 3 ´ 1 7 4 1 7 ´1 9 1 3 ´ 1 9 5 1 9 ´ 1 11 1 3 ´ 1 11 6 1 11 ´ 1 13 1 3 ´ 1 13 7 1 13 ´ 1 15 1 3 ´ 1 15 .
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After adding terms n = 2 through n = N, the partial sum is sN = 1 3 ´ 1 2N + 1 630 because all the terms except the first part of the n = 2 term, and the last part of the n = N term, cancel. Then: 8 ÿ n=2 1 2n ´ 1 ´ 1 2n + 1 = lim NÑ8 sN = lim NÑ8 1 3 ´ 1 2N + 1 = 1 3 All together, 8 ÿ n=2 2n+1 3n + 1 2n ´ 1 ´ 1 2n + 1 = 8 ÿ n=2 2n+1 3n + 8 ÿ n=2 1 2n ´ 1 ´ 1 2n + 1 = 8 3 + 1 3 = 3 S-25: We split the sum into two parts.
8 ÿ n=1 1 3 n + ´ 2 5 n´1 = 8 ÿ n=1 1 3 n + 8 ÿ n=1 ´ 2 5 n´1 Both are geometric series.
= 8 ÿ n=0 1 3 n+1 + 8 ÿ n=0 ´ 2 5 n = 1 3 8 ÿ n=0 1 3 n + 8 ÿ n=0 ´ 2 5 n We use Equation 3.2.2 in the CLP-2 text with a1 = 1 3 and r1 = 1 3, then with a2 = 1 and r2 = ´2 5.
= 1 3 ¨ 1 1 ´ 1 3 + 1 1 + 2 5 = 1 2 + 5 7 = 17 14 S-26: We split the sum into two parts.
8 ÿ n=0 1 + 3n+1 4n = 8 ÿ n=0 1 4n + 8 ÿ n=0 3n+1 4n = 8 ÿ n=0 1 4n + 3 8 ÿ n=0 3 4 n 631 Using Equation 3.2.2 in the CLP-2 text, = 1 1 ´ 1 4 + 3 1 ´ 3 4 = 4 3 + 12 = 40 3 S-27: Using logarithm rules, we see 8 ÿ n=5 log n ´ 3 n = 8 ÿ n=5 log(n ´ 3) ´ log n which looks like a telescoping series. Let’s make a table to figure out the partial sums.
n log(n ´ 3) ´ log n sn 5 log 2 ´ log 5 log 2 ´ log 5 6 log 3 ´ log 6 log 2 + log 3 ´ log 5 ´ log 6 7 log 4 ´ log 7 log 2 + log 3 + log 4 ´ log 5 ´ log 6 ´ log 7 8 log 5 ´ log 8 log 2 + log 3 + log 4 ´ log 6 ´ log 7 ´ log 8 9 log 6 ´ log 9 log 2 + log 3 + log 4 ´ log 7 ´ log 8 ´ log 9 10 log 7 ´ log 10 log 2 + log 3 + log 4 ´ log 8 ´ log 9 ´ log 10 11 log 8 ´ log 11 log 2 + log 3 + log 4 ´ log 9 ´ log 10 ´ log 11 .
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There is a “lag” before the terms cancel, which is why they “build up” more than we saw in past examples. Still, we can clearly see the Nth partial sum: N ÿ n=5 log(n ´ 3) ´ log(n) = log 2 + log 3 + log 4 ´ log(N ´ 2) ´ log(N ´ 1) ´ log(N) = log 24 N(N ´ 1)(N ´ 2) when N ě 7. So, 8 ÿ n=5 log(n ´ 3) ´ log(n) = lim NÑ8 sN = lim NÑ8 log 24 N(N ´ 1)(N ´ 2) = ´8 632 S-28: This is a telescoping series. Let’s investigate it in the usual way. To make the pattern of cancellation clearer, we express 2 n ´ 1 n+1 ´ 1 n´1 = 1 n + 1 n ´ 1 n+1 ´ 1 n´1, and leave the fractions in the middle of the table unsimplified. Then every fraction has numerator one and two terms with the same denominator and opposite sign cancel.
n 1 n 1 n ´ 1 n + 1 ´ 1 n ´ 1 sn 2 1 2 1 2 ´1 3 ´1 1 1 2 + 1 2´1 3 ´ 1 1 3 1 3 1 3 ´1 4 ´1 2 1 2 + 1 3´1 4 ´ 1 1 4 1 4 1 4 ´1 5 ´1 3 1 2 + 1 4´1 5 ´ 1 1 5 1 5 1 5 ´1 6 ´1 4 1 2 + 1 5´1 6 ´ 1 1 6 1 6 1 6 ´1 7 ´1 5 1 2 + 1 6´1 7 ´ 1 1 7 1 7 1 7 ´1 8 ´1 6 1 2 + 1 7´1 8 ´ 1 1 8 1 8 1 8 ´1 9 ´1 7 1 2 + 1 8 ´ 1 9 ´ 1 1 .
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Concentrate on any row n, except the very first row and the very last row. The first 1 n in that row cancels the ´ 1 n in the middle of the row above it, and the second 1 n in that row cancels the ´ 1 n at the end of the row below it. As far as the first (n = 2) row is concerned, the first 1 2 and the last ´1 1 never get cancelled out because there is no row above the first one. And as far as the very last row is concerned, the two middle terms never get cancelled out because there is no row after the last one. So the partial sum is sN = N ÿ n=2 2 n ´ 1 n + 1 ´ 1 n ´ 1 = from the first row hk kik kj 1 2 ´ 1 1 + from the last row hkkkkkikkkkkj 1 N ´ 1 N + 1 There is another purely algebraic way to find the same sN, motivated by the above discussion.
sN = N ÿ n=2 2 n ´ 1 n + 1 ´ 1 n ´ 1 = N ÿ n=2 1 n ´ 1 n + 1 + N ÿ n=2 1 n ´ 1 n ´ 1 633 The first half N ÿ n=2 1 n ´ 1 n + 1 = 1 2´1 3 + 1 3´1 4 + 1 4´1 5 + ¨ ¨ ¨ + 1 N ´ 1 N + 1 = 1 2 ´ 1 N + 1 and the second half N ÿ n=2 1 n ´ 1 n ´ 1 = 1 2 ´ 1 1 + 1 3´1 2 + 1 4´1 3 + ¨ ¨ ¨ + 1 N ´ 1 N ´ 1 = " ´1 1 + 1 N So sN = 1 2 ´ 1 N + 1 + " ´1 1 + 1 N = ´1 2 + 1 N ´ 1 N + 1 and the limit 8 ÿ n=2 2 n ´ 1 n + 1 ´ 1 n ´ 1 = lim NÑ8 sN = lim NÑ8 ´1 2 + 1 N ´ 1 N + 1 = ´1 2 S-29: The stone at position x has mass 1 4x kg, and we have to pull it a distance of 2x metres, so the work involved in moving that one stone is 1 4x kg (2x m) 9.8 m sec2 = 9.8 2x J Therefore, the work to move all the stones is: 8 ÿ x=1 9.8 2x = 8 ÿ x=0 9.8 2x+1 = 8 ÿ x=0 9.8 2 ¨ 1 2x = 9.8 2 ¨ 1 1 ´ 1 2 = 9.8 J 634 S-30: The volume of a sphere of radius 1 πn is vn = 4 3π 1 πn 3 = 4π 3 1 π3 n So, the volume of all the spheres together is: 8 ÿ n=1 vn = 8 ÿ n=1 4π 3 1 π3 n = 8 ÿ n=0 4π 3 1 π3 n+1 = 8 ÿ n=0 4 3π2 1 π3 n We use Equation 3.2.2 in the CLP-2 text with a = 4 3π2 and r = 1 π3.
= 4 3π2 ¨ 1 1 ´ 1 π3 = 4π 3 (π3 ´ 1) S-31: Let’s make a table. Keep in mind cos2 θ + sin2 θ = 1.
n sin2 n 2n cos2(n + 1) 2n+1 sn 3 sin2 3 23 cos2 4 24 sin2 3 23 + cos2 4 24 4 sin2 4 24 cos2 5 25 sin2 3 23 + 1 24 + cos2 5 25 5 sin2 5 25 cos2 6 26 sin2 3 23 + 1 24 + 1 25 + cos2 6 26 6 sin2 6 26 cos2 7 27 sin2 3 23 + 1 24 + 1 25 + 1 26 + cos2 7 27 7 sin2 7 27 cos2 8 28 sin2 3 23 + 1 24 + 1 25 + 1 26 + 1 27 + cos2 8 28 .
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This gives us an equation for the partial sum sN, when N ě 4: sN = N ÿ n=3 sin2 n 2n + cos2(n + 1) 2n+1 !
= sin2 3 23 + N ÿ n=4 1 2n !
+ cos2(N + 1) 2N+1 635 Using Definition 3.2.3 in the CLP-2 text, our series evaluates to: lim NÑ8 sN = lim NÑ8 " sin2 3 23 + N ÿ n=4 1 2n !
+ cos2(N + 1) 2N+1 # = sin2 3 8 + lim NÑ8 cos2(N + 1) 2N+1 + 8 ÿ n=4 1 2n We evaluate the limit using the squeeze theorem; the series is geometric.
= sin2 3 8 + 0 + 8 ÿ n=0 1 2n+4 = sin2 3 8 + 1 24 8 ÿ n=0 1 2n Using Equation 3.2.2 in the CLP-2 text, = sin2 3 8 + 1 24 1 1 ´ 1 2 = sin2 3 8 + 1 8 « 0.1275 S-32: Since tSMu is the sequence of partial sums of 8 ÿ N=1 SN, we can find tSNu from tSMu as in Question 3: SN = SN ´ SN´1 = N + 1 N ´ N N ´ 1 = ´ 1 N(N ´ 1) if N ě 2, S1 = S1 = 2 Similarly, we find tanu from tSNu. Do be careful: SN only follows the formula we found above when N ě 2. In the next line, we use an expression containing Sn´1; in order for the subscript to be at least two (so the formula fits), we need n ě 3.
an = Sn ´ Sn´1 = ´ 1 n(n ´ 1) ´ ´1 (n ´ 1)(n ´ 2) = 2 n(n ´ 1)(n ´ 2) if n ě 3, a2 = S2 ´ S1 = ´ 1 2(2 ´ 1) ´ 2 = ´5 2, a1 = S1 = 2 All together, an = $ ’ & ’ % 2 n(n´1)(n´2) if n ě 3, ´5 2 if n = 2, 2 if n = 1 636 S-33: We consider a circle of radius R, with an “inner ring” from R 3 to 2R 3 and an “outer ring” from 2R 3 to R.
The area of the outer ring is: πR2 ´ π 2R 3 2 = 5 9πR2 The area of the inner ring is: π 2R 3 2 ´ π R 3 2 = 3 9πR2 So, the ratio of the inner ring’s area to the outer ring’s area is 3 5.
In our bullseye diagram, if we pair up any red ring with the blue ring just inside it, the blue ring has 3 5 the area of the red ring. So, the blue portion of the bullseye has 3 5 the area of the red portion.
Since the circle has area 1, if we let the red portion have area A, then 1 = A + 3 5 A = 8 5 A So, the red portion has area 5 8.
Solutions to Exercises 3.3 — Jump to TABLE OF CONTENTS S-1: (A) lim nÑ8 1 n = 0, so the divergence test is inconclusive. It’s true that this series diverges, but we can’t show it using the divergence test.
(B) lim nÑ8 n2 n + 1 = 8, which is not zero, so the divergence test tells us this series diverges.
(C) We’ll show below that lim nÑ8 sin n does not exist at all. In particular, it is not zero.
Therefore, the divergence test tells us this series diverges.
Now we’ll show that lim nÑ8 sin n does not exist. Suppose that it does exist and takes the value S. We will now see that this assumption leads to a contradiction. Add together the two trig identities (see Appendix A.8 in the CLP-2 text) sin(n + 1) = sin(n) cos(1) + cos(n) sin(1) sin(n ´ 1) = sin(n) cos(1) ´ cos(n) sin(1) This gives sin(n + 1) + sin(n ´ 1) = 2 sin(n) cos(1) 637 Taking the limit n Ñ 8 gives 2S = 2S cos(1). Since cos(1) ‰ 1, this forces S = 0.
Now the first trig identity above gives cos(n) = sin(n + 1) ´ sin(n) cos(1) sin(1) Taking the limit as n Ñ 8 of that gives lim nÑ8 cos(n) = S ´ S cos(1) sin(1) = 0 But that provides the contradiction. Because sin2(n) + cos2(n) = 1, we can’t have both sin(n) and cos(n) converging to zero. So lim nÑ8 sin n does not exist.
(D) For all whole numbers n, sin(πn) = 0, so lim nÑ8 sin(πn) = 0 and the divergence test is inconclusive.
S-2: Let f (x) be a function with f (n) = an for all whole numbers n. In order to apply the integral test (Theorem 3.3.5 in the CLP-2 text) we need f (x) to be positive and decreasing for all sufficiently large values of n.
(A) f (x) = 1 x, which is positive and decreasing for all x ě 1, so the integral test does apply here.
(B) f (x) = x2 x+1, which is not decreasing–in fact, it goes to infinity. So, the integral test does not apply here. (The divergence test tells us the series diverges, though.) (C) f (x) = sin x, which is neither consistently positive nor consistently decreasing, so the integral test does not apply. (The divergence test tells us the series diverges, though.) (D) f (x) = sin x+1 x2 is positive for all whole numbers n. To determine whether it is decreasing, we consider its derivative.
f 1(x) = x2(cos x) ´ (sin x + 1)(2x) x4 = x cos x ´ 2 sin x ´ 2 x3 This is sometimes positive, and sometimes negative. (For example, if x = 100π, f 1(x) = 100π´0´2 (100π)3 ą 0, but if x = 101π then f 1(x) = 101π(´1)´0´2 (101π)3 ă 0.) Then f (x) is not a decreasing function, so the integral test does not apply.
S-3: (a) If Olaf is old, and I am even older, then I am old as well.
(b) If Olaf is old, and I am not as old, then perhaps I am old as well (just slightly less so), or perhaps I am young. There is not enough information to tell.
(c) If Yuan is young, and I am older, then perhaps I am much older and I am old, or perhaps I am only a little older, and I am young. There is not enough information to tell.
(d) If Yuan is young, and I am even younger, then I must also be young.
Another way to think about this is with a timeline of birthdates. People born before the threshold are old, and people born after it are young.
638 threshold Yuan Olaf If I’m born before (older than) Olaf, I’m born before the threshold, so I’m old.
If I’m born after (younger than) Yuan I’m born after the threshold, so I’m young.
threshold Yuan definitely young Olaf definitely old If I’m born after Olaf or before Yuan, I don’t know which side of the threshold I’m on. I could be old or I could be young.
threshold Yuan older than Yuan also older than Yuan S-4: The comparison test is Theorem 3.3.8 in the CLP-2 text. However, rather than trying to memorize which way the inequalities go in all cases, we use the same reasoning as Question 3.
If a sequence has positive terms, it either converges, or it diverges to infinity, with the partial sums increasing and increasing without bound. If one sequence diverges, and the other sequence is larger, then the other sequence diverges–just like being older than an old person makes you old.
If ř an converges, and tanu is the red (larger) series, then ř bn converges: it’s smaller than a sequence that doesn’t add up to infinity, so it too does not add up to infinity.
If ř an diverges, and tanu is the blue (smaller) series, then ř bn diverges: it’s larger than a sequence that adds up to infinity, so it too adds up to infinity.
In the other cases, we can’t say anything. If tanu is the red (larger) series, and ř an diverges, then perhaps tbnu behaves similarly to tanu and ř bn diverges, or perhaps tbnu is much, much smaller than tanu and ř bn converges.
Similarly, if tanu is the blue (smaller) series, and ř an converges, then perhaps tbnu behaves similarly to tanu and ř bn converges, or perhaps tbnu is much, much bigger than tanu and ř bn diverges.
if ř an converges if ř an diverges and if tanu is the red series then ř bn CONVERGES inconclusive and if tanu is the blue series inconclusive then ř bn DIVERGES 639 S-5: (a) Since ř 1 n is divergent, we can only use it to prove series with larger terms are divergent. This is the case here, since 1 n´1 ą 1 n. So, the direct comparison test is valid.
For the limit comparison test, we calculate: lim nÑ8 an bn = lim nÑ8 1 n´1 1 n = lim nÑ8 1 1 ´ 1 n = 1 Since the limit exists and is not zero, the limit comparison test is also valid.
(b) Since the series ř 1 n2 converges, we can only use the direct comparison test to show the convergence of a series if its terms have smaller absolute values. Indeed, ˇ ˇ ˇ ˇ sin n n2 + 1 ˇ ˇ ˇ ˇ = | sin n| n2 + 1 ă 1 n2 so the series are set for a direct comparison.
To check whether a limit comparison will work, we compute: lim nÑ8 an bn = lim nÑ8 sin n n2+1 1 n2 = lim nÑ8 n2 n2 + 1 sin n = lim nÑ8(1) sin n The limit does not exist, so the limit comparison test is not a valid test to compare these two series.
(c) Since the series ř 1 n3 converges, we can only use the direct comparison test to conclude something about a series with smaller terms. However, n3 + 5n + 1 n6 ´ 2 ą n3 n6 = 1 n3.
Therefore the direct comparison test does not apply to this pair of series.
For the limit comparison test, we calculate: lim nÑ8 an bn = lim nÑ8 n3+5n+1 n6´2 1 n3 = lim nÑ8 n3 + 5n + 1 n3 ´ 2 n3 1 n3 1 n3 !
= lim nÑ8 1 + 5 n2 + 1 n3 1 ´ 2 n6 = 1 Since the limit is a nonzero real number, we can use the limit comparison test to compare this pair of series.
(d) Since the series ř 1 4 ?n diverges, we can only use the direct comparison test to show that a series with larger terms diverges. However, 1 ?n ă 1 4 ?n 640 so the direct comparison test isn’t valid with this pair of series.
For the limit comparison test, we calculate: L = lim nÑ8 an bn = lim nÑ8 1 ?n 1 4 ?n = lim nÑ8 1 4 ?n = 0 Since the series ř 1 4 ?n diverges, it is part (b) of the limit comparison test, Theorem 3.3.11 in the CLP-2 text, that is appropriate here. As the limit L = 0, the limit comparison test doesn’t apply.
S-6: It diverges by the divergence test, because lim nÑ8 an ‰ 0.
S-7: The divergence test (Theorem 3.3.1 in the CLP-2 text) is inconclusive when lim nÑ8 an = 0. We cannot use the divergence test to show that a series converges.
S-8: The integral test does not apply because f (x) is not decreasing.
S-9: The inequality goes the wrong way, so the direct comparison test (with this comparison series) is inconclusive.
S-10: Although the terms of (A) are sometimes negative and sometimes positive, they are not strictly alternating in a positive-negative-positive-negative pattern. For instance, sin 1 and sin 2 are both postive. So, (A) is not an alternating series.
When n is a whole number, cos(πn) = (´1)n, so (B) is alternating.
Since the exponent of (´n) in (C) is even, the terms are always positive. Therefore (C) is not alternating.
(D) is an alternating series.
S-11: One possible answer: 8 ÿ n=1 1 n2. This series converges (it’s a p–series with p = 2 ą 1), but if we take the ratio of consecutive terms: lim nÑ8 an+1 an = lim nÑ8 n2 (n + 1)2 = 1 The limit of the ratio is 1, so the ratio test is inconclusive.
S-12: By the divergence test, for a series ř an to converge, we need lim nÑ8 an = 0. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If lim nÑ8 ˇ ˇ ˇ ˇ an an+1 ˇ ˇ ˇ ˇ ă 1 or 641 (equivalently) lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ ą 1, then |an+1| ą |an| for sufficiently large n, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.
S-13: The terms of the series only see a small portion of the domain of the integral. We can try to think of a function f (x) that behaves “nicely” when x is a whole number (that is, it produces a sequence whose sum converges), but is more unruly when x is not a whole number.
For example, suppose f (x) = sin(πx). Then f (x) = 0 for every integer x, but this is not representative of the function as a whole. Indeed, our corresponding series has terms tanu = t0, 0, 0, . . .u.
• ż 8 1 sin(πx) dx = lim RÑ8 ´ 1 π cos(πx) R 1 = lim RÑ8 h ´ cos(πR) i ´ 1 π Since the limit does not exist, the integral diverges.
• 8 ÿ n=1 sin(πn) = 8 ÿ n=1 0 = 0. The series converges.
S-14: When n is very large, the term 2n dominates the numerator, and the term 3n dominates the denominator. So when n is very large an « 2n 3n. Therefore we should take bn = 2n 3n . Note that, with this choice of bn, lim nÑ8 an bn = lim nÑ8 2n + n 3n + 1 3n 2n = lim nÑ8 1 + n/2n 1 + 1/3n = 1 as desired.
S-15: (a) In general false. The harmonic series 8 ř n=1 1 n diverges by the p–test with p = 1.
(b) Be careful. You were not told that the an’s are positive. So this is false in general. If an = (´1)n 1 n, then 8 ř n=1 (´1)nan is again the harmonic series 8 ř n=1 1 n, which diverges.
(c) In general false. Take, for example, an = 0 and bn = 1.
S-16: First, we’ll check the divergence test. It doesn’t always work, but if it does, it’s likely the easiest path.
lim nÑ8 n2 3n2 + ?n 1 n2 1 n2 !
= lim nÑ8 1 3 + 1 n?n = 1 3 ‰ 0 Since the limit of the terms being added is not zero, the series diverges by the divergence test.
642 S-17: This precise question was asked on a 2014 final exam. Note that the nth term in the series is an = 5k 4k+3k and does not depend on n! There are two possibilities. Either this was intentional (and the instructor was being particularly nasty) or it was a typo and the intention was to have an = 5n 4n+3n. In both cases, the limit lim nÑ8 an = lim nÑ8 5k 4k + 3k = 5k 4k + 3k ‰ 0 lim nÑ8 an = lim nÑ8 5n 4n + 3n = lim nÑ8 (5/4)n 1 + (3/4)n = +8 ‰ 0 is nonzero, so the series diverges by the divergence test.
S-18: We usually check the divergence test first, to look for low-hanging fruit. The limit of the terms being added is zero: lim nÑ8 1 n + 1 2 = 0 so the divergence test is inconclusive. That is, we need to look harder.
Next, we might consider a comparison test–these can also provide us (if we’re lucky) with an easy path. The terms we’re adding look somewhat like 1 n, but our terms are smaller than these terms, which form the terms of the divergent harmonic series. So, a direct comparison seems unlikely. Now we search for more exotic tests.
Let f (x) = 1 x + 1 2 . Note f (x) is positive and decreases as x increases. So, by the integral test, which is Theorem 3.3.5 in the CLP-2 text, the given series converges if and only if the integral ş8 0 1 x+ 1 2 dx converges. Since ż 8 0 1 x + 1 2 dx = lim RÑ8 ż R 0 1 x + 1 2 dx = lim RÑ8 log x + 1 2 x=R x=0 = lim RÑ8 log R + 1 2 ´ log 1 2 diverges, the series diverges.
S-19: The terms of the series tend to 0, so we can’t use the divergence test.
To generate a guess about its convergence, we do the following: ÿ 1 ?
k ?
k + 1 = ÿ 1 ?
k2 + k « ÿ 1 ?
k2 = ÿ 1 k We guess that our series behaves like the harmonic series, and the harmonic series diverges (which can be demonstrated by p-test or integral test). So, we guess that our 643 series diverges. However, in order to directly compare our series to the harmonic series and show our series diverges, our terms would have to be bigger than the terms in the harmonic series, and this is not the case. So, we use limit comparison.
1 k 1 ?
k2+k = ?
k2 + k k = ?
k2 + k ?
k2 = c k2 + k k2 = c 1 + 1 k, so lim kÑ8 1 k 1 ?
k2+k = lim kÑ8 c 1 + 1 k = 1 Since 1 is a real number greater than 0, by the Limit Comparison Test, ř 1 ?
k ?
k+1 diverges, like ř 1 k.
S-20: This is a geometric series with r = 1.001. Since |r| ą 1, it is divergent.
S-21: This is a geometric series with r = ´1 5 . Since |r| ă 1, it is convergent.
We want to use the formula ř8 n=0 rn = 1 1´r, but our series does not start at 0, so we re-write it: 8 ÿ n=3 ´1 5 n = 8 ÿ n=0 ´1 5 n ´ 2 ÿ n=0 ´1 5 n = 1 1 ´ (´1/5) ´ 1 ´ 1 5 + 1 25 = 1 6/5 ´ 1 + 1 5 ´ 1 25 = 5 6 + ´25 + 5 ´ 1 25 = ´ 1 150 S-22: For any integer n, sin(πn) = 0, so ř sin(πn) = ř 0 = 0. So, this series converges.
S-23: For any integer n, cos(πn) = ˘1, so lim nÑ8 cos(πn) ‰ 0.
By the divergence test, this series diverges.
S-24: Factorials grow super fast. Like, wow, really fast. Even faster than exponentials. So the terms are going to zero, and the divergence test won’t help us. Let’s use ratio–it’s a good go-to test with factorials.
ak+1 ak = ek+1 (k+1)!
ek k!
= ek+1 ek ¨ k!
(k + 1)! = e ¨ k(k ´ 1) ¨ ¨ ¨ (1) (k + 1)(k)(k ´ 1) ¨ ¨ ¨ (1) = e ¨ 1 k + 1 = e k + 1 Since e is a constant, lim kÑ8 ak+1 ak = lim kÑ8 e k + 1 = 0 Since 0 ă 1, by the ratio test, the series converges.
644 S-25: This is close to being in the form of a geometric series. First, we should have our powers be k, not k + 2, but we notice 3k+2 = 3k32 = 9 ¨ 32, so: 8 ÿ k=0 2k 3k+2 = 8 ÿ k=0 2k 9 ¨ 3k = 1 9 8 ÿ k=0 2k 3k = 1 9 8 ÿ k=0 2 3 k Now it looks like a geometric series with r = 2 3 = 1 9 1 1 ´ (2/3) = 1 3 In conclusion: this (geometric) series is convergent, and its sum is 1 3.
S-26: Usually with factorials, we want to use the divergence test or the ratio test. Since the terms are indeed tending towards zero, we are left with the ratio test.
an+1 an = (n+1)!(n+1)!
(2n+2)!
n!n!
(2n)!
= (n + 1)!(n + 1)!
n!n!
¨ (2n)!
(2n + 2)!
= (n+1)(n)(n´1)¨¨¨(1) n(n´1)¨¨¨(1) ¨ (n+1)(n)(n´1)¨¨¨(1) n(n´1)¨¨¨(1) ¨ (2n)(2n´1)(2n´2)¨¨¨(1) (2n+2)(2n+1)(2n)(2n´1)(2n´2)¨¨¨(1) = (n + 1)(n + 1) ¨ 1 (2n + 2)(2n + 1), so lim nÑ8 an+1 an = lim nÑ8 (n + 1)(n + 1) (2n + 2)(2n + 1) = lim nÑ8 (n + 1)(n + 1) 2(n + 1)(2n + 1) = lim nÑ8 n + 1 4n + 2 = 1 4 Since the limit is a number less than 1, the series converges by the ratio test.
S-27: We want to make an estimation, when n gets big: n2 + 1 2n4 + n « n2 2n4 = 1 2n2 Since ř 1 2n2 is a convergent series (by p-test, or integral test), we guess that our series is convergent as well. If we wanted to use comparison test, we should have to show n2+1 2n4+n ă 1 2n2, which seems unpleasant, so let’s use limit comparison.
lim nÑ8 n2+1 2n4+n 1 2n2 = lim nÑ8 (n2 + 1)2n2 2n4 + n = lim nÑ8 2n4 + 2n2 2n4 + n 1/n4 1/n4 = lim nÑ8 2 + 2 n2 2 + 1 n3 = 1 Since the limit is a positive finite number, by the Limit Comparison Test, ř n1+1 2n4+n does the same thing ř 1 2n2 does: it converges.
645 S-28: First, we rule out some of the easier tests. The limit of the terms being added is zero, so the divergence test is inconclusive. The terms being added are smaller than the terms of the (divergent) harmonic series, ř 1 n, so we can’t directly compare these two series, and there isn’t another obvious series to compare ours to. However, the terms being added seem like a function we could integrate.
Let f (x) = 5 x(log x)3/2. Then f (x) is positive and decreases as x increases. So the sum 8 ÿ 3 f (n) and the integral ż 8 3 f (x) dx either both converge or both diverge, by the integral test, which is Theorem 3.3.5 in the CLP-2 text. For the integral, we use the substitution u = log x, du = dx x to get ż 8 3 5 dx x(log x)3/2 = ż 8 log 3 5 du u3/2 which converges by the p–test (which is Example 1.12.8 in the CLP-2 text) with p = 3 2 ą 1.
S-29: Let f (x) = 1 x(log x)p. Then f (x) is positive for n ě 3, and f (x) decreases as x increases. So, we can use the integral test, Theorem 3.3.5 in the CLP-2 text.
ż 8 2 1 x(log x)p dx = lim RÑ8 ż R 2 1 (log x)p dx x = lim RÑ8 ż log R log 2 1 up du with u = log x, du = dx x Using the results about p-series, Example 3.3.6 in the CLP-2 text, we know this integral converges if and only if p ą 1, so the same is true for the series by the integral test.
S-30: As usual, let’s see whether the “easy” tests work. The terms we’re adding converge to zero: lim nÑ8 e´?n ?n = lim nÑ8 1 ?ne ?n = 0 so the divergence test is inconclusive. Our series isn’t geometric, and it doesn’t seem obvious how to compare it to a geometric series. However, the terms we’re adding seem like they would make an integrable function.
Set f (x) = e´?x ?x . For x ě 1, this function is positive and decreasing (since it is the product of the two positive decreasing functions e´?x and 1 ?x). We use the integral test with this 646 function. Using the substitution u = ?x, so that du = 1 2?x dx, we see that ż 8 1 f (x) dx = lim RÑ8 ż R 1 e´?x ?x dx = lim RÑ8 ż ?
R 1 e´u ¨ 2 du = lim RÑ8 ´2e´uˇ ˇ ˇ ?
R 1 = lim RÑ8 ´2e´ ?
R + 2e´ ?
1 = 0 + 2e´1, and so this improper integral converges. By the integral test, the given series also converges.
S-31: We first develop some intuition. For very large n, 3n2 dominates 7 so that ?
3n2 ´ 7 n3 « ?
3n2 n3 = ?
3 n2 The series 8 ÿ n=2 1 n2 converges by the p–test with p = 2, so we expect the given series to converge too.
To verify that our intuition is correct, it suffices to observe that 0 ă an = ?
3n2 ´ 7 n3 ă ?
3n2 n3 = ?
3 n2 = cn for all n ě 2. As the series 8 ř n=2 cn converges, the comparison test says that 8 ř n=2 an converges too.
S-32: We first develop some intuition. For very large k, k4 dominates 1 so that the numerator 3 ?
k4 + 1 « 3 ?
k4 = k4/3, and k5 dominates 9 so that the denominator ?
k5 + 9 « ?
k5 = k5/2 and the summand 3 ?
k4 + 1 ?
k5 + 9 « k4/3 k5/2 = 1 k7/6 The series 8 ÿ n=1 1 k7/6 converges by the p–test with p = 7 6 ą 1, so we expect the given series to converge too.
To verify that our intuition is correct, we apply the limit comparison test with ak = 3 ?
k4 + 1 ?
k5 + 9 and bk = 1 k7/6 = k4/3 k5/2 647 which is valid since lim kÑ8 ak bk = lim kÑ8 3 ?
k4 + 1/k4/3 ?
k5 + 9/k5/2 = lim kÑ8 3 a 1 + 1/k4 a 1 + 9/k5 = 1 exists. Since the series 8 ř k=1 bk is a convergent p–series (with ratio p = 7 6 ą 1), the given series converges.
Note: to apply the direct comparison test with our chosen comparison series, we would need to show that 3 ?
k4 + 1 ?
k5 + 9 ď 1 k7/6 for all k sufficiently large. However, this is not true: the opposite inequality holds when k is large.
S-33: Solution 1: Let’s see whether the divergence test works here.
lim nÑ8 n42n/3 (2n + 7)4 1 n4 1 n4 !
= lim nÑ8 2n/3 (2 + 7/n)4 = lim nÑ8 2n/3 (2 + 0)4 = 8 The summands of our series do not converge to zero. By the divergence test, the series diverges.
Solution 2: Let’s develop some intuition for a comparison. For very large n, 2n dominates 7 so that n42n/3 (2n + 7)4 « n42n/3 (2n)4 = 1 162n/3 The series 8 ÿ n=1 2n/3 is a geometric series with ratio r = 21/3 ą 1 and so diverges. (It also fails the divergence test.) We expect the given series to diverge too.
To verify that our intuition is correct, we apply the limit comparison test with an = n42n/3 (2n + 7)4 and bn = 2n/3 which is valid since lim nÑ8 an bn = lim nÑ8 n4 (2n + 7)4 = lim nÑ8 1 (2 + 7/n)4 = 1 24 exists and is nonzero. Since the series 8 ř n=1 bn is a divergent geometric series (with ratio r = 21/3 ą 1), the given series diverges.
(It is possible to use the plain comparison test as well. One needs to show something like an = n42n/3 (2n+7)4 ě n42n/3 (2n+7n)4 = 1 94 bn.) 648 Solution 3: Alternately, one can apply the ratio test: lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (n + 1)42(n+1)/3/(2(n + 1) + 7)4 n42n/3/(2n + 7)4 ˇ ˇ ˇ ˇ = lim nÑ8 (n + 1)4(2n + 7)4 n4(2n + 9)4 2(n+1)/3 2n/3 = lim nÑ8 (1 + 1/n)4(2 + 7/n)4 (2 + 9/n)4 ¨ 21/3 = 1 ¨ 21/3 ą 1.
Since the ratio of consecutive terms is greater than one, by the ratio test, the series diverges.
S-34: (a) For large n, n2 " 1 and so ?
n2 + 1 « ?
n2 = n. This suggests that we apply the limit comparison test with an = 1 ?
n2+1 and bn = 1 n. Since lim nÑ8 an bn = lim nÑ8 1/ ?
n2 + 1 1/n = lim nÑ8 1 a 1 + 1/n2 = 1 and since 8 ř n=1 1 n diverges, the given series diverges.
(b) Since cos(nπ) = (´1)n, the given series converges by the alternating series test. To check that an = n 2n decreases to 0 as n tends to infinity, note that an+1 an = (n + 1)2´(n+1) n2´n = 1 + 1 n 1 2 is smaller than 1 (so that an+1 ď an) for all n ě 1, and is smaller than 3 4 (so an+1 ď 3 4an) for all n ě 2.
S-35: For large k, k4 " 2k3 ´ 2 and k5 " k2 + k so k4 ´ 2k3 + 2 k5 + k2 + k « k4 k5 = 1 k.
This suggests that we apply the limit comparison test with ak = k4´2k3+2 k5+k2+k and bk = 1 k.
Since lim kÑ8 ak bk = lim kÑ8 k4 ´ 2k3 + 2 k5 + k2 + k ¨ k 1 = lim kÑ8 k5 ´ 2k4 + 2k k5 + k2 + k = lim kÑ8 1 ´ 2/k + 2/k4 1 + 1/k3 + 1/k4 = 1 and since 8 ř k=1 1 k diverges (by the p–test with p = 1), the given series diverges.
649 S-36: (a) For large n, n2 " n + 1 and so the numerator n2 + n + 1 « n2. For large n, n5 " n and so the denominator n5 ´ n « n5. So, for large n, n2 + n + 1 n5 ´ n « n2 n5 = 1 n3.
This suggests that we apply the limit comparison test with an = n2+n+1 n5´n and bn = 1 n3.
Since lim nÑ8 an bn = lim nÑ8 (n2 + n + 1)/(n5 ´ n) 1/n3 = lim nÑ8 n5 + n4 + n3 n5 ´ n = lim nÑ8 1 + 1/n + 1/n2 1 ´ 1/n4 = 1 exists and is nonzero, and since 8 ř n=1 1 n3 converges (by the p–test with p = 3 ą 1), the given series converges.
(b) For large m, 3m " | sin ?m| and so 3m + sin ?m m2 « 3m m2 = 3 m.
This suggests that we apply the limit comparison test with am = 3m+sin ?m m2 and bm = 1 m.
(We could also use bm = 3 m.) Since lim mÑ8 am bm = lim mÑ8 (3m + sin ?m)/m2 1/m = lim mÑ8 3m + sin ?m m = lim mÑ8 3 + sin ?m m = 3 exists and is nonzero, and since 8 ř m=1 1 m diverges (by the p–test with p = 1), the given series diverges.
S-37: 8 ÿ n=5 1 en = 8 ÿ n=5 1 e n = 8 ÿ n=0 1 e n ´ 4 ÿ n=0 1 e n = 1 1 ´ 1 e ´ 1 ´ 1 e 5 1 ´ 1 e = 1 e 5 1 ´ 1 e = 1 e5 1 ´ 1 e = 1 e5 ´ e4 650 S-38: This is a geometric series.
8 ÿ n=2 6 7n = 8 ÿ n=0 6 7n+2 = 8 ÿ n=0 6 72 ¨ 1 7n We use Equation 3.2.2 in the CLP-2 text with a = 6 72 and r = 1 7.
= 6 72 ¨ 1 1 ´ 1 7 = 6 42 = 1 7 S-39: (a) Solution 1: The given series is 1 + 1 3 + 1 5 + 1 7 + 1 9 + ¨ ¨ ¨ = 8 ÿ n=1 an with an = 1 2n ´ 1 First we’ll develop some intuition by observing that, for very large n, an « 1 2n. We know that the series 8 ř n=1 1 n diverges by the p–test with p = 1. So let’s apply the limit comparison test with bn = 1 n. Since lim nÑ8 an bn = lim nÑ8 n 2n ´ 1 = lim nÑ8 1 2 ´ 1 n = 1 2 the series 8 ř n=1 an converges if and only if the series 8 ř n=1 bn converges. So the given series diverges.
Solution 2: The series 1 + 1 3 + 1 5 + 1 7 + 1 9 + ¨ ¨ ¨ ě 1 2 + 1 4 + 1 6 + 1 8 + 1 10 + ¨ ¨ ¨ = 1 2 1 + 1 2 + 1 3 + 1 4 + 1 5 + ¨ ¨ ¨ The series in the brackets is the harmonic series which we know diverges, by the p–test with p = 1. So the series on the right hand side diverges. By the direct comparison test, the series on the left hand side diverges too.
(b) We’ll use the ratio test with an = (2n + 1) 22n+1 . Since an+1 an = (2n + 3) 22n+3 22n+1 (2n + 1) = 1 4 (2n + 3) (2n + 1) = 1 4 (2 + 3/n) (2 + 1/n) Ñ 1 4 ă 1 as n Ñ 8 the series converges.
651 S-40: (a) For very large k, k ! k2 so that an = 3 ?
k k2 ´ k « 3 ?
k k2 = 1 k5/3.
We apply the limit comparison test with bk = 1 k5/3. Since lim kÑ8 ak bk = lim kÑ8 3 ?
k/(k2 ´ k) 1/k5/3 = lim kÑ8 k2 k2 ´ k = lim kÑ8 1 1 ´ 1/k = 1 exists and is nonzero, and 8 ř k=1 1 k5/3 converges (by the p–test with p = 5 3 ą 1), the given series converges by the limit comparison test.
(b) The kth term in this series is ak = k1010k(k!)2 (2k)!
. Factorials often work well with the ratio test, because they simplify so nicely in quotients.
ak+1 ak = (k + 1)1010k+1((k + 1)!)2 (2k + 2)!
¨ (2k)!
k1010k(k!)2 = 10 k + 1 k 10 (k + 1)2 (2k + 2)(2k + 1) = 10 1 + 1 k 10 (1 + 1/k)2 (2 + 2/k)(2 + 1/k) As k tends to 8, this converges to 10 ˆ 1 ˆ 1 2ˆ2 ą 1. So the series diverges by the ratio test.
(c) We’ll use the integal test. The kth term in the series is ak = 1 k(log k)(log log k) = f (k) with f (x) = 1 x(log x)(log log x), which is continuous, positive and decreasing for x ě 3.
ż 8 3 f (x) dx = ż 8 3 dx x(log x)(log log x) = lim RÑ8 ż R 3 dx x(log x)(log log x) = lim RÑ8 ż log R log 3 dy y log y with y = log x, dy = dx x = lim RÑ8 ż log log R log log 3 dt t with t = log y, dt = dy y = lim RÑ8 h log t ilog log R log log 3 = 8 Since the integral is divergent, the series is divergent as well by the integral test.
S-41: For large n, the numerator n3 ´ 4 « n3 and the denominator 2n5 ´ 6n « 2n5, so the nth term is approximately n3 2n5 = 1 2n2. So we apply the limit comparison test with an = n3´4 2n5´6n and bn = 1 n2. Since lim nÑ8 an bn = lim nÑ8 (n3 ´ 4)/(2n5 ´ 6n) 1/n2 = lim nÑ8 1 ´ 4 n3 2 ´ 6 n4 = 1 2 652 exists and is nonzero, the given series 8 ř n=1 an converges if and only if the series 8 ř n=1 bn converges. Since the series 8 ř n=1 bn = 8 ř n=1 1 n2 is a convergent p-series (with p = 2), both series converge.
S-42: By the alternating series test, the error introduced when we approximate the series 8 ÿ n=1 (´1)n n ¨ 10n by N ÿ n=1 (´1)n n ¨ 10n is at most the magnitude of the first omitted term, 1 (N + 1)10(N+1) . By trial and error, we find that this expression becomes smaller than 10´6 when N + 1 ě 6. So the smallest allowable value is N = 5.
S-43: The sequence t 1 n2u decreases to zero as n increases to infinity. So, by the alternating series error bound, which is given in Theorem 3.3.14 in the CLP-2 text, π2 12 ´ SN lies between zero and the first omitted term, (´1)N (N+1)2. We therefore need 1 (N+1)2 ď 10´6, which is equivalent to N + 1 ě 103 and N ě 999.
S-44: The error introduced when we approximate S by the Nth partial sum SN = řN n=1 (´1)n+1 (2n+1)2 lies between 0 and the first term dropped, which is (´1)n+1 (2n+1)2 ˇ ˇ ˇ n=N+1 = (´1)N+2 (2N+3)2. So we need the smallest positive integer N obeying 1 (2N + 3)2 ď 1 100 (2N + 3)2 ě 100 2N + 3 ě 10 N ě 7 2 So we need N = 4 and then S4 = 1 32 ´ 1 52 + 1 72 ´ 1 92 S-45: (a) There are plenty of powers/factorials. So let’s try the ratio test with an = nn 9nn!.
lim nÑ8 an+1 an = lim nÑ8 (n + 1)n+1 9n+1(n + 1)!
9nn!
nn = lim nÑ8 (n + 1)n+1 nn 9 (n + 1) = lim nÑ8 (1 + 1/n)n 9 = e 9 Here we have used that lim nÑ8(1 + 1/n)n = e. See Example 3.7.20 in the CLP-1 text, with x = 1 n and a = 1. As e ă 9, our series converges.
(b) We know that the series ř8 n=1 1 n2 converges, by the p–test with p = 2, and also that 653 log n ě 2 for all n ě e2. So let’s use the limit comparison test with an = 1 nlog n and bn = 1 n2.
lim nÑ8 an bn = lim nÑ8 1 nlog n ¨ n2 1 = lim nÑ8 1 nlog n´2 = 0 So our series converges, by the limit comparison test.
S-46: (a) Solution 1: • Our first task is to identify the potential sources of impropriety for this integral.
• The domain of integration extends to +8. On the domain of integration the denominator is never zero so the integrand is continuous. Thus the only problem is at +8.
• Our second task is to develop some intuition about the behavior of the integrand for very large x. When x is very large: – | sin x| ď 1 ! x, so that the numerator x + sin x « x, and – 1 ! x2, so that denominator 1 + x2 « x2, and – the integrand x + sin x 1 + x2 « x x2 = 1 x • Now, since ż 8 2 dx x diverges, we would expect ż 8 2 x + sin x 1 + x2 dx to diverge too.
• Our final task is to verify that our intuition is correct. To do so, we set f (x) = x + sin x 1 + x2 g(x) = 1 x and compute lim xÑ8 f (x) g(x) = lim xÑ8 x + sin x 1 + x2 ˜ 1 x = lim xÑ8 (1 + sin x/x)x (1/x2 + 1)x2 ˆ x = lim xÑ8 1 + sin x/x 1/x2 + 1 = 1 • Since ż 8 2 g(x) dx = ż 8 2 dx x diverges, by Example 1.12.8 in the CLP-2 text10, with p = 1, Theorem 1.12.22(b) in the CLP-2 text now tells us that ż 8 2 f (x) dx = ż 8 2 x + sin x 1 + x2 dx diverges too.
10 To change the lower limit of integration from 1 to 2, just apply Theorem 1.12.20 in the CLP-2 text.
654 Solution 2: Let’s break up the integrand as x + sin x 1 + x2 = x 1 + x2 + sin x 1 + x2. First, we consider the integral ż 8 2 sin x 1 + x2 dx.
• | sin x| 1 + x2 ď 1 1 + x2, so if we can show ż 1 1 + x2 dx converges, we can conclude that ż | sin x| 1 + x2 dx converges as well by the comparison test.
• ż 8 2 1 1 + x2 dx ď ż 8 2 1 x2 dx • ż 8 2 1 x2 dx converges (by the p–test with p = 2) • So the integral ż 8 2 sin x 1 + x2 dx converges by the comparison test, and hence • ż 8 2 sin x 1 + x2 dx converges as well.
Therefore, ż 8 2 x + sin x 1 + x2 dx converges if and only if ż 8 2 x 1 + x2 dx converges. But ż 8 2 x 1 + x2 dx = lim rÑ8 ż r 2 x 1 + x2 dx = lim rÑ8 h 1 2 log(1 + x2) ir 2 = 8 diverges, so ż 8 2 x + sin x 1 + x2 dx diverges.
(b) The problem is that f (x) = x + sin x 1 + x2 is not a decreasing function. To see this, compute the derivative: f 1(x) = (1 + cos x)(1 + x2) ´ (x + sin x)(2x) (1 + x2)2 = (cos x ´ 1)x2 ´ 2x sin x + 1 + cos x (1 + x2)2 If x = 2mπ, the numerator is 0 ´ 0 + 1 + 1 ą 0.
Therefore, the integral test does not apply.
(c) Solution 1: Set an = n+sin n 1+n2 . We first try to develop some intuition about the behaviour of an for large n and then we confirm that our intuition was correct.
• Step 1: Develop intuition. When n " 1, the numerator n + sin n « n, and the denominator 1 + n2 « n2 so that an « n n2 = 1 n and it looks like our series should diverge by the p–test (Example 3.3.6 in the CLP-2 text) with p = 1.
• Step 2: Verify intuition. To confirm our intuition we set bn = 1 n and compute the limit lim nÑ8 an bn = lim nÑ8 n+sin n 1+n2 1 n = lim nÑ8 n[n + sin n] 1 + n2 = lim nÑ8 1 + sin n n 1 n2 + 1 = 1 655 We already know that the series 8 ř n=1 bn = 8 ř n=1 1 n diverges by the p–test with p = 1. So our series diverges by the limit comparison test, Theorem 3.3.11 in the CLP-2 text.
Solution 2: Since ˇ ˇ sin n 1+n2 ˇ ˇ ď 1 n2 and the series 8 ř n=1 1 n2 converges by the p–test with p = 2, the series 8 ř n=1 sin n 1+n2 converges. Hence 8 ř n=1 n+sin n 1+n2 converges if and only if the series 8 ř n=1 n 1+n2 converges. Now f (x) = x 1+x2 is a continuous, positive, decreasing function on [1, 8) since f 1(x) = (1 + x2) ´ x(2x) (1 + x2)2 = 1 ´ x2 (1 + x2)2 is negative for all x ą 1. We saw in part (a) that the integral ş8 2 x 1+x2 dx diverges. So the integral ş8 1 x 1+x2 dx diverges too and the sum 8 ř n=1 n 1+n2 diverges by the integral test. So 8 ř n=1 n+sin n 1+n2 diverges.
S-47: Note that e´?x ?x = 1 ?xe ?x decreases as x increases. Hence, for every n ě 1, e´?x ?x ě e ?n ?n for x in the interval [n ´ 1, n] So, ż n n´1 e´?x ?x dx ě ż n n´1 e´?n ?n dx = " e´?n ?n x #x=n x=n´1 = e´?n ?n Then, for every N ě 1, EN = 8 ÿ n=N+1 e´?n ?n ď 8 ÿ n=N+1 ż n n´1 e´?x ?x dx = ż N+1 N e´?x ?x dx + ż N+2 N+1 e´?x ?x dx + ¨ ¨ ¨ = ż 8 N e´?x ?x dx 656 Substituting y = ?x, dy = 1 2 dx ?x, ż 8 N e´?x ?x dx = 2 ż 8 ?
N e´y dy = ´2e´yˇ ˇ ˇ 8 ?
N = 2e´ ?
N This shows that ř8 n=N+1 e´?n ?n converges and is between 0 and 2e´ ?
N. Since E14 = 2e´ ?
14 = 0.047, we may truncate the series at n = 14.
8 ÿ n=1 e´?n ?n = 14 ÿ n=1 e´?n ?n + E14 = 0.3679 + 0.1719 + 0.1021 + 0.0677 + 0.0478 + 0.0352 + 0.0268 + 0.0209 + 0.0166 + 0.0134 + 0.0109 + 0.0090 + 0.0075 + 0.0063 + E14 = 0.9042 + E14 The sum is between 0.9035 and 0.9535. (This even allows for a roundoff error of 0.00005 in each term as we were calculating the partial sum.) S-48: Let’s get some intuition to guide us through a proof. Since 8 ř n=1 an, converges an must converge to zero as n Ñ 8. So, when n is quite large, an 1´an « an 1´0 = an 1 , and we know ř an converges. So, we want to separate the “large” indices from a finite number of smaller ones.
Since lim nÑ8 an = 0, there must be11 some integer N such that 1 2 ą an ě 0 for all n ą N.
Then, for n ą N, an 1 ´ an ď an 1 ´ 1/2 = 2an From the information in the problem statement, we know 8 ÿ n=N+1 2an = 2 8 ÿ n=N+1 an converges.
So, by the direct comparison test, 8 ÿ n=N+1 an 1 ´ an converges as well.
Since the convergence of a series is not affected by its first N terms, as long as N is finite, we conclude 8 ÿ n=1 an 1 ´ an converges.
11 We could have chosen any positive number strictly less than 1, not only 1 2.
657 S-49: By the divergence test, the fact that 8 ř n=0 (1 ´ an) converges guarantees that lim nÑ8(1 ´ an) = 0, or equivalently, that lim nÑ8 an = 1. So, by the divergence test, a second time, the fact that lim nÑ8 2nan = +8 guarantees that 8 ř n=0 2nan diverges too.
S-50: By the divergence test, the fact that 8 ÿ n=1 nan ´ 2n + 1 n + 1 converges guarantees that lim nÑ8 nan ´ 2n + 1 n + 1 = 0, or equivalently, that 0 = lim nÑ8 n n + 1an ´ lim nÑ8 2n ´ 1 n + 1 = lim nÑ8 an ´ 2 ð ñ lim nÑ8 an = 2 The series of interest can be written ´ log a1 + 8 ř n=1 log(an) ´ log(an+1) which looks like a telescoping series. So we’ll compute the partial sum SN = ´ log a1 + N ÿ n=1 log(an) ´ log(an+1) = ´ log a1 + log(a1) ´ log(a2) + log(a2) ´ log(a3) + ¨ ¨ ¨ + log(aN) ´ log(aN+1) = ´ log(aN+1) and then take the limit N Ñ 8 ´ log a1 + 8 ÿ n=1 log(an) ´ log(an+1) = lim NÑ8 SN = ´ lim NÑ8 log(aN+1) = ´ log 2 = log 1 2 S-51: We are told that ř8 n=1 an converges. Thus we must have that lim nÑ8 an = 0. In particular, there is an index N such that 0 ď an ď 1 for all n ě N. Then: 0 ď a2 n ď an for n ą N By the direct comparison test, 8 ÿ n=N+1 a2 n converges.
Since convergence doesn’t depend on the first N terms of a series for any finite N, 8 ÿ n=1 a2 n converges as well.
658 S-52: The most-commonly used word makes up α percent of all the words. So, we want to find α.
If we add together the frequencies of all the words, they should amount to 100%. That is, 20,000 ÿ n=1 α n = 100 We can approximate the sum (with α left as a parameter) using the ideas behind the integral test. (See Example 3.3.4.) x y 1 2 3 4 α 5 f (x) = α x As we see in the diagram above, N ÿ n=1 α n (which is the sum of the areas of the rectangles) is greater than ż N+1 1 α xdx (the area under the curve). That is, ż N+1 1 α xdx ă N ÿ n=1 α n .
Using the fact that our language’s 20,000 words make up 100% of the words used, we can find a lower bound for α.
100 = 20,000 ÿ n=1 α n ą ż 20,001 1 α xdx = h α log(x) i20,001 1 = α log(20, 001) α ă 100 log(20, 001) We can find an upper bound for α in a similar manner.
x y 1 2 3 4 α 5 f (x) = α x 659 From the diagram, we see N ÿ n=2 α n (which is the sum of the areas of the rectangles, excluding the first) is less than ż N 1 α xdx. (The reason for excluding the first rectangle is to avoid comparing our series to an integral that diverges.) That is, N ÿ n=2 α n ă ż N 1 α xdx .
Therefore, 100 = 20,000 ÿ n=1 α n = α + 20,000 ÿ n=2 α n ă α + ż 20,00 1 α xdx = α + α log(20, 000) = α 1 + log(20, 000) α ą 100 1 + log(20, 000) Using a calculator, we see 9.17 ă α ă 10.01 So, the most-commonly used word makes up about 9-10 percent of the total words.
S-53: Generalizing our work in Question 52, we find the approximations: ż b+1 a 1 xdx ă b ÿ n=a 1 n ă ż b a´1 1 xdx when a ě 2. The inequality b+1 ş a 1 xdx ă b ř n=a 1 n can be read off of the sketch x y a ´ 1 b + 1 a b 1 f (x) = 1 x and the inequality b ř n=a 1 n ă b ş a´1 1 xdx can be read off of the sketch 660 x y a ´ 1 b + 1 a b 1 f (x) = 1 x We will evaluate the total population by writing 2ˆ106 ÿ n=1 2 ˆ 106 n = a´1 ÿ n=1 2 ˆ 106 n + 2ˆ106 ÿ n=a 2 ˆ 106 n and applying the above integral approximations to the second sum. We want our error to be less than one million, so we need to choose a value of a such that: 2 ˆ 106 ż 2ˆ106 a´1 1 xdx looooooooooomooooooooooon upper bound ´ 2 ˆ 106 ż 2ˆ106+1 a 1 xdx loooooooooooomoooooooooooon lower bound ă 106 ż 2ˆ106 a´1 1 xdx ´ ż 2ˆ106+1 a 1 xdx ă 1 2 h log 2 ˆ 106 ´ log(a ´ 1) i ´ h log 2 ˆ 106 + 1 ´ log(a) i ă 1 2 h log 2 ˆ 106 ´ log 2 ˆ 106 + 1 i + [log(a) ´ log(a ´ 1)] ă 1 2 log 2 ˆ 106 2 ˆ 106 + 1 + log a a ´ 1 ă 1 2 The first term is extremely close to 0, so we ignore it.
log a a ´ 1 ă 1 2 a a ´ 1 ă e1/2 = ?e a ă a?e ´ ?e ?e ă a(?e ´ 1) ?e ?e ´ 1 ă a Since ?e ?e ´ 1 « 2.5, we use a = 3. That is, we will approximate the value of 2ˆ106 ÿ n=3 1 n using an integral. Then, we will use that approximation to estimate our total population.
661 ż 2ˆ106+1 3 1 xdx ă 2ˆ106 ÿ n=3 1 n ă ż 2ˆ106 2 1 xdx log 2 ˆ 106 + 1 ´ log(3) ă 2ˆ106 ÿ n=3 1 n ă log 2 ˆ 106 ´ log(2) 1 + 1 2 + log 2 ˆ 106 + 1 ´ log(3) ă 2ˆ106 ÿ n=1 1 n ă 1 + 1 2 + log 2 ˆ 106 ´ log(2) 3 2 + log 2 ˆ 106 + 1 3 ă 2ˆ106 ÿ n=1 1 n ă 3 2 + 6 log(10) 2 ˆ 1063 2 + log 2 3 ˆ 106 + 1 3 ă 2ˆ106 ÿ n=1 2 ˆ 106 n ă 2 ˆ 1063 2 + 6 log (10) 29, 820, 091 ă population ă 30, 631, 021 Solutions to Exercises 3.4 — Jump to TABLE OF CONTENTS S-1: False. For example if bn = 1 n, then 8 ř n=1 (´1)n+1bn = 8 ř n=1 (´1)n+1 1 n converges by the alternating series test, but 8 ř n=1 1 n diverges by the p–test.
Remark: if we had added that tbnu is a sequence of alternating terms, then by Theorem 3.4.2 in the CLP-2 text, the statement would have been true. This is because 8 ÿ n=1 (´1)n+1bn would either be equal to 8 ÿ n=1 |bn| or ´ 8 ÿ n=1 |bn|.
S-2: Absolute convergence describes the situation where ř |an| converges (see Definition 3.4.1 in the CLP-2 text). By Theorem 3.4.2 in the CLP-2 text, this guarantees that also ř an converges.
Conditional convergence describes the situation where ř |an| diverges but ř an converges (see again Definition 3.4.1 in the CLP-2 text).
If ř an diverges, we just say it diverges. The reason is that if ř an diverges, we automatically know ř |an| diverges as well, so there’s no need for a special name.
ř an converges ř an diverges ř |an| converges converges absolutely not possible ř |an| diverges converges conditionally diverges 662 S-3: The series 8 ř n=1 (´1)n 9n+5 converges by the alternating series test. On the other hand the series 8 ř n=1 ˇ ˇ(´1)n 9n+5 ˇ ˇ = ř8 n=1 1 9n+5 diverges by the limit comparison test with bn = 1 n. So the given series is conditionally convergent.
S-4: Note that (´1)2n+1 = (´1) ¨ (´1)2n = ´1. So we can simplify 8 ÿ n=1 (´1)2n+1 1 + n = ´ 8 ÿ n=1 1 1 + n Since 1 1 + n ě 1 n + n = 1 2n, 8 ÿ n=1 1 1 + n diverges by the comparison test with the divergent harmonic series 8 ř n=1 1 n. The extra overall factor of ´1 in the original series does not change the conclusion of divergence.
S-5: Since lim nÑ8 1 + 4n 3 + 22n = lim nÑ8 1 + 4n 3 + 4n = 1 the alternating series test cannot be used. Indeed, lim nÑ8(´1)n´1 1 + 4n 3 + 22n does not exist (for very large n, (´1)n´1 1+4n 3+22n alternates between a number close to +1 and a number close to ´1) so the divergence test says that the series diverges. (Note that “none of the above” cannot possibly be the correct answer — every series either converges absolutely, converges conditionally, or diverges.) S-6: First, we’ll develop some intuition. For very large n ˇ ˇ ˇ ˇ ?n cos(n) n2 ´ 1 ˇ ˇ ˇ ˇ « ˇ ˇ ˇ ˇ ?n cos(n) n2 ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ cos(n) n3/2 ˇ ˇ ˇ ˇ ď 1 n3/2 since |cos(n)| ď 1 for all n. By the p-test, which is in Example 3.3.6 in the CLP-2 text, the series 8 ÿ n=5 1 np converges for all p ą 1. So we would expect the given series to converge absolutely.
Now, to confirm that our intuition is correct, we’ll first try the limit comparison theorem, 663 which is Theorem 3.3.11 in the CLP-2 text, with an = ˇ ˇ ˇ ?n cos(n) n2´1 ˇ ˇ ˇ and bn = 1 n3/2.
lim nÑ8 an bn = lim nÑ8 ˇ ˇ ˇ ?n cos(n) n2´1 ˇ ˇ ˇ 1 n3/2 = lim nÑ8 ?n ¨ ?n3| cos n| n2 ´ 1 = lim nÑ8 n2| cos n| n2 ´ 1 = lim nÑ8 1 1 ´ 1/n2 | cos n| = lim nÑ8 1 ¨ | cos n| Unfortunately, this limit doesn’t exist, so this attempt to use the limit comparison theorem has failed. Fortunately, having seen that the cos n caused the failure, it is not hard to adjust our strategy to get a successful proof of absolute convergence. First, in step 1 below, we use the comparison test to eliminate the cos n and then, in step 2 below, we apply the limit comparison test.
Step 1: Since | cos n| ď 1, we have ˇ ˇ ˇ ˇ ?n cos(n) n2 ´ 1 ˇ ˇ ˇ ˇ ď ?n n2 ´ 1 for all n ą 1. So, by part (a) of the comparison test, which is Theorem 3.3.8 in the CLP-2 text, if the series 8 ÿ n=5 ?n n2 ´ 1 converges, then we will have that the series 8 ÿ n=5 ˇ ˇ ˇ ˇ ?n cos(n) n2 ´ 1 ˇ ˇ ˇ ˇ also converges, and hence that the series 8 ÿ n=5 ?n cos(n) n2 ´ 1 converges absolutely.
Step 2: Now, to prove that the series 8 ÿ n=5 ?n n2 ´ 1 converges, we apply the limit comparison test with an = ?n n2´1 and bn = 1 n3/2 (for n ě 5). Since lim nÑ8 an bn = lim nÑ8 ?n n2´1 1 n3/2 = lim nÑ8 ?n ¨ ?n3 n2 ´ 1 = lim nÑ8 n2 n2 ´ 1 = lim nÑ8 1 1 ´ 1/n2 = 1 and since 8 ÿ n=5 1 n3/2 converges by the p-test, the limit comparison test tells us that the series 8 ÿ n=5 ?n n2 ´ 1 converges. So, by step 1, 8 ÿ n=5 ?n cos(n) n2 ´ 1 converges absolutely.
S-7: We first develop some intuition about 8 ÿ n=1 ˇ ˇ ˇ ˇ n2 ´ sin n n6 + n2 ˇ ˇ ˇ ˇ, where we take the absolute value of the summands to consider whether the series converges absolutely. For very 664 large n, n2 dominates sin n and n6 dominates n2 so that ˇ ˇ ˇ ˇ n2 ´ sin n n6 + n2 ˇ ˇ ˇ ˇ « n2 n6 = 1 n4 The series 8 ÿ n=1 1 n4 converges by the p–test with p = 4 ą 1. We expect the given series to converge too.
To verify that our intuition is correct, we apply the limit comparison test with an = n2 ´ sin n n6 + n2 and bn = 1 n4 which is valid since lim nÑ8 an bn = lim nÑ8 ˇ ˇ ˇ ˇ (n2 ´ sin n) n6 + n2 ˇ ˇ ˇ ˇ ¨ n4 1 = lim nÑ8 |n6 ´ n4 sin n| n6 + n2 = lim nÑ8 1 ´ n´2 sin n 1 + n´4 = 1 exists and is nonzero. Since the series 8 ř n=1 bn converges, the series 8 ř n=1 |n2 ´ sin n| n6 + n2 converges too. Therefore, the series 8 ř n=1 n2 ´ sin n n6 + n2 converges absolutely.
S-8: You might think that this series converges by the alternating series test. But you would be wrong. The problem is that tanu does not converge to zero as n Ñ 8, so that the series actually diverges by the divergence test. To verify that the nth term does not converge to zero as n Ñ 8 let’s write an = (2n)!
(n2+1)(n!)2 (i.e. an is the nth term without the sign) and check to see whether an+1 is bigger than or smaller than an.
an+1 an = (2n + 2)!
((n + 1)2 + 1)((n + 1)!)2 (n2 + 1)(n!)2 (2n)!
= (2n + 2)(2n + 1) (n + 1)2 n2 + 1 (n + 1)2 + 1 = 2(2n + 1) (n + 1) 1 + 1/n2 (1 + 1/n)2 + 1/n2 = 41 + 1/2n 1 + 1/n 1 + 1/n2 (1 + 1/n)2 + 1/n2 So lim nÑ8 an+1 an = 4 and, in particular, for large n, an+1 ą an. Thus, for large n, an increases with n and so cannot converge to 0. So the series diverges by the divergence test.
S-9: This series converges by the alternating series test. We want to know whether it converges absolutely, so we consider the seris 8 ÿ n=2 ˇ ˇ ˇ ˇ (´1)n n(log n)101 ˇ ˇ ˇ ˇ = 8 ÿ n=2 1 n(log n)101.
We’ve seen similar function before (e.g. Example 3.3.7 in the CLP-2 text, with p = 101 ą 1) and it yields nicely to the integral test. Let f (x) = 1 x(log x)101. Note f (x) is 665 positive and decreasing for n ě 3. Then by the integral test, the series ř8 n=2 1 n(log n)101 converges if and only if the integral ş8 2 1 x(log x)101dx does. We evaluate the integral using the substitution u = log x, du = 1 x dx.
ż 8 2 1 x(log x)101dx = lim bÑ8 ż b 2 1 x(log x)101dx = lim bÑ8 ż log b log 2 1 u101 du = lim bÑ8 ´1 100u100 log b log 2 = 1 100(log 2)100 Since the integral converges, the series 8 ř n=2 1 n(log n)101 converges, and therefore the series 8 ř n=2 (´1)n n(log n)101 converges absolutely.
S-10: The sequence has some positive terms and some negative terms, which limits the tests we can use. However, if we consider the series 8 ÿ n=1 ˇ ˇ ˇ ˇ sin n n2 ˇ ˇ ˇ ˇ, we can use the direct comparison test.
For every n, | sin n| ă 1, so 0 ď ˇ ˇ ˇ ˇ sin n n2 ˇ ˇ ˇ ˇ ă 1 n2. Since 8 ÿ n=1 1 n2 converges, then by the direct comparison test, 8 ÿ n=1 ˇ ˇ ˇ ˇ sin n n2 ˇ ˇ ˇ ˇ converges as well. Then 8 ÿ n=1 sin n n2 converges absolutely– in particular, it converges.
S-11: The terms of this series are sometimes negative (for odd values of n where sin n ă 1 2) and sometimes positive. But, they are not strictly alternating, so we can’t use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.
666 ´1 4 ď sin n 4 ď 1 4 ñ ´1 4 ´ 1 8 ď sin x 4 ´ 1 8 ă 1 4 ´ 1 8 ñ ´3 8 ď sin x 4 ´ 1 8 ă 1 8 ñ 0 ď ˇ ˇ ˇ ˇ sin x 4 ´ 1 8 ˇ ˇ ˇ ˇ ă 3 8 ñ 0 ď ˇ ˇ ˇ ˇ sin x 4 ´ 1 8 nˇ ˇ ˇ ˇ ă 3 8 n Since 8 ÿ n=1 3 8 n converges (it’s a geometric sum with |r| ă 1), by the direct comparison test, 8 ÿ n=1 ˇ ˇ ˇ ˇ sin x 4 ´ 1 8 nˇ ˇ ˇ ˇ converges as well.
Then 8 ÿ n=1 sin x 4 ´ 1 8 n converges absolutely–and so it converges.
S-12: The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can’t use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.
ˇ ˇ ˇ ˇ ˇ sin2 n ´ cos2 n + 1 2 2n ˇ ˇ ˇ ˇ ˇ ď 1 + 1 + 1 2 2n = 5 2n+1 The series 8 ÿ n=1 5 2n+1 converges, because it’s a geometric series with r = 1 2. By the direct comparison test, 8 ÿ n=1 ˇ ˇ ˇ ˇ ˇ sin2 n ´ cos2 n + 1 2 2n ˇ ˇ ˇ ˇ ˇ converges as well. Then 8 ÿ n=1 sin2 n ´ cos2 n + 1 2 2n converges absolutely, so it converges.
S-13: (a) Solution 1: We need to show that 8 ř n=1 24n2e´n3 converges. If we replace n by x in the summand, we get f (x) = 24x2e´x3, which we can integate. (Just substitute u = x3.) So let’s try the integral test. First, we have to check that f (x) is positive and decreasing. It is certainly positive. To determine if it is decreasing, we compute df dx = 48xe´x3 ´ 24 ˆ 3x4e´x3 = 24x(2 ´ 3x3)e´x3 667 which is negative for x ě 1. Therefore f (x) is decreasing for x ě 1, and the integral test applies. The substitution u = x3, du = 3x2 dx, yields ż f (x) dx = ż 24x2e´x3 dx = ż 8e´u du = ´8e´u + C = ´8e´x3 + C.
Therefore ż 8 1 f (x) dx = lim RÑ8 ż R 1 f (x) dx = lim RÑ8 ´8e´x3R 1 = lim RÑ8(´8e´R3 + 8e´1) = 8e´1 Since the integral is convergent, the series 8 ř n=1 24n2e´n3 converges and the series 8 ÿ n=1 (´1)n´124n2e´n3 converges absolutely.
Solution 2: Alternatively, we can use the ratio test with an = 24n2e´n3. We calculate lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ ˇ 24(n + 1)2e´(n+1)3 24n2e´n3 ˇ ˇ ˇ ˇ ˇ = lim nÑ8 (n + 1)2 n2 en3 e(n+1)3 !
= lim nÑ8 1 + 1 n 2 e´(3n2+3n+1) = 1 ¨ 0 = 0 ă 1, and therefore the series converges absolutely.
Solution 3: Alternatively, alternatively, we can use the limiting comparison test. First a little intuition building. Recall that we need to show that 8 ř n=1 24n2e´n3 converges.
The nth term in this series is an = 24n2e´n3 = 24n2 en3 It is a ratio with both the numerator and denominator growing with n. A good rule of thumb is that exponentials grow a lot faster than powers. For example, if n = 10 the numerator is 2400 = 2.4 ˆ 103 and the denominator is about 2 ˆ 10434. So we would guess that an tends to zero as n Ñ 8. The question is “does an tend to zero fast enough with n that our series converges?”. For example, we know that ř8 n=1 1 n2 converges (by the p–test with p = 2). So if an tends to zero faster than 1 n2 does, our series will converge. So let’s try the limiting convergence test with an = 24n2e´n3 = 24n2 en3 and bn = 1 n2.
lim nÑ8 an bn = lim nÑ8 24n2e´n3 1/n2 = lim nÑ8 24n4 en3 668 By l’Hˆ opital’s rule, twice, lim xÑ8 24x4 ex3 = lim xÑ8 4 ˆ 24x3 3x2ex3 by l’Hˆ opital = lim xÑ8 32x ex3 just cleaning up = lim xÑ8 32 3x2ex3 by l’Hˆ opital, again = 0 That’s it. The limit comparison test now tells us that ř8 n=1 an converges.
(b) In part (a) we saw that 24n2e´n3 is positive and decreasing. The limit of this sequence equals 0 (as can be shown with l’Hˆ opital’s Rule, just as we did at the end of the third solution of part (a)). Therefore, we can use the alternating series test, so that the error made in approximating the infinite sum S = 8 ř n=1 an = 8 ř n=1 (´1)n´124n2e´n3 by the sum of its first N terms, SN = N ř n=1 an, lies between 0 and the first omitted term, aN+1. If we use 5 terms, the error satisfies |S ´ S5| ď |a6| = 24 ˆ 36e´63 « 1.3 ˆ 10´91 S-14: The error in our approximation using through term N is at most 1 (2(N+1))!. We want 1 (2(N+1))! ă 1 1000. By checking small values of N, we see that 8! = 40320 ą 1000, so if N = 3, then 1 2(N+1)! = 1 40320 ă 1 1000. So, for our approximation, it suffices to consider the first four terms of our series.
cos(1) « 3 ÿ N=0 (´1)n (2n)! = 1 0! ´ 1 2! + 1 4! ´ 1 6!
= 1 ´ 1 2 + 1 24 ´ 1 720 = 720 ´ 360 + 30 ´ 1 720 = 389 720 When we use a calculator, we see 389 720 = 0.540277 cos(1) « 0.540302 cos(1) ´ 389 720 « 0.000024528 « 1 40770 So, our error is reasonably close to our bound of 1 40320, and far smaller than 1 1000.
669 S-15: The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can’t use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.
If n is prime, then ˇ ˇ ˇan en ˇ ˇ ˇ = ˇ ˇ ˇ ˇ ˇ´en/2 en ˇ ˇ ˇ ˇ ˇ = 1 en/2 = 1 ?e n If n is not prime, then ˇ ˇ ˇan en ˇ ˇ ˇ = ˇ ˇ ˇ ˇ´n2 en ˇ ˇ ˇ ˇ = n2 en For n sufficiently large, n2 ă en/2, so for n sufficiently large, n2 en ď 1 ?e n .
Since e ą 1, then ?e ą 1, so the geometric series ÿ 1 ?e n has |r| = r = 1 ?e ă 1, so it converges. By the direct comparison test, 8 ÿ n=1 ˇ ˇ ˇan en ˇ ˇ ˇ converges as well. Then 8 ÿ n=1 an en converges absolutely, so it converges.
Solutions to Exercises 3.5 — Jump to TABLE OF CONTENTS S-1: f (1) = 8 ÿ n=0 3 ´ 1 4 n = 8 ÿ n=0 1 2 n This is a geometric series with r = 1 2, so we know that it converges and = 1 1 ´ 1 2 = 2 The question does not ask us to find the interval of convergence of the series defining f (x). But we will do so anyway, to get a bit more practice. We may rewrite the nth term of the series defining f (x) as 3 ´ x 4 n = arn with a = 1 and r = 3 ´ x 4 670 That is, for every fixed x, we have a geometric series with r = 3´x 4 . So, by (3.2.2) in the CLP-2 text, the series converges if and only if |r| = ˇ ˇ ˇ ˇ 3 ´ x 4 ˇ ˇ ˇ ˇ ă 1 ð ñ ´ 1 ă 3 ´ x 4 ă 1 ð ñ ´ 4 ă 3 ´ x ă 4 ð ñ ´ 4 ă 3 ´ x and 3 ´ x ă 4 ð ñ ´ 1 ă x ă 7 S-2: By Theorem 3.5.13 in the CLP-2 text, we may differentiate our function term-by-term for all x obeying |x ´ 5| ă R, where R is the radius of convergence of the power series.
The series defining f (x) is reminiscent of the exponential series ř8 n=0 Xn n! of Example 3.5.5 in the CLP-2 text. In that example, we showed that ř8 n=0 Xn n! has radius of convergence 8.
Since ˇ ˇ ˇ ˇ (x ´ 5)n n! + 2 ˇ ˇ ˇ ˇ ď Xn n!
with X = |x ´ 5| the comparison test, Theorem 3.3.8 in the CLP-2 text, tells us that ř8 n=1 (x´5)n n!+2 converges for all x. So we may differentiate our function term-by-term.
f (x) = 8 ÿ n=1 (x ´ 5)n n! + 2 f 1(x) = 8 ÿ n=1 d dx "(x ´ 5)n n! + 2 = 8 ÿ n=1 n(x ´ 5)n´1 n! + 2 Keep in mind that x is our variable, and for each term, n is constant.
S-3: If x = c, then f (x) = Aa(c ´ c)a + Aa+1(c ´ c)a+1 + Aa+2(c ´ c)a+2 + ¨ ¨ ¨ = Aa ¨ 0 + Aa+1 ¨ 0 + Aa+2 ¨ 0 + ¨ ¨ ¨ = 0 So, f (x) converges (to the constant 0) when x = c. (Had we allowed a = 0, it would be possible for f (x) to converge to a nonzero number A0, because we use the convention 00 = 1.) Depending on the sequence tAnu, it’s possible that f (x) diverges for all x ‰ c. For example, suppose An = n!, so f (x) = 8 ÿ n=0 n!(x ´ c)n. If x ‰ c, then the limit 671 lim nÑ8 ˇ ˇ ˇ ˇ (n + 1)!(x ´ c)n+1 n!(x ´ c)n ˇ ˇ ˇ ˇ = lim nÑ8(n + 1)|x ´ c| is infinity, since x ´ c ‰ 0. So, the series diverges.
We’ve now shown that the series definitely converges at x = c, but at any other point, it may fail to converge.
S-4: According to Theorem 3.5.9 in the CLP-2 text, because f (x) diverges somewhere, and because it converges at a point other than its centre, f (x) has a positive radius of convergence R. That is, f (x) converges whenever |x ´ 5| ă R, and it diverges whenever |x ´ 5| ą R.
Since we are told that the series diverges at x = 11, the statement |11 ´ 5| ă R must be false. That is, we must have R ď |11 ´ 5| = 6.
Since we are told that the series converges at x = ´1, the statement | ´ 1 ´ 5| ą R must be false. That is, we must have R ě | ´ 1 ´ 5| = 6.
Therefore, R = 6.
S-5: (a) We give two solutions. For the first solution, we apply the ratio test for the series whose kth term is ak = (´1)k2k+1xk. Then lim kÑ8 ˇ ˇ ˇ ˇ ak+1 ak ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ (´1)k+12k+2xk+1 (´1)k2k+1xk ˇ ˇ ˇ ˇ = lim kÑ8 |2x| = |2x| Therefore, by the ratio test, the series converges for all x obeying |2x| ă 1, i.e. |x| ă 1 2, and diverges for all x obeying |2x| ą 1, i.e. |x| ą 1 2. So the radius of convergence is R = 1 2.
For the second solution we apply (3.5.2) in the CLP-2 text. To do so, we set Ak = (´1)k2k+1 and compute A = lim kÑ8 ˇ ˇ ˇ ˇ Ak+1 Ak ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ (´1)k+12k+2 (´1)k2k+1 ˇ ˇ ˇ ˇ = lim kÑ8 2 = 2 so that R = 1 A = 1 2, again.
(b) The series is 8 ÿ k=0 (´1)k2k+1xk = 2 8 ÿ k=0 (´2x)k = 2 8 ÿ k=0 rkˇ ˇ ˇ r=´2x = 2 ˆ 1 1 ´ r = 2 1 + 2x for all |r| = |2x| ă 1, i.e. all |x| ă 1 2.
S-6: We give two solutions. For the first solution, we apply the ratio test for the series 672 whose kth term is ak = xk 10k+1(k+1)!. Then lim kÑ8 ˇ ˇ ˇ ˇ ak+1 ak ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ xk+1 10k+2(k + 2)! ¨ 10k+1(k + 1)!
xk ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ ˇ 10k+1 10k+2 ˇ ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ (k + 1)!
(k + 2)!
ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ ˇ xk+1 xk ˇ ˇ ˇ ˇ ˇ = lim kÑ8 1 10(k + 2)|x| = 0 ă 1 for all x. Therefore, by the ratio test, the series converges for all x and the radius of convergence is R = 8.
For the second solution we apply (3.5.2) in the CLP-2 text. To do so, we set Ak = 1 10k+1(k + 1)! and compute A = lim kÑ8 ˇ ˇ ˇ ˇ Ak+1 Ak ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ ˇ 10k+1 10k+2 ˇ ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ (k + 1)!
(k + 2)!
ˇ ˇ ˇ ˇ = 0 So, by the bullet just after (3.5.2) in the CLP-2 text, R is again +8.
S-7: We give two solutions. For the first solution, we apply the ratio test with an = (x´2)n n2+1 .
lim nÑ8 ˇ ˇ ˇan+1 an ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (x ´ 2)n+1 (n + 1)2 + 1 ¨ n2 + 1 (x ´ 2)n ˇ ˇ ˇ ˇ = lim nÑ8 n2 + 1 (n + 1)2 + 1|x ´ 2| = lim nÑ8 1 + 1/n2 (1 + 1/n)2 + 1/n2|x ´ 2| = |x ´ 2| So, the series converges if |x ´ 2| ă 1 and diverges if |x ´ 2| ą 1. That is, the radius of convergence is 1.
For the second solution we apply (3.5.2) in the CLP-2 text. To do so, we set An = 1 n2+1 and compute A = lim nÑ8 ˇ ˇ ˇ ˇ An+1 An ˇ ˇ ˇ ˇ = lim nÑ8 n2 + 1 (n + 1)2 + 1 = lim nÑ8 1 + 1/n2 (1 + 1/n)2 + 1/n2 = 1 so that R = 1 A = 1, again.
S-8: We will first find the radius of convergence R of the given series. Once we have found R, we will know that the series converges for |x + 2| ă R and diverges for |x + 2| ą R. Then we will determine what happens at the two end points, x + 2 = R and x + 2 = ´R, of the interval.
673 To determine the radius of convergence, we apply (3.5.2) in the CLP-2 text. To do so, we set An = (´1)n ?n and compute A = lim nÑ8 ˇ ˇ ˇ ˇ An+1 An ˇ ˇ ˇ ˇ = lim nÑ8 ?n ?
n + 1 = lim nÑ8 1 ?
1 + 1/n = 1 so that R = 1 A = 1.
So the series must converge when |x + 2| ă 1 and must diverge when |x + 2| ą 1. When x + 2 = 1, the series reduces to 8 ÿ n=1 (´1)n ?n which converges by the alternating series test. When x + 2 = ´1, the series reduces to 8 ÿ n=1 1 ?n which diverges by the p–series test with p = 1 2. So the interval of convergence is ´1 ă x + 2 ď 1 or (´3, ´1].
S-9: We will first find the radius of convergence R of the given series. Once we have found R, we will know that the series converges for |x + 1| ă R and diverges for |x + 1| ą R. Then we will determine what happens at the two end points, x + 1 = R and x + 1 = ´R, of the interval.
We could determine the radius of convergence by applying (3.5.2), from the CLP-2 text, with An = (´1)n (n+1)3n. As an alternate method, we apply the ratio test for the series whose nth term is an = (´1)n n+1 x+1 3 n .
lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (´1)n+1 n+2 x+1 3 n+1 (´1)n n+1 x+1 3 n ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (´1)n+1 (´1)n ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ n + 1 n + 2 ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ (x + 1)n+1 (x + 1)n ˇ ˇ ˇ ˇ ¨ ˇ ˇ ˇ ˇ 3n 3n+1 ˇ ˇ ˇ ˇ = lim nÑ8 n + 1 n + 2 ¨ ˇ ˇ ˇ ˇ x + 1 3 ˇ ˇ ˇ ˇ = |x + 1| 3 Therefore, by the ratio test, the series converges when |x+1| 3 ă 1 and diverges when |x+1| 3 ą 1. In particular, it converges when |x + 1| ă 3 ð ñ ´3 ă x + 1 ă 3 ð ñ ´4 ă x ă 2 674 and the radius of convergence is R = 3.
Next, we consider the endpoints 2 and ´4. At x = 2, i.e. x + 1 = 3, the series is simply ř8 n=0 (´1)n n+1 , which is an alternating series: the signs alternate, and the unsigned terms decrease to zero. Therefore the series converges at x = 2 by the alternating series test.
At x = ´4 the series is 8 ÿ n=0 (´1)n n + 1 ´4 + 1 3 n = 8 ÿ n=0 (´1)n n + 1 (´1)n = 8 ÿ n=0 1 n + 1, since (´1)n ¨ (´1)n = (´1)2n = (´1)2n = 1. This series diverges, either by comparison or limit comparison with the harmonic series (the p-series with p = 1). (For that matter, it is exactly equal to the standard harmonic series ř8 n=1 1 n, re-indexed to start at n = 0.) In summary, the interval of convergence is ´4 ă x ď 2, or simply (´4, 2].
S-10: We will first find the radius of convergence R of the given series. Once we have found R, we will know that the series converges for |x ´ 2| ă R and diverges for |x ´ 2| ą R. Then we will determine what happens at the two end points, x ´ 2 = R and x ´ 2 = ´R, of the interval.
To determine the radius of convergence, we apply (3.5.2) in the CLP-2 text. To do so, we set An = 1 n4/5(5n´4) and compute A lim nÑ8 ˇ ˇ ˇ ˇ An+1 An ˇ ˇ ˇ ˇ = lim nÑ8 n4/5(5n ´ 4) (n + 1)4/5(5n+1 ´ 4) = lim nÑ8 (1 ´ 4/5n) (1 + 1/n)4/5(5 ´ 4/5n) = 1 5 so that R = 1 A = 5.
Therefore the series converges if |x ´ 2| ă 5 and diverges if |x ´ 2| ą 5. When x ´ 2 = +5, i.e. x = 7, the series reduces to 8 ř n=1 5n n4/5(5n´4) = 8 ř n=1 1 n4/5(1´4/5n) which diverges by the limit comparison test with bn = 1 n4/5. When x ´ 2 = ´5, i.e. x = ´3, the series reduces to 8 ř n=1 (´5)n n4/5(5n´4) = 8 ř n=1 (´1)n n4/5(1´4/5n) which converges by the alternating series test. So the interval of convergence is ´3 ď x ă 7 or [´3, 7).
S-11: We determine the radius of convergence by applying (3.5.2) in the CLP-2 text. To do 675 so, we set An = 1 n2 and compute A = lim nÑ8 ˇ ˇ ˇ ˇ An+1 An ˇ ˇ ˇ ˇ = lim nÑ8 n2 (n + 1)2 = lim nÑ8 1 (1 + 1/n)2 = 1 So we have convergence for |x + 2| ă 1 ð ñ ´1 ă x + 2 ă 1 ð ñ ´3 ă x ă ´1 and divergence for |x + 2| ą 1. For |x + 2| = 1, i.e. for x + 2 = ˘1, i.e. for x = ´3, ´1, the series reduces to 8 ř n=1 (˘1)n n2 , which converges absolutely, because 8 ř n=1 1 np converges for p = 2 ą 1. So the given series converges if and only if ´3 ď x ď ´1.
S-12: (a) We could determine the radius of convergence by applying (3.5.2), from the CLP-2 text, with An = 4n n . As an alternate method, we apply the ratio test for the series whose nth term is an = 4n n (x ´ 1)n. Since lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ 4n+1(x ´ 1)n+1/(n + 1) 4n(x ´ 1)n/n ˇ ˇ ˇ ˇ = lim nÑ8 4|x ´ 1| n n + 1 = 4|x ´ 1| lim nÑ8 n n + 1 = 4|x ´ 1| ¨ 1.
the series converges if 4|x ´ 1| ă 1 ð ñ ´1 ă 4(x ´ 1) ă 1 ð ñ ´1 4 ă x ´ 1 ă 1 4 ð ñ 3 4 ă x ă 5 4 and diverges if 4|x ´ 1| ą 1. Checking the right endpoint x = 5 4, we see that 8 ÿ n=1 4n n 5 4 ´ 1 n = 8 ÿ n=1 1 n is the divergent harmonic series. At the left endpoint x = 3 4, 8 ÿ n=1 4n n 3 4 ´ 1 n = 8 ÿ n=1 (´1)n n converges by the alternating series test. Therefore the interval of convergence of the original series is 3 4 ď x ă 5 4, or 3 4, 5 4 .
676 (b) We could determine the radius of convergence by applying (3.5.2), from the CLP-2 text. But to do so, we would first have to rewrite the series as ř8 n=1 8n n x + 1 2 n to get the nth term into the form An(x ´ c)n required by (3.5.2). Instead, we use the alternate method that consists of applying the ratio test with an = 4n n (2x + 1)n. Repeating the computation of part (a), just with x ´ 1 replaced by 2x + 1, lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ 4n+1(2x + 1)n+1/(n + 1) 4n(2x + 1)n/n ˇ ˇ ˇ ˇ = 4|2x + 1| lim nÑ8 n n + 1 = 4|2x + 1| ¨ 1.
So the series converges if 4|2x + 1| ă 1 ð ñ ´1 ă 4(2x + 1) ă 1 ð ñ ´1 4 ă 2x + 1 ă 1 4 ð ñ ´5 4 ă 2x ă ´3 4 ð ñ ´5 8 ă x ă ´3 8 and diverges if 4|2x + 1| ą 1. At the right endpoint x = ´3 8, the series becomes 8 ÿ n=1 4n n ´ 3 4 + 1 n = 8 ÿ n=1 1 n which is the divergent harmonic series. At the left endpoint x = ´5 8, 8 ÿ n=1 4n n ´ 5 4 + 1 n = 8 ÿ n=1 (´1)n n converges by the alternating series test. Therefore the interval of convergence of the original series is ´5 8 ď x ă ´3 8, or ´ 5 8, ´3 8 .
(c) This time we cannot determine the radius of convergence by applying (3.5.2), from the CLP-2 text, because the nth term is not in the form An(x ´ c)n required by (3.5.2). But we may still apply the ratio test with an = 4n n (2x + 1)2n. Repeating the computation of part (a), just with x ´ 1 replaced by (2x + 1)2, lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ 4n+1(2x + 1)2(n+1)/(n + 1) 4n(2x + 1)2n/n ˇ ˇ ˇ ˇ = 4|2x + 1|2 lim nÑ8 n n + 1 = 4|2x + 1|2 ¨ 1.
677 So the series converges if 4|2x + 1|2 ă 1 ð ñ ´1 ă 2(2x + 1) ă 1 ð ñ ´1 2 ă 2x + 1 ă 1 2 ð ñ ´3 2 ă 2x ă ´1 2 ð ñ ´3 4 ă x ă ´1 4 and diverges if 4|2x + 1|2 ą 1. At the right endpoint x = ´1 4, the series becomes 8 ÿ n=1 4n n ´ 1 2 + 1 2n = 8 ÿ n=1 1 n which is the divergent harmonic series. At the left endpoint x = ´3 4, 8 ÿ n=1 4n n ´ 3 2 + 1 2n = 8 ÿ n=1 1 n which is again the divergent harmonic series. Therefore the interval of convergence of the original series is ´3 4 ă x ă ´1 4, or ´ 3 4, ´1 4 .
S-13: We determine the radius of convergence by applying (3.5.2) in the CLP-2 text. To do so, we set An = (´1)n 2n(n+2) and compute A = lim nÑ8 ˇ ˇ ˇ ˇ An+1 An ˇ ˇ ˇ ˇ = lim nÑ8 2n(n + 2) 2n+1(n + 3) = lim nÑ8 1 2 n + 2 n + 3 = 1 2 lim nÑ8 1 + 2/n 1 + 3/n = 1 2 So the radius of convergence is R = 1 A = 2 and the series converges if |x ´ 1| ă 2 ð ñ ´2 ă (x ´ 1) ă 2 ð ñ ´1 ă x ă 3 and diverges if |x ´ 1| ą 2. Checking the left endpoint x = ´1, we see that 8 ÿ n=0 (´1)n (´1 ´ 1)n 2n(n + 2) = 8 ÿ n=0 1 n + 2 is the divergent harmonic series. At the right endpoint x = 3, so that 8 ÿ n=0 (´1)n (3 ´ 1)n 2n(n + 2) = 8 ÿ n=0 (´1)n n + 2 678 converges by the alternating series test. Therefore the interval of convergence of the original series is ´1 ă x ď 3, or ´ 1, 3 .
S-14: We cannot determine the radius of convergence by applying (3.5.2), from the CLP-2 text, because the nth term is not in the form An(x ´ c)n required by (3.5.2). But we may still apply the ratio test with an = (´1)nn2(x ´ a)2n. Since lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (´1)n+1(n + 1)2(x ´ a)2(n+1) (´1)nn2(x ´ a)2n ˇ ˇ ˇ ˇ = lim nÑ8 |x ´ a|2(n + 1)2 n2 = |x ´ a|2 lim nÑ8 1 + 1/n 2 = |x ´ a|2 ¨ 1.
the series converges if |x ´ a|2 ă 1 ð ñ |x ´ a| ă 1 ð ñ ´1 ă x ´ a ă 1 ð ñ a ´ 1 ă x ă a + 1 and diverges if |x ´ a| ą 1. Checking both endpoints x ´ a = ˘1, we see that 8 ÿ n=1 (´1)nn2(x ´ a)2n ˇ ˇ ˇ ˇ x´a=˘1 = 8 ÿ n=1 (´1)nn2 fails the divergence test — the nth term does not converge to zero as n Ñ 8. Therefore the interval of convergence of the original series is a ´ 1 ă x ă a + 1, or a ´ 1, a + 1 .
S-15: (a) We could determine the radius of convergence by applying (3.5.2), from the CLP-2 text, with An = 1 n29n. As an alternate method, we apply the ratio test for the series whose kth term is ak = (x+1)k k29k . Then lim kÑ8 ˇ ˇ ˇ ˇ ak+1 ak ˇ ˇ ˇ ˇ = lim kÑ8 ˇ ˇ ˇ ˇ (x + 1)k+1 (k + 1)29k+1 k29k (x + 1)k ˇ ˇ ˇ ˇ = lim kÑ8 |x + 1| 1 9 k2 (k + 1)2 = lim kÑ8 |x + 1| 1 9 1 (1 + 1/k)2 = |x + 1| 9 So the series must converge when |x + 1| ă 9 and must diverge when |x + 1| ą 9. When x + 1 = ˘9, the series reduces to 8 ÿ k=1 (˘9)k k29k = 8 ÿ k=1 (˘1)k k2 which converges (since, by the p–test, ř8 k=1 1 kp converges for any p ą 1). So the interval of convergence is |x + 1| ď 9 or ´10 ď x ď 8 or [´10, 8].
679 (b) The partial sum N ÿ k=1 ak ak+1 ´ ak+1 ak+2 = a1 a2 ´ a2 a3 + a2 a3 ´ a3 a4 + ¨ ¨ ¨ + aN aN+1 ´ aN+1 aN+2 = a1 a2 ´ aN+1 aN+2 We are told that 8 ÿ k=1 ak ak+1 ´ ak+1 ak+2 = a1 a2 . This means that the above partial sum converges to a1 a2 as N Ñ 8, or equivalently, that lim NÑ8 aN+1 aN+2 = 0 and hence that lim kÑ8 |ak+1(x ´ 1)k+1| |ak(x ´ 1)k| = |x ´ 1| lim kÑ8 |ak+1| |ak| is infinite for any x ‰ 1. So, by the ratio test, this series converges only for x = 1.
S-16: Using the geometric series 8 ř n=0 xn = 1 1´x, x3 1 ´ x = x3 8 ÿ n=0 xn = 8 ÿ n=0 xn+3 = 8 ÿ n=3 xn S-17: We can find f (x) by differentiating its integral, or antidifferentiating its derivative.
In the latter case, we’ll have to solve for the arbitrary constant of integration; in the former case, we do not. (Remember that many different functions have the same derivative, but a single function has only one derivative.) To avoid the necessity of finding the arbitrary constant, we can ignore the given equation for f 1(x), which makes the problem much simpler. This is the method used in Solution 1.
Solution 1 Using the Fundamental Theorem of Calculus Part 1: d dx "ż x 5 f (t)dt = f (x) So, f (x) = d dx # 3x + 8 ÿ n=0 (x ´ 1)n+1 n(n + 1)2 + = 3 + 8 ÿ n=1 (n + 1)(x ´ 1)n n(n + 1)2 = 3 + 8 ÿ n=1 (x ´ 1)n n(n + 1) 680 Solution 2 Suppose we had used f 1(x) instead. We would antidifferentiate to find: f (x) = ż 8 ÿ n=0 (x ´ 1)n n + 2 !
dx = 8 ÿ n=0 (x ´ 1)n+1 (n + 1)(n + 2) !
+ C = 8 ÿ n=1 (x ´ 1)n n(n + 1) !
+ C Notice f (1) = 0 + C. So, to find C, we must find f (1). We can’t get that information from f 1(x), so our only option is to consider the given formula for şx 5 f (t)dt. Using the Fundamental Theorem of Calculus Part 1: f (1) = d dx "ż x 5 f (t)dt ˇ ˇ ˇ ˇ x=1 = d dx # 3x + 8 ÿ n=1 (x ´ 1)n+1 n(n + 1)2 +ˇ ˇ ˇ ˇ ˇ x=1 = " 3 + 8 ÿ n=1 (n + 1)(x ´ 1)n n(n + 1)2 # x=1 = " 3 + 8 ÿ n=1 (x ´ 1)n n(n + 1) # x=1 = 3 + 8 ÿ n=1 0n n(n + 1) = 3 So, f (x) = 3 + 8 ÿ n=1 (x ´ 1)n n(n + 1).
Note that in Solution 2, we did the same calculation as Solution 1, and more.
S-18: We’ll start by applying the ratio test twice, • once for the series whose nth term is an = xn 32n log n, and • once for the series whose nth term is an = ˇ ˇ xn 32n log n ˇ ˇ.
681 For both series, i.e. for both an’s, lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ xn+1 32(n+1) log(n + 1) 32n log n xn ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ x log n 32 log(n + 1) ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ x log n 32[log(n) + log(1 + 1/n)] ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ x 32[1 + log(1 + 1/n)/ log(n)] ˇ ˇ ˇ ˇ = |x| 9 Therefore, by the ratio test, when |x| ă 9, the series ř8 n=2 ˇ ˇ xn 32n log n ˇ ˇ converges and the series ř8 n=2 xn 32n log n converges absolutely, and when |x| ą 9 the series ř8 n=2 xn 32n log n diverges. That just leaves x = ˘9.
For x = ´9, 8 ÿ n=2 xn 32n log n = 8 ÿ n=2 (´1)n log n which converges by the alternating series test.
For x = +9, 8 ÿ n=2 xn 32n log n = 8 ÿ n=2 1 log n which is the same series as 8 ÿ n=2 ˇ ˇ ˇ(´1)n log n ˇ ˇ ˇ. We shall shortly show that n ě log n, and hence 1 log n ě 1 n for all n ě 1. This implies that the series 8 ÿ n=2 1 log n diverges by comparison with the divergent series 8 ÿ n=2 1 np ˇ ˇ ˇ ˇ p=1 . This yelds both divergence for x = 9 and also the failure of absolute convergence for x = ´9.
Finally, we show that n ´ log n ą 0, for all n ě 1. Set f (x) = x ´ log x. Then f (1) = 1 ą 0 and f 1(x) = 1 ´ 1 x ě 0 for all x ě 1 So f (x) is (strictly) positive when x = 1 and is increasing for all x ě 1. So f (x) is (strictly) positive for all x ě 1.
S-19: (a) Applying 1 1 + r = 8 ÿ n=0 (´1)nrn with r = x3 gives ż 1 1 + x3 dx = 8 ÿ n=0 (´1)n ż x3n dx = 8 ÿ n=0 (´1)n x3n+1 3n + 1 + C (b) By part (a), ż 1/4 0 1 1 + x3 dx = 8 ÿ n=0 (´1)n x3n+1 3n + 1 ˇ ˇ ˇ ˇ 1/4 0 = 8 ÿ n=0 (´1)n 1 (3n + 1)43n+1 This is an alternating series with successively smaller terms that converge to zero as n Ñ 8. So truncating it introduces an error no larger than the magnitude of the first 682 dropped term. We want that first dropped term to obey 1 (3n + 1)43n+1 ă 10´5 = 1 105 So let’s check the first few terms.
1 (3n + 1)43n+1 ˇ ˇ ˇ ˇ n=0 = 1 4 ą 1 105 1 (3n + 1)43n+1 ˇ ˇ ˇ ˇ n=1 = 1 45 ą 1 105 1 (3n + 1)43n+1 ˇ ˇ ˇ ˇ n=2 = 1 7 ˆ 47 = 1 7 ˆ 214 = 1 7 ˆ 16 ˆ 1024 = 1 112 ˆ 1024 ă 1 105 So we need to keep two terms (the n = 0 and n = 1 terms).
S-20: (a) Differentiating both sides of 8 ÿ n=0 xn = 1 1 ´ x gives 8 ÿ n=0 nxn´1 = 1 (1 ´ x)2 Now multiplying both sides by x gives 8 ÿ n=0 nxn = x (1 ´ x)2 as desired.
(b) Differentiating both sides of the conclusion of part (a) gives 8 ÿ n=0 n2xn´1 = (1 ´ x)2 ´ 2x(x ´ 1) (1 ´ x)4 = (1 ´ x)(1 ´ x + 2x) (1 ´ x)4 = 1 + x (1 ´ x)3 Now multiplying both sides by x gives 8 ÿ n=0 n2xn = x(1 + x) (1 ´ x)3 We know that differentiation preserves the radius of convergence of power series. So this series has radius of convergence 1 (the radius of convergence of the original geometric series). At x = ˘1 the series diverges by the divergence test. So the series converges for ´1 ă x ă 1.
683 S-21: By the divergence test, the fact that 8 ř n=0 (1 ´ bn) converges guarantees that lim nÑ8(1 ´ bn) = 0, or equivalently, that lim nÑ8 bn = 1. So, by equation (3.5.2) in the CLP-2 text, the radius of convergence is R = lim nÑ8 ˇ ˇ ˇbn+1 bn ˇ ˇ ˇ ´1 = 1 1 ´1 = 1 (3.7) S-22: (a) We know that the radius of convergence R obeys 1 R = lim nÑ8 an+1 an = lim nÑ8 n n + 1 (n + 1)an+1 nan = 1C C = 1 because we are told that lim nÑ8 nan = C. So R = 1.
(b) Just knowing that the radius of convergence is 1, we know that the series converges for |x| ă 1 and diverges for |x| ą 1. That leaves x ˘ 1.
When x = +1, the series reduces to 8 ř n=1 an. We are told that nan decreases to C ą 0. So an ě C n . By the comparison test with the harmonic series 8 ř n=1 1 n, which diverges by the p–test with p = 1, our series diverges when x = 1.
When x = ´1, the series reduces to 8 ř n=1 (´1)nan. We are told that nan decreases to C ą 0.
So an ą 0 and an converges to 0 as n Ñ 8. Consequently 8 ř n=1 (´1)nan converges by the alternating series test.
In conclusion 8 ř n=1 anxn converges when ´1 ď x ă 1.
S-23: Equation 2.3.1 in the CLP-2 text tells us the centre of mass of a rod with weights tmnu at positions txnu is ¯ x = ř mnxn ř mn .
We find the combined mass of our weights using Equation 3.2.2 in the CLP-2 text with r = 1 2 and r = 1 3, respectively.
8 ÿ n=1 1 2n + 8 ÿ n=1 1 3n = 8 ÿ n=0 1 2 ¨ 1 2n + 8 ÿ n=0 1 3 ¨ 1 3n = 1 2 ¨ 1 1 ´ 1 2 + 1 3 ¨ 1 1 ´ 1 3 = 1 + 1 2 = 3 2 684 Now, we want to calculate the sum of the products of the masses and their positions.
8 ÿ n=1 1 2n ¨ n + 8 ÿ n=1 1 3n ¨ (´n) We don’t have such a nice formula for this, but we can make one by differentiating.
The following formula is true for any x with |x| ă 1: 8 ÿ n=0 xn = 1 1 ´ x Differentiating both sides with respect to x: 8 ÿ n=0 nxn´1 = 1 (1 ´ x)2 8 ÿ n=1 nxn´1 = 1 (1 ´ x)2 Multiplying both sides by x: 8 ÿ n=1 nxn = x (1 ´ x)2 This allows us to evaluate our series.
8 ÿ n=1 n 2n ´ 8 ÿ n=1 n 3n = 1 2 1 ´ 1 2 2 ´ 1 3 1 ´ 1 3 2 = 2 ´ 3 4 = 5 4 Therefore, ¯ x = 5/4 3/2 = 5 6 = 0.833 Remark: we can check that this makes some sense. Since the weights to the right of x = 0 are heavier than those to the left, but spaced the same, we would expect our rod to balance to the right of x = 0.
685 S-24: First, we differentiate.
f (x) = 8 ÿ n=0 An(x ´ c)n f 1(x) = 8 ÿ n=0 nAn(x ´ c)n´1 = 8 ÿ n=1 nAn(x ´ c)n´1 f 1(c) = 8 ÿ n=1 nAn ¨ 0n´1 = A1 ¨ 1 + 2A2 ¨ 0 + 3A3 ¨ 0 + ¨ ¨ ¨ = A1 So, if A1 = 0, then f 1(c) = 0. That is, f (x) has a critical point at x = c.
To determine the behaviour of this critical point, we use the second derivative test.
f 1(x) = 8 ÿ n=1 nAn(x ´ c)n´1 f 2(x) = 8 ÿ n=1 n(n ´ 1)An(x ´ c)n´2 = 8 ÿ n=2 n(n ´ 1)An(x ´ c)n´2 f 2(c) = 8 ÿ n=2 n(n ´ 1)An ¨ 0n´2 = 2(1)A2 ¨ 00 + 3(2)A3 ¨ 01 + 4(3)A4 ¨ 02 + ¨ ¨ ¨ = 2A2 Following the second derivative test, x = c is the location of a local maximum if A2 ă 0, and it is the location of a local minimum if A2 ą 0. (If A2 = 0, the critical point may or may not be a local extremum.) S-25: We recognize 8 ÿ n=3 n 5n´1 as f (x) = 8 ÿ n=3 n ¨ xn´1, evaluated at x = 1 5. We should figure out what f (x) is in equation form (as opposed to power series form). Notice that this looks similar to the derivative of the geometric series ÿ xn.
686 1 1 ´ x = 8 ÿ n=0 xn when |x| ă 1 d dx " 1 1 ´ x = d dx # 8 ÿ n=0 xn + 1 (1 ´ x)2 = 8 ÿ n=1 nxn´1 = 1x0 + 2x1 + 8 ÿ n=3 nxn´1 = 1 + 2x + 8 ÿ n=3 nxn´1 So, 1 (1 ´ x)2 ´ 1 ´ 2x = 8 ÿ n=3 nxn´1 Setting x = 1 5: 1 (1 ´ 1/5)2 ´ 1 ´ 2 5 = 8 ÿ n=3 n 1 5 n´1 5 4 2 ´ 1 ´ 2 5 = 8 ÿ n=3 n 5n´1 So, our series evaluates to 25 16 ´ 1 ´ 2 5 = 13 80.
S-26: As we saw in in Example 3.5.20 of the CLP-2 text, log(1 + x) = 8 ÿ n=0 (´1)n xn+1 n + 1 which is an alternating series when x is positive. If we use its partial sum SN to approximate log(1 + x), the absolute error involved is no more than x(N+1)+1 (N + 1) + 1 = xN+2 N + 2 We want this error to be at most 10´5 whenever 0 ă x ă 1 10. For this range of x values, xN+2 N + 2 ă 1 (N + 2)10N+2, so we want N that satisfies the inequality: 1 (N + 2)10N+2 ď 1 105 ñ (N + 2)10N+2 ě 105 687 We see N = 3 suffices.
So, the partial sum 3 ÿ n=0 (´1)n xn+1 n + 1 = x ´ x2 2 + x3 3 ´ x4 4 approximates log(1 + x) to within an error of x5 5 .
When x is between 0 and 1 10, that error is at most 1 5 ¨ 105 ă 10´5, as desired.
Now we can approximate log(1.05).
log(1.05) = log 1 + 1 20 « 1 20 ´ 1 20 2 2 + 1 20 3 3 ´ 1 20 4 4 = 12 ˆ 203 ´ 6 ˆ 202 + 4 ˆ 20 ´ 3 12 ˆ 204 = 93677 1920000 We note that a computer approximates 93677 1920000 « 0.04879010 and log(1.05) « 0.04879016.
So, our actual error is around 6 ˆ 10´8.
S-27: As we saw in in Example 3.5.21 of the CLP-2 text, arctan x = 8 ÿ n=0 (´1)n x2n+1 2n + 1 which is an alternating series when x is nonzero. If we use its partial sum SN to approximate arctan x, the absolute error involved is no more than |x|2(N+1)+1 2(N + 1) + 1 = |x|2N+3 2N + 3 We want this error to be at most 10´5 whenever ´1 4 ă x ă 1 4. For this range of x values, |x|2N+3 2N + 3 ă 1 (2N + 3)42N+3, so we want N that satisfies the inequality: 1 (2N + 3)42N+3 ď 1 105 A quick check with a calculator shows that N = 2 suffices.
So, the partial sum 2 ÿ n=0 (´1)n x2n+1 2n + 1 = x ´ x3 3 + x5 5 688 approximates arctan x to within an error of x7 7 .
When x is between ´1 4 and 1 4, that error is at most 1 7 ¨ 47 = 1 114688 ă 1 100000 = 10´5, as desired. (When x = 0, our approximation is 0, the exact value of arctan 0.) Solutions to Exercises 3.6 — Jump to TABLE OF CONTENTS S-1: All functions A, B, and C intersect the function y = f (x) when x = 2. B is a constant function, so this is the constant approximation. A is the tangent line, so A is the linear approximation. C is a tangent parabola, so C is the quadratic approximation.
S-2: Following how a Taylor series is constructed, the Taylor series and the function agree at the point chosen as the centre. So, T(5) = arctan3 e5 + 7 .
If we were evaluating a Taylor series at a point other than its centre, we would generally need to check that (a) the series converges, and (b) it converges to the same value as the function we used to create it.
S-3: These are listed in Theorem 3.6.5 in the CLP–2 text. However, it’s possible to figure out many of them without a lot of memorization. For example, e0 = cos(0) = 1 1´0 = 1, while sin(0) = log(1 + 0) = arctan(0) = 0. So by plugging in x = 0 to the series listed, we can divide them into these two categories.
The derivative of sine is cosine, so we can also look for one series that is the derivative of another. The derivative of ex is ex, so we can look for a series that is its own derivative.
Furthermore, sine and arctangent are odd functions and only II and IV are odd. Cosine is an even function and only III is even.
Alternately, we can find the first few terms of each series using the definition of a Taylor series, and match them up.
In any event, here is what Theorem 3.6.5 in the CLP–2 text gives.
(A) The Taylor series representation of 1 1´x is given in V. The series converges for ´1 ă x ă 1. So by Definition 3.5.3 in the CLP–2 text, the series has radius of convergence 1.
(B) The Taylor series representation of log(1 + x) is given in I. The series converges for ´1 ă x ď 1. In particular, it converges for all |x| ă 1 and diverges for all |x| ą 1. So by Definition 3.5.3 in the CLP–2 text, the series has radius of convergence 1.
(C) The Taylor series representation of arctan x is given in IV. The series converges for ´1 ď x ď 1. In particular, it converges for all |x| ă 1 and diverges for all |x| ą 1. So it has radius of convergence 1.
(D) The Taylor series representation of ex is given in VI. The series converges for all x. So it has infinite radius of convergence.
689 (E) The Taylor series representation of sin x is given in II. The series converges for all x.
So it has infinite radius of convergence.
(F) The Taylor series representation of cos x is given in III. The series converges for all x.
So it has infinite radius of convergence.
S-4: (a) Using the definition of a Taylor series, we know 8 ÿ n=0 n2 (n! + 1)(x ´ 3)n = 8 ÿ n=0 f (n)(3) n!
(x ´ 3)n So, the coefficient of (x ´ 3)20 is f (20)(3) 20!
(using the definition). Using the given series, the coefficient of (x ´ 3)20 is 202 20!+1. So, f (20)(3) 20!
= 202 20! + 1 ñ f (20)(3) = 202 20!
20! + 1 (which is extremely close to 202).
(b) Using the definition of a Taylor series, we know 8 ÿ n=0 n2 (n! + 1)(x ´ 3)2n = 8 ÿ k=0 g(k)(3) k!
(x ´ 3)k So, the coefficient of (x ´ 3)20 is g(20)(3) 20!
(using the definition). Looking at the given series, the coefficient of (x ´ 3)20 occurs when n = 10, so it is 102 10!+1. So, g(20)(3) 20!
= 102 10! + 1 ñ g(20)(3) = 102 20!
10! + 1 (c) With the previous two examples in mind, we find the Maclaurin series for h(x).
(Using the series representation will be much easier than differentiating h(x) directly 690 twenty times.) Recall from the text that we know the Maclaurin series for arctan x.
arctan(x) = 8 ÿ n=0 (´1)n x2n+1 2n + 1 arctan(5x2) = 8 ÿ n=0 (´1)n (5x2)2n+1 2n + 1 = 8 ÿ n=0 (´1)n 52n+1 2n + 1x4n+2 arctan(5x2) x2 = 8 ÿ n=0 (´1)n 52n+1 2n + 1x4n 8 ÿ k=0 h(k)(0) k!
xk = 8 ÿ n=0 (´1)n 52n+1 2n + 1x4n Using the definition of a Maclaurin series, the coefficient of x20 is h(20)(0) 20!
. This occurs in the given series when n = 5, so h(20)(0) 20!
= (´1)5 52ˆ5+1 2 ˆ 5 + 1 = ´511 11 ñ h(20)(0) = ´20! ¨ 511 11 Similarly, the coefficient of x22 in the Maclaurin series is h(22)(0) 22!
. Since no term x22 occurs in our series, that coefficient is 0, so h(22)(0) = 0.
S-5: The definition of a Taylor series tells us we will be computing the coefficients in the series 8 ÿ n=0 f (n)(1) n!
(x ´ 1)n That is, we need a general description of f (n)(1). To find this, we take a few derivatives, and look for a pattern.
f (x) = log(x) f (1) = 0 f 1(x) = x´1 f 1(1) = 1 f 2(x) = (´1)x´2 f 2(1) = ´1 f (3)(x) = (´2)(´1)x´3 f (3)(1) = 2!
f (4)(x) = (´3)(´2)(´1)x´4 f (4)(1) = ´3!
f (5)(x) = (´4)(´3)(´2)(´1)x´5 f (5)(1) = 4!
f (6)(x) = (´5)(´4)(´3)(´2)(´1)x´6 f (6)(1) = ´5!
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f (n)(x) = (´1)n´1(n ´ 1)! x´n f (n)(1) = (´1)n´1(n ´ 1)!
691 Using the convention 0! = 1, our pattern for f (n)(1) begins when n = 1.
8 ÿ n=0 f (n)(1) n!
(x ´ 1)n = 0 + 8 ÿ n=1 (´1)n´1(n ´ 1)!
n!
(x ´ 1)n = 8 ÿ n=1 (´1)n´1 n (x ´ 1)n S-6: To find the Taylor series for sine, centred at a = π, we’ll need to know the various derivatives of sine at π.
f (x) = sin x f (π) = 0 f 1(x) = cos x f 1(π) = ´1 f 2(x) = ´ sin x f 2(π) = 0 f 3(x) = ´ cos x f 3(π) = 1 f (4)(x) = sin x = f (x) f (4)(π) = 0 Even derivatives are 0; odd derivatives alternate between ´1 and +1. (If you’re following along with the derivation of the Maclaurin series for sine in the text, note f (n)(π) = ´f (n)(0).) In our Taylor series, every even-indexed term will be zero, and we will be left with only odd-indexed terms. If we let n be our index, then the term 2n + 1 will capture all the odd numbers. Since the signs alternate, f (2n+1)(π) = (´1)n+1. So, our Taylor series is: 8 ÿ k=0 f (k)(π) k!
(x ´ π)k = 8 ÿ n=0 f (2n+1)(π) (2n + 1)! (x ´ π)2n+1 (since the even terms are all zero) = 8 ÿ n=0 (´1)n+1 (2n + 1)!(x ´ π)2n+1 S-7: The definition of a Taylor series tells us we will be computing the coefficients in the series 8 ÿ n=0 g(n)(10) n!
(x ´ 10)n That is, we need a general description of g(n)(10). To find this, we take a few derivatives, and look for a pattern.
692 g(x) = x´1 g(10) = 1 10 g1(x) = (´1)x´2 g1(10) = ´1 102 g2(x) = (´2)(´1)x´3 g2(10) = (´1)22!
103 g(3)(x) = (´3)(´2)(´1)x´4 g(3)(10) = (´1)33!
104 g(4)(x) = (´4)(´3)(´2)(´1)x´5 g(4)(10) = (´1)44!
105 g(5)(x) = (´5)(´4)(´3)(´2)(´1)x´6 g(5)(10) = (´1)55!
106 .
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g(n)(x) = (´1)nn!x´(n+1) g(n)(10) = (´1)nn!
10n+1 Using the convention 0! = 1, our pattern for g(n)(10) begins when n = 0.
8 ÿ n=0 g(n)(1) n!
(x ´ 10)n = 8 ÿ n=0 (´1)nn!
n!10n+1 (x ´ 10)n = ´ 8 ÿ n=0 (x ´ 10)n (´10)n+1 = 1 10 8 ÿ n=0 10 ´ x 10 n For fixed x, we recognize this as a geometric series with r = 10´x 10 . So it converges precisely when |r| ă 1, i.e.
ˇ ˇ ˇ ˇ 10 ´ x 10 ˇ ˇ ˇ ˇ ă 1 |10 ´ x| ă 10 ´10 ă x ´ 10 ă 10 0 ă x ă 20 So, its interval of convergence is (0, 20).
S-8: The definition of a Taylor series tells us we will be computing the coefficients in the series 8 ÿ n=0 h(n)(a) n!
(x ´ a)n 693 That is, we need a general description of h(n)(a). To find this, we take a few derivatives, and look for a pattern.
h(x) = e3x h(a) = e3a h1(x) = 3e3x h1(a) = 3e3a h2(x) = 32e3x h2(a) = 32e3a h3(x) = 33e3x h3(a) = 33e3a .
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h(n)(x) = 3ne3x h(n)(a) = 3ne3a The pattern for h(n)(a) holds for all (whole numbers) n ě 0. So, our Taylor series for h(x) is 8 ÿ n=0 3ne3a n! (x ´ a)n To find its radius of convergence, we use the ratio test.
ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ 3n+1e3a(x ´ a)n+1 (n + 1)!
¨ n!
3ne3a(x ´ a)n ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ 3n+1 3n ¨ e3a e3a ¨ n!
(n + 1)! ¨ (x ´ a)n+1 (x ´ a)n ˇ ˇ ˇ ˇ = 3 ¨ 1 n + 1 ¨ |x ´ a| lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 3 n + 1 ¨ |x ´ a| = 0 Our series converges for every value of x, so its radius of convergence is 8.
S-9: Substituting y = 2x into 1 1 ´ y = 8 ÿ n=0 yn (which is valid for all ´1 ă y ă 1) gives f (x) = 1 2x ´ 1 = ´ 1 1 ´ 2x = ´ 8 ÿ n=0 (2x)n = ´ 8 ÿ n=0 2nxn for all ´1 2 ă x ă 1 2 S-10: Substituting first y = ´x and then y = 2x into 1 1 ´ y = 8 ÿ n=0 yn (which is valid for all ´1 ă y ă 1) gives 1 1 ´ (´x) = 8 ÿ n=0 (´x)n = 8 ÿ n=0 (´1)nxn for all ´1 ă x ă 1 1 1 ´ (2x) = 8 ÿ n=0 (2x)n = 8 ÿ n=0 2nxn for all ´1 2 ă x ă 1 2 694 Hence, for all ´1 2 ă x ă 1 2, f (x) = 3 x + 1 ´ 1 2x ´ 1 = 3 1 ´ (´x) + 1 1 ´ 2x = 3 8 ÿ n=0 (´1)nxn + 8 ÿ n=0 2nxn = 8 ÿ n=0 3(´1)n + 2n xn So bn = 3(´1)n + 2n.
S-11: We found the Taylor series for e3x from scratch in Question 8. If we hadn’t just done that, we could easily find it by modifying the series for ex.
Substituting y = 3x into the exponential series ey = 8 ÿ n=0 yn n!
gives e3x = 8 ÿ n=0 (3x)n n!
= 8 ÿ n=0 3n xn n!
so that c5, the coefficient of x5, which appears only in the n = 5 term, is c5 = 35 5!
S-12: Since f 1(t) = d dt log(1 + 2t) = 2 1 + 2t = 2 8 ÿ n=0 (´2t)n if |2t| ă 1 i.e. |t| ă 1 2 and f (0) = 0, we have f (x) = ż x 0 f 1(t) dt = 2 8 ÿ n=0 ż x 0 (´1)n2ntn dt = 8 ÿ n=0 (´1)n2n+1 xn+1 n + 1 for all |x| ă 1 2 S-13: We just need to substitute y = x3 into the known Maclaurin series for sin y, to get the Maclaurin series for sin(x3), and then multiply the result by x2.
sin y = y ´ y3 3! + ¨ ¨ ¨ sin(x3) = x3 ´ x9 3! + ¨ ¨ ¨ x2 sin(x3) = x5 ´ x11 3! + ¨ ¨ ¨ so a = 1 and b = ´ 1 3! = ´1 6.
695 S-14: Recall that ey = 8 ÿ n=0 yn n! = 1 + y + y2 2 + y3 3! + ¨ ¨ ¨ Setting y = ´x2, we have e´x2 = 1 ´ x2 + x4 2 ´ x6 3! + ¨ ¨ ¨ e´x2 ´ 1 = ´x2 + x4 2 ´ x6 6 + ¨ ¨ ¨ e´x2 ´ 1 x = ´x + x3 2 ´ x5 6 + ¨ ¨ ¨ ż e´x2 ´ 1 x dx = C ´ x2 2 + x4 8 ´ x6 36 + ¨ ¨ ¨ S-15: Recall that arctan(y) = 8 ÿ n=0 (´1)n y2n+1 2n + 1 Setting y = 2x, we have ż x4 arctan(2x) dx = ż x4 8 ÿ n=0 (´1)n (2x)2n+1 2n + 1 !
dx = ż 8 ÿ n=0 (´1)n 22n+1x2n+5 2n + 1 !
dx = 8 ÿ n=0 (´1)n 22n+1x2n+6 (2n + 1)(2n + 6) + C = 8 ÿ n=0 (´1)n 22nx2n+6 (2n + 1)(n + 3) + C S-16: Substituting y = ´3x3 into 1 1 ´ y = 8 ř n=0 yn gives df dx = x ¨ 1 1 + 3x3 = x 8 ÿ n=0 ´ 3x3n = 8 ÿ n=0 (´1)n3nx3n+1 Now integrating, f (x) = 8 ÿ n=0 (´1)n3n x3n+2 3n + 2 + C 696 To have f (0) = 1, we need C = 1. So, finally f (x) = 1 + 8 ÿ n=0 (´1)n 3n 3n + 2x3n+2 S-17: We’re given a big hint: that our series resembles the Taylor series for arctangent.
The terms of arctangent are (´1)n x2n+1 2n + 1. Our terms resemble those terms, with x2n+1 replaced by 1 3n .
Since 3n = ?
3 2n = 1 ?
3 ?
3 2n+1: 8 ÿ n=0 (´1)n (2n + 1)3n = ?
3 8 ÿ n=0 (´1)n (2n + 1) ?
3 2n+1 = ?
3 8 ÿ n=0 (´1)n x2n+1 2n + 1 ˇ ˇ ˇ ˇ x= 1 ?
3 = ?
3 arctan 1 ?
3 = ?
3 π 6 = π 2 ?
3 S-18: Recall that ex = 8 ÿ n=0 xn n! . So 8 ÿ n=0 (´1)n n!
= h 8 ÿ n=0 xn n!
i x=´1 = h exi x=´1 = e´1 S-19: Recall that ex = 8 ÿ k=0 xk k! . So 8 ÿ k=0 1 ekk! = h 8 ÿ k=0 xk k!
i x=1/e = h exi x=1/e = e1/e S-20: Recall that ex = 8 ÿ k=0 xk k! . So 8 ÿ k=0 1 πkk! = h 8 ÿ k=0 xk k!
i x=1/π = h exi x=1/π = e1/π This series differs from the given one only in that it starts with k = 0 while the given series starts with k = 1. So 8 ÿ k=1 1 πkk! = 8 ÿ k=0 1 πkk! ´ 1 lo omo on k=0 = e1/π ´ 1 697 S-21: Recall, from Theorem 3.6.5 in the CLP-2 text, that, for all ´1 ă x ď 1, log(1 + x) = 8 ÿ k=0 (´1)k xk+1 k + 1 = 8 ÿ n=1 (´1)n´1 xn n (To get from the first sum to the second sum we substituted n = k + 1. If you don’t see why the two sums are equal, write out the first few terms of each.) So 8 ÿ n=1 (´1)n´1 n 2n = h 8 ÿ n=1 (´1)n´1 xn n i x=1/2 = h log(1 + x) i x=1/2 = log(3/2) S-22: Write 8 ÿ n=1 n + 2 n!
en = 8 ÿ n=1 n n!en + 8 ÿ n=1 2 n!en = 8 ÿ n=1 en (n ´ 1)! + 2 8 ÿ n=1 en n!
= e 8 ÿ n=1 en´1 (n ´ 1)! + 2 8 ÿ n=1 en n!
= e 8 ÿ n=0 en n! + 2 8 ÿ n=1 en n!
Recall that ex = 8 ÿ n=0 xn n! . So 8 ÿ n=1 n + 2 n!
en = e h 8 ÿ n=0 xn n!
i x=e + 2 h 8 ÿ n=1 xn n!
i x=e = e h exi x=e + 2 h ex ´ 1 i x=e = ee+1 + 2(ee ´ 1) = (e + 2)ee ´ 2 S-23: Let’s use the ratio test: lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ ˇ 2n+1 n+1 2n n ˇ ˇ ˇ ˇ ˇ = lim nÑ8 2 n n + 1 = 2 ą 1 So, the series diverges.
Remark: it’s tempting to note that log(1 + y) = 8 ÿ n=0 (´1)n yn+1 n + 1 = ´ 8 ÿ n=1 (´y)n n , and try to substitute in y = ´2. But, the Maclaurin series for log(1 + y) has radius of convergence R = 1, so it doesn’t converge at y = ´2. Furthermore, log(1 + (´2)) = log(´1), but this is undefined.
698 S-24: Our series looks something like the Taylor series for sine, sin x = 8 ÿ n=0 (´1)n (2n + 1)!x2n+1.
8 ÿ n=0 (´1)n (2n + 1)!
π 4 2n+1 1 + 22n+1 = 8 ÿ n=0 (´1)n (2n + 1)!
π 4 2n+1 + π 2 2n+1 = 8 ÿ n=0 (´1)n (2n + 1)!
π 4 2n+1 + 8 ÿ n=0 (´1)n (2n + 1)!
π 2 2n+1 = sin π 4 + sin π 2 = 1 ?
2 + 1 = 1 + ?
2 ?
2 S-25: (a) Solution 1: The naive strategy is to set an = x2n (2n)! and apply the ratio test.
lim nÑ8 ˇ ˇ ˇan+1 an ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ ˇ ˇ x2n+2 (2n+2)!
x2n (2n)!
ˇ ˇ ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ x2n+2 x2n ¨ (2n)!
(2n + 2)(2n + 1)(2n)!
ˇ ˇ ˇ ˇ = lim nÑ8 x2 (2n + 2)(2n + 1) = 0 This is smaller than 1 no matter what x is. So the series converges for all x.
Solution 2: Alternatively, the sneaky way is to observe that both ex = 8 ÿ n=0 xn n! and e´x = 8 ÿ n=0 (´x)n n!
are known to converge for all x. So 1 2 ex + e´x = ÿ n even xn n! = 8 ÿ n=0 x2n (2n)!
also converges for all x.
699 (b) Recall that ex = 8 ÿ n=0 xn n! . Then: e = 8 ÿ n=0 1 n!
e´1 = 8 ÿ n=0 (´1)n n!
e + e´1 = 8 ÿ n=0 1 + (´1)n n!
= 2 8 ÿ n even 1 n! = 2 8 ÿ n=0 1 (2n)!
Hence 8 ÿ n=0 1 (2n)! = 1 2 e + 1 e .
S-26: All three series we’re adding up are alternating, so we can bound the absolute error in the approximation SN (the N-th partial sum) by |aN+1|.
The Taylor series for arctangent is arctan(x) = 8 ÿ n=0 (´1)n x2n+1 2n + 1 for every real x.
(a) Using the Taylor series for arctangent when x = 1, we see π 4 = arctan(1) = 8 ÿ n=0 (´1)n 1 2n + 1 π = 8 ÿ n=0 (´1)n 4 2n + 1 The error involved in approximating π with the partial sum SN is at most |aN+1| = 4 2N+3. In order for this to be at most 4 ˆ 10´5, we need: 4 2N + 3 ď 4 ˆ 10´5 2N + 3 ě 105 N ě 105 ´ 3 2 = 5 ˆ 104 ´ 3 2 = 50, 000 ´ 1.5 Since n must be an integer, we need to add up the terms from n = 0 to n = 49, 999.
That is, we add up the first 50,000 terms.
(b) Using the Taylor series for arctangent: π = 16 arctan 1 5 ´ 4 arctan 1 239 = 16 8 ÿ n=0 (´1)n 1 (2n + 1)52n+1 ´ 4 8 ÿ n=0 (´1)n 1 (2n + 1) ¨ 2392n+1 = 8 ÿ n=0 (´1)n 2n + 1 16 52n+1 ´ 4 2392n+1 700 This is an alternating sum, so the absolute error in using the partial sum SN is at most: |aN+1| = 1 2N + 3 16 52N+3 ´ 4 2392N+3 So, we want to find a value of N that makes this at most 4 ˆ 10´5. Several values of N are given below.
N |aN+1| 1 1 5 16 55 ´ 4 2395 « 0.001 2 1 7 16 57 ´ 4 2397 « 0.000029 ă 4 ˆ 10´5 So, it suffices to add up the first three terms (n = 0, n = 1, and n = 2) of the series.
(c) Again, we use the Taylor series for arctangent.
arctan 1 2 + arctan 1 3 = arctan 3 + 2 2 ¨ 3 ´ 1 = arctan(1) = π 4 π = 4 arctan 1 2 + arctan 1 3 = 4 8 ÿ n=0 (´1)n 1 (2n + 1)22n+1 + 4 8 ÿ n=0 (´1)n 1 (2n + 1)32n+1 = 8 ÿ n=0 (´1)n 4 2n + 1 1 22n+1 + 1 32n+1 If we use the partial sum SN, our absolute error is at most |aN+1| = 4 2N + 3 1 22N+3 + 1 32N+3 .
Several of these values are given below.
N |aN+1| 1 4 5 1 25 + 1 35 « 0.028 2 4 7 1 27 + 1 37 « 0.0047 3 4 9 1 29 + 1 39 « 0.00089 4 4 11 1 211 + 1 311 « 0.00018 5 4 13 1 213 + 1 313 « 0.000038 ă 4 ˆ 10´5 So, it suffices to add the first six terms (n = 0 to n = 5) of the series.
Remark: if we actually wanted to approximate π this way, the series from part (a) is probably not ideal–adding 50,000 terms sounds rough. The series from (b) and (c) seem much more practical.
701 S-27: Using the Taylor series for log(1 + x): log(1 + x) = 8 ÿ n=1 (´1)n+1 xn n log(1.5) = log 1 + 1 2 = 8 ÿ n=1 (´1)n+1 1 n2n Since this is an alternating series, the error involved in using the partial sum SN is at most |aN+1| = 1 (N + 1)2N+1.
We want this to be at most 5 ˆ 10´11.
N |aN+1| 10 1 11 ¨ 211 « 4 ˆ 10´5 15 1 16 ¨ 216 « 9.5 ˆ 10´7 20 1 21 ¨ 221 « 2 ˆ 10´8 25 1 26 ¨ 226 « 6 ˆ 10´10 26 1 27 ¨ 227 « 3 ˆ 10´10 27 1 28 ¨ 228 « 1 ˆ 10´10 28 1 29 ¨ 229 « 6 ˆ 10´11 29 1 30 ¨ 230 « 3 ˆ 10´11 So, it suffices to add up the first 29 terms.
S-28: The Taylor Series for ex is not alternating, so we’ll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SN is EN = f (N+1)(c) (N + 1)! (x ´ a)N+1 for some c strictly between a and x. In our case, a = 0 and x = 1. So, we want to find a value of N such that ˇ ˇ ˇ ˇ ˇ f (N+1)(c) (N + 1)! (1 ´ 0)N+1 ˇ ˇ ˇ ˇ ˇ = ec (N + 1)! ă 5 ˆ 10´11 for all c in (0, 1).
If c is between 0 and 1, then ec is between 1 and e. However, since the purpose of this problem is to approximate e precisely, it doesn’t make much sense to use e in our bound.
Since e is less than 3, then ec ă 3 for all c in (0, 1). Now we can search for an appropriate value of N.
702 N 3 (N + 1)!
10 3 11! = 1 910 « 8 ˆ 10´8 11 3 12! « 6 ˆ 10´9 12 3 13! « 5 ˆ 10´10 13 3 14! « 3 ˆ 10´11 So, it suffices to use the partial sum S13.
S-29: The Taylor Series for log(1 ´ x) is not alternating, so we’ll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SN is EN = f (N+1)(c) (N + 1)! (x ´ a)N+1 for some c strictly between a and x. In our case, a = 0 and x = 1 10. So, we want to find a value of N such that ˇ ˇ ˇ ˇ ˇ f (N+1)(c) (N + 1)!
1 10 N+1ˇ ˇ ˇ ˇ ˇ ă 5 ˆ 10´11 for all c in (0, 1 10).
To find this N, we to know f (N+1)(x). Just like when we create a Taylor polynomial from scratch, we’ll differentiate f (x) several times, and look for a pattern.
f (x) = log(1 ´ x) f (6)(x) = ´2(3)(4)(5) (1 ´ x)6 f 1(x) = ´1 1 ´ x f (7)(x) = ´2(3)(4)(5)(6) (1 ´ x)7 f 2(x) = ´1 (1 ´ x)2 .
.
.
f 3(x) = ´2 (1 ´ x)3 f (N+1)(x) = ´N!
(1 ´ x)N+1 f (4)(x) = ´2(3) (1 ´ x)4 f (5)(x) = ´2(3)(4) (1 ´ x)5 Now we want a reasonable bound on f (N+1)(c), when c is in (0, 1 10). Note that in this 703 range, 1 ´ c ą 0.
0 ă c ă 1 10 ñ 9 10 ă 1 ´ c ă 1 ñ 9 10 N+1 ă (1 ´ c)N+1 ă 1 ñ 1 ă 1 (1 ´ c)N+1 ă 10 9 N+1 ñ N! ă N!
(1 ´ c)N+1 ă N!
10 9 N+1 This bound provides us with a “worst-case scenario” error. We don’t know exactly what c is, but we don’t need to–the bound above holds for all c between 0 and 1 10.
Now we’re ready to choose an N that results in a sufficiently small error bound.
ˇ ˇ ˇ ˇ ˇ f (N+1)(c) (N + 1)!
1 10 N+1ˇ ˇ ˇ ˇ ˇ ă N!
10 9 N+1 (N + 1)!
1 10 N+1 = 1 9N+1 ¨ (N + 1) So, we want: 1 9N+1 ¨ (N + 1) ă 5 ˆ 10´11 To find an appropriate N, we test several values.
N 1 9N+1 ¨ (N + 1) 8 1 9 ¨ 99 = 1 910 « 3 ˆ 10´10 9 1 10 ¨ 910 « 3 ˆ 10´11 So, it suffices to use the partial sum S9.
S-30: We’ll use Theorem 3.6.1-b in the CLP-2 text to bound the error of a partial-sum approximation. The error in the partial-sum approximation SN is EN = f (N+1)(c) (N + 1)! (x ´ a)N+1 for some c strictly between a and x. In our case, a = 0 and x is in (´2, 1). So, we want to find a value of N such that ˇ ˇ ˇ ˇ ˇ f (N+1)(c) (N + 1)! (x)N+1 ˇ ˇ ˇ ˇ ˇ ă 5 ˆ 10´11 704 for all x in (´2, 1), and all c in (´2, 1).
To find this N, we to know f (N+1)(x). Just like when we create a Taylor polynomial from scratch, we’ll differentiate f (x) several times, and look for a pattern.
f (x) = sinh(x) = ex ´ e´x 2 f 1(x) = ex + e´x 2 f 2(x) = ex ´ e´x 2 f 3(x) = ex + e´x 2 That is, even derivatives of f (x) are f (x), and odd derivatives of f (x) are ex+e´x 2 (which, incidentally, is the function called cosh x).
Now we want a reasonable bound on f (N+1)(c), when c is in (´2, 1). Since powers of e are always positive, we begin by noting that 0 ă ex´e´x 2 ă ex+e´x 2 . So, all derivatives of f (x) are bounded above by ex+e´x 2 .
´2 ă c ă 1 ñ e´2 ă ec ă e and e´1 ă e´c ă e2 ñ f (N+1)(c) ă ec + e´c 2 ă e2 + e2 2 = e2 ă 9 This bound provides us with a “worst-case scenario” error. We don’t know exactly what c is, but we don’t need to–the bound above holds for all c between ´2 and 1.
We also don’t know exactly what x will be, only that it’s between ´2 and 1. So, we note |x|N+1 ă 2N+1.
Now we’re ready to choose an N that results in a sufficiently small error bound.
ˇ ˇ ˇ ˇ ˇ f (N+1)(c) (N + 1)! (x)N+1 ˇ ˇ ˇ ˇ ˇ ă 9 ¨ 2N+1 (N + 1)!
So, we want: 9 ¨ 2N+1 (N + 1)! ă 5 ˆ 10´11 To find an appropriate N, we test several values.
705 N 9 ¨ 2N+1 (N + 1)!
10 9 ¨ 211 (11)! « 5 ˆ 10´4 15 9 ¨ 216 (16)! « 3 ˆ 10´8 17 9 ¨ 218 (18)! « 4 ˆ 10´10 18 9 ¨ 219 (19)! « 4 ˆ 10´11 ă 5 ˆ 10´11 So, it suffices to use the partial sum S18.
S-31: We’ll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SN is EN = f (N+1)(c) (N + 1)! (x ´ a)N+1 for some c strictly between a and x. In our case, a = 1 2, x = ´1 3, and we are given the nth derivative of f (x): E6 = f (7)(c) 7!
´1 3 ´ 1 2 7 = 1 7! ¨ 6!
2 h (1 ´ c)´7 + (´1)6 (1 + c)´7i ´5 6 7 = ´57 14 ¨ 67 ¨ h (1 ´ c)´7 + (1 + c)´7i = for some c in (´1 3, 1 2).
We want to provide actual numeric bounds for this expression. That is, we want to find the absolute max and min of E(c) = ´57 14 ¨ 67 ¨ h (1 ´ c)´7 + (1 + c)´7i over the interval ´1 3, 1 2 . Absolute extrema occur at endpoints and critical points. So, we’ll start by differentiating E(c), and finding its critical points (if any) in the interval ´1 3, 1 2 .
706 E(c) = ´57 14 ¨ 67 ¨ h (1 ´ c)´7 + (1 + c)´7i E1(c) = ´57 14 ¨ 67 ¨ h 7 (1 ´ c)´8 ´ 7 (1 + c)´8i = 0 (1 ´ c)´8 = (1 + c)´8 1 ´ c = 1 + c c = 0 Since E1(c) is defined over our entire interval, its only critical point is c = 0.
• E(0) = ´57 14 ¨ 67 • E ´1 3 = ´57 14 ¨ 67 4 3 ´7 + 2 3 ´7 = ´57 14 ¨ 67 h 3 4 7 + 3 2 7i • E 1 2 = ´57 14 ¨ 67 1 2 ´7 + 3 2 ´7 = ´57 14 ¨ 67 h 27 + 2 3 7i We want to decide which of these numbers is biggest, and which smallest. Note that 27 is much, much bigger than (3/2)7, and both (3/4)7 and (2/3)7 are less than one.
Furthermore, (3/2)7 is much larger than 2. So: 27 + (2/3)7 ą (3/2)7 + (3/4)7 ą 2.
Therefore, ´57 14 ¨ 67 " 27 + 2 3 7# ă ´57 14 ¨ 67 "3 4 7 + 3 2 7# ă ´57 14 ¨ 67 We conclude that the error E6 is in the interval ´57 14 ¨ 67 " 27 + 2 3 7# , ´57 14 ¨ 67 !
or, equivalently, ´57 14 ¨ 37 1 + 1 37 , ´57 7 ¨ 67 which is approximately (´0.199, ´0.040).
707 S-32: Using the Maclaurin series expansions of cos x and ex, cos x = 1 ´ x2 2! + x4 4! + ¨ ¨ ¨ 1 ´ cos x = x2 2! ´ x4 4! + ¨ ¨ ¨ ex = 1 + x + x2 2! + x3 3! + ¨ ¨ ¨ 1 + x ´ ex = ´x2 2! ´ x3 3! + ¨ ¨ ¨ 1 ´ cos x 1 + x ´ ex = x2 2! ´ x4 4! + ¨ ¨ ¨ ´ x2 2! ´ x3 3! + ¨ ¨ ¨ = 1 2! ´ x2 4! + ¨ ¨ ¨ ´ 1 2! ´ x 3! + ¨ ¨ ¨ we have lim xÑ0 1 ´ cos x 1 + x ´ ex = lim xÑ0 1 2! ´ x2 4! + ¨ ¨ ¨ ´ 1 2! ´ x 3! + ¨ ¨ ¨ = 1 2!
´ 1 2!
= ´1 S-33: Using the Maclaurin series expansion of sin x, sin x = x ´ x3 3! + x5 5! ´ x7 7! + ¨ ¨ ¨ sin x ´ x + x3 6 = x5 5! ´ x7 7! + ¨ ¨ ¨ sin x ´ x + x3 6 x5 = 1 5! ´ x2 7! + ¨ ¨ ¨ we have lim xÑ0 sin x ´ x + x3 6 x5 = lim xÑ0 1 5! ´ x2 7! + ¨ ¨ ¨ = 1 5! = 1 120 Remark: to solve this using l’Hˆ opital’s rule we would differentiate five times, making series a practical alternative.
S-34: Our limit has the indeterminate form 18; as with l’Hˆ opital’s rule, we can change it to a friendlier form using the natural logarithm.
f (x) = 1 + x + x22/x log( f (x)) = log 1 + x + x22/x = 2 x log 1 + x + x2 Recall log(1 + y) = 8 ÿ n=1 (´1)n+1yn n , and set y = x + x2. The series converges when |y| ă 1, and since we only consider values of x that are very close to 0, we can assume 708 |x + x2| ă 1.
log( f (x)) = 2 x log 1 + (x + x2) = 2 x 8 ÿ n=1 (´1)n+1(x + x2)n n = 2 x (x + x2) ´ (x + x2)2 2 + (x + x2)3 3 ´ ¨ ¨ ¨ = 2 + 2x ´ (x + x2)2 2x + (x + x2)3 3x ´ ¨ ¨ ¨ = 2 + 2x ´ (x2 + x)(1 + x) 2 + (x2 + x)2(1 + x) 3 ´ ¨ ¨ ¨ lim xÑ0 log( f (x)) = lim xÑ0 2 + 2x ´ (x2 + x)(1 + x) 2 + (x2 + x)2(1 + x) 3 ´ ¨ ¨ ¨ = 2 + 0 + 0 ¨ ¨ ¨ = 2 lim xÑ0 f (x) = lim xÑ0 elog f (x) = e2 S-35: We have an indeterminate form 18. We can use a natural logarithm to change this to a friendlier form. Furthermore, to avoid negative powers, we substitute y = 1 2x. As x grows larger and larger, y gets closer and closer to zero, while staying positive.
log 1 + 1 2x x = x log 1 + 1 2x = 1 2y log(1 + y) = 1 2y 8 ÿ n=1 (´1)n+1 n yn = 1 2y y ´ y2 2 + y3 3 ´ y4 4 + ¨ ¨ ¨ = 1 2 ´ y 4 + y2 6 ´ y3 8 + ¨ ¨ ¨ lim xÑ8 log 1 + 1 2x x = lim yÑ0+ 1 2 ´ y 4 + y2 6 ´ y3 8 + ¨ ¨ ¨ = 1 2 lim xÑ8 1 + 1 2x x = e1/2 = ?e 709 S-36: The factor (n + 1)(n + 2) reminds us of a derivative. Start with the geometric series.
1 1 ´ x = 8 ÿ n=0 xn d dx " 1 1 ´ x = d dx # 8 ÿ n=0 xn + 1 (1 ´ x)2 = 8 ÿ n=0 nxn´1 = 8 ÿ n=1 nxn´1 d dx " 1 (1 ´ x)2 = d dx # 8 ÿ n=1 nxn´1 + 2 (1 ´ x)3 = 8 ÿ n=1 n(n ´ 1)xn´2 = 8 ÿ n=2 n(n ´ 1)xn´2 = 8 ÿ n=0 (n + 2)(n + 1)xn Let x = 1 7. Then |x| ă 1, so our series converges.
2 (1 ´ 1/7)3 = 8 ÿ n=0 (n + 2)(n + 1) 1 7 n 2 (6/7)3 = 8 ÿ n=0 (n + 2)(n + 1) 7n S-37: Recall the Taylor series for arctangent is: arctan x = 8 ÿ n=0 (´1)n x2n+1 2n + 1 There are similarities between this and our given series: skipping powers of x, and a denominator that’s not factorial. We’ll try to manipulate it to look like our series. First, we antidifferentiate, to get a factor of (2n + 2) on the bottom.
ż arctan x dx = 8 ÿ n=0 (´1)n x2n+2 (2n + 1)(2n + 2) + C We can find the antiderivative of arctangent using integration by parts. Let u = arctan x and dv = dx; then du = 1 1+x2dx and v = x.
ż arctan x dx = x arctan x ´ ż x 1 + x2 dx + C 710 Now, we use the substitution w = 1 + x2, dw = 2xdx.
= x arctan x ´ 1 2 log(1 + x2) + C So, 8 ÿ n=0 (´1)n x2n+2 (2n + 1)(2n + 2) = x arctan x ´ 1 2 log(1 + x2) + C To find C, we evaluate both sides of the equation at x = 0.
0 = 0 arctan 0 ´ 1 2 log(1) + C = C Therefore, 8 ÿ n=0 (´1)n x2n+2 (2n + 1)(2n + 2) = x arctan x ´ 1 2 log(1 + x2) Multiplying both sides by x2, 8 ÿ n=0 (´1)n x2n+4 (2n + 1)(2n + 2) = x3 arctan x ´ x2 2 log(1 + x2) S-38: (a) We’ll start, as we usually do, by finding a pattern for f (n)(0).
f (x) = (1 ´ x)´1/2 f 1(x) = 1 2(1 ´ x)´3/2 f 2(x) = 1 ¨ 3 22 (1 ´ x)´5/2 f 3(x) = 1 ¨ 3 ¨ 5 23 (1 ´ x)´7/2 f (4)(x) = 1 ¨ 3 ¨ 5 ¨ 7 24 (1 ´ x)´9/2 .
.
.
f (n)(x) = 1 ¨ 3 ¨ 5 ¨ . . . ¨ (2n ´ 1) 2n (1 ´ x)´(2n+1)/2 f (n)(0) = 1 ¨ 3 ¨ 5 ¨ . . . ¨ (2n ´ 1) 2n We could leave it like this, but we simplify, to make our work cleaner later on.
= 1 2n ¨ (2n)!
2 ¨ 4 ¨ 6 ¨ . . . ¨ (2n) = 1 2n ¨ (2n)!
2n ¨ n!
= (2n)!
22n n!
711 This pattern holds for n ě 0. Now, we can write our Maclaurin series for f (x).
(1 ´ x)´1/2 = 8 ÿ n=0 f (n)(0) n!
xn = 8 ÿ n=0 (2n)!
22n (n!)2 xn To find the radius of convergence, we use the ratio test.
ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = (2n + 2)!
22n+2((n + 1)!)2 ¨ 22n (n!)2 (2n)!
¨ |x| = (2n + 2)!
(2n)!
n!
(n + 1)!
2 ¨ 22n 22n+2|x| = (2n + 2)(2n + 1) 1 n + 1 2 ¨ 1 4|x| = 4n2 + 4n + 2 4n2 + 8n + 4|x| lim nÑ8 ˇ ˇ ˇ ˇ an+1 an ˇ ˇ ˇ ˇ = lim nÑ8 4n2 + 4n + 2 4n2 + 8n + 4|x| = |x| So, the radius of convergence is R = 1.
(b) We note the derivative of the arcsine function is 1 ?
1 ´ x2 = f (x2). With this insight, we can manipulate our Taylor series for f (x) into a Taylor series for arcsine.
1 ?
1 ´ x = 8 ÿ n=0 (2n)!
22n (n!)2 xn 1 ?
1 ´ x2 = 8 ÿ n=0 (2n)!
22n (n!)2 x2n ż 1 ?
1 ´ x2 dx = ż 8 ÿ n=0 (2n)!
22n (n!)2 x2n !
dx arcsin x = 8 ÿ n=0 (2n)!
22n (n!)2(2n + 1)x2n+1 + C arcsin x = 8 ÿ n=0 (2n)!
22n (n!)2(2n + 1)x2n+1 where we found the value of C by setting x = 0. Its radius of convergence is also 1, by Theorem 3.5.13.
S-39: We use that log(1 + y) = 8 ÿ n=1 (´1)n´1 yn n for all ´ 1 ă y ď 1 712 with y = x ´ 2 2 to give log(x) = log(2 + x ´ 2) = log 2 1 + x ´ 2 2 = log 2 + log 1 + x ´ 2 2 = log 2 + 8 ÿ n=1 (´1)n´1 n 2n (x ´ 2)n It converges when ´1 ă y ď 1, or equivalently, 0 ă x ď 4.
S-40: (a) Using the geometric series expansion with r = ´t4, 1 1 ´ r = 8 ÿ n=0 rn ù ñ 1 1 + t4 = 8 ÿ n=0 (´t4) n = 8 ÿ n=0 (´1)nt4n Substituting this into our integral, I(x) = ż x 0 1 1 + t4 dt = ż x 0 8 ÿ n=0 (´1)nt4n !
dt = " 8 ÿ n=0 (´1)n t4n+1 4n + 1 #t=x t=0 = 8 ÿ n=0 (´1)n x4n+1 4n + 1 (b) Substituting in x = 1 2, I(1/2) = 8 ÿ n=0 (´1)n 1 (4n + 1)24n+1 = 1 2 ´ 1 5 ˆ 25 + 1 9 ˆ 29 ´ 1 13 ˆ 213 + ¨ ¨ ¨ = 0.5 ´ 0.00625 + 0.000217 ´ 0.0000094 + ¨ ¨ ¨ = 0.493967 ´ 0.0000094 + ¨ ¨ ¨ See part (c) for the error analysis.
(c) The series for I(x) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is between zero and the first omitted term. In this case, the first omitted term was negative (´0.0000094). So the exact value of I(1/2) is the approximate value found in part (b) plus a negative number whose magnitude is smaller than 0.00001 = 10´5. So the approximate value of part (b) is larger than the true value of I(1/2).
713 S-41: Expanding the exponential using its Maclaurin series, I = ż 1 0 x4e´x2 dx = 8 ÿ n=0 ż 1 0 x4(´x2)n n!
dx = 8 ÿ n=0 (´1)n n!
ż 1 0 x2n+4 dx = 8 ÿ n=0 (´1)n n!(2n + 5) = 1 5 lo omo on n=0 ´ 1 7 lo omo on n=1 + 1 18 lo omo on n=2 ´ 1 3!(11) loomoon n=3 + ¨ ¨ ¨ The signs of successive terms in this series alternate. Futhermore the magnitude of the nth term decreases with n. Hence, by the alternating series test, I lies between 1 5 ´ 1 7 + 1 18 and 1 5 ´ 1 7 + 1 18 ´ 1 3!(11). So |I ´ a| ď 1 3!(11) = 1 66 S-42: Expanding the exponential using its Taylor series, I = ż 1 2 0 x2e´x2 dx = 8 ÿ n=0 ż 1 2 0 x2(´x2)n n!
dx = 8 ÿ n=0 (´1)n n!
ż 1 2 0 x2n+2 dx = 8 ÿ n=0 (´1)n n!(2n + 3) 1 22n+3 The signs of successive terms in this series alternate. Futhermore the magnitude of the nth term decreases with n. Hence, by the alternating series test, I lies between N ř n=0 (´1)n n!(2n+3) 1 22n+3 and N+1 ř n=0 (´1)n n!(2n+3) 1 22n+3, for every N. The first few terms are, to five decimal places, n 0 1 2 3 (´1)n n!(2n + 3) 1 22n+3 0.04167 -0.00625 0.00056 -0.00004 Allowing for a roundoff error of 0.000005 in each of these, I must be between 0.04167 ´ 0.00625 + 0.00056 + 0.000005 ˆ 3 = 0.035995 and 0.04167 ´ 0.00625 + 0.00056 ´ 0.00004 ´ 0.000005 ˆ 4 = 0.035920 where the multiples of 0.000005 are the maximum possible accumulated roundoff errors in the added terms.
S-43: (a) Using the Taylor series expansion of ex with x = ´t, e´t = 8 ÿ n=0 (´t)n n!
ù ñ e´t ´ 1 = 8 ÿ n=1 (´1)n tn n! ù ñ e´t ´ 1 t = 8 ÿ n=1 (´1)n tn´1 n!
714 Substituting this into our integral, I(x) = ż x 0 e´t ´ 1 t dt = 8 ÿ n=1 (´1)n ż x 0 tn´1 n! dt = 8 ÿ n=1 (´1)n xn n ¨ n!
(b) Substituting in x = 1, I(1) = 8 ÿ n=1 (´1)n 1 n ¨ n!
= ´1 + 1 2 ¨ 2! ´ 1 3 ¨ 3! + 1 4 ¨ 4! ´ 1 5 ¨ 5! + ¨ ¨ ¨ = ´1 + 0.25 ´ 0.0556 + 0.0104 ´ 0.0017 + ¨ ¨ ¨ = ´0.80 See part (c) for the error analysis.
(c) The series for I(x) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is no larger than the first omitted term. So the magnitude of ´ 1 5 5! + ¨ ¨ ¨ is no larger than 0.0017. Allowing for a roundoff error of at most 0.0001 in each of the two terms ´0.0556 + 0.0104 I(1) = ´1 + 0.25 ´ 0.0556 + 0.0104 ˘ 0.0019 = ´0.7952 ˘ 0.0019 S-44: (a) Using the Taylor series expansion of sin x with x = t, sin t = 8 ÿ n=0 (´1)n t2n+1 (2n + 1)! ù ñ sin t t = 8 ÿ n=0 (´1)n t2n (2n + 1)!
So Σ(x) = ż x 0 sin t t dt = 8 ÿ n=0 (´1)n ż x 0 t2n (2n + 1)! dt = 8 ÿ n=0 (´1)n x2n+1 (2n + 1)(2n + 1)!
(b) The critical points of Σ(x) are the solutions of Σ1(x) = 0. By the fundamental theorem of calculus Σ1(x) = sin x x , so the critical points of Σ(x) are x = ˘π, ˘2π, ¨ ¨ ¨ . The absolute maximum occurs at x = π.
(c) Substituting in x = π, Σ(π) = 8 ÿ n=0 (´1)n π2n+1 (2n + 1)(2n + 1)!
= π ´ π3 3 ¨ 3! + π5 5 ¨ 5! ´ π7 7 ¨ 7! + ¨ ¨ ¨ = 3.1416 ´ 1.7226 + 0.5100 ´ 0.0856 + 0.0091 ´ 0.0007 + ¨ ¨ ¨ 715 The series for Σ(π) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is no larger than the first omitted term. So Σ(π) = 3.1416 ´ 1.7226 + 0.5100 ´ 0.0856 + 0.0091 = 1.8525 with an error of magnitude at most 0.0007 + 0.0005 (the 0.0005 is the maximum possible accumulated roundoff error in all five retained terms).
S-45: (a) Using the Taylor series expansion of cos t, cos t = 1 ´ t2 2! + t4 4! ´ t6 6! + ¨ ¨ ¨ = 8 ÿ n=0 (´1)n t2n (2n)!
cos t ´ 1 t2 = ´ 1 2! + t2 4! ´ t4 6! + ¨ ¨ ¨ = 8 ÿ n=1 (´1)n t2n´2 (2n)!
I(x) = ż x 0 cos t ´ 1 t2 dt = ´ x 2! + x3 4!3 ´ x5 6!5 + ¨ ¨ ¨ = 8 ÿ n=1 (´1)n x2n´1 (2n)!(2n ´ 1) (b), (c) Substituting in x = 1, I(1) = ´1 2 + 1 4!3 ´ 1 6!5 + ¨ ¨ ¨ = ´0.5 + 0.0139 ´ 0.0003 ´ ¨ ¨ ¨ = ´0.486 ˘ 0.001 The series for I(1) is an alternating series with decreasing successive terms that converge to zero. So approximating I(1) by ´1 2 + 1 4!3 introduces an error between 0 and ´ 1 6!5.
Hence I(1) ă ´1 2 + 1 4!3.
S-46: (a) Using the Taylor series expansions of sin x and cos x with x = t, sin t = 8 ÿ n=0 (´1)n t2n+1 (2n + 1)!
=t ´ t3 3! + t5 5! ´ t7 7! + ¨ ¨ ¨ t sin t = 8 ÿ n=0 (´1)n t2n+2 (2n + 1)!
= t2 ´ t4 3! + t6 5! ´ t8 7! + ¨ ¨ ¨ = ´ 8 ÿ n=1 (´1)n t2n (2n ´ 1)!
cos t = 8 ÿ n=0 (´1)n t2n (2n)!
=1 ´ t2 2! + t4 4! ´ t6 6! + t8 8! + ¨ ¨ ¨ cos t ´ 1 = 8 ÿ n=1 (´1)n t2n (2n)!
= ´ t2 2! + t4 4! ´ t6 6! + t8 8! + ¨ ¨ ¨ 716 cos t + t sin t´1 = 8 ÿ n=1 (´1)n t2n (2n)! ´ 8 ÿ n=1 (´1)n t2n (2n ´ 1)!
= 8 ÿ n=1 (´1)nt2n 1 (2n)! ´ 1 (2n ´ 1)!
= 1 ´ 1 2!
t2 ´ 1 3! ´ 1 4!
t4 + ¨ ¨ ¨ = 8 ÿ n=1 (´1)nt2n 1 (2n)! ´ 2n (2n)!
= 8 ÿ n=1 (´1)nt2n 1 ´ 2n (2n)!
= 2 2! ´ 1 2!
t2 ´ 4 4! ´ 1 4!
t4 + ¨ ¨ ¨ = 8 ÿ n=1 (´1)n+1t2n 2n ´ 1 (2n)!
= 1 2!t2 ´ 3 4!t4 + 5 6!t6 ´ 7 8!t8 + ¨ ¨ ¨ cos t + t sin t ´ 1 t2 = 8 ÿ n=1 (´1)n+1t2n´2 2n ´ 1 (2n)!
= 1 2!t ´ 3 4!t2 + 5 6!t4 ´ 7 8!t6 + ¨ ¨ ¨ Now, we’re ready to integrate.
I(x) = ż x 0 cos t + t sin t ´ 1 t2 = ż x 0 8 ÿ n=1 (´1)n+1t2n´2 2n ´ 1 (2n)!
!
dt = " 8 ÿ n=1 (´1)n+1 t2n´1 (2n)!
x 0 = 8 ÿ n=1 (´1)n+1 x2n´1 (2n)!
(b) I(1) = 1 2! ´ 1 4! + 1 6! ´ 1 8! + ¨ ¨ ¨ = 0.5 ´ 0.041˙ 6 + 0.00139 ´ 0.000024 + ¨ ¨ ¨ = 0.460 . The error analysis is in part (c).
(c) The series for I(1) is an alternating series with decreasing successive terms that convege to zero. So approximating I(1) by 1 2! ´ 1 4! + 1 6! introduces an error between 0 and ´ 1 8!. So I(1) ă 1 2! ´ 1 4! + 1 6! ă 0.460.
S-47: (a) Substituting x = ´t into the known power series ex = 1 + x + x2 2! + x3 3! + x4 4! + ¨ ¨ ¨ , we see that: e´t = 1 ´ t + t2 2! ´ t3 3! + t4 4! ´ ¨ ¨ ¨ 1 ´ e´t = t ´ t2 2! + t3 3! ´ t4 4! + ¨ ¨ ¨ 1 ´ e´t t = 1 ´ t 2! + t2 3! ´ t3 4! + ¨ ¨ ¨ ż 1 ´ e´t t dt = C + x ´ x2 2 ¨ 2! + x3 3 ¨ 3! ´ x4 4 ¨ 4! + ¨ ¨ ¨ 717 Finally, f (0) = 0 (since f (0) is an integral from 0 to 0) and so C = 0. Therefore f (x) = ż x 0 1 ´ e´t t dt = x ´ x2 2 ¨ 2! + x3 3 ¨ 3! ´ x4 4 ¨ 4! + ¨ ¨ ¨ .
We can also do this calculation entirely in summation notation: ex = 8 ÿ n=0 xn n! , and so e´t = 8 ÿ n=0 (´t)n n!
= 1 + 8 ÿ n=1 (´1)ntn n!
1 ´ e´t = ´ 8 ÿ n=1 (´1)ntn n!
= 8 ÿ n=1 (´1)n´1tn n!
1 ´ e´t t = 8 ÿ n=1 (´1)n´1tn´1 n!
f (x) = ż x 0 1 ´ e´t t dt = 8 ÿ n=1 (´1)n´1xn n ¨ n!
(b) We set an = Anxn = (´1)n´1 n ¨ n!
xn and apply the ratio test.
lim nÑ8 ˇ ˇ ˇan+1 an ˇ ˇ ˇ = lim nÑ8 ˇ ˇ ˇ ˇ (´1)nxn+1/((n + 1) ¨ (n + 1)!) (´1)n´1xn/(n ¨ n!) ˇ ˇ ˇ ˇ = lim nÑ8 |x|n+1 |x|n n ¨ n!
(n + 1) ¨ (n + 1)!
= lim nÑ8 |x| n (n + 1)2 since (n + 1)! = (n + 1) n!
= 0 This is smaller than 1 no matter what x is. So the series converges for all x.
S-48: ex = 1 + x + x2 2! + x3 3! + ¨ ¨ ¨ ě 1 + x for all x ě 0 ù ñ ex ´ 1 ě x ù ñ x3 ex ´ 1 ď x3 x = x2 ù ñ ż 1 0 x3 ex ´ 1 dx ď ż 1 0 x2 dx = 1 3 718 S-49: (a) We know that ex = 8 ř n=0 xn n! for all x. Replacing x by ´x, we also have e´x = 8 ř n=0 (´x)n n!
for all x and hence cosh(x) = 1 2 ex + e´x = 1 2 h 8 ÿ n=0 xn n! + 8 ÿ n=0 (´x)n n!
i = 8 ÿ n=0 n even xn n! = 8 ÿ n=0 x2n (2n)!
for all x. In particular, the interval of convergence is all real numbers.
(b) Using the power series expansion of part (a), cosh(2) = 1 + 22 2! + 24 4! + 8 ÿ n=3 22n (2n)! = 32 3 + 8 ÿ n=3 22n (2n)!
So it suffices to show that ř8 n=3 22n (2n)! ď 0.1. Let’s write bn = 22n (2n)!. The first term in ř8 n=3 22n (2n)! is b3 = 26 6! = 26 6 ˆ 5 ˆ 4 ˆ 3 ˆ 2 = 4 45 The ratio between successive terms in ř8 n=3 22n (2n)! is bn+1 bn = 22n+2/22n (2n + 2)!/(2n)! = 4 (2n + 2)(2n + 1) ď 4 8 ˆ 7 = 1 14 for all n ě 3 Hence 8 ÿ n=3 22n (2n)! ď b3 hk kik kj 4 45 + b4ď hkkkikkkj 4 45 ˆ 1 14 + b5ď hkkkkikkkkj 4 45 ˆ 1 142 + b6ď hkkkkikkkkj 4 45 ˆ 1 143 + ¨ ¨ ¨ = 4 45 1 1 ´ 1 14 = 4 45 14 13 = 56 585 ă 1 10 (c) Comparing cosh(t) = 8 ÿ n=0 t2n (2n)! = 8 ÿ n=0 (t2)n (2n)!
and e 1 2t2 = 8 ÿ n=0 (1 2t2) n n!
= 8 ÿ n=0 (t2)n 2nn!
we see that it suffices to show that (2n)! ě 2nn!. Now. for all n ě 1, (2n)! = n factors hkkkkkkkikkkkkkkj 1 ˆ 2 ˆ ¨ ¨ ¨ ˆ n n factors hkkkkkkkkkkkkkkkkkikkkkkkkkkkkkkkkkkj (n + 1) ˆ (n + 2) ˆ ¨ ¨ ¨ ˆ 2n ě n factors hkkkkkkkikkkkkkkj 1 ˆ 2 ˆ ¨ ¨ ¨ ˆ n n factors hkkkkkkkikkkkkkkj 2 ˆ 2 ˆ ¨ ¨ ¨ ˆ 2 = 2n n!
719 S-50: (a) For Newton’s method, recall we approximate a root of the function g(x) in iterations: given an approximation xn, our next approximation is xn+1 = xn ´ g(xn) g1(xn). In our case, xn+1 = xn ´ x3 n ´ 2 3x2 n = 2 3 xn + 1 x2 n .
We want to start somewhere reasonably close to the actual root we want, so let’s set x0 = 1. (Your starting point may vary.) x0 = 1 ù ñ x1 = 2 3 1 + 1 1 = 4 3 «1.3333 x1 = 4 3 ù ñ x2 = 2 3 4 3 + 9 16 = 91 72 «1.2639 x2 = 91 72 ù ñ x3 = 2 3 91 72 + 722 912 = 1126819 894348 «1.2599 x3 = 1126819 894348 ù ñ x4 = 2 3 1126819 894348 + 8943482 11268192 «1.2599 So, 3 ?
2 « 1.26.
(b) We’ll evaluate the given series at x = 2. This yields the series 3 ?
2 = 1 + 1 6 + 8 ÿ n=2 (´1)n´1(2)(5)(8) ¨ ¨ ¨ (3n ´ 4) 3n n!
.
This series is alternating, so if we use the partial sum SN, our absolute error is at most |aN+1| = (2)(5)(8) ¨ ¨ ¨ (3N ´ 1) 3N+1 (N + 1)!
(if N ě 2). We want to know which value of N makes this at most 0.01. We test several values.
720 N |aN+1| 3 (2)(5)(8) 34 ¨ 4!
« 0.04 4 (2)(5)(8)(11) 35 5!
« 0.03 5 (2)(5)(8)(11)(14) 36 6!
« 0.023 6 (2)(5)(8)(11)(14)(17) 37 7!
« 0.019 7 (2)(5)(8)(11)(14)(17)(20) 38 8!
« 0.016 8 (2)(5)(8)(11)(14)(17)(20)(23) 39 9!
« 0.013 9 (2)(5)(8)(11)(14)(17)(20)(23)(26) 310 10!
« 0.012 10 (2)(5)(8)(11)(14)(17)(20)(23)(26)(29) 311 11!
« 0.0103 11 (2)(5)(8)(11)(14)(17)(20)(23)(26)(29)(32) 312 12!
« 0.009 So, the approximation S11 has a sufficiently small error. That is, we would add up the first twelve terms.
S-51: Our plan is as follows: • Make a Taylor series for f (x) • Calculate the tenth derivative of the Taylor series of f (x).
• Decide how many terms we need to add to achieve the desired accuracy.
• Approximate f (10) 1 5 with a partial sum.
We know that the Taylor series for arctan x is 8 ÿ n=0 (´1)n x2n+1 2n + 1, which converges for ´1 ď x ď 1. So, the Taylor series for arctan(x3) is f (x) = arctan(x3) = 8 ÿ n=0 (´1)n (x3)2n+1 2n + 1 = 8 ÿ n=0 (´1)n x6n+3 2n + 1 It is much easier to differentiate this series many times than it is to differentiate 721 arctan(x3) directly many times.
f 1(x) = 8 ÿ n=0 (´1)n (6n + 3)x6n+2 2n + 1 f 2(x) = 8 ÿ n=0 (´1)n (6n + 3)(6n + 2)x6n+1 2n + 1 f 3(x) = 8 ÿ n=0 (´1)n (6n + 3)(6n + 2)(6n + 1)x6n 2n + 1 .
.
.
f (10)(x) = 8 ÿ n=0 (´1)n (6n + 3)(6n + 2)(6n + 1) ¨ ¨ ¨ (6n ´ 6)x6n´7 2n + 1 = 8 ÿ n=2 (´1)n (6n + 3)(6n + 2)(6n + 1) ¨ ¨ ¨ (6n ´ 6)x6n´7 2n + 1 = 8 ÿ n=2 (´1)n (6n + 3)!
(2n + 1)(6n ´ 7)!x6n´7 f (10) 1 5 = 8 ÿ n=2 (´1)n (6n + 3)!
(2n + 1)(6n ´ 7)! ¨ 56n´7 (Notice, after ten differentiations, the terms a0 and a1 are both zero.) Since this is an alternating series, the absolute error involved in using the approximation SN is at most |aN+1| = (6N + 9)!
(2N + 3)(6N ´ 1)! ¨ 56N´1 By testing a few values of N, we find |a6| = |a5+1| = 39!
(13)(29!) ¨ 529 « 0.00000095 ă 10´6 So, S5 is a sufficient approximation. That is, f (10) 1 5 « 5 ÿ n=2 (´1)n (6n + 3)!
(2n + 1)(6n ´ 7)! ¨ 56n´7 = (´1)2 15!
5 ¨ 5! ¨ 55 + (´1)3 21!
7! ¨ 11! ¨ 511 + (´1)4 27!
9! ¨ 17! ¨ 517 + (´1)5 33!
11! ¨ 23! ¨ 523 = 15!
5! ¨ 56 ´ 21!
7! ¨ 11! ¨ 511 + 27!
9! ¨ 17! ¨ 517 ´ 33!
11! ¨ 23! ¨ 523 Remark: if we had calculated f (10)(1/5) directly, using derivative rules instead of series, we would have found an exact value; however, our value here is easier to find, and is highly accurate (if not exact).
722 S-52: (a) To sketch y = f (x), we note the following: • f (x) is never negative.
• lim xÑ˘8 f (x) = e0 = 1, so the curve has horizontal asymptotes in both directions at y = 1.
• lim xÑ˘0 f (x) = lim xÑ˘0 1 e1/x2 = lim uÑ+8 1 eu = 0 = f (0), so the curve is continuous at x = 0.
• For x ‰ 0, f 1(x) = 2 x3 e´1/x2, so our curve is decreasing on (´8, 0) and increasing on (0, 8) • For x ‰ 0, f 2(x) = 2x´6(2 ´ 3x2)e´1/x2, so our curve is concave up on (´ ?
2/3, ?
2/3), and concave down elsewhere.
x y 1 ´ ?
2/3 ?
2/3 y = f (x) (b) Since f (n)(0) = 0 for all whole n (that is, the graph is really quite flat at the origin), and since f (0) = 0, the Maclaurin series for f (x) is 8 ÿ n=0 0 n!xn = 0.
(c) The Maclaurin series converges for all real values of x (to the constant 0).
(d) Since ey ą 0 for any real y, we see f (x) = 0 only when x = 0. So, f (x) is only equal to its Maclaurin series at the single point x = 0.
Remark: the function f (x) is an example of a function whose Maclaurin series converges, but not to f (x)! To describe this behaviour, we say f (x) is non-analytic.
S-53: Solution 1: Since f (x) is odd, f (´x) = ´f (x) for all x in its domain. We plug this into our power series, then consider the even-indexed terms and the odd-indexed terms 723 separately.
f (´x) = ´f (x) 8 ÿ n=0 f (n)(0) n!
(´x)n = ´ 8 ÿ n=0 f (n)(0) n!
xn 8 ÿ n=0 f (2n+1)(0) (2n + 1)! (´x)2n+1 + 8 ÿ n=0 f (2n)(0) (2n)! (´x)2n = ´ 8 ÿ n=0 f (2n+1)(0) (2n + 1)! x2n+1´ 8 ÿ n=0 f (2n)(0) (2n)! x2n ´ 8 ÿ n=0 f (2n+1)(0) (2n + 1)! x2n+1 + 8 ÿ n=0 f (2n)(0) (2n)! x2n = ´ 8 ÿ n=0 f (2n+1)(0) (2n + 1)! x2n+1´ 8 ÿ n=0 f (2n)(0) (2n)! x2n 8 ÿ n=0 f (2n)(0) (2n)! x2n = ´ 8 ÿ n=0 f (2n)(0) (2n)! x2n 2 8 ÿ n=0 f (2n)(0) (2n)! x2n = 0 8 ÿ n=0 f (2n)(0) (2n)! x2n = 0 Solution 2: Alternately, we could note the following: • Since all derivative of f (x) exist, all its derivatives are continuous.
• The derivative of an odd function is even, and the derivative of an even function is odd.
• So, the even-indexed derivatives of f (x) are continuous, odd functions.
• Every continuous, odd function passes through the origin. That is, f (2n)(0) = 0.
• So, every term in the series is 0.
724 |
6736 | https://www.pbs.org/video/good-know-partial-products-grade-4/ | Good To Know | Partial Products | Grade 4 | PBS
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Partial Products | Grade 4
Special | 1m 58s
A concrete method for multiplication.
03/07/2017 | Rating NR
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Partial Products | Grade 4
Special | 1m 58s
A concrete method for multiplication.
When your child first learns to multiply two two-digit numbers, she will use the area model. This visual tool illustrates how to decompose numbers and find four different products. As her skills improve, she will move from this pictorial model into a concrete method called partial products.
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When your child first learns to multiply two two-digit numbers, she will use the area model. This visual tool illustrates how to decompose numbers and find four different products. As her skills improve, she will move from this pictorial model into a concrete method called partial products.
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6737 | https://www.youtube.com/watch?v=caAu2yGccRg | Geometry of Complex numbers | Lecture 4 | Equilateral Triangle | Important results
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Posted: 30 Jun 2021
Geometry of Complex numbers | Lecture 4 | Equilateral Triangle | Important results
00:00 #SE1 If z1, z2 and z3 are vertices of an equilateral triangle then prove that z1^2+z2^2+z3^2=z1z2+z2z3+z3z1. If z0 is the circumcenter then prove that z1^2+z2^2+z3^2=3z0^2
05:12 #SE2 Ifz1 and z2 are roots of the equation z^2+az+b=0 and O, z1 and z2 are vertices of an equilateral triangle then prove a^2=3b
06:37 #SE3 If the roots of the equation z^3+3az^2+3bz+c=0 correspond to vertices of an equilateral triangle then prove a^2=b
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SE1 If z1, z2 and z3 are vertices of an equilateral triangle then prove that z1^2+z2^2+z3^2=z1z2+z2z3+z3z1. If z0 is the circumcenter then prove that z1^2+z2^2+z3^2=3z0^2 ए लेदर कंपलेक्स नंबर्स डोनेट टो इन्वेस्ट इन बीजों बर्ड्स उसकी लाइफ इन सब्सक्राइब सीबीसी ए फ्रेंड यह दो एक्जेक्टली ओम ने Bigg Boss रिजल्ट प्रूव्ड थिस ए डिफरेंस को यह लड्डू स्क्वेयर प्लस बैट्री स्क्वेयर फिट जून 2017 टू गर्ल्स इन लर्च में जुटी डेक पलट वीडियो सिंह प्रिपरेशन आफ कंपलेक्स नंबर्स विल कंसीडर ए टू वेक्टर्स नज़ीर व्यक्ति मैंने सैट 2018 - बैक टू लोटस आयुर्वेद इंडिया को सब्सक्राइब करें - 121 एक व्यक्ति - बिट्टू अली गुड्डू बेबी अपऑन बीसीडी इंप्रूव डिफाल्टर फाइबर शेडस पे लुटा लुटा है का प्रतीक कृष्ण-राधा विल कंसीडर एवरी स्टेट आफ्टर सपोच कि चीफ निदेशक जयपुर - 302 अगेन विल रोटेट एंटी क्लॉकवाइज सब्सक्राइब टो ए विजिट टो - रैफरी अपऑन थे गवर्नमेंट लेटर ई लव बीइंग गुड बुक्स ए बी सी अपऑन ए सी एक बड़ा यह प्राइवेसी नवीन ठाकुर ट्रायंगल डिसाइड बिल्कुल ग्रेटर नोएडा सब्सक्राइब विड ऊ एक व्यक्ति - टैटू पिंपल टू बैक टू - रील पॉर्न का रेट - कि जेटली ने Bigg Boss मलिक वेब पेज ऐड्रेसेस एंड स्क्वेयर - 2013 - बैक टू जेड प्लस जेड 273 लिए बिल्कुल 2016 और मर्डर-2 सर थे माइंड व्यक्ति स्क्वायर प्लस एप कि वे टू व्यक्ति अवर बैक टू साथ ही कैंसल वीरवार के विध्वंस को प्लस J2 स मतलब स्क्वेयर मेडिकल टूट गठबंधन टू प्लस बैक टू जेड प्लस जेड डी जे डैन टर्नर अधिवेशन इफिजी आफ बर्ड्स पेंसिल ड्रॉइंग रूम आ अजय देवगन स्क्वेयर लेग टैटू स्क्वेयर प्लस जॉब स्क्वैड गुड्डू स्टूडेंट्स वेयर नॉट वर्क वेल नोन एस कि फिल्म इक्विलैटरल ट्रायंगल इस Android सरकमसेंटर और तू सेंटर इंटरेस्ट रिसीव्ड पॉइंट मिडनाइट बिल्कुल टू जेड प्लस जेड टू प्लस बी स्क्वायर नोट्स ही टू-डू लिस्ट को सबस्क्राइब कर अजय देवगन J2 J2 सेठ है जेट रिज़र्व व्होल ट्रुथ लैब्स के लड्डू सब्सक्राइब subscribe and subscribe the Channel को subscribe Video subscribe and subscribe है पिंपल टू डेकोरेट वे नोटिस सरकमसेंटर इन सेंटर सेंट्रल प्वाइंट सेंटर mid-2012 SE2 Ifz1 and z2 are roots of the equation z^2+az+b=0 and O, z1 and z2 are vertices of an equilateral triangle then prove a^2=3b प्लेसेस page-09 पिंपल्स टू डू द फॉलोइंग इज नॉट अ को जेड प्लस जेड 2012 - से एग प्रोडक्ट डेवलपमेंट उसे ज़ोर सुनो थे इफिजी डेक्सजेट 2131 तय व्हाट्सएप ओपन इक्विलैटरल ट्रायंगल दिन जेम्स को सबस्क्राइब प्लस यौवन 2012 प्लस ए लेटर f3 प्लस कर देती थी जड़वाल ओम ने खुद इस ट्रायंगल वृद्धि पर टेक्स्ट और जिन सीमेंट फैक्ट्री इस और एजेंट डेविड कि Bigg Boss कंडीशन है जब स्कूल एक्टिविटीज उर्दू नेटवर्क डेविड 2012 बोथ साइड्स डिसकनेक्ट कि जर्मन प्रेसिडेंट to goals रेमिडी कुल्फी टाइम्स कि J2 तेरे इश्क़ वे इस वक्त बिल्कुल टू चीफ हंद्स्फ्री ऑप्शन मैं नागिन क्वेश्चसं के थ्रू सबसीक्वेंट SE3 If the roots of the equation z^3+3az^2+3bz+c=0 correspond to vertices of an equilateral triangle then prove a^2=b कर पॉइंट टू वर्ड्स सिर्फ निकला कि ट्रायंगल इंप्रूव दैट इज क्वेश्चन बीमर सपोर्ट जरूर सा जज्बात जून 2016 जेड प्लस जेड टू प्लस टू है लैटर टू जहरीले प्लस कि जब 301 पिंपल टू थ्री कि अ सीबीसी रूट सैमसंग J2 इन एवरी दर वर्ड्स आफ न इक्विलैटरल ट्रायंगल धुंध में लिपटी नॉट फॉर दिस चैनल को सबस्क्राइब सब्सक्राइब टो के विरुद्ध टू प्लस से जूझती प्लस पिंपल एक अपने बी कंपलीट्स परफेक्ट वेडिंग टू मैं जरूर थे टू प्लस टू जेड 3 प्लस हु इज द प्रेसिडेंट वेबसाइट लुट लो कि फीचर नहीं है को जेड प्लस जेड टू प्लस जतिन है उसको या फिर गिलहरी डांस 2012 प्लस कि J2 जैसी प्लस के पिंपल्स नजरबंद सब्सक्राइब एंड हिस्ट्री लेमन जूस टू द टूरिस्ट प्लेसेस पेज न एडिट बिल्कुल टू टू टू टू [संगीत] |
6738 | https://artofproblemsolving.com/wiki/index.php/Root_(polynomial)?srsltid=AfmBOoomi_qNlstxZrAdQEa8sHR3l5bVnIEvvfS8RPKNV64aekvNfIOH | Art of Problem Solving
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Root (polynomial)
A root of a function (often a polynomial) whose range is the real, the complex numbers or any abstract field is a value in the domain of the function such that .
Contents
1 Finding roots
1.1 General techniques
1.2 Rational roots
1.3 Real roots
1.4 Complex roots
1.5 In algebraic form
Finding roots
Let with all and , a general degree- polynomial. The degree of is , so has at most complex roots. In fact, the sum of the multiplicities of all distinct complex roots is exactly ; that is, counting any double roots twice, triple roots three times, and so on, there are in fact exactly complex roots of .
General techniques
Multiplying or dividing all of the coefficients of a polynomial by a nonzero constant does not change its roots. Thus, there will always be a monic polynomial with degree having the same roots as , given by
Once a root of is found, the Factor Theorem gives that is a factor of . Therefore, can be divided out of using synthetic division, and the roots of the resulting quotient will be the remaining roots of . Once another root is found, the process can be repeated. Dividing through by is most practical when the root and all the coefficients of are rational.
Rational roots
The three simplest values to test are and .
is a root of if and only if the constant term is .
is a root of if and only if the sum of the coefficients is .
is a root of if and only if the alternating sum of the coefficients is .
If all of the coefficients of are integers, then the Rational Root Theorem applies; namely, if is a root of , with and relatively prime, then is a (positive or negative) divisor of and is a divisor of .
If all of the are rational, but not necessarily integers, then multiplying through by the least common multiple of the denominators yields a polynomial with the same roots as and integer coefficients. The Rational Root Theorem can then be applied to the new polynomial to search for rational roots of .
In some cases the search may be simplified by substituting , where is a nonconstant linear polynomial with rational coefficients. If is a rational root of , then is a rational root of . Conversely, if is a rational root of , then the inverse of at must be a rational root of .
Real roots
Evaluating at chosen values can point to the location of roots. If and have opposite signs, then the Intermediate Value Theorem states that has at least one root in .
For single roots and other roots of odd multiplicity (triple roots, quintuple roots, etc.), will always change sign on opposite sides of the root, but for roots of even multiplicity (such as double roots), the sign of will be the same on either side of the root, so the Intermediate Value Theorem will not detect even-multiplicity roots.
Descartes Rule of Signs yields information on the number of positive real roots and negative real roots of . Writing the coefficients of in descending order of degree and excluding any that equal , the number of positive real roots of is equal to the number of sign changes between adjacent coefficients, minus some even nonnegative integer. The number of negative real roots of is equal to the number of such sign changes after reversing the sign of every odd-degree coefficient, again minus some even nonnegative integer. Here roots are counted according to multiplicity, so double roots are counted twice, triple roots three times, and so on.
More broadly, counting roots according to their multiplicity as before the number of real roots always has the same parity as the degree . Specifically, if is odd then must have at least one real root.
Rolle's Theorem guarantees that if has two roots and , then its derivative has at least one root in the interval . In particular, if has no roots in an interval , then has at most one root in .
Newton's method generates arbitrarily close approximations of the value of a real root of a polynomial. Use of Newton's method generally requires an educated guess for the location of the root based on the above criteria. We let the guess equal and compute the approximations recursively:
Complex roots
Complex roots of polynomials with real coefficients always exist in conjugate pairs. That is, if , , and all coefficients of are real and is a root of then is also a root of . In particular, must both have an even number of distinct nonreal roots and an even sum of the multiplicities of all distinct nonreal roots.
The product of the factors corresponding to and is a quadratic polynomial with real coefficients. As such, dividing through by the product leaves a polynomial which still has real coefficients and all roots of the original except and .
In algebraic form
A root of a polynomial with integer coefficients will always be an algebraic number. General formulas give the algebraic form of the roots of polynomials with degree at most .
For a quadratic equation , the roots are given by the quadratic formula
The cubic formula and quartic formula also exist, but they are quite lengthy and not very practical for computation by hand.
No similar formula exists for quintics or polynomials of any higher degree. Although the roots of such polynomials are algebraic numbers, they cannot be expressed in terms of the coefficients using only addition, subtraction, multiplication, division, powers, and radicals.
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6739 | https://www.quora.com/How-do-you-make-b-the-subject-of-a-the-square-root-of-b+6 | How to make b the subject of a = the square root of b+6 - Quora
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5
How do you make b the subject of a = the square root of b+6?
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Chris Zhou
Just loved it · Author has 76 answers and 208.1K answer views
·7y
A2A
a=√b+6 a=b+6
a 2=(√b+6)2 a 2=(b+6)2
a 2=b+6 a 2=b+6
a 2−6=b+6−6 a 2−6=b+6−6
a 2−6=b a 2−6=b
b=a 2−6 b=a 2−6
Upvote ·
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More answers below
What is b the subject of a=\sqrt {b+6}?
Algebra: What is the square root of 6?
What is the method to calculate a square root by hand?
Who discovered the square and square roots?
What is the square root of 49?
Girija Warrier
Studied at Sufficiently Educated · Author has 5.9K answers and 13.9M answer views
·6y
That is, write ‘b’ in terms of ‘a’
a = √(b+6)
On squaring both sides
=> a² = b+6
=> b = a² - 6
Upvote ·
9 2
9 1
9 1
Assistant
Bot
·
Sep 6
Step 1 — square both sides to eliminate the square root:
a = √(b + 6) ⇒ a^2 = b + 6
Step 2 — isolate b:
b = a^2 − 6
Check domain (implicit step): the original equation requires √(b + 6) to be defined and equal to a, so b + 6 ≥ 0 and a ≥ 0. If a is known to be nonnegative and b = a^2 − 6, that b satisfies the original equation.
Upvote ·
Akan Okon
Studied Petroleum Engineering at University of Ibadan, Nigeria (Graduated 2012)
·6y
a = square root of (b + 6)
a^2 = b + 6
b = a^2 - 6
Upvote ·
9 5
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Who invented square root?
What is the square root of 64? (A) 6 (B) 8 (C) 21 (D) 1 trillion?
How we can prove that S n=n 2(2 a+(n−1)d)S n=n 2(2 a+(n−1)d)?
Who invented square roots and cube roots?
What is the square root 9 plus square root 16?
A Richard Manner
MIT Chemcal Engineer BS/MS 1979 · Author has 1.2K answers and 2.9M answer views
·7y
You probably want to square both sides. a^2 = b+6; b= a^2–6
that assumes you meant a=sqrt(b+6) if you meant sqrt(b) + 6
then b = (a-6)^2 = a^2 -12a +36
Upvote ·
9 3
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Related
How do I make x the subject of the formula ax²+bx+c=0?
I have TWO ways to do this…
This way is very clever but a little harder to follow…
Continue Reading
I have TWO ways to do this…
This way is very clever but a little harder to follow…
Upvote ·
99 54
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Upvoted by
Kobe Ye
, M.S Mathematics, The University of British Columbia (2007) and
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 6.8K answers and 52.8M answer views
·3y
Related
How do you make the "b" subject of the quadratic formula?
I just laughed out loud when I read this question because the quadratic formula is just the original equation transformed to make x the subject!
This is the original quadratic…
All the above equations are the SAME EQUATION.
Continue Reading
I just laughed out loud when I read this question because the quadratic formula is just the original equation transformed to make x the subject!
This is the original quadratic…
All the above equations are the SAME EQUATION.
Upvote ·
99 99
99 11
Nicholas McConnell
Studied Mathematics at Rutgers University (Graduated 2021) · Upvoted by
Aditya Garg
, M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) and
Utkarsh Priyam
, studied Mathematics at Harker School (2021) · Author has 1.1K answers and 1.8M answer views
·Updated 6y
Related
If a+b=3 a+b=3 and a 2+b 2=7 a 2+b 2=7, what is a 4+b 4 a 4+b 4?
There are many ways to tackle this problem. One is simply tackle it through algebraic means.
a+b=3 a+b=3 (Given)
a 2+b 2=7 a 2+b 2=7 (Given)
a 2+2 a b+b 2=9 a 2+2 a b+b 2=9 (square #1)
2 a b=2 2 a b=2 (subtract #2 from #3)
a b=1 a b=1 (divide #4 by 2)
(a 2+b 2)2=49(a 2+b 2)2=49 (square #2)
a 4+2 a 2 b 2+b 4=49 a 4+2 a 2 b 2+b 4=49 (expand #6 via the Binomial Theorem)
(a 4+b 4)+2(a b)2=49(a 4+b 4)+2(a b)2=49(rewrite general polynomial on left side of #6)
(a 4+b 4)+2⋅1 2=49(a 4+b 4)+2⋅1 2=49 (substitute #5 into left side of #8)
(a 4+b 4)+2=49(a 4+b 4)+2=49 (simplify left side of #9)
a 4+b 4=47 a 4+b 4=47 (subtract 2 from #10)
And hence the answer to your question is 47 47.
Another way is
Continue Reading
There are many ways to tackle this problem. One is simply tackle it through algebraic means.
a+b=3 a+b=3 (Given)
a 2+b 2=7 a 2+b 2=7 (Given)
a 2+2 a b+b 2=9 a 2+2 a b+b 2=9 (square #1)
2 a b=2 2 a b=2 (subtract #2 from #3)
a b=1 a b=1 (divide #4 by 2)
(a 2+b 2)2=49(a 2+b 2)2=49 (square #2)
a 4+2 a 2 b 2+b 4=49 a 4+2 a 2 b 2+b 4=49 (expand #6 via the Binomial Theorem)
(a 4+b 4)+2(a b)2=49(a 4+b 4)+2(a b)2=49(rewrite general polynomial on left side of #6)
(a 4+b 4)+2⋅1 2=49(a 4+b 4)+2⋅1 2=49 (substitute #5 into left side of #8)
(a 4+b 4)+2=49(a 4+b 4)+2=49 (simplify left side of #9)
a 4+b 4=47 a 4+b 4=47 (subtract 2 from #10)
And hence the answer to your question is 47 47.
Another way is to use computational algebraic geometry. And by that, I mean:
Take an ideal in the polynomial ring Q[a,b]Q[a,b], presumably of all general polynomials we are declaring to be zero. (In this problem it’s I=⟨a+b−3,a 2+b 2−7⟩I=⟨a+b−3,a 2+b 2−7⟩)
Pick a useful monomial ordering. (In this case let’s go with the lexicographical ordering: a i b j<a i′b j′a i b j<a i′b j′ iff either j<j′j<j′, or else j=j′j=j′ and i<i′i<i′. That is, 1<a<a 2<⋯<b<a b<a 2 b<…1<a<a 2<⋯<b<a b<a 2 b<…) With that, the leading term of a polynomial in Q[a,b]Q[a,b] is defined as the term with the largest monomial in this ordering.
Use Buchburger’s Algorithm to find a Gröbner basis for the ideal. [I.e., a finite generating set where the leading term of any polynomial in the ideal whatsoever is divisible by the leading term of a polynomial in the generating set.] You can also arrange for the basis to be reduced; i.e., no two elements of the basis have leading terms, one of which is divisible by the other.
Once you have a Gröbner basis, write the generators in decreasing order of leading term. Then taking any polynomial and reducing it modulo every basis element from left to right will give you a residue modulo the ideal, and every polynomial of the same congruence class mod the ideal will land at this same residue; in particular doing this to any element of the ideal gives you zero. (This isn’t true for arbitrary bases.)
In this case, we have I=⟨a+b−3,a 2+b 2−7⟩I=⟨a+b−3,a 2+b 2−7⟩ as our ideal. However, this is not a reduced basis, because the leading terms of the polynomials are b b and b 2 b 2, and b∣b 2 b∣b 2. We thus may reduce it by changing the second polynomial to its remainder when directly divided by the first polynomial, and we still have an ideal:
(a 2+b 2−7)−b(a+b−3)=a 2−a b+3 b−7(a 2+b 2−7)−b(a+b−3)=a 2−a b+3 b−7
The leading term of that polynomial is −a b−a b, which is still divisible by b b, so we may do this again:
(a 2−a b+3 b−7)+a(a+b−3)=2 a 2+3 b−3 a−7(a 2−a b+3 b−7)+a(a+b−3)=2 a 2+3 b−3 a−7
Finally, the leading term of that polynomial is 3 b 3 b, so subtract 3 3 times (a+b−3)(a+b−3) as follows:
(2 a 2+3 b−3 a−7)−3(a+b−3)=2 a 2−6 a+2(2 a 2+3 b−3 a−7)−3(a+b−3)=2 a 2−6 a+2
Thus we may rewrite I=⟨a+b−3,2 a 2−6 a+2⟩I=⟨a+b−3,2 a 2−6 a+2⟩, or better yet, I=⟨a+b−3,a 2−3 a+1⟩I=⟨a+b−3,a 2−3 a+1⟩ (since multiplication by a nonzero constant rational does not change the ideal). Effectively, this means that in the context of your question, we must have a 2−3 a+1=0 a 2−3 a+1=0. (However in the context of the polynomial ring, a 2−3 a+1 a 2−3 a+1 is a nonzero element of the ideal I I.) Let us keep note:
I=⟨a+b−3,a 2−3 a+1⟩I=⟨a+b−3,a 2−3 a+1⟩.
The leading terms of I I above are b b and a 2 a 2 respectively, hence I I is reduced and its leading terms are in descending order. In fact, I I is a Gröbner basis: the leading terms of the polynomials are relatively prime, so there is no need to find their Syzygy polynomial.
Now we have enough information to reduce a 4+b 4 a 4+b 4 modulo I I. Any polynomial we get must have its value equal to a 4+b 4 a 4+b 4 in the context that a+b−3=a 2−3 a+1=0 a+b−3=a 2−3 a+1=0. We start by observing that a 4+b 4 a 4+b 4 has leading term b 4 b 4, which is divisible by b b, so we subtract to get (a 4+b 4)−b 3(a+b−3)=a 4−a b 3+3 b 3(a 4+b 4)−b 3(a+b−3)=a 4−a b 3+3 b 3. The leading term of this polynomial is −a b 3−a b 3, which is again divisible by b b, so we cancel out that leading term: (a 4−a b 3+3 b 3)+a b 2(a+b−3)=a 4+3 b 3+a 2 b 2−3 a b 2(a 4−a b 3+3 b 3)+a b 2(a+b−3)=a 4+3 b 3+a 2 b 2−3 a b 2. Continuing to cancel the leading term until it is no longer divisible by b b, you wind up with 2 a 4−12 a 3+54 a 2−108 a+81 2 a 4−12 a 3+54 a 2−108 a+81. Now divide this by the second polynomial a 2−3 a+1 a 2−3 a+1 and the remainder is 47 47.
The algorithm mentioned above can be programmed into a computer, and then used to simplify the solving of any system of polynomial equations. Depending on what you want to get out of the equations, some monomial orderings are better than others. For instance, if you order all powers of a variable x x at the lowest, a Gröbner basis will explicitly state all polynomial relations x x must satisfy alone.
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How do I make b subject of formula in the equation a=√s(s-a) (s-b) (s-c)?
One of the great things about Quora is that folks can answer the question the OP was trying to ask, or should have asked, rather than the one actually asked.
I’m going to mostly ignore this form, with a a on the left and right, and s s left unspecified. This is presumably meant to be Heron’s Formula for the area of a triangle, which requires the stipulation that s s is the semiperimeter, and the result is the area Δ.Δ. Let’s change the question to:
Solve for b b in Δ=√s(s−a)(s−b)(s−c)where s=1 2(a+b+c)Δ=s(s−a)(s−b)(s−c)where s=1 2(a+b+c)
I’ve done this so many times already. Here we go again.
Δ=\sq Δ=\sq
Continue Reading
One of the great things about Quora is that folks can answer the question the OP was trying to ask, or should have asked, rather than the one actually asked.
I’m going to mostly ignore this form, with a a on the left and right, and s s left unspecified. This is presumably meant to be Heron’s Formula for the area of a triangle, which requires the stipulation that s s is the semiperimeter, and the result is the area Δ.Δ. Let’s change the question to:
Solve for b b in Δ=√s(s−a)(s−b)(s−c)where s=1 2(a+b+c)Δ=s(s−a)(s−b)(s−c)where s=1 2(a+b+c)
I’ve done this so many times already. Here we go again.
Δ=√s(s−a)(s−b)(s−c)Δ=s(s−a)(s−b)(s−c)
16 Δ 2=2 s(2 s−2 a)(2 s−2 b)(2 s−2 c)16 Δ 2=2 s(2 s−2 a)(2 s−2 b)(2 s−2 c)
16 Δ 2=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)16 Δ 2=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)
That form is already superior to Heron’s formula, with the spurious semiperimeter ousted and the fractions moved to the end. The next step is the crucial one; we want to aim for the form with b b isolated, so I think we associate this thus:
16 Δ 2=(((a+c)+b)((a+c)−b))((b−(a−c))(b+(a−c)))16 Δ 2=(((a+c)+b)((a+c)−b))((b−(a−c))(b+(a−c)))
16 Δ 2=((a+c)2−b 2)(b 2−(a−c)2)16 Δ 2=((a+c)2−b 2)(b 2−(a−c)2)
16 Δ 2=(2 a c+(a 2−b 2+c 2))(2 a c−(a 2−b 2+c 2))16 Δ 2=(2 a c+(a 2−b 2+c 2))(2 a c−(a 2−b 2+c 2))
16 Δ 2=4 a 2 c 2−(a 2−b 2+c 2)2 16 Δ 2=4 a 2 c 2−(a 2−b 2+c 2)2
That’s Archimedes’ Theorem, for the square of the area in terms of the squares of the sides. It tells us many things, including that, surprisingly, Heron’s Formula only depends on the squares of the sides, and ignores any sign on the sides.
We’re after b,b, which is downhill from here:
4 a 2 c 2−16 Δ 2=(a 2−b 2+c 2)2 4 a 2 c 2−16 Δ 2=(a 2−b 2+c 2)2
±√4 a 2 c 2−16 Δ 2=a 2−b 2+c 2±4 a 2 c 2−16 Δ 2=a 2−b 2+c 2
b 2=a 2+c 2±√4 a 2 c 2−16 Δ 2 b 2=a 2+c 2±4 a 2 c 2−16 Δ 2
Answer: b=√a 2+c 2±√4 a 2 c 2−16 Δ 2 b=a 2+c 2±4 a 2 c 2−16 Δ 2
We don’t have to worry about the negative sign for the outer square root, because geometry.
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Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Jhevon Smith
, M.S. Mathematics (2014) · Author has 8.8K answers and 173.8M answer views
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If a+b=3 a+b=3 and a 2+b 2=7 a 2+b 2=7, what is a 4+b 4 a 4+b 4?
a 4+b 4 a 4+b 4 begs us to consider (a+b)4(a+b)4. Why? Because (a+b)4(a+b)4, expanded, is going to contain a 4+b 4 a 4+b 4, and at the same time it’s a known quantity because we got told what a+b a+b is.
So, trusty binomial theorem:
(a+b)4=a 4+4 a 3 b+6 a 2 b 2+4 a b 3+b 4(a+b)4=a 4+4 a 3 b+6 a 2 b 2+4 a b 3+b 4
We’d like to rearrange this into something that plays nice with a+b a+b and a 2+b 2 a 2+b 2. One natural thing comes to mind:
(a+b)4=a 4+b 4+4 a b(a 2+b 2)+6(a b)2(a+b)4=a 4+b 4+4 a b(a 2+b 2)+6(a b)2
We simply collected the terms in symmetric pairs (first and last, second and next-to-last, leave the center alone.)
Almost everything here is either a known quantity or the thing we’re trying to fi
Continue Reading
a 4+b 4 a 4+b 4 begs us to consider (a+b)4(a+b)4. Why? Because (a+b)4(a+b)4, expanded, is going to contain a 4+b 4 a 4+b 4, and at the same time it’s a known quantity because we got told what a+b a+b is.
So, trusty binomial theorem:
(a+b)4=a 4+4 a 3 b+6 a 2 b 2+4 a b 3+b 4(a+b)4=a 4+4 a 3 b+6 a 2 b 2+4 a b 3+b 4
We’d like to rearrange this into something that plays nice with a+b a+b and a 2+b 2 a 2+b 2. One natural thing comes to mind:
(a+b)4=a 4+b 4+4 a b(a 2+b 2)+6(a b)2(a+b)4=a 4+b 4+4 a b(a 2+b 2)+6(a b)2
We simply collected the terms in symmetric pairs (first and last, second and next-to-last, leave the center alone.)
Almost everything here is either a known quantity or the thing we’re trying to find. The only exception is a b a b. Can we determine a b a b?
Why, sure, using the exact same trick.
(a+b)2=a 2+b 2+2 a b(a+b)2=a 2+b 2+2 a b
9=7+2 a b 9=7+2 a b
a b=1 a b=1
Hooray, nice and simple. So back to our equation:
81=a 4+b 4+4⋅7+6 81=a 4+b 4+4⋅7+6
and we’ve determined that a 4+b 4 a 4+b 4 must be 81−(4⋅7+6)=47 81−(4⋅7+6)=47. ■◼
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How do I rearrange this equation to make b the subject 3 (4b - a) = 6a (2 - b)?
How do I rearrange this equation to make b the subject 3 (4b - a) = 6a (2 - b)?
First I multiply out on both sides:
3(4 b−a)=6 a(2−b)∣z(x±y)=x z±y z 3(4 b−a)=6 a(2−b)∣z(x±y)=x z±y z
12 b−3 a=12 a−6 a b 12 b−3 a=12 a−6 a b
Then I bring everything with b b to the left side and everything with out it to the right:
12 b−3 a=12 a−6 a b∣+3 a+6 a b 12 b−3 a=12 a−6 a b∣+3 a+6 a b
12 b+6 a b=15 a 12 b+6 a b=15 a
Third I extract b b on the left side:
12 b+6 a b=15 a∣x z±y z=z(x±y)12 b+6 a b=15 a∣x z±y z=z(x±y)
b(12+6 a)=15 a b(12+6 a)=15 a
Finally I divide by this parentheses expression and get b b by itself. Done.
b(12+6 a)=15 a∣÷(12+6 a)b(12+6 a)=15 a∣÷(12+6 a)
b=15 a 12+6 a=5 a 4+2 a=5 a 2(2+a)b=15 a 12+6 a=5 a 4+2 a=5 a 2(2+a)
That is true for a≠−2 a≠−2.
For a=−2 a=−2 there is no solution as tryin
Continue Reading
How do I rearrange this equation to make b the subject 3 (4b - a) = 6a (2 - b)?
First I multiply out on both sides:
3(4 b−a)=6 a(2−b)∣z(x±y)=x z±y z 3(4 b−a)=6 a(2−b)∣z(x±y)=x z±y z
12 b−3 a=12 a−6 a b 12 b−3 a=12 a−6 a b
Then I bring everything with b b to the left side and everything with out it to the right:
12 b−3 a=12 a−6 a b∣+3 a+6 a b 12 b−3 a=12 a−6 a b∣+3 a+6 a b
12 b+6 a b=15 a 12 b+6 a b=15 a
Third I extract b b on the left side:
12 b+6 a b=15 a∣x z±y z=z(x±y)12 b+6 a b=15 a∣x z±y z=z(x±y)
b(12+6 a)=15 a b(12+6 a)=15 a
Finally I divide by this parentheses expression and get b b by itself. Done.
b(12+6 a)=15 a∣÷(12+6 a)b(12+6 a)=15 a∣÷(12+6 a)
b=15 a 12+6 a=5 a 4+2 a=5 a 2(2+a)b=15 a 12+6 a=5 a 4+2 a=5 a 2(2+a)
That is true for a≠−2 a≠−2.
For a=−2 a=−2 there is no solution as trying to solve it with this assumption leads to a contradiction:
3(4 b+2)=−12(2−b)3(4 b+2)=−12(2−b)
Expand
12 b+6=12 b−24 12 b+6=12 b−24
Subtract 12 b 12 b
6=−24 6=−24
⚡️
Similar rearranging for a a we get:
a=12 b 15−6 b=4 b 5−2 b a=12 b 15−6 b=4 b 5−2 b
For b≠5 2=2+1 2=2.5 b≠5 2=2+1 2=2.5
And again assuming it is leads to a contradiction (in this case 30=0 30=0).
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Simon Tsai
Doing maths for fun. · Author has 4.9K answers and 2M answer views
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How do you make x the subject of the formula a=b+x/b-x?
Mr Straube has answered your question already. For the sake of variety, I am offering an alterantive. We are given the equation:
a=b+x b−x.a=b+x b−x.
Let x=−b u x=−b u and see
a=1−u 1+u.a=1−u 1+u.
Manipulate both sides:
a=u−1/2−u 1/2 u−1/2+u 1/2.a=u−1/2−u 1/2 u−1/2+u 1/2.
Let u=e−2 s u=e−2 s:
a=e s−e−s e s+e−s.a=e s−e−s e s+e−s.
Recognise the hyperbolic tangent on the right side:
a=tanh(s).a=tanh(s).
So, accordingly,
\DeclareMathOperator{\artanh}{ar\DeclareMathOperator{\artanh}{ar
Continue Reading
Mr Straube has answered your question already. For the sake of variety, I am offering an alterantive. We are given the equation:
a=b+x b−x.a=b+x b−x.
Let x=−b u x=−b u and see
a=1−u 1+u.a=1−u 1+u.
Manipulate both sides:
a=u−1/2−u 1/2 u−1/2+u 1/2.a=u−1/2−u 1/2 u−1/2+u 1/2.
Let u=e−2 s u=e−2 s:
a=e s−e−s e s+e−s.a=e s−e−s e s+e−s.
Recognise the hyperbolic tangent on the right side:
a=tanh(s).a=tanh(s).
So, accordingly,
s=artanh(a).s=artanh(a).
Undo the substitutions. Finally,
x=−b e−2 artanh(a).x=−b e−2 artanh(a).
The formula above is valid for all |a|≤1|a|≤1. For |a|≥1|a|≥1,
x=−b e−2 arcoth(a).x=−b e−2 arcoth(a).
See the graphs below:
This is unnecessary complication. I did it just for fun.
Upvote ·
9 1
9 1
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Related
If a√5+b√2 is the square root of 95-30√10, what are the values of a and b respectively?
I will just square a√5+b√2 and equate the answer
to 95 – 30√10
Now I need to solve a QUARTIC...
Continue Reading
I will just square a√5+b√2 and equate the answer
to 95 – 30√10
Now I need to solve a QUARTIC...
Upvote ·
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HomeMath WorksheetsNumber WorksheetsSkip Counting WorksheetsMath Number Skip Counting by 20, worksheet 2
Math Number Skip Counting by 20, worksheet 2
Math skip counting worksheets to practice and master skip counting by 20. Write the missing numbers in the number sequence by doing skip counting by20. Download free printable for 2nd graders to practice and test skipping number by 20. Grade 2 math worksheet to supplement�math�lessons on�skip counting.�
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trade fixture
trade fixture
A trade fixture is something attached to property rented by the lessee which they are entitled to take with them after the lease ends. Trade fixtures can be many things such as a machine or shelves which require the object to be fastened to the building. Often disputes will arise surrounding trade fixtures in commercial real estate when a tenant makes many improvements on a building and the lease does not make clear who gets to keep what items. For example, if a tenant replaces something essential like windows, it is most likely that the owner of the property will keep the windows, but if the tenant replaced a ceiling light with a chandelier, it may be less clear whether the light is something the tenant may take with them.
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6742 | https://www.vcalc.com/equation/?uuid=5742adc5-0d73-11e4-b7aa-bc764e2038f2 | Deuteron - Mass
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Deuteron - Mass
vCalc Reviewed
Last modified by
MichaelBartmess
on
Jul 24, 2020, 6:28:07 PM
Created by
MichaelBartmess
on
Jul 17, 2014, 5:29:36 AM
m d=3.34358348⋅10−27 k g m d=3.34358348⋅10-27 k g
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To Full Decimal
3.34358348E-27 (kg)kilogram (mg)milligram (g)gram (oz)ounce (lb)pound (ton_us)ton(U.S.) (t)metric ton (ton_uk)ton(U.K.) (me)electron mass (u)atomic mass (gr)grain (ct)carat (dwt)penny weight (troy_ounce)troy ounce (Earth_Mass)Earth Mass (Jupiter_Mass)Jupiter Mass (Solar_Mass)Solar Mass
| | | |
---
TagsAtomicMasselementperiod tableVerified
UUID 5742adc5-0d73-11e4-b7aa-bc764e2038f2
Advertise Here
The Mass of a Deuteron (Deuterium nucleus) is 2.01355345 atomic masses (u). This constant defines the mass of deuteron, the nucleus of Deuterium. The mass of Deuteron can be automatically converted to other mass units, such as kilograms (kg) or mass of electrons, via the pull-down menu.
Notes
Deuterium (symbol D or 2H, also known as heavy hydrogen) is one of two stable isotopes of hydrogen. It has a natural abundance in Earth's oceans of about one atom in 6,420 of hydrogen.
This constant, m d m d, is specified with a standard uncertainty (standard deviation) of: 0.00000015E-27 kg
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6743 | https://math.stackexchange.com/questions/4723657/criterion-for-factorability-of-cubic-polynomials | number theory - Criterion for factorability of cubic polynomials - Mathematics Stack Exchange
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Criterion for factorability of cubic polynomials
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It is well-known that, over an arbitrary field k k, a quadratic polynomial f(x)=a x 2+b x+c f(x)=a x 2+b x+c with a,b,c∈k a,b,c∈k factorises over k k if and only if its discriminant, Δ=b 2−4 a c Δ=b 2−4 a c, is a square in k k. From an algorithmic standpoint, checking that Δ Δ is a square is much faster than checking that the polynomial f(x)f(x) factorises (especially when you have a whole bunch of polynomials to attempt to factor).
Was wondering if there is a similar condition that can be found for cubic polynomials. Namely, a cubic polynomial f(x)=a x 3+b x 2+c x+d f(x)=a x 3+b x 2+c x+d if and only if some condition holds. I know that when k=R k=R, there is a simple criterion again involving the discriminant of the cubic polynomial. However, a bit like in the quadratic setting, would like a more general criterion that applies to any field (including when k=Q k=Q). This general condition is proving a little elusive for me to find.
If anyone has any suggestions for this, please let me know.
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edited Jun 22, 2023 at 19:17
J. W. Tanner
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If a cubic polynomial is irreducible in K[x]K[x], then it cannot be factored further. It is known that a cubic polynomial is irreducible in K[x]K[x] if and only if it has no root over K K. We can test this, for example, for K=Q K=Q by the Rational Root Test. Over R R a cubic has always a real root, so it factorizes into a linear factor and a quadratic factor (which you have studied before, and which may factor further).
Edit: It is easy to see that the cubic polynomial from the comments,
47 x 3−1222 x 2−3928 x+209664 47 x 3−1222 x 2−3928 x+209664
is irreducible over Q Q, since it is already irreducible over F 5 F 5, which is easy to see (it has no root there).
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edited Jun 23, 2023 at 11:51
answered Jun 22, 2023 at 18:47
Dietrich BurdeDietrich Burde
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Very true, but the rational root test is not necessarily the most practical method to test for factorability if the number of factors of a a and d d is large. Unless I'm mistaken I don't this method will be better than straight up factoring it as the polynomial.Brudogopus –Brudogopus 2023-06-22 19:06:27 +00:00 Commented Jun 22, 2023 at 19:06
Right, but often the polynomial has small coefficients (like in all exercises), like f=x 3−9 x 2+20 x−12 f=x 3−9 x 2+20 x−12. Then we see quickly (also without the rational root test) if there is a rational root. In the same way I could say, I rarely use the Δ Δ for finding a rational root of a quadratic equation.Dietrich Burde –Dietrich Burde 2023-06-22 19:59:49 +00:00 Commented Jun 22, 2023 at 19:59
I think my point is very specific to the circumstances I find myself in whereby I want to find the most efficient algorithm to testing factoring of such cubic polynomials that could have large coefficients or large numbers of factors. I could apply the rational root test to factor quadratic polynomials in the exact same way you mentioned but I've found that doing a discriminant computation is more effective (especially for the sorts of polynomials I'm encountering). Not only does the discriminant idea work for factoring quadratics but it is also quite effective at factoring certain quartics Brudogopus –Brudogopus 2023-06-22 21:54:35 +00:00 Commented Jun 22, 2023 at 21:54
I don't want to discredit your suggestion since for degree 3 polynomials it is a very very good option. I just want to see if there's a better and potentially more algorithmically friendly option Brudogopus –Brudogopus 2023-06-22 21:58:01 +00:00 Commented Jun 22, 2023 at 21:58
I see. In this case, you should add a typical example with large coefficients to your post, and explain where they come from. This will help to provide more context for your question. I was expecting the typical exercise cubic polynomials.Dietrich Burde –Dietrich Burde 2023-06-23 08:57:52 +00:00 Commented Jun 23, 2023 at 8:57
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6744 | https://www.ahajournals.org/doi/10.1161/CIRCULATIONAHA.120.045708 | The Universal Definition of Myocardial Infarction | Circulation
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Reference #1
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Originally Published 4 May 2020
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The Universal Definition of Myocardial Infarction: Present and Future
Yader Sandoval, MD, Kristian Thygesen, MD, DSc, and Allan S.Jaffe, MD Info & Affiliations
Circulation
Volume 141, Number 18
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Acute myocardial infarction (MI) historically is defined as a clinical syndrome that meets a certain set of criteria, usually a combination of symptoms, electrocardiographic changes, and cardiac biomarkers in the proper clinical context. These criteria have evolved and have been interdigitated with noninvasive and invasive diagnostic imaging and biomarkers. Before 2000, clinicians used a variety of MI definitions. To provide consistency, the Task Force for the Universal Definition of Myocardial Infarction (UDMI) developed international, collaborative, multidisciplinary, consensus definitions to facilitate standardization and a clear nomenclature based on pathophysiology.1 Although 2 of the authors of the present article were involved in those efforts, our advocacy now reflects our individual concerns with the suggestions made to change the UDMI rather than a feeling that the document cannot be improved. Accordingly, we provide clarification about issues recently raised concerning the UDMI.2
The UDMI document provides clinical definitions. It is neither a clinical practice guideline nor a clinical decision pathway. Three elements are key. First, it is a pathophysiologic-based scheme. It predicates the diagnosis of acute MI as a clinical syndrome based on common signs and symptoms of acute myocardial ischemia. Finally, recognizing the variability in diagnostic resource availability across the globe, it endorses a definition based on criteria that can be implemented widely.
Proposals have been made to modify the UDMI to emphasize the presence or absence of acute coronary obstruction.2 In comparison with the UDMI recommendations,1 which favor the diagnosis of acute MI as a clinical syndrome, such proposals dichotomize patients with or without acute coronary obstruction and advocate the preeminent use of coronary angiography to determine MI subtypes. Significant problems exist with relying on coronary angiography as the gate keeper. Foremost, it is not routinely available in all hospitals, constraining the ability to diagnose and treat MI, and its broader use among patients with a lower pretest probability for atherothrombosis could lead to more invasive diagnostic procedures and coronary interventions with inherent procedural risks that may not be justified, especially in those with possible type 2 MIs who have comorbidities that increase such risks. Even when available, angiographically complex atherosclerotic lesions and even plaque disruption occur in patients with stable coronary artery disease with intracoronary imaging,3 which makes unclear what angiographic criteria might be used to define diagnoses. The ACT-2 trial (Appropriateness of Coronary Investigation in Myocardial Injury and Type 2 Myocardial Infarction; URL: https//www.anzctr.org.au; Unique identifier: ACTRN1261800378224) should provide insights into the broader use of coronary angiography,4 but pending more data, its routine use to determine MI subtypes should not be advised.
Some researchers have advocated that conditions leading to acute coronary obstruction other than atherothrombosis, such as coronary embolism, spasm, microvascular dysfunction, or spontaneous coronary artery dissection, should be classified as type 1 MI events because their diagnosis requires coronary angiography.2 These conditions, however, are not related to acute atherothrombotic plaque disruption. The therapeutic approach to type 1 MI presently emphasizes therapies targeting acute atherothrombosis, whereas the diagnoses of spontaneous coronary artery dissection, embolism, spasm, or microvascular dysfunction require individualized diagnostic and treatment approaches, which are often difficult to deploy, and very different than the approaches advocated for patients with acute atherosclerotic plaque disruption. The fact that coronary angiography may be useful for diagnosis should not mean that it is necessary to establish an MI diagnosis or MI subtypes. We favor the present UDMI model and are against subdividing type 1 MI, a homogeneous entity, into a heterogeneous entity of various mechanisms and disease subtypes.
Misunderstandings exist about how the UDMI relates to other terms used in clinical practice and research. Clinical practice guidelines have long used the term acute coronary syndrome to define patients presenting with symptoms and signs of myocardial ischemia, whether attributable to non–ST-segment–elevation MI and ST-segment–elevation MI or unstable angina. When used in this manner, the term does not equate to acute atherothrombosis. It is appropriate to use the term initially in undifferentiated patients to identify those in whom additional diagnostic approaches and therapies for myocardial ischemia are needed. Once a more precise diagnosis is confirmed, however, the term acute coronary syndrome is broad and ambiguous. We suggest at that point that the specific MI subtypes that elaborate on pathophysiology should be used. The Figure illustrates our concept of the relationship between this nomenclature and MI subtypes.
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Figure. Application of universal definition of myocardial infarction (MI) and relationship to other nomenclature/classification schemes. Acute coronary syndrome (ACS) is a term that can be used when patients with possible acute ischemic heart disease present. However, once a more specific diagnosis is made, the more specific term should be used. Type 1 MI is a homogeneous condition that relates to acute atherothrombotic coronary artery disease for which standardized, evidence-based management guidelines should be used. Type 2 MI is a heterogeneous syndrome, which can be a primary or secondary illness, and is caused by either acute nonatherothrombotic coronary artery disease or other acute noncoronary trigger/illness. The term myocardial infarction with no obstructive coronary artery disease (MINOCA), and the ECG terminology (ST-segment–elevation myocardial infarction [STEMI]/non–ST-segment–elevation myocardial infarction[NSTEMI]), as well, apply for both type 1 and 2 MIs. CAD indicates coronary artery disease; and CPG, Clinical Practice Guidelines.
New terms, such as MI with no obstructive coronary artery disease (MINOCA), are increasingly used as a working diagnosis to raise awareness that the absence of obstructive disease (≤50% in major epicardial vessels) should not lead to the false reassurance that MI has not occurred when other clinical criteria are met. MINOCA does not have a solitary pathophysiology. MINOCA can potentially occur because of acute atherothrombotic (type 1 MI) or nonatherothrombotic (type 2 MI) etiologies. The proposal cited claims that the suggested classification would eliminate the need for the MINOCA term because spontaneous coronary artery dissection, embolism, and macrovascular and microvascular obstruction would all fit the construct.2 However, given their proposed requisite for acute coronary obstruction to define type 1 MI, this would lead to all MINOCAs being categorized as type 2 MI. In addition, this proposal that requires acute coronary obstruction to define type 1 MI may disfavor certain patient subsets, like women, whose MIs often are attributable to nonobstructive mechanisms.
We share the idea that MI diagnostic and management pathways are needed and have proposed such algorithms.5 Our advocacy for the present classification system does not prevent the tabulation of the data necessary for patient management; it simply suggests that the present system is a better universal standard. It is also clear that, for type 2 MI, more data are needed on the management and prognosis of the distinct type 2 MI phenotypes, including distinguishing those that occur from coronary and noncoronary causes.5 Given the adverse prognosis of patients who have type 2 MI, we especially urge trials to develop treatment pathways. This would be far more effective than altering a well-established system of definitions that are based on pathophysiologic mechanisms.
References
1.
Thygesen K, Alpert JS, Jaffe AS, Chaitman BR, Bax JJ, Morrow DA, White HD; Executive Group on behalf of the Joint European Society of Cardiology (ESC)/American College of Cardiology (ACC)/American Heart Association (AHA)/World Heart Federation (WHF) Task Force for the Universal Definition of Myocardial Infarction. Fourth universal definition of myocardial infarction (2018). Circulation. 2018;138:e618–e651. doi: 10.1161/CIR.0000000000000617
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a [...] nomenclature based on pathophysiology.
b [...] comparison with the UDMI recommendations,
2.
de Lemos JA, Newby LK, Mills NL. A proposal for modest revision of the definition of type 1 and type 2 myocardial infarction. Circulation. 2019;140:1773–1775. doi: 10.1161/CIRCULATIONAHA.119.042157
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a [...] issues recently raised concerning the UDMI.
b [...] or absence of acute coronary obstruction.
c [...] diagnosis requires coronary angiography.
d [...] obstruction would all fit the construct.
3.
Maehara A, Mintz GS, Bui AB, Walter OR, Castagna MT, Canos D, Pichard AD, Satler LF, Waksman R, Suddath WO, et al. Morphologic and angiographic features of coronary plaque rupture detected by intravascular ultrasound. J Am Coll Cardiol. 2002;40:904–910. doi: 10.1016/s0735-1097(02)02047-8
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Lambrakis K, French JK, Scott IA, Briffa T, Brieger D, Farkouh ME, White H, Chuang AM, Tiver K, Quinn S, et al. The appropriateness of coronary investigation in myocardial injury and type 2 myocardial infarction (ACT-2): a randomized trial design. Am Heart J. 2019;208:11–20. doi: 10.1016/j.ahj.2018.09.016
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Sandoval Y, Jaffe AS. Type 2 myocardial infarction: JACC Review Topic of the Week. J Am Coll Cardiol. 2019;73:1846–1860. doi: 10.1016/j.jacc.2019.02.018
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a [...] needed and have proposed such algorithms.
b [...] occur from coronary and noncoronary causes.
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Circulation
Volume 141 • Number 18 • 5 May 2020
Pages: 1434 - 1436
PubMed: 32364775
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Keywords
acute coronary syndrome
clinical decision-making
coronary angiography
myocardial infarction
myocardial ischemia
Subjects
Myocardial Infarction
Authors
Affiliations Expand All
Yader Sandoval, MD
Departments of Cardiovascular Diseases (S.Y., A.S.J.), Mayo Clinic, Rochester, MN.
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Kristian Thygesen, MD, DSc
Department of Cardiology, Aarhus University Hospital, Denmark (K.T.).
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Allan S.Jaffe, MD
Departments of Cardiovascular Diseases (S.Y., A.S.J.), Mayo Clinic, Rochester, MN.
Laboratory Medicine and Pathology (A.S.J.), Mayo Clinic, Rochester, MN.
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Notes
The opinions expressed in this article are not necessarily those of the editors or of the American Heart Association.
Allan S. Jaffe, MD, Department of Cardiovascular Diseases, Gonda 5, Mayo Clinic, 200 First St SW, Rochester, MN 55905. Email jaffe.allan@Mayo.edu
Disclosures
Dr Sandoval has served on an advisory board/speaker for Abbott Diagnostics and an advisory board for Roche Diagnostics, all without personal financial compensation. Dr Jaffe has consulted or is presently consulting for most of the major diagnostic companies, including Beckman, Abbott, Siemens, ET Healthcare, Roche, Quidel, Brava, and Sphingotec. He also consults for Blade and Novartis. Dr Thygesen reports no conflict.
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Figure. Application of universal definition of myocardial infarction (MI) and relationship to other nomenclature/classification schemes. Acute coronary syndrome (ACS) is a term that can be used when patients with possible acute ischemic heart disease present. However, once a more specific diagnosis is made, the more specific term should be used. Type 1 MI is a homogeneous condition that relates to acute atherothrombotic coronary artery disease for which standardized, evidence-based management guidelines should be used. Type 2 MI is a heterogeneous syndrome, which can be a primary or secondary illness, and is caused by either acute nonatherothrombotic coronary artery disease or other acute noncoronary trigger/illness. The term myocardial infarction with no obstructive coronary artery disease (MINOCA), and the ECG terminology (ST-segment–elevation myocardial infarction [STEMI]/non–ST-segment–elevation myocardial infarction[NSTEMI]), as well, apply for both type 1 and 2 MIs. CAD indicates coronary artery disease; and CPG, Clinical Practice Guidelines.
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1.
Thygesen K, Alpert JS, Jaffe AS, Chaitman BR, Bax JJ, Morrow DA, White HD; Executive Group on behalf of the Joint European Society of Cardiology (ESC)/American College of Cardiology (ACC)/American Heart Association (AHA)/World Heart Federation (WHF) Task Force for the Universal Definition of Myocardial Infarction. Fourth universal definition of myocardial infarction (2018). Circulation. 2018;138:e618–e651. doi: 10.1161/CIR.0000000000000617
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a [...] needed and have proposed such algorithms.
b [...] occur from coronary and noncoronary causes.
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2004 AIME I Problems/Problem 2
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (Sketchy Solution for Speedrunners)
6 Solution 5
7 See also
Problem
Set consists of consecutive integers whose sum is , and set consists of consecutive integers whose sum is The absolute value of the difference between the greatest element of and the greatest element of is . Find
Solution 1
Note that since set has consecutive integers that sum to , the middle integer (i.e., the median) must be . Therefore, the largest element in is .
Further, we see that the median of set is , which means that the "middle two" integers of set are and . Therefore, the largest element in is . if , which is clearly not possible, thus .
Solving, we get
Solution 2
Let us give the elements of our sets names: and . So we are given that so and (this is because so plugging this into yields ). Also, so so and .
Then by the given, . is a positive integer so we must have and so .
Solution 3
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set
Where and represent the first and last terms of . This comes from the sum of an arithmetic sequence.
Solving for , we find the sum of the two terms is .
Doing the same for set B, and setting up the equation with and being the first and last terms of set ,
and so .
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set has half the number of elements as set , and the difference between the greatest terms of the two two sequences is (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where is the last term of set A:
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to and respectively (add and to see what i mean).
Solving this equation we find . We know the first and last terms have to sum to so we find the first term of the sequence is . Now, the solution is in clear sight, we just find the number of integers between and , inclusive, and it is .
Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.
Solution 4 (Sketchy Solution for Speedrunners)
First, calculate the average of set and set . It's obvious that they are and respectively. Let's look at both sets. Obviously, there is an odd number of integers in the set with being in the middle, which means that is an odd number and that the number of consecutive integers on each side of are equal. In set , it is clear that it contains an even number of integers, but since the number in the middle is , we know that the range of the consecutive numbers on both sides will be to and to .
Nothing seems useful right now, but let's try plugging an odd number, , for in set . We see that there are consecutive integers and on both sides of . After plugging this into set , we find that the set equals . From there, we find the absolute value of the difference of both of the greatest values, and get 0.
Let's try plugging in another odd number, . We see that the resulting set of numbers is to , and to . We then plug this into set , and find that the set of numbers is to which indeed results in the average being . We then find the difference of the greatest values to be 26.
From here, we see a pattern that can be proven by more trial and error. When we make equal to , then the difference is whearas when we make it , then the difference is . equals to and is just . We then see that increases twice as fast as the difference. So when the difference is , it increased from when it was , which means that increased by which is . We then add this to our initial of , and get as our answer.
Solution 5
Let the first term of be and the first term of be . There are elements in so is . Adding these up, we get . Set contains the numbers . Summing these up, we get . The problem gives us that the absolute value of the difference of the largest terms in and is . The largest term in is and the largest term in is so . From the first two equations we get, we can get that . Now, we make a guess and assume that (if we get a negative value for , we can try ). From here we get that . Solving for , we get that the answer is
-Heavytoothpaste
See also
2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1Followed by
Problem 3
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
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6746 | https://www.rxfiles.ca/rxfiles/uploads/documents/SADMANS-Rx.pdf | • When throwing up, having diarrhea, running a fever, or not eating / drinking enough for about 24 hours, there is a risk of becoming dehydrated.
• When dehydrated, some medications can harm the kidneys. These medications should be paused (for a few days) to protect the kidneys. It is important to restart the medications once feeling better.
PROTECTING THE KIDNEYS when unwell and dehydrated © September 2025 RxFiles Academic Detailing Reminders for people with DIABETES Reminders for people with HEART FAILURE Adjust insulin Insulin needs may change when unwell. For example, if not eating, meal time insulin is not needed.
Ask about pausing diuretic Please contact your healthcare team before stopping a diuretic (water pill) such as furosemide.
Stay hydrated Drink lots of fluids with minimal sugar. Limit caffeine. Consider electrolyte replacement drinks (such as Hydralyte). Stay hydrated Keep total fluids (such as soup, water, tea) to 2 litres per day. Replace fluids lost from throwing up, diarrhea, or fever.
Check blood sugar Check blood sugar more often while unwell. Medications can lower blood sugar too much. If blood sugar is too low, contact a healthcare provider.
Track weight and blood pressure Contact a healthcare provider if body weight goes up or down by more than 1.5 kilograms (3 pounds) in 2 days or if blood pressure is lower than usual.
Be aware of salt Canned soup and packaged foods are high in salt and can cause your body to hold on to extra fluid.
Medication Family Medication Name Healthcare provider to fill out. Date :____ ACE inhibitor / ARB / ARNI Diuretic SGLT2 inhibitor NSAID (anti-inflammatory) Sulfonylurea and / or metformin Avoid over-the-counter products and cold medications that have ibuprofen (Advil, Motrin), naproxen (Aleve), or decongestants (pseudoephedrine, phenylephrine). Ask a pharmacist what can be taken instead.
Signs of Dehydration q Peeing Less q More Tired Than Usual q Dizzy q Confused q Dry / Cool Skin q Thirsty, Dry Mouth q Irritable q Headache CHECK for dehydration. If unwell and showing signs of dehydration, continue to step 2.
1 PAUSE some medications. Stop the medications listed below for 1 to 3 days until feeling better.
2 3 Page 1 RESTART medications when feeling well and eating again.
Talk to a healthcare provider if not feeling better after 3 days or with any questions.
TAKE NOTE!
To protect your kidneys when unwell and dehydrated, follow the steps below: Medication Family Generic Name Brand Name Risk if Taken when Dehydrated ACE inhibitors (ACEi) Angiotensin receptor blockers (ARB) Angiotensin receptor blocker neprilysin inhibitor (ARNI) benazepril Lotensin Increased risk of kidney damage or affecting how well the kidneys work.
captopril Capoten cilazapril Inhibace enalapril Vasotec fosinopril Monopril lisinopril Zestril perindopril Coversyl quinapril Accupril ramipril Altace trandolapril Mavik candesartan Atacand irbesartan Avapro losartan Cozaar olmesartan Olmetec telmisartan Micardis valsartan Diovan sacubitril / valsartan Entresto Diuretics (water pills) chlorthalidone Hygroton If using a diuretic (water pill) for heart failure, check with a healthcare provider before pausing (especially furosemide). Increased risk of kidney damage or affecting how well the kidneys work.
eplerenone Inspra finerenone Kerendia furosemide Lasix hydrochlorothiazide HCTZ indapamide Lozide metolazone Zaroxolyn spironolactone Aldactone SGLT2 inhibitors canagliflozin Invokana Increased risk of kidney damage or affecting how well the kidneys work.
dapagliflozin Forxiga empagliflozin Jardiance Non-steroidal anti-inflammatory drugs (NSAID) acetylsalicylic acid (ASA) Aspirin / Entrophen Increased risk of kidney damage or affecting how well the kidneys work. It is okay to continue low dose 81mg ASA.
Be aware: over-the-counter cough, cold and flu products often contain these medications.
celecoxib Celebrex diclofenac Voltaren ibuprofen Advil / Motrin indomethacin Indocid ketorolac Toradol naproxen Aleve / Naprosyn Sulfonylureas & secretagogues gliclazide Diamicron MR Increased risk of low blood sugar.
glimepiride Amaryl glyburide Diabeta repaglinide Gluconorm Metformin metformin Glucophage / Glumetza Increased risk of side effects. Consider restarting at a lower dose for a few days if having nausea and / or diarrhea. Medications to Pause when Dehydrated For references visit www.rxfiles.ca/kidney Disclosures: No conflicts of interest are reported.
Disclaimer: RxFiles Academic Detailing is part of the College of Pharmacy and Nutrition at the University of Saskatchewan. The content of this work represents the research, experience and opinions of the authors and not those of the University of Saskatchewan. Neither the authors nor the University of Saskatchewan nor any other party who has been involved in the preparation or publication of this work warrants or represents that the information contained herein is accurate or complete, and they are not responsible for any errors or omissions or for the result obtained from the use of such information. Any use of the materials will imply acknowledgment of this disclaimer and release any responsibility of the University of Saskatchewan, its employees, servants or agents. Readers are encouraged to confirm the information contained herein with other sources.
© September 2025 RxFiles Academic Detailing Page 2 References for Sick Day Management Tool: 1. Kidney Disease: Improving Global Outcomes (KDIGO) CKD Work Group. KDIGO 2024 Clinical Practice Guideline for the Evaluation and Management of Chronic Kidney Disease. Kidney Int. 2024 Apr;105(4S):S117-S314. 2. McDonald M, Virani S, Chan M, et al. CCS/CHFS Heart Failure Guidelines Update: Defining a New Pharmacologic Standard of Care for Heart Failure With Reduced Ejection Fraction. Can J Cardiol. 2021 Apr;37(4):531-546. 3. Diabetes Canada Clinical Practice Guidelines Expert Working Group. The User's Guide to the Pharmacologic Glycemic Management of Type 2 Diabetes in Adults---2024 Update. Can J Diabetes. 2024 Oct;48(7):425-430. 4. Faber SJ, Scherpbier ND, Peters HJG, Uijen AA. Preventing acute kidney injury in high-risk patients by temporarily discontinuing medication - an observational study in general practice. BMC Nephrol. 2019 Dec 4;20(1):449. 5. Watson KE, Dhaliwal K, McMurtry E, et al. Sick Day Medication Guidance for People With Diabetes, Kidney Disease, or Cardiovascular Disease: A Systematic Scoping Review. Kidney Med. 2022 May 28;4(9):100491. 6. Watson KE, Dhaliwal K, Robertshaw S, et al; PAUSE (Preventing Medication Complications During Acute Illness Through Symptom Evaluation and Sick Day Guidance) Medication Safety Advisory Panel. Consensus Recommendations for Sick Day Medication Guidance for People With Diabetes, Kidney, or Cardiovascular Disease: A Modified Delphi Process. Am J Kidney Dis. 2023 May;81(5):564-574. 7. Fink JC, Maguire RM, Blakeman T, et al. Medication Holds in CKD During Acute Volume-Depleting Illnesses: A Randomized Controlled Trial of a "Sick-Day" Protocol. Kidney Med. 2022 Jul 31;4(9):100527. 8. Kado K, Okada H, Suzuki S, et al. Study of assessment of knowledge and understanding for coping with sick days among patients with diabetes in community pharmacy: a cluster randomized controlled trial (SAKURA trial). J Pharm Policy Pract. 2023 Sep 25;16(1):104. 9. Bayoumi I, Dolovich L, Hutchison B, Holbrook A. Medication-related emergency department visits and hospitalizations among older adults. Can Fam Physician. 2014 Apr;60(4):e217-22. 10. Martindale AM, Elvey R, Howard SJ, et al. Understanding the implementation of 'sick day guidance' to prevent acute kidney injury across a primary care setting in England: a qualitative evaluation. BMJ Open. 2017 Nov 8;7(11):e017241. |
6747 | https://math.stackexchange.com/questions/2888137/what-is-the-correct-way-to-obtain-the-position-of-a-quartile-and-its-value | probability - What is the correct way to obtain the position of a quartile and its value? - Mathematics Stack Exchange
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What is the correct way to obtain the position of a quartile and its value?
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I'm studying statistics, specifically position measurements.
The problem is that I have seen many ways to calculate the quartile, quintile, percentile, decile, etc.
The main formulas that I have seen to calculate the POSITION of a quartile are:
(n+1)i 4(n+1)i 4 and n i 4 n i 4, where n=n= number of data, i=i= number of quartile(1,2,3)(1,2,3)
Some websites claim that one formula is to calculate odd data and the other in pairs. However, in many exercises, they use the formula (n + 1) q / 4 for a number of even data and odd numbers of data, without distinction.
What has given me success in the exercises (so far), is to obtain the POSITION of the quartile with the formula (n+1)i 4(n+1)i 4, without distinguishing between odd and even data, and to obtain the VALUE of the quartile I have two options:
If the position of the quartile is a decimal, the VALUE will be the average between the two closest integers.
If the position of the quartile is an INTEGER, the VALUE will be the data that is in that position.
And I have the same doubt, with the formulas for quintile, decile and percentile.
But my recently exposed methodology has worked well for me, however I come to ask, Is the way I am getting the POSITION and VALUE of the position measurements correct or not correct at all? Thanks in advance.
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edited Aug 20, 2018 at 2:43
BruceET
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asked Aug 19, 2018 at 20:53
MattiuMattiu
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There is no one 'correct way'. Definitions differ from textbook to textbook. If you are taking a course, try to get a clear idea what you textbook and class notes say to do. Then after the course, you can choose your favorite method. See my answer for several methods in common use.BruceET –BruceET 2018-08-20 02:39:14 +00:00 Commented Aug 20, 2018 at 2:39
BruceET has given an excellent answer. Adapting a remark from Numerical Recipes (from a related subject), if you care about the difference between these definitions you are probably up to no good. In his example the lower quartile is somewhere in the range 95−96 95−96 depending on the exact definition you use. How much does that matter? We could add 1,000,000 1,000,000 to all the observations 96 96 and above in his dataset and the range of answers would grow enormously, but they would still be between 95 95 and 1,000,096 1,000,096 Ross Millikan –Ross Millikan 2018-08-20 02:58:52 +00:00 Commented Aug 20, 2018 at 2:58
Related question, answer, and discussion at the stats.stackexchange.com question Finding Quartiles in R.ruffin –ruffin 2020-10-26 18:39:31 +00:00 Commented Oct 26, 2020 at 18:39
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Data: If you are finding the lower quartile of a sample, you should know that various textbooks and software programs use slightly different definitions. Roughly speaking, the lower quartile separates the lower quarter of the sorted data from the upper three quarters. Differences among the specific definitions arise most clearly when the sample size is not divisible by 4.4.
For example, consider the sample of n=25 n=25 observations shown below sorted from smallest to largest.
x
66 82 86 91 95 95 96 96 96 97 101 104
104 106 106 107 108 108 109 111 111 112 112 120
125
The median is at 104 (the 13th observation in order). The 7th observation is 96. it has 'not more than 25% of the observations below': precisely, 6/25=24%;6/25=24%; it has 'not more than 75% of the observations above': precisely, 16/25=60%16/25=60% (taking tied values into account). So many textbooks would say the lower quartile is 96.
According to the default method in R statistical software the lower quartile (25th percentile) is 96 as shown below:
quantile(x, .25)
25%
96
However, for compatibility with other software, R has nine different 'types' of definitions that you can request. They make different compromises when the sample size is not divisible by 4 and when there are ties in the vicinity of the lower quartile. Here are a few that give different answers:
quantile(x, .25, type=3)
25%
95
quantile(x, .25, type=4)
25%
95.25
quantile(x, .25, type=5)
25%
95.75
quantile(x, .25, type=8)
25%
95.66667
Differences among the definitions of quantiles (including quartiles) may seem important in small datasets, but in practical applications, quantiles are most often used for very large datasets where the differences are usually not consequential.
Continuous distributions: If you are dealing with continuous probability distributions, then the 25th percentile is usually unique. Here are the lower quartiles of the distributions N o r m(μ=0,σ=1)N o r m(μ=0,σ=1) and E x p(r a t e=1):E x p(r a t e=1):
```
qnorm(.25, 0, 1)
-0.6744898
qexp(.25, 1)
0.2876821
```
Discrete distributions: For discrete distributions such as binomial and Poisson, it is not usually possible to find an integer value that cuts off exactly the lower quarter of the probability in the distribution. Below are results for B i n o m(n=5,p=1/3)B i n o m(n=5,p=1/3) and P o i s(λ=5).P o i s(λ=5). The R functions pbinom and ppois designate CDFs of binomial and Poisson distributions respectively.
```
qbinom(.25, 5, 1/3)
1 # 25th percentile is 1
pbinom(1, 5, 1/3)
0.4609053 # P(X <= 1) = .4609 > .25
pbinom(0, 5, 1/3)
0.1316872 # P(X <= 0) = .1317 < .25
qpois(.25, 5)
3
ppois(3, 5)
0.2650259
ppois(2, 5)
0.124652
```
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edited Aug 20, 2018 at 2:54
answered Aug 20, 2018 at 2:36
BruceETBruceET
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Thanks for your answer, In short, there are different correct forms. However, I want to know if mine is one of the correct ones?Mattiu –Mattiu 2018-08-21 20:35:13 +00:00 Commented Aug 21, 2018 at 20:35
1 In short, I'm not absolutely sure I believe you have specified any reasonable method. Suppose n=5 n=5 and observations are x=(1,3,6,11,100).x=(1,3,6,11,100). Looking for the first quartile, one of your formulas seems to give 6/4=1.5 6/4=1.5 and the other 5/4=1.25.5/4=1.25. I guess that means the first quartile is somewhere between 1 and 3. The default method in R gives 3. It would be difficult to defend 2 2 because then 4/5>75%4/5>75% of data lie above 2.2. \ The strategy of my answer was to try to give you a realistic view of how quartiles are found in practice. // Perhaps re-read @RossMillikan's Comment.BruceET –BruceET 2018-08-21 20:49:48 +00:00 Commented Aug 21, 2018 at 20:49
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Cayley-Menger Determinant
Definition
The Cayley-Menger determinant is a determinant that gives the volume of a simplex in j dimensions. If S is a j-simplex in R^n with vertices v_1, ..., v_(j + 1) and B = (β_(i k)) denotes the (j + 1)×(j + 1) matrix given by β_(i k) = left bracketing bar v_i - v_k right bracketing bar _2^2, then the content V_j is given by V_j^2(S) = (-1)^(j + 1)/(2^j (j!)^2) det(B^^),
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6750 | http://www.seas.ucla.edu/~vandenbe/ee236b/lectures/problems.pdf | L. Vandenberghe ECE236B (Winter 2025) 4. Convex optimization problems • standard form (convex) optimization problem • linear optimization • quadratic optimization • geometric programming • semidefinite optimization • quasiconvex optimization • vector and multicriterion optimization 4.1 Optimization problem in standard form minimize 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 • 푥∈R푛is the optimization variable • 푓0 : R푛→R is the objective or cost function • 푓푖: R푛→R, for 푖= 1, . . . , 푚, are the inequality constraint functions • ℎ푖: R푛→R, for 푖= 1, . . . , 푝, are the equality constraint functions Convex optimization problems 4.2 Feasible and optimal points Feasible point: 푥is feasible if 푥∈dom 푓0 and it satisfies all constraints Optimal value 푝★= inf { 푓0(푥) | 푓푖(푥) ≤0, 푖= 1, . . . , 푚, ℎ푖(푥) = 0, 푖= 1, . . . , 푝} • 푝★= ∞if the problem is infeasible (set of feasible 푥is empty) • 푝★= −∞if the problem is unbounded below Optimal solution • a feasible 푥is optimal if 푓0(푥) = 푝★ • the set of optimal points will be denoted by 푋opt • ˆ 푥is locally optimal if there is an 푅> 0 such that ˆ 푥is optimal for the problem minimize (over 푥) 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 ∥푥−ˆ 푥∥2 ≤푅 Convex optimization problems 4.3 Examples (푛= 1, 푚= 푝= 0) 푓0(푥) dom 푓0 graph 푝★ 푋opt 1/푥 R++ 0 empty −log 푥 R++ −∞ empty 푥log 푥 R++ −1/푒 {1/푒} max{0, |푥| −1} R 0 [−1, 1] 푥3 −3푥 R −∞ empty Convex optimization problems 4.4 Implicit constraints the standard form optimization problem has an implicit constraint 푥∈D = 푚 \ 푖=0 dom 푓푖∩ 푝 \ 푖=1 dom ℎ푖, • we call D the domain of the problem • the constraints 푓푖(푥) ≤0, ℎ푖(푥) = 0 are the explicit constraints • a problem is unconstrained if it has no explicit constraints (푚= 푝= 0) • the distinction will be important when we diccuss duality Example minimize 푓0(푥) = − 푘 X 푖=1 log(푏푖−푎푇 푖푥) this is an unconstrained problem with implicit constraints 푎푇 푖푥< 푏푖 Convex optimization problems 4.5 Feasibility problem find 푥 subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 can be considered a special case of the general problem with 푓0(푥) = 0: minimize 0 subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 • 푝★= 0 if constraints are feasible; any feasible 푥is optimal • 푝★= ∞if constraints are infeasible this formulation is not meant as a practical method for solving feasibility problems Convex optimization problems 4.6 Convex optimization problem in standard form minimize 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 푎푇 푖푥= 푏푖, 푖= 1, . . . , 푝 • objective and inequality constraint functions 푓0, 푓1, ..., 푓푚are convex • equality constraints are linear, often written as 퐴푥= 푏 • feasible set is convex: the intersection of several convex sets dom 푓0, sublevel sets {푥| 푓푖(푥) ≤0}, the affine set {푥| 퐴푥= 푏} • optimal set is convex: any convex combination of optimal 푥1, 푥2 is feasible, with 푓0(휃푥1 + (1 −휃)푥2) ≤ 휃푓0(푥1) + (1 −휃) 푓(푥2) = 푝★ hence, 푓0(휃푥1 + (1 −휃)푥2) = 푝★, so the convex combination is optimal Convex optimization problems 4.7 Example minimize 푓0(푥) = 푥2 1 + 푥2 2 subject to 푓1(푥) = 푥1/(1 + 푥2 2) ≤0 ℎ1(푥) = (푥1 + 푥2)2 = 0 • 푓0 is convex • feasible set {(푥1, 푥2) | 푥1 = −푥2 ≤0} is convex • not a convex problem (according to our definition): 푓1 not convex, ℎ1 not affine • the problem is equivalent (but not identical) to the convex problem minimize 푥2 1 + 푥2 2 subject to 푥1 ≤0 푥1 + 푥2 = 0 Convex optimization problems 4.8 Local and global optima any locally optimal point of a convex problem is (globally) optimal • suppose 푥is locally optimal: there is an 푅> 0 such that 푧feasible, ∥푧−푥∥2 ≤푅 =⇒ 푓0(푧) ≥푓0(푥) • suppose 푥is not globally optimal: there exists a feasible 푦with 푓0(푦) < 푓0(푥) • convex combinations of 푥and 푦are feasible • cost function at convex combination of 푥and 푦with 0 < 휃≤1 satisfies 푓0((1 −휃)푥+ 휃푦) ≤ (1 −휃) 푓0(푥) + 휃푓0(푦) < (1 −휃) 푓0(푥) + 휃푓0(푥) = 푓0(푥) • for 0 < 휃≤푅/∥푦−푥∥2 this contradicts the assumption of local optimality of 푥 Convex optimization problems 4.9 Optimality criterion for differentiable 푓0 푥is optimal if and only if it is feasible and ∇푓0(푥)푇(푦−푥) ≥0 for all feasible 푦 −∇푓0(푥) 푋 푥 if nonzero, ∇푓0(푥) defines a supporting hyperplane to feasible set 푋at 푥 Convex optimization problems 4.10 Proof (necessity) • consider feasible 푦≠푥and define line segment 퐼= {푥+ 푡(푦−푥) | 0 ≤푡≤1} • by convexity of 푋, points in 퐼are feasible • let 푔(푡) = 푓0(푥+ 푡(푦−푥)) be the restriction of 푓0 to 퐼 • derivative at 푡is 푔′(푡) = ∇푓0(푥+ 푡(푦−푥))푇(푦−푥), so 푔′(0) = ∇푓0(푥)푇(푦−푥) • if 푔′(0) = ∇푓0(푥)푇(푥−푦) < 0, the point 푥is not even locally optimal Proof (sufficiency) if 푦is feasible and ∇푓0(푥)푇(푦−푥) ≥0, then, by convexity of 푓0, 푓0(푦) ≥ 푓0(푥) + ∇푓0(푥)푇(푦−푥) ≥ 푓0(푥) Convex optimization problems 4.11 Examples Unconstrained problem: 푥is optimal if and only if 푥∈dom 푓0, ∇푓0(푥) = 0 (recall our assumption that dom 푓0 is an open set if 푓0 is differentiable) Minimization over nonnegative orthant minimize 푓0(푥) subject to 푥⪰0 푥is optimal if and only if 푥∈dom 푓0, 푥⪰0, ∇푓0(푥)푖≥0 푥푖= 0 ∇푓0(푥)푖= 0 푥푖> 0 Convex optimization problems 4.12 Equality constrained problem minimize 푓0(푥) subject to 퐴푥= 푏 푥is optimal if and only if there exists a 휈such that 푥∈dom 푓0, 퐴푥= 푏, ∇푓0(푥) + 퐴푇휈= 0 • first two conditions are feasibility of 푥 • gradient ∇푓0(푥) can be decomposed as ∇푓0(푥) + 퐴푇휈= 푤with 퐴푤= 0 • if 푤= 0, the optimality condition on page 4.10 holds: ∇푓0(푥)푇(푦−푥) = −휈푇퐴(푦−푥) = 0 for all 푦with 퐴푦= 푏 • if 푤≠0, condition on p. 4.10 does not hold: 푦= 푥−푡푤is feasible for small 푡> 0, ∇푓0(푥)푇(푦−푥) = −푡(푤−퐴푇휈)푇푤= −푡∥푤∥2 2 < 0 Convex optimization problems 4.13 Linear program (LP) minimize 푐푇푥+ 푑 subject to 퐺푥⪯ℎ 퐴푥= 푏 • convex problem with affine objective and constraint functions • feasible set is a polyhedron P 푥★ −푐 Convex optimization problems 4.14 Examples Diet problem: choose quantities 푥1, ..., 푥푛of 푛foods • one unit of food 푗costs 푐푗, contains amount 푎푖푗of nutrient 푖 • healthy diet requires nutrient 푖in quantity at least 푏푖 to find cheapest healthy diet, minimize 푐푇푥 subject to 퐴푥⪰푏, 푥⪰0 Piecewise-linear minimization minimize max 푖=1,...,푚(푎푇 푖푥+ 푏푖) equivalent to an LP minimize 푡 subject to 푎푇 푖푥+ 푏푖≤푡, 푖= 1, . . . , 푚 Convex optimization problems 4.15 Chebyshev center of a polyhedron Chebyshev center of P = {푥| 푎푇 푖푥≤푏푖, 푖= 1, . . . , 푚} is center of largest inscribed ball B = {푥c + 푢| ∥푢∥2 ≤푟} 푥c 푥c • 푎푇 푖푥≤푏푖for all 푥∈B if and only if sup{푎푇 푖(푥c + 푢) | ∥푢∥2 ≤푟} = 푎푇 푖푥c + 푟∥푎푖∥2 ≤푏푖 • hence, 푥c, 푟can be determined by solving the LP maximize 푟 subject to 푎푇 푖푥c + 푟∥푎푖∥2 ≤푏푖, 푖= 1, . . . , 푚 Convex optimization problems 4.16 Quadratic program (QP) minimize 1 2푥푇푃푥+ 푞푇푥+ 푟 subject to 퐺푥⪯ℎ 퐴푥= 푏 • 푃∈S푛 +, so objective is convex quadratic • minimize a convex quadratic function over a polyhedron P 푥★ −∇푓0(푥★) Convex optimization problems 4.17 Examples Least squares minimize ∥퐴푥−푏∥2 2 • analytical solution 푥★= 퐴†푏(퐴† is pseudo-inverse) • can add linear constraints, e.g., 푙⪯푥⪯푢 Linear program with random cost minimize ¯ 푐푇푥+ 훾푥푇Σ푥= E 푐푇푥+ 훾var(푐푇푥) subject to 퐺푥⪯ℎ 퐴푥= 푏 • 푐is random vector with mean ¯ 푐and covariance Σ • hence, 푐푇푥is random variable with mean ¯ 푐푇푥and variance 푥푇Σ푥 • 훾> 0 is risk aversion parameter • 훾controls trade-offbetween expected cost and variance (risk) Convex optimization problems 4.18 Quadratically constrained quadratic program (QCQP) minimize 1 2푥푇푃0푥+ 푞푇 0푥+ 푟0 subject to 1 2푥푇푃푖푥+ 푞푇 푖푥+ 푟푖≤0, 푖= 1, . . . , 푚 퐴푥= 푏 • 푃푖∈S푛 +; objective and constraints are convex quadratic • if 푃1, . . . , 푃푚∈S푛 ++, feasible set is intersection of 푚ellipsoids and an affine set Convex optimization problems 4.19 Second-order cone programming minimize 푓푇푥 subject to ∥퐴푖푥+ 푏푖∥2 ≤푐푇 푖푥+ 푑푖, 푖= 1, . . . , 푚 퐹푥= 푔 (퐴푖∈R푛푖×푛, 퐹∈R푝×푛) • inequalities are called second-order cone (SOC) constraints: (퐴푖푥+ 푏푖, 푐푇 푖푥+ 푑푖) ∈second-order cone in R푛푖+1 • for 푛푖= 0, reduces to an LP; if 푐푖= 0, reduces to a QCQP • more general than QCQP and LP Convex optimization problems 4.20 Robust linear programming the parameters in optimization problems are often uncertain, e.g., in an LP minimize 푐푇푥 subject to 푎푇 푖푥≤푏푖, 푖= 1, . . . , 푚 there can be uncertainty in 푐, 푎푖, 푏푖 two common approaches to handling uncertainty (in 푎푖, for simplicity) • deterministic model: constraints must hold for all 푎푖∈E푖 minimize 푐푇푥 subject to 푎푇 푖푥≤푏푖for all 푎푖∈E푖, 푖= 1, . . . , 푚, • stochastic model: 푎푖is random variable; constraints must hold with probability 휂 minimize 푐푇푥 subject to prob(푎푇 푖푥≤푏푖) ≥휂, 푖= 1, . . . , 푚 Convex optimization problems 4.21 Deterministic approach via SOCP choose an ellipsoid as E푖: E푖= { ¯ 푎푖+ 푃푖푢| ∥푢∥2 ≤1} ( ¯ 푎푖∈R푛, 푃푖∈R푛×푛) center is ¯ 푎푖, semi-axes determined by singular values/vectors of 푃푖 SOCP formulation minimize 푐푇푥 subject to 푎푇 푖푥≤푏푖 ∀푎푖∈E푖, 푖= 1, . . . , 푚 this is equivalent to the SOCP minimize 푐푇푥 subject to ¯ 푎푇 푖푥+ ∥푃푇 푖푥∥2 ≤푏푖, 푖= 1, . . . , 푚 (follows from sup ∥푢∥2≤1 ( ¯ 푎푖+ 푃푖푢)푇푥= ¯ 푎푇 푖푥+ ∥푃푇 푖푥∥2) Convex optimization problems 4.22 Stochastic approach via SOCP • assume 푎푖∼N ( ¯ 푎푖, Σ푖)): Gaussian with mean ¯ 푎푖, covariance Σ푖 • 푎푇 푖푥is Gaussian random variable with mean ¯ 푎푇 푖푥, variance 푥푇Σ푖푥 • if we denote the CDF of N (0, 1) by Φ(푥) = 1 √ 2휋 ∫푥 −∞푒−푡2/2 푑푡, prob(푎푇 푖푥≤푏푖) = Φ 푏푖−¯ 푎푇 푖푥 ∥Σ1/2 푖 푥∥2 !
SOCP formulation of robust LP minimize 푐푇푥 subject to prob(푎푇 푖푥≤푏푖) ≥휂, 푖= 1, . . . , 푚 for 휂≥1/2, this is equivalent to the SOCP minimize 푐푇푥 subject to ¯ 푎푇 푖푥+ Φ−1(휂)∥Σ1/2 푖 푥∥2 ≤푏푖, 푖= 1, . . . , 푚 Convex optimization problems 4.23 Example prob(푎푇 푖푥≤푏푖) ≥휂, 푖= 1, . . . , 5 feasible set for three values of 휂 휂= 10% 휂= 50% 휂= 90% Φ−1(휂) < 0 Φ−1(휂) = 0 Φ−1(휂) > 0 Convex optimization problems 4.24 Geometric programming Monomial function 푓(푥) = 푐푥푎1 1 푥푎2 2 · · · 푥푎푛 푛, dom 푓= R푛 ++ with 푐> 0; exponent 푎푖can be any real number Posynomial function: sum of monomials 푓(푥) = 퐾 X 푘=1 푐푘푥푎1푘 1 푥푎2푘 2 · · · 푥푎푛푘 푛, dom 푓= R푛 ++ Geometric program (GP) minimize 푓0(푥) subject to 푓푖(푥) ≤1, 푖= 1, . . . , 푚 ℎ푖(푥) = 1, 푖= 1, . . . , 푝 with 푓푖posynomial, ℎ푖monomial Convex optimization problems 4.25 Geometric program in convex form change variables to 푦푖= log 푥푖, and take logarithm of cost, constraints • monomial 푓(푥) = 푐푥푎1 1 · · · 푥푎푛 푛transforms to log 푓(푒푦1, . . . , 푒푦푛) = 푎푇푦+ 푏 (푏= log 푐) • posynomial 푓(푥) = P퐾 푘=1 푐푘푥푎1푘 1 푥푎2푘 2 · · · 푥푎푛푘 푛 transforms to log 푓(푒푦1, . . . , 푒푦푛) = log( 퐾 X 푘=1 푒푎푇 푘푦+푏푘) (with 푏푘= log 푐푘) • geometric program transforms to convex problem minimize log( 퐾 P 푘=1 exp(푎푇 0푘푦+ 푏0푘)) subject to log( 퐾 P 푘=1 exp(푎푇 푖푘푦+ 푏푖푘)) ≤0, 푖= 1, . . . , 푚 퐺푦+ 푑= 0 Convex optimization problems 4.26 Design of cantilever beam 퐹 segment 4 segment 3 segment 2 segment 1 • 푁segments with unit lengths, rectangular cross-sections of size 푤푖× ℎ푖 • given vertical force 퐹applied at the right end Design problem minimize total weight subject to upper & lower bounds on 푤푖, ℎ푖 upper bound & lower bounds on aspect ratios ℎ푖/푤푖 upper bound on stress in each segment upper bound on vertical deflection at the end of the beam variables: 푤푖, ℎ푖for 푖= 1, . . . , 푁 Convex optimization problems 4.27 Objective and constraint functions • total weight 푤1ℎ1 + · · · + 푤푁ℎ푁is posynomial • aspect ratio ℎ푖/푤푖and inverse aspect ratio 푤푖/ℎ푖are monomials • maximum stress in segment 푖is given by 6푖퐹/(푤푖ℎ2 푖), a monomial • vertical deflection 푦푖and slope 푣푖of central axis at the right end of segment 푖: 푣푖 = 12(푖−1/2) 퐹 퐸푤푖ℎ3 푖 + 푣푖+1 푦푖 = 6(푖−1/3) 퐹 퐸푤푖ℎ3 푖 + 푣푖+1 + 푦푖+1 for 푖= 푁, 푁−1, . . . , 1, with 푣푁+1 = 푦푁+1 = 0 (퐸is Young’s modulus) 푣푖and 푦푖are posynomial functions of 푤, ℎ Convex optimization problems 4.28 Formulation as a GP minimize 푤1ℎ1 + · · · + 푤푁ℎ푁 subject to 푤−1 max푤푖≤1, 푤min푤−1 푖 ≤1, 푖= 1, . . . , 푁 ℎ−1 maxℎ푖≤1, ℎminℎ−1 푖 ≤1, 푖= 1, . . . , 푁 푆−1 max푤−1 푖ℎ푖≤1, 푆min푤푖ℎ−1 푖 ≤1, 푖= 1, . . . , 푁 6푖퐹휎−1 max푤−1 푖ℎ−2 푖 ≤1, 푖= 1, . . . , 푁 푦−1 max푦1 ≤1 note • we write 푤min ≤푤푖≤푤max and ℎmin ≤ℎ푖≤ℎmax 푤min/푤푖≤1, 푤푖/푤max ≤1, ℎmin/ℎ푖≤1, ℎ푖/ℎmax ≤1 • we write 푆min ≤ℎ푖/푤푖≤푆max as 푆min푤푖/ℎ푖≤1, ℎ푖/(푤푖푆max) ≤1 Convex optimization problems 4.29 Minimizing spectral radius of nonnegative matrix Perron–Frobenius eigenvalue 휆pf(퐴) • exists for (elementwise) positive 퐴∈R푛×푛 • a real, positive eigenvalue of 퐴, equal to spectral radius max푖|휆푖(퐴)| • determines asymptotic growth (decay) rate of 퐴푘: 퐴푘∼휆푘 pf as 푘→∞ • alternative characterization: 휆pf(퐴) = inf{휆| 퐴푣⪯휆푣for some 푣≻0} Minimizing spectral radius of matrix of posynomials • minimize 휆pf(퐴(푥)), where the elements 퐴(푥)푖푗are posynomials of 푥 • equivalent geometric program: minimize 휆 subject to 푛 P 푗=1 퐴(푥)푖푗푣푗/(휆푣푖) ≤1, 푖= 1, . . . , 푛 variables 휆, 푣, 푥 Convex optimization problems 4.30 Conic linear optimization minimize 푐푇푥 subject to 퐹푥+ 푔⪯퐾0 퐴푥= 푏 • 퐾is a proper convex cone in R푚 • 퐹is an 푚× 푛matrix, 푔is a 푚-vector • constraint means −(퐹푥+ 푔) ∈퐾 • linear programming is special case with 퐾= R푚 + • same properties as standard convex problem (local optimum is global, etc.) Convex optimization problems 4.31 Semidefinite program (SDP) minimize 푐푇푥 subject to 푥1퐹1 + 푥2퐹2 + · · · + 푥푛퐹푛+ 퐺⪯0 퐴푥= 푏 with 퐹푖, 퐺∈S푘 • inequality constraint is called linear matrix inequality (LMI) • includes problems with multiple LMI constraints: for example, 푥1 ˆ 퐹1 + · · · + 푥푛ˆ 퐹푛+ ˆ 퐺⪯0, 푥1 ˜ 퐹1 + · · · + 푥푛˜ 퐹푛+ ˜ 퐺⪯0 is equivalent to single LMI 푥1 ˆ 퐹1 0 0 ˜ 퐹1 + 푥2 ˆ 퐹2 0 0 ˜ 퐹2 + · · · + 푥푛 ˆ 퐹푛 0 0 ˜ 퐹푛 + ˆ 퐺 0 0 ˜ 퐺 ⪯0 Convex optimization problems 4.32 LP and SOCP as SDP LP and equivalent SDP LP: minimize 푐푇푥 subject to 퐴푥⪯푏 SDP: minimize 푐푇푥 subject to diag(퐴푥−푏) ⪯0 (note different interpretation of generalized inequality ⪯) SOCP and equivalent SDP SOCP: minimize 푓푇푥 subject to ∥퐴푖푥+ 푏푖∥2 ≤푐푇 푖푥+ 푑푖, 푖= 1, . . . , 푚 SDP: minimize 푓푇푥 subject to (푐푇 푖푥+ 푑푖)퐼 퐴푖푥+ 푏푖 (퐴푖푥+ 푏푖)푇 푐푇 푖푥+ 푑푖 ⪰0, 푖= 1, . . . , 푚 Convex optimization problems 4.33 Eigenvalue minimization minimize 휆max(퐴(푥)) where 퐴(푥) = 퐴0 + 푥1퐴1 + · · · + 푥푛퐴푛(with given 퐴푖∈S푘) Equivalent SDP minimize 푡 subject to 퐴(푥) ⪯푡퐼 • variables 푥∈R푛, 푡∈R • equivalence follows from 휆max(퐴) ≤푡 ⇐⇒ 퐴⪯푡퐼 Convex optimization problems 4.34 Matrix norm minimization minimize ∥퐴(푥)∥2 = 휆max(퐴(푥)푇퐴(푥))1/2 where 퐴(푥) = 퐴0 + 푥1퐴1 + · · · + 푥푛퐴푛(with given 퐴푖∈R푝×푞) Equivalent SDP minimize 푡 subject to 푡퐼 퐴(푥) 퐴(푥)푇 푡퐼 ⪰0 • variables 푥∈R푛, 푡∈R • constraint follows from ∥퐴∥2 ≤푡 ⇐⇒ 퐴푇퐴⪯푡2퐼, 푡≥0 ⇐⇒ 푡퐼 퐴 퐴푇 푡퐼 ⪰0 Convex optimization problems 4.35 Quasiconvex optimization minimize 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 퐴푥= 푏 • 푓0 is quasiconvex • 푓1, ..., 푓푚are convex can have locally optimal points that are not (globally) optimal (푥, 푓0(푥)) Convex optimization problems 4.36 Linear-fractional program minimize 푓0(푥) subject to 퐺푥⪯ℎ 퐴푥= 푏 (1) where 푓0(푥) = 푐푇푥+ 푑 푒푇푥+ 푓, dom 푓0 = {푥| 푒푇푥+ 푓> 0} • a quasiconvex optimization problem • also equivalent to the LP (variables 푦, 푧) minimize 푐푇푦+ 푑푧 subject to 퐺푦⪯ℎ푧 퐴푦= 푏푧 푒푇푦+ 푓푧= 1 푧≥0 (2) Convex optimization problems 4.37 Exercise assume the linear-fractional program (1) is feasible • show how to obtain the solution of (1) from the solution of the LP (2) • what do solutions (푦, 푧) of (2) with 푧= 0 mean for (1)?
Solution: denote the optimal values of (1) and (2) by 푝★ lfp and 푝★ lp, respectively 1. for every feasible 푥in (1), there is a corresponding feasible (푦, 푧) in (2): 푦= 푥 푒푇푥+ 푓, 푧= 1 푒푇푥+ 푓, 푐푇푦+ 푑= 푐푇푥+ 푑 푒푇푥+ 푓 2. for every feasible (푦, 푧) in (2) with 푧> 0, there is a feasible 푥in (1): 푥= 푦 푧, 푓0(푥) = 푐푇푥+ 푑 푒푇푥+ 푓= 푐푇푦+ 푑푧 푒푇푦+ 푓푧= 푐푇푦+ 푑 Convex optimization problems 4.38 3. suppose (푦, 푧) is feasible for (2) with 푧= 0: 퐺푦⪯0, 퐴푦= 0, 푒푇푦= 1 let ˆ 푥be a feasible point for (1): 퐺ˆ 푥⪯ℎ, 퐴ˆ 푥= 푏, 푒푇ˆ 푥+ 푓> 0 all points on the half-line {ˆ 푥+ 훼푦| 훼≥0} are feasible for (1), 퐺( ˆ 푥+ 훼푦) ⪯ℎ, 퐴( ˆ 푥+ 훼푦) = 푏, 푒푇( ˆ 푥+ 훼푦+ 푓) > 0, and the cost function at ˆ 푥+ 훼푦tends to 푐푇푦as 훼→∞: 푓0( ˆ 푥+ 훼푦) = 푐푇ˆ 푥+ 푑+ 훼푐푇푦 푒푇ˆ 푥+ 푓+ 훼 −→푐푇푦 • 1 shows that 푝★ lfp ≥푝★ lp and 2, 3 show that 푝★ lp ≥푝★ lfp; therefore 푝★ lp = 푝★ lfp • if (푦, 푧) is an optimal solution of (2) and 푧> 0, then 푥= 푦/푧is optimal for (1) • (푦, 0) of (2) indicates the optimal value of (1) is finite but not attained Convex optimization problems 4.39 Generalized linear-fractional program 푓0(푥) = max 푖=1,...,푟 푐푇 푖푥+ 푑푖 푒푇 푖푥+ 푓푖 , dom 푓0(푥) = {푥| 푒푇 푖푥+ 푓푖> 0, 푖= 1, . . . , 푟} • a quasiconvex optimization problem • LP reformulation of page 4.37 does not extend to generalized problem Example: Von Neumann model of a growing economy maximize (over 푥, 푥+) min 푖=1,...,푛푥+ 푖/푥푖 subject to 푥+ ⪰0, 퐵푥+ ⪯퐴푥 • 푥, 푥+ ∈R푛: activity levels of 푛sectors, in current and next period • (퐴푥)푖: amount of good 푖produced in current period • (퐵푥+)푖: amount consumed in next period, cannot exceed (퐴푥)푖 • 푥+ 푖/푥푖: growth rate of sector 푖 allocate activity to maximize growth rate of slowest growing sector Convex optimization problems 4.40 Convex representation of sublevel sets of 푓0 if 푓0 is quasiconvex, there exists a family of functions 휙푡such that: • 휙푡(푥) is convex in 푥for fixed 푡 • 푡-sublevel set of 푓0 is 0-sublevel set of 휙푡, i.e., 푓0(푥) ≤푡 ⇐⇒ 휙푡(푥) ≤0 Example 푓0(푥) = 푝(푥) 푞(푥) with 푝convex, 푞concave, and 푝(푥) ≥0, 푞(푥) > 0 on dom 푓0 can take 휙푡(푥) = 푝(푥) −푡푞(푥): • for 푡≥0, 휙푡convex in 푥 • 푝(푥)/푞(푥) ≤푡if and only if 휙푡(푥) ≤0 Convex optimization problems 4.41 Quasiconvex optimization via convex feasibility problems 휙푡(푥) ≤0, 푓푖(푥) ≤0, 푖= 1, . . . , 푚, 퐴푥= 푏 (3) • for fixed 푡, a convex feasibility problem in 푥 • if feasible, we can conclude that 푡≥푝★; if infeasible, 푡≤푝★ Bisection method given: 푙≤푝★, 푢≥푝★, tolerance 휖> 0 repeat 1. 푡:= (푙+ 푢)/2 2. solve the convex feasibility problem (3) 3. if (3) is feasible, 푢:= 푡 else 푙:= 푡 until 푢−푙≤휖 requires exactly l log2 푢−푙 휖 m iterations Convex optimization problems 4.42 Vector optimization General vector optimization problem minimize (w.r.t. 퐾) 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 vector objective 푓0 : R푛→R푞, minimized with respect to proper cone 퐾∈R푞 Convex vector optimization problem minimize (w.r.t. 퐾) 푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 퐴푥= 푏 where 푓1, ..., 푓푚are convex and 푓0 is “퐾-convex”, i.e., 푓0(휃푥+ (1 −휃)푦) ⪯퐾휃푓0(푥) + (1 −휃) 푓0(푦) for all 푥, 푦∈dom 푓0 and 휃∈[0, 1] Convex optimization problems 4.43 Multicriterion optimization vector optimization problem with 퐾= R푞 + 푓0(푥) = (퐹1(푥), . . . , 퐹푞(푥)) • 푞different objectives 퐹푖; roughly speaking we want all 퐹푖’s to be small • feasible 푥★is optimal if 푦feasible =⇒ 푓0(푥★) ⪯푓0(푦) if there exists an optimal point, the objectives are noncompeting • feasible 푥po is Pareto optimal if 푦feasible, 푓0(푦) ⪯푓0(푥po) =⇒ 푓0(푥po) = 푓0(푦) if Pareto optimal values are not unique, there is a trade-offbetween objectives • 푓0 is 퐾-convex if 퐹1, ..., 퐹푞are convex (in the usual sense) Convex optimization problems 4.44 Optimal and Pareto optimal points set of achievable objective values O = { 푓0(푥) | 푥feasible} • feasible 푥is optimal if 푓0(푥) is the minimum value of O • feasible 푥is Pareto optimal if 푓0(푥) is a minimal value of O O 푓0(푥★) 푥★is optimal O 푓0(푥po) 푥po is Pareto optimal Convex optimization problems 4.45 Regularized least-squares minimize (w.r.t. R2 +) (∥퐴푥−푏∥2 2, ∥푥∥2 2) 0 10 20 30 40 50 0 5 10 15 20 25 퐹1(푥) = ∥퐴푥−푏∥2 2 퐹2(푥) = ∥푥∥2 2 O example for 퐴∈R100×10; heavy line is formed by Pareto optimal points Convex optimization problems 4.46 Risk–return trade-offin portfolio optimization minimize (w.r.t. R2 +) (−¯ 푝푇푥, 푥푇Σ푥) subject to 1푇푥= 1, 푥⪰0 • 푥∈R푛is investment portfolio; 푥푖is fraction invested in asset 푖 • return is 푟= 푝푇푥where 푝∈R푛is vector of relative asset price changes • 푝is modeled as a random variable with mean ¯ 푝, covariance Σ • ¯ 푝푇푥= E 푟is expected return; 푥푇Σ푥= var 푟is return variance (risk) Example mean return standard deviation of return 0% 10% 20% 0% 5% 10% 15% standard deviation of return allocation 푥 푥(1) 푥(2) 푥(3) 푥(4) 0% 10% 20% 0 0.5 1 Convex optimization problems 4.47 Scalarization to find Pareto optimal points: choose 휆≻퐾∗0 and solve scalar problem minimize 휆푇푓0(푥) subject to 푓푖(푥) ≤0, 푖= 1, . . . , 푚 ℎ푖(푥) = 0, 푖= 1, . . . , 푝 • solutions 푥of scalar problem are Pareto-optimal for vector optimization problem 푥not Pareto-optimal ⇓ ∃feasible 푦: 푓0(푦) ⪯퐾푓0(푥), 푓0(푦) ≠푓0(푥) ⇓ 휆푇푓0(푦) < 휆푇푓0(푥) for 휆≻퐾∗0 O 푓0(푥1) 휆1 푓0(푥2) 휆2 푓0(푥3) • partial converse for convex vector optimization problems (see later in duality): can find (almost) all Pareto optimal points by varying 휆≻퐾∗0 • objective of scalar problem is convex if 푓0 is 퐾-convex Convex optimization problems 4.48 Scalarization for multicriterion problems to find Pareto optimal points, minimize positive weighted sum 휆푇푓0(푥) = 휆1퐹1(푥) + · · · + 휆푞퐹푞(푥) • regularized least squares problem of page 4.46 take 휆= (1, 훾) with 훾> 0 minimize ∥퐴푥−푏∥2 2 + 훾∥푥∥2 2 for fixed 훾, a LS problem 0 5 10 15 20 0 5 10 15 20 ∥퐴푥−푏∥2 2 ∥푥∥2 2 훾= 1 • risk–return trade-offof page 4.47: with 훾> 0, minimize −¯ 푝푇푥+ 훾푥푇Σ푥 subject to 1푇푥= 1, 푥⪰0 Convex optimization problems 4.49 |
6751 | https://www.ahajournals.org/doi/pdf/10.1161/circulationaha.109.885137 | Poor R-Wave Progression in the Precordial Leads in Left-Sided Spontaneous Pneumothorax
Wataru Mitsuma, MD; Masahiro Ito, MD; Tadayuki Honda, MD; Makoto Kodama, MD; Hiroshi Endoh, MD; Yoshifusa Aizawa, MD A 30-year-old man without any cardiovascular history was transferred to our hospital because of chest pain. On arrival, his blood pressure was 166/93 mm Hg, with a pulse rate of 58 bpm. His oxygen saturation level was 98%. His ECG showed normal sinus rhythm, with poor R-wave pro-gression in the precordial leads (Figure, A). The white blood cell count was 7990/mm 3 , and the level of creatine kinase was 93 IU/L (normal range 163). Left ventricular wall motion was normal on echocardiography. The chest x-ray revealed left-sided spontaneous pneumothorax (Figure, A). After sim-ple aspiration with cannula, the left lung was reexpanded, and the poor R-wave progression in the precordial leads was completely resolved (Figure, B). Pneumothorax and myocardial infarction are common diseases presenting chest pain, and ECG is one of the most important diagnostic tools for them. Here, we describe ECG findings with left-sided pneumothorax mimicking anterior myocardial infarction. The mechanism of poor precordial R-wave progression in this patient seemed to be rotation of the heart due to intrathoracic air, because the ECG findings immediately improved after simple aspiration. Left-sided pneumothorax should be considered in patients with chest pain and suspected anterior myocardial infarction on ECG.
Disclosures
None.
Figure. ECG and chest radiograph of patient on arrival (A) and after aspiration (B). A, ECG showed normal sinus rhythm with poor precordial R-wave progression, and the collapse of the left lung was seen on the chest radiograph (arrows). B, After simple aspiration, both abnormalities as demon-strated by ECG and chest radiograph improved.
From the Division of Cardiology (W.M., M.I., M.K., Y.A.) and the Advanced Disaster Medical and Emergency Critical Care Center (W.M., T.H., H.E.), Niigata University Graduate School of Medical and Dental Sciences, Niigata 951– 8510, Japan. Correspondence to Wataru Mitsuma, MD, Division of Cardiology, Niigata University Graduate School of Medical and Dental Sciences, 1–754 Asahimachi dori, Chuoku, Niigata 951– 8510, Japan. E-mail mwataru@med.niigata-u.ac.jp
(Circulation. 2009;120:2122.)
© 2009 American Heart Association, Inc.
Circulation is available at DOI: 10.1161/CIRCULATIONAHA.109.885137
2122
Images in Cardiovascular Medicine Downloaded from by on September 29, 2025 |
6752 | http://qed.econ.queensu.ca/pub/faculty/clintonk/econ223/3%20Solow%20growth%20model.pdf | Kevin Clinton Winter 2005 Lecture notes 3 Economic growth: Solow model 1. Introduction Solow’s classic model is a superb piece of work, everything you could ask of a theory. It takes on the biggest questions—e.g., what determines standards of living, why some countries are rich and others poor. The argument is based on standard assumptions, yet it arrives at not-at-all obvious implications. It fits the facts well. So much so that Solow’s model sets the framework for all serious empirical studies of growth and productivity.
Solow highlights technical change—i.e. productivity growth—as the key to long-run growth of per capita income and output. Accumulation of capital creates growth in the long run only to the extent that it embodies improved technology.
To develop the model, we start with the artificial situation of constant population and constant technology, and then, in steps, allow population to grow, and technology to improve.
2. The steady state Production function The aggregate production function is: Y = F(K,L) With constant returns to scale we can transform this into a function relating output per worker to capital per worker.
y = f(k) where y = Y/L, and k = K/L.
2 Figure 3.1: Per worker production function Accumulation of capital The change in the capital stock per worker (known as capital deepening) is equal to per worker gross investment minus depreciation: ∆k = i - δk.
Ignore government for present purposes, so that investment is equal to private sector saving: i = S/L = s Y/L = sy.
where s is the saving ratio (the MPS is for simplicity the same as the APS). This we can write in terms of the production function: i = s f(k).
The proportional saving-income relationship implies that this investment function is like a scaled-down production function.
Figure 3.2 Investment and production functions y, i k y= f(k) i = sf(k) Diminishing MPK y k 3 Thus, both output per worker and investment per worker are an increasing function (at a decreasing rate, because of diminishing MPK) of capital per worker.
To show capital accumulation on the graph, we focus on the i = s f(k) curve, and introduce depreciation.
Figure 3.3 Investment and depreciation Depreciation is a straight-line function of k. At some point, call it k, the depreciation line slices through the flattening investment curve. To the left of k, net investment is positive (gross greater than depreciation), to the right negative. Investment along the straight line just keeps capital worker constant, so we can regard the line as a break-even investment schedule.
In other words, to the left of k, k is increasing, to the right, k is decreasing. Therefore k is the steady state level of capital per worker—the long-run equilibrium of the economy.
Figure 3.5: Steady-state equilibrium i k i = sf(k) δ δ δ δk k i k i = sf(k) δ δ δ δk: break-even line net investment k negative net 4 Accumulation and growth In equilibrium, with a given saving rate, there is no net accumulation of capital, and no growth of output. What if the saving rate goes up?
Figure 3.6 Transition to a higher-saving steady state The capital stock rises eventually to a new steady state equilibrium, at k2. During the transition output as well as capital grows, both at a diminishing rate. Growth tapers off to nothing in the new steady state.
Implications A permanent increase in the saving ratio will raise the level of output permanently, but not its rate of growth. During the transition period, which might last decades, growth will be higher. But the increased investment eventually results in an offsetting increase in depreciation, and hence capital per worker levels off. Saving and capital accumulation on its own, with given technology, cannot explain long-run economic expansion.
Evidence Current affairs The US has a very large budget deficit, about 5% of GDP, for as far as the eye can see, with the Administration offering more tax cuts, and some high-priced new programs (medicine for seniors, Americans on Mars,). Budget deficits reduce national savings, which might not be too bad if the private saving rate were high, but in the US it is low. The Solow model warns that such a policy is likely to reduce income growth over an extended period. i k δ δ δ δk k1 k2 i = s2 f(k) i = s1 f(k) 5 3. Population growth Accumulation to stand still Population growth, of course, affects accumulation of capital per worker. To see how, start with the approximation that the proportional change in a ratio is equal to the proportional change in the numerator minus the proportional change in the denominator yields. The approximation is good for modest-sized changes, such those for macroeconomic aggregates from year to year. Here the ratio of concern is k = K/L, for which the rate of change is: ∆k/k = ∆Κ/K – ∆L/L.
Since ∆Κ is equal to investment (I) minus depreciation (δK), rewrite this as: ∆k/k = I/K - δK/K – n.
For the change in the capital stock per worker, as opposed to the rate of change, multiply each side by k, or K/L, as convenient: ∆k = (I/K - δK/K)K/L – nk = I/L - δK/L – nk, this simplifies to: ∆k = i – (δ + n)k.
The change in capital per worker is given by net investment less the investment required to provide newly arriving workers with the same capital as existing workers, nk. The break-even line rotates upward.
Figure 3.7 Negative impact of population growth i k δ δ δ δk k2 k1 i = s f(k) (δ+ (δ+ (δ+ (δ+n) ) ) )k 6 Implications Population growth, in itself, reduces the steady-state level of capital per worker. Via the production function, this translates directly to lower per capita output and income.
Steady-state per capita income is constant; total output grows at the rate of population growth.
So far, the model does not explain permanently increasing per capita income (ironic, given the title of Ch 4)—for this we need improving productivity.
Evidence What does the model explain?
Solow’s model, even in a rudimentary version without technical change, explains • positive correlation of investment rates and per capita income • negative correlation of population growth and per capita income It also helps explain these remarkable phenomena: • 2-3 decade growth miracles following wartime destruction • China and Asian tigers • ultimate collapse of Soviet heavy industry expansion 4. Technological progress Equilibrium with increasing productivity Y = F(K, L×E) We can measure labour in efficiency units, E. Technological progress (improved equipment, education, skills, health, infrastructure, etc.) increases the productivity of labour. Let this improvement be at a steady rate, g. Redefine k to stand for capital per effective worker, i.e. k = K/( L×E), and likewise y = Y/( L×E). The equation for capital deepening in terms of effective workers, ∆k, is ∆k = i – (δ + n + g)k.
The additional term gk represents investment that merely keeps the capital stock per effective worker constant as efficiency increases.
Figure 3.8: Equilibrium with technological progress 7 Capital per effective worker is in equilibrium at k, for the same reasons as in the constant technology case. An increase in g, just like an increase in δ or n, rotates the break-even line upward.
Capital per actual worker grows at rate g, as does output per worker (the capital/output ratio is therefore stable).
Table 3.1: Steady state with technological progress Growth rate Capital per effective worker k = K/ (E × L) 0 Output per effective worker y = Y/ (E × L) = f(k) 0 Capital & output per worker Y/L & K/L g Total output Y n + g Implication Technological progress explains long-run expansion of income per capita. i per effective worker k i = s f(k) k (δ+ (δ+ (δ+ (δ+n+g) ) ) ) k 8 Income and factor shares The distribution of income between capital and labour remains constant along the steady-state growth path. The return on capital (in this model, the interest rate) is constant, while the stock grows at rate n+g. The wage rate grows at g, the labour force at n, so the wage bill also grows at n+g.
Table 3.2: Steady-state distribution of income Golden Rule Level Growth rate Total income Y n + g Return on capital (interest rate) MPK = n + g 0 Total return to capital MPK K n + g Wage rate MPL g Total return to labour MPL L n + g Capital share α = MPK K / Y 0 Labour share 1 − α 0 Evidence Factor shares have remained roughly stable, over long periods of time. In Canada and the US the labour share has been about 70%.
9 Golden rule As we have seen, the equilibrium value of capital per effective worker increases with the saving ratio. In steady state, the per capita income path is higher for a greater savings ratio.
Is more saving always better? No. We want to maximize consumption, not income. A greater capital stock requires more break-even investment—i.e. investment just to keep capital per effective worker constant. At some point, because of decreasing MPK, capital increases start to cannibalize capital itself, at the margin leaving nothing for consumption.
Figure 3.8: Consumption possibilities To maximize consumption, we want to hold the capital ratio at the point where further increases in capital cease to yield a marginal gain in consumption. This point in Figure 3.8 is G. Beyond that point increases in capital per effective worker create decreases in consumption, despite continued increases in output.
The Golden Rule describes the consumption-maximizing path of capital accumulation corresponding to G. As a matter of geometry, at G the slope of the production function (the gross marginal product of capital) is equal to the slope of the break-even line. That is: MPK = δ + n + g or MPK - δ = n + g.
Under the Golden Rule, the net marginal product of capital is equal to the growth rate of total output. y etc.
per effective worker y k Consumption G (δ+ (δ+ (δ+ (δ+n+g) ) ) ) k: break-even line 10 Real interest and growth rates In equilibrium, the interest rate (the return on saving) is equal to the net marginal product of capital after depreciation. If the interest rate is less than the growth rate, the economy is saving too much—i.e. consuming more provides a free lunch.
Figure 3.9: Consumption possibilities For example, at point P, the interest rate is less than the rate of growth. A reduction in the saving rate, from sP to sG, reduces steady state output (OP), but reduces break-even investment even more (OQ). [Magnify.] An efficient economy therefore would have the long-run equilibrium real rate of interest (or natural rate) equal to, or above, the growth rate. The Golden Rule has equality.
To apply this to the real world one has to recognize that • actual interest rates are many and varied, with different terms and risks and other features, depending on the asset and the borrower • returns on capital may be highly uncertain, and therefore contain a risk premium over and above the interest rate on financial assets Bearing in mind these qualifications, one can say that if the economy is operating efficiently, then in the long run the real yield on long-term government bonds (credit-risk-free assets) should be about equal, or slightly above, the growth rate.
Evidence In Canada, over time, the average long-term bond yield has been in the same range as the rate of growth (3-4%). But the overall return on capital has been above this (M&S estimate 7 ½%, p 11), reflecting risk premiums.
i, y, etc.
per effective worker y k G (δ+ (δ+ (δ+ (δ+n+g) ) ) ) k sG y P kG kP sP y O Q |
6753 | https://www.bioblast.at/index.php/Chinese_numerals | Chinese numerals
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Chinese numerals
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Chinese numerals
The Arabic numeral system used today in China was introduced to China by the Europeans in the early 17th century. But the Chinese character-based number systems are still in use. The financial numerals are used only when writing an amount on a form for remitting money at a bank. They function as anti-fraud numerals.
The character 零 (zero) appeared very early in ancient Chinese writing. However, at that time, it did not mean "nothing", but "bits and pieces", "not much". 一百零五(105) means in Chinese: In addition to a hundred, there is a fraction of five. With the introduction of the Arabic numerals, 105 is exactly pronounced “one hundred zero five”, the character 零 corresponds exactly to the symbol 0. Thus, the character 零has the meaning of 0. But the character 〇 was one of the Chinese characters created and promulgated by the only empress (with greater achievements than countless emperors) in the history of China in 690 AD (much later than the invention of 0 in India) for the purpose of demonstrating her power. At that time the character 〇 meant “star”, representing a round planet. It is now used as a synonym for the 零 (zero).
Contents
1 Chinese numerals
```
Communicated by Zhang Feiyuan 2020-07-21
```
Chinese numerals
| Arabic | Normal Chinese | Financial Chinese |
---
| | 小寫／小写 | 大寫／大写 |
| 0 | 〇 | 零 |
| 1 | 一 | 壹 |
| 2 | 二 | 貳／贰 |
| 3 | 三 | 參／叁 |
| 4 | 四 | 肆 |
| 5 | 五 | 伍 |
| 6 | 六 | 陸／陆 |
| 7 | 七 | 柒 |
| 8 | 八 | 捌 |
| 9 | 九 | 玖 |
| 10 | 十 | 拾 |
| 102 | 百 | 佰 |
| 103 | 千 | 仟 |
| 104 | 萬／万 | |
| 108 | 億／亿 | |
| 1012 | 兆 (Taiwan) | |
| 1012 | 萬億／万亿 (Mainland China) | |
In my opinion, the Normal Chinese characters may not be used by non-Mandarin speaking Chinese since this requires the knowledge of the Chinese language. However, these Chinese characters or Chinese character-based numerals are shared with countries influenced by the Chinese culture in the past, e.g. Japan, Korea, Vietnam. So non-Chinese speakers in certain countries use the Chinese character-based numerical system, too. But they pronounce them differently.
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Arthritis Rheumatol
. Author manuscript; available in PMC: 2024 Mar 18.
Published in final edited form as: Arthritis Rheumatol. 2023 May 25;75(8):1299–1311. doi: 10.1002/art.42507
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2022 American College of Rheumatology (ACR) Guideline for Exercise, Rehabilitation, Diet, and Additional Integrative Interventions for Rheumatoid Arthritis
Bryant R England
Bryant R England, MD, PhD
1 University of Nebraska Medical Center and VA Nebraska-Western Iowa Health Care System, Omaha, NE
Find articles by Bryant R England
1,, Benjamin J Smith
Benjamin J Smith, DMSc, PA-C
2 Florida State University, Tallahassee, FL
Find articles by Benjamin J Smith
2,, Nancy A Baker
Nancy A Baker, ScD, MPH, OTR/L
3 Tufts University, Boston, MA
Find articles by Nancy A Baker
3, Jennifer L Barton
Jennifer L Barton, MD, MCR
4 VA Portland Health Care System and Oregon Health & Science University, Portland, OR
Find articles by Jennifer L Barton
4, Carol A Oatis
Carol A Oatis, PT, PhD
5 Arcadia University, Glenside, PA
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5, Gordon Guyatt
Gordon Guyatt, MD
6 McMaster University, Hamilton, ON, Canada
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6, Allen Anandarajah
Allen Anandarajah, MD, MS
7 University of Rochester Medical Center, Rochester, NY
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7, Kristine Carandang
Kristine Carandang, PhD, OTR/L
8 San Diego, CA
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8, Deb Constien
Deb Constien
9 Sun Prairie, WI
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9, Karmela Kim Chan
Karmela Kim Chan, MD
10 Hospital for Special Surgery, New York, NY
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10, Eileen Davidson
Eileen Davidson
11 Burnaby, BC, Canada
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11, Carole V Dodge
Carole V Dodge, OTR, CHT
12 University of Michigan Hospital and Health System, Ann Arbor, MI
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12, Anita Bemis-Dougherty
Anita Bemis-Dougherty, PT, DPT, MAS
13 American Physical Therapy Association, Alexandria, VA
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13, Sotiria Everett
Sotiria Everett, EdD, RD, CDN, CSSD
14 Department of Family, Population, Preventive Medicine, Stony Brook Renaissance School of Medicine, Stony Brook, NY
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14, Nadine Fisher
Nadine Fisher, EdD
15 University of Buffalo, Buffalo, NY
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15, Liana Fraenkel
Liana Fraenkel, MD, MPH
16 Yale School of Medicine, New Haven, CT
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16, Susan M Goodman
Susan M Goodman, MD
10 Hospital for Special Surgery, New York, NY
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10, Janet Lewis
Janet Lewis, MD
17 University of Virginia, Charlottesville, VA
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17, Victoria Menzies
Victoria Menzies, PhD, APRN
18 University of Florida, Gainesville, FL
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18, Larry W Moreland
Larry W Moreland, MD
19 University of Colorado Anschutz Medical Campus, Aurora, CO
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19, Iris Navarro-Millan
Iris Navarro-Millan, MD, MSPH
20 Weill Cornell Medicine, New York, NY
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20, Sarah Patterson
Sarah Patterson, MD
21 UCSF Osher Center for Integrative Medicine, San Francisco, CA
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21, Lawrence “Rick” Phillips
Lawrence “Rick” Phillips, EdD
22 Noblesville, IN
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22, Neha Shah
Neha Shah, MD
23 Stanford Health Care, Palo Alto, CA
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23, Namrata Singh
Namrata Singh, MD, MSCI
24 University of Washington, Seattle, WA
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24, Daniel White
Daniel White, PT, ScD, MSc
25 University of Delaware, Newark, DE
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25, Rawan AlHeresh
Rawan AlHeresh, MSOT, PhD, OTR/L
26 MGH Institute of Health Professions
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26, Kamil E Barbour
Kamil E Barbour, PhD, MPH
27 Centers for Disease Control and Prevention, Atlanta, GA
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27, Thomas Bye
Thomas Bye, PT, DPT, MS
25 University of Delaware, Newark, DE
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25, Dana Guglielmo
Dana Guglielmo, MPH
28 Research Consultant, Los Angeles, CA
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28, Rebecca Haberman
Rebecca Haberman, MD
29 NYU Langone Health, New York, NY
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29, Tate Johnson
Tate Johnson, MD
1 University of Nebraska Medical Center and VA Nebraska-Western Iowa Health Care System, Omaha, NE
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1, Anatole Kleiner
Anatole Kleiner, MD
7 University of Rochester Medical Center, Rochester, NY
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7, Chris Y Lane
Chris Y Lane, PT, DPT
30 University of North Carolina at Chapel Hill, Chapel Hill, NC
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30, Linda C Li
Linda C Li, PT, PhD
31 University of British Columbia and Arthritis Research Canada, Vancouver, BC, Canada
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31, Hiral Master
Hiral Master, PT, PhD, MPH
32 Vanderbilt University Medical Center, VICTR, Nashville, TN
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32, Daniel Pinto
Daniel Pinto, PT, PhD
33 Marquette University, Milwaukee, WI
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33, Janet L Poole
Janet L Poole, PhD, OTR/L
34 University of New Mexico, Albuquerque, NM
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34, Kimberly Steinbarger
Kimberly Steinbarger, PT, MHS, DHSc
35 Husson University, Bangor, ME
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35, Daniel Sztubinski
Daniel Sztubinski
36 ECRI Institute, Plymouth Meeting, PA
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36, Louise Thoma
Louise Thoma, PT, PhD
30 University of North Carolina at Chapel Hill, Chapel Hill, NC
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30, Vlad Tsaltskan
Vlad Tsaltskan, MD
37 University of California, San Diego, La Jolla, CA
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37, Marat Turgunbaev
Marat Turgunbaev, MD, MPH
38 American College of Rheumatology, Atlanta, GA
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38, Courtney Wells
Courtney Wells, PhD, MPH, MSW
39 University of Wisconsin – River Falls, River Falls, WI
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39, Amy S Turner
Amy S Turner
35 Husson University, Bangor, ME
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35, Jonathan R Treadwell
Jonathan R Treadwell, PhD
36 ECRI Institute, Plymouth Meeting, PA
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36
Author information
Article notes
Copyright and License information
1 University of Nebraska Medical Center and VA Nebraska-Western Iowa Health Care System, Omaha, NE
2 Florida State University, Tallahassee, FL
3 Tufts University, Boston, MA
4 VA Portland Health Care System and Oregon Health & Science University, Portland, OR
5 Arcadia University, Glenside, PA
6 McMaster University, Hamilton, ON, Canada
7 University of Rochester Medical Center, Rochester, NY
8 San Diego, CA
9 Sun Prairie, WI
10 Hospital for Special Surgery, New York, NY
11 Burnaby, BC, Canada
12 University of Michigan Hospital and Health System, Ann Arbor, MI
13 American Physical Therapy Association, Alexandria, VA
14 Department of Family, Population, Preventive Medicine, Stony Brook Renaissance School of Medicine, Stony Brook, NY
15 University of Buffalo, Buffalo, NY
16 Yale School of Medicine, New Haven, CT
17 University of Virginia, Charlottesville, VA
18 University of Florida, Gainesville, FL
19 University of Colorado Anschutz Medical Campus, Aurora, CO
20 Weill Cornell Medicine, New York, NY
21 UCSF Osher Center for Integrative Medicine, San Francisco, CA
22 Noblesville, IN
23 Stanford Health Care, Palo Alto, CA
24 University of Washington, Seattle, WA
25 University of Delaware, Newark, DE
26 MGH Institute of Health Professions
27 Centers for Disease Control and Prevention, Atlanta, GA
28 Research Consultant, Los Angeles, CA
29 NYU Langone Health, New York, NY
30 University of North Carolina at Chapel Hill, Chapel Hill, NC
31 University of British Columbia and Arthritis Research Canada, Vancouver, BC, Canada
32 Vanderbilt University Medical Center, VICTR, Nashville, TN
33 Marquette University, Milwaukee, WI
34 University of New Mexico, Albuquerque, NM
35 Husson University, Bangor, ME
36 ECRI Institute, Plymouth Meeting, PA
37 University of California, San Diego, La Jolla, CA
38 American College of Rheumatology, Atlanta, GA
39 University of Wisconsin – River Falls, River Falls, WI
These authors contributed equally.
✉
Correspondence: Bryant R. England, MD, PhD, University of Nebraska Medical Center, Address: 986270 Nebraska Medical Center, Omaha, NE 68198-6270, Fax: 404-559-6788, Bryant.england@unmc.edu
Issue date 2023 Aug.
PMC Copyright notice
PMCID: PMC10947582 NIHMSID: NIHMS1972855 PMID: 37227071
The publisher's version of this article is available at Arthritis Rheumatol
Abstract
Objective:
To develop initial American College of Rheumatology (ACR) guidelines on the use of exercise, rehabilitation, diet, and additional integrative interventions in conjunction with disease-modifying anti-rheumatic drugs (DMARDs) as part of an integrative management approach for people with rheumatoid arthritis (RA).
Methods:
An interprofessional guideline development group constructed clinically relevant Population, Intervention, Comparator, and Outcome (PICO) questions. A literature review team then completed a systematic literature review and applied the Grading of Recommendations Assessment, Development and Evaluation (GRADE) approach to rate the certainty of evidence. An interprofessional voting panel (n=20 participants) that included 3 persons with RA achieved consensus on the direction (for or against) and strength (strong or conditional) of recommendations.
Results:
The voting panel achieved consensus on 28 recommendations for the use of integrative interventions in conjunction with DMARDs for the management of RA. Consistent engagement in exercise received a strong recommendation. Of 27 conditional recommendations, 4 pertained to exercise, 13 to rehabilitation, 3 to diet, and 7 to additional integrative interventions. These recommendations are specific to RA management, recognizing that other medical indications and general health benefits may exist for many of these interventions.
Discussion:
This guideline provides initial ACR recommendations on integrative interventions for the management of RA to accompany DMARD treatments. The broad range of interventions included in these recommendations illustrates the importance of an interprofessional team-based approach to RA management. The conditional nature of most recommendations requires clinicians to engage patients in shared decision making when applying these recommendations.
Keywords: Rheumatoid Arthritis, Integrative Medicine, Physical Activity, Diet, Dietary Supplements, Rehabilitation, Physical Therapy, Occupational Therapy
INTRODUCTION
Rheumatoid arthritis (RA) is a chronic, systemic, inflammatory condition, and improved outcomes occur with the early diagnosis, evaluation, and management. The American College of Rheumatology (ACR) has previously published pharmacologic guidelines to aid clinicians and people with RA (1–4). In addition to pharmacological interventions, people with RA and their clinicians consider how exercise, rehabilitation, diet, and additional adjunctive therapies can benefit and be integrated into their disease management. Using the Grading of Recommendations Assessment, Development and Evaluation [GRADE] methodology, the ACR developed this first guideline to support decision making when using specific integrative interventions in the management of RA. The interventions considered in this guideline are defined in Table 1. Although people with RA may have other indications for these interventions (e.g., comorbidities), this guideline focuses specifically on managing RA.
Table 1.
Descriptions and examples of interventions included in the integrative management of rheumatoid arthritis guideline.
| Intervention | Description and/or examples |
:--- |
| Exercise |
| Physical activity | Movement of the body requiring energy expenditure. |
| Exercise | Performance of physical activity in regular and structured manner to improve fitness and health. |
| Aerobic exercise | Exercise intended to improve cardiorespiratory fitness and muscular endurance. Examples include walking, biking or cycling, running, hiking, aerobics, rowing, swimming, using an elliptical. |
| Aquatic exercise | Exercise performed in water, containing elements of both aerobic and resistance exercise. Examples include swimming, water aerobics, water walking or jogging. |
| Resistance exercise | Exercise intended to increase muscular strength. Examples include free weights, weight machines, resistance bands, Pilates. |
| Mind-body exercise | Exercise that combines movement, mental focus, and controlled breathing. Examples include yoga, Tai Chi, Qigong. |
| Rehabilitation |
| Comprehensive occupational therapy | Evaluation and treatment by Occupational Therapist with the goal of increasing physical function and participation. Receives patient-centered individualized treatment. Components of OT services vary and may include arthritis education, ADL evaluation and training, joint protection, activity pacing, work simplification and fatigue management, exercise (particularly for the hand and arm), splinting/orthotics, provision of assistive/adaptive devices, environmental adaptations, work and leisure counselling/rehabilitation, work-site assessment, sexual advice, relaxation, and pain and stress management training. |
| Comprehensive physical therapy | Evaluation and treatment by a physical therapist. Components of PT services will vary and should include exercise. May also include functional training and physical activity, energy conservation, workplace accommodations, mobility and gait training, manual therapy, self-management education, electrotherapy, application of orthoses, instruction in assistive devices, pain-management including thermal therapy. |
| Hand therapy exercises | Exercises of the hand to improve mobility and strength. |
| Bracing and orthoses | Devices to correct and support musculoskeletal function, improve joint alignment, or protect the joint. Examples include wrist and finger splints, foot or knee orthoses, compression gloves, taping. |
| Joint protection techniques | Self-management approach that aims to maintain function by providing people with ways to alter work methods and movement patterns of affected joints to reduce pain, inflammation, and joint stress. Examples include changing the way of performing activities to avoid pain, resting, using alternative muscle groups. |
| Activity pacing | Balancing activity and rest to accomplish activities. Includes activity pacing, energy conservation, activity modification, fatigue management techniques. |
| Assistive devices | Devices to assist with mobility. Examples include crutches, canes, walkers, wheelchairs, tricycles, scooters. |
| Adaptive equipment | Devices to assist with activities of daily living. Examples include built up and/or long handled equipment, sock aide, button hook, reachers, pill cutters, cell phone holders. |
| Environmental adaptations | Adapting environment to improve safety. Examples include adaptations for toileting (raised toilet seat, commode, toilet safety rail), showering (tub seat, handheld shower, walk in bath), grab bars, ramps, stair lifts, home modification. |
| Vocational rehabilitation | Training programs to overcome barriers preventing successful employment. |
| Work site evaluation and modifications | Evaluating and adjusting work-site conditions and duties for safety and well-being. |
| Dietary |
| Formally defined diet | Specific formally defined diets evaluated in this guideline were: anti-inflammatory, Mediterranean-style, ketogenic, paleo, gluten-free, vegetarian, vegan, intermittent fasting, elemental, elimination, raw foods, whole food plant-based. |
| Mediterranean-style diet | Diet pattern that emphasizes intake of vegetables, fruits, whole grains, nuts, seeds, and olive oil; moderate amounts of low-fat dairy and fish; and limits added sugars, sodium, highly processed foods, refined carbohydrates, and saturated fats. |
| Dietary supplement | Substances used to add nutrients, botanicals, herbs, or microbials (probiotics) to the diet. Specific supplements evaluated in this guideline were vitamin D, probiotics, fish oil and omega fatty acids, antioxidants (selenium, zinc, vitamin A, vitamin C, vitamin E), turmeric, glucosamine, γ-linolenic acid, borage seed oil, evening primrose oil, black currant seed oil, selenium, Boswellia, ginger. |
| Weight loss | Intentional loss of body weight. Examples include lifestyle modification through diet and/or exercise, support groups, health coaching, medically-supervised weight loss programs, branded dietary weight loss programs, weight loss surgery. |
| Additional adjunctive therapies |
| Self-management program | Standardized program to guide self-management. Examples include Arthritis Self-Management Program, Chronic Disease Self-Management Program, Better Choices Better Health, Tomando Control de su Salud, RA Self-Management Intervention, OPERAS (an On-demand Program to EmpoweR Active Self-management). |
| Cognitive behavioral therapy | Psychological therapy to identify and change thought and behavior patterns. |
| Mind-body approaches | Practices engaging both mind and body functions. Examples include biofeedback, goal setting, meditation, mindfulness, breathing exercises, progressive muscle relaxation, guided imagery. |
| Acupuncture | Stimulation of specific body points through insertion of thin needles. |
| Massage therapy | Rubbing and kneading of muscles and joints with the hands. Examples include Swedish, Deep Tissue, Trigger Point. |
| Thermal modalities | Use of heat and cold for medical treatment. Examples include cryotherapy, heat, therapeutic ultrasound, infrared sauna, paraffin therapy, and laser therapy. |
| Electrotherapy | Use of electrical energy for medical treatment. Examples include transcutaneous electrical nerve stimulation (TENS), neuro-muscular electrical nerve stimulation (NMES). |
| Vagal nerve stimulation | Implantation of a device to stimulate the vagus nerve with electrical impulses. |
| Chiropractic | Diagnosis and manipulation of malaligned joints, particularly the spine. |
| Tobacco cessation | Counseling on tobacco cessation, tobacco cessation programs (phone, mobile applications), nicotine replacement therapies, tobacco cessation medications without nicotine. |
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Abbreviations: ADL, activities of daily living; OT, occupational therapy; PT, physical therapy
METHODS
This guideline follows the ACR guideline development process and ACR policy guiding management of conflicts of interest and disclosures ( which includes GRADE methodology (5, 6) and adheres to AGREE criteria (7). Supplementary Appendix 1 includes a detailed description of the methods. Briefly, the core leadership team (BE, BS, NB, JB, CO, GG) drafted clinical population, intervention, comparator, and outcome (PICO) questions (See Supplementary Appendix 2). For most questions, the critical outcomes were physical function, which refers to the ability to perform both basic and instrumental activities of daily living, and pain. Disease activity was an additional critical outcome for questions pertaining to diet and dietary supplements. Work outcomes were additional critical outcomes for questions pertaining to vocational rehabilitation and work site evaluation and modification. The Literature Review Team performed a systematic literature review for all PICO questions, extracted relevant study data, graded the quality of evidence (high, moderate, low, very low) and produced the evidence report (see Supplementary Appendix 3). A Patient Panel of 12 patients with varying manifestations of RA and varying experiences with the considered interventions for RA management met virtually. This panel was moderated by a member of the core team (JB) and literature review team (LT). The panel reviewed the evidence report (along with a summary and interpretation by the moderator) and provided patient perspectives and preferences for consideration by the Voting Panel. At a separate Voting Panel meeting held virtually, the resulting evidence was reviewed, patient perspectives considered, and recommendations formulated and voted on. Three members of the Patient Panel were also members of the Voting Panel, to ensure the Patient Panel’s perspective was considered when final decisions on the recommendations were made. Rosters of the Core Leadership Team, Literature Review Team, Voting Panel, and Patient Panel are included in Supplementary Appendix 4. These teams included individuals with expertise in epidemiology, exercise physiology, GRADE methodology, integrative medicine, nursing, nutrition, occupational therapy, physical therapy, rheumatology, and social work.
Consensus among the Voting Panel members required ≥70% agreement on both direction (for or against) and strength (strong or conditional) of each recommendation, as per ACR practice. According to GRADE, a recommendation is categorized as strong if the panel is very confident that the benefits of an intervention clearly outweigh the harms (or vice versa); a conditional recommendation denotes uncertainty regarding the balance of benefits and harms, such as when the evidence quality is low or very low, or when the decision is particularly sensitive to individual patient preferences, or when costs are expected to affect the decision. Thus, conditional recommendations refer to decisions in which incorporation of patient preferences and values is an essential element of shared decision making.
Guiding Principles
Eight guiding principles (Table 2) were established by the Core Leadership Team to aid in the preparation of this guideline. These guiding principles specify that integrative interventions considered in this guideline should complement pharmacologic treatments, an interprofessional approach for the management of RA should be used, and shared decision making is needed when caring for people with RA.
Table 2.
Guiding principles.
| Integrative Rheumatoid Arthritis Guideline Guiding Principles |
| Rheumatoid arthritis is a chronic, systemic, inflammatory condition that requires early diagnosis, evaluation, and management to achieve optimal outcomes. Persons with chronic diseases, like rheumatoid arthritis, seek many available therapies to maintain physical function, reduce pain, and improve their quality of life. |
| Rheumatoid arthritis should be treated with disease-modifying anti-rheumatic drugs and follow a treat-to-target management strategy, as detailed in the 2021 American College of Rheumatology (ACR) Rheumatoid Arthritis Pharmacologic Treatment Guidelines (1). |
| Treatment decisions should follow a shared decision-making process. Persons with rheumatoid arthritis present with a variety of manifestations and experiences. |
| Optimum rheumatoid arthritis treatment outcomes are achieved through interprofessional teams providing expert patient-centered care. |
| Recommendations assume no contraindications to listed management strategies. |
| Recommendations pertain to rheumatoid arthritis management. Recommendations do not pertain to clinical situations when patients have alternative indications for listed treatments. Other general health benefits may exist for listed treatments. |
| Surgical interventions are not included in this guideline because there are other guideline efforts that address large joint replacement, and small joint surgeries are not frequently a part of the current management of rheumatoid arthritis. |
| Disease activity and disease activity levels refer to those calculated using an ACR-endorsed RA disease activity measure (2). |
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1.
Fraenkel L, Bathon JM, England BR, St. Clair EW, Arayssi T, Carandang K, et al. 2021 American College of Rheumatology guideline for the treatment of rheumatoid arthritis. Arthritis Care Res 2021;73:924–939.
2.
England BR, Tiong BK, Bergman MJ, Curtis JR, Kazi S, Mikuls TR, et al. 2019 Update of the American College of Rheumatology Recommended Rheumatoid Arthritis Disease Activity Measures. Arthritis Care. 2019;71:1540–1555.
RESULTS/RECOMMENDATIONS
Twenty-eight recommendations were made based on a set of 28 PICO questions. The systematic literature review initially identified 8,994 manuscripts. After screening, 275 manuscripts were mapped to ≥1 PICO question (see flow diagram in Supplementary Appendix 5). The literature review did not identify any evidence fulfilling eligibility criteria for 29% (8/28) of the PICO questions.
Exercise recommendations (Table 3)
Table 3.
Exercise recommendations for the management of rheumatoid arthritis.
| Recommendation | Certainty of Evidence | Based on the Evidence Report of the Following PICO(s) | Page no(s) of Evidence Table(s) in Suppl. App. 3 |
:---: :---: |
| Consistent engagement in exercise is strongly recommended over no exercise. | Moderate | 4–7 | 205–354 |
| Consistent engagement in aerobic exercise is conditionally recommended over no exercise. | Very low to Low | 4 | 205–252 |
| Consistent engagement in aquatic exercise is conditionally recommended over no exercise. | Low | 5 | 253–270 |
| Consistent engagement in resistance exercise is conditionally recommended over no exercise. | Very low | 6 | 271–327 |
| Consistent engagement in mind-body exercise is conditionally recommended over no exercise. | Very low to low | 7 | 328–354 |
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Insert link to Supplementary Appendix 3 on journal website – TBD
Intervention definitions and examples are provided in Table 1.
Consistent engagement in exercise is strongly recommended over no exercise.
This recommendation is based on moderate certainty evidence suggesting improved physical function and pain. Aerobic, resistance, aquatic, and mind-body exercise were considered together in the evidence supporting this recommendation. The exercise type, frequency, intensity, and duration were not formally defined, though the Voting and Patient Panels emphasized “moving regularly”. The specific elements of an exercise intervention should be tailored to each patient at the given time in their disease trajectory, considering their capabilities, access, and other health conditions. National physical activity guidelines can aid such instruction (8).
Consistent engagement in aerobic exercise is conditionally recommended over no exercise.
This recommendation is based on very low to low certainty evidence suggesting improved physical function but moderate certainty evidence of no difference in pain. The recommendation is conditional because of the certainty of evidence and recognizing that patient preferences may vary due to RA disease activity level, the presence of joint damage or deformities, comorbidities, and the cost, access, or burden of engaging in consistent aerobic exercise.
Consistent engagement in aquatic exercise is conditionally recommended over no exercise.
This recommendation is based on low certainty evidence of improvement in physical function but no difference in pain. The recommendation is conditional because of the certainty of evidence, variability in patient preferences related to comfort in water, cost, access, and burden.
Consistent engagement in resistance exercise is conditionally recommended over no exercise.
This recommendation is based on very low to low certainty evidence of improvement in physical function (inferred from performance measures) and pain. The recommendation is conditional because of the certainty of evidence, variability in patient preferences related to joint damage or deformities that may limit participation, access, cost, and burden. The Voting Panel and patient panel emphasized the importance of appropriate prescription and supervision of resistance exercise by physical therapists or other qualified exercise professionals to prevent harm.
Consistent engagement in mind-body exercise (yoga, Tai Chi, qigong) is conditionally recommended over no exercise.
This recommendation is based on very low to low certainty evidence of improvement in physical function but no difference in pain. The recommendation is conditional because of the certainty of evidence, variability in patient preferences, cost, access, and burden.
Rehabilitation recommendations (Table 4)
Table 4.
Rehabilitation interventions for the management of rheumatoid arthritis.
| Recommendation | Certainty of Evidence | Based on the Evidence Report of the Following PICO(s) | Page no(s) of Evidence Table(s) in Suppl. App. 3 |
:---: :---: |
| Participation in comprehensive occupational therapy is conditionally recommended over no comprehensive occupational therapy. | Very low | 17 | 419–437 |
| Participation in comprehensive physical therapy is conditionally recommended over no comprehensive physical therapy. | Very low | 18 | 438–453 |
| For patients with hand involvement, performing hand therapy exercises is conditionally recommended over no hand therapy exercises. | Low | 8 | 355–378 |
| For patients with hand and/or wrist involvement and/or deformity, use of splinting, orthoses, and/or compression is conditionally recommended over no splinting, orthoses, and/or compression. | Very low | 9 | 379–386 |
| For patients with foot and/or ankle involvement, use of bracing, orthoses, and/or taping is conditionally recommended over no bracing, orthoses, and/or compression. | Very low | 10 | 387–408 |
| For patients with knee involvement, use of bracing and/or orthoses is conditionally recommended over no bracing and/or orthoses. | No studies met eligibility criteria | 11 | 409 |
| Use of joint protection techniques is conditionally recommended over no joint protection techniques. | Low | 12 | 410–413 |
| Use of activity pacing, energy conservation, activity modification, and/or fatigue management is conditionally recommended over no activity pacing, energy conservation, activity modification, and/or fatigue management. | No studies met eligibility criteria | 13 | 415 |
| Use of assistive devices is conditionally recommended over no assistive devices. | No studies met eligibility criteria | 14 | 416 |
| Use of adaptive equipment is conditionally recommended over no adaptive equipment. | No studies met eligibility criteria | 15 | 417 |
| Use of environmental adaptations is conditionally recommended over no environmental adaptations. | No studies met eligibility criteria | 16 | 418 |
| For patients who are currently employed or desire to become employed, use of vocational rehabilitation is conditionally recommended over no work interventions. | No studies met eligibility criteria | 21 | 510 |
| For patients who are currently employed or desire to become employed, work site evaluations and/or modifications are conditionally recommended over no work site evaluations and/or modifications. | Low | 22 | 511–517 |
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Insert link to Supplementary Appendix 3 on journal website – TBD
Intervention definitions and examples are provided in Table 1.
Participation in comprehensive occupational therapy (OT) is conditionally recommended over no comprehensive OT.
Participation in comprehensive physical therapy (PT) is conditionally recommended over no comprehensive PT.
These recommendations are conditional based on very low certainty evidence of improvement in pain and physical function, expected variability in patient preferences, burden, access, and cost. In these recommendations, comprehensive refers to the numerous different approaches and interventions that occupational therapists and physical therapists utilize in the assessment and management of people with RA. The comprehensive nature of these interventions also highlights the importance of expertise and experience in tailoring these interventions to the management of RA through a shared decision-making approach. This recommendation applies throughout the RA disease course. Clinicians should discuss the opportunity to refer to OT and/or PT early in the RA disease course with the recognition that OT and/or PT interventions can be tailored to unique patient needs throughout the patient’s experience with RA. Access to OT and PT services (e.g., availability, insurance coverage) may be a barrier to care. Interventions in the subsequent recommendation statements are often included in comprehensive OT and/or PT services.
For patients with hand involvement, performing hand therapy exercises is conditionally recommended over no hand therapy exercises.
This recommendation is conditional based on low certainty evidence of pain reduction and improvement in physical function. Evaluation of the unique needs of the person with RA with hand involvement is best performed by an experienced hand therapist (e.g., a certified hand therapist (CHT) who is typically an occupational or physical therapist with additional training) who can guide the specific design and intensity of the intervention.
For patients with hand and/or wrist involvement and/or deformity, use of splinting, orthoses, and/or compression is conditionally recommended over no splinting, orthoses, and/or compression.
For patients with foot and/or ankle involvement, use of bracing, orthoses, and/or taping is conditionally recommended over no bracing, orthoses, and/or compression.
For patients with knee involvement, use of bracing and/or orthoses is conditionally recommended over no bracing and/or orthoses.
These recommendations are conditional based on very low (no studies met eligibility criteria for this PICO) certainty evidence of improvement in pain and physical function at the respective anatomic sites. Although the Patient Panel discussed the discomfort and burden accompanying the periodic and regular use of these interventions, the Patient Panel and Voting Panel also recognized their potential to reduce pain and improve physical function. In addition, although these interventions are available without a prescription, the Voting Panel recommends their prescription and use under the guidance of an experienced occupational therapist or physical therapist to ensure appropriate item selection and fit.
Use of joint protection techniques is conditionally recommended over no joint protection techniques.
This recommendation is conditional based on low certainty evidence of improvement in pain and function. Experienced healthcare professional guidance in joint protection techniques at various stages of a patient’s experience with RA is vital for this intervention to aid the patient in maintaining physical function. The Voting Panel also stressed the importance of proper patient education in joint protection techniques by occupational or physical therapists.
Use of activity pacing, energy conservation, activity modification, and/or fatigue management is conditionally recommended over no activity pacing, energy conservation, activity modification, and/or fatigue management.
There was no evidence found for this PICO question. However, these interventions are generally safe and may help preserve physical function and manage fatigue. Proper instruction in these approaches by occupational or physical therapists as well as periodic reminders to employ them were suggested by the Patient Panel and Voting Panel.
Use of assistive devices is conditionally recommended over no assistive devices.
Use of adaptive equipment is conditionally recommended over no adaptive equipment.
Use of environmental adaptations is conditionally recommended over no environmental adaptations.
In the absence of evidence addressing these PICO questions, a conditional recommendation was made in favor of using assistive devices/equipment because of the potential for meaningfully improving function and quality of life and the lack of known harms. The timing of the use of interventions, guidance on intervention selection, and education on how to use these interventions should be considered. Involving an occupational or physical therapist can aid these processes and ensure patient safety. The Voting Panel recognized cost and burden as barriers to the use of these interventions.
For patients who are currently employed or want to become employed, use of vocational rehabilitation (training programs to support employment) is conditionally recommended over no vocational rehabilitation.
For patients who are currently employed or want to become employed, work site evaluations and/or modifications are conditionally recommended over no work site evaluations and/or modifications
These recommendations were conditional based on the absence of evidence for vocational rehabilitation and low certainty evidence for work site evaluations and modifications. The Voting Panel recognized the following considerations in implementing worksite evaluations and modifications: 1) the employee/employer relationship regarding health-specific variables and confidentiality, 2) comfort with disclosure of RA to the employer, 3) the requirements of the Americans with Disabilities Act and the Family and Medical Leave Act, 4) the heterogeneity of employer resources and employee job responsibilities, and 5) the variable availability of experienced work and ergonomics specialists.
Diet recommendations (Table 5)
Table 5.
Diet recommendations for the management of rheumatoid arthritis.
| Recommendation | Certainty of Evidence | Based on the Evidence Report of the Following PICO(s) | Page no(s) of Evidence Table(s) in Suppl. App. 3 |
:---: :---: |
| Adherence to a Mediterranean-style diet is conditionally recommended over no formally defined diet. | Low to moderate | 1 | 8–86 |
| Adherence to a formally defined diet, other than Mediterranean-style, is conditionally recommended against. | Very low to moderate | 1 | 8–86 |
| Following established dietary recommendations without dietary supplements is conditionally recommended over adding dietary supplements. | Very low to moderate | 2 | 87–201 |
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Intervention definitions and examples are provided in Table 1.
Adherence to a Mediterranean-style diet is conditionally recommended over no formally defined diet.
The Mediterranean-style diet pattern emphasizes the intake of vegetables, fruits, whole grains, nuts, seeds, and olive oil; moderate amounts of low-fat dairy and fish; and limits added sugars, sodium, highly processed foods, refined carbohydrates, and saturated fats. This recommendation is based on low to moderate certainty of evidence of improvement in pain and no difference in physical function or disease activity. The recommendation is conditional because of the evidence certainty, patient preferences, costs, access, and burden. The Voting Panel recognized the potential benefits of a Mediterranean-style diet for long-term health outcomes (e.g., longevity and cardiovascular disease) that are affected by RA disease activity and the evidence from studies in the general population (9, 10). The expert role of a registered dietician as a member of the interprofessional team is recognized.
Adherence to a formally defined diet, other than a Mediterranean-style diet, is conditionally recommended against.
This recommendation is based on very low to moderate certainty evidence demonstrating no consistent, clinically meaningful benefit on physical function, pain, or disease activity specific to RA for formally defined diets other than a Mediterranean-style diet (listed in Table 1). In addition to the certainty of evidence, this recommendation is conditional because of the burden and costs that accompany adhering to a formally defined diet and patient preferences are expected to differ.
Following established dietary recommendations without dietary supplements is conditionally recommended over adding dietary supplements.
This recommendation for RA management pertains to all dietary supplements considered (listed in Table 1) and is based on very low to moderate certainty evidence demonstrating no consistent, clinically meaningful benefit on physical function, pain, or disease activity specific to RA. The recommendation is conditional because of the certainty of evidence, expected variation in patient preferences, adequacy of nutrient intake through diet, lack of regulation (e.g., U.S. Food and Drug Administration), possibility of harm (e.g., interactions with medications, side effects), and costs. The Voting Panel supported a “food first” approach but recognized the role dietary supplements may serve for bone (e.g., Vitamin D) and cardiovascular (e.g., fish oil) health, which are particularly important in people with RA (11). In this recommendation, established dietary recommendations refer to those produced by the U.S. Department of Agriculture and U.S. Department of Health & Human Services (12) and the American Heart Association (13). Recommendations on folic acid supplementation in the setting of treatment with methotrexate are included in the pharmacologic treatment guidelines (1).
Body weight and weight loss
Given the broad spectrum of weight loss interventions, including lifestyle modification, commercial weight loss programs, pharmacologic therapies, and surgical interventions, the Voting Panel did not vote on recommendations regarding specific weight loss interventions in overweight or obese people with RA specifically for RA management. However, the Voting Panel was unanimous in its support of clinicians engaging in discussion about maintaining a healthy body weight for people with RA to optimize long-term RA and general health outcomes. In RA, obesity is associated with higher disease activity, impairments in physical function, and poorer treatment response, in addition to poor long-term health outcomes (14). General population recommendations on body weight classification and weight loss strategies, for those who are overweight or obese, can serve as a guide for these discussions (15, 16).
Additional integrative intervention recommendations (Table 6)
Table 6.
Additional integrative interventions for the management of rheumatoid arthritis.
| Recommendation | Certainty of Evidence | Based on the Evidence Report of the Following PICO(s) | Page no(s) of Evidence Table(s) in Suppl. App. 3 |
:---: :---: |
| Use of a standardized self-management program is conditionally recommended over no standardized self-management program. | Low | 19 | 454–466 |
| Use of cognitive behavioral therapy and/or mind-body approaches is conditionally recommended over no cognitive behavioral therapy and/or mind-body approaches. | Very low to low | 20 | 467–510 |
| Use of acupuncture is conditionally recommended over no acupuncture. | Low | 23 | 518–538 |
| Use of massage therapy is conditionally recommended over no massage therapy. | Very low | 24 | 539–543 |
| Use of thermal modalities is conditionally recommended over no thermal modalities. | Very low | 25 | 544–563 |
| Using electrotherapy is conditionally recommended against. | Low | 26 | 564–572 |
| Using chiropractic therapy is conditionally recommended against. | No studies met eligibility criteria | 27 | 573 |
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Insert link to Supplementary Appendix 3 on journal website – TBD
Intervention definitions and examples are provided in Table 1.
Use of a standardized self-management program is conditionally recommended over no standardized self-management program.
This recommendation is conditional based on low certainty evidence of improvement in physical function and pain. The Patient Panel described how these programs can be “life changing” and can provide motivation related to several factors that contribute to quality of life including mental wellness and psychological adaptation to disease experience. The availability of and access to these programs as well as their costs were noted as potential barriers.
Use of cognitive behavioral therapy and/or mind-body approaches is conditionally recommended over no cognitive behavioral therapy and/or mind-body approaches
This conditional recommendation is based on very low to low certainty evidence of no consistent improvement in pain and physical function (critical outcomes), but low to moderate certainty evidence of improvement in depression, anxiety, fatigue, and sleep (important outcomes). Although these interventions are beneficial for chronic disease management, access to experienced healthcare professionals, cost, and the burden of using these interventions were recognized barriers.
Use of acupuncture is conditionally recommended over no acupuncture.
This recommendation for using acupuncture is conditional based on low certainty evidence of inconsistent improvements in pain and function. The Patient Panel generally found acupuncture to be of lower value than other considered interventions for RA management based on their disease experiences. For people with RA, the burden, cost, access, and invasiveness may impact the choice to use this intervention.
Use of massage therapy is conditionally recommended over no massage therapy.
This recommendation for using massage therapy is conditional based on very low certainty evidence of improvement of pain. Massage therapy intensity and technique may affect a patient’s experience; therefore, it is best delivered by a provider (e.g., massage therapist, physical therapist) with knowledge and experience of treating people with RA. Burden, cost, access, and short-term duration of benefit should be considered.
Use of thermal modalities is conditionally recommended over no thermal modalities.
This recommendation for using thermal modalities, such as cryotherapy, heat, and therapeutic ultrasound, is conditional based on very low certainty evidence of improvement for pain and physical function. People with RA receive varying levels of benefit from thermal modalities, and patient preferences are expected to vary regarding the choice of a thermal modality. Patients can control and administer many of these modalities at home though others may benefit from guidance from an occupational or physical therapist.
Using electrotherapy is conditionally recommended against.
This recommendation for “not” using electrotherapy modalities, such as transcutaneous electrical nerve stimulation (TENS) and neuro-muscular electrical nerve stimulation, for RA management is conditional based on low certainty evidence of no improvement of pain and physical function specific to RA. While some people with RA may receive benefit from these interventions (e.g., in the setting of comprehensive PT or OT), the Voting Panel recommended against electrotherapy because the evidence was not felt to outweigh the burden and costs.
No recommendation was made by the Voting Panel on the use of vagus nerve stimulation because this invasive procedure is not currently approved by the FDA for RA.
Using chiropractic therapy is conditionally recommended against.
In the absence of evidence, this recommendation for “not” using chiropractic therapy (i.e., chiropractic spinal adjustment) directly for the management of RA is conditional because of the potential cervical spine complications that can occur in people with RA (17), Voting and Patient Panels’ perceived lack of benefit specific to RA, burden, and costs.
Tobacco cessation
Due to existing clinical quality measures for tobacco use screening and cessation (18) and the absence of studies on tobacco cessation in RA that met eligibility criteria, the Voting Panel did not make further recommendations on individual tobacco cessation interventions for the specific management of RA beyond the clinical quality measures. The Voting Panel recognized the well-established harms of tobacco including detrimental effects on RA that include higher disease severity, poorer treatment response, and increased risk of poor long-term disease outcomes (19). Because of the trust that is frequently developed between people with RA and their clinicians and the low success rate of individual tobacco cessation counseling efforts (20), there was unanimous agreement that clinicians caring for people with RA serve an integral role in counseling on tobacco cessation (21).
DISCUSSION
This is the first guideline produced by the ACR on the use of exercise, rehabilitation, diet, and additional integrative interventions in conjunction with DMARDs for the management of RA. This guideline highlights the importance of an interprofessional healthcare team in providing optimal care to people with RA. The recommended interventions do not replace DMARD treatments in accordance with existing ACR pharmacologic treatment guidelines (1) but are intended to be integrated into the comprehensive management of people with RA. Additionally, the recommended interventions were considered for their efficacy for the management of RA outcomes specifically, rather than other general health benefits or alternative medical indications. The guideline is meant to increase patient and clinician awareness of these interventions, provide evidence to inform shared decision making, improve access to the recommended interventions, and inspire much-needed future research into the integrative management of RA to generate higher-certainty evidence for the management of RA.
The one strong recommendation in this guideline was for consistent engagement in exercise. Most people with RA are expected to benefit from engaging in exercise. However, the specific exercise modality should be determined by patient preference and values, considering the potential burden on and capacity of each patient (22). Recommendations for exercise include multiple types (aerobic, aquatic, resistance, mind-body), which is consistent with physical activity guidelines produced by the U.S. Department of Health and Human Services (8). The frequency, intensity, and duration of exercise were not specified in the recommendation statement, recognizing that these specifics will need to be tailored to each person. The U.S. recommendations on exercise and physical activity can serve as a guide to clinicians counseling patients (8). Benefits of exercise and being physically active extend beyond management of RA symptoms, with evidence from the general population suggesting improved longevity, a lower risk of developing chronic diseases that are overrepresented in RA (e.g., cardiovascular disease), and improved mental health (8). Because symptoms and consequences of RA may impact participation (23), more personalized exercise prescription and monitoring may be needed with the assistance of physical therapists and/or clinical exercise physiologists.
Several rehabilitation interventions as well as comprehensive OT and PT were recommended for their benefits on pain, physical function, preserving independence, remaining in work, and safety, although the certainty of evidence was low or very low. Consistently, the Patient Panel emphasized the importance of receiving these interventions from occupational and/or physical therapists to ensure proper use, exemplifying the unique and substantial contributions these professions provide to the interprofessional RA care team. Patient panel members also reported wishing they had been referred to occupational and physical therapists earlier in the disease course. Clinicians should consider discussing rehabilitation interventions and involving occupational and physical therapists early in the disease course. Early referral to these services can provide education to people with RA on ways to independently use these interventions (e.g., exercise, joint protection, energy conservation, assistive and adaptive devices) to self-manage their disease. Ensuring a sufficient workforce of occupational and physical therapists well-versed in the management of RA and access to this care are high priorities.
Dietary patterns and quality have been associated with RA risk and severity in many, though not all, epidemiologic studies (24). Of several diets evaluated in this guideline (e.g., vegan, anti-inflammatory, elimination), only a Mediterranean-style diet had sufficient evidence to be recommended, given the burden and costs that accompany adhering to a formally defined diet. Dietary supplements were heavily debated by the Voting Panel. Ultimately, there was not sufficient evidence to recommend use of dietary supplements for RA management. The Voting Panel supported a “food first” approach, which emphasizes using high-quality foods to obtain necessary nutrients. Although no recommendation was made on weight loss interventions, the Voting Panel was unanimous in their support for maintaining a healthy body weight. Rheumatology clinicians should consider involving registered dieticians as part of the interprofessional care team to assist people with RA who desire to modify their diet as part of their RA management plan.
Additional integrative interventions that were conditionally recommended included standardized self-management programs, CBT and mind body approaches, acupuncture, massage therapy, and thermal modalities. Although the evidence supporting these interventions was of very low to low certainty, these interventions possess few harms and a modest burden for many people. The Patient Panel favored standardized self-management, CBT, mind body approaches, and thermal modalities because they felt empowered by the interventions, were better able to cope with the chronic disease aspects of RA, and/or it provided them a management option they could control independently, often in their own home.
People with RA who are currently using tobacco should be supported in their tobacco cessation journey. The limited efficacy of counseling on tobacco cessation (20) illustrates why it is critical for all members of the interprofessional care team to engage in this practice, which is an existing clinical quality measure (25). There was not sufficient evidence to establish any formal recommendations for RA beyond the existing quality measures.
A broad range of interventions was considered in this guideline. It is highly unlikely that one clinician will possess the necessary expertise in all these areas, which illustrates the importance of assembling an interprofessional healthcare team to best support people with RA. The Patient Panel emphasized that rheumatology clinicians (e.g., physicians, physician assistants, nurse practitioners) are most often the first contact for therapeutic decisions. Thus, it was important to patients that their rheumatology clinician(s) be knowledgeable about these integrative therapies and help guide patients to other members of the interprofessional care team with the relevant expertise. These discussions should take place early in the disease course and provide patients with options of therapies/interventions and the clinicians who could support them through their disease experience. In addition to the rheumatology clinician, potential members of the interprofessional team include physical and occupational therapists, dieticians, clinical exercise physiologists, psychologists, massage therapists, acupuncturists, and others. The timing in the disease course when a patient might need to involve each care team member will depend on their specific situation and preferences.
RA is a chronic disease that imposes considerable costs to affected persons and society (26, 27). The recommended interventions in this guideline are variably covered by health insurance, and many of the costs become the responsibility of the person with RA. We encourage health policymakers to support the coverage of these interventions to support an integrative and comprehensive approach to the management of RA. In addition to cost concerns, the availability of and access to these interventions was a concern of both the Patient Panel and Voting Panel, particularly for underserved populations. Improving access to and ensuring high-quality delivery of these interventions across diverse settings are important endeavors to immediately support. The Voting Panel acknowledged that patients and/or clinicians may have implicit and/or explicit biases regarding the considered interventions that may make them reluctant to recommend or use these interventions (28). While the evidence-based approach used in this guideline can help overcome such biases, clinicians should consider whether such biases may exist and work to reduce them.
The majority of recommendations were conditional because of low certainty evidence for each intervention. Several factors contributed to the low-certainty evidence, including the limited number of studies evaluating relevant interventions, lack of blinding and study attrition, small sample sizes resulting in imprecision, and heterogeneity of study designs (e.g., various interventions, comparators, and outcomes) that prevented pooling results through a formal meta-analysis. Many of these problems resulted from inherent difficulties in conducting clinical trials evaluating the considered interventions. These conditional recommendations indicate that clinicians should engage in shared decision making with patients when deciding whether to use these interventions. The low or very low certainty evidence supporting most recommendations calls for prioritizing research into these interventions and prompted the Core Team and Voting Panel to propose a research agenda (Table 7). Key research agenda items include determining the efficacy, safety, optimal timing, mode of delivery, and personalization of these interventions.
Table 7.
Research agenda for the integrative management of rheumatoid arthritis.
Evaluate the efficacy and safety of integrative therapies for the management of RA. Initial evidence is needed in the areas of activity pacing, energy conservation, activity modification, fatigue management, and vocational rehabilitation. Additional strong evidence from larger, well-designed studies is needed in all other areas.
Determine the appropriate timing of different integrative interventions in the RA disease course.
Examine the delivery, education, and implementation of integrative interventions. For example, evaluating various methods of instruction and training of joint protection and activity pacing interventions.
Establish the cost-effectiveness of different integrative interventions and develop approaches for cost-effective delivery.
Identify barriers to the adoption and implementation of integrative therapies. These may include variability in access, costs, and implicit and/or explicit biases.
Describe the assembly of, communication between, and role delineation among the interprofessional care team delivering pharmacologic and integrative interventions.
Improve access to experienced healthcare professionals who provide integrative interventions.
Determine efficacy and safety of integrative interventions based on RA disease manifestations and pharmacologic therapies, e.g., modifying exercise interventions based on disease activity level or diet based on disease-modifying anti-rheumatic drugs utilized.
Tailor interventions (and their delivery) studied in the general population to persons with RA, e.g., tobacco cessation programs, exercise programs, and weight loss.
Define efficacy and safety of therapies not included in this guideline such as cannabidiol, vaping, and occupational exposures and protections as well as emerging therapies such as vagal nerve stimulation.
Evaluate dietary supplements (especially Vitamin D and Omega-3 fatty acids such as “fish oil”) in the context of dietary intake. The assessment and implementation of diets should focus on being inclusive of different cultures.
Develop vocational rehabilitation programs and work site modifications that improve the ability of persons with RA to work without negative stigma in the workplace.
Establish the efficacy and safety of integrative therapies on extra-articular manifestations and long-term RA-related outcomes. In addition to disease activity, physical function, pain, and quality of life outcomes considered in this guideline, other outcomes that should be considered are longevity, cardiovascular disease, lung disease, cancer, osteoporosis, and infection.
Develop research methodology to study integrative therapies, e.g., defining an adequate control intervention. Publication of research standards for integrative therapies relevant to RA to guide research efforts.
Establish dedicated funding from organizations to study integrative RA therapies and their implementation.
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There are additional limitations to the development of these guidelines beyond the generally low certainty evidence. Studies that were conducted prior to more recent treatment eras (characterized by early diagnosis and a treat-to-target approach) were included in the evidence report and may be less generalizable than more recently completed studies. In addition, although broad expertise was recruited and an extensive list of interventions considered in this guideline, we could not ensure expertise in every area of integrative RA management or consider all possible integrative interventions. Additional integrative interventions can be considered for inclusion in future guidelines. For example, members of the Patient Panel inquired about cannabinoids given the rising prevalence of their use in rheumatic diseases (29, 30). Cannabinoids were not included in this guideline, and emerging evidence for cannabidiol, a pharmacologic therapy that is not FDA-approved for RA, is being synthesized in a living systematic review through a joint U.S. Department of Veterans Affairs and Center for Evidence-Based Policy at Oregon Health & Sciences University (31). In addition, different modes of delivering interventions (e.g., telehealth vs. in-person) were not assessed, as this was beyond the scope of this project.
In summary, we have provided initial recommendations on the management of RA with exercise, rehabilitation, diet, and additional integrative interventions. These recommendations complement existing pharmacologic treatment guidelines that instruct on the use of DMARDs and, together, can guide a shared decision-making approach between patients and clinicians. Interprofessional treatment teams are crucial to implementing these recommendations. The generally low-quality evidence highlights the need for well-designed studies in the area of integrative management of RA. Policy efforts are needed to ensure access to recommended interventions for people with RA from diverse backgrounds and settings. Together, the integrative and pharmacologic guidelines support the comprehensive management of RA in pursuit of optimal outcomes for people living with RA.
Supplementary Material
Appendix 1
NIHMS1972855-supplement-Appendix_1.docx (37.8KB, docx)
Appendix 2
NIHMS1972855-supplement-Appendix_2.docx (36.2KB, docx)
Appendix 3
NIHMS1972855-supplement-Appendix_3.docx (2MB, docx)
Appendix 4
NIHMS1972855-supplement-Appendix_4.docx (32KB, docx)
Appendix 5
NIHMS1972855-supplement-Appendix_5.docx (37.2KB, docx)
Appendix 6
NIHMS1972855-supplement-Appendix_6.docx (66.5KB, docx)
Guidelines and recommendations developed and/or endorsed by the American College of Rheumatology (ACR) are intended to provide guidance for patterns of practice and not to dictate the care of a particular patient. The ACR considers adherence to the recommendations within this guideline to be voluntary, with the ultimate determination regarding their application to be made by the clinician in light of each patient’s individual circumstances. Guidelines and recommendations are intended to promote beneficial or desirable outcomes but cannot guarantee any specific outcome. Guidelines and recommendations developed and endorsed by the ACR are subject to periodic revision as warranted by the evolution of medical knowledge, technology, and practice. ACR recommendations are not intended to dictate payment or insurance decisions, and drug formularies or other third-party analyses that cite ACR guidelines should state this. These recommendations cannot adequately convey all uncertainties and nuances of patient care.
The American College of Rheumatology is an independent, professional medical, and scientific society that does not guarantee, warrant, or endorse any commercial product or service.
SIGNIFICANCE.
This is the first American College of Rheumatology clinical practice guideline addressing integrative interventions for the management of rheumatoid arthritis (RA).
Use of integrative interventions for the management of RA should occur in concert with disease-modifying anti-rheumatic drugs according to ACR guidelines for pharmacologic treatment of RA (1).
This guideline highlights the vital role expert members of interprofessional healthcare teams serve in providing optimal care to people with RA.
People with RA desire awareness of different management options early in their disease course and seek counsel from their interprofessional healthcare team to decide when to employ the use of these management options.
ACKNOWLEDGEMENTS
We thank the patients who (along with authors Deb Constien, Eileen Davidson, and Lawrence “Rick” Phillips) participated in the Patient Panel meeting: Grace M. Becker, Denise Cedar, Judith Flanagan, Carolyn R. Mason, Eileen Julie O’Rourke, Catherine Simons, Sharon A. Sharp, and Sumayya Spencer. We thank the ACR staff, including Regina Parker for assistance in coordinating the administrative aspects of the project and Cindy Force for assistance with manuscript preparation. We thank Janet Waters for her assistance in developing the literature search strategy, as well as performing the initial literature search and update searches. We thank Theresa Wampler Muskardin and Karen Smarr for their thoughtful review and feedback during the project.
Grant Support:
This guideline project was supported by the American College of Rheumatology.
Footnotes
Disclaimer: The findings and conclusions in this report are those of the authors and do not necessarily represent the official position of the Centers for Disease Control and Prevention, the National Institutes of Health, or the Department of Veterans Affairs.
Financial Conflict: Forms submitted as required.
IRB Approval: Approval from Human Studies Committees was not required.
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6756 | https://web.ma.utexas.edu/users/rusin/Putnam/2018/2018-8.pdf | Problems for 2018 University of Texas Putnam Prep Session, week 8 (Nov 8) Functional Equations 1. Suppose f : R →R satisfies the equation f(x + 1) = 1 2 + p f(x) −(f(x))2 for all x. Prove that f is periodic.
2. Let G be a set of functions f : R →R of the form f(x) = ax + b where a ̸= 0 and b are real numbers, such that (a) if f, g ∈G then f ◦g ∈G (b) If f ∈G then f −1 ∈G (c) For each f ∈G there is a point xf ∈R where f(xf) = xf.
Prove that there exists a point x∗∈R with f(x∗) = x∗for all f ∈G.
3. Let P1(x) = x2 −2 and for j > 1 let Pj(x) = P1(Pj−1(x)). Show that for each positive integer n, the solutions of the equation Pn(x) = x are real and distinct.
4. Suppose f : R →R is (n + 1)-times differentiable and that f (k)(a) = f (k)(b) = 0 for every non-negative k < n. Prove that there is a point c ∈[a, b] where f (n+1)(c) = f(c).
5. (Putnam 1959) Find all functions f : C →C for which f(z) + zf(1 −z) = 1 + z for all points z ∈C.
6. Suppose f(m, n) is a real-valued function defined for all pairs of natural numbers, such that (a) f(0, n) = n + 1 (b) f(m + 1, 0) = f(m, 1) (c) f(m + 1, n + 1) = f(m, f(m + 1, n)) for all m, n ≥0. Compute f(4, 2018).
7. Find all differentiable functions f : R →R for which f(x + y) + f(x −y) = 2f(x)f(y) for every x, y ∈R.
8. Find all analytic functions f : [0, 1) →R for which f(x) + f(x2) = x for every x ∈[0, 1). What if the domain of the function must include the point 1 as well?
9. (Putnam 1972) The series X k≥0 xk(x −1)2k k!
may be expanded into a power series P i≥0 aixi. Show that for no value of i do we have ai−1 = ai = ai+1 = 0.
10. Show that there is no function f : R →R which satisfies f(x) + f(y) 2 ≥f x + y 2 + |x −y| for all real x, y.
11. (Putnam 1988) Prove that there is a unique function f : [0, ∞) →[0, ∞) such that f(f(x)) = 6x −f(x) for all x > 0.
12. (Putnam 1997) Suppose g(x) : R →[0, ∞) is given. Show that any solution to the differential equation f ′′(x) + f(x) = −x g(x) f ′(x) is bounded.
13. (Putnam 1979) First, find a solution (not identically zero) of the homogeneous linear differential equation (3x2 + x −1)y′′ −(9x2 + 9x −2)y′ + (18x + 3)y = 0.
(Intelligent guessing of the form of a solution may be helpful.) Then let y = f(x) be the solution of the inhomogeneous differential equation (3x2 + x −1)y′′ −(9x2 + 9x −2)y′ + (18x + 3)y = 6(6x + 1) which has f(0) = 1 and (f(−1) −2)(f(1) −6) = 1. Find integers a, b, c such that (f(−2) −a)(f(2) −b) = c . |
6757 | https://www.math.lsu.edu/~madden/Resources2011/M6302Madden/Lecture-1.pdf | Lecture 1. The Number Line The Number Line is the most important picture in mathematics.
-3 -2 -1 0 1 2 3 -52 -32 -12 12 32 52 To make a number line, one decorates a naked line. Follow these steps: • STAGE I (setting things up) ◦Choose a unit of length.
◦Choose a point on the line.
◦Choose a direction from the point.
Unit Chosen point Chosen direction • STAGE II (labeling the integer points) ◦Lay copies of the unit end-to-end, starting from the chosen point and going in the chosen direction.
◦Label the endpoints of the units with the number of units one crosses to reach them.
0 1 2 3 ◦Lay copies of the unit, end-to-end, starting from the chosen point and going in the opposite of the chosen direction.
◦Label the endpoints of the units with the negative of the number of units one crosses to reach them.
0 1 2 3 -3 -2 -1 • STAGE III (labeling rational points) ◦Divide the unit into 2 equal parts and use this to label the halves.
-3 -2 -1 0 1 2 3 -62 -52 -42 -32 -22 -12 02 12 22 32 42 52 62 ◦Divide the unit into 3 equal parts and use this to label the thirds.
-3 -2 -1 0 1 2 3 -52 -32 -12 12 32 52 -93 -83 -73 -63 -53 -43 -33 -23 -13 03 13 23 33 43 53 63 73 83 93 ◦Etc.
1 How do you divide the unit into n equal parts?
• Draw more than n parallel lines that are spaced so close that the unit can cross all of them.
• Put the one endpoint of the unit on the first line, and tilt it until the other endpoint lies on the (n+1)th line.
Problems Divide into groups of 3 to 5 people. Each group will consider one problem. Before talking, each participant will take 5 minutes to write and answer to his or her group’s question. Discussion begins with each group member reading his/her written answer. After 10 more minutes, we will pick group representatives to report on the group’s work to the class.
1. Why do we need a common denominator when we add fractions?
2. Using the number line, show how we add 2/5 and 7/6. How does the common denominator 30 come into the picture?
3. Why don’t we need a common denominator when we multiply fractions?
4. Use a rectangle to illustrate the meaning of multiplying 2/5 and 7/6.
5. When we divide one fraction by another, why do we invert and multiply?
6. Draw a picture to illustrate what it means to divide one fraction by another. (Compare with question 4.) Advanced discussion and questions for later in the day. Stage II requires making infinitely many marks, so no person could ever complete it. Nonetheless, we can imagine a number line with all the integers labelled.
Now, each substage of STAGE III requires infinitely many steps, but we can also imagine a number line with all the units and all the halves marked. If we can imagine to first two substages complete, the we can also imagine the first hundred substages complete.
1. Suppose I have marked all the multiples of 1/2, of 1/3, . . . etc. and so on to 1/100. What is the smallest distance between neighboring marks?
2. I have marked all the multiples of 1/2, of 1/3, . . . etc. and so on to 1/100. Suppose that I also mark all the positions that I can get to by adding and subtracting these positions. Now, what is the smallest distance between neighboring marks?
3. If we can imagine the first 100 substages complete, then we can imagine all the substages complete.
After completing them all, can we create additional marks on the line by adding and subtracting existing marks? How about multiplying and dividing?
4. After completing STAGE III, are there any positions on the line that are not yet marked?
2 |
6758 | https://en.wikipedia.org/wiki/Complex_logarithm | Jump to content
Complex logarithm
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From Wikipedia, the free encyclopedia
Logarithm of a complex number
In mathematics, a complex logarithm is a generalization of the natural logarithm to nonzero complex numbers. The term refers to one of the following, which are strongly related:
A complex logarithm of a nonzero complex number , defined to be any complex number for which . Such a number is denoted by . If is given in polar form as , where and are real numbers with , then is one logarithm of , and all the complex logarithms of are exactly the numbers of the form for integers . These logarithms are equally spaced along a vertical line in the complex plane.
A complex-valued function , defined on some subset of the set of nonzero complex numbers, satisfying for all in . Such complex logarithm functions are analogous to the real logarithm function , which is the inverse of the real exponential function and hence satisfies eln x = x for all positive real numbers x. Complex logarithm functions can be constructed by explicit formulas involving real-valued functions, by integration of , or by the process of analytic continuation.
There is no continuous complex logarithm function defined on all of . Ways of dealing with this include branches, the associated Riemann surface, and partial inverses of the complex exponential function. The principal value defines a particular complex logarithm function that is continuous except along the negative real axis; on the complex plane with the negative real numbers and 0 removed, it is the analytic continuation of the (real) natural logarithm.
Problems with inverting the complex exponential function
[edit]
For a function to have an inverse, it must map distinct values to distinct values; that is, it must be injective. But the complex exponential function is not injective, because for any complex number and integer , since adding to has the effect of rotating counterclockwise radians. So the points
equally spaced along a vertical line, are all mapped to the same number by the exponential function. This means that the exponential function does not have an inverse function in the standard sense. There are two solutions to this problem.
One is to restrict the domain of the exponential function to a region that does not contain any two numbers differing by an integer multiple of : this leads naturally to the definition of branches of , which are certain functions that single out one logarithm of each number in their domains. This is analogous to the definition of on as the inverse of the restriction of to the interval : there are infinitely many real numbers with , but one arbitrarily chooses the one in .
Another way to resolve the indeterminacy is to view the logarithm as a function whose domain is not a region in the complex plane, but a Riemann surface that covers the punctured complex plane in an infinite-to-1 way.
Branches have the advantage that they can be evaluated at complex numbers. On the other hand, the function on the Riemann surface is elegant in that it packages together all branches of the logarithm and does not require an arbitrary choice as part of its definition.
Principal value
[edit]
Definition
[edit]
For each nonzero complex number , the principal value is the logarithm whose imaginary part lies in the interval . The expression is left undefined since there is no complex number satisfying .
When the notation appears without any particular logarithm having been specified, it is generally best to assume that the principal value is intended. In particular, this gives a value consistent with the real value of when is a positive real number. The capitalization in the notation is used by some authors to distinguish the principal value from other logarithms of
Calculating the principal value
[edit]
The polar form of a nonzero complex number is , where is the absolute value of , and is its argument. The absolute value is real and positive. The argument is defined up to addition of an integer multiple of 2π. Its principal value is the value that belongs to the interval , which is expressed as .
This leads to the following formula for the principal value of the complex logarithm:
For example, , and .
The principal value as an inverse function
[edit]
Another way to describe is as the inverse of a restriction of the complex exponential function, as in the previous section. The horizontal strip consisting of complex numbers such that is an example of a region not containing any two numbers differing by an integer multiple of , so the restriction of the exponential function to has an inverse. In fact, the exponential function maps bijectively to the punctured complex plane , and the inverse of this restriction is . The conformal mapping section below explains the geometric properties of this map in more detail.
The principal value of as an analytic continuation
[edit]
On the region consisting of complex numbers that are not negative real numbers or 0, the function is the analytic continuation of the natural logarithm. The values on the negative real line can be obtained as limits of values at nearby complex numbers with positive imaginary parts.
Properties
[edit]
Not all identities satisfied by extend to complex numbers. It is true that for all (this is what it means for to be a logarithm of ), but the identity fails for outside the strip . For this reason, one cannot always apply to both sides of an identity to deduce . Also, the identity can fail: the two sides can differ by an integer multiple of ; for instance,
but
The function is discontinuous at each negative real number, but continuous everywhere else in . To explain the discontinuity, consider what happens to as approaches a negative real number . If approaches from above, then approaches which is also the value of itself. But if approaches from below, then approaches So "jumps" by as crosses the negative real axis, and similarly jumps by
Branches of the complex logarithm
[edit]
Is there a different way to choose a logarithm of each nonzero complex number so as to make a function that is continuous on all of ? The answer is no. To see why, imagine tracking such a logarithm function along the unit circle, by evaluating as increases from to . If is continuous, then so is , but the latter is a difference of two logarithms of so it takes values in the discrete set so it is constant. In particular, , which contradicts .
To obtain a continuous logarithm defined on complex numbers, it is hence necessary to restrict the domain to a smaller subset of the complex plane. Because one of the goals is to be able to differentiate the function, it is reasonable to assume that the function is defined on a neighborhood of each point of its domain; in other words, should be an open set. Also, it is reasonable to assume that is connected, since otherwise the function values on different components of could be unrelated to each other. All this motivates the following definition:
: A branch of is a continuous function defined on a connected open subset of the complex plane such that is a logarithm of for each in .
For example, the principal value defines a branch on the open set where it is continuous, which is the set obtained by removing 0 and all negative real numbers from the complex plane.
Another example: The Mercator series
converges locally uniformly for , so setting defines a branch of on the open disk of radius 1 centered at 1. (Actually, this is just a restriction of , as can be shown by differentiating the difference and comparing values at 1.)
Once a branch is fixed, it may be denoted if no confusion can result. Different branches can give different values for the logarithm of a particular complex number, however, so a branch must be fixed in advance (or else the principal branch must be understood) in order for "" to have a precise unambiguous meaning.
In some literature, the notation is used to explicitly denote the -th branch of the complex logarithm. This notation is particularly useful when working with multi-valued logarithms in complex analysis and topology. It was first introduced in the paper Unwinding the Branches of the Lambert W function and was later referenced in the work of David Jeffrey.
Branch cuts
[edit]
The argument above involving the unit circle generalizes to show that no branch of exists on an open set containing a closed curve that winds around 0. One says that has a branch point at 0. To avoid containing closed curves winding around 0, is typically chosen as the complement of a ray or curve in the complex plane going from 0 (inclusive) to infinity in some direction. In this case, the curve is known as a branch cut. For example, the principal branch has a branch cut along the negative real axis.
If the function is extended to be defined at a point of the branch cut, it will necessarily be discontinuous there; at best it will be continuous "on one side", like at a negative real number.
The derivative of the complex logarithm
[edit]
Each branch of on an open set is the inverse of a restriction of the exponential function, namely the restriction to the image . Since the exponential function is holomorphic (that is, complex differentiable) with nonvanishing derivative, the complex analogue of the inverse function theorem applies. It shows that is holomorphic on , and for each in . Another way to prove this is to check the Cauchy–Riemann equations in polar coordinates.
Constructing branches via integration
[edit]
The function for real can be constructed by the formula
If the range of integration started at a positive number other than 1, the formula would have to be
instead.
In developing the analogue for the complex logarithm, there is an additional complication: the definition of the complex integral requires a choice of path. Fortunately, if the integrand is holomorphic, then the value of the integral is unchanged by deforming the path (while holding the endpoints fixed), and in a simply connected region (a region with "no holes"), any path from to inside can be continuously deformed inside into any other. All this leads to the following:
If is a simply connected open subset of not containing 0, then a branch of defined on can be constructed by choosing a starting point in , choosing a logarithm of , and defining for each in .
The complex logarithm as a conformal map
[edit]
Any holomorphic map satisfying for all is a conformal map, which means that if two curves passing through a point of form an angle (in the sense that the tangent lines to the curves at form an angle ), then the images of the two curves form the same angle at .
Since a branch of is holomorphic, and since its derivative is never 0, it defines a conformal map.
For example, the principal branch , viewed as a mapping from to the horizontal strip defined by , has the following properties, which are direct consequences of the formula in terms of polar form:
Circles in the z-plane centered at 0 are mapped to vertical segments in the w-plane connecting to , where is the real log of the radius of the circle.
Rays emanating from 0 in the z-plane are mapped to horizontal lines in the w-plane.
Each circle and ray in the z-plane as above meet at a right angle. Their images under Log are a vertical segment and a horizontal line (respectively) in the w-plane, and these too meet at a right angle. This is an illustration of the conformal property of Log.
The associated Riemann surface
[edit]
Construction
[edit]
The various branches of cannot be glued to give a single continuous function because two branches may give different values at a point where both are defined. Compare, for example, the principal branch on with imaginary part in and the branch on whose imaginary part lies in . These agree on the upper half plane, but not on the lower half plane. So it makes sense to glue the domains of these branches only along the copies of the upper half plane. The resulting glued domain is connected, but it has two copies of the lower half plane. Those two copies can be visualized as two levels of a parking garage, and one can get from the level of the lower half plane up to the level of the lower half plane by going radians counterclockwise around 0, first crossing the positive real axis (of the level) into the shared copy of the upper half plane and then crossing the negative real axis (of the level) into the level of the lower half plane.
One can continue by gluing branches with imaginary part in , in , and so on, and in the other direction, branches with imaginary part in , in , and so on. The final result is a connected surface that can be viewed as a spiraling parking garage with infinitely many levels extending both upward and downward. This is the Riemann surface associated to .
A point on can be thought of as a pair where is a possible value of the argument of . In this way, R can be embedded in .
The logarithm function on the Riemann surface
[edit]
Because the domains of the branches were glued only along open sets where their values agreed, the branches glue to give a single well-defined function . It maps each point on to . This process of extending the original branch by gluing compatible holomorphic functions is known as analytic continuation.
There is a "projection map" from down to that "flattens" the spiral, sending to . For any , if one takes all the points of lying "directly above" and evaluates at all these points, one gets all the logarithms of .
Gluing all branches of log z
[edit]
Instead of gluing only the branches chosen above, one can start with all branches of , and simultaneously glue every pair of branches and along the largest open subset of on which and agree. This yields the same Riemann surface and function as before. This approach, although slightly harder to visualize, is more natural in that it does not require selecting any particular branches.
If is an open subset of projecting bijectively to its image in , then the restriction of to corresponds to a branch of defined on . Every branch of arises in this way.
The Riemann surface as a universal cover
[edit]
The projection map realizes as a covering space of . In fact, it is a Galois covering with deck transformation group isomorphic to , generated by the homeomorphism sending to .
As a complex manifold, is biholomorphic with via . (The inverse map sends to .) This shows that is simply connected, so is the universal cover of .
Applications
[edit]
The complex logarithm is needed to define exponentiation in which the base is a complex number. Namely, if and are complex numbers with , one can use the principal value to define . One can also replace by other logarithms of to obtain other values of , differing by factors of the form . The expression has a single value if and only if is an integer.
Because trigonometric functions can be expressed as rational functions of , the inverse trigonometric functions can be expressed in terms of complex logarithms.
In electrical engineering, the propagation constant involves a complex logarithm.
Generalizations
[edit]
Logarithms to other bases
[edit]
Just as for real numbers, one can define for complex numbers and
with the only caveat that its value depends on the choice of a branch of log defined at and (with ). For example, using the principal value gives
Logarithms of holomorphic functions
[edit]
If f is a holomorphic function on a connected open subset of , then a branch of on is a continuous function on such that for all in . Such a function is necessarily holomorphic with for all in .
If is a simply connected open subset of , and is a nowhere-vanishing holomorphic function on , then a branch of defined on can be constructed by choosing a starting point a in , choosing a logarithm of , and defining
for each in .
Notes
[edit]
^ a b c d e f g Ahlfors, Section 3.4.
^ a b c d e f g h Sarason, Section IV.9.
^ Conway, p. 39.
^ Another interpretation of this is that the "inverse" of the complex exponential function is a multivalued function taking each nonzero complex number z to the set of all logarithms of z.
^ Jeffrrey, D.J.; Hare, D.E.G.; Corless, Robert M. (1996). "Unwinding the branches of the Lambert W function" (PDF). The Mathematical Scientist. 21: 1–7.
^ Calkin, Neil J.; Chan, Eunice Y. S.; Corless, Robert M. (2023). Computational Discovery on Jupyter. Society for Industrial and Applied Mathematics. ISBN 978-1-61197-749-3.
^ Lang, p. 121.
^ Strictly speaking, the point on each circle on the negative real axis should be discarded, or the principal value should be used there.
^ Ahlfors, Section 4.3.
^ The notations R and logR are not universally used.
^ Kreyszig, p. 640.
References
[edit]
Calkin, Neil J.; Chan, Eunice Y. S.; Corless, Robert M. (2023). Computational Discovery on Jupyter. Society for Industrial and Applied Mathematics. ISBN 978-1-61197-749-3.
Jeffrey, D. J.; Hare, D. E. G.; Corless, Robert M. (1996). Unwinding the branches of the Lambert W function. The Mathematical Scientist, 21, 1–7.
Ahlfors, Lars V. (1966). Complex Analysis (2nd ed.). McGraw-Hill.
Conway, John B. (1978). Functions of One Complex Variable (2nd ed.). Springer. ISBN 9780387903286.
Kreyszig, Erwin (2011). Advanced Engineering Mathematics (10th ed.). Berlin: Wiley. ISBN 9780470458365.
Lang, Serge (1993). Complex Analysis (3rd ed.). Springer-Verlag. ISBN 9783642592737.
Moretti, Gino (1964). Functions of a Complex Variable. Prentice-Hall.
Sarason, Donald (2007). Complex Function Theory (2nd ed.). American Mathematical Society. ISBN 9780821886229.
Whittaker, E. T.; Watson, G. N. (1927). A Course of Modern Analysis (Fourth ed.). Cambridge University Press.
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6759 | https://www.sciencedirect.com/science/article/abs/pii/S1473309918301245 | Mass chemoprophylaxis for control of outbreaks of meningococcal disease - ScienceDirect
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Cited by (24)
The Lancet Infectious Diseases
Volume 18, Issue 9, September 2018, Pages e272-e281
Review
Mass chemoprophylaxis for control of outbreaks of meningococcal disease
Author links open overlay panel Lucy A McNamara PhD a, Jessica R MacNeil MPH a, Amanda C Cohn MD b, Prof David S Stephens MD c
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Summary
Although vaccination is the main strategy used to control meningococcal disease outbreaks, mass chemoprophylaxis has also been used as an immediate response to outbreaks, either to supplement vaccination or when vaccination is not possible. However, public health guidelines regarding the use of mass chemoprophylaxis for outbreak control vary by country, partly because the impact of mass chemoprophylaxis on the course of an individual outbreak is difficult to assess. We have reviewed data for the use of mass chemoprophylaxis during 33 outbreaks that occurred both in military populations and in communities and non-military organisations. In most outbreaks, no additional cases of meningococcal disease occurred after mass chemoprophylaxis, or cases occurred only in individuals who had not received prophylaxis. A delay of several weeks was common before cases occurred among prophylaxis recipients. Overall, the outbreak reports that we reviewed suggest that mass chemoprophylaxis might provide temporary protection to chemoprophylaxis recipients during outbreaks.
Introduction
Meningococcal disease outbreaks occur when multiple cases are caused by the same meningococcal strain in a community or institution over a short period and the cases are not linked by direct close contact. Depending on the size of the institution and specific circumstances, having just two cases of the same strain can be considered an outbreak, whereas in other circumstances, particularly in community settings, two cases can instead be considered a cluster.1, 2 Although outbreaks account for only 2–5% of meningococcal disease cases in the USA each year (unpublished),2 each outbreak requires an immediate public health response to help prevent additional cases. Vaccination of the population at increased risk against the serogroup responsible for the outbreak is the recommended response for a meningococcal disease outbreak and is the best method to provide individuals with protection for the duration of the outbreak.1, 2
In addition to polysaccharide–protein conjugate vaccines against serogroups A, C, W, and Y, the US licensure of two serogroup B vaccines means that meningococcal vaccines are now available for protection against the three most common disease-causing serogroups in the USA—B, C, and Y—and against serogroups A and W. However, the serogroup B vaccines require multiple doses to achieve maximum protection and might not protect against all serogroup B strains.3, 4, 5 Furthermore, immunity following vaccination can take up to 2 weeks to develop, leaving even vaccinated people susceptible to meningococcal disease during this period. For these reasons, mass antibiotic chemoprophylaxis for the population at risk has been proposed as an additional measure for outbreak control, either to supplement a vaccination campaign or as an independent measure when vaccination is not appropriate. Vaccination might not be appropriate when the vaccine is not expected to help protect against the meningococcal outbreak strain, or when no vaccine is licensed or recommended for the affected age group.
Antibiotic chemoprophylaxis is routinely recommended for close contacts of patients with meningococcal disease, who are estimated to have an up to 1600-fold increased risk of contracting meningococcal disease compared with the general population.6, 7 The aim of chemoprophylaxis for intimate contacts is to eliminate colonisation with Neisseria meningitidis before the bacterium causes invasive disease or is transmitted to other contacts. By contrast, mass chemoprophylaxis is the expansion of chemoprophylaxis beyond close contacts of patients with meningococcal disease to interrupt meningococcal transmission on a broader scale. A key characteristic of meningococcal disease outbreaks is that outbreak cases are not all directly linked by close contact. Instead, an outbreak occurs when transmission and acquisition are continuing throughout a population via asymptomatic carriage of virulent N meningitidis. By reducing asymptomatic carriage in the population, administration of mass antibiotic chemoprophylaxis to the population at risk during an outbreak could, in theory, reduce transmission of the outbreak strain and prevent additional cases of disease. However, mass chemoprophylaxis is logistically challenging and could also have negative consequences, including the development of antibiotic resistance and occurrence of adverse drug reactions. Public health guidelines from various countries differ in their recommendations of mass chemoprophylaxis for outbreak control (panel 1).
The existing data for the effectiveness of mass chemoprophylaxis as an outbreak response have not previously been compiled into a single reference. To provide a resource for future decisions around mass chemoprophylaxis, we have summarised previous reports of the use of mass chemoprophylaxis as a response to meningococcal disease outbreaks.
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Section snippets
Antibiotics for mass chemoprophylaxis
Rifampicin, ceftriaxone, and ciprofloxacin are the primary antibiotics recommended for chemoprophylaxis of meningococcal disease for the close contacts of patients with meningococcal disease in the USA. These antibiotics are most likely to be choices for a mass chemoprophylaxis campaign. However, several other antibiotics have also been used for mass chemoprophylaxis in response to meningococcal disease outbreaks, such as minocycline and ofloxacin (table). The table summarises when antibiotics
Sulfadiazine chemoprophylaxis in military populations and the development of resistance
Mass chemoprophylaxis for meningococcal disease was first used in the 1940s in populations of US military recruits, who frequently had a high incidence of meningococcal disease.13 After an initial report 13 by Kuhns and colleagues showed that mass administration of sulfadiazine was highly effective both in reducing meningococcal carriage and in preventing cases of meningococcal disease, sulfadiazine was used routinely for mass chemoprophylaxis in military populations for the next 20 years.21 In
Rifampicin and minocycline chemoprophylaxis in military populations
After sulfadiazine prophylaxis became ineffective, different antibiotics were introduced for mass chemoprophylaxis of meningococcal disease in military populations. Rifampicin32, 33, 34 or minocycline 35 were used for mass chemoprophylaxis in response to three outbreaks of meningococcal disease reported in military populations (appendix). We excluded one study 36 from this Review in which ciprofloxacin was used in response to a meningococcal disease outbreak, because insufficient detail was
Community and organisational outbreaks
By contrast with military populations, outbreak responses in non-military organisations or in the community are more likely to have challenges with regard to achieving rapid and complete administration of antibiotics or vaccines. Thus, the data from community and organisational outbreaks show substantial variability in how mass chemoprophylaxis was implemented and the extent of population coverage that was achieved.
Conclusions
In 19 of the 33 outbreaks that were reviewed,18, 33, 34, 39, 40, 41, 42, 43, 44, 45, 53, 54, 55, 56, 62 no meningococcal disease cases occurred after mass chemoprophylaxis was implemented. Although additional cases occurred in the remaining outbreaks, these cases often occurred only in individuals who had not received the initial round of chemoprophylaxis, either because they refused the chemoprophylaxis or because they were outside the targeted population.47, 49, 50, 57, 59 Among prophylaxis
Search strategy and selection criteria
We identified references for this Review through searches of Google Scholar and PubMed using the terms “chemoprophylaxis”, “mass chemoprophylaxis”, “mass prophylaxis”, “antibiotic prophylaxis”, “mass antibiotic chemoprophylaxis”, “preventive antibiotic”, “meningococcal”, “meningococcal disease”, “Neisseria meningitidis”, “carriage”, “carrier”, “carrier state”, “nasopharyngeal carriage”, “outbreak”, “quinolones”, “anti-bacterial agents”, “sulfadiazine”, “rifampin”, “minocycline”, “ceftriaxone”,
Recommended articles
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Cited by (24)
MenACWY-CRM conjugate vaccine booster dose given 4–6 years after priming: Results from a phase IIIb, multicenter, open label study in adolescents and adults
2019, Vaccine Citation Excerpt :
Considerable increases in the percentage of participants with hSBA titers ≥1:8 after 6 and 8 days post-booster vaccination have been reported previously [22,23]. Thus, as has already been established , this time interval necessary to observe a rise in protective antibody titers is of particular importance from a public health perspective in cases of outbreak, as antibiotic prophylaxis may still need to be administered to close contacts of patients with IMD, in addition to a booster dose of the MenACWY-CRM vaccine. Prevailing antibody titers 4–6 years after priming with MenACWY-CRM and MenACWY-D vaccines were also evaluated. Show abstract Vaccination strategies against bacterial meningitis vary across countries. In the United States, a single dose of quadrivalent meningococcal conjugate vaccine (MenACWY) is recommended at 11–12 years of age, with a booster dose approximately 5 years later. We assessed immune responses to a booster dose of MenACWY-CRM vaccine after priming with MenACWY-CRM or MenACWY-D vaccines in adolescents and adults. In this phase IIIb, multicenter, open-label study, healthy 15–55-year-olds, who received MenACWY-CRM (N = 301) or MenACWY-D (N = 300) 4–6 years earlier or were meningococcal vaccine-naïve (N = 100), received one MenACWY-CRM vaccine dose. Immunogenicity was evaluated pre-vaccination, 3 or 5 days post-vaccination (sampling subgroups), and 28 days post-vaccination by serum bactericidal activity assay using human complement (hSBA). After vaccination, participants were monitored for 7 days for reactogenicity, 29 days for unsolicited adverse events (AEs), and 181 days for serious AEs and medically-attended AEs. Sufficiency of the immune response to a MenACWY-CRM booster dose was demonstrated; the lower limit of the 1-sided 97.5% confidence interval for percentages of participants with hSBA seroresponse at 28 days post-vaccination was >75% for each serogroup in those primed with either the MenACWY-CRM or MenACWY-D vaccine. Seroresponse was observed in ≥93.24% of primed participants and ≥35.87% of naïve participants 28 days post-vaccination. At 5 days post-booster, among primed participants, hSBA titers ≥1:8 were achieved in ≥47.14% of participants for MenA and in ≥85.52% of participants for MenC, MenW and MenY, and 3.25- to 8.59-fold increases in hSBA geometric mean titers against each vaccine serogroup were observed. No safety concerns were raised throughout the 6-month follow-up period. A booster dose of the MenACWY-CRM vaccine induced a robust and rapid anamnestic response in adolescents and adults, irrespectively of either MenACWY-CRM or MenACWY-D vaccine administered 4–6 years earlier, with an acceptable clinical safety profile. ClinicalTrials.gov registration: NCT02986854. An Audio Summary linked to this article that can be found on Figshare
### Selection of Antibiotics as Prophylaxis for Close Contacts of Patients with Meningococcal Disease in Areas with Ciprofloxacin Resistance - United States, 2024
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2022, International Journal of STD and AIDS
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2020, Frontiers in Pharmacology
### Lessons from mass vaccination response to meningococcal B outbreaks at US universities
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6760 | https://sites.millersville.edu/bikenaga/calculus1/related-rates/related-rates.html | Related Rates
Related rates problems deal with situations in
which several things are changing at rates which are
related. The way in which the rates are related often arises
from geometry, for example.
Example. The radius of a circle increases at 2
light-years per fortnight. At what rate is the area increasing when
the radius is 3 light-years?
The area of a circle is , where r is the radius.
Differentiate this equation with respect to t:
I want to know the rate of change of the area, which is .
The rate at which the radius increases is , so . I want to find when . Plug in:
The units should be square light-years per fortnight.
Notice that I don't plug into the equation . I differentiate first, then plug in.
In most related rates problems, you will perform the steps above:
Differentiate a starting equation with respect to t (time), then plug
in and answer the question. In most cases, the difficult part of the
problem is obtaining the starting equation.
Before I give some examples involving real-world situations, I'll do
an example which just illustrates the kind of mathematics that is
involved.
Example. x and y are functions of t, and are
related by
Find the rate at which x is changing when and , if y is decreasing at 3 units
per second.
I differentiate the given equation with respect to t. To save
writing, I'll use to denote and to denote .
Here's how I got some of the terms.
is a product, so I applied the Product Rule to get
. Note that the derivative of y
is , and by the Chain Rule, the derivative of
is
I differentiated using the Chain Rule:
I differentiated using the Chain Rule:
I plug in , , and into (). (Note that is negative, because I was told that y is decreasing.)
Simplify and solve for :
Example. Calvin and Phoebe elope in a hot air
balloon, which rises at a constant rate of 3 meters per second. Five
seconds after they cast off, Phoebe's jilted suitor Bonzo McTavish
races up in his Porsche. He parks 50 meters from the launch pad, and
runs toward the pad at 2 meters per second. At what rate is the
distance between Bonzo and the balloon changing when the balloon is
30 meters above the ground?
Here's a picture:
The triangle is a right triangle, and I've labelled the three sides
with variables. x is the distance from Bonzo to the launch
pad, y is the height of the balloon, and s is the distance from Bonzo
to the balloon. I use variables because the three sides are changing
with time: The balloon rises (y changes), Bonzo runs toward the
launch pad (x changes), and the distance between Bonzo and the
balloon (s) changes.
There is an obvious relationship among the variables --- Pythagoras'
theorem says
Differentiate with respect to t:
Cancel the 2's:
Now I plug in the numbers. Bonzo's velocity is --- negative, because the distance from Bonzo to the
launch pad is decreasing. is the rate
at which the balloon rises. I want to know what is when . I see that I still need x and s.
The balloon was already 15 ( ) meters high when Bonzo
drove up. It has risen an additional meters. Since
it's rising at 3 meters per second,
additional seconds elapse.
During these 5 seconds, Bonzo is running at 2 meters per second, so
he travels meters. That leaves him meters from the launch pad. In other words, .
Finally, using the equation ,
Plug all this stuff into the derivatives equation:
Here are some rules of thumb for setting up a related rates problem:
Draw a picture. Be sure you've drawn it at an arbitrary
instant. (In the previous example, it would be a bad idea to draw the
picture with the balloon on the ground, or with Bonzo just getting
out of his car.)
Label the quantities that are relevant to the problem. Quantities
that are changing should be labelled with
variables.
Don't put numbers on your picture unless they represent
things which aren't changing.
Look for a relationship among the variables. Think of Pythagoras'
theorem, trigonometry, area or volume formulas, and similar
triangles.
Once you have a starting equation, differentiate with respect to
t, Then plug in the given values and solve for the appropriate rate.
In the next example, I'll need the following facts from physics. The
velocity of an object dropped from rest is feet per second after t seconds. During that t
seconds, it will have fallen feet.
Example. A bowl of petunias is dropped from a
point 12 feet above the ground, so that it will land 4 feet from the
base of a 12 foot lamppost. At what rate is the shadow of the bowl
moving along the ground when the bowl is 8 feet above the ground?
Here's a picture:
x is the distance from the shadow to the post. (In order to measure
the rate at which the shadow is moving, I have to measure its
distance to some fixed object.) h is the height of the bowl above the
ground.
By similar triangles,
(Notice that if you label the picture properly, it's usually obvious
what mathematical formula you should use.)
Clear denominators:
Differentiate with respect to t, being careful to use the Product
Rule on the right side:
Now I plug the numbers in. I'm trying to find , the rate at which the shadow is moving.
When the bowl is 8 feet above the ground, , so the bowl has fallen 4 feet. By the physics fact
I mentioned earlier, the bowl falls feet in t seconds;
setting , I get . The
bowl has been falling for second.
By the other physics fact I mentioned, its velocity is feet per second. Thus, --- negative, because h is decreasing.
The only other thing I need is x. Set in :
Substitute x, h, and :
Example. Calvin and Bonzo drive on long
parallel roads in opposite directions at 20 feet per second and 80
feet per second, respectively. The roads are 200 feet apart. How
rapidly is the distance between them changing 5 seconds after they
pass one another?
Here's a picture:
x is the horizontal separation between Calvin and Bonzo. s is the
distance between them.
By Pythagoras,
Differentiate with respect to t:
I want .
The rate of change of the distance between them is the same as it
would be if Bonzo was moving at feet per
second and Calvin was stationary. So , and 5
second after they pass each other, .
To find s, plug into :
Now plug everything in:
Example. A ferris wheel 50 feet in diameter
makes 2 revolutions per minute. Assume that the wheel is tangent to
the ground and let P be the point of tangency. At what rate is the
distance between P and a rider changing, when she is 25 feet above
the ground and going up?
Here's a picture:
is the angle between the vertical and a radius out to
the rider. s is the distance between the bottom of the wheel and the
rider.
By the Law of Cosines,
Differentiate with respect to t:
Assuming that the wheel is turning counterclockwise in the picture,
when the rider is 25 feet above the ground, she's on the far right
hand side. At this point, , and (since the triangle is then an isosceles right
triangle).
is the rate at which the wheel is turning.
I must be careful to express this in radians per minute:
Plug in:
Example. A spotlight on the top of a police
cruiser makes one revolution per second. The spotlight is 40 feet
from a long straight wall. At what rate is the spot of light moving
across the wall at the instant when the beam makes a angle with the wall?
Here's a picture:
x is the distance from the spot to the point on the wall nearest the
cruiser. is the angle between the beam and the
altitude to the wall from the light.
By basic trigonometry,
Differentiate with respect to t:
When the beam makes a angle with the wall, . The rate at which the beam is
turning is
Plug in:
Example. A helium balloon is released from
ground level and rises at a constant rate of 2 feet per second.
Calvin stands at the top of a 100-foot observation tower located 60
feet from the point where the balloon was released. At what rate is
the distance between Calvin and the balloon changing when the balloon
is 55 feet above the ground?
Let y be the height of the balloon above the ground, and let r be the
distance from Calvin to the balloon.
By Pythagoras,
Differentiate with respect to t:
I want . Now feet per
second; when ,
Thus,
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6761 | https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Notes/CAIE/A2-Inorganic-Chemistry/Detailed/28.%20Chemistry%20of%20Transition%20Elements.pdf | CAIE Chemistry A-level 28: Chemistry of Transition Elements (A-level only) Notes This work by PMT Education is licensed under CC BY-NC-ND 4.0 General Physical Properties of the First Row of Transition Elements, Titanium to Copper Transition Elements Transition elements are found in the d-block of the periodic table. Transition elements form one or more stable ions with incomplete d orbitals. The d subshell can hold up to 10 electrons. A d orbitals looks like this: CK-12 Foundation, CC BY-SA 3.0 Scandium and zinc Scandium and zinc are in the d block of the periodic table but they are not transition elements because their ions do not have an incomplete d subshell. Scandium forms one stable ion, Sc3+, which has an empty d sub shell. Electron configurations Sc: [Ar]3d14s2 → Sc3+: [Ar] Zinc also only forms one stable ion, Zn2+, which has a full d sub shell. Electron configurations Zn: [Ar]3d104s2 → Zn2+: [Ar]3d10 Transition element Atom electron configuration Ion electron configuration Titanium Ti: [Ar]3d24s2 Ti2+: [Ar]3d2 Vanadium V: [Ar]3d34s2 V2+: [Ar]3d3 also common: V3+, V4+ and V5+ Chromium Cr: [Ar]3d54s1 Cr3+: [Ar]3d3 also common: Cr2+ and Cr6+ Manganese Mn: [Ar]3d54s2 Mn2+: [Ar]3d5 also common: Mn3+, Mn4+, Mn6+ and Mn7+ Iron Fe: [Ar]3d64s2 Fe2+: [Ar]3d6 also common: Fe3+ Cobalt Co: [Ar]3d74s2 Co2+: [Ar]3d7 also common Co3+ Nickel Ni: [Ar]3d84s2 Ni2+: [Ar]3d8 also common: Ni+ Copper Cu: [Ar]3d104s1 Cu2+: [Ar]3d10 also common: Cu+ It is important to remember that the 4s orbital is filled before the 3d so the 4s electrons are removed before the 3d electrons. Also remember that the electron configurations of copper and chromium are different than expected as the 4s orbital only has one electron. This is so that the 3d subshell is filled for copper and half filled for chromium, as these are more stable electron configurations. Melting Points and Densities All transition metals have similar physical properties such as high melting points and high densities. Melting point Transition metals have higher melting points that metals found in the s-block, like calcium. This is because the transition metals have the extra 3d electrons so metallic bonding is stronger, requiring more energy to overcome. Density As you go across the periodic table, nuclear charge increases. This draws the electrons inwards slightly, decreasing the atomic radius. As a result, transition metals have higher densities than metals in the s-block because the atoms are smaller so atoms fit into a given volume compared with s-block metal atoms. Density is also greater because atomic mass increases across the period. Variable Oxidation States Transition metals can exist in variable oxidation states. For example, iron forms two common ions, Fe2+ (+2 oxidation state) and Fe3+ (+3 oxidation state). Variable oxidation states form because the electrons in the 4s and 3d orbitals have very similar energies. This means that a relatively similar amount of energy is required to gain or lose different numbers of electrons. General Characteristic Chemical Properties of the First Set of Transition Elements, Titanium to Copper Transition elements and their complexes function as catalysts because of their ability to have variable stable oxidation states and they have vacant d orbitals that are energetically accessible and can form dative bonds with ligands. Transition elements form coloured compounds due to the way that the part filled d orbitals split to different energies. Reactions with Ligands to Form Complexes ● Ligand: a species that has a lone pair of electrons that forms a dative covalent bond with a central metal atom or ion. Monodentate ligands form one coordinate bond, bidentate ligands form two and polydentate ligands form more than two. ● Complex: a molecule containing a central metal atom or ion surrounded by one or more ligands. ● Coordination number: the number of dative bonds formed with the central metal ion. Reactions with water: Transition metal ions react with monodentate water ligands to form complexes. Cu2+ + 6H2O → [Cu(H2O)6]2+ Observation: Blue solution formed Co2+ + 6H2O → [Co(H2O)6]2+ Observation: Pink solution formed Reactions with ammonia: Metal aqua ions react with aqueous ammonia to form precipitates. Ammonia molecules are monodentate ligands. Copper and cobalt undergo two reactions when ammonia is added to their metal aqua ion. The second reaction only occurs when there is an excess of ammonia. Aqua ion Equation Observation sol=solution / ppt=precipitate Cu2+ [Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4 + Excess: [Cu(H2O)4(OH)2]+4NH3 → [Cu(H2O)2(NH3)4]+2H2O+2OH- Blue sol → light blue ppt Light blue ppt → dark blue sol Co2+ [Co(H2O)6]2+ + 2NH3 → [Co(H2O)4(OH)2] + 2NH4 + Excess: [Co(H2O)4(OH)2] + 6NH3 → [Co(NH3)6]2+ + 4H2O + 2OH- Pink sol → blue ppt Blue ppt → yellow brown sol The ammonia ligands make Co(II) unstable so Co(II) is oxidised to Co(III) in air. [Co(NH3)6]2+ becomes [Co(NH3)6]3+ (red-brown solution). Reactions with hydroxide ions Metal aqua ions react with hydroxide ions to form precipitates. Hydroxide ions are monodentate ligands. Aqua ion Equation Observation sol=solution / ppt=precipitate Cu2+ [Cu(H2O)6]2+ + 2OH-→ [Cu(H2O)4(OH)2] + 2H2O Blue sol → light blue ppt Co2+ [Co(H2O)6]2+ + 2OH-→ [Co(H2O)4(OH)2] + 2H2O Pink sol → blue ppt Reactions with chloride ions Chloride ions act as monodentate ligands when they react with metal aqua ions. They are charged and larger so only four Cl- ions are able to fit around the metal ion. This means the coordination number decreases from 6 to 4. Aqua ion Equation Observation sol=solution / ppt=precipitate Cu2+ [Cu(H2O)6]2+ + 4Cl-→ [CuCl4]2- + 6H2O Blue sol → yellow sol Co2+ [Co(H2O)6]2+ + 4Cl-→ [CoCl4]2- + 6H2O Pink sol → blue sol Geometry of Transition Metal Complexes Transition metal complexes can be described as linear, octahedral, tetrahedral or square planar, depending on the ligands and the metal ions in the complex. Linear A linear complex is most commonly formed by a silver metal ion with two coordinate bonds. The bond angles are 180o. Octahedral An octahedral complex is formed when a metal ion has 6 coordinate bonds. This happens with water and ammonia because they are small so 6 ligands are able to surround the central metal ion. Two octahedral complexes are shown below, one with a single type of ligand and the second with a mixture of ligands. The bond angles are 90o. Tetrahedral If there are four coordinate bonds in a complex, like [CuCl4]2-, the shape is usually tetrahedral. The bond angles are 109.5o. Square planar Sometimes a square planar shape when there are four coordinate bonds instead of a tetrahedral shape. The bond angle is 90o. The compound shown is cisplatin, an anti cancer drug. Ligand Exchange When ligands swap places with one another, this is called ligand exchange or ligand substitution. The coordination number and shape of the ion stays the same if the ligands are a similar size and charge. Some examples of ligand substitution reactions are shown below: [Cu(H2O)6]2+ + 4Cl- → [CuCl4]2- + 6H2O ● Blue solution → yellow solution ● Octahedral → tetrahedral ● Coordination number reduces from 6 to 4 because of the change in size of the ligands. [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)(H2O)2]2+ + 4H2O ● This exchange only occurs with excess ammonia. If it is not in excess then [Cu(H2O)4(OH)2] precipitate forms. ● Blue solution → dark blue solution. ● Shape stays as an octahedral. ● Coordination number remains at 6. Redox Systems Redox reactions can be used in titrations. Potassium manganate(VII) Potassium manganate(VII) contains the MnO4 - ion and is an oxidising agent commonly used in redox titrations. The half equation for this redox system is shown below: MnO4 - + 8H+ + 5e- → Mn2+ + 4H2O The titration: ● Purple potassium manganate(VII) is placed in a burette. The sample being analysed is placed in the conical flask with excess dilute sulfuric acid. ● As potassium manganate(VII) is added from the burette, a reaction occurs, forming colourless Mn2+. The end point is when the solution is first tinted pink. Potassium dichromate(VI) Potassium dichromate(VI) solution can be acidified with dilute sulfuric acid. Potassium dichromate is an oxidising agent. The half equation for this redox system is shown below: Cr2O7 2- + 14H+ + 6e- → 2Cr3+ + 7H2O The titration: ● Orange potassium dichromate(VI) is placed in a burette. The sample to be analysed is placed in the conical flask with excess dilute sulfuric acid. ● As potassium dichromate(VI) is added from the burette, a reaction occurs, forming green Cr3+. The end point is when the Cr2O7 2- stops reacting and doesn’t turn green. This can be seen when the solution in the flask starts becoming discoloured due to the orange solution being added. Conditions Acidic conditions are required for both oxidising agents to work so that H+ ions are present. Dilute sulfuric acid is chosen for several reasons: -Hydrochloric acid can not be used because the oxidising agent will cause the formation of Cl2. As Cl2 is a gas, this will affect the volume of the oxidising agent in the reaction mixture. -Concentrated sulfuric acid or concentrated nitric acid can not be used because they are only weak oxidising agents. -Ethanoic acid can not be used because it is a weak acid so not enough H+ ions would be released. Fe2+/Fe3+ can be analysed with acidified potassium manganate(VII) The half equations: Fe2+ → Fe3+ + e- MnO4 - + 8H+ + 5e- → Mn2+ + 4H2O To write an overall equation, the first equation must be multiplied by 5 so that there are equal numbers of electrons in each equation. Combine the equations, cancelling out any terms which are the same on both sides: 5Fe2+ + MnO4 - + 8H+ → 5Fe3+ + Mn2+ + 4H2O C2O4 2- can be analysed with acidified potassium manganate(VII) The half equations: MnO4 - + 8H+ + 5e- → Mn2+ + 4H2O C2O4 2-→ 2CO2 + 2e- To write an overall equation, the first equation must be multiplied by 2 and the second multiplied by 5 so that there are equal numbers of electrons in each equation. Combine the equations, cancelling out any terms which are the same on both sides: 2MnO4 - +16H+ +5C2O4 2-→ 10CO2 +2Mn2+ +8H2O Predicting the Likelihood of Redox Reactions The redox potential of an ion tells you how easily it can be reduced to a lower oxidation state. The smaller the redox potential, the more stable the ion is so the less likely it will be reduced. E.g. Half equation Standard electrode potential, E (V) Cu2+ + e- → Cu+ +0.15 Cr3+ + e- → Cr2+ -0.41 For the table above, the Cr3+ has a smaller redox potential so it is more stable. This means Cu2+ will be reduced because it is more unstable since it is a larger value of E . The species with the smaller/ more negative E value will always be oxidised so Cr2+ is oxidised in the example above. This means that Cu2+ will react with Cr2+ but Cr3+ will not react with Cu+. Colour of Complexes Splitting of Degenerate d Orbitals Degenerate orbitals are electron orbitals that have the same energy level. In an isolated metal atom or ion the five d orbitals are degenerate because they all have the same energy level. When ligands join to a metal ion, the electric fields of the ligands repel the d orbitals, causing the energy levels of the d orbitals to split. They gain/ lose different amounts of energy as some d orbitals are closer to the ligands so they experience more repulsion meaning their energy level is increased more than those further away from the ligands. In a tetrahedral complex none of the orbitals point directly to the ligands so the splitting energy is less than that of octahedral complexes. The diagram on the right is a simple representation of how the d orbitals, originally at the same level in the centre, split for both tetrahedral and octahedral complexes. Another example of this can be seen in the diagram below for Ni2+. Origin of Coloured Complexes When visible light hits a transition metal ion, electrons are excited to higher energy levels. Some frequencies of the visible light are absorbed when electrons jump up to higher orbitals. The rest of the frequencies of visible light are transmitted or reflected. These frequencies combine to make the complement of the colour of the absorbed frequencies. This causes the colour of the complex seen. Effect of Ligands on Absorption Different ligands cause different amounts of visible light to be absorbed. This means that different ligands in metal complexes cause different colours to be seen. The frequencies of light absorbed depend on the size of the energy gap, E. Each ligand produces an energy gap with a different size, which depends on the repulsion that the d orbitals experience from the ligands. This can be seen with coloured complexes containing copper ions. For the example of [Cu(H2O)6]2+, ions absorb light from the red part of the visible light spectrum, so blue is the colour seen as it is the complementary colour formed from the remaining frequencies. [Cu(H2O)6]2+ → Blue solution [CuCl4]2- → Yellow solution [Cu(NH3)4(H2O)2]2+ → Dark blue solution Stereoisomerism in Transition Metal Complexes Stereoisomerism Cis-trans isomerism Square planar complexes with two pairs of ligands show cis-trans isomerism. A trans isomer is when the two paired ligands are opposite each other in the complex. A cis isomer is when the pair of ligands are next to each other. The image below shows cis-platin and trans-platin: CISPLATIN TRANSPLATIN Octahedral complexes with two monodentate ligands of one type and four monodente ligands of another type also show cis-trans isomerism: Trans-[NiCl2(H2O)4] Cis-[NiCl2(H2O)4] Optical Isomerism Optical isomerism is where an ion can exist in two forms which are non-superimposable mirror images. This occurs in octahedral complexes with three bidentate ligands. For example when Ni2+ bonds with the bidentate ligand NH2CH2CH2NH2 . ‘25.3: Coordination Compounds’, Chemistry LibreTexts CC BY-NC-SA 3.0 US Cisplatin Cisplatin is a complex of platinum(II) containing two ammonia ligands and two chloride ion ligands. It has a square planar shape, shown on the previous page. Cisplatin is used as an anticancer drug. Cancer is caused by cells dividing uncontrollably, forming tumours. Cisplatin binds to DNA, causing kinks in the DNA helix. This stops the DNA being copied properly which in turn stops cancerous cells reproducing. How it works: ● DNA contains bases, one of which is guanine. A nitrogen atom in guanine forms a coordinate bond with the platinum ion in cisplatin, replacing a chloride ion ligand. ● A second nitrogen from a different guanine base bonds to the platinum ion, replacing the other chloride ion ligand. ● The cisplatin has bonded to the DNA strands, creating a kink in the strands and preventing correct cell replication. Issues: ● Cisplatin can also bind to DNA in healthy cells. This is a problem for cells which replicate frequently, like hair cells and blood cells. This is why the drugs can cause hair loss and why it can suppress the immune system. ● The drug can also cause kidney damage. The issues with the drug are minimised by giving patients small dosages and by targeting the tumour directly. Stability Constants, Kstab Ligand Exchange in Terms of Competing Equilibria When there is competing equilibria in ligand exchange reactions, the reaction which forms the most stable complex will be prioritised. Example The two equilibria reactions below are being carried out in the same system: [Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O [Cu(NH3)4(H2O)2]2+ is the most stable product so the solution will eventually turn dark blue. The [Cu(H2O)6]2+ ions will react with the ammonia to form the more stable product. The Stability Constant, Kstab The stability constant, Kstab, is the equilibrium constant (Kc) for the formation of the complex ion in a solvent from its constituent ions or molecules. ● Calculating overall stability: [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O As can be seen from the stability expression on the left, it has a similar layout to the equilibrium constant, Kc. The square brackets of the complex have been removed so the square brackets in the expression indicate concentrations in mol dm-3. The concentration of water also is not included on the top of the equation as one of the products. This is because everything is dissolved in water so water is in excess. The water produced as a product is minimal by comparison. ● Calculating individual stability constants: This is the overall equation: [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O However, this reaction takes part in four steps: [Cu(H2O)6]2+ + NH3 ⇌ [Cu(NH3)(H2O)5]2+ + H2O [Cu(NH3)(H2O)5]2+ + NH3 ⇌ [Cu(NH3)2(H2O)4]2+ + H2O [Cu(NH3)2(H2O)4]2+ + NH3 ⇌ [Cu(NH3)3(H2O)3]2+ + H2O [Cu(NH3)3(H2O)3]2++ NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + H2O The stability constant for the first reaction can be called K1: K2, K3 and K4 can also be found for the reactions that follow. The larger the value of Kstab, the more stable the complex ion formed. The larger the value of Kstab, the further right of the equilibrium the reaction lies. |
6762 | https://www.ck12.org/book/ck-12-algebra-i-concepts-honors/r9/section/3.9/ | Vertex Form of a Quadratic Function | CK-12 Foundation
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3.9 Vertex Form of a Quadratic Function
Difficulty Level: At Grade | Created by: CK-12
Last Modified: Jan 30, 2013
Read Resources Details Attributions
Given the equation y=3(x+4)2+2, list the transformations of y=x 2.
Watch This
James Sousa: Find the Equation of a Quadratic Function from a Graph
Guidance
The equation for a basic parabola with a vertex at (0,0) is y=x 2. You can apply transformations to the graph of y=x 2 to create a new graph with a corresponding new equation. This new equation can be written in vertex form. The vertex form of a quadratic function is y=a(x−h)2+k where:
|a| is the vertical stretch factor. If a is negative, there is a vertical reflection and the parabola will open downwards.
k is the vertical translation.
h is the horizontal translation.
Given the equation of a parabola in vertex form, you should be able to sketch its graph by performing transformations on the basic parabola. This process is shown in the examples.
Example A
Given the following function in vertex form, identify the transformations of y=x 2.
y=−1 2(x−2)2−1
Solution:
a – Is a negative? YES. The parabola will open downwards.
a – Is there a number in front of the squared portion of the equation? YES. The vertical stretch factor is the absolute value of this number. Therefore, the vertical stretch of this function is 1 2.
k – Is there a number after the squared portion of the equation? YES. The value of this number is the vertical translation. The vertical translation is -1.
h – Is there a number after the variable ‘x’? YES. The value of this number is the opposite of the sign that appears in the equation. The horizontal translation is +2.
Example B
Given the following transformations, determine the equation of the image of y=x 2 in vertex form.
Vertical stretch by a factor of 3
Vertical translation up 5 units
Horizontal translation left 4 units
Solution:
a – The image is not reflected in the x-axis. A negative sign is not required.
a – The vertical stretch is 3, so a=3.
k – The vertical translation is 5 units up, so k=5.
h – The horizontal translation is 4 units left so h=−4.
The equation of the image of y=x 2 is y=3(x+4)2+5.
Example C
Using y=x 2 as the base function, identify the transformations that have occurred to produce the following image graph. Use these transformations to write the equation in vertex form.
[Figure 1]
Solution:
a – The parabola does not open downward so a will be positive.
a – The y-values of 1 and 4 are now up 3 and up 12. a=3.
k – The y−coordinate of the vertex is -5 so k=−5.
h – The x−coordinate of the vertex is +3 so h=3.
The equation is y=3(x−3)2+5.
Example D
When the equation of the basic quadratic function is written in vertex form, the function can also be expressed in mapping notation form. This form describes how to obtain the image of a given graph by using the changes in the ordered pairs.
The standard base table of values for the base quadratic function y=x 2 is given by:
X−3−2−1 0 1 2 3 Y 9 4 1 0 1 4 9
When these ordered pairs are plotted, we get the base parabola. The mapping rule used to generate the image of a quadratic function is (x,y)→(x′,y′) where (x′,y′) are the coordinates of the image graph. The resulting mapping rule from the equation y=a(x−h)2+k is (x,y)→(x+h,a y+k). A mapping rule details the transformations that were applied to the coordinates of the base function y=x 2.
Given the following quadratic equation, y=2(x+3)2+5 write the mapping rule and create a table of values for the mapping rule.
Solution:
The mapping rule for this function will tell exactly what changes were applied to the coordinates of the base quadratic function.
y=2(x+3)2+5:(x,y)→(x−3,2 y+5)
[Figure 2]
These new coordinates of the image graph can be plotted to generate the graph.
Concept Problem Revisited
Given the equation y=3(x+4)2+2, list the transformations of y=x 2.
a=3 so the vertical stretch is 3. k=2 so the vertical translation is up 2. h=−4 so the horizontal translation is left 4.
Vocabulary
Horizontal translation The horizontal translation is the change in the base graph y=x 2 that shifts the graph right or left. It changes the x−coordinate of the vertex. Mapping Rule The mapping rule is another form used to express a quadratic function. The mapping rule defines the transformations that have occurred to the base quadratic function y=x 2. The mapping rule is (x,y)→(x′,y′) where (x′,y′) are the coordinates of the image graph. Transformation A transformation is any change in the base graph y=x 2. The transformations that apply to the parabola are a horizontal translation, a vertical translation, a vertical stretch and a vertical reflection. Vertex form of y=x 2 The vertex form ofy=x 2 is the form of the quadratic base function y=x 2 that shows the transformations of the image graph. The vertex form of the equation is y=a(x−h)2+k. Vertical Reflection The vertical reflection is the reflection of the image graph in the x-axis. The graph opens downward and the y-values are negative values. Vertical Stretch The vertical stretch is the change made to the base function y=x 2 by stretching (or compressing) the graph vertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the original base graph of y=x 2. Vertical Translation The vertical translation is the change in the base graph y=x 2 that shifts the graph up or down. It changes the y−coordinate of the vertex.
Guided Practice
Identify the transformations of y=x 2 for the quadratic function −2(y+3)=(x−4)2
List the transformations of y=x 2 and graph the function =−(x+5)2+4
Graph the function y=2(x−2)2+3 using the mapping rule method.
Answers:
a – a is negative so the parabola opens downwards.
a – The vertical stretch of this function is 1 2.
k – The vertical translation is -3.
h – The horizontal translation is +4.
[Figure 3]
a a k h→n e g a t i v e→1→+4→−5
Mapping Rule(x,y)→(x+2,2 y+3)
Make a table of values:
| x→x+2 | | y→2 y+3 | |
--- --- |
| −3 | -1 | 9 | 21 |
| -2 | 0 | 4 | 11 |
| -1 | 1 | 1 | 5 |
| 0 | 2 | 0 | 3 |
| 1 | 3 | 1 | 5 |
| 2 | 4 | 4 | 11 |
| 3 | 5 | 9 | 21 |
Draw the Graph
[Figure 4]
Practice
Complete the following table to identify the transformations of y=x 2 in each of the given functions:
| Number | a | k | h |
--- --- |
| 1. | | | |
| 2. | | | |
| 3. | | | |
| 4. | | | |
| 5. | | | |
y=4(x−2)2−9
y=−1 6 x 2+7
=−3(x−1)2−6
y=1 5(x+4)2+3
y=5(x+2)2
Graph the following quadratic functions using the mapping rule method:
y=2(x−4)2−5
y=−1 3(x−2)2+6
y=−2(x+3)2+7
y=−1 2(x+6)2+9
y=1 3(x−4)2
Using the following mapping rules, write the equation, in vertex form, that represents the image of y=x 2.
(x,y)→(x+1,−1 2 y)
(x,y)→(x+6,2 y−3)
(x,y)→(x−1,2 3 y+2)
(x,y)→(x+3,3 y+1)
(x,y)→(x−5,−1 3 y−7)
Notes/Highlights
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Description
This concept explores writing equations for quadratic functions in vertex form.
Learning Objectives
Here you will learn to write the equation for a parabola that has undergone transformations.
Difficulty Level
At Grade
author
Brenda Meery,Kaitlyn Spong
Tags
Transformation,vertex form,vertical stretch,horizontal translation,vertical translation,vertical reflection,mapping rule, (4 more)
Subjects
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Periodic Trends
Barium Hydroxide
Strong Bases
Nitric Acid
Acidic Strength
Sodium Hydroxide
Weak and Strong Acids and...
5
Is there a trend to follow for ranking the strength of strong acids and bases? (acids: HClO4, HCl, HBr, HI, HNO3, H2SO4; bases: LiOH, NaOH, KOH, CaOH2, SrOH2, BaOH2)
·
Yes, there are general trends for ranking the strength of strong acids and strong bases.
Strong Acids
For the strong acids you listed (HClO₄, HCl, HBr, HI, HNO₃, H₂SO₄), the trend in strength is largely determined by the stability of the conjugate base and the bond strength of the H-X bond (where X is the conjugate base). The general order of strength for these acids is as follows:
HClO₄ (Perchloric acid) - Strongest
H₂SO₄ (Sulfuric acid) - Strong, but depends on concentration
HBr (Hydrobromic acid)
HI (Hydroiodic acid) - Stronger than HBr
HCl (Hydrochloric acid)
HNO₃ (Nitric acid) - Weaker than HCl,
Yes, there are general trends for ranking the strength of strong acids and strong bases.
Strong Acids
For the strong acids you listed (HClO₄, HCl, HBr, HI, HNO₃, H₂SO₄), the trend in strength is largely determined by the stability of the conjugate base and the bond strength of the H-X bond (where X is the conjugate base). The general order of strength for these acids is as follows:
HClO₄ (Perchloric acid) - Strongest
H₂SO₄ (Sulfuric acid) - Strong, but depends on concentration
HBr (Hydrobromic acid)
HI (Hydroiodic acid) - Stronger than HBr
HCl (Hydrochloric acid)
HNO₃ (Nitric acid) - Weaker than HCl, HBr, and HI
Strong Bases
For the strong bases you mentioned (LiOH, NaOH, KOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂), the strength of bases generally increases down the group in the alkali and alkaline earth metals due to the increasing solubility and dissociation in water. The order of strength is:
Ba(OH)₂ (Barium hydroxide) - Strongest
Sr(OH)₂ (Strontium hydroxide)
Ca(OH)₂ (Calcium hydroxide) - Moderately strong, depending on concentration
KOH (Potassium hydroxide)
NaOH (Sodium hydroxide)
LiOH (Lithium hydroxide) - Weakest of the group
Strong Acids (from strongest to weakest): HClO₄ > H₂SO₄ > HI > HBr > HCl > HNO₃
Strong Bases (from strongest to weakest): Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > KOH > NaOH > LiOH
These rankings are based on general chemical principles, and while there may be exceptions in specific contexts, this is the accepted trend for these acids and bases.
Related questions
Is LiOH a strong base or weak?
How do strong acids react to bases, e.g., HCl + NaOH or HNO3 + KOH?
Which is the strongest acid: HCl or HNO3 or HI or HBr or HF/HClO4/HIO4/HI(aq)?
Which are very strong acids among HCl, HNO3, H2SO4, and H2ClO4?
What is the strongest base among the following, LiOH, NaOh, KOH, and CsOH?
Afranio Torres-Filho
Post-Doc in Polymer Science and Engineering & Physical Chemistry, University of Massachusetts, Amherst (Graduated 1991)
·
Author has 85 answers and 488.3K answer views
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5y
I believe it would make more sense, certainly as a first step, to compare the strength of the acids using aspects of their Chemical Structures. Let´s start with the HYDRACIDS.
I)STRENGTH OF HYDRACIDS: HF x HCl x HBr x HI.
They have Hydrogen as a common component and the properties of the Halogen atom determine the strength of the acid. When thinking about a PROPERTY that would correlate well with the acid strength trend of this group, the first one that came to mind was the ELECTRONEGATIVITY of the Halogen atoms. Electronegativity is the ability of an atom to attract electrons. The larger its te
I believe it would make more sense, certainly as a first step, to compare the strength of the acids using aspects of their Chemical Structures. Let´s start with the HYDRACIDS.
I)STRENGTH OF HYDRACIDS: HF x HCl x HBr x HI.
They have Hydrogen as a common component and the properties of the Halogen atom determine the strength of the acid. When thinking about a PROPERTY that would correlate well with the acid strength trend of this group, the first one that came to mind was the ELECTRONEGATIVITY of the Halogen atoms. Electronegativity is the ability of an atom to attract electrons. The larger its tendency to attract electrons, the larger its Electronegativity.
The correlation would work well with either the individual Electronegativities as well as with the difference between the individual values and the one for Hydrogen, En(H) = 2.2.
Table 1, and the graph displayed in the picture (Fig. 1), both shown below, present the results when correlating the log Ka (proportional to the strength) and the difference between the Electronegativities, En(X)–En(H).
TABLE I:ELECTRONEGATIVITIES OF HALOGENS(X) x HX ACID STRENGTH
FIG. 1: Log Ka x D[En(X)–En(H)] WITH X = F, Cl, Br, I.
It clearly shows a good correlation, great trend: the smaller the Electronegativy difference, the greater the acid strength. Observe that Ka(HI) >> Ka(HBr) >> Ka(HCl) >> Ka(HF)
The explanation involves a more detailed explanation of the Covalent H–X bond. Two electrons are shared by the H and the X atom, forming what is known as a Covalent Bond. Because of the difference in their tendency to attract electrons (Electronegativity) the bonding electrons are not going to be equally shared by the two atoms.
If you have Classical Physics in mind, you may think: “the electron pair will be closer to the atom with larger Electronegativity”, or “the electrons are moving, but they will stay longer close to the atom with larger Electronegativity”. These are not accurate, from the Quantum point of view, but they may be useful for your initial understanding of the bond.
From the Quantum point of view you may say: “the electron cloud (with the 2 bonding electrons) will be shifted towards the atom with larger Electronegativity”. Or, even better, more accurate: “the wave function that describes the pair (if using a function which is the sum of the individual atomic wave functions), has a larger component from the atom with larger Electronegativity”. As an example, for a Covalent Bond between atoms A and B, if the individual atomic wave funtions participating in bond are Fa and Fb, and atom A has a larger Electronegativity, the linear combination function MO (Molecular Orbital) will be:
MO = a Fa + b Fb
with a > b, since atom A has a higer Electronegativity. Another Quantum view of the bonding: “there will be a higher probability of finding the electron pair closer to the atom with higher Electronegativity”.
However, here´s what is really important on a basic level: there will be an asymmetry in the spatial distribution of the electric charge. A larger NEGATIVE electric charge will be closer to the atom with larger Electronegativity. And, just in relative term, the other atom (smaller En) will have a “POSITIVE” charge (in reality, it will be less negative). This spatial separation of electric charges forms what is called an ELECTRIC DIPOLE: two regions, close by, with opposing electric charges. And is this electric charge “unbalance” that is responsible for the Covalent bonding: attraction between opposite charges.
The picture below (Fig. 2) shows all four bonds with the corresponding electric dipole highlighted.
FIG 2: BONDS AND DIPOLE MOMENTS OF THE HYDROGEN HALIDES
The Greek letter DELTA
represents both the negative and the positive electric charges close to the bonding atoms. They are different in signs but identical in terms of absolute values:
The absolute DELTA value is proportional to the difference in Electronegativity. The Dipole Moment is a vectorial physical property. In the case of a diatomic molecule it is represented by:
This bond dipole is interpreted as the dipole from a charge separation over a distance r between the partial charges Q+ and Q− (or the more commonly used terms, δ+ and δ−). It is a vector quantity and the orientation of the dipole is along the axis of the bond. By convention, the vector points towards the more Electronegative atom, as shown for HF and HCl on Figure 2 above.
And here comes the line of reasoning (especially applied to diatomic species):
1)The greater the Electronegativity difference, THE LARGER THE DIPOLE MOMENT (comparing chemical species with similar internuclear distances). Figure 3 shows the Dipole Moments of the Hydrogen Halides (HX).
FIG 3: DIPOLE MOMENT OF THE HYDROGEN HALIDES
2) This dipole moment is the “responsible” for the chemical bond strength: attraction of electric charges with oposite signs.
3) The greater the charge, the larger the attraction between the poles. Therefore, THE LARGER THE DISSOCIATION ENERGY!
4) During the dissolution and ionization of the acid, the H will separate from the Halogen.
HX(g) + H2O(l) === > H3O+(aq) + X–(aq)
It is obvious that the larger de Dissociation Energy the harder it will be for the atoms to separate.
CONCLUSION:
LARGER DIPOLE === > LARGER DISSOCIATION ENERGY === > LOWER ACID STRENGTH (proportional to Ka).
Table II, below, shows the correlation between the Dipole Moments and the Dissociation Energies, besides the Ka of the hydracids.
TABLE II: SOME PROPERTIES x ACID STRENGTH OF HYDRACIDS
There are several properties that also correlate with this difference in En, the acid strength of the hydracids, and some properties of the individual Elements. Some have a direct relationship with the acid strength, others is an inverse relationship. Table III displays a few of them, including the Ka of the hydracids.
TABLE III: DIRECT AND INVERSE RELATIONSHIP BETWEEN SEVERAL PROPERTIES OF HALOGEN HYDRACIDS (HX) AND ACID STRENGTH.
Now lets move to a different set of acids: Oxyacids (or Oxoacids), all containing only one type of Halogen atom.
II) STRENGTH OF OXYACIDS CONTAINING A SINGLE HALOGEN ATOM: HClO, HClO2, HClO3, HClO4.
These are the hipochlorous, chlorous, chloric and hiperchloric acids.
Actually, I found a great link where the correlation between the strength of these acids, their structure, and some properties of the Halogens is well described and discussed. It belongs to the Chemistry LibreTexts links, entitled:
The Acidity of an Oxoacid is Determined by the Electronegativity and Oxidation State of the Oxoacid's Central Atom
Here are some of their highlights on the subject:
For oxoacids of a given central atom the acidity increases with the central element's oxidation state or, in other words, the number of oxygens bound to the central atom.
Here we are looking at the trend for acids in which there are variable numbers of oxygen bound to a given central atom. An examples is the perchloric (ClO4−), chloric (ClO3−), chlorous (ClO2−), and hypochlorous (ClO−) acid series. In such series the greater the number of oxygens the stronger the acid. This can be explained in several ways. From the viewpoint of the acid itself the key factor is again the inductive effect, in this case involving the ability of the oxygens attached to the central atom to pull on electron density across the OH bond. This is seen from the charge density diagram for the chlorine oxoacids shown in Figure 4, in which the partial positive charge on the acidic hydrogen increases with the number of oxygens present.
Figure 4: CHARGE DENSITY AND THE RELATIVE STRENGTH OF OXYACIDS OF CHLORINE
In other words: Increasing number of oxygens increases Ka, which means, increase the acid strength. This is evidenced by the decreased electron density on the acidic hydrogen (which is most blue in HClO4 ).
NOTE, AND VERY IMPORTANT:
The pKa represented is:
pKa = - log Ka
or
Ka =10E(-pKa)
and the larger pKa , the smaller Ka, the weaker the acid.
The increase in oxoacid acidity with the number of oxygens bound to the central atom may also be seen by considering the stability of the conjugate oxyanion. That the stability of the conjugate base increases with the number of oxygens may be seen from the charge distribution diagrams and Lewis bonding models for the chlorine oxyanions shown in Figure 5. As the negative charge is spread over more oxygen atoms it becomes increasingly diffuse.
FIGURE 5: CHARGE DISTRIBUTION IN THE CHLORINE CONTAINING OXYACIDS SERIES
The Figure above shows an increased diffusion of charge in chlorine oxyanions with increasing number of oxygens. The larger the ion the more dispersed the charge and thus the less the charge density, making the perchlorate the most stable anion in the series. Even the simplistic treatment of bonding depicted in the resonance structures correctly show an increased dispersion of the charge density.
The trend shown above also works when the main element in the acid (the one besides H and O) is a non–Halogen element.
II) STRENGTH OF OXYACIDS WITH SIMILAR STRUCTURES BUT DIFFERENT HALOGEN ATOMS: HClO, HBrO, HIO
Now, comparing Oxyacids containing different Halogen central atoms (a homologous series), the acidity increases with the electronegativity of the central atom, X.
Elements in the same group frequently form oxyacids of the same general formula. For example, chlorine, bromine, and iodine all form oxyacids of formula HOX: hypochlorous, hypobromous and hypoiodous acids. In the case of these homologous oxyacids the acidity is largely determined by the electronegativity of the central element. Central atoms, which are better able to inductively pull electron density towards themselves, make the oxygen-hydrogen bond (that is to be ionized) more polar, and stablilze the conjugate base, OX−.
Thus, the acid strength in such homologous series increases with the electronegativity of the central atom.
This may be seen from the data for the hypohalous acids in Table IV, in which the acid stength increases with the electronegativity of the halogen.
TABLE IV: ELECTRONEGATIVITY OF X AND THE STRENGTH OF HOX OXYACIDS
The order of acidity is: HClO > HBrO > HIO
III) COMPARING OXYACIDS WITH HALOGENS, OR OTHER ELEMENTS, IN THE STRUCTURE: HClO4, H2SO4, HNO3
(from the link Oxyacid | chemical compound)
The strength of an oxyacid is defined by the extent to which it dissociates in water (i.e., its ability to form H+ ions).
In general, the relative strength of oxyacids can be predicted on the basis of the ELECTRONEGATIVITY and OXIDATION NUMBER of the CENTRAL NONMETAL ATOM.
The acid strength increases as the electronegativity of the central atom increases. For example, because the electronegativity of chlorine (Cl), En(Cl) = , is greater than that of sulfur (S), En(S) = , which is in turn greater than that of phosphorus (P), En(P) = , it can be predicted that perchloric acid, HClO4, is a stronger acid than sulfuric acid, H2SO4, which should be a stronger acid than phosphoric acid, H3PO4.
For a given nonmetal central atom, the acid strength increases as the oxidation number of the central atom increases. For example, nitric acid, HNO3, in which the nitrogen (N) atom has an oxidation number of +5, is a stronger acid than nitrous acid, HNO2, where the nitrogen oxidation state is +3.
In the same manner, sulfuric acid, H2SO4, with sulfur in its +6 oxidation state, is a stronger acid than sulfurous acid, H2SO3, where a +4 oxidation number of sulfur exists.
IV) COMPARING BASES OF ALKALI AND ALKALINE EARTH METALS
IV.A)BASES OF ALKALI METALS:
On moving down the group of alkali metals, the basic character of hydroxides increases from LiOH to CsOH. As the size of the metal cation increases, the internuclear distance betwen metal cation and the oxygen of the hydroxide group increases. This increases the ease of ionization of the hydroxide ion. Hence, the BASICITY INCREASES.
• The basic strength of these hydroxides increases as we move down the group Li to Cs.
• The hydroxides of alkali metals behave as strong bases due to their low ionization energies which decrease down the group.
• The decrease in ionization energies leads to weakening of the bond between metal and hydroxide ion and M–O bond in M–O–H can easily break giving M+ and OH- .
• This results in the increased concentration of hydroxyl ions in the solution, i.e., increased basic characters.
Other properties of the alkali metals and their hydroxides can be correlated to the strength (basicity) of their bases.
Table V, below, presents some of the properties that correlate, directly or inversely, with strength of their bases. These include, the Ionization Energy of the atoms corresponding to the Alkali Metals, the size of their ions, their Lattice Energy, the heat released during the dissolution process, and the solubility. The latter one results in the strength of the corresponding bases, shown here by their Kb.
TABLE V: M AND MOH PROPERTIES x BASE STRENGTH CORRELATIONS
CORRELATION BETWEEN SOME PROPERTIES OF THE ALKALI METALS AND THEIR HYDROXIDES, AND THE CORRESPONDING STRENGTH OF THEIR BASES
IV.B) BASES OF ALKALINE EARTH METALS:
Considering oxides containing Alkaline Earth Metals, they become more basic down the Group, namely Be to Ba. BeO and Be(OH)2 are amphoteric, and react with acids and strong bases, such as NaOH. MgO is basic and Mg(OH)2 is weakly basic and do not dissolve in NaOH solution. The oxides of calcium, strontium, and barium are basic and the hydroxides are strongly basic.
The solubilities of the hydroxides in water, and their basic strength follow the order:
Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
In resume, the reasoning for explaining the increase in basic strength as one moves down in a periodic table group, goes as follows:
1)A large value of the Lattice Energy will make the M+(n)–(OH)n harder to break. Therefore, there should be an inverse relationship between the Lattice Energy and Basic Strength.
2) However, Lattice Energy is not the only factor to take in consideration when establishing a trend in Basic Strength of Hydroxides (or Solubility of Ionic Chemical compounds in general). As the íons go into solution they´ll interact with water molecules, through their electric dipole moment. The pictures below (Figs 6A and 6B) depict the interaction of the Sodium (Na) and the Magnesium (Mg) cations with water molecules.
Fig. 6A: SODIUM CATION (Na+) INTERACTION WITH WATER MOLECULES.
Fig.6B : MAGNESIUM CATION (Mg2+) INTERACTION WITH WATER MOLECULES.
When a Chemical Bond is formed heat is released. The formation of these
(M+)– (OH)2 bonds also release energy. The final Solubility of a compound (in the case of the Alkaline Earth metals hydroxides, and Hydroxides in general, the Solubility is also related to their Basic Strength) is determined by Thermodynamics. There must be a balance between the Energy needed to break the bonds in the solid state (Lattice Energy) and the Heat released during the hydration of their íons, both the metal (Mn+) and the hydroxyl (OH–) íons. Even if the Lattice Energy is large, the compound may be soluble (and the Base strong) if enough heat (energy) is released during the íons hydration. Formally, it is said that the process must have a negative Gibbs Free Energy, given by:
Where DH is the total energy released, or absorbed, during the process, T is the temperature, and DS is the change in Entropy of the system. For dissolution of a substance, DS is always positive, since there will be a decrease in the order of the system, passing from an organized solid to íons dispersed in a “sea of water”.
The Table below presents the correlation between several properties of the Alkaline Earth metals, and of their Hydroxides, and the Solubility, Basic Strength of their corresponding Bases. The values of Kb are directly proportional to the Base Strength, and pKb = - log (Kb), is inversely related to the Base Strength.
Some trends are immediately observed:
1)INVERSE CORRELATION: The LARGER the Lattice, and the Ionization Energy of the Alkaline Earth metal, the LOWER the Solubility, and the Basic Strength of the corresponding Hydroxide.
2) DIRECT CORRELATION: The LARGER the Ionic Radius of the corresponding Cations, the LARGER the Solubility and Basic Strength of the Hydroxide.
3) DIRECT CORRELATION: Another direct correlation is observed between the Heat of Hydration, the Solubility and Basic Strength: the LARGER the ENERGY RELEASED during the Hydration of the íons, the LARGER the Solubility and the Basic Strength. The numbers on the Table may be misleading because of a convention normally used in Thermodynamics: Energy released (Exothermic process) is expressed as a NEGATIVE VALUE. Mathematically the highest heat released (~ –47 kJ/mol) of Ba(OH)2 would be the smallest number, but this hydroxide is the one in the group which releases more energy, top Solubility, larger Basic Strength!
PS: Although the correlations presented on the last Table are correct, the individual numbers for some of the properties may not be very accurate. Some variation was observed when looking through the web for values of some properties. In particular, Heat of Dissolution (or Enthalpy of Solution) values for the Hydroxides were calculated, based on the hydration values for each íons and the Lattice Energy.
For example:
Calculation of the Heat of Dissolution for the Ca(OH)2:
1)Lattice Energy for Ca(OH)2 === > E(L)= + 2637 kJ/mol
2) Hydration Energy fpr the Ca2+ ion === > E(Hyd-Ca) = 1591 kJ/mol
3) Hydration Energy for the (OH –) íon:== > E(Hyd-OH) = 530 kJ/mol
E[Dissolution of M(OH)2] = E(L) + [(Heat Hyd (M+2) + 2 x (Heat Hyd OH–)]
The value of the Heat of Hyd for (OH–) was multiplied by 2 since 2 OH– íons are released during the dissolution process.
E(Dissolution of Ca(OH)2) = 2637 + [ (–15991) + 2 x (–530) ] = – 14 kJ/mol
I do hope the whole discussion will be useful to many!
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Master's Degree from Harvard University (Graduated 2011)
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Mike Jones
MAEd in Chemistry & Physics, Western Carolina University (Graduated 1974)
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Author has 6.2K answers and 8.4M answer views
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5y
In water, there is no ranking of the strong acids and bases. All the strong acids are strong …. they have the same strength. They all ionize to form a hydrogen ion. Sulfuric is the tricky one since it is only strong for the first H+. The remaining HSO4^- ion is a weak acid.
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2 … (fixed the formulas) are strong bases and ionize completely. The group 2 hydroxides are not very soluble in water, but that which does dissolve ionizes completely.
Ysterklou Botha
Former Retired (1976–2014)
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Author has 4.5K answers and 1.7M answer views
·
Updated 1y
Related
Why is aqueous NaOH not used to neutralize strong acids like HCl, HBr, HI and HF?
I once experienced the explosion of a lead-acid battery, spilling fairly concentrated sulfuric acid all over the place. Also remember that it necessarily contained lead ions. A fairly obnoxious situation.
I used sodium carbonate (fly ash) to neutralize the spill before cleaning up. Why not concentrated NaOH? Or even NaOH pearls.
Who wants to prepare and store concentrated NaOH? Not me. Concentrated NaOH is a hazardous solution. If it is inadvertently spilled while storing it, you have a very hazardous situation to contend with. And if it could be stored safely, you still have to handle it while
I once experienced the explosion of a lead-acid battery, spilling fairly concentrated sulfuric acid all over the place. Also remember that it necessarily contained lead ions. A fairly obnoxious situation.
I used sodium carbonate (fly ash) to neutralize the spill before cleaning up. Why not concentrated NaOH? Or even NaOH pearls.
Who wants to prepare and store concentrated NaOH? Not me. Concentrated NaOH is a hazardous solution. If it is inadvertently spilled while storing it, you have a very hazardous situation to contend with. And if it could be stored safely, you still have to handle it while neutralizing the acid, and this handling constitutes a dangerous operation.
The use of slightly less dangerous NaOH pellets or pearls? To protect yourself it is better to avoid even that.
What is the most important reason for not using NaOH? Consider the case of a battery explosion. Or even the breakage of an acid container. The acid spreads and splashes all over the place. And now you apply a NaOH. The strong acid reacts violently with the strongly basic NaOH. Spreading the mixture containg still unreacted acid and base even further, including all over the person that applied it.
In addition, applying a strong base haphazardly everywhere is going to cause lots of damage everywhere that it touches where there is no acid. Then you have to neutralize the NaOH and not the acid spill only.
Weak bases are easy to store, safe to apply, and much less damaging to materials that it contacts where there is no acid.
Why don't we use aqueous NaOH to neutralize acid spills. It is a stupid thing to do.
Related questions
Why is aqueous NaOH not used to neutralize strong acids like HCl, HBr, HI and HF?
Is HF the weakest acid among HCl, HNO3, and H2SO4?
Why are HCL, HNO3, and H2SO4 known as mineral acids?
Which is a strong acid HCl, H2SO4, HNO3, or HClO4?
Which of following acids is not strong? A) HCl B) HBr C) HNO3 D) H3PO3
Sambhuti Vijay
Studied at DAV Public School, Unit-8, Bhubaneswar
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Author has 115 answers and 13.1K answer views
·
Mar 5
Related
H2SO4 is a strong acid but HSO4- is a weak acid. What is the difference in the strength of these two related species?
The difference in acid strength between H₂SO₄ (sulfuric acid) and HSO₄⁻ (hydrogen sulfate ion) is due to their ability to donate protons () and their degree of ionization in water.
H₂SO₄ (Sulfuric Acid)
Complete ionization: The reason Sulfuric acid is very strong is because it’s Ka value almost tends to 1 thereby making it easily donate a proton.:This means almost every molecule of donates a proton, making it very strong.
HSO₄⁻ (Hydrogen Sulfate Ion)
Partial ionization: The hydrogen sulfate ion, , is a weak acid, meaning it only partially di
The difference in acid strength between H₂SO₄ (sulfuric acid) and HSO₄⁻ (hydrogen sulfate ion) is due to their ability to donate protons () and their degree of ionization in water.
H₂SO₄ (Sulfuric Acid)
Complete ionization: The reason Sulfuric acid is very strong is because it’s Ka value almost tends to 1 thereby making it easily donate a proton.:This means almost every molecule of donates a proton, making it very strong.
HSO₄⁻ (Hydrogen Sulfate Ion)
Partial ionization: The hydrogen sulfate ion, , is a weak acid, meaning it only partially dissociates in water:The equilibrium does not lie completely toward the right, meaning not all molecules lose their protons. This makes it a much weaker acid than . It’s Ka values are much much smaller than H₂SO₄.
Now we come to reason why is a much stronger acid than
Charge on : is a neutral molecule, so losing is relatively easy. is already negatively charged, meaning losing another (which is also positively charged) is harder due to electrostatic attraction.
Stability of Conjugate Bases:When loses a proton, it forms , which is relatively stable due to resonance.When loses a proton, it forms , which has a higher negative charge, making it less stable and less likely to form.
And generally a fact is seen that dissociation of HSO₄⁻ is the second step in total dissociation of H₂SO₄ and by the time it dissociates, a lot of is already there in water (because of the first dissociation of H₂SO₄ and thus by Le Chatelier Principle the reaction will shift towards reactants side because a lot of product is already there in the form of & hence reaction will try to balance itself. It also makes HSO₄⁻ VERY DIFFICULT tO dissociate and hence low Ka.
This difference in strength is due to electrostatic charge,Le Chatelier Principle & stability of the conjugate base.
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Sunil Banerjee
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Is NaOH an acid or a base?
NaOH is a base because when dissolved in water it dissociates into Na+ and OH- ions. It is the OH- (hydroxyl ion) which makes NaOH a base. In classical term a base is defined as a compound which reacts with an acid to form salt and water as depicted by the following equation. NaOH+HCl=NaCl+H2O. This equation means a molecule of sodium hydroxide reacts with a molecule of hydrochloric acid to form a molecule of sodium chloride(salt) and a molecule of water. In terms of Lewis theory a base is defined as a compound which donates a pair of electrons and an acid is defined as the one which accepts t
NaOH is a base because when dissolved in water it dissociates into Na+ and OH- ions. It is the OH- (hydroxyl ion) which makes NaOH a base. In classical term a base is defined as a compound which reacts with an acid to form salt and water as depicted by the following equation. NaOH+HCl=NaCl+H2O. This equation means a molecule of sodium hydroxide reacts with a molecule of hydrochloric acid to form a molecule of sodium chloride(salt) and a molecule of water. In terms of Lewis theory a base is defined as a compound which donates a pair of electrons and an acid is defined as the one which accepts the pair of electrons. OH- ion (anion) from NaOH would donate its pair of electrons to the proton (H+) of acid to from water as shown below. OH- + H+ = H2O. This proves that NaOH (caustic soda) is a base. Not only is NaOH a base but also in reality it is a strong base and is an alkali (giving an alkaline solution in water). In terms of pH if 40 g of NaOH is dissolved in water to make a litre of its aqueous solution (1 N NaOH) the pH of the resulting solution will be 14. Similarly the pH of 4g of NaOH in a litre of aqueous solution is 13. A solution having pH 7 is neutral and a solution with a pH lower than 7 is acidic and a solution with a pH higher than 7 is basic. The pH of 1N HCl is 0.
Shubhayu Das
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Why are all of the hydrohalic acids, except HF, strong acids?
The strength of an acid depends on how easily it can donate the hydrogen atom in the form of H^+ ion. This is in the case of Arhennius acid concept only.
Now looking at the halo-acids, we must first observe the nature of the bonding orbitals. Hydrogen uses a 1s orbital for forming a bond. The halogens use 2p, 3p, 4p and 5p orbitals respectively for flourine, chlorine, bromine and iodine. Due to the size difference of these orbitals as compared to 1s, the bond strength decreases. Therefore iodine having.teb largest orbital 5p cannot effectively overlap with 1s. Hence this is a weak bond.
In case
The strength of an acid depends on how easily it can donate the hydrogen atom in the form of H^+ ion. This is in the case of Arhennius acid concept only.
Now looking at the halo-acids, we must first observe the nature of the bonding orbitals. Hydrogen uses a 1s orbital for forming a bond. The halogens use 2p, 3p, 4p and 5p orbitals respectively for flourine, chlorine, bromine and iodine. Due to the size difference of these orbitals as compared to 1s, the bond strength decreases. Therefore iodine having.teb largest orbital 5p cannot effectively overlap with 1s. Hence this is a weak bond.
In case of flourine however, 2p can form a strong bond with 1s of hydrogen ( bond strength is 565kJ/mol). Hence this is a very strong bond. Thus it can't loose the hydrogen atom easily.
Using this argument we can explain why HF is much less acidic as compared to HI.
Note: If you haven't learnt about orbitals yet, then simply understand the fact that due to the large size of iodine, it can't form a strong bond with hydrogen.
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Anant Verma
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Among HClO4, HClO3, HClO2 and HClO, which one is a strong acid, and why?
The acidity is decided on the basis of oxidation state on the central atom.
Higher the oxidation state more the acidic nature.
So, in HClO4 the oxidation state of Cl is +7
Similarly in HClO3 it's +5
In HClO2 it's +3
In HClO it's +1
So, amongst the given compounds most acidic is HClO4.
Remember the key here is to get the most positive charge on the central atom, so if you're even to compare between acidity of HClO4 and HIO4, the more acidic will be HClO4.
Upvote if helped.
If any doubts, feel free to comment/message. :)
Uxleumas
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Some say acid with pH 2-3 is weak. Does it give hint that any acid with only pH 1 or lower is considered strong?
A more accurate definition would be that an acid is considered weak if it doesn’t disassociate in water completely, and strong if it does. Disassociation basically means how much of HA (H being the acidic hydrogen, and A being the rest of the acid) turns into H+ and A-. An acid with the same concentration as another acid but disassociates more would be considered to be stronger. A strong acid can be diluted to a pH of 2–3, yet it is still a strong acid. A weak acid could be concentrated to a pH of 1 (depending on the acid and the solubility), yet still be a weak acid. As a general rule of thum
A more accurate definition would be that an acid is considered weak if it doesn’t disassociate in water completely, and strong if it does. Disassociation basically means how much of HA (H being the acidic hydrogen, and A being the rest of the acid) turns into H+ and A-. An acid with the same concentration as another acid but disassociates more would be considered to be stronger. A strong acid can be diluted to a pH of 2–3, yet it is still a strong acid. A weak acid could be concentrated to a pH of 1 (depending on the acid and the solubility), yet still be a weak acid. As a general rule of thumb, if an acid is fairly concentrated, but still has a pH of 3 or above, it’s likely weak, and if the acid has a ph of below 1, it’s likely strong.
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Antonello Frau
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The increasing order of acidic acids is HClO₄>HI>HCN. Why?
The stronger acid in aqueous solutions among HClO₄ (perchloric acid), HI (hydroiodic acid) and HCN (hydrocyanic acid) is indeed the first one.
Have a look at its atom sequence (disregard its Lewis structure with double bonds, lone pairs, and such):
HClO₄ is composed of a chlorine atom (an electronegative one) surrounded by four oxygen atoms (even more electronegative than Cl). Each of the five atoms exert a remarkable pull on their shared electron pairs and this gets reflected on the terminal H atom bonded to an O: its O-H bond will be affected by the strong electron-withdrawing pull of both O a
The stronger acid in aqueous solutions among HClO₄ (perchloric acid), HI (hydroiodic acid) and HCN (hydrocyanic acid) is indeed the first one.
Have a look at its atom sequence (disregard its Lewis structure with double bonds, lone pairs, and such):
HClO₄ is composed of a chlorine atom (an electronegative one) surrounded by four oxygen atoms (even more electronegative than Cl). Each of the five atoms exert a remarkable pull on their shared electron pairs and this gets reflected on the terminal H atom bonded to an O: its O-H bond will be affected by the strong electron-withdrawing pull of both O and Cl on it and turn out to be very weak; in other words, HClO₄ is very eager to get rid of its H atom as a proton (H⁺).
This way, the chlorate anion (ClO₄¯) becomes very stable: the more stable a base, the stronger its conjugate acid (and vice versa).
Hence, HClO₄ is a very strong acid (pK𝖺 ca. -15). HI and HCN, on the other hand, are smaller molecules wherein there is a long, weak H-I bond (in HI) and a short, kind of stable H-C bond (in HCN).
As a result, HI is a strong acid (pK𝖺 ca. -9) whilst HCN is a very weak acid (pK𝖺 ca. 10). Remember, the more negative the pK𝖺, the stronger the acid.
Toby Block
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In what order do these acids rank HBr, HCl, HI, H2Se, H2S, CH4, and H2O?
HBr, HCl, and HI are all strong acids in aqueous solution. You can’t really say one is stronger than another (no molecule can do better than 100% dissociation).
H2S and H2Se are weak acids. I couldn’t find Ka values (I don’t have access to textbooks at the moment). One site indicated that H2Se is a bit stronger than H2S because the larger Se atom makes HSe- more stable than HS-.
Water itself is amphoteric, producing both hydrogen ions and hydroxide ions. Kw=[H+][OH-] = 1E-14 at 25 deg C
I wouldn’t consider CH4 an acid at all. It is a non-polar gas and however much can be pushed into water at high
HBr, HCl, and HI are all strong acids in aqueous solution. You can’t really say one is stronger than another (no molecule can do better than 100% dissociation).
H2S and H2Se are weak acids. I couldn’t find Ka values (I don’t have access to textbooks at the moment). One site indicated that H2Se is a bit stronger than H2S because the larger Se atom makes HSe- more stable than HS-.
Water itself is amphoteric, producing both hydrogen ions and hydroxide ions. Kw=[H+][OH-] = 1E-14 at 25 deg C
I wouldn’t consider CH4 an acid at all. It is a non-polar gas and however much can be pushed into water at high pressure and low temperature will not produce any hydrogen ions.
Clive Holloway
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Are there any strong bases with no Hydroxide ion?
Depends on the ssolvent you are using. The pH system is based on the self dissociation of water into H+ and OH- ions, respectively acid and base. But in a non-aqueous solution which can self dissociate then the base is different. For example in pure liquid ammonia, the ions formed are NH4+ and NH2-, respectively acid and base, so ammonium salts would be acidic and amides would be basic. Of course some salts in water can behave as bases also, even without an OH- ion, if their anion (-ve) is able to sufficiently disturb the water self ionisation by removing the H+and leaving excess OH-.
Sudipta Saha
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Is H2SO4 considered a strong acid? What is its conjugate base? How does its dissociation constant and Ka value compare to other strong acids such as HCl and HNO3?
Sulfuric acid is a colourless oily liquid. It is soluble in water with release of heat. It is corrosive to metals and tissue. It will char wood and most other organic matter on contact, but is unlikely to cause a fire. Density =15 lb / gal. Long term exposure to low concentrations or short term exposure to high concentrations can result in adverse health effects from inhalation. It is used to make fertilizers and other chemicals, in petroleum refining.
Conjugate base: The conjugate base of H2 SO4 is HSO4 -.
Ka values:
1.3 10^6=Hydrochloric acid
1.0 10^3=Sulfuric acid
2.4 10^1=Nitric acid
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Which are very strong acids among HCl, HNO3, H2SO4, and H2ClO4?
What is the strongest base among the following, LiOH, NaOh, KOH, and CsOH?
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6764 | https://www.cs.ucr.edu/~elenas/CS111/asymptotic_notation.pdf | Asymptotic Notation Basics (Updated April 16, 2013) In computer science it is often difficult to precisely measure certain quantities, and even when they can be measured accurately their exact values are not essential. Accurate, but simple looking approximations could be more useful than complex exact formulas.
A typical example is that of a running time of an algorithm. The actual running time depends on the implementation, the compiler, on the machine, and on the conditions under which the program is executed.
Thus it is simply not possible to tell what is the running time of an algorithm based only on its description.
But in fact the exact formula is not even necessary. For example, if an algorithm makes 4n3 + 3n + 7 steps on inputs of length n, for large values of n the terms 3n and 7 are insignificant. It makes more sense to say that the running time is approximately 4n3. In fact, it makes sense to even ignore the constant 4, since this constant is implementation- and platform-dependent. So we can say that the running time will be proportional to n3. In other words, it will be roughly cn3, for some constant c. This observation motivates the definition of asymptotic notation.
Definition.
For two functions f(x) and g(x), we say that f(x) is of order g(x), and write f(x) = O(g(x)), if there are constants c and x0 such that |f(x)| ≤cg(x) for all x ≥x0.
In all useful contexts the functions we will deal with will be non-negative, in which case we can ignore the absolute value (we will in fact do this in most examples). Occasionally though we may encounter functions that may take some negative values for small values of x, so we need to take this into account in the definition.
Note that the above definition forces g(x) to be non-negative if x is large enough.
Example 1.
We show that 7x + 20 = O(x). Indeed, say, for x ≥10 we have 7x + 20 ≤7x + 2x ≤9x, so the above definition applies with x0 = 10 and c = 9.
But these choices are not unique. We could as well say that for x ≥20 we have 7x + 20 ≤7x + x ≤8x, etc.
Example 2.
Let’s show now that 2n3 + 6n2 −2 = O(n3). We have 2n3 + 6n2 −2 ≤2n3 + 6n3 = 8n3 for n ≥0, so we can conclude that 2n3 + 6n2 −2 = O(n3).
Example 3.
As a first more exciting example, we’ll look at the harmonic numbers: H(n) = n X i=1 1 i = 1 + 1 2 + 1 3 + ... + 1 n.
This function grows with n, but how fast? We claim that its growth rate is the same as that of logarithmic functions, that is H(n) = O(log n).
Let’s now prove it. The idea of the proof is to divide the sequence 1, 1/2, 1/3, .. into about log n blocks, so that the sum of each block is between 1 2 and 1. More specifically, we divide the sequence 1, 1/2, 1/3, ...
into blocks 1 1 2 + 1 3 1 4 + 1 5 + 1 6 + 1 7 . . .
1 2i + 1 2i + 1 + ... + 1 2i+1 −1 . . .
1 2k + 1 2k + 1 + ... + 1 n where k is chosen to be the integer such that 2k ≤n < 2k+1. (Thus k = ⌊log n⌋.) So, for any i except k, the i-th block starts at 1/2i and ends right before 1/2i+1. The last block is exceptional, since it ends at 1/n which could be smaller than 1/(2k+1 −1).
Thus the sum of the i-th block (except the last) is 1 2j + 1 2j + 1 + 1 2j + 2... + 1 2j+1 −1 In this sum, each term is at most 1/2j and is greater than 1/2j+1. This block starts at the 2j-th term and ends right before 2j+1-st term, so it has 2j+1 −2j = 2j terms. Therefore the sum of this block is at most 1 2j · 1 2j = 1 and at least 2j · 1 2j+1 = 1 2. (Except the last block’s sum, which is at most 1 but could be less than 1 2.) Putting it all together, we proceed as follows. We have k + 1 blocks, with each block adding to at most 1, so Hn ≤(k + 1) · 1 ≤log n + 1. On the other hand, all blocks except last add up to at least 1 2, so Hn ≥k · 1 2 ≥1 2(log n −1). Summarizing, we have proved the following theorem.
Theorem 0.1 1 2(log n −1) ≤Hn ≤log n + 1, for all positive integers n.
How can we express this using the big-Oh notation? Bounding Hn from above, we get that for n ≥2, Hn ≤log n+1 ≤log n+log n = 2 log n. So Hn ≤2 log n for n ≥2, and we can conclude that Hn = O(log n).
Much better estimates are known for Hn. It is known that Hn ∼ln n + γ, where γ ≈0.57 is called Euler’s constant. More precisely, the difference Hn −ln n converges to γ with n →∞. We can express this better approximation using the big-Oh notation as well, by using the big-Oh notation for the approximation error rather than to the function itself: Hn = ln n + O(1).
An even more accurate approximation is Hn = ln n + γ + O(1/n), as it shows that the approximation error Hn −ln n −γ vanishes when n →∞.
Example 4.
The asymptotic notation has not been invented by computer scientists – it has been used in mathematics for over a 100 years, for describing approximation errors using various expansions (like Taylor series), and in number theory, to estimate the growth of some functions. Consider for example the following question: for a number n, how many prime numbers are between 2 and n? This value is denoted π(n). It is known that π(n) = O(n/ log n). In fact, more accurate estimates for π(n) exist; for example π(n) = n/ ln n + O(n/ log2 n). This result is often called the Prime Number Theorem.
Example 5.
We now consider an example that involves sequences defined by recurrence equations. Suppose we have a sequence {ai} defined by a0 = 3, a1 = 8, an = 2an−1 + an−2. We claim that an = O(2.75n).
In order to prove this, we show by induction that an ≤3(2.75)n. Indeed, in the base cases, a0 = 3 = 3(2.75)0 and a1 = 8 < 3(2.75)1. In the inductive step, assume the claim holds for numbers less than n. For n, by the inductive assumption, we get an = 2an−1 + an−2 ≤ 2 · 3(2.75)n−1 + 3(2.75)n−2 ≤ 3(2.75)n−2(2 · 2.75 + 1) ≤ 3(2.75)n−2(2.75)2 ≤ 3(2.75)n, and thus the claim holds for n as well. Therefore it holds for all values of n.
Example 6.
Ok, now let’s talk about algorithms. What’s the running time of the algorithm below?
Algorithm WhatsMyRunTime1 (n : integer) for i ←1 to 6n do z ←2z −1 for i ←1 to 2n2 do for j ←1 to n + 1 do z ←z2 −z The first loop makes 6n iterations, the second double loop makes 2n2 · (n + 1) = 2n3 + 2n2 iterations, for the total of 2n3 + 2n2 + 6n iterations. For n ≥1 this is at most 2n3 + 2n3 + 6n3 ≤10n3, so the running time is O(n3).
2 Reference functions.
The main reason for using the asymptotic notation is that it allows us to estimate a complicated or difficult to determine function in terms of a simple looking function, for example: 197n5 + 13n3 −n2 + 21 = O(n5).
So when we write f(n) = O(g(n)), we typically use some simple functions g(n), whose behavior is well-understood. Thus, although it would be perfectly correct to write n5 = O(197n5 + 13n3 −n2 + 21), it would also be completely pointless, as n5 is a much simpler function than 197n5 + 13n3 −n2 + 21.
These are four types of functions that are most commonly used in the big-O notation: 1, log n, nb, cn, where b > 0 and c > 1. Occasionally powers of logarithms are used as well, that is functions loga n, for a > 0. In almost all applications, the asymptotic values are expressed in terms of these functions. For a warmup, we first compare functions n and 2n.
Claim: 2n ≥n + 1 for all integers n ≥1.
Indeed, this can be easily shown by induction. For n = 1, both sides are equal 2. For n > 1, assume the claim holds, that is 2n ≥n + 1. Then 2n+1 = 2 · 2n ≥2(n + 1) ≥n + 2. By the principle of induction, the claim holds.
This claim can be generalized to real numbers x, that is 2x ≥x + 1 for all real numbers x ≥1. To see this, note first that it holds for x = 1. The derivative of the right-hand side is 1. The derivative of the left-hand side is (2x)′ = 2xln 2 > 1 for x ≥1. So the left-hand side grows faster than the right-hand side, and the inequality follows.
As a conclusion, by taking logarithms of both sides, we also get that n > log n for n ≥1. So we have n = O(2n) and log n = O(n).
Logarithms.
Consider first logarithmic functions, f(n) = logr n, where r > 1. We know that any p > 1 logr n = logp n logp r = 1 logp r logp n.
What this formula implies is that all logarithmic functions are within a constant factor of each other, so for the purpose of asymptotic notation it does not matter what base we use. In particular, we can as well use base 2. So we have logr n = O(log n) for all r > 1.
Polynomials.
Polynomials are among the most common functions that appear in the analysis of algo-rithms. So we deal with polynomial functions first.
Theorem 0.2 Let f(x) = Pk i=0 aixi. Then f(x) = O(xk).
Proof: Let A = max |ai|, be the maximum absolute value of the coefficient in f(x). We can estimate f(x) as follows. For x ≥1 we have f(x) = akxk + ak−1xk−1 + ... + a1x + a0 ≤ A(xk + xk−1 + ... + x + 1) ≤ A(k + 1)xk.
Thus f(x) ≤cxk for c = A(k + 1) and x ≥1. The theorem follows. 2 Theorem 0.3 For all a, b > 0 and c > 1, we have: (a) 1 = O(loga n). (b) loga n = O(nb). (c) nb = O(cn).
3 Proof: Part (a) is obvious. We’ll skip the proof of (b). We will now prove part (c). Take d = c1/b. Since c > 1 and b > 0, we have d > 1. We then have n ≤ 1 + d + d2 + ... + dn−1 = dn −1 d −1 ≤ Adn, where A = 1/(d −1). The first inequality follows from d > 1 (which implies that di > 1 for all i). The middle equation uses the summation of the geometric sequences, and the in the last inequality we simply dropped the −1 in the numerator.
Now, recall that d = c1/b. The derivation above implies then that n ≤Ac(1/b)n, which is equivalent (raising both sides to the power of b) to nb ≤Bcn for B = Ab. We conclude that nb = O(cn), as claimed in (c). 2 In particular, for example, we have n100 = O(1.0000001n). The left-hand side is a fast-growing polyno-mial. The right-hand side is a slow-growing exponential function. Still, eventually, the second function will overtake the first one.
Combinations of functions.
Many functions can be expressed as sums or products of other more ele-mentary functions. To determine the growth rate of some functions, we can often do this by finding the growth rate of some of their components and using the theorem below.
Theorem 0.4 Suppose that f1(x) = O(g1(x)) and f2(x) = O(g2(x)). Then (a) f1(x) + f2(x) = O(max(g1(x), g2(x))). In fact, we also have f1(x) + f2(x) = O(g1(x) + g2(x)).
(b) f1(x)f2(x) = O(g1(x)g2(x)).
As a corollary, we get the following: if f1(x), f2(x) = O(g(x)) then f1(x) + f2(x) = O(g(x)) as well.
Proofs and examples in the book.
Exercise 1: Prove that the O-relation is transitive, in the following sense: if f(x) = O(g(x)) and g(x) = O(h(x)) then f(x) = O(h(x)).
Exercise 2: We showed that all logarithms have the same rate of growth. What about exponential functions?
More specifically, for any a, b > 1, is it always true that ax = O(bx)? Justify your answer.
Example 7.
Determine the asymptotic value of function f(n) = 5n2 log6 n + n3 log n + 2√n + 1. We can ignore all constant coefficients. Then we eliminate low order terms one by one. First, we can discard 1, because 1 = O(n3 log n). Next, let’s look at √n. We have √n = n1/2 = O(n3), so √n = O(n3 log n) as well.
Finally, we compare n2 log6 n and n3 log n. Since log5 n = O(n), multiplying both sides by n2 log n we get n2 log6 n = O(n3 log n). Putting it all together, we get f(n) = O(n3 log n).
Example 8.
Determine the asymptotic value of function f(n) = 7n52n + 3n. Again, we can ignore the constant coefficients. Comparing the two terms, the intuition is that 3n grows much much faster than 2n, even if we multiply 2n by a polynomial. For a rigorous proof, divide both terms by 2n, getting n52n/2n = n5 and 3n/2n = (1.5)n.
We know that n5 = O(1.5n). Now, multiplying it back by 2n, we obtain n52n = O(3n). Therefore f(n) = O(3n).
4 Big-Ωand Θ notations.
Notation f(x) = O(g(x)) means that f(x) does not grow faster than g(x), but it could be the case that f(x) actually grows slower. For example, we have n2 + 3n + 1 = O(n3), or n2 + 3n + 1 = O(2n), etc. But these bounds are “wasteful”, because we also have n2 + 3n + 1 = O(n2), which is a more accurate and, in fact, the best possible estimate. This motivates the introduction of other notations for asymptotic growth of functions.
Definition.
We say that f(x) is Ω(g(x)), and write f(x) = Ω(g(x)), if there are constants C and k such that |f(x)| ≥C|g(x)| for all x > k.
We have f(x) = O(g(x)) if and only if g(x) = Ω(f(x)). (Easy exercise.) We write f(x) = Θ(g(x)) if f(x) = O(g(x)) and at the same time f(x) = Ω(g(x)). Thus the Θ-notation provides the “exact” (asymptotic) estimate for function growth.
In other words, f(x) = Θ(g(x)) if there are constants C1, C2, and k such that C1|g(x)| ≤f(x) ≤C2|g(x)| for all x > k.
See examples in the book.
Example 9.
Consider function f(n) = 3n2 +2n−7. We start with an upper bound: f(n) = 3n2 +2n−7 ≤ 3n2 + 2n2 = 5n2, so f(n) = O(n2). Next, we get a lower-bound estimate. For n ≥4 we have 2n ≥7, so f(n) = 3n2 + 2n −7 ≥3n2. Therefore f(n) = Ω(n2). Putting these two bounds together, we obtain that f(n) = Θ(n2).
Example 10.
Consider harmonic numbers again, Hn = 1 + 1 2 + ... + 1 n. We proved earlier an upper bound Hn ≤log n + 1, which implies Hn = O(log n).
But we also have a lower bound Hn ≥1 2 log n −1 2. Notice that 1 4 log n ≥1 2 for n ≥4. So for n ≥4 we can estimate Hn as follows: Hn ≥1 2 log n −1 2 = 1 4 log n + ( 1 4 log n −1 2) ≥1 4 log n. This implies that Hn = Ω(log n), and we can thus conclude that Hn = Θ(log n).
Example 11.
Let us consider again an algorithm consisting of a number of loops. Our goal is to determine its running time, using the asymptotic notation Θ. In the previous example, Algorithm WhatsMyRun-Time1, the running time was O(n3), and in fact this was the best estimate, so the running time was actually Θ(n3).
To make it more concrete, let us determine the number of words (rather than the running time) printed by the following pseudo-code: Algorithm HowManyHellos (n : integer) for i ←1 to 6n do print(“HELLO”) for i ←1 to 4n + 1 do for j ←1 to 3i + 2 do print(“HELLO”) The first loop will print 6n words, that’s easy. What about the second nested loop? Unlike in Algo-rithm WhatsMyRunTime1, we cannot simply multiply the ranges of the two loops, because the number of iterations in the inner loop depends on i; it changes from one iteration to next. For any given i, it will print 3i + 2 words. Therefore we can write the total number of words using the summation notation as 6n + P4n+1 i=1 (3i + 2). Simplifying, and using the formula for the sum of an arithmetic sequence, we get 6n + 4n+1 X i=1 (3i + 2) = 6n + 4n+1 X i=1 (3i) + 4n+1 X i=1 2 = 6n + 3 4n+1 X i=1 i + 2(4n + 1) = 6n + 3(4n + 1)(4n + 2)/2 + 2(4n + 1) = 24n2 + 32n + 5 = Θ(n2).
5 |
6765 | https://ntrs.nasa.gov/api/citations/19660011808/downloads/19660011808.pdf | SHOCK TUBE DETERMINATION OF THE DRAG COEFFICIENT OF SMALL SPHERICAL PARTICLES by Bmce P. Selberg Prepared under Grant No. NsG-86/23-05-003 by UNIVERSITY OF MICHIGAN East Lansing, Mich. for NATIONAL AERONAUTICS AND SPACE ADMINISTRATION WASHINGTON, D. C. APRIL 1966 I TECH LIBRARY KAFB, NM SHOCK TUBE DETERMINATION OF THE DRAG COEFFICIENT OF SMALL SPHERICAL PARTICLES By Bruce P. Selberg Distribution of this report is provided in the interest of information exchange. Responsibility for the contents resides in the author o r organization that prepared it. Prepared under Grant No. NsG-86/23-05-003 by UNIVERSITY OF MICHIGAN East Lansing, Mich. for NATIONAL AERONAUTICS AND SPACE ADMINISTRATION For sale by the Clearinghouse for Federal Scientific and-Technical%formation Springfield, Virginia 22151 - Price $1.50 I ACKNOWLEDGEMENTS This investigation was supported by the National Aeronautics and Space Administration in the form o f a grant, NsG-86-60. The author wishes to thank Professors J.A. Nicholls and Martin Sichel for their helpful suggestions and criticisms. Special appreciation is given to Professor J.A. Nicholls and Stuart W. Bowen for their advice, dis- cussions, and encouragement throughout the study. Further appreciation is extended to Philip Malte who wrote the computer programs and helped with experiments, to Cletus Iott who pro- vided assistance with the electronic instrumentation and the optical system, to Pai-Lien Lu who took the photomicrographs, and to other members of the Aircraft Propulsion Laboratory who contributed in various ways to this work. iii TABLE OF CONTENTS ACKNOWLEDGEMENTS LIST OF TABLES LIST OF FIGURES NOMENCLATURE ABSTRACT I. 1 1 . 1 1 1 . Page iii vii viii X xiii INTRODUCTION 1 1.1 ORIGIN OF THE PROBLEM 1 FICIENT OF SPHERES 3 1.3 PURPOSE OF THIS STUDY 8 PARTICLE EQUATIONS OF MOTION 9 EXPERIMENTAL APPARATUS 13 3 .
1 CHOICE OF EXPERIMENTAL FACILITY 13 3.2 DESCRIPTION OF FACILITY 13 3,2-1 Shock Tube and Driver Section 14 3.2-2 Optical Equipment 16 3.2-3 Particle Injector System 21 3.2-4 Shock Speed Measurement 22 3.2-5 Pressurization and Vacuum System 22 3.2-6 Sequential Timing During Experiment 24 3 .
2 -7 ' Experimental Procedure 24 3.3 CALIBRATION OF EQUIPMENT 28 1.2 REVIEW OF THE STUDIES ON THE DRAG COEF- 3.3-1 3.3-2 3.3-3 3.3-4 3.3-5 Shock Velocity Measurement 28 Schlieren Photographs of Shock Front 31 Calibration of Optical Equipment 31 Shock Tube Attenuation 33 Shock Tube Test Time 34 V TABLE OF CONTENTS (cont) Page IV. DATA REDUCTION AND ERROR ANALYSIS 4.1 SCOPE OF EXPERIMENTAL ANALYSIS 4.2 DATA REDUCTION EQUATIONS 4.3 REDUCTION OF EXPERIMENTAL DATA 4.4 TYPICAL EXAMPLE OF DATA REDUCTION 4.5 GENERAL EQUATION FOR DETERMINING 4.6 TYPICAL EXAMPLE OF PROBABLE ERROR PROBABLE ERROR V. RESULTS AND DISCUSSION OF RESULTS 5.1 EXPERIMENTAL RESULTS 5.2 SURFACE ROUGHNESS 5.3 UNSTEADINESS I N THE BOUNDARY LAYER 5.4 FREE STREAM TURBULENCE 5.5 PARTICLE ROTATION 5.6 PARTICLE ACCELERATION AND WAKE VI. CONCLUSIONS 36 36 37 38 43 47 49 55 55 59 71 74 74 75 77 APPENDIX 79 REFERENCES 87 vi LIST OF TABLES Table Page 1 PARTICLE CHARACTERISTICS 37 2 DATA SUMMARY - GLASS BEADS 0 < MR < - .15 3 DATA SUMMARY - GLASS BEADS .15 < MR < .30 80 84 4 DATA SUMMARY - WINCHESTER WESTERN HP 295 BALL POWDER 0 < MR < .125 85 5 DATA SUMMARY - SAPPHIRE BALLS .14 < MR < .313 86 vii LIST OF FIGURES Figure Page 1 Mach Number - Reynolds Number Flow Regimes Encountered by a Five-Micron Particle in a Rocket Nozzle 2 2 Data on the Drag Coefficient of Spheres 4 3 Shock Tube Driver Section 4 Optical System 15 17 5 Schematic Diagram s f Microflash System 19 6 First Switching Unit and Xenon Flash Tube 7 Particle Injector 20 21 8 Schematic Diagram of the Pressurization and Vacuum System 23 9 Schematic Diagram of Time Delay Sequencing System 25 10 Circuit Diagram of Time Delay Unit 26 11 General Shock Tube Facility 27 12 Pressure Transducer Outputs 13 Shock Tube Test Time 30 35 14 Typical Example of x versus t Data of a Particle 44 15 Displacement versus Time Curve for Third Order Polynomial 46 16 Drag Coefficients versus Reynolds Number for Third Order Polynomial Data Reduction Technique 57 17 Drag Coefficient versus Reynolds Number for Mean Drag Coefficient Method 58 viii LIST OF FIGURES (cont) Figure Page 18 Drag Coefficient versus Reynolds Number - Glass Par- ticles, Ball Powder, and Sapphire Balls 60 19 Photomicrograph of Glass Particles, Magnification = 200 61 20 Photomicrograph of Glass Particle, Magnification = 1840 62 21 Photomicrograph of Glass Particle, Magnification = 1840 63 22 Photomicrograph of HP 295 Ball Powder, Magnification = 110 65 23 Photomicrograph of HP 295 Ball Powder, Magnifica- tion = 190 66 24 Photomicrograph of HP 295 Ball Powder, Magnifica- tion = 1840 67 25 Photomicrograph of Sapphire Balls, Magnification = 202 68 26 Photomicrograph o f Sapphire Balls, Magnification = 1850 69 27 Photomicrcgraph of a Washed Sapphire Ball, Magnification = 17 60 70 ix I ’ NOMENCLATURE A Projected characteristic area based on particle diameter A Acceleration modulus (ad/ UR ) 2 C At m At.mosphere a cD cf d F I SP m M MR P P V Re T t Speed of sound r) Drag Coefficient (FD/ 1/ 2 p URy A) Skin friction coefficient Sphere diameter Force vector Acceleration due to gravity Mercury Specific impulse Mass Mach number Relative Mach number (UR/a) Camera magnification Pressure Evacuated pressure (P - PI) at m Reynolds number (pURd/ p ) Temperature Time X NOMENCLATURE (cont) r Non-dimensional time U Axial velocity u2 Convective flow velocity uR Relative velocity (U = U2 - Vp) R U Sphere rotation speed V Particle velocity with respect to inertial space S P X CY 6 6 P V Displacement of particle Particle acceleration with respect to inertial space Boundary layer thickness Displacement thickness Density Viscosity Microns Kinematic viscosity (v = p / p ) Axial velocity ratio (u /u ) P g U Relative turbulent intensity SUBSCRIPTS g Gas conditions m Microsecond timer xi NOMENCLATURE (cont) SUBSCRIPTS (cont) P Particle conditions R Relative conditions W Wake conditions 1 Initial conditions at test section 2 Convective flow conditions xii ABSTRACT An experimental study was conducted to determine the drag coef- ficient of inert spherical particles accelerating in a laminar, non-reacting, incompressible continuum flow. The Reynolds number range which was covered in the study was from 150 to 1700, and particle sizes ranged from 150 iI to 450 p. The convective flow behind the shock wave in a shock tube was used to accelerate the particles. The particle's diameter and the displacement versus time measurements were obtained using a rotating drum camera in conjunction with an oscillating light source. The photographic data, the particle density, the shock speed, and the initial pressure and tempera- ture in conjunction with the normal shock relations were combined to cal- culate the drag coefficient. The drag coefficient is usually considered to be a function only of Reynolds number and acceleration modulus, however, C varies consider- ably because of particle roughness. Experiments with H P 295 ball powder, whose surface is relatively rough, produced results which were as much as 85 per cent higher than the steady state curve, with the increase depen- dent upon the relative Mach number of the flow about the particle. Similar drag coefficient experiments with smooth sapphire balls did not produce the scatter, the higher values, nor the dependence on relative Mach number. D xiii I. INTRODUCTION 1.1 ORIGIN OF THE PROBLEM In order to improve the density and specific impulse characteristics of solid propellant rocket fuels, small metal particles are mixed into the propellant. During combustion, condensed liquid and solid metal oxide particles are formed from the combustion products. These particles com- prise 30-40 per cent by weight of the combustion products in current solid propellant rocket motors. Because of inertia effects, these particles leave the nozzle at lower velocity which means a loss of momentum and hence a loss in specific impulse. In order to calculate the loss in specific im- pulse due to the velocity and temperature lags, the drag coefficient of the particles is one of the variables which must be known. Presently most of the specific impulse loss calculations are made using the "standard drag coefficient curve" for spheres. This curve is only valid for a single smooth sphere in a steady, incompressible, laminar, non-reacting, and continuum flow field, conditions which are certainly not satisfied in a rocket nozzle. Thus, in a rocket engine a particle is moving in an accelerating, turbulent, compressible stream whose temperature is dsferent than that of the particle. The particle also moves from the con- tinuum regime to the slip flow regime and possibly to the free molecule regime depending on its size and the flow in the rocket nozzle. Figure 1 1 1 0 1 .
0 0 z = .IO .o I PC= 7.35X10-5 Ibm/ft-sec . . . . . . . , , . . . . . . . . . , . , , , , , . . . 0. I 1 .
0 IO 100 1 , 0 0 0 l0,OOO REYNOLDS NO. P g D(Ug-"p) P v 9 Figure 1. Mach Number-Reynolds Number Flow Regimes Encountered by a Five-Micron Particle in a Rocket Nozzle . represents the path in the Mach number Reynolds number regime which a five-micron particle, produced under chamber conditions typical of solid propellant rocket motors, can experience"). In some cases enough par- ticles are present such that the particles and the gas must be analyzed as a two-phase flow. The particles may remain hot enough, due to the tem- perature lag, to emit electrons by means o f thermionic emission or to change the drag due to heat transfer. Some of the particles which come through the nozzle are the original metal particles which are mixed into the propellant. The drag coefficient of these particles will be a function of all the above effects plus burning. In order then to determine the correct drag force on a particle in a solid rocket motor nozzle, the influence of all of the above parameters must be studied. 1.2 REVIEW OF STUDIES ON THE DRAG COEFFICIENT OF PARTICLES Some theoretical and experimental work has been done to determine the drag coefficients of particles accelerating in the incompressible con- tinuum flow regime. The results of these studies differ appreciably from one another as may be seen in Fig. 2. Ingebo(2) conducted experimental studies on the vaporization rates and drag coefficients €or isotane sprays accelerating in turbulent air streams. He injected the liquid drops into air streams moving at 140 3 too IO 0. I .01 0. I I .o IO 1 0 0 Re 1000 10000 Figure 2. Data on the Drag Coefficient of Spheres I and 180 ft/ sec. A specially designed camera was used to obtain drop-size distributions and drop-velocity data. He was then able to obtain vaporiza- tion rates and drag coefficients for the liquid drops. He found his results could be represented by a single curve given by 27 c = D Re.84 Fledderman and Han~on'~) performed similar experiments by photographing spray droplets accelerating in streams moving from 50 to 75 ft/sec. Their results are one hundredth the steady state value for a sphere. Experimental drag coefficient studies of burning kerosene drops have been done by Bolt and W 0 d 4 ) for Re < 1. These results indicate a de- crease in the drag coefficient due to burning. Habin et al. (5) determined the drag coefficient o f burning and non- burning liquid fuel droplets accelerating due to the convective flow behind a shock wave. For Re > 200 Rabin's drag coefficients were greater than those of a sphere in steady flow. Rabin's data also indicates a de- crease in drag coefficient due to burning. Rudinger@) also used the convective flow behind a shock wave to determine the drag Coefficient of accelerating glass beads, which had an average diameter of 29 microns. The x versus t motion was recorded by streak photography. The Reynolds number range of the experiment 5 was from 40 to 300. Rudinger found that all his data could be correlated by the expression 6000 D Rel.7 c = This relationship deviates widely from the steady state curve for spheres and Rudinger suspects that electric charges on the particles may be the cause of the deviation. A very thorough literature survey on "The Fundamental Aspects of Solids-Gas Flow" was made by Torobin and G a ~ v i n ( ~ - ' ~ ) . They listed and discussed such problems as the sphere wake in laminar fluids; ac- celerated motion of a particle in a fluid; the effects of particle rotation; roughness and shape; and the effect of fluid turbulence on the particle drag coefficient. They also made an experimental study of the drag coefficients of single spheres moving in steady and accelerated motion in a turbulent fluid. Small radioactive smooth spheres were fired into a turbule.nt flow wind tunnel of known turbulence intensity and the motion of the sphere was recorded by means of a radioactive sensing device. Using this technique, continuous x versus t data could be obtained for the spheres. It was found that by increasing the turbulent intensity level, the critical Reynolds number could be shifted to Reynolds numbers as low as 400 and that drag coefficient was independent of acceleration in turbulent flow. 6 Recently Crowe''') did both an analytical and experimental study to determine the effects o f burning, evaporation, and acceleration on the drag coefficients of particles accelerating in gas streams. The Reynolds num- ber range extended from 250 to 1600. Both the analytical and experimental portions of the study were confined to the incompressible continuum flow regime. For the analytical study Crowe chose a spherical model with mass flux through the surface to simulate burning and evaporation. The tangential equation of motion was used as the governing equation in the analysis of the boundary layer flow about the sphere. The velocity distribution outside the boundary layer was assumed to be that corresponding to inviscid flow. With the proper boundary conditions,the equations were solved indicating that for burning or evaporating particles the skin friction coefficient is reduced. A constant form drag coefficient was assumed and used for a11 the results. The total drag coefficient is then just the sum of the skin friction and form drag coefficients. For the experimental portion of his study, Crowe used the convective flow behind a shock wave to accelerate the particles. A high speed framing camera was used to record the particle diameters and x versus t history. With the shock speed, the particle density, and the local temperature and pressure known, Crowe was able to calculate the drag coefficient. Crowe's particle size varied from 100 to 350 microns. 7 Crowe's best fit curve to his experimental non-burning data is approxi- mately 15 per cent higher than the steady drag coefficient curve. His ana- lytical results are another 15 per cent above his experimental data. How- ever, each experimental point has approximately a 40 per cent probable error. 1 . 3 PURPOSE OF THIS STUDY The present study is part of an overall investigation to study the dynamics of inert, reacting, and charged particles in solid rocket motor nozzles, and is a continuation of Crowe's work. Basically it is desired to experimentally determine the relation between CD and the prime variables Re, MR/ & , MR, and a non-dimensional burning rate parameter. In addition the effect of the secondary variables, relative turbulent intensity, unsteady effects in the particles wake, particle rotation, and particle roughness are also to be studied. Due to the relatively large experimental error and scatter in Crowe's data which would obscure slip and compressibility effects, it was necessary to repeat the incompressible continuum flow regime before conducting experiments in the slip flow regime with compressibility effects. In order to determine C more accurately, a new and better instrumented shock tube was constructed. This study is primarily concerned with a more accurate determination of the drag coefficient of small spherical particles D accelerating in the incompressible continuum regime. 8 II. PARTICLE EQUATIONS OF MOTION Newton's second law of motion states that the rate of change of mo- mentum of a particle is equal to the sum of the forces which act on the particle and is in the direction in which the sum of the forces acts. In mathematical form Newton's law is = m a 4 (1) The forces are the viscous and pressure forces which act on the particle surface and the body forces which act on the particle mass. J f the pres- sure and viscous forces are expressed in terms o f a drag coefficient, then Eq. (1) can be written as where C , , = drag coefficient of the particle cy = acceleration o f the particle with respect to inertial space p = density of the fluid U = relative velocity between the particle and the fluid R A = projected characteristic area based on particle diameter m = mass of particle f = body force per unit mass. For the case of burning particles Eq. (2) would contain an additional term for the momentum flux from the particle's surface. The body force term 9 in'this analysis is the gravitational force term that acts on the body. In the present study the particle is accelerated by the convective flow behind a shock wave so that the acceleration due to the viscous and pressure forces is much greater than t.hat due to gravity, i. e., Thus the gravitational force term will be neglected. Equation (2) may be written as 4 -c cD!j IuRI uR A = m a Since the flow velocity and the acceleration vector are in the same direc- tion,the vector notation may be dropped and Eq. (3) becomes For spherical particles o f uniform density, 4p. c u d P D 2 c = 3P UR where p. = particle density P d = particle diameter. 10 I Dimensional analysis indicates that the other similarity parameters which are needed to properly determine the drag coefficient of a non- burning smooth spherical particle in a laminar flow field subject to com- pressibility and non-continuum effects are Reynolds number, Re, Mach number, M M /6, and acceleration modulus, Ac. In terms of the present notation Reynolds number and Mach number are R’ R P U R d Re = I-1 uR M = - R a The particle flow regime is determined by the value of MR/ f i e . The different flow regimes are defined as follows : (14) < _ I MR < 10-1 6 - 10-1 < - a - MR < 3 s - MR > 3 Re continuum regime slip flow regime transition regime free molecular flow regime 11 As seen in Fig. 1, a five-micron particle in a typical solid propellent rocket nozzle experiences all of the above flow regimes as it travels through the nozzle. Equatioqs (5) and (6), along with experimental data, can be used to calculate the C and Re of a non-reacting spherical particle. Al- D though this report is concerned only with the incompressible continuum regime, the calculation of M and M R / 6 are needed to insure that R’ the particle is in the desired flow regime. 12 III. EXPERIMENTAL APPARATUS 3.1 CHOICE OF EXPERIMENTAL FACILITY In order to study the dynamics of solid particles in rocket nozzles, an experimental facility must be able to produce a particle environment under which the following parameters can be studied: 1. Acceleration modulus 2. Mach number 3. Mach number/ dReynolds number 4. Burning rate parameter 5. Electric charges on particles A shock tube was chosen because by using the convective flow behind the shock front in connection with the small spherical particles (50-500 microns in diameter), it is possible to produce the flow conditions under which the above parameters can be studied. This type of facility was used successfully by both Crowe(13) in studying the drag coefficient of solid particles and by in studying the shattering of liquid drops. 3.2 DESCRIPTION OF FACILITY The experimental equipment consists of a horizontal shock tube into which the particles are injected and then accelerated by the convective flow field behind the shock front. The distance, x, versus time, t, of the particles is recorded by a modified AVCO rotating drum camera in conjunction 13 I with a high voltage switching circuit which supplies energy to a Xenon flash tube. By combining the x versus t data, the shock strength, and the initial conditions, the drag coefficient of the particle can be obtained. 3.2-1 Shock Tube and Driver Section The 1 3/ 8 inch square shock tube consists of a three foot stainless steel driver, a six foot stainless section between the driver and the test section, a one and one-half foot aluminum test section, and a six foot stainless section downstream of the test section. The diaphragm material is ruptured by means of a long rod which is inside the driver section and is driven by a Saval 24 volt D. C. solenoid. A photograph of the driver section appears in Fig. 3. Two different types of material have been used as diaphragms for the weakest shock waves. First, Dupont 220 MD-31 cellophane was used. This material tended to shatter into small pieces which necessitated frequent swabbing of the shock t.ube. The second type diaphragm material that was used was Dupont mylar, 0015 inches thick. The mylar was a tougher material and was used for the stronger shock runs. Unlike the cellophane, the mylar did not shatter when punctured but just folded back. However, more energy was lost in this folding process and it took a higher driver pressure using mylar to achieve the same shock Mach number than it did when cellophane was used. 14 Figure 3. Shock Tube Driver Section I 3.2-2 Optical Equipment The test section windows, through which the x versus t history of the particles is recorded, are 1 1/2 inches high and 4 inches long o f optically- flat glass A modified AVCO rotating drum camera is used in conjunction with an oscillating light source to take shadowgraph pictures of the particle's trajectory. The physical layout s f the optical system is reproduced in Fig. 4. The drum camera is mounted on a lathe bed with two compound rests, with rotating drum and main camera body on one compound rest and the camera lens mounted on the other. A flexible bellows connects the lens and the camera body. This arrangement permits accurate and independent movement of the lens and body both perpendicular and parallel to the test section. Thus the magnification can be changed as desired and the length of the test section can be traversed with the lathe setup. The lens is a Goerz Red Dot Art= process lens, which is a highly corrected lens de- signed for applications where the image to film distance and the object to film distance are o f the same order. The light source is a PEK XE-9 Xenon flash tube. The energy is switched to the Xenon tube via a modified version of a Edgerton, Germeshausen, and Grier LS-10 multiple microflash system. This system 16 tr includes a low voltage power supply, a pulse shaper unit, a time delay unit, and five high voltage discharge units. The switching circuits con- trol the discharge to the Xenon flash tube of one . 001 microfarad capacitor bank and four 005 microfarad capacitor banks, all charged to 12 kilovolts. The switching circuit is activated by a signal generator pulse-shaper unit. By varying the frequency of the signal generator, it is possible to vary the time interval between flashes from 10 milliseconds to 10 microseconds. A schematic diagram of the modified EG and G unit is in Fig. 5. To achieve the shortest possible rise time, it is necessary to reduce the circuit inductance to a minimum since the resistance of the ionized Xenon gas is only 2 52 This was accomplished by modifying the high voltage discharge units of the multiple microflash system. The high voltage section of the first switching unit was installed as close to the Xenm t.ube as possible as is in Fig. 6. A spark gap is used in the first unit as the switch for the energy. This was done since the spark gap will switch the current faster than the mercury diode of the other four units. With the above arrangement it was possible to obtain a 480 nanosecond width for the light pulse to decay to 10 per cent of its peak value. The Xenon flash lamp is mounted on a drill press milling vise, per- mitting motion bcth perpendicularly and parallel to the test section. Both collimated and quasi-collimated light have been used in the system. 18 TIME DELAY UNIT - I \ I c b 1 Low out High Signal 1 L Voltage Pulse Output Voltage POWER SUPPLY PULSE SHAPER AUDIO OSCILLATOR Sine wave Sine out Output - Input m Figure 5. Schematic Diagram of Microflash System 19 First high voltage switching unit Xenon flash tube Figure 6. First Switching Unit and Zenon Flash Tube 20 Two types o f film have been used in the experiment: Kodak Royal Pan and Kodak Plus X film. 3.2-3 Particle Injector System Within the particle injector, shown in Fig. 7, the particles are placed on a circular platform which is rotated on its axis by means of a solenoid thus injecting the particles into the test section. The particle injector is designed so that its pressure is the same as that in the test section and the hole in the test section is .059 inches in diameter. Particle Solenoid Figure 7 . Particle Injector 21 3,2-4 Shock Speed Measurement The shock speed is measured with two flush-mounted Kistler model 701A quartz pressure transducers, 1.1667 feet apart, in conjunction with a microsecond timer, The transducers have a rise time of ten micro- seconds. Two Kistler model 566 multi-range electrostatic charge amplifiers are used to amplify the signals from the transducers. These signals are in turn fed to the start and stop channels of a Transistor Specialties Incorporated microsecond Timer. The above transducers are very sensitive and thus allow the recording of very weak shock waves. Desiring the best possible measurement of the wave speed, the transducers were mounted equally upstream and downstream o f the test section. 3.2-5 Pressurization and Vacuum System A schematic diagram of the pressurization and vacuum system is shown in Fig. 8. The system was designed so that the shock tube and driver section could both be evacuated to any desired pressure. The valve arrangement is such that the initial pressure can be controlled in the driver section and shock tube independently. In a similar fashion both the driver section and the shock tube can be pressurized in an inde- pendent controlled fashion. With this flexibility it is possible to maintain certain desired shock strengths while varying the Re number and M R / 6 . 22 I Driver Atm -@- Regulator r c I Air at 25 psi Tube f Diaphragm Test Section I r( d - - n I B Shock 3 HQ Manometer Measures AP I -@ Atm c 3 i Vacuum At m I Hg Manometer Measures PV At m Figure 8. Schematic Diagram of the Pressurization and Vacuum System 23 I 3.2-6 Sequential Timing During Experiment To have the particles in correct position and to begin the photographing sequence at that time, an accurate timing sequence is necessary, Figure 9 is a schematic diagram of this sequencing system. Once the particles are dropped, there must be an accurate delay before the solenoidal drive rod punctures the driver diaphragm. This is accomplished by means of a thyra- tron time delay unit. A circuit diagram of this circuit appears in Fig. 10. A second sequencing unit is needed to start the photographic process, This is accomplished by having the signal from the first pressure trans- ducer sent to a second time delay unit which, after a sufficient delay al- lowing time for the shock to travel to and interact with the particles,sends another signal to the first high voltage switching circuit. The high voltage switching circuit then dumps its energy into the XE-9 flash tube and the picture-taking sequence is begun. Figure 11 is a view of the overall facility . 3.2-7 Experimental Procedure The normal experimental procedure is as follows: 1. Load camera 2. Record atmospheric pressure 3. Record ,atmospheric temperature 4 . Install diaphragm 24 I Diaphragm Solenoid Plunger I - Time Delay i Unit - 1 J f Particle Dropper Solenoid L Start switch Figure 9. Schematic Diagram of Time Delay Sequencing System 25 - - l40Kil Lu b , x x - - - - - - - Figure 10. Circuit Diagram o f Time Delay Unit Figure 11. General Shock Tube Facility 5. 6. 7 . 8. 9. 1 0 . 11. 12. 13. Load injector Evacuate, pressurize, or both Adjust sensitivity levels Read test section pressure Run rotating drum camera up to speed Engage start switch Record microsecond timer reading Mark and remove film from camera Develop film. 3 . 3 CALIBRATION OF EQUIPMENT All of the parameters except p which are needed to calculate C are P D obtained from data that is recorded during an experiment. Thus it is es- sential that the experimental equipment has been properly calibrated, and this calibration procedure is described below. 3. 3-1 Shock Velocity Measurement The convective flow velocity, U and the gas density in the convective 2’ flow region, p are obtained from the measurements o f shock velocity, C initial temperature, and initial pressure and application of the normal shock relations. For weak shock waves U is very sensitive to changes in C or the time for the shock wave to travel the distance between the two pressure pickups. This time measurement must therefore be very accurate. 2’ S’ 2 s9 28 The linearity and slope of the signals from the pressure transducers were checked by putting the output from the amplifiers of the transducers into a Tektronix oscilloscope and photographing the trace with a Polaroid attachment. Photographs of the traces were taken of the start transducer and its amplifier and of the stop transducer and its amplifier at equal shock strengths. The amplifiers were then interchanged and the procedure was repeated at the same shock strengths. Upon comparing the results it was found that the four traces are virtually indistinquishable. Two pictures of such traces appear in Fig. 12a and 1%. The sensitivity adjustments on the start and stop channels of the microsecond timer were calibrated so that both channels could be set at the same level. The gain on the amplifier and the sensitivity on the microsecond timer were always adjusted so that both channels of the timer would be activated on about the first 10 per cent of the signal from the pressure transducers. Since the slopes and linearity of the two transducer amplifier com- binations were the same and the sensitivities on the timer were adjusted to the same level, the shock speed that was measured by the transducer microsecond timer combination would be the actual shock speed if shock 29 Start transducer Stop transducer 5,uSEC/DIV Figure 12. Pressure Transducer Outputs 30 speed decay was unimportant. By setting the same driver to test section pressure ratio and making a series of shock runs, the time interval read on the microsecond timer would consistently be within plus or minus one microsecond out of 900 microseconds. 3.3-2 Schlieren Photographs of Shock Front A Schlieren system was set up to verify the existence of a plane shock front and to make sure that the injector hole did not cause the front to bend locally. A spark gap triggered by the first unit of the multi-microflash was used as the light source for the Schlieren system. By incorporating the spark gap with the microflash system and varying the time delay in the microflash system, it was possible to take photographs of the shock front at various locations along the viewing area of the test section. . In all cases the shock front was plane and perpendicular to the direction of flow. The fact that the shock front is plane and is perpendicular to the flow di- rection is justification for using the normal shock-tube relationships. 3 . 3-3 Calibration of Optical Equipment The calibration of the optical equipment is extremely important for both the particle diameter, and the distance versus time data are obtained photographically. The camera was focused and the magnification was determined in the following way. A cylindrical rod, .043 inches in diameter, was inserted through the particle injector hole parallel to the test section walls. The rotating drum camera was then focused on the rod, 31 and a picture of the rod was taken with the drum stationary. The film was developed and the negative was then viewed through a Bausch and Lomb microscope with graduated eyepiece to determine the size of the image. The magnification of the microscope had previously been determined using the grid from an Edmnd comparator. Thus the magnification of the camera was known, Some doubt remained as to whether this magnification was correct in that the calibration probe was six times larger in diameter than the spherical particles and the test object was not spherical like the particles. Accordingly, synthetic sapphire spheres, .015 inches in dia- meter ('t . 0001 inches), were dropped through the particle injector and photographed. The photographic images of the sapphire balls yielded sphere diameters which were within . 1 per cent of the actual sapphire sphere diameter. Thus, the method for determining the magnification factor proved acceptable. Another optical test was made to determine whether the lines on a grid pattern remained straight and undistorted so that the particle dfsplace- lnent measurements would be true. A grid network on a clear plastic was inserted into the test section and photographed. By viewing the re- sulting image on the film no distortion could be detected. Thus the dis- placement measurements on the film would be a true representation of the actual movement o f the particle. 32 " To determine the influence of development time on image size, a final optical test was made. Three wires, 0.0036 inches, 0.0070 inches, and 0.0104 inches in diameter, were photographed nine times under the same conditions. Using Kodak D-11 developer at 6 8 ' F , each piece of exposed film was developed at times ranging from five to nine minutes at half- minute invervals- seven minutes being normal. Upon microscopic examination, no apparent diameter change could be detected in any of the three wires. Thus in actual experimental runs, it was determined that the development time could be varied plus and minus one minute without affecting the validity of the experimental results. 3. 3-4 Shock Tube Attenuation Ideally a shock wave propagates at a constant velocity in a shock tube. In actuality, however, the shock wave attenuates. The amount of this attenuation depends on several things: the strength of the original wave at the diaphragm, the distance o f the measuring device from the dia- phragm, and the size of the shock tube. An experimental study of at- tenuation in shock tubes was conducted by R. J. Emrich and C. W. Curtis . They conducted their experiments in shock tubes of various sizes. For a shock tube with the same hydraulic radius as in the present study, they found the following relationship to be valid within a factor of two. (1 5) 2 dM -4M - 1 dx - 5.08 x 10 33 The shock Mach number was always less than 1.25. For M = 1.25 dM/dx is ” AM - 2.3 x Ax For x = 14 inches A M =. 003 Thus the attenuation over the interval in which the velocity is measured is negligible and the convective flow conditions can be calculated directly without any correction for attenuation. 3.3-5 Shock Tube Test Time The shock tube test time depends on many factors: driver section length, the distance between the driver section and the test section, the distance between the test section: and. end of the shock tube, the particular gases or gas being used, the strength of the shock, and the local speed of sound. The test time can be determined analytically, however, the actual test time is usually less and thus a series of experiments were performed at various shock strengths to determine the actual test time. This was accomplished by putting the output of the pressure pickups into a Tektronix oscilloscope and photographing the traces. The first pressure pickup started the sweep of the oscilloscope. A typical trace is shown in Fig. 13. The test time extends from the instant of pressure increase to 34 0 0 cu 2 mSEC / DIV Figure 13. Shock Tube Test Time the onset of the rarefaction wave and from the above figure is 4.4 milli- seconds. The second pressure pulse is from the reflected shock wave. The shortest run time that was recorded over the desired range of shock strengths was 4.2 milliseconds. Thus all of the experimental runs were made in 4.2 milliseconds or less. 35 IV. DATA REDUCTION AND ERROR ANALYSIS 4.1 RANGE OF EXPERIMENTAL MEASUREMENTS The experimental study consisted of determining the drag coef- ficient of non-burning particles under unsteady flow conditions. For all the experimental runs, the relative Mach number of the flow was always less than 3. The M /& was such that the flow was always in the con- tinuum regime. The Reynolds number ranged from 150 to 1700. R Three different types of particles were used in the study: glass beads, Winchester Western H P 295 ball powder, and synthetic sapphire balls. The superbrite glass beads were obtained from Reflective Products Division of the Minnesota Mining and Manufacturing Company. The glass beads for the most part were spherical; however, some discretion was needed in examining the photographs o f the glass beads to sort out the few beads which were not spherical. The Winchester Western HP 295 ball powder likewise was mostly spherical, but again, some selection of data was necessary to eliminate the non-spherical powder. The preci- sion-lapped sapphire balls, obtained from Industrial Tectonics, Inc., were very spherical (within 0.000010). The balls are also uniform in their diameter ( 0.0001 inches), and their surface finish is precise, 1. 5 microinches. A table listing some o f the particle characteristics appears below: 36 TABLE I Type of Particle Density Size Range or Exact Size (gm/ cc) , Superbrite Glass 2.49 150 j i " 2501, Beads Winchester -Western 1.67 280 c- 3 5 0 F HP 295 Ball Powder Sapphire Balls 3.978 396.8 The density of the Superbrite glass beads were obtained experimentally using a Beckman Air Comparison Pycnometer. 4.2 REDUCTION OF EXPERIMENTAL DATA From each experimental run the following variablea .are. needed for the determination of C and Re D 1, particle diameter 2. convective flow velocity 3. convective gas density 4. gas viscosity in the convective flow regime 5. particle velocity 6. particle acceleration. As mentioned previously, the particle diameter is obtained by viewing its image on the film under a microscope. Knowing the camera and micro- scope magnification factors, the particle diameter is obtained directly. 37 The convective flow velocity and gas density are obtained from the normal shock relations knowing the shock speed, the temperature of the undis- turbed gas, and the pressure of the undisturbed gas in the test section, The temperature in the convective region is likewise given by the normal shock relations. The viscosity values for air at various temperatures were obtained from air viscosity tables of the National Bureau of Standards . For small temperature ranges, i. e. , AT - 25 R, the viscosity versus temperature curve is linear. Thus, a series of curves of viscosity versus tem- perature were plotted for different temperature ranges from the above table. Linear viscosity relationships versus temperature were obtained from these curves for each temperature range so that the viscosity could be solved for analytically. The remaining two variables which must be found for each particle are the velocity and acceleration. (1 6) 0 The particle distance versus time history is recorded by taking five photographs on the rotating drum camera at equal time intervals. From these photographs, the particle velocity and acceleration must be obtained. 4.3 METHOD OF DATA REDUCTION The simplest technique for the reduction o f the x versus t data would be to set up a difference table. If this method was used, one would obtain the velocity of the particle from the first difference and the particle's acceleration from the second difference. The main disadvantage of the 38 t measurements, and is very sensitive to any errors in these measure- I ments. Another disadvantage is that an average acceleration and velocity is obtained rather than an instantaneous value. It would be more accurate to fit the position data with a polynominal. Since there are five position; points, the highest order polynominal which can be fitted to the points is a fourth order polynominal which would pass through each position point. Thus, any measurement error in the position data would be ampli- fied when the acceleration was obtained by differentiating the analytic x versus t expression twice, since a fourth order polynomial would pass through every point. A second order polynomial could be used to fit the position data. However, this implies that the force on the particle is a constant. The velocity rela.tive to the particle is actually changing with time and thus the force on the particle is changing with time. The remaining choice is a third order polynominal. By using the third order fit, the final curve would not be forced to pass exactly through the position data and thus some mea- surement error could exist without affecting the final result drastically. A least square method was used to fit a third order curve through the five position points. Once the analytic expression was obtained, the velo- city and acceleration of the particle with respect to inertial space was obtained by differentiating the x versus t expression. The time which was 39 substituted in the analytic expression for x was that which corresponded to the third picture. It was felt that the second derivative was more cor- rect near the center point of the x versus t expression than at the times corresponding to the first or fifth pictures. The data reduction process for the most part is handled by an IBM 7090 computer program. The data which is put into the computer program for each run is as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Run number Particle number Diameter of particle on film Position data; xo, xl, x2, x3, x4 Atmospheric pressure Initial temperature Initial test section pressure Oscillator frequency Microsecond timer reading Magnification Density of particles With the above data the computer calculates p2, T2, U2, and p2; solves for the coefficients of the -third order polynomial; then differentiates this 40 expression to obtain the particle velocity and acceleration. Finally, it solves for C and Re and prints this and other pertinent data on the output page. D A second data reduction technique which calculates a mean CD was used to check the third order method results as described below. The particle acceleration and velocity can be written as 2 d x dt a=-”-= dx 2 v = - - P dt - X The equation for CD can be written as 2 B(U2 - X) = X where 3p2 ‘ D 4Pp d B = Let X = p si = dp/ dt. Equation (8) becomes If we assume that over the time interval of interest B is a constant we may integrate Eq. (9). The values of p and x at limits of the integral are t = O P = Po x = x 0 t = t P = P x = x 41 - . . . . . . .
. . . .
. . . . . - I Equation (9) becomes Performing the integration and substituting the limits 1 dx 1 dt p = u - - B t + " u2 - Po The above may be rewritten as X t 1 d[B[U2 - po) t + 1 1 I d x = ' 2 dt 'E [BIU2 - po)t + 11 X 0 0 0 Integrating and substituting in for the limits one has A least squares technique is used on the position data to solve for the coefficients of Eq. (ll), i. e., x B, and V Once these have been evaluated, c, can be obtained from B. The particle velocity, V was calculated from Eq. (10) using the time corresponding to the third picture. 0 ' P O P9 Knowing V U2, d, p2, and p the Reynolds number, Re2, can be cal- culat.ed, These calculations have also been programmed for the computer. P' 2' 42 The cD method assumes that the CD is constant over the time interval of the test, an approximately true assumption in the actual case since for a typical run AC / C .” 1.1 x 10 over the test-time interval. -2 D D’ 4.4 TYPICAL EXAMPLE OF DATA REDUCTION Figure 14 shows five sequential pictures of a typical run (Run 50E) where the particles are being accelerated by the convective flow behind a shock. The x versus t measurements were made using a ruler graduated in 0.01 of an inch in conjunction with an Edmund 12-power comparator. With this system it was possible to estimate readings on the scale to 0.003 of an inch. Various information obtained from Run 50E for the data reduction is listed below: = 3.10 t = 6.812 x 10 sec P d = 5.166 x 10-4ft -4 Tat m = 76. 40°F P = 29.10 inches Hg at m 1 P = 18.28 inches Hg t = 945. o x 10e6sec m The above data when fed into the two computer programs gave the following results 43 Figure 14. Typical Example of x Versus t Data of a Particle 44 M1 = 1.088 U2 = 159.14 ft/sec 3 = .0519 lbs/ f t = 1135.0 ft/ sec = 1166.9 ft/sec P , = 1568.8 Ibs/ft pz “1 a2 2 L p2 = 1 . From this point grams differed. As 287 x lb/ft, sec O n the method of data reduction by the computer pro- mentioned previously, the first program used a third order fit to the x versus t data. A graph of the third order fit for Run 50E appears in Fig. 15. The particlek velocity and acceleration are obtained by differentiating the analytic expression for x versus t once and twice respectively. This expression for the particular particle in question is x = 4.605 t + 4.223 x lo3 t2 - 7.045 x 10 t 4 3 The time which was substituted in for the velocity and acceleration is that which corresponds to the third picture and is t2 = 1.36 x 10 seconds -3 The velocity and acceleration corresponding to t2 is V = 15.72 ft/sec P (Y = 7869.35 ft/sec 2 45 El Data X = 4.61t + 4,222.6t2 - 70.454 f I 21 t3 I 0 0 .4 .€I 1.2 1 .
6 2.0 2.4 2.8 TIME IN MILLISECONDS Figure 15. Displacement Versus Time Curve for Third Order Polynomial 46 When all of the above values are put into the equations for C and Re one has D CD = .7926 Re = 298.81 The second computer program solves by the method of least squares for. the unknown constants x B, and V in Eq. (11). Once these quantities are known ED and Re2 are determined. For this sample run one gets 0’ P O - CD = .7892 Re2 = 299.39 In the present use, and in the evaluation of other data, the two methods of data reduction were found to be in substantial agreement. 4.5 GENERAL EQUATIONS FOR DETERMINING PROBABLE ERROR In order to assess the validity of the experimental data an error ana- lysis must be performed. As, can be seen from Eq. ( 5 ) , CD is a function of the following independent variables; p , CY, d, p2, U2, and V The P P’ probable error in CD can be written as (17) + - ( a c D r 2 (acDja 2 r 2 a u2 (6 u2) + - a vP (6 VP’ 47 Performing the necessary differentiation and substituting into Eq. (12) the probable error in CD becomes The probable deviation or errors in p c y , d, and V can all be estimated or obtained directly from the data. However, the errors in U2 and p 2 must be evaluated from the normal shock relationships in conjunction with Cs, al, and P1, i. e., P' P In all the calculations y = 1.4, 6 U2 and 6 p2 are obtained by taking dif- ferentials of Eqs. (14) and (15). The actual values of 6 U2 and 6 p for a particular run are then obtained by substituting in the numerical values 2 for Cs, al, and PI. These values can be put into Eq. (13) with the other deviations and the probable error in drag coefficient, P(C ), for a parti- D cular case can be obtained. 48 From Eq. (6) the probable error in Re, P(Re), can be written After differentiating Eq. (6) and substituting the results into Eq. (16) P(Re), becomes As in the case of CD,the deviations 6 V 6 d, and 6 p2 can be estimated or obtained directly from the data. The deviations 6 p2 and 6 U have been determined. Thus for a particular run the probable error in Reynolds P' 2 number can be obtained using Eq. (17). 4.6 TYPICAL EXAMPLE OF PROBABLE ERROR The same run which was used in the data reduction section as an example run will be used in determining the probable error in C and Re, namely Run. 50E. The necessary data from Run 50E is given below: D 49 - = 2.475 gm/cc pP V = 15.72 ft/ sec P 2 cy = 7869.3 ft/ sec C = 1234.5 ft/sec d = 5.166 x f t = 1135.0 ft/ sec 2 S "1 P1 = 1292.8 lb/ft p2 = 0519 lb/ft u2 = 159.14 ft/sec Time = 945.0 x 10 sec -6 U = 143.43 ft/ sec p = 1.28681~ low5 lb/ ft sec R 2 The probable error in C will be computed first. The first term in Eq. (13) is 6 pp/pp. The volume o f the glass beads was found by using a Beckman model 930 air comparison pycnometer. A 10 c. c. or larger sample of beads is needed for an accurate volume determination. Once the volume was determined the sample was weighed on a beam balance in order to ob- tain the average density. The 6 p and average density technique was D P 9 7jp using the above 6 p = .050 gm/cc P from which Consider the second term in Eq. (13), 6 c y / c y . The probable error in determining acceleration was attributed only to incorrect particle dis- placement. measurements. Typical displacement errors ( .003 inches) 50 I were incorporated into the original measurements of x versus t. This new data was then run through the computer program for data reduction to find the new value for acceleration. Doing this the deviation in ac- celeration for Run 50E is 2 6 CY = 274.5 ft/ sec The next term is 6 d/d. The deviation in particle diameter was con- sidered due only to the error in estimating the exact location of the particle’s edge. The deviation for the diameter was 6 d = 151 x lom4 f t lyj” = 8.52 x loe4 The deviation in the particle’s velocity, 6 V was calculated as being due to incssrrect displacement measurements as in the acceleration case. The method for determining 6 V was entirely analogous to that of 6 CY as mentioned above, and for Run 50E 6 V i s P’ P P 6 V = 24 ft/sec P 51 6 p2 depends on the deviations o f Cs, al, and Pl. The initial test section pressure, P is read off a mercury manometer and thus 6 P1 is rather small. 1’ 2 6 P1 = 5.17 lb/ft The deviation in a is that caused by a change in the temperature and is for Run 50E 1 6 a1 = 2 ft/ sec The deviation in shock speed, Cs, is considered due only to the error in the microsecond timer. It was shown in a previous section that attenua- tion affects could be neglected. It was also mentioned that since the outputs of the two transducers had the same slopes and both channels of the micro- second timer were calibrated, the measured time would be correct. The deviation due to the error in the microsecond timer is 6 c = 1.2 ft/ sec S Substituting in the proper values in 6 p2 we have 6 p2 = - .416 X 10 -4 52 2 (2,) = .645 x Similarly 6 U2 is obtained and is u2 = - 1.226 ft/sec 2 = 2.92 X 10 -4 All o f the terms which appear in Eq. (14) have been computed. Putting these terms in Eq. (13) the probable error in drag coefficient is for Run 50E PED) cD = 5.28 X 10 -2 In calculating the probable error in Reynolds number Eq. (17) must be used. This equation contains the same deviations as those just cal- culated for C plus 6 p2. Thus for the Re number case, it is necessary only to find 6 p2. 6 p is caused only by an error in the temperature measurement. This gives 6 p as D 2 2 6 p2 = .005 x lom5 lb/ft, sec 16 p2I2 = 1.51 x 10 -5 53 2 Putting (6 p2/ p2) and the other values into Eq. (17) for the probable error in Re for Run 50E we have 54 V. RESULTS AND DISCUSSION OF RESULTS 5.1 EXPEFUMENTAL RESULTS The experiments involving the glass beads were all carried out using a magnification of 3.1. Before a particular particle was used, the film record of the x versus t history was checked against the following condi- t ions. 1. 2. 3. 4. 5. Any particle which was closer than five diameters to another particle perpendicular to the direction o f motion was not used. Any particle which was within twelve diameters of another particle wake or had another particle within twelve diameters of its wake was not used in the data reduction. If any diaphragm material appeared in any of the five frames, the run was discarded. Only spherical particles were used in the data reduction. Only particles which were sufficiently sharp were used for the data reduction. Because of the above restrictions, many particles could not be used to obtain CD versus Re data. Table 2, in the Appendix, is a summary o f experimental data for glass beads in incompressible continuum flow. C and Re results obtained using both methods of data reduction appear in the table. I€ the CD versus Re points are plotted on log-log paper, D 55 they appear to fall in nearly a straight line. Thus, a least squares technique was used to fit the C versus Re data with an equation of the following form: D loglo D o C = A + A 1 loglo Re + + (loglo Re)2 Other analytical forms were tried for CD versus Re data, however, none of these other forms fit the data nearly as well as that of Eq. (18). The results which appear in Table 2 are plotted in Figs. 16 and 17 for the two methods of data reduction. The best fit curve for these points is also plotted. A second series of experimental runs were made with the glass beads at higher shock strengths to begin a study of the influence of compressi- bility. The results of these runs appear in Table 3, The relative Mach number, M for the flow field about the particles is 15 < MR < .30. R’ - - The C has increased as much as 50 per cent with respect to similar data which appeared in Table 2. The scatter in C for a particular Re has increased approximately 40 per cent. This shift in C cannot be at- tributed to compressibility effects since the relative Mach number, M R’ is still quite low. Likewise the scatter in CD cannot be explained by dif- D D D ferences in experimental technique since the results which appear in Table 3 were obtained in precisely the same manner as the data which appears in Table 2. Thus two more series of experimental runs were made in an 56 1 .
0 7 .9 4 i- I I I I I I I I I I Best fit curve for incompressible continuum non- burning spherical glass particles .6 .5 .4 200 300 400 500 600 700 800 IO00 Re Figure 1 6 . Drag Coefficients Versus Reynolds Number for Third Order Polynomial Data Reduction Technique El Glass particles 200 300 400 500 600 700 800 1000 Re Figure 17. Drag Coefficient Versus Reynolds Number for Mean Drag Coefficient Method attempt to explain the above results. They were made using the same experimental technique that was used for the data of Tables 2 and 3. The only change was the types of particles that were used, Winchester-Western HP 295 ball powder and sapphire balls. The data summaries for the HP 295 ball powder and sapphire balls appears in Tables 4 and 5 res- pectively. The results of Tables 3, 4, and 5 are illustrated in Fig. 18. 5.2 SURFACE ROUGHNESS Figure 18 shows that the scatter in the H P 295 ball powder data is of the same level as that obtained for the higher relative Mach number runs using glass beads. The relative Mach number of the flow about the ball powder, however, was M < .124. Both for the HP 295 ball powder and the higher relative Mach number glass bead data, the average C for a particular Reynolds number fell above the best fit curve of the glass beads. In. contrast to this, the average C for the sapphire runs for a particular Reynolds number falls below the best fit curve of glass beads and does not seem to be a function of the M R - D D R' As an attempt to explain the varied results which appear in Fig. 18, photomicrographs were taken of the particles. Photomicrographs of 200 pglass beads appear in Figs. 19, 20, and 21. Under the relatively low power magnification of Fig. 19, the glass beads appear to have fairly smooth surfaces except for a few bubbles or craters which appear on 59 Q) 0 I .o .9 . 0 c, .7 .6 .5 ,4 I T I 3 Best fit curve for incompressible continuum non - burning spherical glass . particles Experimental drag coefficient of spheres -+\I in steady incompressible flow (Ref. 1 8 ) 1 1 6% I / -t IO0 200 300 500 700 1000 Re El Glass balls .I5 S M R 4.30 A Gunpowder balls 0 Sapphire balls - 0 < MR< .
I 2 5 - -140 4 MR .320 I 2000 4000 Figure 18. Drag Coefficient Versus Reynolds Number-Glass Particles, Ball, Powder, and Sapphire Balls Figure 19. Photomicrograph of Glass Particles, Magnification = 200 61 Figure 20. Photomicrograph of Glass Particle, Magnification = 1840 62 Figure 21. Photomicrograph of Glass Particle, Magnification = 1840 63 most of the beads. Figure 20 shows a detailed view of these bubbles or craters. These protrusions and craters are quite pronounced relative to the particle diameter. Figure 21 is a detailed view of a portion of the glass bead which is free of the larger surface blemishes. However, this portion of the surface still possesses a considerable amount of rough- ness relative to the particle diameter. Photomicrographs of H P 295 ball powder appear in Figs. 22, 23, and 24. Figure 22 is a low-power photo- micrograph and shows only that the ball powder is spherical with a few surface protrusions. Figure 23 is of intermediate magnification, and it appears from this figure that the entire surface of the ball powder is rough relative to the similar photomicrograph of the glass beads in Fig. 19. Finally, Fig. 24 is a detailed view of the ball powder. Again the surface appears to be rougher than that of glass beads. Photomicro- graphs of sapphire balls appear in Fig. 25, 26, and 27. Figure 25 is a low-power photomicrograph and indicates only that the sapphire ball is spherical. Figure 26 is a detailed view of a sapphire ball. From this figure it is evident that the sapphire balls have less surface roughness than even the portions of glass that were free of the bubbles or craters. Figure 27 is a photomicrograph of a sapphire ball which was washed in trichloroethylene before the photomicrograph was taken. Almost all of the roughness relative Mach which appeared in Fig. 26 is missing number range for the sapphire was, e in Fig. 27. The 14 < M < .313 - R - 64 Figure 22. Photomicrograph of HP 295 Ball Powder, Magnification = 110 65 Figure 23. Photomicrograph of HP 295 Ball Powder, Magnification = 190 66 Figure 24. Photomicrograph of HP 295 Ball Powder, Magnification = 1840 67 Figure 25. Photomicrograph of Sapphire Balls, Magnification = 202 68 " Figure 26. Photomicrograph of Sapphire Ball, Magnification = 1850 69 . . .
. : . . Figure 27. Photomicrograph of a Washed Sapphire Ball, Magnification = 1760 70 yet none of the violent scatter occurred that was found using both the glass particles and the H P 295 ball powder. As noted earlier, the only dif- ferences between the various types of particles and the experimental runs involving these particles is the particle's surface finish. 5.3 UNSTEADINESS IN THE BOUNDARY LAYER AND WAKE The interaction of the shock wave with a particle is analogous to the impulsive motion of a sphere in a fluid in that the particle impulsively sees and is acted upon by the convective flow velocity behind the shock front. In impulsive motion a certain amount of time is needed for the boundary layer and the wake to reach a quasi-steady condition. In order to determine the time it takes for the boundary layer to become quasi- steady, a diffusion time may be defined as the time required for a sudden change to spread by the process .of molecular or turbulent diffusion . The diffusion time may be taken as 6 / v where 6 is the boundary layer thickness and v is the kinematic viscosity. The experimental displace- ment thickness at the 90 point on a circular cylinder is (19) 2 0 (18) 6 UR " - R v - .8 I f we assume the following values for R, Urn, and v 71 R = 3.28 x ft v = 1.611 x 10 ft /sec -4 2 uOc = 100 ft/ sec Then 6 becomes 900 = I. 841 x ft The diffusion time may be approximated using the displacement thickness for a cylinder and is 2 " - 2.12 x sec V The minimum time from shock front passage over the particles until the first picture was taken was always greater than 100 microseconds. The diffusion time as computed above is only two per cent of this time and thus by the time the experimental data is taken, the boundary layer should be quasi-steady. Experimental work on impulsive flow about cylinders has been done by Schwabe(20) and very recently by Sarpkaya(21). Sarpkaya's experi- ments involved the impulsive flow of water about circular cylinders. He found that the drag coefficient initially rises above the steady state case due to the formation of the vortices and then decays back to the steady 72 state case once the vortex flow has become established. He found that his C data correlated with a non-dimensional time 7 . The drag co- efficient had returned to the steady state result when 7 28. r is defined as D u t R c o w r = - where Uco = flow field velocity t =time W R = radius of cylinder For U = 100, R = 3.28 x ft, and r = 28 then tw becomes co t = 91.8 x sec W It might be noted that since the wake formation time is an order of magni- tude larger than the boundary layer formation time, then in unsteady flow problems, the wake formation time dominates. Since t is less than 100 microseconds, which is the time between when the shock passes the par- ticles to when the first picture is taken, the wake flow field has reached a quasi-steady condition before the data is taken. Thus both the boundary layer and the wake have become quasi-steady before any experimental data is taken. W 73 5.4 TURBULENCE IN THE FREE STREAM Free stream turbulence can cause a considerable shift in C for a particular Re as can be seen, from Torobin's and Gauvin's(12) results which appear in Fig. 2. A literature survey yielded no information on turbulence behind a shock wave. Since the shock waves are weak and the convective flow Mach numbers are low, any turbulence which is generated in the boundary layer will not be transmitted into the convective flow. Turbulence will not be generated by the shock front since Schlieren photographs indicated the shock front was plane and perpendicular to the walls of the shock tube. Further, the fact that smooth sapphire balls yielded CD data close to the steady state curve tends to substantiate a turbulent-free flow field assumption. D 5.5 PARTICLE ROTATION Particle rotation could cause a shift in the drag coefficient of a particle in that on one side of the particle the separation point will move rearward whereas on the other side it will move forward. This will cause the skin friction and the form drag cont.ributions to the drag co- efficient to change. Macco11(22) conducted some experiments to deter- mine the effects of sphere rotation around an axis perpendicular to the flow. These experiments ranged from Reynolds numbers of 6.15 x 10 to 10.7 x 10 . CD fell from . 52 to .48 as the ratio of equatori peed, 4 ' 4 74 U to the relative velocity, U increased from 0 to 1. Davies (23) has conducted experiments in the same Reynolds number regime as Maccoll and his results agree with Maccoll's. Pasternak (24) made observations of freely moving spheres and suggests that the ratio of U / U is on the order of five per cent. L~thander'~~) has measured the drag coefficient of spheres rotating around an axis parallel to the flow direction in the region of the critical Reynolds number. Below the critica1,Reynolds number, a U /U ratio up to two had very little effect on C R' S S S D' Garstang(26) and Drazin'") studied analytically the effects of rotation about the axis parallel to the flow direction for the Stokesian drag regime and found that the drag coefficient remained unaffected by the rotation. Due to the above results that rotation does not change the drag coefficient significantly and since spherical particles are in- jected in the present study in a manner which does not initiate rota- tional motion, it is felt that rotational effects are negligible in the present study. 5.6 ACCELERATION EFFECTS A considerable amount o f experimental work has been done on the ef- fects of acceleration on both sphere and cylinder drag. Keim(28)7 did experiments with accelerating cylinders, He found that the drag coef- ficient versus Reynolds number data could be correlated by using a n 75 acceleration modulus defined as He found that if A was on the order of . 2 or greater then the effect of acceleration upon the drag coefficient was substantial. Bugliarello (2 9) C determined experimentally the drag coefficient for accelerating spheres. He found his data also correlated with the acceleration modulus, Ac. Crowe(13) showed analytically that (4) rr e= < rr Ac " 9 where C = skin friction coefficient. The acceleration moduli for the present study were of the order of 10 to ACf/ Cf becomes f -4 ef Thus acceleration effects do not seem to be important in the present study. 76 VI. CONCLUSIONS Observations on the drag coefficient of small spherical particles in an incompressible, laminar, non-reacting, continuum flow regime are sum- marized below: 1. The drag coefficient of small spherical particles in laminar, non- reacting, incompressible flow regime was found to be consistently higher for a particular Reynolds number than the generally accepted steady state value. 2. The C of the HP 295 ball powder, for MR < .125, increased as much as 85 per cent and the CD of glass beads increased significantly, for . 15 < M < . 30, over the steady state value R - respectively. D 3. The CD and the scatter in the CD data of the HP 295 ball powder and the glass beads substantially increased as the relative Mach number increased, even though the relative Mach number was still in a region which is normally considered incompressible. 4. The CD of smooth sapphire balls fell closer to the steady state curve and did not depend on the relative Mach number nor did the scatter in the CD data increase as the relative Mach number increased. 77 5. Photomicrographs of HP 295 ball powder, glass beads, and sapphire balls indicated that the sapphire balls were relatively smooth, the glass beads somewhat rougher, and the HP 295 ball powder quite rough. It is concluded that surface roughness can cause considerable shift in C for small spherical particles in an incompressible, laminar, non- reacting, continuum flow regime. Since one group of particles in solid propellant rocket exhaust is relatively rough, i. e. , those metalic particles which are originally cast into the solid propellant, their C may be several hundred per cent higher than the steady state results. The other particles in the rocket exhausts are those which are condensed from the com- bustion products. The C of the condensed particles could also deviate appreciably from the steady state value depending upon the relative sur- face roughness and deformation of the particles. Thus, the use of the steady state C curve for velocity lag calculations of particles in rocket nozzles possibly causes considerable error. D D D D This study has only touched upon the regimes a particle encounters in a solid propellant rocket exhaust. Much work is left to be done on the influence of compressibility, burning, and electric charges in the slip- flow, transition, and free molecular regimes. 78 "... .-. ...- . .. . .. -.-.- APPENDIX 79 I RUN 3D- 1 3D- 2 4D- 1 4D- 2 4D- 3 5D- 1 5D- 2 6D- 1 7D- 1 7D- 2 9D- 1 9D-2 10D-1 10D-2 11D-1 11D- 2 11D-3 12D- 1 17D- 1 18D-1 27D- 1 3 5D- 1 41D- 1 0 03 cD .828 . 827 .774 .775 .774 . 769 .802 .835 .754 .742 .841 .805 .869 .847 .763 .778 .751 .756 . 741 . 778 .779 . 642 . 644 Re 344. 5 351.0 343.7 341.8 330.8 351. 4 362.8 351. 2 362.7 396. 4 313.7 329. 5 329. 3 357.8 389. 4 349. 2 394.1 407. 4 508. 1 442. 6 424. 3 543.8 672.7 M1 1.067 1. 067 1.067 1. 067 1.067 1. 067 1. 072 1.070 1.074 1. 075 1.065 1.065 1. 070 1.071 1.077 1.077 1.077 1.064 1. 071 1. 067 1.061 1.081 1. 095 TABLE 2 DATA SUMMARY - GLASS BEADS O < M R < . 1 5 - MR .096 ,096 .095 . 095 ,095 . 101 . 101 .097 . 105 . 106 .091 .092 .096 . 097 . l o 6 . 105 . 107 .094 , 103 . 095 .090 . 116 .133 d ft 4.99 x 10-4 5.09 x 10-4 5.04 x 10-4 5.04 x 10-4 4.89 x 10-4 4.84 x 10-4 4.99 x 10-4 5.04 x 10-4 4.79 x 10-4 5.19 x 10-4 4.84 x 10-4 5.04 x 10-4 4.79 x 10-4 5.14 x 10-4 5.04 x 10-4 4. 59 x 10-4 5.09 x 10-4 6.10 x 6.90 x 6. 55 x 10-4 6.70 x 10-4 6. 30 x 10-4 6.70 x C Y ft/sec 2 7886. 3 7699. 2 7125.7 7064. 2 7229. 2 8340.0 8451. 1 8014. 5 8985.1 8 288.8 7313.8 6831.8 8569.9 7970.9 8998.0 9783.3 8801. 5 5587.8 5869. 4 5489.4 4775. 1 7388.9 9322. 5 uR f t/ sec 111.7 111. 6 110.7 110.1 109.8 117. 2 117.3 112.8 122.1 123.0 105. 6 106. 5 111.4 112.7 124.3 122. 5 124. 6 109.7 120. 3 111.0 105.3 135.8 155. 5 - cD .824 .823 . 769 ,772 .771 .764 .796 .831 . 749 .738 .836 ,801 ,863 .843 . 759 ,773 .747 .755 . 741 . 7 7 4 .777 . 639 . 641 344.3 350.8 342.9 341. 6 331.1 351.0 361.8 351. 4 362.1 395.9 313.1 329. 2 329. 2 357.8 389. 5 348.9 393.6 407. 6 508.8 442.2 424.1 543. 2 672. 4 TABLE 2 (cont) RUN 43D- 1 43D- 2 47D- 1 49D-2 51D-1 56D-1 58D- 1 58D- 2 68D-1 69D-1 7 5D- 1 75D-2 75D-3 76D-1 77D-1 85D-1 87D-1 89D-1 91D-1 97D-1 98D-1 99D-1 99D-2 6E-1 10E-2 11E-1 14E-1 cD . 621 . 612 . 585 . 603 . 614 . 610 . 646 . 625 . 656 . 642 . 652 .702 . C41 . 645 . 654 . 635 . 639 . 678 . 666 . 641 . 633 . 644 . 662 ,731 .822 .899 .861 Re 778.4 766. 6 800.1 709.0 723.9 699. 3 616. 4 650. 3 605. 2 579. 6 656.7 687.9 685. 3 611. 5 587. 1 665.1 642. 1 692. 0 568. 6 596. 6 619.7 546.7 598.7 384.0 295. 2 270.7 235. 2 M1 1.105 1.105 1.110 1.100 1.110 1.100 1,090 1.090 1.090 1.082 1.097 1.097 1. 097 1.086 1.084 1. 092 1. 094 1. 091 1. 083 1. 090 1. 089 1.084 1. 084 1.060 1. 045 1. 041 1.065 MR d ,147 6.90 x ,147 6.80 x 10-4 ft . 156 6. 60 x . 141 6. 60 x . 146 6. 48 x . 140 6. 60 x 10-4 . 131 6. 25 x . 132 6. 55 x . 122 6. 65 x . 117 6.70 x . 137 6.35 x 10-4 . 137 6. 65 x . 137 6. 60 x . 119 6. 65 x . 130 6.85 x . 131 6. 55 x . 130 7.03 x . 116 6. 55 x . 125 6.30 x . 125 6. 55 x . 116 6. 25 x . 117 6.80 x .085 6. 30 x . 066 6. 40 x ,062 6.30 x .122 6.75 x 10-4 .095 4.91 x 10-4 CY ft/sec 2 10977. 3 10961. 1 1237 5.9 10108.8 11350. 4 9982. 2 9732.3 9096. 6 7932.7 6979. 2 10621.9 10912. 6 10123. 4 7637. 7 7481. 1 8492. 5 9094.9 9006.7 7413. 3 8768. 1 8311.8 7600.9 7265.9 4284.7 2735. 6 2642. 4 5637.9 uR ft/sec 173.1 172.9 183.7 165. 5 171. 4 163.9 153. 5 154. 5 142.7 136.2 160. 6 160. 6 161. 2 142.8 139. 5 152. 4 153.3 152.4 135. 5 146. 3 146. 3 136.1 136.9 99. 0 76. 0 71. 1 109.7 . 619 . 610 . 582 . 600 . 612 . 607 . 643 . 622 . 652 . 641 . 648 . 695 . 637 . 643 . 651 . 632 . 637 . 675 . 664 . 638 . 630 . 641 . 659 .728 .821 .894 .858 Re2 778.7 767. 6 799.3 708.7 724.7 699.1 615. 6 649.7 604. 2 580.0 655.8 685. 1 684.1 611. 5 586. 7 664. 4 642.3 691. 1 568. 4 596.9 619. 1 546. 2 598. 0 384.0 295. 3 270. 3 235.0 TABLE 2 (cont) RUN 24E - 1 25E-1 27E-1 34E-1 35E-1 49E-1 50E-1 51E-1 52E-1 53E - 1 533-2 55E-1 58E-1 60E- 1 62E - 1 74E-1 77E-1 78E-1 80E-1 81E-1 93E-1 94E-1 943-2 99E-1 4G-1 6G-1 6G- 2 6G-3 cD .879 ,842 . &42 ,794 .863 ,847 .792 ,875 .795 .811 ,804 .899 .816 ,921 .841 .880 .857 .836 .789 ,790 . 677 . 676 . 670 .726 . 652 . 675 . 683 . 728 Re 291. 1 277. 4 253. 6 305.9 282. 6 277. 1 298.8 280.9 296. 1 278. 5 291.9 270.7 273. 6 268.8 279. 5 280. 7 273.8 279. 6 295.9 299.8 523. 2 563.8 553. 2 486. 1 536. 6 551. 3 553. 4 549. 6 M1 1.079 1.083 1.079 1.089 1.090 1.082 1.088 1.090 1.093 1. 095 1. 095 1. 076 1.079 1.081 1.085 1.089 1.091 1.093 1.097 1. 096 1.098 1.097 1. 097 1.094 1. 092 1.099 1.099 1.099 MR ,112 . 118 . 113 . 128 , 126 . 116 . 123 . 125 . 130 . 131 . 132 . 110 . 114 ,114 . 121 . 124 . 129 . 129 . 133 ,135 . 135 . 135 .135 . 131 . 132 . 139 . 139 . 138 d ft 5.37 x 10-4 4.94 x 10-4 4. 69 x 10-4 5. 24 x 10-4 4.94 x 10-4 5.04 x 10-4 5.17 x 10-4 4.89 x 10-4 4.99 x 10-4 4.69 x 10-4 4.89 x 10-4 5.04 x 10-4 4.96 x 10-4 4.99 x 10-4 4.99 x 10-4 5.01 x 10-4 4. 69 x 10-4 4.81 x 10-4 4.99 x 10-4 4.96 x 10-4 6.10 x 6.35 x 6.22 x 6. 42 x 6. 50 x 10-4 6. 58 x 10-4 6.60 x 6. 60 x 10-4 CY ft/sec 2 7221. 1 8166.7 7935. 6 8215.9 9200.9 7707. 3 7869. 3 9279.8 8933.1 9823. 4 9420. 3 7592. 3 7440. 3 8151. 4 8250.8 8876.3 10105. 4 9536. 3 9140. 3 9520.3 9493. 6 9360. 4 9488.0 8289. 5 8006.1 8781. 6 8856. 6 9303.0 uR ft/sec 131.4 137.7 131.9 149. 5 147.9 135.3 143. 2 145. 6 152.1 153. 6 154.3 128. 5 133.1 132.7 140.9 145.1 150.9 151. 4 155.9 158. 6 159.3 158. 5 158. 7 154.7 155.1 163.1 163. 1 162.0 - cD .876 .838 ,836 .790 .859 .843 .789 ,873 ,792 .808 .800 ,898 .813 .916 ,837 .876 .852 .832 .787 ,785 . 674 . 672 . 667 .724 . 650 , 672 . 680 ,724 Re2 290.8 277. 1 252.8 305.6 282. 5 276.8 298. 6 281. 3 296.0 278. 4 291.7 270.9 273. 6 268. 6 279. 2 280.7 273. 5 279. 5 296. 1 299.0 523.0 562. 6 552.8 486. 0 536. 4 550.8 552. 5 549. 1 TABLE 2 (cont) RUN 12G-1 12G-2 14G-1 15G- 1 15G-2 16G-1 23G - 1 236-2 26G- 1 31G-1 33G-1 37G-1 37G-2 43G- 1 56G-1 66G- 1 72G-1 72G-2 72G-3 726-4 72G-5 77G-1 79G-1 81G-1 81G-2 84G-1 876-2 92G-1 cD . 658 . 691 ,718 .717 ,734 . 685 . 6 1 9 . 642 . 651 . 642 . 667 ,713 . 673 .946 .742 .752 .?lo .735 .741 . 730 . 758 . 6 8 9 . 697 ,709 .720 .715 . 7 2 9 . 685 Re 5 2 5 .
9 537.7 515. 4 5 6 6 . 5 5 2 9 . 3 553.9 605.3 645.0 512.9 5 7 3 .
1 6 0 9 . 6 576.9 551. 2 2 0 0 . 3 413.8 412.3 4 4 9 . 1 4 2 1 . 4 463.0 4 5 1 . 3 436.7 4 1 8 . 5 4 1 9 . 6 3 9 4 . 2 3 7 7 . 5 500.7 4 9 4 . 0 496.7 M1 1.089 1.089 1.085 1.088 1 . 088 1 . 089 1.102 1.102 1.092 1.106 1.105 1.101 1.101 1.055 1 . 0 5 9 1.063 1.065 1.065 1.065 1.065 1.065 1.063 1.064 1 . 0 5 9 1. 059 1 . 069 1 . 074 1 . 073 MR . 127 , 127 . 121 . 125 . 125 , 126 . 143 . 143 . 132 . I47 . 147 . 143 . 143 . 081 .088 .091 .094 .093 . 094 .094 .094 .089 .091 .087 . 086 . 100 . 104 , 105 d ft 6 . 4 0 x 6 . 55 x 6 . 35 x 6.98 x 10-4 6 . 55 x 7.21 x 10-4 7.01 x 10-4 6 . 6 0 x 6.63 x 6 . 55 x 6.98 x 10-4 6.75 x 6.45 x 6 . 55 x 10-4 6.30 x 10-4 6 . 6 0 x 10-4 6 . 25 x 6.80 x lom4 6 . 6 5 x 10-4 6 . 4 5 x 6 . 5 0 x 10-4 6 . 4 0 x 6.35 x 6.10 x 6.85 x 6 . 5 0 x 6 . 55 x 10-4 4 . 6 9 x 10-4 (Y f t / s e c 2 7875. 4 8 0 5 9 . 2 7950.0 7525.0 8132.8 6607.7 8957.8 8831.8 7294. 1 9222.0 8958.0 9350.0 9242.0 4899.0 4 4 3 9 . 2 5060.0 4 9 2 5 . 4 5 2 9 5 . 3 4 9 9 3 . 2 5 0 0 2 . 1 5 3 3 4 . 2 4 3 4 7 . 2 4613.8 4238. 2 4453.8 5445.0 6 3 7 3 . 6 5 9 6 9 . 3 uR f t / s e c 1 4 9 . 5 1 4 9 . 3 1 4 0 . 5 1 4 6 . 3 1 4 5 . 6 1 4 6 . 5 1 6 7 . 3 167.9 154.8 172.8 172.6 166.9 167.0 9 3 . 2 1 0 1 . 5 1 0 5 . 4 1 0 9 . 5 1 0 8 . 6 1 0 9 . 6 1 0 9 . 2 109.0 103.7 1 0 5 . 4 100.1 9 9 . 8 1 1 6 . 5 121. 4 121.9 - cD . 656 . 688 ,714 .715 .731 . 687 . 615 . 638 . 648 . 6 3 9 . 664 .?lo . 670 .942 .739 ,750 , 7 0 7 .732 .738 . 727 ,752 . 689 . 694 .706 ,716 ,712 ,725 . 682 Re2 525.8 5 3 7 . 3 5 1 4 . 4 5 6 6 . 5 5 2 8 .
9 553.3 604.0 6 4 3 . 6 512.8 572.8 6 0 9 . 3 5 7 6 . 4 551.0 2 0 0 . 1 4 1 3 . 5 4 1 2 . 2 448.7 4 2 1 . 3 4 6 2 . 5 450.9 435.0 418.9 4 1 9 . 5 393.8 377.1 5 0 0 . 4 493.8 4 9 6 . 5 RUN 127-1 123-1 135-1 130-1 126-1 154-1 158-1 164-1 165-1 166-1 167 -1 169-1 2 0 3 - 2 2 0 4- 2 205-2 225- 1 226-1 231-1 232-1 238-1 242-1 244- 1 2 4 7 - 1 250-1 2 8 3 - 1 284- 1 286-1 290-1 300-1 300-2 301 -1 301 -3 302-1 304-1 304-2 cD . 728 . 5 9 1 . 614 .781 . 681 .921 . 572 . 701 . 857 . 857 . 639 . 656 . 653 . 686 . 652 ,705 . 584 . 594 .736 . 638 . 719 ,712 . 655 . 631 .788 .834 .739 . 524 .766 .838 . 585 . 5 7 1 .889 . 6 2 9 . 657 TABLE 3 DATA SUMMARY - GLASS BEADS .15 < MR < .30 Re 787.9 894.7 811. 5 6 8 9 . 8 869.8 5 6 0 . 4 707. 6 5 6 4 . 2 6 2 9 . 8 6 1 5 . 0 778.9 751. 4 1378 1290 1577 1269 1331 1304 1150 1239 13 51 1214 1158 1296 803.1 608.8 783.0 896.9 5 7 1 . 5 5 6 6 . 6 866. 2 853.7 656.8 804.1 7 7 4 . 6 M1 1.179 1 . 208 1,199 1 . 168 1.194 1 . 1 6 7 1.177 1.155 1.163 1 . 157 1.197 1 . 1 8 1 1 . 130 1.120 1.138 1.126 1.135 1 . 123 1 . 1 0 5 1.121 1 . 128 1.110 1 . 118 1.134 1.150 I. 1 1 9 1 . 147 1 . 163 1.142 1.142 1 . 207 1 . 2 0 7 1 . 161 1.193 1.194 MR . 2 3 9 . 270 . 258 . 225 . 255 . 216 . 2 4 1 . 213 . 223 . 214 . 258 . 244 . 180 . 166 . 188 , 173 . 185 . 170 . 150 . 167 . 175 . 159 . 163 . 184 . 2 0 9 . 170 , 206 . 230 . 187 . 185 . 258 . 258 . 208 . 246 . 244 84 d f t 6.70 x 10-4 6 . 6 5 x 6 . 2 0 x 6.30 x 6.80 x 10-4 5.29 x 10-4 6.10 x loe4 6.30 x 6.35 x 6 . 55 x 6 . 2 0 x 6.35 x 9 . 5 8 x 10-4 9 . 8 3 x 10-4 I . 04 x 10-3 9.42 x 10-4 9.12 x 10-4 I . 01 x 10-3 9. 63 x 10-4 I. 00 x 10-4 9 . 2 7 x 10-4 9.22 x 10-4 8.92 x 10-4 9.17 x 10-4 9.17 x 10-4 6.56 x 10-4 6.67 x 10-4 9.88 x 9.88 x 8.97 x 6 . 5 6 x 6 . 5 6 x 6 . 5 6 x 6 . 5 6 x 6.37 x C Y ft/ sec 2 23793. 7 25814. 3 26337. 3 23590. 2 25725. 8 30660. 3 20499. 7 16345. 1 2 3 1 8 0 . 4 20300. 4 26551.7 23458. 8 13173. 6 11265.9 13456.0 13130.3 13132. 2 10177. 5 9242. 1 10803. 2 13196. 3 10089. 5 10735. 1 1 4 1 6 3 . 7 11821. 9 8071. 0 10736. 5 9822. 3 14508.9 15594. 7 2 3 4 1 1 . 8 23317. 1 2 1 7 3 7 . 3 2 2 7 0 4 . 7 2 4 0 8 2 . 6 uR f t/ sec 2 8 7 . 5 3 2 8 . 3 311.9 269.9 307.9 2 5 8 . 5 289.9 254.9 2 6 8 . 2 256.9 312.8 294.9 2 1 2 . 8 195.8 222.9 205.0 220.3 201.9 177.7 199.0 2 0 8 . 2 1 8 7 . 4 1 9 2 . 5 2 1 9 . 2 2 4 8 . 9 200.9 244.9 2 7 4 . 5 2 2 3 . 1 221. 2 312. 7 313. 2 2 4 9 . 4 2 9 6 . 8 2 9 4 . 6 TABLE 4 DATA SUMMARY - WINCHESTER WESTERN HP 295 BALL POWDER RUN 389-1 390- 1 390- 2 391-1 577 - 1 577-2 578-1 578-2 581-1 587-1 591-1 593- 1 596- 1 607 - 2 610-1 612- 1 613-1 614-1 cD .939 .969 1.004 . 927 . 685 .758 .784 . 691 .783 . 690 .919 ,774 . 655 . 677 .701 .777 . 530 . 692 R e 596.9 603. 1 595.3 614.5 662. 6 716.7 692.4 752.7 665. 3 683.4 689.0 668.8 744. 3 734.4 724.0 688.2 737.7 722.8 M1 1.101 1.100 1.100 1.101 1.071 1. 071 1.068 1.068 1. 067 1.068 1.069 1.068 1.068 1. 070 1. 070 1.069 1.069 1.069 MR . 121 . 122 . 124 . 124 .099 .099 .096 .097 .094 .094 .096 .095 .097 . 097 .098 .098 . 100 .098 d ft 6.56 x 6.56 x 6.56 x 6.56 x 10 9.32 x 1.01 x 1.02 x 1.09 x 9. 92 x 10 1.03 x 9.78 x 9.73 x 1.05 x 10 1.03 x 101; 1.00 x 9.57 x 1.00 x 1.00 x 10 -4 -4 -3 CY ft/sec 2 11678. 3 12299. 6 12426. 3 12227. 2 5681. 3 5771.7 5499. 8 4618. 3 5472. 3 4596. 2 6832.7 5557.4 4591. 2 4831. 5 5280.0 6060. 1 4160. 1 5218. 2 uR ft/sec 142. 6 144. 1 142. 3 146.8 116.5 116. 2 112. 1 113. 5 110. 8 110.0 112.1 110.1 113. 2 112.0 113. 6 113. 2 116. 1 113. 8 85 TABLE 5 DATA SUMMARY - SAPPHIRE BALLS RUN cD Re M1 5 (Y ft/sec 2 701-1 .581 1477 1.101 . 149 3445. 3 720-1 . 530 1462 1.100 . 146 3048.7 723- 1 .548 1479 1.100 . 148 3229.0 725-1 . 572 1474 1.099 . 148 3369.9 726-1 .556 1487 1.101 . 149 3334. 6 728-1 . 561 1467 1.099 . 147 3294. 2 728-2 . 571 1468 1.099 . 147 3356.4 735- 1 . 624 855.9 1.099 . 149 2181. 2 740- 1 .473 956.0 1. 107 . 160 1958. 2 741-1 .533 968. 8 1.108 . 162 2259. 9 743- 1 .521 955.2 1.107 . 160 2160. 1 7 46- 1 .546 949. 4 1. 108 . 161 2266.0 751-1 . 465 1178 1. 183 . 257 4016. 1 756-1 .521 933.3 1.104 . 156 2048. 6 756-2 . 531 932.9 1. 104 . 156 2083. 8 758-1 .525 923. 3 1.103 . 154 2010.0 761-1 .534 920.8 1.102 . 154 2035.0 767-1 . 501 1068 1. 138 . 202 2988.7 767-2 . 525 1069 1.138 . 202 3134.0 775-1 .522 1036 1.135 . 193 2887. 4 77 6- 1 . 484 1116 1.151 . 217 3268. 1 777- 1 . 520 1088 1. 140 . 205 3209. 1 778-1 . 492 1106 1. 147 . 213 3213. 3 779- 1 . 520 1083 1. 138 . 202 3132. 5 781-1 . 549 1061 1. 136 . 200 3199. 2 783- 1 . 585 1038 1. 135 . 198 3313. 1 820-1 . 488 1181 1.175 .250 4061. 5 821-1 . 527 1158 1.177 . 250 4301. 4 828- 1 .506 1224 1.196 . 276 4895. 5 833- 1/ . 492 1234 1. 227 . 313 5590. 5 uR ft/sec 174. 1 170. 8 172. 9 173.0 174. 4 173.0 173.0 175. 3 187.8 189.9 188.0 189.0 307. 5 183.0 182. 9 180.6 184.0 238.8 238. 9 228. 5 257. 6 242. 5 252. 2 238.4 235.8 233.9 297. 8 297. 5 330. 5 379. 6 Diameter, d, is always 1. 3021 x 10 ft for sapphire balls. -3 86 REFERENCES I 1. "Dynamics of Two-Phase Flow In Rocket Nozzles, '' Fourth Quarterly Technical Progress Report, 26 May 1962, United Technology Corpora- tion, Contract No. NOW-61-0760-C. 2. Ingebo, R.D., "Drag Coefficients for Droplets and Solid Spheres in Clouds Accelerating in Air Streams, '' NACA TN 3762, 1956. 3. Hanson, A. R. , "The Effects of Relative Velocity on Evaporation of a Liquid Fuel Spray, '' University of Michigan, Ph. D. Thesis, 1951. 4. Bolt, J. A. and Wolf, L. W. , "Drag Coefficients for Burning Kerosene Drops,'' University of Michigan ERI Project No. 2253: 3-6-P, 1954. 5. Rabin, E. , Schallenmuller, A. R., and Lawhead, R.B., "Displace- ment and Shattering of Propellent Droplets,'' Rocketdyne, AFOSR- TR-60-75, 1960. 6. Rudinger, G. , "Experiments on Shock Relaxation in Particle Suspensions in a Gas and Preliminary Determination of Particle Drag Coefficients, '' Cornel1 Aeronautical Laboratory, Project Squid Tech. Report CAL-90-P, July 1963. 7. Torobin, L. B. and Gauvin, W. H., "Introductory Concepts and Idealized Sphere Motion in Viscous Regime," The Canadian Journal of Chemical Engineering, 37:129, 1959. - 8. Torobin, L. B. and Gauvin, W. H., "The Sphere Wake in Steady Laminar Fluids, 7 f The ~ - " " Canadian ~ Journal of Chemical Engineering, 37:157, 1959. 9. Torobin, L. B. and Gauvin, W. H. , "Accelerated Motion of a Particle in a Fluid, '' The . . . . - . . - Canadian - Journal of Chemical Engineering, 37:224, 1959. 10. Torobin, L.B. and Gauvin, W.H., "The Effects of Particle Rota-. tion, Roughness, and Shape, ?' The Canadian Journal of Chemical Engineering, 38:142, 1960. REFERENCES (cont) 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. Torobin, L. B. and Gauvin, W. H., "The Effects of Fluid Turbulence on the Particle Drag Coefficient, l 1 The Canadian Journal of Chemical Engineering, 38:189, 1960. - 1__1 Torobin, L. B. and Gauvin, W. H. , "The Drag Coefficients of Single Spheres Moving in Steady and Accelerated Motion in a Turbulent Fluid, '' Pulp and Paper Institute o f Canada and Department of Chemical Engineering, McGill University, Tech. Report No. 193, 1960. Crowe, C. T. , "Drag Coefficients of Inert, Burning, or Evaporat - ing Particles Accelerating in Gas Streams, '' University of Michigan, Ph.D. Thesis, 1961. "Fundamentals of Gas Dynamics, H. W. Emmons, Ed., High Speed Aerodynamics and Jet Propulsion, Vol. 3, Princeton University Press, 1958. Emrich, R. J. and Curtis, C. W., "Attenuation in the Shock Tube, Journal of Applied Physics, Vol. 24, No. 3, 1953. Hilsenrath, J. , Beckett, C. W. , et al. , Tables of Thermal Properties of Gases, National Bureau of Standards Circular 564, 1955, Beers, Y. , Introduction to the Theory of Error, Addison-Wesley Publishing Company, 1957. Schlichting, H., Boundary Layer Theory, Trans. by J. Kestin, McGraw-Hill Book Co., Inc., 1960. ~ Moore, F. K. , "Aerodynamic Effects of Boundary Layer Uristeadi- ness, Sixth Anglo-American Aeronautical Conference, 1957. Rosenhead, L., Laminar Boundary Layers, Oxford Press, 1963. Sarpkaya, T . , "Separated Flow About Lifting Bodies and Impulsive Flow About Cylinders, Paper presented at the Second Annual AIAA Meeting, AIAA Paper No. 65-396, 1965. 88 REFERENCES (cont) 22. 23. 2 4 . 2 5 . 26. 27. 28. 29. 3 0 . 31 e 3 2 . 33. 34. Maccoll, J. W. , Journal of the Royal Aeronautical Society, 32:777, 1928. "_ -~ - ~~ ~~~~ Davies, J. M., Journal .~ of Applied Physics, 20:821, 1949. Pasternak, I.S., Ph.D. Thesis, McGill University, 1959. Luthander, S. and Rydberg, A , , Physik. Z. 36:552, 1935. Garstang, T. E. , Proceedings ~ of the Royal Society, 142A;491,1933. Drazin, M. Po , Proceeding of the Cambridge - -. Philosophical Society, 47:142, 1951. Keim, S. R. , "Fluid Resistance to Cylinders in Accelerated Motion, '' Proceedings of the American Society of Civil Engineerg 82 :HY6, (J. Hri-~Div. ), Paper 1113, 1956. ___ _ _ ~ - " " ~~ - i -~ ~ - = - " " . .
. Bugliarello, G. , "La Resistenza a1 Mot0 Accelerato di Sfere in Acqua, 'I La Richerca.Scientifica, 26:437, 1956. Hoerner, S., "Tests of Spheres with Reference to Reynolds Number, Turbulence, and Surface Roughness, '' Luftfahrtforschung, Vsl. 12, No. 1, 1935. Lunnon, R. G., "Fluid Resistance to Moving Spheres, 'I Proceedings " of the Royal Society, - Series A., Vol. 110, 1926. Carlson, D. J. and Hoglund, R. F. "Particle Drag and Heat Transfer in Rocket Nozzles, '' AIAA Journal, Vol. 2, No. 11, 1964. Iversen, H. W. and Balent, R. "A Correlating Modulus for Fluid Resistance in Accelerated Motion, '' Journal Of Applied Physics, Vol. 22, No. 3, 1951. Hoglund, R. F. , "Recent Advances in Gas Particle Nozzle Flows, '' American Rocket Society Solid Propellent Rocket Conference, Baylor University, 1962. 89 REFERENCES (cont) 35. Gilbert, M. , Davies, L., and Altman, D. , Velocity .Lag of Particles in Linearly Accelerated Combustion Gases, '' Jet Propulsion, Vol. 25 1955. 36. Liepmann, H.W. and Roshko, A., Elements of G a s Dynamics, John Wiley and Sons, Inc., 1957. 37. Spokes, G. N. , "The Role of Aluminum and its Oxides as Sources or Moderators of Electrons in Aluminized Solid Propellent Rocket Exhausts, '' Final Report, Part 2 , August 1964, Stanford Research Institutes Contract No. AF 04(694)-128. 38. Dryden, H. L. , "Review of Published Data on the Effect of Roughness on Transition from Laminar to Turbulent Flow, '' Journal of the Aeronautical Sciences, Vol. 20, No. 7, pp. 477-482, July 1953. 39. Klebanoff, P.S., Schubauer, G.B., and Tidstrom, K. D., "Mea- surements of the Effect of Two-Dimensional and Three Dimensional Roughness Elements on Boundary Layer Transition, " Journal of the Aeronautical Sciences, Vol. 22, No. 11, pp. 803-804, Nov. 1955. 40. Smith, A. M. 0. and Clutter, D. W. , "The Smallest Height of Rough- ness Capable of Affecting Boundary-Layer Transition, (' Journal of the Aeronautical Sciences, Vol. 26, No. 4 , pp. 229-245, April 1959. 41. Potter, J.A. and Whitfield, J. D. , "Effects of Unit Reynolds Number, Nose Bluntness, and Roughness on Boundary Layer Transition, AEDC-TR-60-5, March 1960. 90 NASA-Langley, 1966 CR-418 |
6766 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15?srsltid=AfmBOopCO4nQumet7ntiAbDDRd0K_uqdmPZi-0GHvd4D-IHy_qPR5OqQ | Art of Problem Solving
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2021 AMC 10B Problems/Problem 15
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6(Non-rigorous for little time)
8 Solution 7
9 Video Solution (Super Fast. 2 min and 9 seconds)
10 Video Solution (Easiest and Fastest BASIC UNDERSTANDING ONLY Required)
11 Video Solution by OmegaLearn
12 Video Solution by Interstigation (Simple Silly Bashing)
13 Video Solution by TheBeautyofMath
14 See Also
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can factor out from our original expression of to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , and notice that this is the golden mean , which is well-known to be a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),
~Lcz
Solution 3
We can immediately note that the exponents of are an arithmetic sequence, so they are symmetric around the middle term. So, . We can see that since , and therefore . Continuing from here, we get , so . We don't even need to find what is! This is since is evidently , which is our answer.
~sosiaops
Solution 4
We begin by multiplying by , resulting in . Now we see this equation: . The terms all have in common, so we can factor that out, and what we're looking for becomes . Looking back to our original equation, we have , which is equal to . Using this, we can evaluate to be , and we see that there is another , so we put substitute it in again, resulting in . Using the same way, we find that is . We put this into , resulting in , so the answer is .
~purplepenguin2
Solution 5
The equation we are given is Yuck. Fractions and radicals! We multiply both sides by square, and re-arrange to get Now, let us consider the expression we wish to acquire. Factoring out we have Then, we notice that Furthermore, Thus, our answer is ~peace09
Solution 6(Non-rigorous for little time)
Multiplying by x and solving, we get that Note that whether or not we take or we take our answer has to be the same. Thus, we take . Since this number is small, taking it to high powers like , , and will make the number very close to , so the answer is ~AtharvNaphade
Solution 7
We know that . Multiply both sides by to get Squaring both sides: Subtract from both sides: Squaring both sides: Subtract from both sides: Multiply on both sides: ~sid2012
Video Solution (Super Fast. 2 min and 9 seconds)
~Education, the Study of Everything
Video Solution (Easiest and Fastest BASIC UNDERSTANDING ONLY Required)
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution by Interstigation (Simple Silly Bashing)
~ Interstigation
Video Solution by TheBeautyofMath
Not the most efficient method, but gets the job done.
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14Followed by
Problem 16
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Comprehensive compilation of conic section formulae
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Modified 8 years, 8 months ago
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15
$\begingroup$
My frustration started after hours of searching failed to turn up a formula for the vertex of a parabola in the general form
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
As is already well known, the discriminant $\Delta=B^2-4AC$ can be used to diagnose the type of conic one has if it it given in the general form. Other useful formulae I found in the midst of searching include the formula for the slope of the principal axis:
$$\tan\;\theta=\frac{B}{A-C+(\mathrm{sgn}\;B)\sqrt{B^2+(A-C)^2}}$$
the eccentricity (intended only for elliptic or hyperbolic cases)
$$\varepsilon=\left(\sqrt{\frac12-\frac{(\mathrm{sgn}\;\Delta)(A+C)}{2\sqrt{B^2+(A-C)^2}}}\right)^{-1}$$
and the coordinates of the center for a central conic (I won't list it here so you can figure it out or find it yourself ;) ), but there were no formulae given for finding the coordinates of the vertex of a parabola.
Of course, for the parabola case, I could rotate axes, find the vertex through completing the square, and then rotate back, but I was hoping somebody already went through the trouble of deriving a formula so I don't have to reinvent the wheel.
On to my query: is there a comprehensive compilation (book, article) somewhere of formulae related to dealing with conic sections? The implicit Cartesian form is what I'm currently dealing with, but lists relating to other forms (polar, parametric) are welcome as well.
geometry
reference-request
conic-sections
plane-curves
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asked Sep 14, 2010 at 23:15
J. M. ain't a mathematicianJ. M. ain't a mathematician
76.7k88 gold badges222222 silver badges347347 bronze badges
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$\begingroup$ My suggestion is the book "Analytic geometry: a vector approach", by Charles Wexler. There is a very good chapter on conic sections, proving each passage. It's very clear. $\endgroup$
Ronaldo
– Ronaldo
2010-09-14 23:32:32 +00:00
Commented Sep 14, 2010 at 23:32
2
$\begingroup$ I remember there used to be a book by S.L.Loney which was filled with cartesian geometry results about conic sections. Don't recollect the name of the book though. $\endgroup$
Aryabhata
– Aryabhata
2010-09-15 00:18:43 +00:00
Commented Sep 15, 2010 at 0:18
1
$\begingroup$ I don't know what the principal axis is but if you know its slope you should be able to find the vertex by finding the point whose tangent line has the appropriate slope. $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2010-09-15 03:58:11 +00:00
Commented Sep 15, 2010 at 3:58
1
$\begingroup$ Qiaochu: it's the line of symmetry for a conic (one of the lines of symmetry for a central conic). From $\tan\;\theta$ you can compute the sine and cosine, and thus rotate the conic to a form where the more usual formulae apply (algebraically, it removes the cross term $xy$). Done by hand, I can easily find the vertex of a parabola, but this is meant for computer implementation. $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2010-09-15 04:15:48 +00:00
Commented Sep 15, 2010 at 4:15
1
$\begingroup$ The thing with the parabolic case... as I said, I'd prefer something that doesn't involve rotate, complete the square, undo rotation on vertex thus found since this is intended for a dumb computer (without geometric insight) to do. $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2010-09-15 04:19:05 +00:00
Commented Sep 15, 2010 at 4:19
| Show 7 more comments
2 Answers 2
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5
$\begingroup$
Appendix 1 of Keith Kendig's Conics seems to carry a lot of the formulae I want. On the other hand, I'd sure be interested in an even more extensive listing...
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answered Aug 25, 2011 at 3:16
J. M. ain't a mathematicianJ. M. ain't a mathematician
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An old rich treatise is
John Casey: A treatise on the analytical geometry of the point, line, circle, and conic sections. Second edition 1893. Hodges, Figgis, & Co. (Ltd.) Dublin; Longmans, Green & Co. London; Dublin University Press.
That one also contains background for the complex tangent-through-focus topic which I find so useful in determining foci.
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edited Jan 22, 2017 at 16:26
answered Jan 22, 2017 at 16:18
ccornccorn
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$\begingroup$ "1893" - I don't think any library easily accessible to me has something of that vintage, but I'll see what I can manage. Thanks! $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2017-01-22 16:26:30 +00:00
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Can we always find such a permutation?
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Let p p be an odd prime number. Does there always exist integers {a 1,a 2,...,a p−1}{a 1,a 2,...,a p−1} which is a permutation of {1,2,3,...,p−1}{1,2,3,...,p−1} such that a 1 1+a 2 2+⋯+a p−1 p−1 a 1 1+a 2 2+⋯+a p−1 p−1 is divisible by p p?
Ideas so far:
I have tried taking a primitive root modulo p and ranging it and considering the sum of it with the other p−2 p−2 terms. Another idea I had was to somehow find a permutation so that we got 1,2,...,p−1 1,2,...,p−1 modulo p p. Both of these have failed.
I do not even know if the answer to the problem statement is positive. Does anyone have any suggestions?
elementary-number-theory
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edited Nov 20, 2023 at 23:07
Bill Dubuque
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asked Nov 16, 2023 at 13:38
LulunLulun
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An easier (and much more likely to be true) variant would be to show that there are infinitely many p such that it satisfies the problem statement (instead of trying to show that it is true for all p).Lulun –Lulun 2023-11-16 13:38:40 +00:00 Commented Nov 16, 2023 at 13:38
This is an interesting question: After running a bf(until prime 1489) up to 1487 the first permutation(lexicographically) fixes all numbers up to p-8 for p>8. In the case of 1489, the first permutation only fixes the first 8 numbers. I tried bf using a permutation generated with a primitive element, it doesn't work for p=101 for example.Phicar –Phicar 2023-11-16 15:41:45 +00:00 Commented Nov 16, 2023 at 15:41
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Note: This problem has been solved already by someone else a while ago. I am posting it since it appears that a few people are still visiting this thread and I hope to provide a complete solution. The proof is surprisingly quite elegant. [Solution by u\ComfortableOwl2322 is as follows]
Yes, you always can.
Proof: We think of the powers as being set by the permutation instead, i.e. 1^{a_{1}} + (p-1)^{a_{p-1}}
Fix a primitive root A. It suffices to show that there is a {a_1,...,a_{p-1}} such that the part of the sum for the odd powers of A is equal to the part of the sum for even powers of A --- i.e.
(A^1)^a_1 + (A^3)^a_3 + (A^5)^(a_5) + ... + (A^p-2)^{a_{p-2}) = (A^2)^a2 + .... + (A^{p-1})^a_{p-1} (mod p).
To see this, note that we can replace {a_1, ..., a_{p-1}} by the shift of adding (p-1)/2 to each a_i (mod p-1). This has the effect of negating the left side above, but preserving the right side, so, if the left and right sides were originally equal, the sum would be zero after this shift.
But actually we can explicitly give a_1,..., a_{p-1} such that the above equation holds ---- even term for term! Namely, take a_1 = 2, a_2 = 1, a_3 = 4, a_4 = 3, ..., a_{p-2} = p-1, a_{p-1} = p-2.
Note then that (A^{2i-1})^{a_{2i-1}} = (A^{2i-1}) ^{2i} = {A^{2i}} ^{2i-1} = {A^{2i}}^{a_{2i}}.
Construction: Unwinding the proof details, we have the following procedure to obtain a setting of the original powers which gives zero sum:
Choose a primitive root A mod p.
For each A^2i, i.e. even power of A, choose its power to be (2i-1+ (p-1)/2) mod (p-1).
For each A^{2i-1}, i.e. odd power of A, choose its power to be (2i + (p-1)/2) mod (p-1).
For example, take p=11. A primitive root is 2, with powers (2,4,8,5,10,9,7,3,6,1), and we take our exponents to be (2,1,4,3,6,5,8,7,10,9) shifted by (p-1)/2 = 5 mod 10, so (7,6,9,8,1,10,3,2,5,4).
And one can check that 2^7+4^6+8^9+...+1^4 = 0 (mod 11).
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edited Nov 21, 2023 at 14:19
answered Nov 21, 2023 at 14:01
LulunLulun
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Though I haven't proved yet, it seems like this construction works
let g g be a primitive root modulo p p . then atleast one of ∑p−1 i=1(g i)i∑i=1 p−1(g i)i or ∑p−1 i=1(g i−1)i∑i=1 p−1(g i−1)i is divisible by p.
Edit : proof :-
if p≡1(m o d 4)p≡1(m o d 4) .
We show that the second summation works
as g p−1 2≡−1(m o d p)g p−1 2≡−1(m o d p)
g k(m o d p)g k(m o d p) depends on k(m o d p−1)k(m o d p−1) .
We show that each i can be paired with j such that j(j−1)=i(i−1)+p−1 2(m o d p−1)j(j−1)=i(i−1)+p−1 2(m o d p−1) So that the terms cancel out in the sum.
if p−1 p−1 is factorized as 2 α k 2 α k where k is odd. we set i≡j(m o d k)i≡j(m o d k) . And j≡2 α−1+i(m o d 2 α)j≡2 α−1+i(m o d 2 α), u can easily verify that it works.
if p≡3(m o d 4)p≡3(m o d 4)
The first sum works we show there exists j such that
j 2=i 2+p−1 2(m o d p−1)j 2=i 2+p−1 2(m o d p−1)
If p−1=2 k p−1=2 k where k is odd. set j≡i(m o d k)j≡i(m o d k) , and j=i+1(m o d 2)j=i+1(m o d 2) and use CRT .
I have not really shown why pairing is unique for each but hopefully, you can complete that part yourself.
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edited Nov 22, 2023 at 9:53
answered Nov 21, 2023 at 5:46
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I will put a proof if I get it Lucid –Lucid 2023-11-21 05:46:47 +00:00 Commented Nov 21, 2023 at 5:46
Thanks for your interest. It turns out it's easier to switch perspectives and fix the bases instead of the exponents. You may refer to the solution above if you're interested in seeing the solution.Lulun –Lulun 2023-11-21 14:17:18 +00:00 Commented Nov 21, 2023 at 14:17
I see but i think i proved it so i will just put it in Lucid –Lucid 2023-11-22 09:39:04 +00:00 Commented Nov 22, 2023 at 9:39
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6769 | https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/Metric_Imperial_Conversion_Errors | Skip to main content
Metric/Imperial Conversion Errors
Last updated
: Aug 29, 2023
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Units of Measure
Metric Prefixes - from yotta to yocto
Page ID
: 349
( \newcommand{\kernel}{\mathrm{null}\,})
To learn more about systems of measurement, visit the SI Unit page and the Non-SI Unit page. As the four examples below can attest, small errors in these unit systems can harbor massive ramifications.
The Mars Climate Orbiter: A Multimillion Dollar Mistake
Although NASA declared the metric system as its official unit system in the 1980s, conversion factors remain an issue. The Mars Climate Orbiter, meant to help relay information back to Earth, is one notable example of the unit system struggle. The orbiter was part of the Mars Surveyor ’98 program, which aimed to better understand the climate of Mars. As the spacecraft journeyed into space on September 1998, it should have entered orbit at an altitude of 140-150 km above Mars, but instead went as close as 57km. This navigation error occurred because the software that controlled the rotation of the craft’s thrusters was not calibrated in SI units. The spacecraft expected newtons, while the computer, which was inadequately tested, worked in pound forces; one pound force is equal to about 4.45 newtons. Unfortunately, friction and other atmospheric forces destroyed the Mars Climate Orbiter. The project cost $327.6 million in total. Tom Gavin, an administrator for NASA's Jet Propulsion Laboratory in Pasadena, stated, "This is an end-to-end process problem. A single error like this should not have caused the loss of Climate Orbiter. Something went wrong in our system processes in checks and balances that we have that should have caught this and fixed it."
The Mars Climate Orbiter, image courtesy NASA/JPL-Caltech
NASA's Constellation Program: A Possible Casualty of Metric/Imperial Conversions
Another NASA related conversion concern involves the Constellation project, which is focused mainly on manned spaceflight. Established in 2005, it includes plans for another moon landing. The Constellation project is partially based upon decades-old projects such as the Ares rocket and the Orion crew capsule. These figures and plans are entirely in British Imperial units; converting this work into metric units would cost approximately $370 million.
Work on the Constellation Project, image courtesy NASA/Kim Shiflett
Disneyland Tokyo: A Bumpy Blunder
Tokyo Disneyland’s Space Mountain roller coaster came to a sudden halt just before the end of a ride on December 5, 2003. This startling incident was due a broken axle. The axle in question fractured because it was smaller than the design’s requirement; because of the incorrect size, the gap between the bearing and the axle was over 1 mm – when it should have been a mere 0.2 mm (the thickness of a dime vs. the thickness of two sheets of common printer paper.) The accumulation of excess vibration and stress eventually caused it to break. Though the coaster derailed, there were no injuries. Once again, unit systems caused the accident. In September 1995, the specifications for the coaster’s axles and bearings were changed to metric units. In August 2002, however, the British-Imperial-unit plans prior to 1995 were used to order 44.14 mm axels instead of the needed 45 mm axels.
Air Canada Flight 143: Unit-Caused Fuel Shortage
A Boeing 767 airplane flying for Air Canada on July 23, 1983 diminished its fuel supply only an hour into its flight. It was headed to Edmonton from Montreal, but it received low fuel pressure warnings in both fuel pumps at an altitude of 41,000 feet; engine failures followed soon after. Fortunately, the captain was an experienced glider pilot and the first officer knew of an unused air force base about 20 kilometers away. Together, they landed the plan on the runway, and only a few passengers sustained minor injuries. This incident was due partially to the airplane’s fuel indication system, which had been malfunctioning. Maintenance workers resorted to manual calculations in order to fuel the craft. They knew that 22,300 kg of fuel was needed, and they wanted to know how much in liters should be pumped. They used 1.77 as their density ratio in performing their calculations. However, 1.77 was given in pounds per liter, not kilograms per liter. The correct number should have been 0.80 kilograms/liter; thus, their final figure accounted for less than half of the necessary fuel.
The Air Canada craft, image courtesy Akradecki
Example 1
If Jimmy walks 5 miles, how many kilometers did he travel?
Solution
: 5miles×(1.6kilometers1mile)=8kilometers
Example 2
A solid rocket booster is ordered with the specification that it is to produce a total of 10 million pounds of thrust. If this number is mistaken for the thrust in Newtons, by how much, in pounds, will the thrust be in error? (1 pound = 4.5 Newtons)
Solution
: 10,000,000 Newtons x (1 pound / 4.448 Newtons) = 2,200,000 pounds.
10,000,000 pounds - 2,200,000 pounds = 7.800,000 pounds.
The error is a missing 7,800,000 pounds of thrust.
Example 3
The outer bay tank at the Monterey Bay Aquarium holds 1.3 million gallons. If NASA takes out all the fish in this tank and sends them to swim around in space, what is the theoretical volume of all the fish in liters? Assume there are 3,027,400 liters of water left in the tank after the fish are removed.
Solution
: 3,027,400liters×(0.264gallons1liter)=800,000gallons remaining in tank
The volume of the space fish is 1,300,000 - 800,000 = 500,000 gallons, which converts to 1,892,100 liters worth of fish swimming around the solar system.
Example 4
A bolt is ordered with a thread diameter of 1.25 inches. What is this diameter in millimeters? If the order was mistaken for 1.25 centimeters, by how many millimeters would the bolt be in error?
Solution
: 1.25inches×25.4millimeters1inch=31.75millimeters
Since 1.25 centimeters x (10 millimeters / 1 centimeter) = 12.5 millimeters, the bolt would delivered 31.75 - 12.5 = 19.25 millimeters too small.
Example 5
The Mars Climate Orbiter was meant to stop about 160 km away from the surface of Mars, but it ended up within 36 miles of the surface. How far off was it from its target distance (in km)? If the Orbiter is able to function as long as it stays at least 85 km away from the surface, will it still be functional despite the mistake?
Solution
: 36miles×1.6kilometers1mile=57.6km kilometers from surface
The difference then is (in kilometers): 160 - 57.6 kilometers = 102.4 kilometers away from targeted distance. Hence, the Orbiter is unable to function due to this mistake since it is beyond the 85 km error designed into its function.
References
Nelson, Wade. “Gimli Glider.” Mobility Forum: The Journal of the Air Mobility Command’s Magazine. 10.1. (2001): 24-29.
Lawler, Andrew. “Obama Backs Big Launcher and Bigger NASA Budget.” Science. 327.5961. (2010): 18.
Cowen, Ron. "NASA Loses Climate Obiter." Science News. 156.14. (1999): 214.
Units of Measure
Metric Prefixes - from yotta to yocto |
6770 | https://ocw.mit.edu/courses/14-30-introduction-to-statistical-methods-in-economics-spring-2009/resources/mit14_30s09_lec11/ | MIT OpenCourseWare
14.30 Introduction to Statistical Methods in Economics
Spring 200 9
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1
14.30 Introduction to Statistical Methods in Economics
Lecture Notes 11
Konrad Menzel
March 17, 2009
Order Statistics
Let X1, . . . , X n be independent random variables with identical p.d.f.s fX1 (x) = . . . = fXn (x) - we’ll generally call such a sequence ”independent and identically distributed”, which is typically abbreviated as iid . We are interested in the function
Yn = max {X1, . . . , X n}
i.e. Yn is the largest value in the sample. We can derive the c.d.f. of Yn using independence
FYn (y) = P (Yn ≤ y) = P (X1 ≤ y, X 2 ≤ y, . . . , X n ≤ y)
= P (X1 ≤ y)P (X2 ≤ y) . . . P (Xn ≤ y)
= FX1 (y)FX2 (y) . . . F Xn (y)
= [FX (y)] n
Using the chain rule, we can obtain the p.d.f. of the maximum,
d n−1
fYn (y) = FYn (y) = n [FX (y)] fX (y)
dy
Example 1 An old painting is sold at an auction. n identical bidders submit their bids B1, . . . , B n
independently, and the marginal c.d.f. of the bids is FB (b). The potential buyer who submitted the highest bid gets to buy the painting and has to pay his bid (this type of auction is called Dutch, or first price auction). Then the revenue of the seller of the painting has p.d.f. is
n−1
fY (y) = fmax {B1 ,...,B n }(y) = n [FB (y)] fB (y)
Now we can generalize this to other ranks in the sample, e.g.
Yn−1 = ”The 2nd highest value in X1, . . . , X n ”
This random variable is called the ( n − 1) th order statistic of X1, . . . , X n, and we can state its p.d.f.
Proposition 1 Let X1, . . . , X n be an iid sequence of random variables with p.d.f. fX (f ) and c.d.f.
FX (x). Then the kth order statistic Yk has p.d.f.
fYk (y) = k n [FX (y)] k−1[1 − FX (y)] n−kfX (y)
k
1�� �� �� � � ���� �
Proof: We can split the experiment into two parts, (a) one of the Xi’s has to take the value y according to the density fX (y), and (b) given the value y, the other draws in the sequence have to be grouped around y in a way that makes y the kth smallest value in the sample. Part (b) is a binomial experiment in which the n trials correspond to the n draws of X1, . . . , X n, and we define a ”success” in the ith round as the event ( Xi ≤ y). Since draws are independent and correspond to the same p.d.f. the parameter p in the binomial distribution is equal to FX (y). y being smaller or equal to the kth smallest value corresponds to at least k ”successes” in the binomial experiment, and therefore the corresponding c.d.f. is
n
� �
FYk (y) = n [FX (y)] l[1 − FX (y)] n−l
l
l=k
We can now obtain the p.d.f. by differentiating the c.d.f. with respect to y, using the product and the chain rule
d
fYk (y) = FYk (y)
dy
n
� � n � �
= n l[FX (y)] l−1[1 − FX (y)] n−lfX (y) − n (n − l)[ FX (y)] l[1 − FX (y)] n−l−1fX (y) = T1 − T2
l l
l=kl=k
This expression looks complicated, but it turns out to be essentially a telescopic sum, so that most summands will drop out. Noting that for the second term, the summand corresponding to l = n is zero, we can rewrite it as
n−1
� � n � �
T2 = n
l (n−l)[ FX (y)] l[1 −FX (y)] n−l−1fX (y) = l −
n
1 (n−l+1)[ FX (y)] l−1[1 −FX (y)] n−lfX (y)
l=kl=k+1
replacing the running index l with l − 1. Since for the first term,
n n!l n! n
l · l = l!( n − l)! = (l − 1)!( n − l + 1)! (n − l + 1) = l − 1 (n − l + 1)
The first term becomes
n
� �
T1 = l −
n
1 (n − l + 1)[ FX (y)] l−1[1 − FX (y)] n−lfX (y)
l=k
so that the density equals the l = k term of the sum defining T1 since it is the only one which doesn’t get canceled out when we subtract T2. Therefore
k
� �
fYk (y) = T1 − T2 = l −
n
1 (n − l + 1)[ FX (y)] l−1[1 − FX (y)] n−lfX (y)
l=k
= k!( n −
n
k
!
1)! (n − k + 1)[ FX (y)] k−1[1 − FX (y)] n−kfX (y)
= k n [FX (y)] k−1[1 − FX (y)] n−kfX (y)
k
which is the result we were going to prove �
2� �� � � �
2
Example 2 We could now think of an auction which is different from the first one in that the buyer submitting the highest bid still gets to buy the painting, but only has to pay the amount offered by the second-highest bidder (this auction format is more common than the first, and it’s known as an English, or second-price auction). If the submitted bids are random variables C1, . . . , C n, the revenue Y of the seller now has p.d.f.
fY (y) = 2 n FC (y)[1 − FC (y)] n−2fC (y)
2
Notice that I used different letters for the bids, since it is known from economic theory that the same bidders should submit different bids under the two different auction formats.
Digression: The Distribution of the ”First Digit of Anything” (skipped in lecture)
Here’s a somewhat cute but non-standard problem, for which we are not going to use the methods we saw earlier. So this is definitely something you can skip for problem sets or exam preparation, but I’d still like to go over it.
What is the distribution of the first digit of a number we basically don’t know anything about? To be more precise, let X be a measurement of any kind, for which we don’t know what it represents, nor in what units 1 it is measured - e.g. we could leaf through a newspaper and collect any numbers Y
representing a measurement of some kind (incomes, stock indices, population etc.). What is the p.d.f. of the first digit of a number of that kind if we don’t know anything else about where it comes from? I.e. how do we derive the p.d.f. of the first decimal of the random variable Z = X Y if all we know that X·
and Y are positive, but can be anything? A first guess might be that the distribution of the first digit is (discrete) uniform, since this distribution intuitively does not seem to contain much ”information” about the numbers nor their units. However, if we take a uniform distribution and change the units (e.g. double or quadruple all numbers in the assumed underlying distribution), the distribution of first digits doesn’t remain uniform. E.g. if the true numbers are X ∼ U [1 , 10], 4 X ∼ U [0 , 40], so that for the first digit Y of 4 X has p.d.f.
⎧ 1 1
⎨ 40 + 4 if y ∈ { 1, 2, 3}
fY (y) = 1
40
if y ∈ { 4, 5, 6, 7, 8, 9}
⎩ 0 otherwise
Or, in pictures, So a minimal requirement for the distribution of the first digits should be that it doesn’t change if we change the units of measurement. What we are looking for is in fact a random variable X for which the distribution of X doesn’t change for changes of the scale, i.e. aX for a > 0. This is true if we assume that Z = log( X) ∼ U [log(1) , log(10)], since for a scale shift,
� � z z+1
− log z z + 1 log a a
P (z ≤ aX ≤ z + 1) = P a ≤ X ≤ a log(10 a) − log( a)
=
= log( z + 1) − log( z) = z + 1)
log(10) − log(1) P (z ≤ X ≤
Then the first digit Y of Z has p.d.f.
�
fY (y) =
log( z+1) −log( z)
log(10)
0
if y ∈ { 1, 2, 3, . . . , 9}
otherwise
30 .2 .4 .6 1 2 3 4 5 6 7 8 9uniform density old units times 2 old units times 4
Figure 1: Effect of a Change of Measurement Units on the Uniform
This invariance idea may seem like a very artificial way of obtaining a distribution, since there is no obvious connection to the measurements X nor the measurement units. However, the resulting p.d.f. seems to give a good approximation to ”real-world data” which falls into the category - as e.g. numbers representing measurements of some kind which appear in the New York Times. The following graph shows the ”theoretical” density together with a histogram of the first digit of GDP measured in local currency units (i.e. Yen for Japan, C$ for Canada, etc.) for the 77 countries included in ”World in Figures 2007” pocket book from The Economist. Summarizing, this example took a radically different approach
0 .1 .2 .3 123456789gdp density
Figure 2: Distribution of First Digit of GDP in Local Currency Units and Theoretical Density (Numbers from The Economist, Pocket World in Figures 2007)
to determining the distribution of a given function of two random variables X and Y : here we started out not knowing the p.d.f.s of X and Y , but then imposed that whatever the resulting distribution was going to be, it had to be invariant with respect to a change of units - i.e. the realization of Y . The concept of ”invariance” plays a major role in advanced statistics, but for the purpose of this class, we
4����
won’t go beyond this example.
3 Expected Values and Median
Given a random variable X with a p.d.f. fX (x), we’d like to summarize the most important characteristics of the entire distributions without having to give the entire density function. The expected value tells us basically where the distribution of X is centered.
3.1 Definitions
Definition 1 If X is a discrete random variable, the expected value of X, denoted E[X] is given by
E[X] := xf X (x)
x
if this sum is finite. If X is continuous, the expected value is defined as
� ∞
E[X] = xf X (x)dx
−∞
if the integral is finite.
Example 3 What is the expectation of a binomial random variable, i.e. X ∼ B(n, p )?
n
� �
E[X] = x n px(1 − p)n−x
x
x=0
n
= �
x!( n
n!
−
x
x)! px(1 − p)n−x
x=1
n
(n − 1)! x−1(1 − p)(n−1) −(x−1)
= np p
(x − 1)!(( n − 1) − (x − 1))! x=1
n−1
� �
n − 1
= np x − 1 px(1 − p)n−1−x
x=0
= np 1·
where in the second row, we can ignore the summand corresponding to x = 0 since it’s zero, in the third row, we pull n out of the binomial coefficient, and in the following step, we switch the summation index from x to x−1. Finally, if we pull np out, the summands are the binomial probabilities for X˜ ∼ B(n−1, p ),and therefore they sum to one.
Note that a random variable which takes infinitely many different value may not have a finite ex pectation, in which case the expectation is not defined. You should also notice that even though the expectation gives a sense of the ”location” of a distribution, it is in general not a ”typical” value for the random variable: e.g. the expected value of a die roll is 1
6
(1 + 2 + 3 + 4 + 5 + 6) = 3 2
1
, which is not a possible outcome.
An alternative measure of the position of a distribution is the median :
Definition 2 The median m(X) of a random variable X is a real number such that
1
P (X < m ) = 2
5��
The median and the expected value of a random variable X coincide if the distribution of X is symmetric around m(X), i.e. fX (m(X) − x) = fX (m(x) + x), but need not be the same in general.
Example 4 Say X has p.d.f.
12
fX (x) = 9 x if 0 ≤ x ≤ 30 otherwise
The expected value is
� 3 � 3 � �3
1 1 1 4 81 9
t t2dt = t3dt = t = = = 2 .25
0
· 9 9 0 36 0 36 4
In order to obtain the median, let’s first calculate the c.d.f. of X
� x �x 3
1 2 1 3 x
FX (x) = t dt = t =
9 27 27 0 0
Therefore, solving FX (m) = 2
1
for m gives
27 33
m = 2 ⇔ m = √2 ≈ 2.38 > 2.25 3
Therefore, the median of this distribution is greater than the mean.
Note that the median may not be unique:
Example 5 Let X be the result of rolling a fair die once. Then for any number m ∈ (3 , 4] , P (X < m ) =
P (X ≤ 3) = 1 . Therefore any value in that interval is a median. 2
3.2 Properties of Expectations
Property 1 If X = c, where c is a constant, then
E[X] = c
Property 2 If Y = aX + b, then
E[Y ] = aE[X] + b
Proof: Let’s only look at the continuous case: if X is a continuous random variable with p.d.f. fX (x), then the expectation of Y is given by
� ∞ � � ∞
E[Y ] = (ax + b)fX (x)dx = a xf X (x)dx + b fX (x)dx = aE[X] + b 1·
−∞ −∞ −∞
so as we can see, the linearity of integrals translates directly to linearity of expectations �
Property 3 For Y = a1X1 + a2X2 + . . . + anXn + b,
E[Y ] = a1E[X1] + a2E[X2] + . . . + anE[Xn] + b
This is the most general statement of the linearity of expectations, and we will use this property over and over in the remainder of this class.
6� �� � �� � ���� �
Example 6 Above, we calculated the expectation of X B(n, p ), E[X] = np by summing over the ∼
possible outcomes of X. But from the last result we can see that there is an easier way of obtaining the same result: since X is the number of successes from a sequence of n trials, we can code the outcome of each trial as Z1, Z 2, . . . , Z n, where Zi = 1 if the ith trial was a success and Zi = 0 otherwise.
E[Zi] = 1 p + 0 (1 − p) = p· ·
and therefore, n n n
E[X] = E Zi = E[Zi] = p = np
i=1 i=1 i=1
Property 4 IF X and Y are independent , then
E[XY ] = E[X]E[Y ]
If X and Y are not independent, this will generally not be true.
3.3 Expectations of Functions of Random Variables
Let Y = r(X). Last week, we saw how we could derive the p.d.f. of Y if we knew fX (x). For expectations, this problem is much simpler since we are only looking at one single characteristic of the distribution. The expectation of Y = r(X) is given by
r(x)fX (x) if X discrete
E[Y ] = E[r(X)] = � ∞
x
−∞
r(t)fX (t)dt if X continuous
Example 7 Suppose Y = X1/2 where the p.d.f. of X is given by
2x if 0 < x < 1
fX (x) = 0 otherwise
� 1 � 1 � �1
5/2
E[Y ] = t1/2fX (t)dt = 2 t3/2dt = 2 2 4
x =
5 50 0 0
The same principle works for functions of 2 or more random variables
Example 8 Suppose we are interested in the function Z = X2 + Y 2 of two random variables X and Y
with joint p.d.f.
fXY (x, y ) = 1 if 0 ≤ x, y ≤ 10 otherwise
Then � ∞ � ∞
E[Z] = (x2 + y 2)fXY (x, y )dxdy
−∞ −∞
� 1 1
= (x2 + y 2)dxdy
00
� 1 � �1
1
= x3 + y 2 x dy
30 0
� 1 1
= + y 2 dy
30
� �1
1 1 3 1 1 2
= y + y = + =
3 3 3 3 30
7For linear functions of a random variable X of the type Y = aX + b, we saw above that E[aX + b] =
aE[X] + b. This doesn’t work for nonlinear functions of a random variable. A particulary important result on this is Jensen’s Inequality :
u(X)
x
X2X1E(X)
E[u(X)]
u(E[X])
Figure 3: Example with a discrete distribution
Image by MIT OpenCourseWare.
over {x1, x 2}
Proposition 2 (Jensen’s Inequality) Let X be a random variable, and u(x) be a convex function. Then
E[u(X)] ≥ u (E[X])
The inequality is strict if u( ) is strictly convex and X takes at least two different values with positive ·
probability.
Proof: We can define a linear function
r(x) = u(E[X]) + u�(E[X])( x − E[X]) which is tangent to u(x) which passes through the point ( E[X], u (E[X])). Since u( ) is convex, ·
u(x) ≥ r(x) for all x
In particular, � ∞ � ∞
E[u(X)] = u(t)fX (t)dt ≥ r(t)fX (t)dt = E[r(X)]
−∞ −∞
Since r(x) is linear by construction, we can use property 2 on expectations of linear functions with
a = u�(E[X]) and b = u(E[X]) − u�(E[X]) E[X] to obtain
E[r(X)] = E[aX + b] = aE[X] + b = u�(E[X]) E[X] + u(E[X]) − u�(E[X]) E[X] = u(E[X]) Putting this together with the inequality derived before,
E[u(X)] ≥ E[r(X)] = u(E[X]) which completes the proof �
Notice that, since for a concave function v(x), its negative −v(x) is convex, Jensen’s Inequality also implies that for a concave function v(·),
E[v(X)] ≤ v(E[X])
8u (x) x
E [x] r (x)
�� �� �
Figure 4: r(x) is always less than u(x)
Example 9 (Risk Aversion): Suppose you buy a laptop for 1,200 dollars which comes with a limited warranty for the first year. During that first year, there is a probability p = 10% that you spill a cup of coffee over the laptop (or have a similar accident, which is all your own fault) and have to replace the entire motherboard, which will cost 1100 dollars. This repair is not covered by the limited warranty, but you may purchase an extended service plan for 115 dollars. Should you buy this additional ”insurance”? Without the additional insurance, we can think of the total cost of your laptop as a random variable which takes values X = 1 , 200 with probability 1 − p, and X = 1 , 200 + 1 , 100 = 2 , 300 with probability p (there are different ways of setting up this problem, but let’s keep things simple for now). With the extended service plan, your laptop will cost you Y = 1 , 200 + 115 = 1 , 315 dollars for sure. If you only care about the expected value of the laptop, then E[X] = 2 , 300 p+1 , 200(1 −p) = 1 , 200+1 , 100 p.This is greater than E[Y ] = 1 , 315 if p ≥ 10 .45% . But since we said that p = 10% , would it still be a good idea to buy the service plan? - Economists typically assume that when people take decisions under uncertainty, they do not care about the expected amount of money W they can spend, but that they experience a utility u(W ) over a dollar amount, for which the additional value of an additional dollar increases in the total amount consumed. That means that we assume that u( ) is a concave function in ·
costs, say
u(c) = 4, 800 − c
where I assume that our initial wealth is 4, 800 , and we can spend 4, 800 − C, where C is the total ”cost” of the laptop. Then the expected utility from not having the service plan is
E[u(C1)] = 0 .9 4, 800 − 1, 200 + 0 .1 4, 800 − 2, 300 = 0 .9 60 + 0 .1 50 = 59 · ·
whereas with the insurance plan,
E[u(C2)] = 4, 800 − 1, 315 > 3, 481 = 59
In fact, you would be willing to spend up to 4, 800 − 3, 481 = 119 dollars for the insurance, whereas the expected additional cost is only 1, 100 p = 110 dollars. This 9-dollar difference in what we are willing to pay for the insurance is referred to as the risk-premium comes from the fact that u( ) is concave. By ·
Jensen’s inequality, this risk-premium is positive if u( ) is concave, and we say that this type of preferences ·
exhibits risk aversion .
Example 10 The following example is known as the St. Petersburg Paradox , and it gives an example of a random variable which doesn’t have a finite expectation. We are offered the following gamble: suppose a fair coin is tossed over and over until the first head
9
Image by MIT OpenCourseWare.
�� �
appears. You get 2 dollars when heads appears on the first flip, 22 dollars if it appears on the second, and, more generally, 2k dollars if the first head appears on the xth flip. How much would you pay to play this game? - in principle, you should be willing to pay your expected winnings, so let’s do the calculation: the probability that exactly x flips are required equals
1 1
fX (x) = P (x − 1 tails , 1heads) = = 2 −x
2x−1 2
Therefore, expected winnings Y can be calculated as
∞∞
� �x ∞� � 2 �
E[Y ] = 2xfX (x) = 2 = 1 = ∞
x=1 x=1 x=1
Therefore, there is no upper bound on expected winnings. Does this mean that we’d see people willing to pay an infinite amount for this type of bet? - certainly not: typically people would offer at most around 25 dollars to play the game. This paradox can be resolved in different ways:
• people do not actually care about the expected amount of money, but their valuation of money decreases in the total amount they already have, i.e. people maximize some concave function u( ) ·
of winnings, as in the previous example.
• with very small probabilities, the amount you win is extremely high - i.e. in the trillions, quadrillions etc. of dollars, and we would not believe that in this case, our opponent could live up to his promise, so in fact there would be some ceiling as to how much we could at best hope to win from this bet.
In order to make the link back to Jensen’s inequality, let’s also calculate the expected number of coin flips in the game for a coin which comes up heads with probability p = a
1
:
∞∞
� �x−1 � �
G�
a a a a
E[X] = � xa −x = 1 � 1 1 1
x =:
x=1 x=1
where G�(a) is, as you can easily check, the first derivative with respect to a
1
of
� � ∞ � �x
1 1 1
G = =
a ax=1 1 − a
1
Therefore, by taking the derivative of the new expression for G �
a
1
�, we get that
1 1 1 1
E[X] = G� = � �2a a a 1
1 − a
and since in our example, 1
1
= 2
1
, the expected number of flips is 2. Hence, the 25 dollars many people are still willing to bet are far more than 2E[X] = 2 2 = 4 dollars you’d get from the average number of flips. The explanation for this is once more Jensen’s inequality, and the fact that u(x) = 2 x is an (extremely) convex function of x.
Example 11 Suppose you can choose between two assets: the first is the stock of an obscure online start up which makes it possible for people to search web sites for free. It is very risky in that with probability 90% dividends are constant at e0t = 1 , and with 10% probability, the company’s name is Google, and the
10 dividends it pays at any instant in time t develop according to e0.1T , i.e. there is a random growth rate
G1 which takes values 0% with probability 90%, and 10% with probability 10%, respectively. Alternatively, you can hold a government bond which will e0.02 t interest at every instant t in the future, i.e. G2 = 2%
for sure. You value one dollar you receive t periods from now as much as e−0.15 t dollars you can have right now, but invest in one of the two assets anyway. I.e. in general you value an asset whose returns grow at rate
g for sure as � ∞
V (g) = e(g−r)tdt = 0 − 1 = 1
0
g − r r − g
The expected growth rate of dividends for the risky stock is
E[G1] = 0 .9 0% + 0 .1 10% = 1% < 2% = E[G2]· ·
However, your valuation for the stock is
1 1 90 10 40
E[V (G1)] = 0 .9 + 0 .1 = + = = 8
0.15 − 0 0.15 − 0.1 15 5 5
whereas you value the bond as
1 100 200 200
E[V (G2)] = = = < = E[V (G1)]
0.15 − 0.02 13 26 25
Intuitively, uncertainty over the growth rates means that even though 90% of the start-ups don’t develop (or even fail), the 10% which are successful do so spectacularly as to compensate for the investments which turn bad. Formally, the function V (g) is convex in growth rates, so that by Jensen’s inequality investors should value risk in growth rates - though typically not in levels .
11 |
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Esophagus - Clinical Vignettes/Case Reports
S3389 Inconsistency With CT Scan and X-ray Esophagogram Findings: A Rare Case of Boerhaave Syndrome
Kumar, Naresh MD 1,; Karishma, Fnu MBBS 2; Aakash, Aakash MD 3; Ginjupalli, Manasa MBBS, MD 4; Al-bustami, Iyad MD, MPH(c)5; Kumar, Rahul MD 6; Kajal, Diksha MD 7; Moparty, Hamsika MD 1; Forlemu, Arnold N. MD, MPH 8; Kumar, Vikash MD 9
Author Information
1 The Brooklyn Hospital Center, Brooklyn, NY;
2 Liaquat University of Medical and Health Science, Brooklyn, NY;
3 Florida State University Cape Coral Hospital, Cape Coral, FL;
4 The Brooklyn Hospital Center, New York, NY;
5 Brooklyn Hospital Center, Houston, TX;
6 Jacobi/ North Central Bronx Hospital Bronx, Bronx, NY;
7 Florida State University College of Medicine, Cape Coral, FL;
8 Brooklyn Hospital Center, Athens, GA;
9 Creighton University School of Medicine, Brooklyn, NY.
Presenter
The American Journal of Gastroenterology 119(10S):p S2260-S2261, October 2024. | DOI: 10.14309/01.ajg.0001042924.85137.b6
Free
Introduction:
Boerhaave syndrome results from the rupture of the esophagus due to a sudden increase in intraesophageal pressure combined with negative intrathoracic pressure (e.g., severe straining or vomiting). Diagnosis is made by CT with contrast and esophagogram. Management relies on rapid recognition and intervention, as a lack of therapeutic interventions can be fatal. We present a rare case of Boerhaave syndrome that has discordant findings between the CT scan and X-ray esophagogram.
Case Description/Methods:
A 29-year-old healthy man presented with nausea, abdominal pain, and multiple episodes of vomiting for one day after binge drinking. On admission, labs showed acute kidney injury (AKI), leukocytosis (17.1), and normal liver function tests (LFTs) with unremarkable vitals but notable crepitus around the neck. A chest X-ray (CXR) showed a band-like opacity paralleling and abutting the mediastinum and subcutaneous emphysema. Persistent intractable epigastric abdominal pain led to a CT of the abdomen, which showed pneumomediastinum, pneumopericardium, and some gas penetrating through the distal esophageal wall, perhaps indicative of Boerhaave syndrome consistent with CT chest findings. A CT of the neck showed extensive subcutaneous emphysema with extension into the superior mediastinum. Interestingly, an esophagogram with Gastrografin showed no evidence of esophageal perforation on the same day. The patient was managed conservatively with broad-spectrum antibiotics and fluids. Due to the lack of evidence of perforation on the esophagogram and improvement in clinical condition, surgery was deferred. A repeat CT of the chest a few days later revealed improvement in pneumomediastinum and pneumopericardium. The patient was started on a diet and discharged on the 10th day with a short course of antibiotics
Discussion:
Boerhaave syndrome is a medical emergency, and diagnosis is usually made by esophagogram and CT of the chest and abdomen with contrast. However, Gastrografin esophagogram can possibly miss small and cervical perforations, which can be better visualized on a barium esophagogram or CT scan of the chest and abdomen. Similarly, in this case, the Gastrografin contrast wasn't able to detect the perforation, possibly due to its small size, despite its distal location. This case underscores the complexity of diagnosing such conditions and suggests the need for research on possible alternative water-soluble contrasts to detect smaller esophageal perforations (see Figures 1 and 2).
Figure 1.:
XR Esophagogram showing no evidence of esophageal perforation.
Figure 2.:
CT chest showing suspected tear of the distal esophagus along the right anterolateral aspect.
© 2024 by The American College of Gastroenterology
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S3389 Inconsistency With CT Scan and X-ray Esophagogram Findings: A Rare Case of Boerhaave Syndrome
Official journal of the American College of Gastroenterology | ACG119(10S):S2260-S2261, October 2024.
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6772 | https://www.cuemath.com/ncert-solutions/bisectors-of-angles-b-and-c-of-a-triangle-abc-intersect-each-other-at-the-point-o-prove-that-angle-boc-equal-to-90-1-2-angle-a/ | Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + 1/2 ∠A
Solution:
Given, bisectors of angles B and C of a triangle ABC intersect each other at the point O.
We have to prove that ∠BOC = 90° + 1/2 ∠A.
BO is the bisector of angle B such that ∠OBC = (∠1)
CO is the bisector of angle C such that ∠OCB = (∠2)
Considering the triangle BOC,
By angle sum property,
∠OBC + ∠BOC + ∠OCB = 180°
∠1 + ∠BOC + ∠2 = 180° -------------------------- (1)
Considering triangle ABC,
By angle sum property,
∠A + ∠B + ∠C = 180°
∠A + 2(∠1) + 2(∠2) = 180°
Dividing by 2 on both sides,
∠A/2 + ∠1 + ∠2 = 180°/2
∠A/2 + ∠1 + ∠2 = 90°
∠1 + ∠2 = 90° - ∠A/2 ------------------------------- (2)
Substituting (2) in (1),
∠BOC + 90° - ∠A/2 = 180°
∠BOC - ∠A/2 = 180° - 90°
∠BOC - ∠A/2 = 90°
Therefore, ∠BOC = 90° + ∠A/2
✦ Try This: In fig., if l∥m, find the value of a and b.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6
NCERT Exemplar Class 9 Maths Exercise 6.4 Sample Problem 3
Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + 1/2 ∠A
Summary:
An angle bisector or the bisector of an angle is a ray that divides an angle into two equal parts. Bisectors of angles B and C of a triangle ABC intersect each other at the point O. It is proven that ∠BOC = 90° + 1/2 ∠A
☛ Related Questions:
If two lines intersect, prove that the vertically opposite angles are equal
Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. Prove that ∠BTC = 1/ . . . .
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding a . . . .
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6773 | https://k12.libretexts.org/Bookshelves/Mathematics/Trigonometry/02%3A_Trigonometric_Ratios/2.05%3A_Radians/2.5.06%3A_Length_of_an_Arc | 2.5.6: Length of an Arc - K12 LibreTexts
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2.5: Radians
Unit 2: Trigonometric Ratios
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Sun, 27 Mar 2022 16:40:45 GMT
2.5.6: Length of an Arc
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Contents
1. Home
2. Bookshelves
3. Mathematics
4. Trigonometry
5. Unit 2: Trigonometric Ratios
6. 2.5: Radians
7. 2.5.6: Length of an Arc
Expand/collapse global location
Trigonometry
Front Matter
Unit 1: Right Triangles and the Pythagorean Theorem
Unit 2: Trigonometric Ratios
Unit 3: Trigonometric Identities
Unit 4: Non-Right Triangle Trigonometry
Unit 5: Polar System and Complex Numbers
Back Matter
2.5.6: Length of an Arc
Last updated Mar 27, 2022
Save as PDF
2.5.5: Angular Velocity
2.5.7: Area of a Sector
Page ID 4239
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Example 2.5.6.1
2. Example 2.5.6.2
3. Example 2.5.6.3
4. Example 2.5.6.4
5. Review
6. Review (Answers)
7. Vocabulary
8. Additional Resources
1. Interactive Element
Radius times the angle in radians.
You have taken your little cousin to the amusement park for the day. While there, she decides she would like a ride on the carousel. After the ride, she excitedly bounces over to you. She is amazed that she went around, but in a way ‘‘didn't go anywhere’’, since she ended up where she started.
‘‘How far did I go when I was halfway around the turn?’’, she asks.
You know that the radius of the carousel is 7 meters. Can you tell your little cousin how far she went in one half of a turn around the ride?
The length of an arc on a circle depends on both the angle of rotation and the radius length of the circle. If you recall from the last lesson, the measure of an angle in radians is defined as the length of the arc cut off by one radius length. What if the radius is 4 cm? Then, the length of the half-circle arc would be π multiplied by the radius length, or 4π cm in length.
Figure 2.5.6.1
This results in a formula that can be used to calculate the length of any arc.
s=rθ,
where s is the length of the arc, r is the radius, and \theta is the measure of the angle in radians.
Solving this equation for θ will give us a formula for finding the radian measure given the arc length and the radius length.
Let's look at some problems involving arc length.
Find the length of the arcs
The free-throw line on an NCAA basketball court is 12 ft wide. In international competition, it is only about 11.81 ft. How much longer is the half circle above the free-throw line on the NCAA court?
Figure 2.5.6.2
Find both arc lengths.
NCAA INTERNATIONAL s 1=rθ s 2=rθ s 1=12 2(π)s 2≈11.81 2(π)s 1=6π s 2≈5.905π
So the answer is approximately 6π−5.905π≈0.095π
Figure 2.5.6.3
This is approximately 0.3 ft, or about 3.6 inches longer.
Two connected gears are rotating. The smaller gear has a radius of 4 inches and the larger gear’s radius is 7 inches. What is the angle through which the larger gear has rotated when the smaller gear has made one complete rotation?
Figure 2.5.6.4
Because the blue gear performs one complete rotation, the length of the arc traveled is:
s=rθ s=4×2π
So, an 8πarc lengthon the larger circle would form an angle as follows:
θ=s r θ=8π 7 θ≈3.6
So the angle is approximately 3.6 radians.
3.6×180π≈206∘
The radius of a standard car tire is 27.94 cm. How far does a car go in one revolution of the tire?
Figure 2.5.6.5
Since the distance traveled by the tire is equal to the distance around the tire, we can use the circumferenceof the tire to answer the question.
s=rθ s=(27.94)(2π)s=175.46
Example 2.5.6.1
Earlier, you were asked can you tell your little cousin how far she went in one half of a turn around the ride.
Solution
Since you now know that you can measure an arc length using s=rθ, you can use this to find a solution to your cousin's question. Since your cousin wants to know how far she went around when she went 1 2 of a rotation, and the radius of the ride is 7 meters, you can calculate her arc length:
s=rθ=7π≈21.98 meters
Example 2.5.6.2
You are trying to push your car after it has broken down. Unfortunately, you aren't very strong, and so the car is just rocking back and forth instead of rolling as you push. If the radius of your car's tire is 14 inches, and the change in the tire's angle is π 2 radians, how far did the tire move?
Solution
Since the distance the tire moved is equal to the length of the arc the tire rolled, you can use the equation s=r\theta to determine how far the tire went:
s=rθ s=(14)(π 2)s=7π s≈21.98 in
Example 2.5.6.3
If an object with a radius of 10 cm spins so that its arc covers 54 cm, what is the change in angle of the object?
Solution
You can again use the equation s=rθ to solve this problem:
s=rθ θ=s r=54 10=5.4
The disk moves 5.4 radians, which is a little less than a complete rotation, since a complete rotation is approximately 6.28 radians.
Example 2.5.6.4
If your DVD has a radius of 4.5 inches, how far does a point on the disk spin if the player turns it \dfrac{\pi}{2} radians?
Solution
Using s=rθ,
s=rθ s=(4.5)(π 2)s=2.25π≈7.065
A point on the disk turns 7.065 inches.
Review
The radius of a carousel is 8 meters. Use this information to answer questions 1-3.
You are half way around the carousel. How far did you travel?
You are all the way around the carousel. How far did you travel?
You have now traveled all the way around the carousel twice. How far did you travel?
A pizza has a radius of 10 in. Use this information to answer questions 4-6.
A slice is removed. The length of the crust of the missing slice is 3 in. What is the central angle of the missing slice?
You eat three pieces with a central angle of 4π 5. What is the length of the crust you ate?
A large pizza has a radius of 12 in. What is the length of the crust of half of the large pizza?
The diameter of a tire is 35 in. Use this information to answer questions 7-10.
What is the length around the whole tire?
The tire travels one mile (5280 ft). How many revolutions did the tire make?
You roll the tire so it rotates 7π radians. How far did it move?
The tire travels half a mile. How many radians did the tire rotate?
Consider a standard 12 hour clock like the one below with a radius of 5 inches. Use this to answer questions 11-15.
Figure 2.5.6.6
What is the length of the arc between the 3 and the 7?
What is the length of the arc between the 3 and the 2?
It is 12:30. What is the length of the arc between the minute and hour hands?
It is 7:20. What is the length of the arc between the minute and hour hands?
It is 1:25. What is the length of the arc between the minute and hour hands?
Review (Answers)
To see the Review, open this PDF file and look for section 2.5.
Vocabulary
| Term | Definition |
--- |
| Arc | An arc is a section of the circumference of a circle. |
Additional Resources
Interactive Element
Video: Arc Length, Area of Sector, Linear Speed, and Angular Speed - Example 5
Practice: Length of an Arc
This page titled 2.5.6: Length of an Arc is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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6776 | https://www.ck12.org/section/transformations-of-functions/ | Elementary Math
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HomeMathematicsTransformations of functions
Transformations of functions
Difficulty Level: Basic | Created by: CK-12
Last Modified: Dec 29, 2014
Learning objectives
Graph functions by shifting basic functions.
Graph functions by stretching or compressing basic functions.
Graph functions that are reflections of basic functions.
Combine transformations in order to graph more complicated functions.
Introduction
In the previous lesson we examined the key characteristics of different families of functions. In this lesson we will focus on the relationships among functions within a given family. In particular, we will analyze how a function is related to the parent function in the family (e.g., y = x2, y = x3) so that we can efficiently identify the key characteristics of a particular function and sketch its graph. We will consider several kinds of relationships between functions and their parents: shifts, stretches or compressions, and reflections. We will begin with shifts.
Vertical shifts
Consider the graphs of three functions shown below: y = x2, y = x2 - 3, and y = x2 + 4
At first glance, it may seem that the graphs have different widths. For example, it might look like y = x2 + 4, the uppermost of the three parabolas, is thinner than the other two parabolas. However, this is not the case. The parabolas are congruent. That is, if we shifted the graph of y = x2 up four units, we would have the exact same graph as y = x2 + 4. Similarly, if we shifted y = x2 down three units, we would have the graph of y = x2 - 3.
Numerically, we can determine that for any value of x, the value of y = x2 + 4 is 4 more than the value of y = x2, and the value of y = x2 - 3 is 3 less than the value of y = x2. The table below shows several function values:
| x | y = x2 | y = x2 + 4 | y = x2 - 3 |
--- --- |
| 1 | 5 |
| 0 | 0 | 4 |
| 1 | 1 | 5 |
| 2 | 4 | 8 | 1 |
| 5 | 25 | 29 | 22 |
| 10 | 100 | 104 | 97 |
We can formally express this relationship using function notation: for any function f(x), the function g(x) = f(x) + c has a graph that is the same as f(x), shifted c units vertically. If c is positive, the graph is shifted up. If c is negative, the graph is shifted down. We can then use this relationship to graph functions by hand.
Example 1. Use the graph of y = x2 to graph the function y = x2 - 5.
Solution:
The graph of y = x2 is a parabola with vertex at (0, 0). The graph of y = x2- 5 is therefore a parabola with vertex (0, -5). To quickly sketch y = x2 - 5, you can sketch several points on y = x2, and then shift them down 5 units.
In sum, when we add or subtract a constant from the equation of a function, the graph of the new function is a vertical shift from the original function. We can use similar reasoning to graph functions that are horizontal shifts of parent graphs.
Horizontal shifts
Consider the functions f(x) = |x| and g(x) = |x - 3|. From the examples of vertical shifts above, you might think that the graph of g(x) is the graph of f(x), shifted 3 units to the left. However, this is not the case. The graph of g(x) is the graph of f(x), shifted 3 units to the right. The direction of the shift makes sense if we look at specific function values.
| x | g(x) = abs(x - 3) |
--- |
| 0 | 3 |
| 1 | 2 |
| 2 | 1 |
| 3 | 0 |
| 4 | 1 |
| 5 | 2 |
| 6 | 3 |
From the table we can see that the vertex of the graph is the point (3, 0). The function values on either side of x = 3 are symmetric, and greater than 0.
We can formalize horizontal shifts in the same way we formalized vertical shifts: given a function f(x), and a constant a > 0, the function g(x) = f(x - a) represents a horizontal shift a units to the right from f(x). The function h(x) = f(x + a) represents a horizontal shift a units to the left.
Notice that in the case of a horizontal shift, the constant is part of the “input” of the function. For example, y = (x + 7)2 represents a horizontal shift of the original function y=x2, while y = x2 + 7 represents a vertical shift. It is important to keep this distinction in mind as you interpret equations.
Example 2: What is the relationship between f(x) = x2 and g(x) = (x - 2)2 - 6?
Solution: The graph of g(x) is the graph of f(x), shifted 2 units to the right, and down 6 units.
In sum, adding or subtracting constants within the equation of a function creates a vertical or horizontal shift in the graph. When the constant is “inside” the parentheses, the shift will be horizontal. Now we will consider the results of multiplying a function or the input of a function (x) by a coefficient.
Stretching and compressing graphs
If we multiply a function by a coefficient, the graph of the function will be stretched or compressed. Consider the graph of three parabolas below.
The widest parabola is the function y = (1/5) x2. The middle parabola is the function y = x2. The thinnest parabola is the function y = 10x2. From these graphs, we can see that multiplying by different coefficients has different effects on the graph of y = x2. The graph of y = 10x2 represents a vertical stretch of the graph of y = x2. Every function value of y = 10x2 is 10 times the equivalent function value of y = x2. For example, y = x2 contains the point (2, 4), while y = 10x2 contains the point (2, 40).
The coefficient of 1/5 has an opposite effect. We describe this as a vertical compression. Every function value of y = (1/5) x2 is 1/5 the value of y = x2. For example, y = x2 contains the point (10, 100), while y = (1/2) x2 contains the point (10, 20).
: We can formalize vertical stretching and compressing as follows:
A function g(x) = cf(x) represents a vertical stretch if c > 1.
A function h(x) = cf(x) represents a vertical compression if 0 < c < 1.
Notice that the graph of y = (1/5) x2 is wider than the other two parabolas. While 1/5 has the effect of compressing the graph of y = x2 vertically, we can also describe the graph as having a horizontal stretch. Formally, we characterize a transformation as a horizontal stretch or compression if the function is of the form g(x) = f(cx). That is, instead of multiplying the function by a coefficient, the variable x is multiplied by c. However, if we analyze a quadratic equation, we can see why a vertical compression looks like a horizontal stretch. (The parabola gets shorter/wider, or taller/thinner.)
Consider for example the function y = (3x)2. We have multiplied x by 3. This should affect the graph horizontally. However, if we simplify the equation, we get y = 9x2. Therefore the graph if this parabola will be taller/thinner than y = x2. Multiplying x by a number greater than 1 creates a horizontal compression, which looks like a vertical stretch.
Now consider the function y = ((1/2)x)2. If we simplify this equation, we get y = (1/4) x2. Therefore multiplying x by a number between 0 and 1 creates a horizontal stretch, which looks like a vertical compression. That is, the parabola will be shorter/wider.
We can see similar behavior in other functions. For example, consider the functions y=x−−√, y=4x−−√, and y=2x−−√, shown in the graph below. If we simplify the second equation, we get y=4x−−√=4–√x−−√=2x−−√. Therefore the second and third equations are actually the same function, and we can describe the function as a vertical stretch or as a horizontal compression.
Given a function f(x), we can formalize compressing and stretching the graph of f(x) as follows:
A function g(x) represents a vertical stretch of f(x) if g(x) = cf(x) and c > 1.
A function g(x) represents a vertical compression of f(x) if g(x) = cf(x) and 0 < c < 1.
A function h(x) represents a horizontal compression of f(x) if h(x) = f(cx) and c > 1.
A function h(x) represents a horizontal stretch of f(x) if h(x) = f(cx) 0 < c < 1.
Notice that a vertical compression or a horizontal stretch occurs when the coefficient is a number between 0 and 1. Now we will consider the effects of a coefficient less than 0.
Reflecting graphs over the y-axis and x-axis.
Consider the graphs of the functions y = x2 and y = -x2, shown below.
The graph of y = -x2 represents a reflection of y = x2, over the x-axis. That is, every function value of y = -x2 is the negative of a function value of y = x2. In general, g(x) = -f(x) has a graph that is the graph of f(x), reflected over the x-axis.
Example 3: Sketch a graph of y = x3 and y = -x3 on the same axes.
Solution:
At first the two functions might look like two parabolas. If you graph by hand, or if you set your calculator to sequential mode (and not simultaneous), you can see that the graph of y = -x3 is in fact a reflection of y = x3 over the x-axis.
However, if you look at the graph, you can see that it is a reflection over the y-axis as well. This is the case because in order to obtain a reflection over the y-axis, we negate x. That is, h(x) = f(-x) is a reflection of f(x) over the y-axis. For the function y = x3, h(x) = (-x)3 = (-x) (-x) (-x) = -x3. This is the same function as the one we have already graphed.
It is important to note that this is a special case. The graph of y = x2 is also a special case. If we want to reflect y = x2 over the y-axis, we will just get the same graph! This can be explained algebraically: y = (-x)2 = (-x) (-x) = x2.
: Now let’s consider a different function.
Example 4: Graph the functions y=x−−√ and y=−x−−−√.
Solution:
The equation y=−x−−−√ might look confusing because of the -x under the square root. It is important to keep in mind that - x means the opposite of x. Therefore the domain of this function is restricted to values ≤ 0. For example, if x = - 4, y=−(−4)−−−−−√=4–√=2. It is this domain, which includes all real numbers not in the domain of y=x−−√ plus zero, that gives us a graph that is a reflection over the y-axis.
In sum, a graph represents a reflection over the x-axis if the function has been negated (i.e. the y has been negated if we think of y = f(x)). The graph represents a reflection over the y-axis if the variable x has been negated. Now that we have considered shifts, stretches/compressions, and reflections, we will look at graphs that combine these transformations.
Combining transformations
Consider the equation y = 2(x - 3)2 + 1. We can compare the graph of this function to the graph of the parent y = x2: the graph represents a vertical stretch by a factor of 2, a horizontal shift 3 units to the right, and a vertical shift of 1 unit.
We can use this relationship to graph the function y = 2(x - 3)2 + 1. You can start by sketching y = x2 or y = 2x2. Then you can shift the graph 3 units to the right, and up 1 unit.
It is slightly more complicated to graph functions that represent reflections of the parent graphs. Next we will examine reflections combined with other transformations.
Example 5. Graph each function using your knowledge of the function y = |x| and your knowledge of transformations.
: a. f(x) = -|x| + 3
: b. g(x) = |-x + 3|
Solution:
a. f(x) = -|x| + 3
The parent graph of this function is the graph of y = |x|, reflected over the x-axis, and shifted up 3 units. The question is: which transformation do you perform first?
We can answer this question if we consider a few key function values. The table below shows several function values for f(x) = -|x| + 3:
| x | f(x) = - abs(x) + 3 |
--- |
| -3 | -abs(-3) + 3 = -(+3) + 3 = -3 + 3 = 0 |
| -2 | -abs(-2) + 3 = -(+2) + 3 = -2 + 3 = 1 |
| -1 | 2 |
| 0 | 3 |
| 1 | 2 |
| 3 | 0 |
From the function values in the table we can see that the function increases until a vertex at (0, 3), and then it decreases again. This tells us that we can obtain the graph if we first reflect y = |x| over the x-axis (turn the “v” upside down), and then shift the graph up 3 units.
We can also justify this ordering of the transformations of we think about the order of operations. To find any function value we take an x value, find its absolute value, find the negative of that number, and then add 3. This is the same as the order of the transformation: reflection comes before shifting up.
b. g(x) = |-x + 3|
This function represents a horizontal shift of y = |x|, and a reflection over the x-axis. Before graphing, consider a few function values:
| x | g(x) = abs(−x+3) |
--- |
| abs(-(-3) + 3) = abs(3 + 3) = abs(6) = 6 |
| abs(-(-2) + 3 ) = (2 + 3) = abs(5) = 5 |
| 4 |
| 0 | 3 |
| 1 | 2 |
| 2 | 1 |
| 3 | 0 |
| 4 | 1 |
| 5 | 2 |
| 6 | 3 |
From the values in the table, we can see that the vertex of the graph is at (3, 0). The graph is shown below.
The graph looks the same as the graph of y = |x - 3|. This is the case because y = |-x + 3| = |-(x - 3)|, and because |- a| = |a| for all values of a, then |-(x - 3)| = |x - 3|. So the original function is equal to |x - 3|. We can still think of this graph as a reflection: if we reflect y = |x| over the x-axis, the graph remains the same, as it is symmetric over the x-axis. Then we shift the graph 3 units to the right. What is important to note here is that in order to “read” the equation as a horizontal shift, the entire expression inside the function (in this case, inside the absolute value) must be negated.
Lesson Summary
In this lesson we have explored transformations of functions. We have related different functions to their parent graphs, in terms of vertical and horizontal shifts, vertical and horizontal stretches and compressions, and reflections over the y-axis and the x-axis. By analyzing the functions, we can relate the graphs of functions to the graphs of their parent functions, and we can use these relationships to sketch graphs. The issue of “inside” and “outside”, raised throughout the lesson, and seen specifically in example 5b, will be examined more closely in the next lesson.
Points to Consider
How can you analyze an equation and determine if the graph represents a vertical or horizontal shift?
How can you analyze an equation and determine if the graph represents a stretch or compression?
How is it that different equations might yield the same graphs?
Review Questions
How is the graph of f(x) = x3 + 5 related to the graph of g(x) = x3?
How is the graph of f(x) = (x - 3)2 related to the graph of f(x) = (x + 3)2? Sketch these functions together.
Write an equation that represents a vertical stretch of the graph of y=x−−√ by a factor of 5.
Describe the function f(x) = (5x)2 as both a stretch and a compression of y = x2.
Graph the function g(x)=|x−1| and −g(x) on the same grid. What is the equation of −g(x)?
Describe the relationship between the graphs of f(x) = 4(x + 8)3 - 3 and g(x) = x3.
Graph the functions f(x)=2x and h(x)=2x+1 on the same graph. State the asymptotes of both functions.
Graph the functions f(x) = 3x + 1 and f(-x) on the same grid. What is the equation of f(-x)?
Explain how a function can be its own reflection over an axis.
Give an example of two equations that can be represented by the same graph.
Notes/Highlights
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Difficulty Level
Basic
Tags
range, Function, domain, vertical line test, relation, compression, composition of functions, stretch, vertical compression, asymptote, horizontal stretch, horizontal compression, vertical stretch, limit, end behavior, extrema, minimize, maximize, average rate of change, CK.MAT.ENG.SE.1.Math-Analysis.1, (17 more)
Subjects
mathematics
Grades
11, 12
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Last Modified
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6777 | https://en.wikipedia.org/wiki/Interquartile_range | Jump to content
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Contents
(Top)
1 Use
2 Algorithm
3 Examples
3.1 Data set in a table
3.2 Data set in a plain-text box plot
4 Distributions
4.1 Interquartile range test for normality of distribution
5 Outliers
6 See also
7 References
8 External links
Interquartile range
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From Wikipedia, the free encyclopedia
Measure of statistical dispersion
"IQR" redirects here. For other uses, see IQR (disambiguation).
In descriptive statistics, the interquartile range (IQR) is a measure of statistical dispersion, which is the spread of the data. The IQR may also be called the midspread, middle 50%, fourth spread, or H‑spread. It is defined as the difference between the 75th and 25th percentiles of the data. To calculate the IQR, the data set is divided into quartiles, or four rank-ordered even parts via linear interpolation. These quartiles are denoted by Q1 (also called the lower quartile), Q2 (the median), and Q3 (also called the upper quartile). The lower quartile corresponds with the 25th percentile and the upper quartile corresponds with the 75th percentile, so IQR = Q3 − Q1.
The IQR is an example of a trimmed estimator, defined as the 25% trimmed range, which enhances the accuracy of dataset statistics by dropping lower contribution, outlying points. It is also used as a robust measure of scale It can be clearly visualized by the box on a box plot.
Use
[edit]
Unlike total range, the interquartile range has a breakdown point of 25% and is thus often preferred to the total range.
The IQR is used to build box plots, simple graphical representations of a probability distribution.
The IQR is used in businesses as a marker for their income rates.
For a symmetric distribution (where the median equals the midhinge, the average of the first and third quartiles), half the IQR equals the median absolute deviation (MAD).
The median is the corresponding measure of central tendency.
The IQR can be used to identify outliers (see below). The IQR also may indicate the skewness of the dataset.
The quartile deviation or semi-interquartile range is defined as half the IQR.
Algorithm
[edit]
The IQR of a set of values is calculated as the difference between the upper and lower quartiles, Q3 and Q1. Each quartile is a median calculated as follows.
Given an even 2n or odd 2n+1 number of values
: first quartile Q1 = median of the n smallest values
: third quartile Q3 = median of the n largest values
The second quartile Q2 is the same as the ordinary median.
Examples
[edit]
Data set in a table
[edit]
The following table has 13 rows, and follows the rules for the odd number of entries.
| i | x[i] | Median | Quartile |
--- --- |
| 1 | 7 | Q2=87 (median of whole table) | Q1=31 (median of lower half, from row 1 to 6) |
| 2 | 7 |
| 3 | 31 |
| 4 | 31 |
| 5 | 47 |
| 6 | 75 |
| 7 | 87 | |
| 8 | 115 | Q3=119 (median of upper half, from row 8 to 13) |
| 9 | 116 |
| 10 | 119 |
| 11 | 119 |
| 12 | 155 |
| 13 | 177 |
For the data in this table the interquartile range is IQR = Q3 − Q1 = 119 - 31 = 88.
Data set in a plain-text box plot
[edit]
+−−−−−+−+ |−−−−−−−−−−−| | |−−−−−−−−−−−| +−−−−−+−+ +−−−+−−−+−−−+−−−+−−−+−−−+−−−+−−−+−−−+−−−+−−−+−−−+ Number line 0 1 2 3 4 5 6 7 8 9 10 11 12
For the data set in this box plot:
Lower (first) quartile Q1 = 7
Median (second quartile) Q2 = 8.5
Upper (third) quartile Q3 = 9
Interquartile range, IQR = Q3 - Q1 = 2
Lower 1.5IQR whisker = Q1 - 1.5 IQR = 7 - 3 = 4. (If there is no data point at 4, then the lowest point greater than 4.)
Upper 1.5IQR whisker = Q3 + 1.5 IQR = 9 + 3 = 12. (If there is no data point at 12, then the highest point less than 12.)
Pattern of latter two bullet points: If there are no data points at the true quartiles, use data points slightly "inland" (closer to the median) from the actual quartiles.
This means the 1.5IQR whiskers can be uneven in lengths. The median, minimum, maximum, and the first and third quartile constitute the Five-number summary.
Distributions
[edit]
The interquartile range of a continuous distribution can be calculated by integrating the probability density function (which yields the cumulative distribution function—any other means of calculating the CDF will also work). The lower quartile, Q1, is a number such that integral of the PDF from -∞ to Q1 equals 0.25, while the upper quartile, Q3, is such a number that the integral from -∞ to Q3 equals 0.75; in terms of the CDF, the quartiles can be defined as follows:
where CDF−1 is the quantile function.
The interquartile range and median of some common distributions are shown below
| Distribution | Median | IQR |
---
| Normal | μ | 2 Φ−1(0.75)σ ≈ 1.349σ ≈ (27/20)σ |
| Laplace | μ | 2b ln(2) ≈ 1.386b |
| Cauchy | μ | 2γ |
Interquartile range test for normality of distribution
[edit]
The IQR, mean, and standard deviation of a population P can be used in a simple test of whether or not P is normally distributed, or Gaussian. If P is normally distributed, then the standard score of the first quartile, z1, is −0.67, and the standard score of the third quartile, z3, is +0.67. Given mean = and standard deviation = σ for P, if P is normally distributed, the first quartile
and the third quartile
If the actual values of the first or third quartiles differ substantially[clarification needed] from the calculated values, P is not normally distributed. However, a normal distribution can be trivially perturbed to maintain its Q1 and Q2 std. scores at 0.67 and −0.67 and not be normally distributed (so the above test would produce a false positive). A better test of normality, such as Q–Q plot would be indicated here.
Outliers
[edit]
The interquartile range is often used to find outliers in data. Outliers here are defined as observations that fall below Q1 − 1.5 IQR or above Q3 + 1.5 IQR. In a boxplot, the highest and lowest occurring value within this limit are indicated by whiskers of the box (frequently with an additional bar at the end of the whisker) and any outliers as individual points.
See also
[edit]
Interdecile range – Statistical measure
Midhinge
Probable error – Measure of statistical dispersion
Robust measures of scale – Statistical indicators of the deviation of a sample
References
[edit]
^ a b c d e Dekking, Frederik Michel; Kraaikamp, Cornelis; Lopuhaä, Hen Paul; Meester, Ludolf Erwin (2005). A Modern Introduction to Probability and Statistics. Springer Texts in Statistics. London: Springer London. doi:10.1007/1-84628-168-7. ISBN 978-1-85233-896-1.
^ Upton, Graham; Cook, Ian (1996). Understanding Statistics. Oxford University Press. p. 55. ISBN 0-19-914391-9.
^ Zwillinger, D., Kokoska, S. (2000) CRC Standard Probability and Statistics Tables and Formulae, CRC Press. ISBN 1-58488-059-7 page 18.
^ Ross, Sheldon (2010). Introductory Statistics. Burlington, MA: Elsevier. pp. 103–104. ISBN 978-0-12-374388-6.
^ a b Kaltenbach, Hans-Michael (2012). A concise guide to statistics. Heidelberg: Springer. ISBN 978-3-642-23502-3. OCLC 763157853.
^ Rousseeuw, Peter J.; Croux, Christophe (1992). Y. Dodge (ed.). "Explicit Scale Estimators with High Breakdown Point" (PDF). L1-Statistical Analysis and Related Methods. Amsterdam: North-Holland. pp. 77–92.
^ Yule, G. Udny (1911). An Introduction to the Theory of Statistics. Charles Griffin and Company. pp. 147–148.
^ a b c Bertil., Westergren (1988). Beta [beta] mathematics handbook : concepts, theorems, methods, algorithms, formulas, graphs, tables. Studentlitteratur. p. 348. ISBN 9144250517. OCLC 18454776.
^ Dekking, Kraaikamp, Lopuhaä & Meester, pp. 235–237
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6778 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10762924/ | Carrier frequency estimation of pathogenic variants of autosomal recessive and X-linked recessive mendelian disorders using exome sequencing data in 1,642 Thais - PMC
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BMC Med Genomics
. 2024 Jan 2;17:9. doi: 10.1186/s12920-023-01771-w
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Carrier frequency estimation of pathogenic variants of autosomal recessive and X-linked recessive mendelian disorders using exome sequencing data in 1,642 Thais
Wanna Chetruengchai
Wanna Chetruengchai
1 Excellence Center for Genomics and Precision Medicine, King Chulalongkorn Memorial Hospital, the Thai Red Cross Society, Bangkok, 10330 Thailand
2 Center of Excellence for Medical Genomics, Medical Genomics Cluster, Department of Pediatrics, Faculty of Medicine, Chulalongkorn University, Bangkok, 10330 Thailand
Find articles by Wanna Chetruengchai
1,2, Prasit Phowthongkum
Prasit Phowthongkum
1 Excellence Center for Genomics and Precision Medicine, King Chulalongkorn Memorial Hospital, the Thai Red Cross Society, Bangkok, 10330 Thailand
3 Division of Medical Genetics and Genomics, Department of Medicine, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand
Find articles by Prasit Phowthongkum
1,3,✉, Vorasuk Shotelersuk
Vorasuk Shotelersuk
1 Excellence Center for Genomics and Precision Medicine, King Chulalongkorn Memorial Hospital, the Thai Red Cross Society, Bangkok, 10330 Thailand
2 Center of Excellence for Medical Genomics, Medical Genomics Cluster, Department of Pediatrics, Faculty of Medicine, Chulalongkorn University, Bangkok, 10330 Thailand
Find articles by Vorasuk Shotelersuk
1,2
Author information
Article notes
Copyright and License information
1 Excellence Center for Genomics and Precision Medicine, King Chulalongkorn Memorial Hospital, the Thai Red Cross Society, Bangkok, 10330 Thailand
2 Center of Excellence for Medical Genomics, Medical Genomics Cluster, Department of Pediatrics, Faculty of Medicine, Chulalongkorn University, Bangkok, 10330 Thailand
3 Division of Medical Genetics and Genomics, Department of Medicine, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand
✉
Corresponding author.
Received 2023 Aug 6; Accepted 2023 Dec 11; Collection date 2024.
© The Author(s) 2023
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated in a credit line to the data.
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PMCID: PMC10762924 PMID: 38167091
Abstract
Background
People with autosomal recessive disorders often were born without awareness of the carrier status of their parents. The American College of Medical Genetics and Genomics (ACMG) recommends screening 113 genes known to cause autosomal recessive and X-linked conditions in couples seeking to learn about their risk of having children with these disorders to have an appropriate reproductive plan.
Methods
We analyzed the exome sequencing data of 1,642 unrelated Thai individuals to identify the pathogenic variant (PV) frequencies in genes recommended by ACMG.
Results
In the 113 ACMG-recommended genes, 165 PV and likely PVs in 60 genes of 559 exomes (34%, 559/1642) were identified. The carrier rate was increased to 39% when glucose-6-phosphate dehydrogenase (G6PD) was added. The carrier rate was still as high as 14.7% when thalassemia and hemoglobinopathies were excluded. In addition to thalassemia, hemoglobinopathies, and G6PD deficiency, carrier frequencies of > 1% were found for Gaucher disease, primary hyperoxaluria, Pendred syndrome, and Wilson disease. Nearly 2% of the couples were at risk of having offsprings with the tested autosomal recessive conditions.
Conclusions
Based on the study samples, the expanded carrier screening, which specifically targeted common autosomal recessive conditions in Thai individuals, will benefit clinical outcomes, regarding preconception/prenatal genetic carrier screening.
Supplementary Information
The online version contains supplementary material available at 10.1186/s12920-023-01771-w.
Keywords: Carrier screening, WES, Thais
Background
Although each Mendelian disease is generally uncommon, with > 5,000 known heritable disorders, they are accounted for ~ 20% of infant mortality, ~ 18% of pediatric hospitalizations, and substantial numbers in adults in the United States . At least 2,000 autosomal recessive diseases and 500 X-linked diseases have been identified. The preconception carrier screening for these disorders and proper reproductive option counseling have been shown to be cost-effective measures that reduce the burden of Mendelian diseases [4–7]. The studies of the carrier rate of these disorders are highly varied with the number of tested genes and the genes included in the studies among different populations with no consensus agreement [8–11]. Until recently, the American College of Medical Genetics and Genomics (ACMG) released a set of 113 genes, both autosomal recessive and X-linked conditions, proposed as a standard list of genes for carrier screening . Because X-linked glucose-6-phosphate dehydrogenase (G6PD) deficiency is the most common genetic disorder in Thailand and Southeast Asia , we added G6PD into the list and determined the carrier frequency and variant distribution of 114 recessive genes in the Thai population using the exomes of 1,642 unrelated Thais.
Methods
Study samples
In our study, we collected a sample of 1642 exomes from unrelated healthy Thai individuals, consisting of 811 men and 831 women. Prior to participation, informed consent was obtained from each individual. The study was conducted in accordance with ethical guidelines and was approved by the Institutional Review Board of the Faculty of Medicine, Chulalongkorn University (IRB No. 264/62). Our research with human subjects conforms with the Declaration of Helsinki (1964). The data analyzed in our study were part of the Thai Reference Exome database, although not all data from the database were used for our analysis . It is important to note that for the patient cohort, we specifically excluded variants in genes known to be responsible for the patients' diagnosed conditions. This ensured that our analysis focused on carriers of pathogenic variants unrelated to the diagnosed diseases in the patient cohort.
Sequencing and bioinformatics analysis
Genomic DNA was extracted from peripheral blood leukocytes. Exomes were captured using either a TruSeq® Exome Kit (Illumina, San Diego, CA) on the NextSeq 500 system or SureSelect Human All Exons (Agilent Inc., Santa Clara, CA) on the Hiseq 4000 and Novaseq 6000 Systems. The reagents are according to the manufacturer's standard protocol. Sequences were aligned to the human reference genome (GRCh37) using the Burrows–Wheeler Aligner package version 0.7.15 . Variant calling was performed using the Genome Analysis Tool Kit (GATK Best Practice V3.7; Broad Institute), which is called by HaplotypeCaller . ANNOVAR was used to annotate genetic variants. The variant interpretation was limited to the 113 ACMG-recommended genes and G6PD. We included variants listed as PV and LPV according to VarSome's ACMG classification with a final manual curation by the authors.
Statistical analysis
The frequency and 95% confidence interval were calculated using the proportion of carriers in total individuals tested (n/1642) and the Wilson method, respectively.
Results
Using 1,642 exome sequencing data of unaffected parents whose children had rare diseases, we first excluded pathogenic and likely pathogenic variants (PV and LPV, respectively) found in these exomes, which caused the diseases in their children. We excluded 13 PV and LPV in 18 exomes (1%, 18/1642) (S1 Table).
Of the 113 ACMG-recommended genes, 165 PV and LPV in 60 genes of 559 exomes were found (559/1642: 34%). For G6PD, 7 PV and LPV of 127 exomes (female) were identified. In the analysis of a total of 114 genes, 172 PV and LPV in 61 genes of 640 exomes (640/1642: 39%) were identified (Table1). Although we excluded HBA1, HBA2, and HBB genes causing thalassemia and hemoglobinopathies, the carrier rate of at least one screened disorder was still as high as 14.7% (241/1642) in our cohort. Of the samples, 7.4% (121/1642) carried more than one PV and LPV (Fig.1 and S2 Table).
Table 1.
Carrier frequencies of genetic diseases identified in 114 genes of 1,642 unrelated Thai individuals (Bold text: carrier frequency ≥ 1/250)
| Disease group (Number of individuals) | Gene | Phenotype disease | Mode of inheritance | Number of Individual | 1 in | Carrier frequency (%) (Wilson 95% Interval) |
:--- :--- :---
| 1.Hemolytic disorders with positive selection from malaria (518) | HBB | β-thalassemia | AR | 321 | 6 | 19.55 (17.7–21.54) |
| G6PD | Glucose-6-phosphate dehydrogenase deficiency | X-linked | 127 | 13a | 7.73 (6.5–9.1)a |
| HBA2 | ɑ-Thalassemia1 | AR | 65 | 26 | 3.96 (3.12–5.01) |
| HBA1 | ɑ-Thalassemia2 | AR | 5 | 329 | 0.3 (0.13–0.7) |
| 2.Inherited metabolic disorders (142) | GBA | Gaucher disease, type I | AR | 27 | 61 | 1.64 (1.13–2.38) |
| AGXT | Hyperoxaluria, primary type I, II | AR | 18 | 92 | 1.1 (0.7–1.73) |
| ATP7B | Wilson disease | AR | 17 | 97 | 1.04 (0.65–1.66) |
| PAH | Phenylketonuria | AR | 13 | 127 | 0.79 (0.46–1.35) |
| CYP27A1 | Cerebrotendinous xanthomatosis | AR | 7 | 235 | 0.43 (0.21–0.88) |
| ABCD1 | Adrenoleukodystrophy (ALD) | X-linked | 5 | 329 a | 0.3 (0.13–0.71)a |
| GAA | Glycogen storage disease, type II (Pompe disease) | AR | 4 | 411 | 0.24 (0.09–0.62) |
| PMM2 | Carbohydrate-deficient glycoprotein syndrome type Ia | AR | 4 | 411 | 0.24 (0.09–0.62) |
| TYR | Oculocutaneous albinism type 1A and 1B | AR | 4 | 411 | 0.24 (0.09–0.62) |
| ALPL | Hypophosphatasia, adult, childhood and infantile | AR | 3 | 548 | 0.18 (0.06–0.53) |
| FMO3 | Trimethylaminuria | AR | 3 | 548 | 0.18 (0.06–0.53) |
| MMACHC | Methylmalonic aciduria with homocystinuria cblC type | AR | 3 | 548 | 0.18 (0.06–0.53) |
| ARSA | Metachromatic leukodystrophy | AR | 2 | 821 | 0.12 (0.03–0.44) |
| CBS | Homocystinuria, B6 responsive and nonresponsive | AR | 2 | 821 | 0.12 (0.03–0.44) |
| DHCR7 | Smith–Lemli–Opitz syndrome | AR | 2 | 821 | 0.12 (0.03–0.44) |
| DLD | Dihydrolipoamide dehydrogenase deficiency | AR | 2 | 821 | 0.12 (0.03–0.44) |
| FAH | Tyrosinemia type I | AR | 2 | 821 | 0.12 (0.03–0.44) |
| FKRP | Muscular dystrophy–dystroglycanopathy, type A, 5, type B, 5 | AR | 2 | 821 | 0.12 (0.03–0.44) |
| FKTN | Cardiomyopathy;dilated;1X, Walker–Warburg congenital muscular dystrophy | AR | 2 | 821 | 0.12 (0.03–0.44) |
| GALT | Galactosemia | AR | 2 | 821 | 0.12 (0.03–0.44) |
| GBE1 | Glycogen storage disease;type IV, GBE1-related disorders | AR | 2 | 821 | 0.12 (0.03–0.44) |
| MMUT | Methylmalonic aciduria–methylmalonyl–CoA mutase deficiency | AR | 2 | 821 | 0.12 (0.03–0.44) |
| POLG | Mitochondrial DNA depletion syndrome 4A,4B | AR | 2 | 821 | 0.12 (0.03–0.44) |
| ACADM | Medium-chain acyl-coenzyme A dehydrogenase deficiency | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| ACAT1 | ɑ-Methylacetoacetic aciduria | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| ALDOB | Hereditary fructosuria | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| BTD | Biotinidase deficiency | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| CPT2 | Carnitine palmitoyltransferase II deficiency, infantile, lethal neonatal | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| GNPTAB | Mucolipidosis type II alpha/beta | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| HEXA | Tay–Sachs disease | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| HPS1 | Hermansky Pudlak S. 1 | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| IDUA | Mucopolysaccharidosis, Ih (Hurler S), Ih/s (Hurler–Scheie S) | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| MCCC2 | 3-methylcrotonyl CoA carboxylase 2 deficiency | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| OTC | Ornithine transcarbamylase deficiency | X-linked | 1 | 1642 a | 0.06 (0.01–0.34)a |
| RARS2 | Pontocerebellar hypoplasia type 6 | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| 3.Other (123) | SLC26A4 | Deafness autosomal recessive 4, Pendred syndrome | AR | 18 | 92 | 1.1 (0.7–1.73) |
| CFTR | Cystic fibrosis | AR | 16 | 103 | 0.97 (0.6–1.58) |
| USH2A | Usher syndrome, type 2A | AR | 16 | 103 | 0.97 (0.6–1.57) |
| NEB | Nemaline myopathy 2 | AR | 12 | 137 | 0.73 (0.42–1.27) |
| GJB2 | Nonsyndromic hearing loss recessive 1A, Nonsyndromic hearing loss dominant 3A | AR | 9 | 183 | 0.55 (0.29–1.04) |
| MCPH1 | Primary microcephaly 1, recessive | AR | 8 | 206 | 0.49 (0.25–0.96) |
| CEP290 | Joubert syndrome 5, Leber congenital amaurosis 10 | AR | 7 | 235 | 0.43 (0.21–0.88) |
| CLCN1 | Congenital myotonia, autosomal recessive form | AR | 7 | 235 | 0.43 (0.21–0.88) |
| PKHD1 | Autosomal recessive polycystic kidney disease | AR | 7 | 235 | 0.43 (0.21–0.88) |
| AIRE | Autoimmune polyendocrinopathy syndrome type I | AR | 5 | 329 | 0.3 (0.13–0.7) |
| BBS2 | Bardet–Biedl syndrome 2, Retinitis pigmentosa 74 | AR | 3 | 548 | 0.18 (0.06–0.53) |
| EVC2 | Chondroectodermal dysplasia | AR | 3 | 548 | 0.18 (0.06–0.53) |
| ABCC8 | Diabetes mellitus, permanent neonatal 3 | AR | 2 | 821 | 0.12 (0.03–0.44) |
| CC2D2A | Joubert syndrome 9, Meckel syndrome 6 | AR | 2 | 821 | 0.12 (0.03–0.44) |
| DYNC2H1 | thoracic dysplasia 3 with or without polydactyly | AR | 2 | 821 | 0.12 (0.03–0.44) |
| ERCC2 | Cerebrooculofacioskeletal syndrome 2, Trichothiodystrophy 1;photosensitive | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| FANCC | Fanconi anemia, complementation group C | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| NPHS1 | Finnish congenital nephrotic syndrome | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| PCDH15 | Deafness, autosomal recessive 23,Usher syndrome;type 1F | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| PRF1 | Hemophagocytic lymphohistiocytosis, familial, 2 | AR | 1 | 1642 | 0.06 (0.01–0.34) |
| XPC | Xeroderma pigmentosum | AR | 1 | 1642 | 0.06 (0.01–0.34) |
Open in a new tab
a For X-linked disorders, the carrier frequencies are calculated based on the total number of female individuals which is 831 out of 1,642 individuals in this study
Fig.1.
Open in a new tab
Carrier frequencies of one or more genetic disorders in 114 genes of 1,642 unrelated Thais. Orange color represents carrier frequencies in 113 ACMG recommended genes. Blue color represents those in 114 genes, which are the 113 genes and G6PD. Gray color represents those in 110 genes, which are the 113 genes excluding three genes underlying thalassemia and hemoglobinopathies (HBB, HBA1, and HBA2)
Gene carrier rate
In our cohort, all top three gene carrier rates were in hemolytic disorders with positive selection from malaria. The gene with the highest carrier rate is HBB, with 50.2% (321/640) and 19.6% (291/1642) of all carriers and participants, respectively. The second most common carrier gene is G6PD, with 19.8% (127/640) and 7.7% (127/1642) of all carriers and participants, respectively. HBA2 is the third most common gene, with 10.2% (65/640) and 4% (65/1642) of all carriers and participants, respectively.
There were 17 autosomal recessive genes with carrier frequencies of at least 1/250, an expected prevalence at birth of at least 1 in 250,000. As shown in Fig.2, three, five, and nine of these genes are related to hemolytic disorders (G6PD, HBB, and HBA2), inborn errors of metabolism (AGXT, ATP7B, CYP27A1, GBA, and PAH), and in the miscellaneous group (CEP290, CFTR, CLCN1, GJB2, MCPH1, NEB, PKHD1, SLC26A4, and USH2A), respectively.
Fig.2.
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The 17 genes underlying autosomal recessive disorders with carrier frequencies of ≥ 1/250 (0.4%) in the 1,642 unrelated Thai individuals
Variant carrier rate
The most common PV is HBB; c.79G > A (p.Glu27Lys) responsible for hemoglobin E (HbE), representing 45.5% (291/640) and 17.7% (291/1642) of all carriers and participants, respectively. HBA2; c.427 T > C (p. Ter143Glnext31, also known as Hb Constant Spring) is the second most common variant, representing 9.2% (59/640) and 3.6% (59/1642) of all carriers and participants, respectively. The third most common variant is G6PD; c.961G > A (p.Val321Met) causing G6PD deficiency, with 8.6% (55/640) and 3.4% (55/1642) of all carriers and participants, respectively. The variant carrier rates are shown in S3 Table.
Thalassemia and hemoglobinopathies
Although, the most common type of α-thalassemia mutation is the deletion of one or more of the α-globin genes, HBA1 and HBA2 , we did not attempt to analyze this type of genetic variant due to the inherent limitation of exome sequencing. The carrier rates of nondeletional α-thalassemia were detected in 5.1% (71/1642; 6 in HBA1 and 65 in HBA2) of our cohort. The most common variant is c.427 T > C 3.6% (59/1642) in Hb Constant Spring, followed by Hb Paksé c.429A > T 0.3% (5/1642). The frequency rate of β-thalassemia carriers is 19.6% (321/1642). The most prevalent variant, c.79G > A (p.Glu27Lys [HbE]), accounts for 17.7% (291/1642) of cases in the HBB genes. The next two most prevalent variants are c.52A > T (0.79%, 13/1642) and c.126_129del (0.67%, 11/1642).
G6PD deficiency
The carrier rate of G6PD variants was 7.7% (127/1642). The prevalence of heterozygous/carrier G6PD variant and homozygous females in our cohort is 15.28% (127/831) and 0.8% (7/831), respectively (S4 Table). The percentage of symptomatic or biochemical G6PD deficiency in these samples was not known. The prevalence of hemizygous G6PD variant males in our cohort is 8% (65/811). The female carrier rate corresponds with the predicted prevalence calculated with male hemizygous prevalence using the Hardy–Weinberg formula. c.961G > A (p.Val321Met) (G6PD Viantien) and c.577G > A (p.Gly193Ser) (G6PD Mahidol) are the two most common G6PD variants found in our cohort.
Other autosomal recessive disorder carrier rate and common variants
The carrier frequency of PV and LPV in GBA causing Gaucher disease is 1.6% (27/1642) with c.605G > A (p.Arg202Gln) (0.54%, 9/1642) and c.1448 T > C (p.Leu483Pro) (0.42%, 7/1642) being the two most common variants.
The carrier frequency of PV and PLV in genes associated with hereditary syndromic hearing loss SLC26A2 is 1.1% (18/1642). c.919-2A > G was the most common variant (0.3%, 5/1642). The second most common gene causing syndromic hearing loss carriers was USH2A (Usher syndrome) (0.97%, 16/1642). The carrier frequency of PV and PLV in GJB2, one of the most common genes causing nonsyndromic deafness is 0.6% (9/1642). All cases identified carried c.235delC (p.Leu79CysfsTer3) in GJB2.
Besides Gaucher disease, the most common inherited metabolic disease gene carriers in our cohort are hyperoxaluria (AGXT; 1.1%, 18/1642), Wilson disease (ATP7B; 1%, 17/1642), and phenylketonuria (PAH; 0.79%, 13/1642).
Discussion
Every approach possesses both advantages and limitations. In our study, we chose to engage parents whose children were afflicted with rare diseases rather than employing a population-based screening methodology. This strategy, akin to "killing two birds with one stone," aimed to optimize the benefits derived from trio exome testing. Specifically, children with rare diseases stood a higher chance of obtaining molecular diagnoses through trio exomes as opposed to singleton testing. Furthermore, the exomes of the parents provided valuable insights into a spectrum of other biological and medical issues. To mitigate bias, we deliberately excluded genes identified as causative for rare diseases in the children. Notably, our approach is characterized by a lower ethical burden. The apparently healthy parents, whose children were affected by rare diseases in our cohort, received genetic counseling and were well-informed about the implications of these tests. They made a conscious choice to permit the use of their genomic information for purposes beyond the primary focus of the testing.Conversely, a population-based approach is not without its constraints. Recent evidence [19, 20], suggests the presence of participation bias. Additionally, ascertainment bias may manifest, with individuals referred or participating in the project having a known family history of rare diseases showing heightened interest. The ethical complexities surrounding population screening pose a challenge. Carrier screening currently remains optional and not a standard mandatory test. While couples undergoing population screening could potentially benefit from counseling, extending genetic counseling to research subjects who are apparently normal may be cumbersome and miss the opportunity to support those genuinely in need. Moreover, the practical medical advantages of counseling may be constrained by the exorbitant costs associated with reproductive choices such as in vitro fertilization (IVF) and preimplantation genetic diagnosis (PGD).
Prior investigations into the carrier rates of these disorders in various countries have exhibited substantial variability in terms of the number of genes tested and the specific genes encompassed within the studies across diverse populations. Notably, within the Thai population, genetic disorders that manifest with a comparatively higher prevalence than in other populations encompass thalassemia, hemoglobinopathies, and glucose-6-phosphate dehydrogenase (G6PD) deficiency . Carriers of these hematological disorders in Thais demonstrate a relative resistance to malaria, affording them a selective advantage. Since HBA and HBB are already incorporated into the ACMG 113-gene list, we have augmented our gene panel to encompass G6PD deficiency, resulting in a total of 114 genes. This expansion aligns with our commitment to comprehensively address the genetic landscape relevant to carrier screening, particularly in the context of the prevalent hematological disorders observed in the Thai population.
Next-Generation Sequencing (NGS) has been increasingly used for expanded genetic carrier screening for multiple genes in a highly efficient manner in clinical laboratories . When screening for carriers in individuals with an average risk, it is deemed acceptable or advisable to report solely (likely) pathogenic variants . The carrier rate ranges from as high as 43.6% in Ashkenazi Jewish individuals to 8.5% in East Asians , although the number of genes tested and which genes are included are varied among these studies. In our study using 1642 exomes of a Thai population, 39% of Thai individuals carried a PV and an LPV in at least one of 114 genes (113 genes have been recommended by ACMG ) plus G6PD, causing G6PD deficiency, which is most common in the Thai population. We observed that 23.8% of individuals were carriers of thalassemia and hemoglobinopathies (HBB, HBA1, and HBA2), followed by 7.7% who were carries of G6PD deficiency. The carrier rate observed in our cohort is considered high, comparable to that of the Ashkenazi Jewish group. The majority of these variants are found in the Hb genes and G6PD, which are believed to be the result of heterozygous advantage in regions with a high prevalence of malaria . This differs from the Ashkenazi Jewish population, where the high carrier rate is primarily attributed to in-group mating.
Consanguineous marriage is generally considered taboo in the majority of the Thai population, with exceptions observed in certain minor communities such as Muslims and hill tribes. Our assertions are substantiated by evidence gleaned from our analysis of inheritance patterns among patients referred for genetic testing at our institution. Notably, our findings reveal that only 25% of positive tests exhibit inheritance in an autosomal recessive pattern, in stark contrast to Middle Eastern countries where the autosomal recessive pattern can reach up to 80%, particularly in regions where consanguinity is commonplace . It is imperative to underscore that thalassemia and hemoglobinopathies are highly prevalent in our region, and the pseudoautosomal dominant pattern observed in our study is reflective of this prevalence rather than indicative of prevailing marriage patterns.
Even after excluding thalassemia and hemoglobinopathies, the carrier rate in our cohort remained significantly high at 14.7%. We observed 14 couples (1.7%) who were carriers of PV and LPV in the same gene (S5 Table). Of these couples, seven (0.9%, 7/821) were non thalassemia carrier couples at risk of having children affected with recessive disorders. Thalassemia carrier screening is already included and available in the screening program of Thai pregnant women; therefore, it is estimated that nearly 1% of the couples could benefit from expanded carrier screening.
The prevalence of common recessive disorders in our cohort can be estimated and, in general, is comparable to previous prevalence rates reported for common autosomal recessive disorders, such as Thalassemia and G6PD deficiency [18, 26–31]. This suggests that our approach of excluding samples carrying genetic variants responsible for the diseases in their offspring during ascertainment and analysis helps eliminate bias and ensures that the frequency estimates are comparable to those of non-disease or population cohorts. We have previously demonstrated the effectiveness of this strategy in our analysis of the prevalence of secondary findings, particularly since the majority of reportable diseases are inherited in an autosomal dominant pattern.
The most common carriers of inherited metabolic disease identified in our cohort are individuals with Gaucher disease caused by mutations in the GBA gene. However, interpreting the results can be challenging due to the presence of a pseudogene associated with GBA. The GBA pseudogene exhibits the highest homology with GBA between exons 8 and 11, which is likely a result of recombination events . Many clinical exome sequencing laboratories do not analyze or report the variants in GBA. However, within the exome, there exists a mappable region of GBA that encompasses known recurrent pathogenic variants, including the c.1448 T > C mutation. This region can be analyzed with a high degree of sensitivity and specificity. This methodology has been detailed in our recent publication on early-onset Parkinson’s disease in Thai cohorts . In addition, the variants commonly identified in our cohort are outside these highly homologous areas and are therefore less likely to be affected in our analysis. Conversely, the prevalence of the carrier of GBA variant is likely to be underestimated in our cohort or a similar cohort using exome sequencing. Nevertheless, it is crucial to acknowledge the limitation associated with utilizing exome sequencing for GBA carrier screening. While this approach offers high sensitivity and specificity for certain regions, its effectiveness in reducing the posttest probability of being a GBA carrier for other areas is comparatively low. This limitation underscores the importance of considering alternative testing methods for a comprehensive and accurate assessment of GBA carrier status.
In our study, carriers of primary hyperoxaluria and Wilson disease PV and PLV were found to be the next most common. Founder variants in East Asian populations (c.2 T > C (p.Met1Thr) in AGXT [34–37] and c.2333G > T (p.Arg778Leu) in ATP7B [38–43] were also identified in our cohort. Our study further supports the existing newborn screening program in Thailand for Phenylketonuria (PKU). The estimated carrier rate of 1 in 127 individuals is comparable to rates reported in other countries [44–46]. With an annual live birth rate of 600,000, our findings suggest that approximately 8–10 children will be born with PKU each year in Thailand. These findings emphasize the importance of continued implementation and monitoring of the newborn screening program for PKU in the country.
Hearing loss is one of the major disability problems in childhood, with hereditary causes, both syndromic and nonsyndromic, being the primary etiologies in affected children. In our study, we identified carriers of Pendred syndrome, Usher syndrome, and GJB2 (nonsyndromic hereditary neurosensory hearing loss) to be common in the Thai cohort, aligning with previous estimations [47–58]. These findings highlight the relevance of these hereditary conditions as significant contributors to hearing loss in the Thai population.
Compared to the common recommendation of screening for cystic fibrosis carrier status in Caucasians [59–61], our study revealed that only 0.97% of our sample carried pathogenic variants (PV) and likely pathogenic variants (LPV) in the CFTR gene. This suggests that the estimated prevalence of cystic fibrosis among Thais is lower than that among Caucasians (1 in 2,000 in Northern European descent vs. 1 in 400,000). Interestingly, we observed three females (0.18%) who carried the 5T variant (c.1210–12 ) in CFTR. While this variant is associated with mild cystic fibrosis, it can result in male infertility due to congenital bilateral absence of the vas deferens .
Limitations
Our study had some limitations. Whole Exome Sequencing (WES) cannot reliably detect structural variants that are common in α-thalassemia. Spinal muscular atrophy, a common muscular disorder, most commonly caused by a copy number change in the SMN1 gene and modified by the copy numbers of SMN2, is also missed by WES. The genome analysis with structural variant detection will help estimate these common disorders. Additionally, the frequency of variants based on NGS analysis may be underestimated without validation using Sanger sequencing, particularly for complex recombinant alleles of GBA pseudogenes. The other common PVs that will be missed by WES are trinucleotide repeat expansion. PV in FMR1 causing fragile X syndrome, the most common cause of intellectual disability especially in males, cannot be detected with exome sequencing data. It is worth mentioning that our sample size provides over 80% power to detect alleles with a frequency of 1%. However, for the detection of uncommon variants, a larger sample size would be necessary .
Conclusion
Among the 114 genes analyzed, which included the 113 ACMG-recommended genes along with G6PD, our study revealed that 39% (640/1642) of Thai individuals in our cohort were carriers of pathogenic variants (PVs). Through a combination of computational filtering and manual curation, we were able to unveil the landscape of these PVs that are commonly observed in the Thai population. This information will serve as a foundation for designing gene lists for individual-level carrier screening and future public health plans aimed at addressing inherited genetic disorders.
Supplementary Information
12920_2023_1771_MOESM1_ESM.docx (58.4KB, docx)
Additional file 1: S1 Table. Genetic variants of the 18 parents which were responsible for the presenting symptoms in their children. S2 Table. Carrier frequencies of one or more genetic disorders in 114 genes of 1,642 unrelated Thais. S3 Table. Variant carrier rate (VCR) of each variant. S4 Table. Characteristics of G6PD phenotypes in 1,642 unrelated Thais. S5 Table. Pathogenic or likely pathogenic variants in the same genes which were harbored by the couples.
Acknowledgements
Not applicable.
Abbreviations
ACMG
The American College of Medical Genetics and Genomics
LPV
Likely pathogenic variants
PV
Pathogenic variants
NGS
Next-Generation Sequencing
WES
Whole exome sequencing
Authors’ contributions
WC Data curation, analysis, and writing the original draft. PP Conceptualization and editing the manuscript. VS Conceptualization, funding acquisition, and editing the manuscript. All authors have read and approved the final version of the manuscript.
Funding
This study was supported by the Health Systems Research Institute (65–040).
Availability of data and materials
The datasets used and/or analyzed during the current study available from the corresponding author on reasonable request.
Declarations
Ethics approval and consent to participate
This study was approved by the Institutional Review Board of Faculty of Medicine, Chulalongkorn University, Thailand (IRB No. 264/62). Written informed consent was obtained from parents or legal guardians of the participants.
Consent for publication
Not applicable.
Competing interests
The authors declare that they have no competing interests.
Footnotes
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Additional file 1: S1 Table. Genetic variants of the 18 parents which were responsible for the presenting symptoms in their children. S2 Table. Carrier frequencies of one or more genetic disorders in 114 genes of 1,642 unrelated Thais. S3 Table. Variant carrier rate (VCR) of each variant. S4 Table. Characteristics of G6PD phenotypes in 1,642 unrelated Thais. S5 Table. Pathogenic or likely pathogenic variants in the same genes which were harbored by the couples.
Data Availability Statement
The datasets used and/or analyzed during the current study available from the corresponding author on reasonable request.
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What counts as a hit in baseball?
What counts as a hit in baseball?
By HotBot Updated: September 23, 2024
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Introduction to Hits in Baseball
In baseball, a "hit" refers to the successful act of a batter reaching at least first base after striking the ball into fair territory, without the benefit of an error or a fielder's choice. Hits are a fundamental aspect of the game, contributing significantly to a team's offensive performance. Understanding what constitutes a hit involves exploring various types of hits, the rules governing them, and their strategic importance in the game.
Types of Hits
Single
A single is the most basic type of hit, where the batter safely reaches first base. Singles are often the result of a well-placed ground ball, line drive, or a bunt that evades the infielders. While they may not appear as dramatic as other hits, singles are critical for setting up scoring opportunities and advancing base runners.
Double
A double occurs when the batter hits the ball and safely reaches second base. Doubles usually result from hard-hit balls to the outfield that roll to the fence or find gaps between outfielders. Doubles are significant because they often put a runner in scoring position, increasing the chances of the team scoring a run.
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Triple
A triple is a hit where the batter reaches third base safely. Triples are rare compared to singles and doubles, typically requiring a combination of speed and precise ball placement. They are often the result of hitting the ball into the deep outfield or taking advantage of fielding misplays. Triples are valuable as they place the batter just 90 feet away from scoring a run.
Home Run
A home run is the most celebrated type of hit, where the batter hits the ball out of the playing field in fair territory, allowing them to circle all the bases and score a run. Home runs can be either inside-the-park, where the batter runs all the bases on a ball that stays in play, or over-the-fence, where the ball clears the outfield fence. Home runs are crucial for scoring multiple runs, especially with runners already on base.
Rules Governing Hits
Fair Territory
For a hit to be counted, the ball must land or be touched in fair territory. Fair territory is defined as the area within the two foul lines extending from home plate to the outfield fence and vertically upwards. A ball that lands in foul territory or is first touched in foul territory is not considered a hit.
Exclusion of Errors
If a batter reaches base due to a defensive error, it does not count as a hit. Official scorers are responsible for distinguishing between hits and errors. For example, if a fielder misplays a routine ground ball, allowing the batter to reach base, it is scored as an error, not a hit.
Fielder's Choice
A fielder's choice occurs when a fielder chooses to try to put out another base runner instead of the batter. In such cases, the batter reaching base is not credited with a hit. This distinction helps maintain the integrity of individual and team statistics by ensuring only genuine hitting achievements are recorded as hits.
Statistical Importance of Hits
Batting Average
One of the primary statistics in baseball is the batting average, which is calculated by dividing the number of hits by the number of official at-bats. A higher batting average indicates a batter's effectiveness at getting hits. Maintaining a high batting average is a key indicator of a batter's success and value to their team.
Slugging Percentage
Slugging percentage measures the power of a hitter by calculating the total bases achieved from hits divided by the number of at-bats. Unlike the batting average, slugging percentage accounts for the type of hits (singles, doubles, triples, and home runs), giving more weight to extra-base hits. This statistic provides a more comprehensive view of a batter's offensive capabilities.
On-Base Percentage
On-base percentage (OBP) includes hits, walks, and hit-by-pitches to determine how frequently a batter reaches base. While hits are a significant component of OBP, this statistic also considers a batter's ability to draw walks and endure being hit by a pitch, offering a broader perspective on their overall offensive contributions.
Strategic Importance of Hits
Advancing Runners
Hits play a crucial role in advancing base runners and setting up scoring opportunities. For example, a single with runners on first and second base can load the bases, creating a high-pressure situation for the opposing pitcher and defense. Effective hitting strategies, such as situational hitting and hit-and-run plays, are designed to maximize the potential of advancing runners through hits.
Creating Momentum
Hits can shift the momentum of a game, energizing the batting team and putting pressure on the opposing defense. A series of consecutive hits, known as a "rally," can lead to multiple runs scored in a single inning, dramatically changing the game's outcome. Teams often rely on timely hits to build and sustain offensive momentum.
Psychological Impact
The psychological aspect of hitting cannot be understated. A well-timed hit can boost the morale of the batting team while simultaneously demoralizing the opposing pitcher and defense. Conversely, a slump in hitting can lead to decreased confidence and increased pressure on the batting team. The mental game is an integral part of baseball, and hits are a significant factor in shaping the psychological dynamics of a game.
Rarely Known Details About Hits
Infield Hits
Infield hits occur when a batter reaches base safely on a ball that does not leave the infield. These hits often result from speed, as the batter beats the throw from an infielder. Infield hits are more common among faster players and can catch the defense off guard, contributing to the overall unpredictability of the game.
Ground Rule Double
A ground rule double is a specific type of hit where the ball bounces over the outfield fence in fair territory. The batter is automatically awarded second base, and any base runners advance two bases from their positions at the time of the pitch. While not as common as other types of hits, ground rule doubles add an interesting twist to the game's dynamics.
Hitting for the Cycle
Hitting for the cycle is a rare and notable achievement in baseball, where a batter hits a single, double, triple, and home run in a single game. This feat demonstrates a batter's versatility and ability to hit in various situations. Achieving the cycle is celebrated as a significant milestone and is a testament to a player's hitting prowess.
The Art and Science of Hitting
Hitting in baseball is both an art and a science, requiring a combination of physical skills, mental acuity, and strategic thinking. Players spend countless hours honing their techniques, studying pitchers, and developing game plans to maximize their hitting potential. From the mechanics of a swing to the intricacies of pitch recognition, the journey to becoming a proficient hitter is complex and multifaceted.
The Ever-Evolving Nature of Hits
As baseball continues to evolve, so too do the strategies and technologies related to hitting. Advances in analytics and data-driven approaches have transformed how players and teams approach hitting, leading to new insights and innovations. Whether through the use of advanced metrics, video analysis, or biomechanical studies, the pursuit of excellence in hitting remains a dynamic and ever-changing aspect of the game.
Understanding what counts as a hit in baseball involves delving into the various types of hits, the rules governing them, and their strategic importance. From the simplicity of a single to the excitement of a home run, hits are a central element of the game's offensive strategy. By appreciating the nuances and complexities of hitting, fans and players alike can gain a deeper appreciation for the skill and strategy that make baseball a timeless and captivating sport.
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What is a bb in baseball?
In the lexicon of baseball, "BB" stands for "base on balls," commonly known as a walk. This term plays a significant role in the game, impacting various aspects from a player's statistics to strategic gameplay. To fully appreciate the importance and intricacies of BB, it is essential to delve into its definition, historical context, strategic implications, statistical significance, and its impact on players and games.
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6780 | https://cdn.kutasoftware.com/Worksheets/Precalc/11%20-%20Induction.pdf | ©j J2^0N1M6D tKPuFt\aJ lSeokfVt^wdaTrtex ELlLtC].o V BAQlhly crlidgkhXtYsr srweTsMeZrOv^edt.Y r ZMBahdkeC HwziPtHhu pIfnZfrixnEietWej tPkrXec_aTlVcfuxlYuIsQ.
Worksheet by Kuta Software LLC Kuta Software - Infinite Precalculus Mathematical Induction Name_____ Date____ Period_ -1-Write the induction proof statements P , Pk , and Pk for each conjecture.
1) Pn: nn(n) 2) Pn: nn is divisible by 3) Pn: nn Use mathematical induction to prove that each statement is true for all positive integers 4) (n)n(n) ©o f2N0I1K6q [KbuftaaS iSoLfqtwuaArjeG BLCLXCl.J e zAOlml] HrLiggZhWtes] Br^esLekrgvkeDdy.H C KMEandbeA awgiUtyhW fIPnYfbi]nZiCtTeN tPorheGcgaelecbuVlDuZsy.
Worksheet by Kuta Software LLC -2-5) is a factor of n 6) nn ©M G2X0M1d6k lKAuHtoaF NSWoHfLtzwGaqryeJ DLoLgCE.k G OAKl]lJ xr\ibgMhktEsz sroeQsoetrUvIekdu.` S OMJaIdFei LwViTtTht mIDnVfoi^nAihtRes FPBrAe[cXawllcwuSlkucsX.
Worksheet by Kuta Software LLC Kuta Software - Infinite Precalculus Mathematical Induction Name_____ Date__ Period____ -1-Write the induction proof statements P , Pk , and Pk for each conjecture.
1) Pn: nn(n) P: Pk: kk(k) Pk: k(k)k(k)(k) 2) Pn: nn is divisible by P: is divisible by Pk: kk is divisible by . Therefore, kkr for some integer r.
Pk: (k)(k) is divisible by 3) Pn: nn P: Pk: kk for some positive integer k Pk: k(k) Use mathematical induction to prove that each statement is true for all positive integers 4) (n)n(n) Let Pn be the statement (n)n(n) Anchor Step P is true since ()() Inductive Hypothesis Assume that Pk is true: (k)k(k) Inductive Step We now show that Pk is true: (k)((k)) = k(k)((k)) kk(k) kkkk kkk (kkk) (kkkk) ((kkk)(k)) ((k)(k)) (k)((k)) Conclusion By induction Pn is true for all n.
©P n2^0k1Z6k QKBuktDai bS]offHtmwoaor\es YLWLSCL.P B HAmlHlS nrSiggbh_ttsJ arPeRsOeQrVvJeId].M D KMUaCdves EwviCtrhZ HIxndfjirnXiStVeN qPmrweEcgaplYcIuLlQuTsl.
Worksheet by Kuta Software LLC -2-5) is a factor of n Let Pn be the statement is a factor of n Anchor Step P is true: is a factor of Inductive Hypothesis Assume that is a factor of k. Therefore, kr for some integer r.
Inductive Step We now show that Pk is true: is a factor of (k) k ()k kk kr (kr) Conclusion By induction Pn is true for all n.
6) nn Let Pn be the statement nn Anchor Step Pis true: Inductive Hypothesis Assume that Pk is true: kk for some positive integer k Inductive Step We now show that Pk is true: kk kk kk kkk kk k(k) k k Conclusion By induction Pn is true for all n.
Create your own worksheets like this one with Infinite Precalculus. Free trial available at KutaSoftware.com |
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6782 | https://math.stackexchange.com/questions/2276971/modulus-problem-complex-number | Stack Exchange Network
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Modulus problem (Complex Number)
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$\begingroup$
If complex number $z(z\neq2)$ satisfies the equation : $z^2=4z+\lvert z\rvert^2+\frac{16}{\lvert z\rvert^3}$, then what is the value of $\lvert z\rvert^4$?
My try: I tried to take $z=x+iy$ and solved for the value of $\lvert z\rvert^4$ but everytime ended up getting a value greater than $9$.
Hint- The answer lies between $0$ and $9$ , both included.
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asked May 11, 2017 at 20:04
ZlatanZlatan
66922 gold badges66 silver badges1818 bronze badges
$\endgroup$
6
$\begingroup$ So can you show your work? Are you sure you typed up the problem correct? Please check the signs $\endgroup$
imranfat
– imranfat
2017-05-11 20:15:30 +00:00
Commented May 11, 2017 at 20:15
$\begingroup$ actually i cant at the moment , but on substituting $z=x+iy$, i got $x=2$, so whatever my $y$ would be , my $\lvert z\rvert^4$ would already come out to be greater than $9$. $\endgroup$
Zlatan
– Zlatan
2017-05-11 20:18:09 +00:00
Commented May 11, 2017 at 20:18
$\begingroup$ I need to think about this one, made a mistake myself :) I was actually thinking of writing $z$ as $re^{i\theta}$ and then plug in. The absolute value would be just $r$ $\endgroup$
imranfat
– imranfat
2017-05-11 20:26:18 +00:00
Commented May 11, 2017 at 20:26
$\begingroup$ lol , okay bro, take your time. $\endgroup$
Zlatan
– Zlatan
2017-05-11 20:28:11 +00:00
Commented May 11, 2017 at 20:28
1
$\begingroup$ Since this came up as a Review Audit, let me Comment that it doesn't make sense to ask for a complex number $z$ to be "between $0$ and $9$" unless you actually mean that $z$ is a real number. This turns out to be consistent with the solution offered below. $\endgroup$
hardmath
– hardmath
2017-06-10 15:31:10 +00:00
Commented Jun 10, 2017 at 15:31
| Show 1 more comment
1 Answer 1
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4
$\begingroup$
This question CAN be solved by assuming $z=x+iy$, where $x$ and $y$ are reals.
The given equation can be written as $$z^2-4z= |z|^2+ \frac{16}{|z|^3}$$
Plugging in $z=x+iy$ and
$$x^2-y^2 +2ixy -4x-4iy=x^2+y^2 + \frac{16}{(\sqrt{x^2+y^2})^3}$$
taking the complex terms to one side, we obtain$$2iy(x-2)=2y^2+4x+\frac{16}{(\sqrt{x^2+y^2})^3}$$
But if $x$ and $y$ are real numbers, then the LHS cannot be equal to the RHS, as one is a real number and the other is a complex number . This means the LHS must be zero.
$\implies y=0$ or $x=2$.
First lets check for $y=0$. The equation reduces to $$4x+\frac{16}{|x|^3}=0 \implies x=-\sqrt2$$
Hence $z=-\sqrt2$
$|z|^4 = 4$
NOTE: $x$ cannot be equal to $2$, because then, the expression can have no real value for $y$.(you can check this yourself)
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edited Jun 26, 2017 at 9:28
answered May 12, 2017 at 8:13
Ananth KamathAnanth Kamath
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$\endgroup$
1
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$\begingroup$ +1) Never thought about the $y=0$ possibility, kept focusing on $x=2$ $\endgroup$
imranfat
– imranfat
2017-05-12 16:53:29 +00:00
Commented May 12, 2017 at 16:53
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Spondylitis Plus Archive
An Updated Overview Of Spondyloarthritis,…
An Updated Overview Of Spondyloarthritis,…
An Updated Overview Of Spondyloarthritis, A Family Of Related Diseases
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We hear a few different terms to describe ankylosing spondylitis (AS) and related diseases: Spondyloarthritis, spondyloarthritides and spondyloarthropathy. . . Is there a difference?
Spondyloarthritis and spondyloarthropathy are often used interchangeably. Some experts prefer the term spondyloarthritis rather than spondyloarthropathy because the ending “arthritis” indicates inflammation of the joint, whereas the ending “arthropathy” can refer to any type of joint disease. Inflammation is a key feature that helps distinguish spondyloarthritis from other types of arthritis, including wear-and-tear arthritis, such as osteoarthritis. Spondyloarthritides is the plural form of spondyloarthritis.
How is this group of diseases related? Why is it sometimes called a “family” of conditions?
These diseases look and behave in similar ways because they share overlapping disease features. Common features of spondyloarthritis include inflammation in the spine, pelvis, other joints, intestine, eyes [Editor’s note: please see side bar on Uveitis / Iritis to the right], and heels. This family of diseases is divided into individual categories according to the predominant disease feature(s). For example, inflammation of the intestine can occur with any type of spondyloarthritis, but is most pronounced in patients with IBDassociated arthritis (also called enteropathic arthritis).
What does “seronegative” mean? How does it relate to this group of diseases?
Seronegative means that specific blood tests used to help diagnose rheumatoid arthritis are negative. In some instances, these blood tests are helpful in determining whether a person has rheumatoid arthritis or spondyloarthritis. In most cases, a diagnosis of spondyloarthritis can be made without these blood tests.
Is ankylosing spondylitis (AS) considered the “primary” disease? Why or why not?
For people with ankylosing spondylitis, it is the primary disease. For people with other types of spondyloarthritis, it is not.
In the past, ankylosing spondylitis has been portrayed as the primary type of spondyloarthritis for several reasons including the following:
Ankylosing spondylitis is easier to study than reactive arthritis and IBD-associated arthritis because it is much more common.
Ankylosing spondylitis is often easier for doctors and patients to recognize than undifferentiated spondyloarthritis and reactive arthritis.
Ankylosing spondylitis has been recognized as a unique type of arthritis for hundreds of years, whereas other types of spondyloarthritis were described more recently. For example, psoriatic arthritis was not widely recognized as a distinct form of arthritis until the 1960s.
Can you give us a key symptom or “feature” of each of the conditions in this group? What makes each one distinct or different from the others?
Ankylosing Spondylitis (AS)
Inflammation in the pelvis and/or spine causes inflammatory back pain. Inflammatory back pain usually starts gradually before the age of 40, tends to improve with activity but not rest, and occurs with stiffness in the morning that lasts at least 30 minutes.
Reactive Arthritis (Reiter’s Syndrome – ReA)
An infection in the intestine or urinary tract usually occurs before inflammation in the joints.
Juvenile Spondyloarthritis (JSpA)
Symptoms begin in childhood. JSpA can look like any other type of spondyloarthritis. Enthesitis (inflammation where tendons or ligaments meet bone) is often a dominant disease feature.
Arthritis Associated With Inflammatory Bowel Disease (Enteropathic Arthritis – EnA)
Inflammation of the intestine is a predominant feature. Symptoms may include chronic diarrhea, abdominal pain, weight loss, and/or blood in the stool. The most common types of inflammatory bowel disease are Crohn’s, ulcerative colitis, and undifferentiated colitis.
Psoriatic Arthritis (PsA)
PsA frequently causes pain and swelling in the small joints of the hands and feet. Most people with PsA have a psoriasis skin rash. Some people have a “sausage digit” with a toe or finger that swells between the joints as well as around the joints.
Undifferentiated Spondyloarthritis (USpA)
People with USpA have symptoms and disease features consistent with spondyloarthritis, but their disease doesn’t fit into another category of spondyloarthritis. For example, an adult may have iritis, heel pain (caused by enthesitis), and knee swelling, WITHOUT back pain, psoriasis, a recent infection, or intestinal symptoms. This person’s combination of disease features suggests spondyloarthritis, but he or she doesn’t fit into the categories of ankylosing spondylitis, psoriatic arthritis, reactive arthritis, juvenile spondyloarthritis or IBD-associated arthritis.
Can one of these conditions share symptoms or complications with another one of the conditions? In general terms, do symptoms overlap? If so, how? What are the main similarities – if any?
Absolutely. The main similarities that can occur with any type of spondyloarthritis are:
Inflammation in the pelvis and spine that usually causes inflammatory back pain.
Pain and/or swelling of any other joint in the body (hips, knees, ankles, feet, hands, wrists, elbows, shoulders etc.)
Sudden onset of marked pain and redness in one eye at a time (uveitis/iritis).
Psoriasis skin rash.
Inflammation in the intestine (Crohn’s, ulcerative colitis, undifferentiated colitis).
Inflammation along the tendons of the finger or toes (sausage digits, also called dactylitis).
Inflammation where tendons and ligaments meet the bone (enthesitis). This commonly occurs at the back or bottom of the heel.
Why would a doctor diagnose one form of spondyloarthritis over another?
Doctors classify people as having a certain type of spondyloarthritis according to the predominant disease feature(s). For example, a person with psoriasis and joint swelling in the hands and feet will most likely be classified as having psoriatic arthritis. A person with inflammatory back pain and x-ray changes consistent with inflammation in the sacroiliac joints in the pelvis will likely be classified as having ankylosing spondylitis. A person with Crohn’s and swelling in the knees and ankles most likely has IBD-associated arthritis. Sometimes, disease features are equally dominant and a person may fit into more than one type of spondyloarthritis. For example, a person could have psoriasis, inflammation in the pelvis/spine, and Crohn’s disease. This person could correctly be said to have any of the following:
Psoriatic arthritis with ankylosing spondylitis and Crohn’s.
Ankylosing spondylitis with psoriasis and Crohn’s.
IBD-associated arthritis with ankylosing spondylitis and psoriasis.
Can a diagnosis change from, say, undifferentiated spondyloarthritis (USpA) to ankylosing spondylitis or another one of these conditions? Why would this occur?
Yes. The diseases can evolve or change over time, since not all symptoms occur at once. For example, the previously discussed person with USpA with iritis, enthesitis, and knee swelling could develop back pain and inflammatory changes on x-rays that would lead to the diagnosis of ankylosing spondylitis.
Are men and women affected at different rates between these diseases? Can you give us some details on how these rates were determined?
Similar numbers of men and women are affected with spondyloarthritis. In the past, it was thought that ankylosing spondylitis was more common in men than women. More recent studies suggest that ankylosing spondylitis occurs in similar numbers of men and women. Early estimates of the occurence of ankylosing spondylitis suggested that ankylosing spondylitis occurred 9-10 times more frequently in men than women. However, there were problems with how these early studies were done. More recent studies reported that men are 2 to 3 times more likely than women to have ankylosing spondylitis. These studies use relatively narrow definitions of ankylosing spondylitis that rely on classic manifestations of inflammatory back pain and damage on x-rays. Classic inflammatory back pain may be the initial symptom in men more frequently than in women, and women may have less x-ray damage than men. Despite these differences, the overall disease severity is similar in men and women.
When broader definitions are used to identify people with spondyloarthritis in the pelvis and/or spine (axial spondyloarthritis), the prevalence is similar in men and women.
How are these conditions treated? Are there any notable differences in treatment such as prescribed medications?
There are several treatment options for various types of spondyloarthritis. The treatments for each disease overlap, but they are not identical. For example, certain treatments may simultaneously help with psoriasis, inflammatory bowel disease, enthesitis, and arthritis. Other treatments may help with one or two disease features, but not the others. There are even some treatments that may help with one disease feature, but make another feature worse. Treatments need to be tailored for each individual, according to the type and severity of specific disease features. Many other factors must also be considered when selecting therapies, including other medical conditions, access to therapies, and the preferences of patients.
Please see our medications guide for more information.
Is there a known cause for these diseases?
We know that there are several specific genes that increase the risk of developing spondyloarthritis. HLA-B27 is the best studied gene and it associates most strongly with inflammation in the pelvis (sacroiliac joints) and spine. Most people with HLA-B27 and other high risk genes never develop spondyloarthritis. We don’t yet understand why some people develop disease and others don’t. There are also studies suggesting that things in our environment may cause disease. For example, specific types of infections may trigger disease. However, environmental triggers are not known for most people who develop spondyloarthritis. There is much research that needs to be done to better understand why certain people get these diseases.
Is there a cure?
Not yet…
By: Jessica Walsh
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In a recursive pattern, repetition of a rule or procedure can be used to extend the sequence or to find the values of any terms missing from the sequence.
Given a sequence of numbers, how can we discover the pattern and find the next number in the sequence? There are some common sequences that are straightforward to recognize, such as the sequence of positive integers
1,2,3,4,5,…
or the sequence of odd prime numbers
3,5,7,11,13,17,….
But what if we are given a more complicated sequence? What deductive approaches should we attempt to discover the sequence pattern? First, asking a few basic questions might be helpful. Is the sequence always increasing, or always decreasing, or neither? Is the sequence always positive, or always negative, or neither? Once we have answered these questions, here are a few methods to try to find the underlying sequence pattern:
Look at successive terms and take their differences (or ratios):
is there a clear pattern in the differences (or ratios)?
Try adding another sequence (often constant or linear) to the current sequence.
Divide out by common factors.
Be aware of common patterns in the sequence itself (or in the differences or ratios of successive terms), such as integers, primes, factorial, arithmetic or geometric progressions, etc.
Draw the next term in this pattern.
In drawing the answer, the boxes could be added to the third term in two different locations:
a horizontal row of five squares under the existing squares
or a diagonal line of five squares created by adding one box to the end of each existing row and starting a new row at the top of the image.
In either case, the resulting image is the same, and five squares were added to the pattern.
The same principles can be used to fill in the blanks if a term in a pattern is missing.
Draw the missing image from the following pattern:
Fill in the missing number x in the following pattern: 1,4,9,x,25,36,….
The pattern should read 1,4,9,16,25,36,….
In analyzing this sequence, you may have noticed the values were perfect squares. Depending on how you solved the previous example, you may also have noticed that each value corresponds to the total number of small triangles in the pattern shown above. If you did not know what square numbers were, or did not notice the pattern, you could have counted the number of triangles in the fourth image if you noticed that pattern. Often, there are multiple ways of defining a recursive pattern and solving for missing terms. □
What comes next in the sequence 2,4,6,8,…?
Looking at the sequence, we recognize the even numbers, each of which is 2 more than the one before it. Thus the answer is 10. □
What comes next in the sequence 1,2,5,10,17,26,…?
Looking at the sequence carefully, we might decide to record the difference between terms:
2−15−210−517−1026−17=1=3=5=7=9.
Thus we can see that the differences between adjacent terms in the sequence are the odd numbers and that the next difference must be 11. Therefore the next term in the sequence is 26+11=37. □
What comes next in the sequence 50,49,47,44,40,35,…?
First, notice that this sequence is decreasing. Taking the difference between successive terms, we obtain
50−4949−4747−4444−4040−35=1=2=3=4=5.
From these differences, we see that terms in the sequence differ by increasing integers, and the next difference is 6. Therefore, the next term in the sequence is 35−6=29. □
What comes next in the sequence 1,−3,9,−27,81,…?
This sequence is alternating between positive and negative terms, and the first terms are multiples of 3. This suggests that we might try taking the ratio of successive terms, which gives
1−3−399−27−2781=−3=−3=−3=−3.
From these ratios, we see that the ratio between successive terms are all −3. Therefore, the next term in the sequence is (81)⋅(−3)=−243. □
What comes next in the sequence 1,3,7,15,31,63,…?
This sequence is increasing and the gaps between successive terms is increasing. If we take the difference between successive terms, we obtain
3−17−315−731−1563−31=2=4=8=16=32.
We recognize these differences as powers of 2. Therefore, the next term in the sequence is 63+26=63+64=127.
Alternatively, we can notice that the terms in the sequence are all odd, and adding 1 to each term in the sequence gives 2,4,8,16,32,64…. Therefore, the next term in the sequence is 128−1=127 and, in general, term n in the sequence is given by 2n−1.
□
What comes next in the sequence 1,2,4,8,16,32,64,…?
If you carefully observe, you will see that these numbers are the powers of 2. Therefore, the numbers are 20,21,22,23,24,25,26,…. Hence, the next number in this sequence should be 27, which equals 128. □
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6786 | https://www.statisticshowto.com/probability-and-statistics/confidence-interval/ | Confidence Interval: Definition, Examples - Statistics How To
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Confidence Interval: Definition, Examples
< Probability and Statistics < Confidence Interval
Contents (Click to Skip to Section)
What is a Confidence Interval?
How to Find a Confidence Interval by Hand:
Sample (T-Distribution)
Small sample example 2
Normal Distribution / Z-Distribution
Proportion
Two Populations (Proportions)
How to Find a Confidence Interval using Technology
Excel: Mean
TI 83: Two Populations;
TI 83: Population Proportion, p
TI 83: Population Mean
TI 89: Population Mean
TI 89: Population Proportion
History of Confidence Intervals
What is a Confidence Interval
A confidence interval is an estimate of a statistic, plus and minus a variation in that estimate. For example, a voting poll might give a result of 3% for an unpopular candidate plus or minus 2%. It represents how much uncertainty there is with an estimate.
A confidence interval represents a range of values that are likely to include a population parameter. For instance, if you want to know the mean weight of all adults in the United States, you could take a sample of 250 adults and measure their weights. The sample’s mean height would serve as a “point estimate” of the population average; a point estimate is a number used to estimate a population parameter, such as the mean or the standard deviation; in the above image that statistic is “3”.
However, the sample mean might not precisely match the population average. In this case, a confidence interval can help estimate the range of values that probably include the population average. In the above image, the range of values is from 1 to 5.
A more technical definition
More precisely, a confidence interval is a range of likely values for the population parameter based on a point estimate (e.g., the sample mean), a confidence level (e.g., 95%) and the sampling variability or standard error of the point estimate.
Usually, analysts will include a 95% or 99% confidence level with results. For example, they might report a mean of 100 with a 95% confidence interval of between 95 to 105. This tells us that they are 95% confident that the true mean lies between 95 and 105. Should someone else repeat the experiment, 95% of the time the mean will fall into that interval.
Real life example
A 2008 Gallup survey found that TV ownership may be good for wellbeing. The results from the poll stated that the confidence level was 95% +/-3, which means that if Gallup repeated the poll over and over, using the same techniques, 95% of the time the results would fall within the published results. The 95% is the confidence level and the +/-3 is called a margin of error. The article reported, “For the European data, one can say with 95% confidence that the true population for wellbeing among those without TVs is between 4.88 and 5.26.” The confidence interval here is “between 4.88 and 5.26.“
The fact that televisions are connected with wellbeing may come as a surprise, especially since we often view television-watching as being associated with an unhealthy couch potato lifestyle. But the results of the analysis only tell us that there is something about TV ownership that is good for wellbeing — it doesn’t tell us exactly what that “something” is. It could be an association with wealth, access to entertainment, or some other factor. In addition, it doesn’t say that watching TV is good for us — just that owning one is. Thus, we shouldn’t take too much stock in individual study results — even if those results are highly confident.
Confidence Intervals vs. Confidence Levels
Confidence levels are expressed as a percentage (for example, a 95% confidence level). It means that should you repeat an experiment or survey over and over again, 95% of the time your results will match the results you get from a population (in other words, your statistics would be sound!).
Confidence intervals are usually numbers. For example, you survey a group of pet owners to see how many cans of dog food they purchase a year. You test your statistic at the 99% confidence level and get a confidence interval of (200, 300). That means you think they buy between 200 and 300 cans a year. You’re super confident (99% is a very high level!) that your results are sound, statistically.
| Feature | Confidence interval | Confidence level |
---
| Definition | A range of values that probably contains a population parameter | The probability that the confidence interval contains the population parameter |
| Calculation | Uses sample size, standard deviation of the sample, and desired level of confidence | Usually chosen by the researcher |
| Interpretation | Indicates the range of values that are likely to be true for the population | Indicates the probability that the confidence interval contains the true population parameter |
Table summarizing key differences between confidence interval and confidence level.
Back to Top
Confidence Interval For a Sample: Overview
When you don’t know anything about a population’s behavior (i.e. you’re just looking at data for a sample), you need to use the t-distribution to find the confidence interval. That’s the vast majority of cases: you usually don’t know population parameters, otherwise you wouldn’t be looking at statistics! The confidence interval tells you how confident you are in your results. With any survey or experiment, you’re never 100% sure that your results could be repeated. If you’re 95% sure, or 98% sure, that’s usually considered “good enough” in statistics. That percentage of sureness is the confidence interval.
Steps
If you have one small set of data (under 30 items), you’ll want to use the t-distribution instead of the normal distribution to construct your confidence interval. Example Question:
A group of 10 foot surgery patients had a mean weight of 240 pounds. The sample standard deviation was 25 pounds. Find a confidence interval for a sample for the true mean weight of all foot surgery patients. Use a 95% CL.
The formula you’re going to use is:
where:
t = the t-score (from the t-distribution table, shown in Steps 1 through 3 below),
s = the sample standard deviation,
n = the sample size,
x̄ = the sample mean.
Steps:
Subtract 1 from your sample size. 10 – 1 = 9. This gives you degrees of freedom, which you’ll need in step 3.
Subtract the confidence level from 1, then divide by two. (1 – .95) / 2 = .025
Look up your answers to step 1 and 2 in the t-distribution table. For 9 degrees of freedom (df) and α = 0.025, my result is 2.262.
dfα = 0.10.050.0250.010.0050.0010.0005∞tα=1.2821.6451.9602.3262.5763.0913.29113.0786.31412.70631.82163.656318.289636.57821.8862.9204.3036.9659.92522.32831.60031.6382.3533.1824.5415.84110.21412.92441.5332.1322.7763.7474.6047.1738.61051.4762.0152.5713.3654.0325.8946.86961.4401.9432.4473.1433.7075.2085.95971.4151.8952.3652.9983.4994.7855.40881.3971.8602.3062.8963.3554.5015.04191.3831.8332.262
Divide your sample standard deviation by the square root of your sample size. 25 / √(10) = 7.90569415
Multiply step 3 by step 4. 2.262 × 7.90569415 = 17.8826802
For the lower end of the range , subtract step 5 from the sample mean. 240 – 17.8826802 = 222.117
For the upper end of the range, add step 5 to the sample mean. 240 + 17.8826802 = 257.883
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Small sample example 2
The formula for constructing a CI with the t-distribution.
Example problem: Construct a 98% Confidence Interval based on the following data: 45, 55, 67, 45, 68, 79, 98, 87, 84, 82.
Find the sample mean, μ and standard deviation, σ (try our standard deviation calculator) for the data.
σ: 18.172.
μ: 71Set these numbers aside for a moment.
Subtract 1 from your sample size to find the degrees of freedom (df). We have 10 numbers listed, so our sample size is 10, so our df = 9. Set this number aside.
Subtract the confidence level from 1, then divide by two. This is your alpha level. (1 – .98) / 2 = .01
Look up df (Step 2) and α (Step 3) in the t-distribution table. For df = 9 and α = .01, the table gives us 2.821.
Divide your std dev (step 1) by the square root of your sample size. 18.172 / √(10) = 5.75
Step 6: : Multiply step 4 by step 5. 2.821 × 5.75 = 16.22075
Step 7: For the lower end of the range , subtract step 6 from the mean (Step 1). 71 – 16.22075 = 54.77925
Step 8: For the upper end of the range, add step 6 to the mean (Step 1). 71 + 16.22075 = 87.22075
Back to Top
Confidence Interval with the Normal Distribution / Z-Distribution
If you don’t know your population mean (μ) but you do know the standard deviation (σ), you can find a confidence interval for the population mean, with the formula: x̄ ± z (σ / √n), Where
x̄ = the sample mean,
z = z-score,
σ = standard deviation,
n – sample size.
Example problem: Construct a 95% confidence interval an experiment that found the sample mean temperature for a certain city in August was 101.82, with a population standard deviation of 1.2. There were 6 samples in this experiment.
Subtract the confidence level (Given as 95 percent in the question) from 1 and then divide the result by two. This is your alpha level, which represents the area in one tail. (1 – .95) / 2 = .025
Subtract your result from Step 1 from 1 and then look that area up in the middle of the z-table to get the z-score:
1 – 0.025 = 0.975
z score = 1.96.
Plug the numbers into the second part of the formula and solve: z σ / (√n) = 1.96 1.2/√(6) = 1.96 0.49 = 0.96
For the lower end of the range, subtract step 3 from the mean. 101.82 – 0.96 = 100.86
For the upper end of the range, add step 3 to the mean. 101.82 + 0.96 = 102.78. The CI is (100.86,102.78)
Back to Top
How to Find a Confidence Interval for a Proportion
When we talk about a confidence interval (CI), we’re dealing with data.
For example, let’s say the manager for that job you applied for told you he would get back with you in a “couple of days.” A couple of days could mean two. Or three. Or there might be a paperwork backlog and it could be a week. It definitely doesn’t mean in an hour. So your CI would probably be between 2 and 4 days.
Perhaps the trickiest part of CIs is recognizing the various parts needed for the formula, like z a/2.
Example Question: 510 people applied to the Bachelor’s in Elementary Education program at Florida State College. Of those applicants, 57 were men. Find the 90% CI of the true proportion of men who applied to the program.
Read the question carefully and figure out the following variables:
Find z α/2. You don’t have to look this up in the z-table every time, you can find common ones in the table below.
According to the table, for a 90% CI, z α/2 = 1.645.
p-hat: Divide the proportion given (i.e. the smaller number)by the sample size. 57/510 = 0.112
q-hat: To find q-hat, subtract p-hat (from directly above) from 1. This gives: 1 – 0.112 = 0.888
Common confidence levels and corresponding z α/2.
Multiply p-hat by q-hat (from Step 1). 0.112 x 0.888 = 0.099456
Divide step 2 by the sample size. 0.099456 / 510 = 0.000195011765
Take the square root of step 3: sqrt(0.000195011765) = 0.0139646613
Multiply step 4 by z a/2: 0.0139646613 x 1.645 = 0.023.
For the lower percentage, subtract step 5 from p-hat. 0.112 – 0.023 \= 0.089 = 8.9%.
For the upper percentage, add step 5 to p-hat. 0.112 + 0.023 \= 13.5%. This next method involves plugging in numbers into the actual formula. You’ll get the same results if you use the “formula free” method above or if you use the steps below.
Confidence intervals for a proportion are calculated using the following formula: The formula might look daunting, but all you really need are two pieces of information: the z-score and the P-hat. You should be familiar with looking up z-scores from previous sections on the normal distribution and P-hat is just dividing the number of events by the number of trials. Once you’ve figured those two items out, the rest is basic math.
Example 2: Steps
Example question: Calculate a 95% confidence interval for the true population proportion using the following data: Number of trials(n) = 160 Number of events (x) = 24
Step 1: Divide your confidence level by 2: .95/2 = 0.475.
Step 2: Look up the value you calculated in Step 1 in the z-table and find the corresponding z-value. The z-value that has an area of .475 is 1.96.
Step 3: Divide the number of events by the number of trials to get the “P-hat” value: 24/160 = 0.15. Step 4: Plug your numbers into the formula and solve:
0.15 ± (1.96) √ ((0.15(1 – 0.15) / 160))=
0.15 ± (1.96) √ ((0.15(0.85)/160))=
0.15 ± (1.96) √ ((0.1275)/160))=
0.15 ± (1.96) √ (0.000796875)=
0.15 ± (1.96) 0.0282289744765905\=
0.15 – 0.0553 = 0.0947 ← this is your lower confidence interval for a proportion
0.15 + 0.0553 = 0.2053 ←this is your upper confidence interval for a proportion
Your answer can be expressed as: (0.0947,0.2.053).
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Two Populations (Proportions)
Finding confidence intervals for two populations can look daunting, especially when you take a look at the ugly equation below. It looks a lot worse than it is, because the right side of the equation is actually a repeat of the left! Finding confidence intervals for two populations can be broken down to three steps. Example question: A study revealed that 65% of men surveyed supported the war in Afghanistan and 33% of women supported the war. If 100 men and 75 women were surveyed, find the 90% confidence interval for the data’s true difference in proportions.
Find the following variables from the information given in the question: n1 (population 1)=100
Phat1 (population 1, positive response): 65% or 0.65 Qhat1 (population 1, negative response): 35% or 0.35 n2(population 2)=75 Phat2 (population 2, positive response): 33% or 0.33 Qhat2 (population 2, negative response): 67% or 0.67
Find z__α/2 (If you’ve forgotten how to find α/2, see the directions in: How to Find a Confidence Interval for a Proportion above) zα/2\=0.13
Enter your data into the following formula and solve:
If formulas scare you, here’s the step-by-step to solve the equation (refer back to step 1 for the variables):
multiply phat1 and qhat1 together (.65 x .35 = .2275)
divide your answer to (1) by n1. Set this number aside. (.2275 x 100=.00275)
multiply phat2 and qhat2 together (.33 x .67=.2211).
divide your answer to (3) by n2 (.2211/75=.002948).
Add (3) and (4) together (.00275 + .002948=.005698)
Take the square root of (5): (sqrt.005698=.075485)
Multiply (6) by zα/2 found in Step 2. (.075485 x 0.13=.0098). Set this number aside.
Subtract phat2 from phat1 (.65-.33=.32).
Subtract (8) from (7) to get the left limit (.32-0.0098 = 31.9902)
Add (7) to (8) to get the right limit (.32+.0.0098=32.0098)
That’s it!
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Confidence Interval for the Mean in Excel
CI for the mean formula.
A confidence interval for the mean is a way of estimating the true population mean. Instead of a single number for the mean, a confidence interval gives you a lower estimate and an upper estimate.
For example, instead of “6” as the mean you might get {5,7}, where 5 is the lower estimate and 7 is the upper. The narrower the estimate, the more precise your estimate is. The equations involved in statistics often involve a lot of minor calculations (such as summation), plus you would also need to calculate the margin of error and the mean of the sample.
It’s very easy for errors to slip in if you calculate the confidence interval by hand. However, Excel can calculate the mean of the sample, the margin or error and confidence interval for the mean for you. All you have to do is provide the data —which for this technique must be a sample greater than about 30 to give an accurate confidence interval for the mean.
Example problem: Calculate the 95 percent confidence interval for the mean in Excel using the following sample data: 2, 5, 78, 45, 69, 100, 34, 486, 34, 36, 85, 37, 37, 84, 94, 100, 567, 436, 374, 373, 664, 45, 68, 35, 56, 67, 87, 101, 356, 56, 31.
Type your data into a single column in Excel. For this example, type the data into cells A1:A31.
Click the “Data” tab, then click “Data Analysis,” then click “Descriptive Statistics” and “OK.” If you don’t see Data Analysis, load the Excel data analysis toolpak.
Enter your input range into the Input Range box. For this example, your input range is “A1:A31”.
Type an output range into the Output Range box. This is where you want your answer to appear. For example, type “B1.”
Click the “Summary Statistics” check box and then place your chosen confidence level into the ‘Confidence Level for Mean’ check box. For this example, type “95”.
Click “OK. “
Microsoft Excel will return the confidence interval for the mean and the margin of error for your data. For this sample, the mean (Xbar) is 149.742 and the margin of error is 66.9367. So the mean has a lower limit of 149.742-66.936 and an upper limit of 149.742+66.936. That’s it!
Warning: A 99 percent confidence interval doesn’t mean that there’s a 99 percent probability that the calculated interval has the actual mean. Your sample is either going to contain the actual mean, or it isn’t. Over the long-term, if you ran tests on many, many samples, there is a 99 percent probability that the calculated intervals would contain the true mean.
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TI 83 Confidence Interval: Two Populations
Statistics about two populations is incredibly important for a variety of research areas. For example, if there’s a new drug being tested for diabetes, researchers might be interested in comparing the mean blood glucose level of the new drug takers versus the mean blood glucose level of a control group. The confidence interval(CI) for the difference between the two population means is used to assist researchers in questions such as these.
The TI 83 allows you to find a CI for the difference between two means in a matter of a few keystrokes.
Example problem: Find a 98% CI for the difference in means for two normally distributed populations with the following characteristics: x1 = 88.5 σ1 = 15.8 n1 = 38 x2 = 74.5 σ2\= 12.3 n2 = 48
Step 1: Press STAT, then use the right arrow key to highlight TESTS. Step 2: Press 9 to select 2-SampZInt….
Step 3: Right arrow to Stats and then press ENTER. Enter the values from the problem into the appropriate rows, using the down arrow to switch between rows as you complete them.
Step 4: Use the down arrow to select Calculate then press ENTER.
The answer displayed is (6.7467, 21.253). We’re 98% sure that the difference between the two means is between 6.7467 and 21.253.
That’s it!
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How to Find a Confidence Interval for Population Proportion, p on the TI 83
Example problem: A recent poll shows that 879 of 1412 Americans have had at least one caffeinated beverage in the last week. Construct a 90% confidence interval for p, the true population proportion. Note:
“x” is the number of successes and must be a whole number. Successes in this question is how many Americans have had at least one caffeinated beverage (879). If you are given p̂ instead (the sample proportion), multiply p̂ by n to get x (because x=n p̂).
“n” is the number of trials.
Step 1: Press STAT.
Step 2: Right arrow over to “TESTS.”
Step 3: Arrow down to “A:1–PropZInt…” and then press ENTER.
Step 4: Enter your x-value: 879.
Step 5: Arrow down and then enter your n value: 1412.
Step 6: Arrow down to “C-Level” and enter .90. This is your confidence level and must be entered as a decimal.
Step 7: Arrow down to calculate and press ENTER. The calculator will return the range (.6013, .64374) That means the 98 percent CI for the population proportion is between 0.6013 and .64374.
Tip: Instead of arrowing down to select A:1–PropZInt…, press Alpha and MATH instead. Back to Top
How to Find a Confidence Interval on the TI 83 for the Population Mean
If you don’t know how to enter data into a list, you can find the information in this article on TI 83 cumulative frequency tables. Watch the video for an overview of how to do this, or read on below.
Can’t see the video? Click here to watch it on YouTube.
Example problem: 40 items are sampled from a normally distributed population with a sample mean x̄ of 22.1 and a population standard deviation(σ) of 12.8. Construct a 98% confidence interval for the true population mean.
Press STAT, then right arrow over to “TESTS.”
Press 7 for “Z Interval.”
Arrow over to “Stats” on the “Inpt” line and press ENTER to highlight and move to the next line, σ.
Enter 12.8, then arrow down to x̄.
Enter 22.1, then arrow down to “n.”
Enter 40, then arrow down to “C-Level.”
Enter .98. Arrow down to “calculate” and then press ENTER. The calculator will give you the result of (17.392, 26.808) meaning that your 98% confidence interval is 17.392 to 26.808. This is the same as: 17.392 > μ > 26.808.
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How to Find a Confidence Interval for the Mean on the TI 89
Example problem #1 (known standard deviation): Fifty students at a Florida college have the following grade point averages: 94.8, 84.1, 83.2, 74.0, 75.1, 76.2, 79.1, 80.1, 92.1, 74.2, 64.2, 41.8, 57.2, 59.1, 65.0, 75.1, 79.2, 95.0, 99.8, 89.1, 59.2, 64.0, 75.1, 78.2, 95.0, 97.8, 89.1, 65.2, 41.9, 55.2. Find the 95% confidence interval for the population mean, given that σ = 2.27.
Press APPS and scroll to Stats/List Editor. ENTER.
Press F1 then 8. This clears the list editor.
Press ALPHA ) 9 2 to name the list “CI2.”
Enter your data in a list. Follow each number with the ENTER key: 94.8, 84.1, 83.2, 74.0, 75.1, 76.2, 79.1, 80.1, 92.1, 74.2, 64.2, 41.8, 57.2, 59.1, 65.0, 75.1, 79.2, 95.0, 99.8, 89.1, 59.2, 64.0, 75.1, 78.2, 95.0, 97.8, 89.1, 65.2, 41.9, 55.2.
Press F4 then 1.
Enter “ci” in the “List” box: ALPHA key then ) 9 2.
Enter 1 in the frequency box. Press ENTER. This should give you the mean (xbar, the first in the list) = 75.033.
Press ENTER. Press 2nd F7 1 ENTER. This brings up the z-distribution menu.
Press the right arrow key then the down arrow to select a “Data Input Method” of “Stats.” Press ENTER.
Enter your σ from the question (in our case, 2.27), xbar from Step 7 (75.3033), n = 30 and the Confidence Interval from the question (in our example, it’s .95).
Press ENTER and read the results. The “C Int” is {74.49,76.123}. This means we are 95% confident that the population mean falls between 74.49 and 76.123.
That’s it!
Example problem #2 (unknown standard deviation): A random sample of 30 students at a Florida college has the following grade point averages: 59.1, 65.0, 75.1, 79.2, 95.0, 99.8, 89.1, 65.2, 41.9, 55.2, 94.8, 84.1, 83.2, 74.0, 75.1, 76.2, 79.1, 80.1, 92.1, 74.2, 59.2, 64.0, 75.1, 78.2, 95.0, 97.8, 89.1, 64.2, 41.8, 57.2. What is the 90% confidence interval for the population mean?
Press APPS. Scroll to the Stats/List Editor and press ENTER.
Press F1 8 to clear the editor.
Press ALPHA ) 9 to name the list “CI.”
Enter your data in a list. Follow each number with the ENTER key: 59.1, 65.0, 75.1, 79.2, 95.0, 99.8, 89.1, 65.2, 41.9, 55.2, 94.8, 84.1, 83.2, 74.0, 75.1, 76.2, 79.1, 80.1, 92.1, 74.2, 59.2, 64.0, 75.1, 78.2, 95.0, 97.8, 89.1, 64.2, 41.8, 57.2.
Press F4 1.
Enter “ci” in the List box: Press ALPHA ) 9.
Enter1 in the frequency box. Press ENTER. This should give you the sample standard deviation, sx = 15.6259, n = 30, and x (the sample mean) = 75.033.
Press ENTER. Press 2nd F2 2.
Press the right arrow key then the down arrow to select a “Data Input Method” of “Stats.” Press ENTER.
Enter your x, sx and n from Step 7. In our example, sx = 15.6259. n = 30 and x = 75.033. Enter the Confidence Interval from the question (in our example, it’s .9).
Press ENTER and read the results. The C Int is {70.19,79.88} which means that we are 90% confident that the population mean falls between 70.19 and 79.88.
That’s it!
Tip: If you know σ, use ZInterval instead of TInterval.
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How to find a Confidence Interval for a Proportion on the TI 89
Example problem #1: In a simple random sample of 295 students, 59.4% of students agreed to a tuition increase to fund increased professor salaries. What is the 95% CI for the proportion in the entire student body who would agree?
Press APPS and scroll down to Stats/List Editor. Press ENTER. If you don’t see the Stats/List editor, download it HERE from the TI-website. You’ll need the graphlink cable that came with your calculator to transfer the software.
Press 2nd F2 5 for the 1-PropZInt menu.
Figure out your “successes.” Out of 295 people, 59.4% said yes, so .694 × 295 = 175 people.
Enter your answer from Step 3 into the Successes,x box: 175.
Scroll down to n. Enter 295, the number in the sample.
Scroll down to C Level. Enter the given confidence level. In our example, that’s .95. Press ENTER twice.
Read the result. The calculator returns the result C Int {.5372, .6493}. This means that you are 95% confident that between 54% and 65% of the student body agree with your decision.
Tip: If you are asked for a folder when entering the Stats Editor, just press Enter. It doesn’t matter which folder you use.
Warning: Make sure your round your “success” entries to the nearest integer to avoid a domain error.
Example problem #2: A recent poll in a simple random sample of 986 women college students found that 699 agreed that textbooks were too expensive. Out of 921 men surveyed by the same manner, 750 thought that textbooks were too expensive. What is the 95% confidence interval for the difference in proportions between the two populations?
Step 1: Press APPS, scroll to the Stats/List Editor, and press ENTER.
Step 2: Press 2nd F2 6 to reach 2-PropZint.
Step 3: Enter your values into the following boxes (Use “women” for population 1 (x1 and n1) and “men” for population 2 (x2 and n2)):
Successes, x1: 590
n1: 796
Successes, x2: 548
n2: 800
C Level: 0.95
Step 4: Press ENTER. Step 5: Read the result. The confidence interval is displayed at the top as C Int { .0119,.10053}. This means that your confidence interval is between 1.19% and 10.05%.
Tip: As long as you keep track of which population is x1/n1 and x2/n2, it doesn’t matter which is entered in which box.
You must enter a whole number here, or you’ll get ERR:DOMAIN. You’ll come across this common type of problem in elementary stats: find a confidence interval given a large random sample and the number of “successes” in that sample.
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The 95% Confidence Interval Explained
The terms confidence level and confidence interval are often confused. A 95% confidence level means is that if the survey or experiment were repeated, 95 percent of the time the data would match the results from the entire population. Sometimes you just can’t survey everyone because of time or cost (think about how much it would cost to do a telephone survey of over 300 million Americans!).Therefore, you take a sample of the population.
Having a 95% confidence level means that you’re almost certain your results are the same as if you had surveyed everyone. A 95% confidence interval gives you a very specific set of numbers for your confidence level. For example, let’s suppose you were surveying a local school to see what the student’s state test scores are. You set a 95% confidence level and find that the 95% confidence interval is (780,900). That means if you repeated this over and over, 95 percent of the time the scores would fall somewhere between 780 and 900.
Image: WUSTL.EDU
The above image shows a 95% confidence interval on a normal distribution graph. The red “tails” are the remaining 5 percent of the interval. Each tail has 2.5 percent (that’s .025 as a decimal).
You don’t have to draw a graph when you’re working with confidence intervals, but it can help you visualize exactly what you are doing — especially in hypothesis testing.
If your results fall into the red region, then that’s outside of the 95% confidence level that you, as a researcher, set. If you have a small sample or if you don’t know the population standard deviation which in most real-life cases is true), then you’ll find the 95% Confidence Interval with a t-distribution.
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Asymmetric Confidence Interval
An asymmetric confidence interval just means that the point estimate doesn’t lie in the exact center of the CI. You can end up with asymmetric CIs for many reasons, including:
You transform your data (for example, using log transformations).
You incorporate random error.
You incorporate systematic bias into the interval:
A positive systematic bias will increase the right side of the interval.
A negative systematic bias will increase the left side of the interval.
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History of Confidence Intervals
The idea of confidence intervals originated in the early 1930s, thanks to the work of Jerzy Neyman and Egon Pearson, who were Polish-American and English statisticians, respectively. They introduced confidence intervals as a method to quantify the uncertainty surrounding a statistical estimate. Before confidence intervals, statisticians relied on hypothesis testing for making inferences about populations.
Hypothesis testing involves comparing a sample statistic to an assumed population parameter. If the sample statistic differs significantly from the assumed population parameter, the null hypothesis is rejected. However, if the sample statistic does not deviate significantly, the null hypothesis remains unchallenged.
Neyman and Pearson contended that hypothesis testing was insufficient for drawing conclusions about populations, as it failed to provide information about the statistical estimate’s uncertainty. Confidence intervals offer a means to quantify this uncertainty by specifying ranges of values likely to include the true population parameter.
The confidence level represents the probability of the confidence interval containing the actual population parameter.
Nowadays, confidence intervals are widely used across various fields, such as economics, finance, medicine, and social science, proving invaluable for making population inferences. Key figures in the history of confidence intervals include:
Jerzy Neyman (1894-1980): Polish-American statistician who co-developed the concept of confidence intervals with Egon Pearson.
Egon Pearson (1895-1980): English statistician who co-developed the concept of confidence intervals with Jerzy Neyman.
Ronald Fisher (1890-1962): English statistician who pioneered the use of hypothesis testing in statistics.
William Sealy Gosset (1876-1937): Irish statistician who created the student t-distribution, a commonly used distribution for estimating confidence intervals.
Confidence intervals have significantly impacted the field of statistics as a versatile and powerful tool, applicable in various settings and valuable for making inferences about populations.
More explanations and Definitions:
Asymptotic Confidence Interval
95% Confidence Interval
Asymmetric Confidence Interval
Clopper-Pearson Exact Method
95 Percent Confidence Interval (Part III of Intro to Statistics)
Wald CI
Wilson CI
Binomial Confidence Intervals.
Z Interval
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6787 | https://medium.com/the-innovation/everything-you-need-to-know-about-covariance-and-independence-and-how-to-determine-the-a809ed29bb44 | Everything you need to know about covariance and independence and how to determine the relationship between the two. | by dapanji.eth | Medium
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Everything you need to know about covariance and independence and how to determine the relationship between the two.
dapanji.eth
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Oct 20, 2020
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Covariance and independence are two commonly used concepts in statistics. As many may know, independence implies Cov = 0 while Cov=0 not always imply independence. Today we will go over both of them and give you a concrete understanding of the relationship between the two. The prerequisite of this article is the basic understanding of linear algebra and expectation.
First, let’s talk about Covariance
Definition: The covariance of x and y is Cov(x,y)=E[(x-Ex)(y-Ey)]
Remark: Cov(x,x)=E((x-Ex)²)=Var(x)
Property: Cov(x,y) = Exy-ExEy. Now let’s prove this property.
Proof: Cov(x,y) = E[(x-Ex)(y-Ey)] = E(xy-xEy-yEx+ExEy) = Exy-ExEy-EyEx+ExEy (by linearity of expectations) =Exy-ExEy.
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Now let’s introduce independence. Recall that if x and y are independent, then Exy=ExEy. Therefore if x and y are independent, Cov(x,y) = Exy-ExEy = 0. Thus, we have shown that independence implies 0 Coviance.
Now we have the question: does Cov(x,y)=0 implies x and y are independent? The answer is no except the time if the two variables are bivariate normal random variables. Let’s show how to bivariate normal random variable has this property.
Let x~N(µ1, σ1), y~N(µ2, σ2), Cov(x,y) = σ12. In order show independence, we need to show the joint distribution is the product of the two marginal distribution: fxy(x,y) = fx(x)fy(y). To calculate the joint pdf(probability density function), we need to use the formula of the joint pdf of the bivariate(x,y) formula and the proof is shown in figure 1.
Press enter or click to view image in full size
figure 1. the proof of independence for bivariate r.v.
Note you might need to refer back to the formula for normal distribution to understand the last two steps.
In conclusion, we have
If (x,y) is bivariate normal r.v, then independence <=>Cov=0.
In general, Independence => Cov = 0.
Hope you learned the concepts and please comment below if you have any questions
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6.10: Size of Ions
6: Structures and Energetics of Metallic and Ionic solids
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Wed, 03 May 2023 22:23:39 GMT
6.10A: Ionic Radii
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Contents
1. Home
2. Bookshelves
3. Inorganic Chemistry
4. Map: Inorganic Chemistry (Housecroft)
5. 6: Structures and Energetics of Metallic and Ionic solids
6. 6.10: Size of Ions
7. 6.10A: Ionic Radii
Expand/collapse global location
Map: Inorganic Chemistry (Housecroft)
Front Matter
1: Basic Concepts- Atoms
2: Basic Concepts- Molecules
3: Introduction to Molecular Symmetry
4: Experimental Techniques
5: Bonding in Polyatomic Molecules
6: Structures and Energetics of Metallic and Ionic solids
7: Acids, bases and ions in aqueous solution
8: Reduction and Oxidation
9: Non-aqueous Media
10: Hydrogen
11: Group 1 - Alkali Metals
12: Goup 2- Alkaline Earth Metals
13: The Group 13 Elements
14: The Group 14 Elements
15: The Group 15 Elements
16: The Group 16 Elements
17: The Group 17 Elements
18: The Group 18 Elements
19: d-Block Metal Chemistry - General Considerations
20: d-Block Metal Chemistry - Coordination Complexes
21: d-Block Metal Chemistry - The First Row Metals
22: d-Block Metal Chemistry - The Heavier Metals
23: Organometallic chemistry- s-Block and p-Block Elements
24: Organometallic chemistry- d-block elements
25: Catalysis and some industrial processes
26: d-Block Metal Complexes- Reaction Mechanisms
27: f-Block Metals- Lanthanides and Actinides
28: Inorganic materials and nanotechnology
29: The trace metals of life
Back Matter
6.10A: Ionic Radii
Last updated May 3, 2023
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6.10: Size of Ions
6.10B: Periodic Trends in Ionic Radii
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Table of contents
1. Introduction
2. Hard-Sphere model
3. Periodic Trends
4. References
5. Outside Links
6. Problems
7. Answers
In a crystal lattice, the ionic radius is a measure of the size of the atom's ion.6 When formed, ionic atoms change in size with respect to their orginal atom. Cation radii will decrease and the anion radii will increase in size compared to their neutral atoms. Questions such as: "What methodology is used by chemists to measure ionic radii?" and "Are there any non-experimental ways to estimate the size of ionic radii?" will be answered in this module. Accordingly, there are many ways to determine ionic radii.
Introduction
Hard-Sphere model
Periodic Trends
References
Outside Links
Problems
Answers
Introduction
In the past, after an atom is ionized, X-ray diffraction is used to measure how much the radius of the atom increased or decreased. However, scientists wanted to use another technique, due to the fact, that X-ray diffraction is difficult to distinguish a boundary between two ions. As a result, the hard sphere model can be used.
Hard-Sphere model
The Hard-Sphere model are impenetrable spheres that do not overlap in space.5 The Hard-Sphere model has been tested by well-known scientists; Lande', Pauling and Goldsmidt. The ion radii measured under crystal state of ionic compound which cations and anions are stacking in pattern as shown below.
Figure 1: Schematic of the hard-sphere model
The Hard-Sphere model can be applied to metallic and ionic compounds such as NaCl, which is shown below.
Figure 2: 3-D hard sphere model of Sodium Chloride, taken with permission from en.Wikipedia.org/wiki/File:So...e-3D-ionic.png
In general, scientists uses formula of Internuclear distance to test out the radii of ion then compared with the ion radii had done on X-ray diffraction:
Internuclear distance (d) = r cation+ r anion 2
To calculate ion radii, Lande used ionic compound under solid state (ex: NaCl). This will minize the distribution of electrons.
Find the radii of anion (r-) atom.
Find internuclear distance (d) between anion and cation.
Use Internuclear distance formula to find the r+.
Figure 3: The Hard sphere model can roughly determine the ion radii.
Periodic Trends
As described earlier, cations are smaller in size compared to their neutral atoms while anions are larger in size.Cations are smaller than its neutral atoms because the positive nuclear charge, which holds the electrons in closer, exceeds the negative charge when a metal atom loses an electron. On the contrary, anions are larger because the electrons are not held as tightly, repulsions of electrons increase, and the electrons spread out more due to nonmetal atoms gaining an electron. Refer to the outside link to learn more about the periodic trends for ionic radii (
References
Kotz, Treichel, Weaver,2006, Chemistry and Chemical Reactivity, Thomson Brooks/Cole, Mason, OH, p.358-364
Housecroft C., and Shappe A., 2008, Inorganic Chemistry 3rd edition, Pearson Education Limited, England, p.162-164
Barrera, M, and FZuloagat. "Determination of the ionic radii by means of the Kohn-Sham potential: Identification of the chemical potential." International journal of quantum chemistry 106.9 (2006):2044-2053.
"Ionic radii for Group 1 and Group 2 halide, hydride, fluoride, oxide, sulfide, selenide and telluride crystals." Dalton transactions 39.33 (2010):7786-.
Outside Links
Problems
What is the most general formula that used to determine the ion radii for hard sphere model?
Find radius for Cacium ion in Calcium Chloride (CaCl 2). List out all the steps (numbers are not necessary)
Determine which is larger:
a) K+ or Cs+?
b) La 3+ or Lu 3+?
c) Ca 2+ or Zn 2+?
Answers
1. Internuclear distance (d) = r cation+ r anion
2. Find the radii of anion (r-) atom; Find internuclear distance (d) between anion and cation; Use Internuclear distance formula to find the r+.
a.) Cs+ b.) La 3+ c.) Ca 2+
6.10A: Ionic Radii is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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6789 | https://www.optometrists.org/general-practice-optometry/optical/guide-to-optical-lenses/guide-to-bifocals-and-multifocals/ | Guide to Bifocals and Multifocals
By Dr. Russel Lazarus
Have you noticed the need to hold your phone, books or restaurant menus farther from your eyes to improve their clarity?
Presbyopia is the most common reason most adults begin to wear eyeglasses. The condition generally develops overtime, beginning at around age 40, and is considered a normal part of the aging process.
Presbyopia causes an ability to focus on near objects, such as words on a page, numbers on a phone, or text on a computer. In a young eye, the flexible lens inside your eye, is able to shift focus between close objects and objects at a distance. However, as you age, this lens thickens and begins to harden— losing its flexibility and affecting your near vision.
If you already wear prescription eyeglasses for distance vision, you will likely be prescribed a pair of bifocals or multifocals to continue to see clearly at all distances.
Bifocals and multifocals for eye focusing problems
Though bifocal and multifocal lenses are typically prescribed for adults with presbyopia, in some cases these lenses may be prescribed for eye teaming or focusing problems that may cause eye strain when reading.
In these cases, the bottom segment of the lens serves to reduce the stress on the eyes by supporting the focusing skills necessary for reading.
Additionally, research shows that wearing bifocals or multifocals may help control myopia progression by reducing the stress on the focusing muscles.
What are bifocal lenses?
Bifocal lenses contain two optical powers to accommodate clear vision for both near and far.
The lens is divided into two segments, the top of the lens contains the distance vision prescription, while the bottom of the lens contains the near vision prescription. Some bifocals contain a bisecting horizontal line between the two lens powers.
Types of bifocal lenses
All bifocals lenses function in the same way. However, the lenses differ in appearance, based on the design of the segment that contains your near vision prescription.
A half-moon/ flat-top/ straight-top/ D segment
A round segment
A ribbon segment– a narrow rectangular area
Executive/ Franklin/ E style- the entire bottom half of the lens
What are trifocal lenses?
Trifocals contain three optical powers— distance, intermediate and near vision, with two segmenting lines to delineate the powers. The intermediate segment is located above the segment for near vision and is helpful for viewing objects at arm’s length, such as a computer, or car dashboard.
While trifocal lenses may be practical to those who require three different vision corrections, they are not commonly prescribed— mainly because of their unfavorable segmenting lines.
What are multifocal lenses?
Multifocal lenses are similar to bifocals, but contain multiple lens powers to provide vision at all distances— close up, intermediate, distance, and any other lens power necessary for vision clarity.
Multifocals gradually blend the lens powers together without a bisecting line, making them more attractive than bifocals. The downside of multifocals is that the clear zone for each part of the lens may be limited.
There are three different types of multifocal lenses:
Standard progressive
Premium progressive
Premium personalized progressive
Contact an eye doctor near you to discuss bifocal lens options.
SEE RELATED: What is Presbyopia?
Find an eye doctor near you
Standard progressive multifocals vs. premium progressive multifocals
While standard and premium progressive lenses will both provide clear vision at three distances (near, intermediate, far), the two lenses differ in the quality of improved vision, as well as affordability.
Standard progressive lenses
These lenses contain a large segment at the bottom of the lens for near vision correction. This type of lens is helpful if you have different optical prescriptions for near and distance vision, and are favorable because of the progressive transition it provides between the two viewing areas.
A disadvantage of this type of lens is that it often requires larger frames to accommodate the larger near vision segment.
Benefits of standard progressive lenses:
Prescribed for children and adults
Lowest cost
Large area for near vision
Appropriate for most prescriptions
Premium progressive lenses
Premium progressive lenses are more closely designed for your individual needs.
These lenses provide a wider viewing area and less visual distortions. These lenses can also be fitted to smaller frames.
Premium progressive lenses are recommended if you wear your eyeglasses for work, driving or full-time. However, they can be more expensive than standard progressive lenses.
Benefits of premium progressive lenses:
Less distortions
Wider viewing area
Smaller frames
Easier adaptation
Improved vision for 3 distances
What are premium personalized progressive lenses?
These lenses are custom designed to meet all of the wearer’s individual needs.
The lenses are created based on your primary daily activities, and offer the best natural vision in all situations. Premium personalized progressive lenses provide pupil size optimization to ensure natural vision both day and night.
These lenses also offer enhanced vision through all segments of the lens, with significantly less distortions, and without the prismatic, swaying effects on the periphery of the lens.
Moreover, using special technology, near and intermediate lens segments provide the easiest adaptation, widest visual width and smoothest focus transition, without influencing distance vision.
Benefits of premium personalizes progressive lenses:
Enhanced vision in all situations
Improved visual comfort
Easy adaptation to new lenses
Fast focus transitioning
Unlimited frame choice
Which lenses are right for me?
Each type of lens has both advantages and disadvantages. It is important to understand the differences between each lens. If you are still uncertain about which lenses to pursue, keep these important factors in mind:
Standard progressive lenses are a great option and at an affordable choice.
If you wear your glasses throughout the day or need customized lenses, premium progressive lenses are the way to go.
The personalized premium progressive lenses can offer an even higher quality lens with additional benefits.
LEARN MORE: Optical and Contact Lenses
Schedule an appointment with an eye doctor to determine the most appropriate option for you.
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6790 | https://www.collinsdictionary.com/us/dictionary/english-thesaurus/bombastic | English
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Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar
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Synonyms of 'bombastic' in American English
bombastic
(adjective)
Synonyms
grandiloquent
grandiose
high-flown
inflated
pompous
verbose
wordy
Synonyms of 'bombastic' in British English
bombastic
(adjective)
the bombastic style of his oratory
Synonyms
grandiloquent
She attacked her colleagues for indulging in `grandiloquent' language.
inflated
Some of the most inflated prose is held up for ridicule.
ranting
windy
I have a horror of turning into a windy old bore.
high-flown
Many personnel were put off by such high-flown rhetoric.
pompous
She winced at his pompous phraseology.
grandiose
Not one of his grandiose plans has ever come to anything.
histrionic
wordy
His speech is full of wordy rhetoric.
verbose
When drunk, he becomes pompous and verbose.
declamatory
She has a reputation for making bold, declamatory statements.
fustian
magniloquent
He waxed magniloquent on the cultivation of turnips.
Copyright © 2016 by HarperCollins Publishers. All rights reserved.
Additional synonyms
in the sense of declamatory
She has a reputation for making bold, declamatory statements.
Synonyms
rhetorical,
theatrical,
inflated,
high-flown,
pompous,
turgid,
bombastic,
discursive,
grandiloquent,
fustian,
orotund,
stagy,
magniloquent
in the sense of grandiose
Definition
impressive, or meant to impress
Not one of his grandiose plans has ever come to anything.
Synonyms
pretentious,
ambitious,
extravagant,
flamboyant,
high-flown,
pompous,
showy,
ostentatious,
bombastic
in the sense of high-flown
Definition
extravagant or pretentious
Many personnel were put off by such high-flown rhetoric.
Synonyms
extravagant,
elaborate,
pretentious,
exaggerated,
inflated,
lofty,
grandiose,
overblown,
florid,
high-falutin (informal),
arty-farty (informal),
magniloquent
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6791 | https://researcher.life/blog/article/simple-random-sampling-definition-methods-examples/ | Skip to content
Educational resources and simple solutions for your research journey
Home » R Discovery » Simple Random Sampling: Definition, Methods, and Examples
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Simple Random Sampling: Definition, Methods, and Examples
Dhanya Alex
Researchers often rely on samples to draw conclusions about a larger population. Allowing researchers to study a subset of the population makes the research process more manageable and cost-effective. Here, we discuss Simple Random Sampling, considered one of the most straightforward and unbiased sampling methods.
Table of Contents
Simple random sampling is employed when the researcher believes that each member of the population has an equal chance of being selected, ensuring the sample’s representativeness. This method is particularly suitable for relatively homogeneous populations and minimizes bias, facilitating the generalization of research findings to the larger population. Simple random sampling can be used in various scenarios across different fields. Here are a few examples of cases where simple random sampling can be applied.
Medical Research: A hospital wants to study the average recovery time of patients who have undergone a specific surgery. To ensure the study is unbiased, researchers use simple random sampling to select 100 patients from the hospital’s database of all patients who had this surgery in the past year. This way, every patient has an equal chance of being included in the study.
Market Research: A company wants to understand customer satisfaction with its new product. The marketing team uses simple random sampling to select 200 customers from a list of all customers who purchased the product. This ensures that the sample represents the entire customer base, avoiding any biases that might come from targeting a specific group.
Education Studies: A school district is interested in evaluating the effectiveness of a new teaching method. Researchers randomly select 10 schools from the district’s total of 50 schools. Then, within each selected school, they randomly choose 50 students to participate in the study. This two-stage simple random sampling helps ensure that the sample is representative of the entire district.
Public Health Surveys: A city health department wants to estimate smoking prevalence among adults in the city. They use simple random sampling to select 1,000 residents from the city’s census data. By doing so, they can obtain a representative sample that accurately reflects the smoking habits of the city’s population.
Environmental Studies: Scientists studying the biodiversity of a forest may use simple random sampling to select 100 plots of land from a larger forest area. By analyzing the selected plots, they can make inferences about the entire forest’s biodiversity without studying every plot.
This article will provide a clear understanding of the importance and practicality of simple random sampling in research. We explore what simple random sampling entails, including its definition and how it is conducted. We will also explore when to use simple random sampling in research and highlight the advantages and disadvantages.
What is Simple Random Sampling?
Simple random sampling is defined as “a sampling technique where each member of the population has an equal and independent chance of being selected“.1
In simple random sampling, the selection of each unit is entirely independent of the selection of any other unit. This allows a fair sample selection mechanism, making simple random sampling suitable for many types of quantitative research. The key features of simple random sampling are as follows:
Equal Probability: Each member of the population has an equal chance of being selected.
Independence: The selection of one member does not influence the selection of another.
Unbiased: This method minimizes the risk of bias, making the sample representative of the population.
Simplicity: The process is straightforward to understand and implement.
When to Use Simple Random Sampling in Research?
Simple random sampling is most appropriate when the following conditions are met: 2
Availability of the Complete Population List: A comprehensive and accessible list of the entire population is available, allowing every member an equal chance of being selected.
Homogeneous Population: The population is relatively uniform in characteristics relevant to the study, reducing the risk of bias and ensuring that the sample accurately represents the whole.
Smaller or Manageable Population Size: The population size is sufficiently small to allow for the practical implementation of the sampling process without excessive time and resource demands.
Need for Unbiased Results: The research aims to produce unbiased results, making it crucial to avoid systematic errors in sample selection.
Knowledge of Statistical Analysis: The study requires precise statistical analysis, and the researcher needs to apply techniques that assume random sampling, such as certain inferential statistics.
How to Do Simple Random Sampling (Step by Step)
The steps involved in simple random sampling are as follows:
1. Define the Population:
Identify the population from which you want to sample. This includes determining the total number of units (people, items, etc.) in the population.
Example: If you’re studying the sleep patterns of students in a university, your population might be all students enrolled in that university.
2. Determine the Sample Size:
Decide how many units you need to include in your sample. This depends on factors like the research objective, population size, and desired level of accuracy.
Example: If the university has 10,000 students and you decide to sample 500 students, 500 is your sample size.
3. Assign Numbers to Each Member:
Give each member of the population a unique identifier (usually a number). This is essential for random selection.
Example: Assign numbers 1 to 10,000 to the students.
4. Randomly Select the Sample:
Use a random method to select the desired number of units. This can be done using random number generators, drawing lots (lottery method), or software designed for random sampling. The students corresponding to these numbers form your sample.
Example: Use a random number generator to pick 500 numbers between 1 and 10,000. The students assigned these numbers are then surveyed about their sleeping habits.
In the given example, simple random sampling ensures that every student in the university has an equal chance of being selected for the study. This helps to ensure that the sample is representative of the entire student population, allowing for accurate conclusions about students’ sleeping patterns.
What are the Advantages of Simple Random Sampling?
Simple random sampling is widely used in research and statistics due to its various advantages.
1. Unbiased Representation:
Each individual in the population has an equal probability of being chosen, which minimizes selection bias and promotes fairness and equality.
It ensures that the sample represents the population accurately.
Reduces the likelihood of favoritism or systematic exclusion.
2. Ease of Use:
Simple to understand and implement.
Requires minimal technical knowledge and statistical tools.
3. High Level of Validity:
Results are highly reliable and valid for making inferences about the population.
It helps in achieving the generalizability of the findings.
4. Data Analysis:
Simplifies the process of data analysis as the statistical formulas are straightforward.
Facilitates the use of various statistical techniques to analyze the data.
5. Flexibility:
Can be applied to any known population size, regardless of nature.
Suitable for small and large populations.
6. Foundation for Advanced Techniques:
Serves as a basis for more complex sampling methods.
Provides a solid foundation for stratified, cluster, and systematic sampling.
7. Reduced Sampling Error:
Random selection helps reduce sampling errors.
Enhances the accuracy and reliability of the results.
These advantages make simple random sampling a preferred choice in many research and statistical applications.
What Are the Limitations of Simple Random Sampling?
Despite its many advantages, simple random sampling has some limitations that researchers need to consider.
1. Complexity in Large Populations:
Identifying and listing every member of a large population can be difficult and time-consuming.
It may require considerable resources to manage large datasets.
2. Not Always Practical:
In some cases, a complete list of the population is not available or up-to-date, making it impossible to implement simple random sampling.
Can be impractical for populations that are geographically dispersed.
3. Homogeneity Issues:
If the population is very homogeneous, simple random sampling may not capture the diversity within the population.
May result in samples that do not adequately reflect subgroups within the population.
4. Sample Size Concerns:
Requires a sufficiently large sample size for large populations to ensure accurate representation and reliability of results.
Small sample sizes can lead to high sampling errors and unreliable results.
5. Implementation Costs:
Can be costly and resource-intensive, especially for large-scale studies.
Requires significant effort in terms of time, money, and human resources to implement properly.
6. Data Collection Challenges:
Gathering data from randomly selected individuals may be difficult if they are unwilling or unable to participate.
Non-response or low response rates can affect the validity of the sample.
7. Risk of Sampling Error:
Random selection does not guarantee that the sample will perfectly represent the population.
Although reduced, there is still a possibility of sampling error.
8. Need for Statistical Knowledge:
Proper implementation and analysis require a good understanding of statistical principles.
Errors in design or execution can compromise the validity of the results.
While simple random sampling is a fundamental and widely used method in research, its limitations must be carefully considered before selection. Researchers need to weigh the practicality, cost, and potential implementation challenges against the need for unbiased, representative samples. In some cases, alternative sampling methods may be more suitable to address specific research needs and constraints.
Key Takeaways
Simple random sampling is a powerful technique used to ensure that every member of a population has an equal chance of being included in the sample.
Simple random sampling is ideal for studies where the population is equally accessible, and no sub-groups need specific representation.
Simple random sampling helps to eliminate bias and provides a representative sample, which is crucial for the validity of research findings.
While it is easy to implement, simple random sampling requires a complete and accurate list of the population and can be time-consuming for large populations.
Despite these limitations, simple random sampling remains a powerful tool for obtaining reliable and valid research data when executed correctly.
References
Levy, P. S., & Lemeshow, S. (2013). Sampling of Populations: Methods and Applications. Wiley.
Pandey, P., & Pandey, M. M. (2021). Research methodology tools and techniques. Bridge Center.
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6792 | https://www.ecfr.gov/current/title-40/chapter-I/subchapter-U/part-1065/subpart-A/section-1065.20 | Site Feedback
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Title 40 Protection of Environment
Chapter I Environmental Protection Agency
Subchapter U Air Pollution Controls
Part 1065 Engine-Testing Procedures
Subpart A Applicability and General Provisions
§ 1065.20
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Part 1065
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42 U.S.C. 7401-7671q.
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70 FR 40516, July 13, 2005, unless otherwise noted.
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§ 1065.20 Units of measure and overview of calculations.
(a) System of units. The procedures in this part generally follow the International System of Units (SI), as detailed in NIST Special Publication 811, which we incorporate by reference in § 1065.1010. The following exceptions apply:
(1) We designate angular speed, fn, of an engine's crankshaft in revolutions per minute (r/min), rather than the SI unit of radians per second (rad/s). This is based on the commonplace use of r/min in many engine dynamometer laboratories.
(2) We designate brake-specific emissions in grams per kilowatt-hour (g/(kW · hr)), rather than the SI unit of grams per megajoule (g/MJ). In addition, we use the symbol hr to identify hour, rather than the SI convention of using h. This is based on the fact that engines are generally subject to emission standards expressed in g/kW · hr. If we specify engine standards in grams per horsepower · hour (g/(hp · hr)) in the standard-setting part, convert units as specified in paragraph (d) of this section.
(3) We generally designate temperatures in units of degrees Celsius ( °C) unless a calculation requires an absolute temperature. In that case, we designate temperatures in units of Kelvin (K). For conversion purposes throughout this part, 0 °C equals 273.15 K. Unless specified otherwise, always use absolute temperature values for multiplying or dividing by temperature.
(b) Concentrations. This part does not rely on amounts expressed in parts per million. Rather, we express such amounts in the following SI units:
(1) For ideal gases, µmol/mol, formerly ppm (volume).
(2) For all substances, cm3/m3, formerly ppm (volume).
(3) For all substances, mg/kg, formerly ppm (mass).
(c) Absolute pressure. Measure absolute pressure directly or calculate it as the sum of atmospheric pressure plus a differential pressure that is referenced to atmospheric pressure. Always use absolute pressure values for multiplying or dividing by pressure.
(d) Units conversion. Use the following conventions to convert units:
(1) Testing. You may record values and perform calculations with other units. For testing with equipment that involves other units, use the conversion factors from NIST Special Publication 811, as described in paragraph (a) of this section.
(2) Humidity. In this part, we identify humidity levels by specifying dewpoint, which is the temperature at which pure water begins to condense out of air. Use humidity conversions as described in § 1065.645.
(3) Emission standards. If your standard is in g/(hp · hr) units, convert kW to hp before any rounding by using the conversion factor of 1 hp (550 ft · lbf/s) = 0.7456999 kW. Round the final value for comparison to the applicable standard.
(e) Rounding. You are required to round certain final values, such as final emission values. You may round intermediate values when transferring data as long as you maintain at least six significant digits (which requires more than six decimal places for values less than 0.1), or all significant digits if fewer than six digits are available. Unless the standard-setting part specifies otherwise, do not round other intermediate values. Round values to the number of significant digits necessary to match the number of decimal places of the applicable standard or specification as described in this paragraph (e). Note that specifications expressed as percentages have infinite precision (as described in paragraph (e)(7) of this section). Use the following rounding convention, which is consistent with ASTM E29 and NIST SP 811:
(1) If the first (left-most) digit to be removed is less than five, remove all the appropriate digits without changing the digits that remain. For example, 3.141593 rounded to the second decimal place is 3.14.
(2) If the first digit to be removed is greater than five, remove all the appropriate digits and increase the lowest-value remaining digit by one. For example, 3.141593 rounded to the fourth decimal place is 3.1416.
(3) If the first digit to be removed is five with at least one additional non-zero digit following the five, remove all the appropriate digits and increase the lowest-value remaining digit by one. For example, 3.141593 rounded to the third decimal place is 3.142.
(4) If the first digit to be removed is five with no additional non-zero digits following the five, remove all the appropriate digits, increase the lowest-value remaining digit by one if it is odd and leave it unchanged if it is even. For example, 1.75 and 1.750 rounded to the first decimal place are 1.8; while 1.85 and 1.850 rounded to the first decimal place are also 1.8. Note that this rounding procedure will always result in an even number for the lowest-value digit.
(5) This paragraph (e)(5) applies if the regulation specifies rounding to an increment other than decimal places or powers of ten (to the nearest 0.01, 0.1, 1, 10, 100, etc.). To round numbers for these special cases, divide the quantity by the specified rounding increment. Round the result to the nearest whole number as described in paragraphs (e)(1) through (4) of this section. Multiply the rounded number by the specified rounding increment. This value is the desired result. For example, to round 0.90 to the nearest 0.2, divide 0.90 by 0.2 to get a result of 4.5, which rounds to 4. Multiplying 4 by 0.2 gives 0.8, which is the result of rounding 0.90 to the nearest 0.2.
(6) The following tables further illustrate the rounding procedures specified in this paragraph (e):
| Quantity | Rounding increment |
--- |
| 10 | 1 | 0.1 | 0.01 |
| 3.141593 | 0 | 3 | 3.1 | 3.14 |
| 123,456.789 | 123,460 | 123,457 | 123,456.8 | 123,456.79 |
| 5.500 | 10 | 6 | 5.5 | 5.50 |
| 4.500 | 0 | 4 | 4.5 | 4.50 |
| Quantity | Rounding increment |
--- |
| 25 | 3 | 0.5 | 0.02 |
| 229.267 | 225 | 228 | 229.5 | 229.26 |
| 62.500 | 50 | 63 | 62.5 | 62.50 |
| 87.500 | 100 | 87 | 87.5 | 87.50 |
| 7.500 | 0 | 6 | 7.5 | 7.50 |
(7) This paragraph (e)(7) applies where we specify a limit or tolerance as some percentage of another value (such as ±2% of a maximum concentration). You may show compliance with such specifications either by applying the percentage to the total value to calculate an absolute limit, or by converting the absolute value to a percentage by dividing it by the total value.
(i) Do not round either value (the absolute limit or the calculated percentage), except as specified in paragraph (e)(7)(ii) of this section. For example, assume we specify that an analyzer must have a repeatability of ±1% of the maximum concentration or better, the maximum concentration is 1059 ppm, and you determine repeatability to be ±6.3 ppm. In this example, you could calculate an absolute limit of ±10.59 ppm (1059 ppm à 0.01) or calculate that the 6.3 ppm repeatability is equivalent to a repeatability of 0.5949008498584%.
(ii) Prior to July 1, 2013, you may treat tolerances (and equivalent specifications) specified in percentages as having fixed rather than infinite precision. For example, 2% would be equivalent to 1.51% to 2.50% and 2.0% would be equivalent to 1.951% to 2.050%. Note that this allowance applies whether or not the percentage is explicitly specified as a percentage of another value.
(8) You may use measurement devices that incorporate internal rounding, consistent with the provisions of this paragraph (e)(8). You may use devices that use any rounding convention if they report six or more significant digits. You may use devices that report fewer than six digits, consistent with good engineering judgment and the accuracy, repeatability, and noise specifications of this part. Note that this provision does not necessarily require you to perform engineering analysis or keep records.
(f) Interpretation of ranges. Interpret a range as a tolerance unless we explicitly identify it as an accuracy, repeatability, linearity, or noise specification. See § 1065.1001 for the definition of tolerance. In this part, we specify two types of ranges:
(1) Whenever we specify a range by a single value and corresponding limit values above and below that value (such as X ±Y), target the associated control point to that single value (X). Examples of this type of range include ±10% of maximum pressure, or (30 ±10) kPa. In these examples, you would target the maximum pressure or 30 kPa, respectively.
(2) Whenever we specify a range by the interval between two values, you may target any associated control point to any value within that range. An example of this type of range is (40 to 50) kPa.
(g) Scaling of specifications with respect to an applicable standard. Because this part 1065 is applicable to a wide range of engines and emission standards, some of the specifications in this part are scaled with respect to an engine's applicable standard or maximum power. This ensures that the specification will be adequate to determine compliance, but not overly burdensome by requiring unnecessarily high-precision equipment. Many of these specifications are given with respect to a flow-weighted mean that is expected at the standard or during testing. Flow-weighted mean is the mean of a quantity after it is weighted proportional to a corresponding flow rate. For example, if a gas concentration is measured continuously from the raw exhaust of an engine, its flow-weighted mean concentration is the sum of the products (dry-to-wet corrected, if applicable) of each recorded concentration times its respective exhaust flow rate, divided by the sum of the recorded flow rates. As another example, the bag concentration from a CVS system is the same as the flow-weighted mean concentration, because the CVS system itself flow-weights the bag concentration. Refer to § 1065.602 for information needed to estimate and calculate flow-weighted means. Wherever a specification is scaled to a value based upon an applicable standard, interpret the standard to be the family emission limit if the engine is certified under an emission credit program in the standard-setting part.
[70 FR 40516, July 13, 2005, as amended at 73 FR 37292, June 30, 2008; 76 FR 57438, Sept. 15, 2011; 79 FR 23753, Apr. 28, 2014]
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6793 | https://www.rae.es/libro-estilo-lengua-espa%C3%B1ola/el-acento | Libro de estilo de la lengua española
Pronunciación y elocución
Elocución
El acento
El acento es la sensación perceptiva que en una palabra resalta una sílaba con respecto a las restantes, de modo que la hace más perceptible para los oyentes. Por ejemplo, en es.ta.ción —la segmentación silábica se representa siempre con puntos—, la sílaba ción se pronuncia acentuada y, por ello, se ve resaltada en relación con es y con ta. Conviene no confundir este acento, que es fonético y se debe a las variaciones en el tono (➤ P-73), en la intensidad (➤ P-49) y en la duración que experimenta la vocal acentuada en cada caso, con el acento ortográfico o tilde, sujeto a unas determinadas reglas de acentuación gráfica (➤ O-58 y ss.).
Realización del acento en el habla. Al resultar más perceptible, es decir, al oírse más, podría pensarse que la sílaba acentuada se pronuncia con una mayor intensidad, pero en realidad esto es solo parcialmente correcto, puesto que el acento, efectivamente, entraña cambios en la intensidad, pero sobre todo conlleva una alteración del tono y a veces también de la duración de la vocal sobre la que recae. Son los valores de estas tres variables, en conjunto, los que determinan si una sílaba es o no acentuada.
Acento prosódico frente a acento ortográfico. El acento prosódico confiere mayor prominencia o relieve a una sílaba frente a las demás de una palabra (➤ P-56). El acento ortográfico o tilde es la marca que llevan en la escritura solo algunas de las sílabas con acento prosódico, siguiendo las reglas académicas de colocación de tilde propuestas al efecto (➤ O-58 y ss.). Por ejemplo, sa.lón lleva acento prosódico en la sílaba lón y también acento ortográfico sobre la o; gra.va.men recibe acento prosódico en la sílaba va, pero no presenta acento ortográfico.
Ubicación del acento en la palabra. El español es una lengua de acentuación semilibre, puesto que el acento puede variar de posición dentro de la palabra, pero solo puede recaer en cualquiera de sus tres últimas sílabas. Cuando, en algunos casos, va situado en la anterior a la antepenúltima es porque la palabra lleva adosados a su parte final uno o más pronombres átonos (➤ O-59): cuéntamelo, presentándomelas. Las formas verbales obedecen a unas reglas de acentuación diferentes de las formas no verbales:
Pautas de acentuación de las formas no verbales. El acento recae siempre sobre una de las tres últimas sílabas, de forma que da lugar a la acentuación aguda (en la última sílaba), a la llana o grave (en la penúltima) o a la esdrújula (en la antepenúltima). Lo más frecuente es que las palabras que terminan en vocal sean llanas (ca.sa) y las que terminan en consonante agudas (a.mor); sin embargo, hay palabras que acaban en vocal y son esdrújulas (tí.pi.co) o agudas (ca.fé), y palabras que terminan en consonante y son llanas (ní.quel) o esdrújulas (ré.gi.men). En español, predomina la acentuación llana sobre todas las demás: alrededor del 80 % de las palabras son llanas.
Pautas de acentuación de las formas verbales. En el tiempo presente de todos los verbos de las zonas sin voseo flexivo (➤ G-41, a), el acento prosódico recae sobre la penúltima sílaba, excepto en la forma vosotros, que va acentuada en la última: vosotros so.ñáis. En los restantes tiempos, el acento se sitúa siempre en el mismo componente de la forma verbal, con independencia de cuál sea la posición silábica con la que quede asociado. Así, por ejemplo, en el imperfecto o en el perfecto de indicativo de los verbos regulares (➤ APÉNDICE 1), el acento se coloca en la sílaba que sigue a la raíz verbal, sin importar su situación en el conjunto de la palabra: so.ña.ba (en la penúltima) / so.ñas.te (en la penúltima) / so.ñá.ba.mos (en la antepenúltima).
Palabras tónicas. Todas las palabras, pronunciadas aisladamente, llevan un acento prosódico que se ajusta a las pautas expuestas en P-58, pero, al ir insertas en la cadena hablada, algunas lo mantienen y otras lo pierden (➤ P-60). Aquellas que conservan el acento prosódico se denominan tónicas. En principio, son tónicos los verbos (estábamos, cantar), la mayoría de los adverbios (bastante), los sustantivos (cárcel, arroz), los adjetivos (terrible), los pronombres interrogativos y exclamativos (quién, cuánto), los demostrativos (esto, aquel), los indefinidos (alguien) y, como su propio nombre indica, los posesivos tónicos (mío) y los pronombres personales tónicos (tú, nosotros), así como las interjecciones (ah).
En contextos neutros (sin que intervengan otros factores relacionados con el hablante o la situación), todas las clases de palabras de significado pleno que transmiten contenidos conceptuales (nombres, adjetivos, verbos, adverbios, pronombres) suelen conllevar acento, esto es, se pronuncian como tónicas.
6. Palabras átonas. Las palabras que pierden su acento prosódico al insertarse en la cadena hablada se denominan átonas. En principio, son átonos los artículos determinados (por ejemplo, en el coche, el acento recae en la primera sílaba del sustantivo, aquí en negrita; el artículo, subrayado, es átono), las conjunciones (quiero que vengas) y preposiciones (con suerte; para ti; cabe exceptuar según y vía), los pronombres personales átonos (me caigo), la mayor parte de los pronombres relativos (el niño, que lloraba, no escuchaba) y los posesivos antepuestos (mi libro). Estas palabras átonas constituyen, junto con los vocablos tónicos a los que acompañan, grupos acentuales, es decir, agrupaciones en las que se pueden integrar diversos vocablos, con un número variable de sílabas, pero solo con un acento; en los ejemplos anteriores: que vengas, con suerte, para ti, que lloraba, mi libro.
En contextos neutros (sin que intervengan otros factores relacionados con el hablante o la situación), las clases de palabras que expresan relaciones gramaticales (artículos, conjunciones y preposiciones) no llevan acento prosódico, es decir, son átonas.
7. Palabras átonas que se acentúan ocasionalmente. Determinadas palabras que son normalmente átonas reciben acento en ciertos contextos. Cuando se trata de entornos no enfáticos ni connotados (➤ P-66, f), la acentuación tiene que ver, por lo general, con la presencia de pausas adyacentes. Por ejemplo, si el hablante introduce una pausa —motivada por una coma u otro signo de puntuación— tras una palabra átona, esta suele recibir acento (Será un edificio de, al menos, tres pisos), lo cual se desaconseja en P-46. Sin embargo, otros casos se ajustan plenamente a la norma. Un ejemplo puede ser el vocablo pues, con acento prosódico cuando va entre dos pausas: Así empezó, pues, mi accidentado viaje (frente a Así pues, no hay solución, en donde pues es átono porque no está rodeado por pausas).
8. Palabras tónicas que pierden el acento ocasionalmente. Algunas palabras habitualmente tónicas se pronuncian como átonas en determinados entornos. En estos casos, los vocablos que no se acentúan constituyen junto con el tónico un grupo caracterizado por un único acento:
1. En ciertos compuestos y elementos similares, el miembro que no ocupa la posición final suele perder su acento en la pronunciación: abrebotellas, pelirrojo, José Carlos, bocarriba, afroamericano, tres mil, unos diez, Nochebuena, cada uno, etc. Según se puede apreciar en estos ejemplos, las combinaciones pueden implicar diferentes clases de palabras: sustantivo + adjetivo; verbo + sustantivo; dos nombres propios, etc. Sin embargo, en otros casos tal desacentuación no ocurre: casa cuna, ciudad jardín, vista cansada.
2. No reciben acento las fórmulas de tratamiento como don/doña cuando acompañan al nombre propio: doña María, don Pepe. Lo mismo ocurre con señor/señora, que no se acentúan cuando forman parte de un vocativo: Venga, señor García, que lo espero. Tampoco es tónico Dios en ¡Dios mío!, ni cielo o vida en ¡cielo mío!, ¡vida mía! De igual manera, algunos adjetivos y nombres antepuestos a otros nombres pierden su acento cuando toda la expresión funciona como un vocativo: buena mujer, doctora Flores, capitán Ochoa.
3. El primer elemento de los nombres geográficos que constan de dos miembros a veces pierde su acento y a veces lo mantiene: Nuevo México, Ciudad Juárez, frente a la Riviera Maya, la Costa Blanca.
Incluso en los enunciados neutros que no están cargados de connotaciones dependientes de la situación o del hablante, las palabras tónicas pueden perder su acento y las átonas pue-den convertirse en acentuadas por causas no siempre obvias. Cuando, además, el acto de habla se ve influido por factores que dependen del entorno en que se produce o del usuario, la variabilidad puede ser mucho mayor.
Para evitar errores, es recomendable tener un conocimiento al menos somero de las circunstancias en las que ocurren estos cambios.
9. Palabras que pueden ser átonas o tónicas dependiendo de su función y de su significado. Sucede también que algunas palabras pueden ser átonas o tónicas, pero su función y su significado cambian en cada caso, por lo que su pronunciación se ha de deducir de la grafía —y del contexto (➤ O-62)—: mí (tónico, pronombre personal) / mi (átono, posesivo), como en Es para mí / Es para mi madre; aún (tónico, adverbio de tiempo o ponderativo) / aun (átono, adverbio equivalente a incluso), como en No vengas aún / Aun queriendo, no pude.
10. Palabras que admiten variantes acentuales. Existe un grupo de palabras que presentan variantes acentuales, es decir, que, por lo que respecta a la posición del acento, admiten ser pronunciadas de dos maneras distintas (e incluso, en escasas ocasiones, de tres):
1. Muchas de ellas son cultismos procedentes directamente del griego o del latín, como por ejemplo elegiaco o elegíaco; quiromancia o quiromancía; hemiplejia o hemiplejía; endoscopia o endoscopía.
2. Algunas proceden de otras lenguas: aeróbic o aerobic, Everest o Éverest.
3. Otras varían por razones muy diversas: élite o elite; travesti o travestí; ibero o íbero; omóplato u omoplato.
4. En palabras como vídeo/video, chófer/chofer, pudin/pudín, icono/ícono, entre muchos otros mencionables, la alternancia acentual se corresponde con áreas geográficas distintas del mundo hispanohablante (➤ O-84, e). En los ejemplos aquí mencionados, la primera forma es la típica del español europeo, mientras que la segunda es la más general en América.
Pueden encontrarse más casos en O-84 y en el GLOSARIO.
La acentuación ortográfica como clave para la prosódica. Algunas personas pueden experimentar cierta dificultad a la hora de saber cómo acentuar prosódicamente determinadas palabras, puesto que la casuística es extensa y muy variada.
Conocer las normas que rigen la utilización del acento ortográfico o tilde (➤ O-58) y saber cómo interpretarlas oralmente puede resultar de mucha ayuda para evitar cometer errores de acentuación al hablar.
Por ejemplo, el plural de carácter es caracteres, el de régimen es regímenes, y el de espécimen es especímenes (➤ G-15, a). Muchos hablantes vacilan acerca de cómo deben pronunciar estos plurales, que conllevan un desplazamiento acentual hacia la derecha con respecto a su correspondiente forma en singular. El dominio de las reglas de acentuación gráfica simplifica el problema y resuelve las dudas.
12. Tipos de acento prosódico. Pueden distinguirse varios tipos de acento prosódico, bien sea por el dominio en el que se aplica, bien sea por la función que cumple:
1. Acento léxico. El acento léxico o de palabra es el que recae en una sílaba concreta de una palabra. Por ejemplo, en cá.ma.ra, el acento de palabra va situado sobre la primera sílaba, ca, que en este caso se marca también con tilde.
2. Acento sintáctico o de frase. El acento sintáctico, de frase, oracional o nuclear es el acento dotado de mayor prominencia en el conjunto del enunciado de que se trate; en español es aquel situado más a la derecha de la unidad: Le ofrecieron una amable acogida.
3. Acento primario. El acento primario es el acento léxico al que se ha venido haciendo referencia, es decir, el que recae sobre la sílaba más prominente de la palabra de que se trate.
4. Acento secundario. Algunos vocablos —típicamente, los adverbios en -mente— presentan también un acento secundario, aunque el realce de la sílaba que lo recibe (aquí subrayada) no es nunca tan marcado con el de la sílaba con acento primario: estupendamente, francamente, etc. Los acentos secundarios aparecen a menudo, en otras palabras, como resultado del énfasis con que se articulen los enunciados (➤ P-66, f).
Conviene marcar correctamente tanto los acentos de palabra (léxicos) como los acentos de frase (➤ P-67). Es poco aconsejable, asimismo, introducir demasiados acentos secundarios en el decurso hablado si no se quiere generar una impresión de afectación y de falta de naturalidad. Es un procedimiento del que no hay que abusar para que ni el ritmo natural del habla (➤ P-69) ni la entonación (➤ P-73) ni el carácter intrínseco del idioma se vean alterados indebidamente.
5. Acento distintivo. El acento puede cumplir una función distintiva cuando diferencia o distingue el significado —y en ocasiones la categoría gramatical— de los vocablos. Por ejemplo, (el) término / (él) terminó / (yo) termino; (yo) viajo / (él) viajó; (la) cántara / (yo o él) cantara / él cantará, etc.
6. Acento enfático. En español existe, además, un acento enfático, es decir, un acento que imprime énfasis al elemento sobre el que recae. El acento enfático puede manifestarse de maneras diferentes:
Tomando la forma de un acento secundario (➤ P-66, d, y P-67) que aparece sobre la sílaba inicial de la palabra, o bien, rítmicamente, en sílabas alternantes partiendo de la que comporta el acento léxico primario hacia la izquierda: constitución, autonómica, compromiso, responsabilidad.
Otras veces el acento enfático se materializa haciendo tónica una palabra que era en principio átona (➤ P-61), para, por ejemplo, establecer un contraste (en este sentido, su valor es contrastivo): me lo debía (a mí), no te lo debía (a ti), en la que los dos pronombres me y te, átonos, se acentúan para recalcar la oposición entre las dos personas a las que se refieren.
En muchas ocasiones el hablante resalta el contenido a su juicio más importante o novedoso de un enunciado mediante el cambio de posición del acento de frase o principal (➤ P-66, b), que, en lugar de ubicarse al final de la unidad de que se trate, se desplaza a otra posición anterior: Enrique fue el que causó el problema. Esta otra modalidad de acento enfático adquiere entonces un valor focalizador (puesto que destaca el foco del enunciado, la palabra informativamente más notoria) y marca la novedad de la información o precisa otros contenidos significativos especiales.
El acento enfático no es frecuente en el habla conversacional cotidiana de muchas zonas, pero aparece con regularidad en la elocución expositiva o didáctica de políticos, conferenciantes o profesores, y es sumamente característico del estilo oral de los medios de comunicación y del de algunos comunicadores en particular.
El acento en los medios orales de comunicación. La preocupación de muchos locutores de radio y televisión por que sus mensajes capten y mantengan la atención de la audiencia los aboca a imprimir un énfasis exagerado a su elocución, el cual no solo no es acorde con la estructura fónica del idioma, sino que además provoca el efecto contrario al deseado: enfatizar todo o casi todo es como no enfatizar nada. Para conseguir dotar de una mayor expresividad a sus comunicaciones, los locutores se sirven, fundamentalmente, de los siguientes procedimientos indebidos:
El realce muy marcado del acento primario de la palabra que se quiere destacar, mediante la elevación de su tono, intensidad y duración: El crimen cometido esta mañana…; Los terroristas se reunían en la casa de uno de ellos.
La sobreacentuación de las palabras enfatizadas añadiendo continuamente acentos secundarios, sobre todo en su sílaba inicial: Se iniciaron conversaciones entre los ministros…; La presidencia del Gobierno…; Las autoridades alemanas…
La acentuación indebida de elementos átonos con la consiguiente ruptura de grupos de palabras en principio inseparables (➤ P-43): Son datos de la Agencia de Meteorología; Se espera que las protestas disminuyan.
Si bien resulta comprensible que los locutores recurran a distintas estrategias para conceder desigual importancia a unos contenidos que a otros, es imprescindible que tales recursos se apliquen de modo correcto para que el resultado no sea precisamente el opuesto al buscado, es decir, que los oyentes confundan lo accesorio con lo fundamental o que rechacen una elocución forzada y ajena al idioma.
14. Recomendaciones generales con respecto al acento:
1. Conocer las reglas ortográficas de acentuación es fundamental para conseguir una buena elocución, lo mismo que cabe afirmar en relación con las normas de puntuación o con las que atañen a las otras variables implicadas.
2. Es conveniente huir siempre de la afectación y buscar la naturalidad, incluso en el estilo profesional. Al igual que en lo concerniente a las demás propiedades que dan forma a una buena elocución, en el caso de la acentuación, lo fundamental es no dejar ningún aspecto a la improvisación.
3. Por lo que se refiere a los profesionales de los medios de comunicación, la recomendación básica es que no generalicen el énfasis mediante excesivas dislocaciones, adiciones, etc., del acento, porque confunde a la audiencia e, incluso, puede provocar su rechazo.
4. Teniendo en cuenta que las reglas de acentuación en español no son fáciles, los textos que leen los locutores de los medios deben presentar un uso correcto de la tilde, puesto que, si no es así, surgirán en la lectura muchas vacilaciones o, incluso, errores.
La intensidad
El ritmo
Real Academia Española y Asociación de Academias de la Lengua Española:
Libro de estilo de la lengua española [en línea], [Consulta: 19/07/2025].
Sobre el Libro de estilo de la lengua española
Información general
Presentación institucional
Índice general
Sinopsis
Presentación: para mejorar el estilo
Guía de uso de la obra
Cuestiones gramaticales
Cuestiones ortográficas
Cuestiones de ortotipografía
Pronunciación y elocución
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Luis Mateo Díez gana el Premio Cervantes
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Visita institucional del director de la RAE y presidente de la ASALE a la República Argentina y a la República de Chile
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Diccionario histórico (1960-1996)
Diccionario de la lengua española (2001)
Diccionario esencial (2006)
Unidad Interactiva del Diccionario (UNIDRAE)
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6794 | https://capillaryswitch.com/liquid-expansion-capillary-thermostat/ | Capillary Thermostats: The Science Behind Liquid Expansion
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Bulb and Capillary Thermostats: The Science Behind Liquid Expansion
on February 15, 2018
Bulb and capillary thermostats are advanced devices that leverage the principles of liquid expansion to control temperatures accurately. Whether you’re using them in HVAC systems, food equipment, or pool and spa controls, these thermostats rely on the same underlying physics to function. This guide dives into the science of how bulb and capillary thermostats work and explores their various applications and benefits. As always, if you have any questions, please do not hesitate to contact us.
The Science of Liquid Expansion in Capillary Thermostats
At the heart of bulb and capillary thermostats lies a fundamental physical principle: liquid expansion. Here’s a detailed look at how this principle operates within the device:
Components and Mechanism
Bulb: The bulb is a larger diameter tube typically mounted within the material flow being monitored. This bulb contains a liquid that responds to temperature changes.
Capillary: Connected to the bulb is a slender tube known as the capillary. The capillary transports the liquid from the bulb to the thermostat’s head.
Head: The head of the thermostat contains a diaphragm or bellow. This component moves in response to changes in the liquid volume caused by temperature fluctuations.
When the temperature of the surrounding environment changes, the liquid inside the bulb expands or contracts. For example:
Temperature Increase: The liquid heats up and expands. This expansion causes the liquid to move through the capillary tube and push against the diaphragm or bellow in the head.
Temperature Decrease: The liquid cools down and contracts. This contraction pulls the liquid back from the diaphragm or bellow, allowing it to return to its original position.
This movement of the diaphragm or bellow, in turn, activates a snap-action switch within the thermostat. Depending on the design, this switch can either open or close an electrical circuit, thereby controlling the connected device.
Key Features and Advancements
Bulb and capillary thermostats are not just simple on-off devices. They incorporate several advanced features to enhance performance and reliability:
Ambient Temperature Compensation: This feature offsets the differential in temperature, ensuring that the thermostat maintains accuracy despite fluctuations in the surrounding environment. This compensation is crucial for maintaining stable temperature control in dynamic conditions.
Snap-Action Switch: The use of a snap-action switch ensures a rapid response to temperature changes. This quick action minimizes the lag between detecting a temperature change and triggering the corresponding control action.
Applications and Practical Uses
Due to their precise control and reliability, bulb and capillary thermostats find applications in a variety of fields. Here are some common uses:
Food Equipment: These thermostats serve as limit or control devices in ovens, refrigerators, and other food-processing equipment. They help maintain safe and consistent temperatures crucial for food safety and quality.
HVAC Systems: In heating, ventilation, and air conditioning systems, bulb and capillary thermostats regulate the temperature of the air being conditioned. They ensure that indoor environments remain comfortable and energy-efficient.
Pool and Spa Controls: These thermostats maintain the desired water temperature in pools and spas, providing a consistent and enjoyable experience for users.
Industrial Equipment: In industrial settings, they control temperatures in machinery and processes, ensuring operational safety and efficiency.
Quality and Reliability
The design and manufacturing process of these thermostats are critical to their performance. High-quality materials and rigorous testing protocols ensure that each device operates reliably under various conditions. At Stemco, a division of Senasys, we prioritize quality control, with each thermostat undergoing 100% individual testing before leaving our facility.
Understanding the science behind bulb and capillary thermostats reveals why they are so effective in various temperature control applications. The principle of liquid expansion, coupled with advanced features like ambient temperature compensation and snap-action switches, makes these devices indispensable in many industries. By ensuring precise and reliable temperature control, they contribute significantly to safety, efficiency, and comfort in numerous applications.
Related posts:
Mounting a Capillary ThermostatSTEMCO Capillary SwitchesTemperature Control (Fan) vs. Limit SwitchUnderstanding the Importance of Capillary Thermostat Tube Fill in Temperature Regulation
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Bulb & Capillary Thermostats
Original equipment manufacturers worldwide use thermostat switches designed and manufactured by STEMCO, a Division of Senasys. No matter which style or model you choose, all of the capillary thermostat switches operate using the same physics. A fluid is encapsulated in a metal and when the temperature changes the fluid either expands or contracts. This fluid change moves a diaphragm which is in touch with a snap action switch and either opens or closes a circuit.
Ambient Compensation
Our capillary thermostats can be ambient compensated to offset the differential in temperature which keeps our thermostat controls more accurate as it’s environment changes.
Custom Options
Capillary thermostat switches built by Senasys can also be built to customer specifications. This insures you get the exact control you want without paying for features you don’t need.
Learn more
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6795 | https://habr.com/ru/articles/128787/ | Дерево отрезков / Хабр
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Я расскажу о структуре под названием дерево отрезков и приведу его простую реализацию на языке С++. Эта структура весьма полезна в случаях, когда необходимо часто искать значение какой-то функции на отрезках линейного массива и иметь возможность быстро изменять значения группы подряд идущих элементов.
Типичный пример задачи на дерево отрезков:
Есть линейный массив, изначально заполненный некоторыми данными. Далее приходят 2 типа запросов:
1й тип — найти значение максимального элемента на отрезке массива [a..b].
2й тип — заменить iй элемент массива на x.
Возможен запрос «добавить х ко всем элементам на отрезке [a..b]», но в данной статье я его не рассматриваю.
С помощью дерева отрезков можно искать не только максимум чисел, но и любую функцию, удовлетворяющую свойству ассоциативности.
Это ограничение связано с тем, что используется предпросчет значений для некоторых отрезков.
Так же можно искать, например, не значения, а порядковые номера элементов.
Желательно, что бы функция имела «нейтральный» аргумент, который не оказывает влияния на результат. Например, для суммы это число 0: (a + 0 = a), а для максимума это бесконечность: max(a, -inf) = a.
Итак, поехали.
Самый простой (и медленный) способ решать представленную выше задачу, это завести линейный массив, и покорно делать то, что от нас хотят.
при такой реализации время нахождения ответа на запрос имеет порядок О(n). в среднем, придется пройтись по половине массива что бы найти максимум. Хотя есть и положительные моменты — изменение значения какого-либо элемента требует O(1) времени. Этот алгоритм можно ускорить в 2 раза, если выполнить небольшой предпросчет: для каждой пары элементов найдем значение максимального из них, и запишем в другой массив. Тогда при поиске максимума на отрезке, для каждой пары элементов уже известно значение большего, и сравнивать придется только с ним. остается только аккуратно проверить граничные элементы, так как граница запрашиваемого отрезка не обязательно четная.
На рисунке выделены элементы, которые необходимо проверять.
Понятно, что над этими массивами можно ввести еще один, что бы поиск был еще в 2 раза быстрее, а над ним еще один… и так до тех пор, пока самый верхний массив не будет состоять из одного элемента. Как не трудно догадаться, значение единственного элемента в самом верхнем массиве – это значение максимального элемента.
Некоторые пояснения: число рядом с вершиной дерева — это положение этой вершины в реальном массиве. При такой реализации хранения дерева очень удобно искать предка и потомков вершины: предок вершины i имеет номер i/2, а потомки номера i2 и i2+1. Из рисунка видно, что необходимо, что бы длинна массива была степенью двойки. Если это не так, то массив можно в конце дополнить нейтральными элементами. Расход памяти на хранение структуры от 2n до 4n, (n — количество элементов).
Алгоритм поиска «сверху» (есть еще и «снизу») весьма прост и в понимании и в реализации (хотя тех, кто не знаком с рекурсией, это всё может озадачить).
Алгоритм таков:
Начинаем опрос с вершины 1 (самая верхняя).
1.пусть текущая вершина знает максимум на промежутке l..r.
«пересекаются области [a..b] и [l..r] ?»
возможные варианты:
a. вообще не пересекаются.
что бы не влиять на результат вычисления, вернем нейтральный элемент (-бесконечность).
b. область [l..r] полностью лежит внутри [a..b].
вернуть предпросчитанное значение в текущей вершине.
с. другой вариант перекрытия областей.
спросить то же самое у детей текущей вершины и вычислить максимум среди них (смотрите код, если непонятно).
Как видно, алгоритм короткий, но рекурсивный. Временная сложность O(logN), что намного луче, чем О(N). например, при массиве длинной 10^9 элементов, необходимо примерно 32 сравнения.
Изменить число в этой структуре еще проще — надо пройти по всем вершинам от заданной до 1й, и если значение в текущей меньше чем новое, то заменить его. Это так же занимает O(log N) времени.
Реализация алгоритма.
Предполагается, что количество элементов массива не более 1024 (номера 0..1023).
```
include
include
using namespace std;
define INF 1000000000 // предпологаем, что чисел больше такого не будет.
define TREE_REAL_DATA 1024 // максимальное допустимое количество элементов
int tree_data[TREE_REAL_DATA 2];
// основная функция поиска.
// аргументы: p - текушая вершина(пронумерованы согласно рисунку).
// l, p - левая и правая границы отрезка, для которого tree_data[p] является максимумом.
// вообще можно было обойтись без этих параметров, и определять их исходя из p, но так проще и понятней.
// a, b - левая и правая границы отрезка, для которого необходимо найти минимум.
int __tree_find_max(int p, int l, int r, int a, int b)
{
if (b < l || r < a) return -INF;
if (a <= l && r <= b) return tree_data[p];
int r1 = __tree_find_max(p2 , l, (l+r) / 2, a, b); // опрос левого предка
int r2 = __tree_find_max(p2+1, (l+r)/2 + 1, r, a, b); // опрос правого предка
return max(r1, r2); // нахождение большего из левого и правого поддеревьев
}
// более юзабильная оболочка для функции выше.
int tree_find_max(int a, int b)
{
return __tree_find_max(1, 0, TREE_REAL_DATA - 1, a, b);
}
// обновление элемента № р.
void tree_update(int p, int x)
{
p += TREE_REAL_DATA; // преобразование позиции p к позиции в самом нижнем массве,
// в котором реально находится массив со значениями.
tree_data[p] = x;
for(p/=2; p ; p/=2) {
if (tree_data[p 2] > tree_data[p 2 + 1])
tree_data[p] = tree_data[p 2];
else tree_data[p] = tree_data[p 2 + 1];
}
}
// простейшая инициализация - установка всех значений в -INF
void tree_init()
{
for (int i = 0; i < TREE_REAL_DATA 2; i++)
tree_data[i] = -INF;
}
int main()
{
tree_init();
while ( 1 ){
char c;
scanf("%c", &c);
if (c == 'Q') return 0; // выход
if (c == 'F') { // найти максимум на интервале a..b
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", tree_find_max(a, b));
}
if (c == 'U') { // установить значение элемента p равым x.
int p, x;
scanf("%d%d", &p, &x);
tree_update(p, x);
}
}
}
```
Вот, в общем, и все, что необходимо в первую очередь знать о дереве отрезков.
Для разнообразия можно еще разобраться с алгоритмом вычисления «снизу» (он, кстати, нерекурсивный), хотя я нахожу его менее симпатичным. Ну и, конечно же, стоит разобраться с быстрым добавлением суммы ко всем элементам на отрезке (тоже за O(log N)), но это будет слишком утомительно для человека, который впервые разбирается с деревом отрезков.
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6796 | https://forum.thegamecreators.com/thread/179559 | Determining whether a 3D point is within a square based pyramid - GameCreators Forum
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Thanks, I've read this.
DarkBASIC Professional Discussion / Determining whether a 3D point is within a square based pyramid
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Michael P
19
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Joined: 6th Mar 2006
Location: London (UK)
Posted: 28th Dec 2010 21:08Edited at: 28th Dec 2010 21:08
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I have a square based pyramid, and I know the coordinates of all 5 points of the pyramid (the top of the pyramid and the 4 corners of the base). The pyramid can be rotated at any angle in 3D space and can be of any dimensions.
I want to know how I can determine whether a point is within this pyramid using mathematics.
This pyramid represents the area which the camera can see where the top corner is the camera and the base is at the end of the camera's range. It is generated by (more information on this is here):
Determining the positions of the 4 corners of the base when the camera is positioned at 0,0,0 facing at angles 0,0,0.
Generating a left handed ZYX rotation matrix from the angles that the camera is facing.
Transforming each base corner's position vector by this rotation matrix.
Adding the camera's position vector to each base corner's vector.
Here is a picture that might help:
Once again this is in aid of solving this problem; I am now very close to finding a solution!
DarkNet - multiplayer plugin
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Phaelax
DBPro Master
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Location: Metropia
Posted: 28th Dec 2010 22:01Edited at: 28th Dec 2010 22:02
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Easiest way I can think of is to check each plane that makes up the pyramid against your 3D point to see which side of the plane the point lies. Doing this we can determine if its enclosed by all 4 sides (5 counting the base)
First you need to calculate the normal of each plane. For each of the 4 sides, you have 3 points we'll call A, B, C. The cross product of vectors AB and AC will create a vector perpendicular to the plane. You might need to normalize this vector for the next part to work, I can't remember for sure exactly.
Next, take your 3D and calculate the dot product with the normal of each side. If the dot product is positive on all 5 planes, then the point is outside the pyramid. If its negative for each one, then the point is inside.
I may have the positive and negative switched around, it depends on the rotational order the points, see the attached pic. It's important that each side follows the same order of ABC.
"Only the educated are free" ~Epictetus
"Imagination is more important than knowledge..." ~Einstein
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Sven B
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Location: Belgium
Posted: 28th Dec 2010 22:14Edited at: 28th Dec 2010 22:15
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By using the view matrix and projection matrix of the camera you're using, it's quite easy to do this though.
Let's say V = view matrix and P = projection matrix.
If Y = X V P, where X is the vector you want to test, then what you need to know is right in there.
If the point is in the pyramid, then it means the camera can see it (which is my guess of what you want to do), so you basically need to check whether or not Y.x and Y.y are in the 'screen' (depending on which width/height you pick), and you also need to check the Z-depth after projection.
DBP can build all these matrices for you using the 3D math commands.
Extra info on this:
Sven B
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OldPMan
TGC Store Seller
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Posted: 28th Dec 2010 22:47
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I can only assume that there is such an option.
By checking the distance from the point to all vertices of the pyramid.
can be
.....already beside..... for all
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Diggsey
19
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Location: On this web page.
Posted: 28th Dec 2010 23:32
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The viewing region is actually a frustum rather than a pyramid (as there is a near plane as well as a far plane). Check out this code snippet from LiT which shows how to generate the six planes of the frustum from a camera, and also how to check if different shapes are inside it. (Remember, a point is just a sphere with a radius of zero)
[b]
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Michael P
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Location: London (UK)
Posted: 29th Dec 2010 19:15Edited at: 29th Dec 2010 19:17
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Thanks guys, I used Phaelax's method and it seems to work well!
I formed 4 triangles (left, right, bottom and top) and calculated their distance from the point being checked. If the left triangle distance is positive, right triangle distance is negative, top triangle distance is negative and bottom triangle distance is positive then the point is inside the square based pyramid.
The distance function is (C++ code):
float distance = (GetEquationalRepresenationD() - n->GetInnerProduct(Q)) / n->GetDistance();
GetEquationalRepresenationD refers to converting from the parametric to equational representation of the plane and is the letter d in the formula: ax1 + bx2 + cx3 = d. n is the normal form of the plane and Q is a point within the plane.
@ Diggsey
This is true but when the near value is 1 (which it is in my case) it is pretty much a square based pyramid.
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Neuro Fuzzy
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Posted: 29th Dec 2010 21:07Edited at: 29th Dec 2010 21:29
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svenB... I'm really confused now whether to use a column or a row vector... I had always used column vectors (not that I'm great with 3d matrix math).
[edit]
well this code kinda works
+ Code Snippet Copy
make object sphere 1,1
position object 1,rnd(100),.5,rnd(100)
make matrix 1,100,100,10,10
do
move camera .01((mouseclick()=1)-(mouseclick()=2))
yrotate camera camera angle y()+mousemovex().1
xrotate camera camera angle x()+mousemovey().1
if pointInCamera(object position x(1), object position y(1), object position z(1),50)
color object 1,rgb(0,255,0)
else
color object 1,rgb(255,0,0)
endif
sync
loop
function pointInCamera(x as float, y as float, z as float,range as float)
null=make matrix4(1)
null=make matrix4(2)
null=make matrix4(3)
null=make vector4(4)
null=make vector4(5)
set vector4 5,x,y,z,1
view matrix4 1
projection matrix4 2
multiply matrix4 3,1,2
transform vector4 4,5,3
if x vector4(4)>=0 and x vector4(4)=0 and y vector4(4)0 and z vector4(4)<range
circle x vector4(4),y vector4(4),20
exitfunction 1
endif
endfunction 0
but... i don't get it :S
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Sven B
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Location: Belgium
Posted: 29th Dec 2010 22:20Edited at: 30th Dec 2010 09:35
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Neuro Fuzzy,
When changing between the two, just inverse the order of matrix multiplication. I'm used to working with column vectors too, but Direct3D works with a row vector. I'm also used to working with a Right Handed axis having the Z-axis pointing up, but Direct3D likes a Left Handed axis having the Y-axis pointing up better.
Really, doesn't make things easier .
Anyways, here's an example on how to get if a vector is in the frustrum or not:
+ Code Snippet Copy
sync on
sync rate 0
`These constants apply to your viewing frustrum
constant Zn 1.0 : `Near clipping plain
constant Zf 1000.0 : `Far clipping plain
set camera range Zn, Zf
make object sphere 1,1
position object 1,rnd(100),.5,rnd(100)
make matrix 1,100,100,10,10
do
move camera .01((mouseclick()=1)-(mouseclick()=2))
yrotate camera camera angle y()+mousemovex().1
xrotate camera camera angle x()+mousemovey().1
if pointInCamera(object position x(1), object position y(1), object position z(1), 50)
color object 1,rgb(0,255,0)
else
color object 1,rgb(255,0,0)
endif
sync
loop
function pointInCamera(x as float, y as float, z as float,range as float)
null = make matrix4(1) : View matrix
null = make matrix4(2) :Projection matrix
null = make vector4(3) : `Vector to be transformed and tested
`Get active view/projection matrix
view matrix4 1
projection matrix4 2
`Set vector
set vector4 3, x, y, z, 1.0
`transform that vector (2 transformations is faster than one matrix4 multiplication if you only want to transform 1 vector)
transform vector4 3, 3, 1 : `X = X ViewMatrix
transform vector4 3, 3, 2 : `X = X ProjectionMatrix
`Finally, we test the Z-depth (-Zn < z < range - Zn)
if z vector4(3) < -Zn or z vector4(3) > range - Zn then exitfunction 0
`Homogeneous coordinates! Divide by it's w coordinate to make it right
divide vector4 3, w vector4(3)
`Retrieve data
if x vector4(3) > -1.0 and x vector4(3) < 1.0
if y vector4(3) > -1.0 and y vector4(3) < 1.0
exitfunction 1
endif
endif
endfunction 0
Cheers!
Sven B
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Neuro Fuzzy
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Posted: 29th Dec 2010 23:36Edited at: 30th Dec 2010 00:18
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[edit]
I'm really confused, I think I've been doing some really stupid stuff with DBPro rotation matrices - that's what you get when you educate yourself on the internet xD
Is it true that "multiply matrix4 ret, a, b" actually says ret=AB and not ret=BA? I think that I was... trying... to fool myself into thinking I was using column vectors, and to get the right results, I had to mess up every multiplication operation like that, and fooled myself into thinking multiply matrix4 switches A and B around? (earlier if you had hasked me I would have said ret=BA)
Arrgh, I feel stupid now... Uhh could you please confirm why I think I've been stupid is correct? I'm just not sure any more X_X
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enderleit
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Joined: 30th May 2007
Location: Denmark
Posted: 30th Dec 2010 01:31
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I know nothing of matrix math, but just thought I'd point out that in normal math it doesn't matter which order you multiply two items...
AB = BA
Don't know if there's something different in matrix math...
Martin Enderleit
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Diggsey
19
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Joined: 24th Apr 2006
Location: On this web page.
Posted: 30th Dec 2010 02:04
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There is
Order matters when multiplying matrices, ie. AB ≠ BA
[b]
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Neuro Fuzzy
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Posted: 30th Dec 2010 02:08
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OOh, right, I didn't say it yet... what I thought I had figured out in my post is wrong, "multiply matrix4 ret, a, b", does actually set ret=ba. I've just been thinking some weird stuff and I've had weird consequences in my programs that confused me. Uhhh It's all clear to me now, I've just been confused ._.
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Sven B
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Location: Belgium
Posted: 30th Dec 2010 10:10Edited at: 30th Dec 2010 10:12
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Hi NeuroFuzzy,
First of all: I made a mistake in my code last time. I edited it.
The line
divide vector4 3, z vector4(3) + Zn
had to be
divide vector4 3, w vector4(3)
It was a bit late yesterday...
Quote:"Is it true that "multiply matrix4 ret, a, b" actually says ret=AB and not ret=BA? I think that I was..."
Multiply matrix4 ret, A, B means ret = A B if you're working with row vectors.
Though I advise you to 'think' in row vectors, it is not strictly necessary. Why?
You see, the transposed of a matrix multiplication is the multiplication in reverse order of the transposed matrices.
A B X = X' B' A'
This allows you to have a certain degree of freedom:
So if you were thinking to do X = A B X
X = column vector
B = first transformation
A = second transformation
Then it would translate to (using ret = second first)
+ Code Snippet Copy
multiply matrix4 C, B, A
transform vector4 X, X, C
And if you were thinking X = X B A (which is the same transformation, but with a row vector where B and A are now transposed)
X = row vector
B = first transformation
A = second transformation
Then it would translate to (using ret = first second)
+ Code Snippet Copy
multiply matrix4 C, B, A
transform vector4 X, X, C
which is clearly the same.
So basically, it doesn't have that big of an impact on how you interpret it, but you have to stick with one. However, if you use row vectors, the commands are more logical.
When you want to use column vectors
transform Y, X, A changes to Y = A X
multiply vector4C, A, B changes to C = B A
subtract matrix4C, A, B does not change and stays C = A - B
get matrix4 element(Mat, E) now 'counts' downwards. E = 2 is element (3, 1), E = 4 is element (1, 2)
All other commands stay the same (they're either commutative or the parameters describe what it should be).
When you want to use row vectors, everything is pretty much as you would suspect:
transform Y, X, A is Y = X A
multiply vector4C, A, B is C = A B
subtract matrix4C, A, B is C = A - B
get matrix4 element(Mat, E) counts to the right. E = 2 is the element (1, 3), E = 4 is element (2, 1).
So if you work with row vectors, you basically avoid a few problems, but technically it's possible to work with column vectors as well.
Cheers!
Sven B
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Posted: 30th Dec 2010 12:16Edited at: 30th Dec 2010 12:19
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In case anyone wants a speedier option that won't cull everything exactly but still do a reasonable job you could have a dummy object (a 3D point) in the centre of the pyramid of view (exactly half way between the camera and the camera's range) and check simple vector distances to that point. - This would surely be faster than checking if something lies exactly within a pyramid as that would require 5 checks for each object instead of 1.
The maths being if you had 100 objects to check, that's 500 checks per loop, but with a dummy object in the middle of the camera's view that's only 100 checks per loop. - O(n) instead of O(5n)
Obviously there will be a bigger margin of error the greater the cameras range as the sphere of permitted objects will increase but I'd guess for the saved performance in the checking you could allow some models to creep into the render area outside the edge of the screen.
I drew all over the original diagram to try and illustrate what I mean but it's probably better to just read what I said carefully.
EDIT:
Re-reading the first post I realise this method may actually be better than 5 times faster as it wouldn't need to calculate things like the positions of the points in the pyramid, it's just calculating one point.
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Posted: 30th Dec 2010 15:17
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Quote:"check simple vector distances to that point"
How would that give just one check?
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Posted: 30th Dec 2010 16:11Edited at: 30th Dec 2010 16:15
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!!!
I don't know, I don't do that kind of 3D maths usually but surely it's quicker (regardless of whether it's 1 check) to calculate the distance between 2 points than to calculate whether a point is on 1 side of 5 different planes?
It must be something along the lines of 3D trigonometry between the centre point and every object in the playarea and determining which lengths are within the range cameraview0.5
(I just said vector because I was picturing lines going from that centre point to each object and checking the lengths of those lines to determine if they were inside the view sphere - I don't actually know if it would be practical to use vectors in this kind of application - I'd just use variables and 3D trigonometry) :S
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Posted: 30th Dec 2010 16:25
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It is alot faster (I have just implemented it)!
Firstly we don't have to generate the square based pyramid, but this isn't actually that expensive because it only needs to be done once and can be reused for checking any number of objects.
The main thing is that the distance check is a simple case of Pythagoras which only needs to be done once per point. With the more precise method, checking the distance between a plane and a point is more expensive, and we have to do this 4 times (once for each plane).
DarkNet - multiplayer plugin
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Posted: 30th Dec 2010 16:30
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Quote:"but surely it's quicker (regardless of whether it's 1 check) to calculate the distance between 2 points than to calculate whether a point is on 1 side of 5 different planes?"
I'm sure it is but what does that have to do with the pyramid check? Which two points are you talking about?
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Posted: 30th Dec 2010 16:45
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xD I'm not the worlds best at explaining things, I may have referred to a point but meant a different one.
What I mean overall is:
Checking the distance between each object and the point directly in front of the camera (at half the camera's range away from that camera) would be faster than checking whether each object is on 1 or the other side of 4 or 5 planes.
I did attempt to do some maths to calculate precisely how much faster it would be but given that the maths for checking distance between to 3D points is somewhat different to checking whether a point lies on 1 or the other side of a plane I just gave up and basically said "Yeah it'll be faster but I'm not sure how much faster."
The connection between the pyramid check is that my method will use the same area (if implemented correctly) and a little bit of extra area to achieve much the same effect.
The confusion is going to be somewhere in my explanation as I'll sometimes reference objects as separate points but should really keep calling them objects and only say point when I'm talking about the one positioned in front of the camera.
xD If that explanation still sucks as much as I do I'll code an example or ask Michael P to explain it.
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Posted: 30th Dec 2010 17:23
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My main point is that you are not checking whether the point is inside the pyramid. Depending on the camera's FOV there could be a big difference between the simple sphere check (which is what I think you are doing) and the correct pyramid check.
Your check is certainly faster - but it's not checking the right thing.
Perhaps I've misunderstood what you are doing?
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Posted: 30th Dec 2010 17:32
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Yeah I tried to say that in my first sentence:
Quote:"won't cull everything exactly but still do a reasonable job"
My method isn't doing the same thing but offers a similar and faster alternative.
It gets more inaccurate the larger the camera range but it's checking far less each loop so it's a bit of a compromise.
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Posted: 30th Dec 2010 17:39
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Quote:"it's a bit of a compromise"
A big bit?
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Posted: 30th Dec 2010 17:48Edited at: 30th Dec 2010 18:07
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It depends on how accurate you want it to be.
If you just want to cull stuff that's behind the camera it's fine.
If you do actually need only what is in the camera's view and nothing else, it isn't.
I'm just giving an alternative here for a speed boost in game dev. - You don't need to be as accurate as a pyramid or frustum cull in a game in my experience so a spherical cull of this type would do the job just nicely.
Here's a diagram of the inaccuracies (in green) but the two triangular shaped ones near the base of the pyramid can be removed just by making the sphere bigger and further forward (dotted blue line).
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Posted: 30th Dec 2010 22:05
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Maybe, but that doesn't answer the question Michael P was asking in his first post.
I've just noticed Michael P's last comment that it is indeed faster (despite it's inaccuracies). That's odd because I would have thought the few extra numerical checks would be outweighed by the loss of culling. Perhaps that isn't relevant in what he was doing?
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Posted: 30th Dec 2010 23:07Edited at: 30th Dec 2010 23:14
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Quote:"Maybe, but that doesn't answer the question Michael P was asking in his first post."
I know, I know, but the thread had been more or less solved so I thought I'd change the topic slightly to performance in game development rather than specific maths and offer people a faster alternative for an actual application rather than a maths question. - It wasn't exactly worth me making another thread just to tell people an idea I thought up based on this thread. (And I had no code to put it in the codebase.)
So I just put it up there for people to make of it what they liked.
And it would seems my prediction that Michael P wanted a fast camera culler was correct as he's implemented it and it works faster.
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Posted: 31st Dec 2010 02:56
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You're probably right.
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6797 | https://personal.math.ubc.ca/~njb/not6.pdf | Now we move onto PDEs expressed in spherical coordinates (ρ, θ, ϕ). This will lead us to more Sturm-Liouville problems, to Legendre’s differential equation and polynomials, and then to spherical harmonics. For this task, let’s solve Laplace’s equation inside the unit sphere. i.e. find u(ρ, θ, ϕ) satisfying ∇2u = 1 ρ2 (ρ2uρ)ρ + 1 ρ2 sin θ(sin θ uθ)θ + 1 ρ2 sin2 θuϕϕ = 0, (1) u regular for ρ →0, θ →0, θ →π, u is 2π −periodic in ϕ, u(1, θ, ϕ) = f(θ, ϕ).
(2) The connection between the spherical polar coordinates and our usual Cartesian coordinates is x = ρ sin θ cos ϕ, y = ρ sin θ sin ϕ, z = ρ cos θ; ρ is distance from the centre; θ is angle from the pole (related to latitude, but θ = 0 and π correspond to the poles here); ϕ is the angle of longitude.
Looking ahead, separation of variables will split the PDE up into three ODEs, with two separation constants; the ODEs in ρ and θ will be non-constant-coefficient. Two of the ODEs should boil down to Sturm-Liouville problems so that we can find infinite sets of suitable eigenfunctions; in view of the various boundary conditions, these should correspond to the problems in the two angles. The general solution will then contain two infinite sums, with a bunch of constants to be set by demanding that we satisfy the more complicated, final condition in (2).
1 Solution by separation of variables We set u = R(ρ)Θ(θ)Φ(ϕ). The PDE can then be re-arranged into sin2 θ (ρ2R′)′ R + (Θ′ sin θ)′ Θ sin θ = −Φ′′ Φ .
(3) This function of (ρ, θ) or ϕ must equal our first separation constant. As we’re heading to a problem that should furnish 2π−periodic functions in ϕ, we’ll set this constant to m2, giving the ODE, Φ′′ + m2Φ = 0, with m = 0, 1, ..., and the usual functions (sin mϕ, cos mϕ or a constant with m = 0) present in a Fourier series.
The other side of (3) can now be re-arranged into a function of θ equalling a function of ρ: (Θ′ sin θ)′ Θ sin θ − m2 sin2 θ = −(ρ2R′)′ R ; (4) i.e. we need another separation constant, that we set to −λ.
At this stage, we might be a little stuck in deciding on which sign to to use for the second separation constant - there’s no obvious physical argument, and the form of the solutions are also not apparent, so deciding what sign is needed to satisfy the boundary conditions is tricky. A minus sign is chosen above, and one can attribute this choice to the vast knowledge and infallibility of the 1 professor. As someone pointed out in class, one could also be guided by the Sturn-Liouville problem that one needs to arrive at, which guarantees afterwards that everything works out happily.1 The θ−part now gives the Sturm-Liouville problem, (Θ′ sin θ)′ + λΘ sin θ −m2Θ sin θ = 0 [a, b] = [0, π], p(θ) = σ(θ) = sin θ & q(θ) = −m2 sin θ , (5) given that the regularity conditions we must impose at the poles are of type (ii). Note that 0 ≤θ ≤π, ensuring that p and σ are, indeed, non-negative. The SL ODE looks a bit formidable, though. In fact, it is not quite as bad as it seems, because the transformation x = cos θ and y(x) = Θ(θ) (indicating (d/dθ) →−sin θ(d/dx)) turns it into the simpler-looking, but still non-constant-coefficient, ODE, d dx (1 −x2)dy dx + ν(ν + 1)y −m2y 1 −x2 = 0, λ = ν(ν + 1).
(6) The singular points (i.e. the poles) are now at x = ±1, and the corresponding SL problem in x has [a, b] = [−1, 1], p(x) = 1 −x2, σ(x) = 1 and q(x) = −m2/(1 −x2).
To motivate the last switch (from λ to ν), we look at the ρ−problem, which now can be seen to revolve around the ODE (ρ2R′)′ −ν(ν + 1)R = 0.
(7) This is another Euler equation with solutions R ∝ρα where α(α + 1) = ν(ν + 1) − → α = ν or −ν −1; the α−values would have been a lot uglier expressed in terms of λ. In fact, to be definitive, we may take ν = q 1 4 + λ −1 2 ≥0 (since λ is not negative, from SL theory). The solutions with ρ−ν−1 are therefore ruled out because they are not regular for ρ →0.
The form of the general solution is therefore u(ρ, θ, ϕ) = X n " 1 2a0nρν0ny0n(cos θ) + ∞ X m=1 (amn cos mϕ + bmn sin mϕ)ρνmnymn(cos θ) # , (8) where {λmn, ymn(x)} denote the SL eigensolutions to (6), indexed by n, and we have added m as a subscript as a reminder that it appears as a known parameter in the ODE.
2 Legendre’s ODE To enjoy the solutions to (6), we divide and conquer. We kick offby considering axisymmetrical solutions without any dependence on ϕ. i.e. we put m = 0, corresponding to the boundary condition u(1, θ, ϕ) = f(θ). Equation (6) reduces to d dx (1 −x2)dy dx + ν(ν + 1)y = 0, (9) which is Legendre’s ODE.
1The most direct way to establish the sign of the separation constant is to multiply (5) by Θ and integrate: after an intergration by parts and a little re-arrangement, one obtains λ Z π 0 Θ2 sin θ dθ = Z π 0 (Θ′)2 sin θ + m2Θ2 sin θ dθ (sin θ = 0 at the limits of the integral and Θ must be regular). Since both integrals cannot be negative, neither can λ.
2 2.1 Lucky guesses By inspecting this ODE, you might be able to come up with some solutions: e.g. if λ = ν = 0, then [(1 −x2)y′]′ = 0 − → y′ = A 1 −x2 − → y = A ln 1 + x 1 −x + B, (10) for two integration constants, A and B. The first of these solutions is not regular at x = ±1, but (λ, y) = (0, B) looks perfectly reasonable. The eigenfunction y(x) = B has no zeros over the interval [−1, 1] and therefore looks as though it ought to be the first of the SL sequence (remember the SL oscillation theorem).
You might also be lucky and guess that a second solution is y(x) = Cx, for some constant C.
Plugging this into (9) gives λ = 2 or ν = 1. This solution has exactly one zero over [−1, 1] and might well be the second of the SL sequence. As the ODE is second-order, there must be another independent solution for λ = 2. To see what this is, we may use reduction of order: set y = xC(x).
Then, plugging into (9) and clearing the smoke gives x[(1−x2)C′]′+2(1−x2)C′ = 0 = ⇒ C′ = B x2(1 −x2) or y = Cx = Ax+B 1 2x log 1 + x 1 −x −1 .
The first term returns our original guessed solution; the second term corresponds to the other solution, and is again singular at x = ±1.
On a role, you might try a third solution taking the form of a quadratic polynomial; this does indeed work (after plugging into (9)), provided that y(x) = C(1 −3x2) and λ = 6 or ν = 2. The solution has two zeros over [−1, 1].
Wow, said Jurgen.
Could the SL eigenfunctions actually be polynomials with ν = n = 0, 1, 2, ...?
2.2 Series solution and recurrence relation for coefficients Armed with this insight, let’s try a series solution, y = ∞ X j=0 cjxj.
(11) Plugging this into the ODE gives ∞ X j=0 {(j + 2)(j + 1)cj+2 + [ν(ν + 1) −j(j + 1)]cj} xj = 0.
(12) To arrive at this relation, we used the fact that [(1 −x2)y′]′ = ∞ X j=0 [j(j −1)xj−2 −j(j + 1)xj]cj ≡ ∞ X l=0 (l + 2)(l + 1)xlcl+2 − ∞ X j=0 j(j + 1)xjcj after setting j = l + 2 in the first sum, then discarding l = −2 and l = −1 in view of the factor (l + 2)(l + 1). Switching the integer over which we sum from l back to j in the first term then leads to (12).
Consequently, as (12) must hold for every value of x, cj+2 = j(j + 1) −ν(ν + 1) (j + 1)(j + 2) cj.
(13) 3 This recurrence relation delivers the coefficient cj+2 in terms of cj. Therefore, if we specify c0, we may compute the sequence {c0, c2, c4, ...}. Alternatively, we can specify c1 and compute the sequence {c1, c3, c5, ...}. This means that we can construct two independent series with either even or odd powers of x. i.e. the solutions must be either even or odd (something that could have been predicted from Legendre’s equation, which is invariant under the reflection x →−x). For the former, we take c0 ̸= 0 and c1 = 0; for the latter, we set c1 ̸= 0 and c0 = 0. More dramatically, if ν is an even integer n, the even series beginning with c0 terminates at cnxn (since cn+2 = 0 by (13)), leaving an even polynomial of degree n. Similarly, if ν = n is an odd integer, the odd series beginning with c1 also terminates at cnxn, furnishing an odd polynomial.
In other words, the SL eigensolutions are λ = n(n+1), n = 0, 1, 2, ..., y(x) = Pn(x) = c0+c2x2+...+cnxn or c1x+c3x3+...+cnxn, (14) along with (13).
Note that, for this set and because of the first value for λ, it is convenient to begin our indexing of the SL sequence with n = 0, rather than n = 1, as done previously.
2.3 The first few Legendre polynomials To make the polynomials into special functions, we need a normalization to eliminate the freedom in the choice of either c0 or c1. The tradition is to take Pn(1) = 1, and we then arrive at the Legendre polynomials. Given our earlier “guesses”, we observe that the first three are P0(x) = 1, P1(x) = x, P2(x) = 3 2x2 −1 2.
(15) We can use the recurrence relation and normalization to compute Pn(x) directly for each n. e.g.
for n = 3, we have P3(x) = c1x + c3x3 & c3 = 1×2−3×4 2×3 c1 = −5 3c1. Then P3(1) = c1(1 −5 3) = 1 = ⇒ c1 = −3 2, giving P3(x) = 5 2x3 −3 2x.
For n = 4, similar calculations indicate that c4 = −7 6, c2 = −10c0 and c0 = 3 8, giving P4(x) = 35 8 x4 −15 4 x2 + 3 8.
This obviously gets a little unwieldy for bigger n.
Figure 1 plots some of the Legendre polynomials. Each time n is raised by one, another zero appears in Pn(x); the polynomials become wigglier. There is not a lot else worth saying about their spatial form.
For each λ = n(n + 1), the other independent solution is not regular at x = ±1 and is usually denoted as Qn(x). In view of the earlier guessed solutions, the first two must be Q0 = 1 2 ln 1 + x 1 −x & Q1 = 1 2x log 1 + x 1 −x −1, having imposed a different normalization (which can be viewed as insisting that Qn diverges as −1 2 ln(1 −x) when x →1). In fact, regular solutions to (9) can only be found if λ takes the value n(n + 1) for n = 0, 1, 2, .... Otherwise, with λ ̸= n(n + 1), the best one can do is is to make the solution regular at one of the singular points, x = ±1, but it remains irregular at the other singular point.
4 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 Figure 1: Legendre polynomials for n = 0, 1, ..., 6.
2.4 Orthogonality and expansion From SL theory we know that the Legendre polynomials must satisfy the orthogonality relation, Z 1 −1 PnPm dx = 0 if n ̸= m (16) (the weight function for Legendre’s equation is σ(x) = 1). In other words, they are examples of “orthogonal polynomials”.
Any sensible function F(x) can also be expanded in terms of the polynomnials, via the expansion formulae, F(x) = ∞ X n=0 dnPn(x), dn = R 1 −1 F(x)Pn(x) dx R 1 −1[Pn(x)]2 dx .
(17) Had F(x) = f(θ), as in our original axisymmetrical PDE problem, this would have corresponded to f(θ) = ∞ X n=0 dnPn(cos θ), dn = R π 0 f(θ)Pn(cos θ) sin θ dθ R π 0 [Pn(cos θ)]2 sin θ dθ .
(18) In other words, the solution to (1)-(2) with u(1, θ) = f(θ) must be u(ρ, θ) = ∞ X n=0 dnPn(cos θ)ρn (19) (given that we now see that νmn →n), with dn prescribed as above.
2.5 Some other charming properties Generating function: 1 √ 1 −2xt + t2 = ∞ X n=0 Pn(x)tn.
(20) Rodrigues’ formula: Pn(x) = 1 2nn!
dn dxn (x2 −1)n.
(21) A useful integral (appearing in the expansion formulae): Z 1 −1 [Pn(x)]2 dx = 2 2n + 1.
(22) 5 Recurrence relation for the polynomials themselves: (n + 1)Pn+1(x) = (2n + 1)xPn(x) −nPn−1(x) (23) All these, and much, more more, can be found in textbooks or online at Wikipedia or DLMF.
The usual route to establishing them involves playing around with some combo of Legendre’s ODE, (13) and the orthogonality condition, or one of the preceding properties. Although one can easily get lost admist all these charming details, we will avoid straying too far from our path of solving PDEs in Math 400.
3 Associated Legendre functions Our original interest was in the SL ODE in (6) rather than the simpler one in (9). So coming up with Legendre polynomials to solve (9) doesn’t look that helpful right now. In fact, this is not the case at all. Amazingly enough, the regular solutions to (6) are CP m n (x), where C is arbitrary, λ is again n(n + 1) with n = 0, 1, 2, ... and P m n (x) is an associated Legendre function that is given by P m n (x) = (−1)m(1 −x2)m/2 dm dxm [Pn(x)].
(24) In other words, νmn is again n, the solutions to (6) can be extracted from suitable derivatives of the Legendre polynomials, and we have another Jurgen Wow moment. Note that conventions are not always followed, and sometimes the (−1)m is omitted in (24). This can cause a bit of a headache at times.
We won’t establish (24) generally, but let’s at least consider m = 1: from Legendre’s ODE we note d dx (1 −x)2P ′′ n −2xP ′ n + n(n + 1)Pn = (1 −x)2 d2 dx2 −4x d dx −2 + n(n + 1) P ′ n = 0.
(25) The first derivative term in the ODE for P ′ n is different from that in the original Legendre ODE (we have −4 d dx rather than −2 d dx), but this can be adjusted with the substitution, P ′ n = Y (x)/ √ 1 −x2.
We then obtain (1 −x2)Y ′′ −2xY ′ + n(n + 1)Y − Y 1 −x2 = 0, (26) which is (6) with m = 1. i.e. we may take P 1 n(x) = − √ 1 −x2P ′ n. In the same way, but with more effort, one can differentiate Legendre’s equation m times, plug in dmPn/dxm = Y/(1 −x2)m/2, to arrive at (6), verifying (24). There is therefore nothing esoteric here, just repeated differentiation and a simple substitution (designed to adjust the first derivative term in the ODE for dmPn/dxm).
As with the Legendre polynomials, there is another, independent solution to (6) for λ = n(n + 1).
That solution is again irregular at x = ±1, and is normally denoted by Qm n (x).
Finally, SL theory implies the orthogonality condition and expansion formulae: Z 1 −1 P m n P m j dx = 0 n ̸= j, 2(n + m)!/[(2n + 1)(n −m)!] n = j, (27) g(x) = ∞ X n=0 dnP m n (x), dn = (2n + 1)(n −m)!
2(n + m)!
Z 1 −1 g(x)P m n (x) dx.
(28) The result for R 1 −1[P m n ]2dx is a generalization of the useful integral in (22) (and can be proved using Rodrigues’ formula).
6 3.1 Some simple Legendre functions and negative order The factor (1 −x2)m/2 in (24) ensures that P m n (x) is still a polynomial if m is even. Conversely, for m odd, P m n (x) consists of a polynomial multiplied by the factor √ 1 −x2 ≡sin θ. For n = 2 or 3 and low m, we have P 1 2 = −3x p 1 −x2, P 2 2 = 3(1 −x2), P 1 3 = 3 2(1 −5x2) p 1 −x2, P 2 3 = 15x(1 −x2), (29) Because Pn(x) is a polynomial of degree n, dnPn/dxn is a constant. Therefore when n = m, the associated Legendre function is just P n n (x) = (−1)n(1 −x2)n/2 (2n)!
2nn! , (30) where the final constant factor follows from Rodrigues’ formula in (21). For example, with n = m = 1, P 1 1 (x) = − p 1 −x2 ≡−sin θ.
(31) For our PDE problem, this implies that the boundary condition f(θ, φ) = sin θ sin ϕ leads to the solution u(ρ, θ, ϕ) = ρ sin θ sin ϕ.
It must also be true that P m n (x) = 0 for m > n.
(32) Later, we see that this result allows a truncation of the eigenfunction expansion.
The parameter m is sometimes called the “order” of the associated Legendre function P m n (x); n is the “degree”. At the moment the order is purely positive and P m n (x) only defined for m > 0.
However, it proves useful to extend this definition so that the order can be negative. In particular, it proves convenient to do this by setting P −m n (x) = (−1)m(n −m)!
(n + m)!
P m n (x).
(33) 4 General solution of the PDE; Spherical harmonics At last, we use the preceding results to write the general solution to the PDE in (1): u = ∞ X n=0 " 1 2a0nρnPn(cos θ) + n X m=1 (amn cos mϕ + bmn sin mϕ)ρnP m n (cos θ) # , (34) where we have used (32). The boundary condition at ρ = 1 can be dealt with by first expanding f(θ, ϕ) as a Fourier series in ϕ: f = 1 2α0(θ) + ∞ X m=1 [αm(θ) cos mϕ + βm(θ) sin mϕ], α0 αm βm = 1 π Z π −π f(θ, ϕ) 1 cos mϕ sin mϕ dϕ, (35) matching up the coefficients with those from (34), and then using the expansion formulae in (28): a0n amn bmn = (2n + 1)(n −m)!
2(n + m)!
Z π 0 P n m(cos θ) α0(θ) αm(θ) βm(θ) sin θ dθ.
(36) 7 This calculation can be cleaned up considerably by defining spherical harmonics. To begin, we note that, by Euler’s formula the cosines and sines in (34) can be replaced by the exponentials e±imϕ.
Therefore, by switching notation for the arbitrary constants (to Am n ), we could just as well write the single double sum, u = ∞ X n=0 n X m=−n Am n ρnY m n (cos θ, ϕ), (37) where we have used the extension to negative order in (33). and defined the expansion functions, Y m n (θ, ϕ) = s (2n + 1)(n −m)!
4π(n + m)!
eimϕP m n (cos θ).
(38) Because of (27) and Z π −π e−im1ϕeim2ϕdϕ = 0 m1 ̸= m2, 2π m1 = m2, (39) we observe Z π 0 Z π −π (Y m1 n1 )∗Y m2 n2 dϕ sin θ dθ = δn1n2δm1m2, (40) where the star denotes complex conjugation and δjk denotes the Kronecker delta (the order of the double integral matters here and the ϕ integral needs to be performed first). Consequently, we can read offimmediately the values of the coefficients in (37): Am n = Z π −π Z π 0 (Y m n )∗f(θ, ϕ) sin θ dθ dϕ (41) (where the integration order no longer matters).
The orthonormal eigenfunctions are the spherical harmonics.
They form a complete set with which one can expand any sensible function of the two angles, θ and ϕ, and satisfy the PDE 1 sin θ ∂ ∂θ sin θ∂Y m n ∂θ + 1 sin2 θ ∂2Y m n ∂ϕ2 = −n(n + 1)Y m n .
(42) 5 The wave equation for a sphere Now let’s consider the wave equation for the ringing of a spherical fluid drop of unit radius. i.e., find u(ρ, θ, ϕ, t), satisfying utt = 1 ρ2 (ρ2uρ)ρ + 1 ρ2 sin θ(sin θ uθ)θ + 1 ρ2 sin2 θuϕϕ, u regular for ρ →0, θ →0, θ →π, u is 2π −periodic in ϕ, u = 0 at ρ = 1.
(43) We also need two initial conditions, but we won’t be interested in any particular initial-value problem below.
We begin using the eigenfunction expansion, u = ∞ X n=0 n X m=−n Rm n (ρ, t)Y m n (cos θ, ϕ).
(44) Plugging in to the wave equation and using (42) furnishes a PDE for each coefficient: ∂2Rm n ∂t2 = 1 ρ2 ∂ ∂ρ ρ2 ∂Rm n ∂ρ −n(n + 1) ρ2 Rm n .
(45) 8 We can separate variables to solve this equation, which leads to solutions with time dependences of cos ωt and sin ωt, for some separation constant −ω2. Imposing the boundary conditions in ρ then determines ω (from another SL problem). Following this route, the general solution for u(ρ, θ, ϕ, t) would involve a triple sum over n, m and a third integer indexing the SL eigensolutions in ρ.
Alternatively, if we are not interested in a specific initial-value problem, but simply the “normal-mode” frequencies at which the drop could ring, we can instead adopt the time dependence in Rm n = Rmn(ρ)e−ωt, to arrive at the ODE R′′ mn + 2 ρR′ mn + ω2Rmn −n(n + 1) ρ2 Rmn = 0.
(46) This is Bessel’s equation in disguise (i.e. it is an example of the more general ODE that leads to Bessel functions noted in Notes V; α = 1 2, β = 1, α2 −ν2 = −n(n + 1)). The solutions that are regular for ρ →0 are Rmn = Cmnρ−1 2 Jn+ 1 2 (ωρ), (47) for some constant Cmn. The half-integer versions of Jν(z) are called “spherical Bessel functions” (precisely because they commonly appear in PDE problems with spherical coordinates like (43)).
The outer boundary condition now demand that the frequencies ω must be dictated by Jn+ 1 2 (ω) = 0.
Spherical Bessel functions can be written more explicitly in terms of trig functions (Bessel functions in general can not). We have already seen this with J 1 2 (z).
Note that ω depends only on n and not m. i.e. all normal modes with the same n, but different m have the same frequency of oscillation. There are (2n + 1) of these modes as m runs from −n to +n.
The frequencies are independent of m because the original drop was spherically symmetric, the PDE (43) has no ϕ−dependent coefficients, and any oscillations cannot tell the difference between east and west in the current coordinate system. Adding any additional effect that does distinguish east from west breaks this symmetry and splits up the frequencies of (2n + 1)−modes that arise for each n. The removal of the (2n + 1)−fold “degeneracy” of the frequencies corresponds to the origin of the Zeeman splitting of spectral lines by a magnetic field, and underlies how astronomers have searched for clues about the rotation rates and magnetic fields of stars by studying pulsation frequencies.
9 |
6798 | https://www.youtube.com/watch?v=0Xoj80MiBLI | DISPLAYING SETS ON A NUMBER LINE, Interval Notation sets
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Transcript:
hello everyone i'm back again with another tutorial uh this time around we're solving question two uh in tutorial sheet one yeah so the question reads let x be equal to uh close brackets these kinds of brackets are normally called close brackets uh close brackets zero and zero comma ten and then when you see these kinds of brackets these are called open brackets yeah so what we're actually talking about uh set intervals yeah so these given uh questions a b c uh be the subset of x meaning if they are subsets of x means that they are all in x find each of the following and display and display them on the real line so we'll quickly begin so we have uh x x is the universal set so if we are now to find a b complement so b complement b complement to just be nothing but equal to uh so you look at the universal set so the universal set is starting from zero so we're trying b complement is we're trying we are trying to list whatever that is not in b but it's in the universal set so big complement means we're listing what is in what is in the universal set but it's not in v all right so make sure those that have not yet subscribed to the channel subscribe so that you're not missing out on how we're going to be solving uh questions on in the in the world in that store sheet all right so big compliment here is uh a big compliment here is is uh we we do what we first look at the universal cell so in this universal set [Music] we have zero yeah the investor state is starting from zero [Music] and then when we go towards b or when we look at b our b is uh two comma eight so these kinds of brackets are implying to say two and eight are not part of the set uh b but we're talking about numbers between two and eight we're talking about real numbers between two and eight so for to to write the b complement here we'll say zero and then comma we write our 2 since 2 is not in b it means that it is in the complement meaning we use this kind of bracket which shows that 2 is part of a b complement these kinds of brackets show that 2 is not part of b and then we combine this set with another set which starts from 8 so i'm going to write 8 this one is starting from 8 up to 10 now is 8 part of b complement yes why because it's not in b so meaning it's part of b complement we're going to use this type of bracket and then is 10 part of b complement yes all right so this is our b complement so now the question is saying we display this on the real line so when displaying this on the real line you write you draw your real line like that so this is your zero if you want you can even include negative one there so zero then you have one you have two you have three four five six seven eight nine then you have ten there like that so we have let me just write important numbers there uh 1 we have 2 we have 10 9 8 yeah but you have to write all the numbers here i'm just writing important numbers that we're going to be using so from there what you do is uh you put this on the real line so when you have these kind of brackets it's mean that it it simply means that you're going to share the circle here so meaning this circle is going to be shaded and then we draw a line up to two you also draw another circle there at two which is shaded okay and then you go on eight you also draw there so using that kind of bracket meaning we're going to shade even at eight uh same applies to this part here we're also going to shade extent like that and then to write now the x so x is the universal set you always have to put it there yeah so x is the inverse edge you need to include it there so this is going to be from 0 to 10 okay so we draw a straight line okay so this is our x then we have our big complement you write you make sure that you show it to say this line is the one that is that is b component yeah so big complement like that so this is how you show them on the real line but make sure that you put all the numbers that are needed today such that i i want to finish first this is why um i'm skipping some of these numbers i'm not writing them all right so this is how you show it on the real line you quickly move on to the next question so the next question is saying we find uh so the next question there is saying we find a intersection b so a intersection b a intersection b so to find the a intersection b we're going to have uh the common elements between a and b so the common elements between a and b we can face that with the smallest number here we have one there do we have one we have one in a now do we have one in b no which is the other smallest number we have two there we have two in b so i'm going to two in b there's also two in a so i'm going to write two there and then we say comma and then what you do there after writing the comma you do what this will be 2 comma 4 this should be 2 comma 4 yeah so once you write 2 comma 4 now which kinds of brackets are we going to put there is it these kinds of brackets or these now you look at uh the two is two part of b no so the two that we got is in b but it's not in b so many we're going to use this kind of brackets on b and then when you look at uh this part here uh where there is uh when you look at this part here uh we are saying uh 4 is in a and 4 is also in b so 4 is in a it's also in b so what you do is you include it you include it in the set the intersection set because when you look at these four it's part of a and this m4 again is is between two and eight meaning you included like that so when drawing the real line what you do is uh you draw the number line like that and then you write your numbers from zero yeah this is one two three four five six seven eight nine ten so you write it in like that so our intersection b is two comma four so we have two and three four we have two up to four so the kinds of um the way we draw this is going to be different from the way we are shading this yeah because on two it's open so meaning will not shade at two and then at four it's closed this kind of brackets is closed so meaning will shed at four yes so this is how it's going to be make sure that you include the the universal set okay so you make sure that you include the universal set say the investor set here is x and then this is our a intersection b all right so to quickly do one to to quickly finish up what you are doing i think let me skip uh okay i'll give you the solutions though i skip them i'll still give you the solutions so the third question here is or let me just finish them up so the third question here is telling us to find a intersection b complement so a intersection b complement so a intersection b complement will simply be equal to this is our a intersection b so now we're going to list the elements that are in the universal set but they are not in a intersection b they are in the universal set but they are not in intersection b in other words we we're trying to list the elements that are not in a intersection b so if this is our a intersection b simply means that our a intersection b complement will start from zero since the investor set our investor set x is starting from zero and then it will end at two first it will end at two but where but at two since two is not part of um the a intersection b when you look at this kind of bracket which has been used at 2 is simply telling us to say 2 is not part of a intersection b so it's going to be part of the complement and then we say union and then we begin at 4 since 4 is part of a intersection with this kind of bracket here which is being used it simply is simply telling us to say 4 is part of intersection b meaning it will not be part of the complement and then you say coma who are ending up to 10 1 or ending up to 10. so if we were not given the universal set or if the universal set was a set of all real numbers instead of writing 10 here would have written a 4 comma infinity now since we have been given x as our investor said meaning we're ending at 10 okay so when it comes to drawing such a such and what a a real line you do the same way the same way i've been drawing the previous one so you just put this on the okay let me just do a sketch this is zero then we have ten so the first one is going to move from zero so this one will be shaded of course up to 2 so at 2 we're also going to shade because 2 is also part of our complement set there then from we're going to start again from 4 to 10 so at 4 it's going to remain open and then at 10 at 10 is going to be closed [Music] it's going to be closed it's going to be closed that thing yeah it's going to be close that's 10 yeah it's going to be close that thing and then now we make sure that we don't forget to put the word the universal set so the inverse of set is going to be x as usual the investor set is going to be x as usual and then this is our a this is our a intersection b complement okay same applies to this one this is our a intersection b complement okay so this is how you solve such questions they are all simple and straightforward so let us now quickly move on to the fourth question uh which says we find okay for this question i think what we should do is uh let us go on top so that at least we should be able to to see what we are to see the the versus the given sets so i'll draw a margin here so we are going to have this so the question is saying uh we find a intersection we open the brackets b slash c so we first start by dealing with what is in the brackets there so we first write our a and then we say intersection b c is simply the same as b uh it's not i mean b intersection c complement yeah so b intersection c complement so the first thing that we can first find is c complement so c complement to c c complement to b uh we we have to stick to the universal set so meaning is going to start from zero because c is here if c is here meaning it's going to start from zero i mean c complement is going to start from zero and end at three and then at three since since three is part of c meaning it's not going to be part of the complement then we say uni union and then we start 6 we write 6 and then 6 c6 is not part of c meaning it's going to be part of the complement and then we say comma we write our 10 like that so our 10 is part of the complement so from there we we we look at b so we find the common elements between b and c complement we're trying to find b intersection c complement because we're trying first to deal with what is in the brackets so b intersection c complement b intersection c complement e is going to be equal to so this is c complement and our b is uh this part here so we find the common elements there you discover to say when you look at 2 we have 2 there so 2 is going to be a part of our b intersection c complement and then it's going to be open because uh the tool that we have is open it's not closed and then it's going to be open and then we write comma this is going to be we're going to end at 3 because we have a 3 there and it's also open yeah and then we say union then we begin from six then end at eight we're trying to find the common elements so begin at six and then end at eight then of course our eight is going to be open since it's open here and then our six is also i mean r6 is going to be closed like that yeah since it's found in both c complement and b so after finding this let us now find let us now intersect this uh this this uh this b intersection c complement with a so let us now find the common elements between a and b intersection c so a is one comma four so we find the common elements between one comma four and this b intersection c complement and then this is what we're going to have as our answer b intersection i mean a intersection b c b c so when you look at this um b intersection c complement and 1 comma 4 there you can agree with me to say 2 comma 3 is between 1 comma 4 so meaning that is the common that is the common set that we have there so 2 comma 3 is our final answer right so this is the answer that we've been looking for so aft having uh found this it is now easier for us to draw it on the real line so on the real line there of course you put your zero up to 10 so what i'm doing i'm just sketching make sure that when you when you are writing it you put all the numbers in between here i'm just trying to do a quick thing so from two to three and at both ends it's going to be open even here it's going to be open what i mean that by sketching i'm not adding the other numbers like one it's not therefore it's not there and other numbers but for you you have to put them when you are writing them when you're writing more especially in your exam on your test so this is our investor set x and then it's shaded on both ends it's closed on both ends so even here it's closed on both ends yeah so this is uh you can name this as a intersection you write your b slash c yeah and then this is your x so let us quickly solve the last question so the last question is uh x slash a union a slash b so um i don't know if i should put it this side okay let me just write it down or just even this side is okay sorry let me just put it down okay so the last question says the last question which is question 5 is requiring us to find x slash a x slash a then union slash b so here what we can do is first we can simplify this uh expression before we we even start finding the answers so let us first simplify the exp expression so how do you simplify this a is just the same as a i mean x slash i mean x intersection a complement okay then union a slash b is just the same as a intersection b complement now so you remember according to the set laws that we learned and any any set intersection uh the universal set because our x is the investment set so any set intersection the investor set is simply that same set then we say union we have a intersection [Music] b complement so a complement union a so i mean we can now use the distributive law here so i'm going to say a complement we distribute a complement with what is in the brackets there say a complement union a [Music] you put it in brackets then you write your symbol there which is the intersection symbol then here you have a complement union b a complement union b complement okay so a complement union a is simply the universal set so a complement and our investor set is denoted by x in this question so air complementary on a is simply the universal set so investor set intersection a so a a complement union so he assessed the amount of simplifying the expression once it's simplified it will even be easier for us to find the answer so a any any set any set meaning this set intersection the investor set will just result to that same set so still remain with this same set and according to de morgan's law this can also be written as a intersection b a intersection b and then complement you just change the sign in between and then remove the sign the the complement outside so this now becomes even easier for us to find yeah if it becomes even easier for us to to add to to solve so we already have a intersection b we found a intersection b in the beginning and we found intersection b2b let us just go back on top all right now i think the solution is even on on this on part three because at one point we were asked to find a intersection b complement and then when you simplify this it's also taking us to the same a intersection b complement so in other words this uh solution the solution for this which is intersection b complement is simply this part this same solution here on top so this is the solution for the last question so instead of me drawing it this is the sketch and this is the solution for our intersection b complement so thank you very much for watching the tutorial video make sure that you don't forget to subscribe to the channel uh so that you don't miss out the uh the other questions that i'm going to be solving in your tutorial sheets yeah i'm not on i'm not only going to solve tutorshit one but i'll soak all the tutorials that that your lecturer is going to be uh releasing including some of your [Music] physics tutorial sheet questions yeah so make sure that you subscribe to the channel and make sure that every time i post the video here you watch it it's really going to help you to clear your mathematics with good grades all right shalom shalom |
6799 | https://physics.info/newton-second/ | Force and Mass – The Physics Hypertextbook
chaos
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facts
get bent
physics
The Physics Hypertextbook
Opus in profectus
…newton-first
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Force and Mass
discuss ion
summary
practice
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[close]
Discussion
introduction
Discuss…
different forces on same object (result?)
different objects with same forces (result?)
different objects with same acceleration (how?)
Consequences…
Newton's second law of motion states that acceleration is directly proportional to net force when mass is constant…
a∝∑F
and that acceleration is inversely proportional to mass when net force is constant…
a∝1
m
and that net force is directly proportional to mass when acceleration is constant…
∑F∝m
This is more compactly written as an equation that combines these relationships…
a=∑F
m
For a variety of reasons, Newton's second law of motion is often written with net force as the subject of the equation like this…
∑F=ma
history
Lex. II.Law II.
Mutationem motus proportionalem eſſe vi motrici impreſſæ, & fieri ſecundum lineaum rectam qua vis illa imprimitur.The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.
Si vis aliqua motum quemvis generet; dupla duplum, tripla triplum generabit, ſive ſimul & ſemel, ſive gradatim & ſucceſſive impreſſa fuerit. Et hic motus quoniam in eandem ſemper plagam cum vi generatrice determinatur, ſi corpus antea movebatur, motui ejus vel conſpiranti additur, vel contrario ſubducitur, vel obliquo oblique adjicitur, & cum eo ſecundum utriuſque determinationem componitur.If any force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion being always directed the same way with the generating force, if the body moved before, is added to or subtracted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both.
Newton also defined what he called "the quantity of matter" and "the quantity of motion". We now call them "mass" and "momentum", respectively.
Definitio. I.Definition I.
Quantitas materiæ est mensura ejusdem orta ex illius densitate et magnitudine conjunctim.The quantity of matter is the measure of the same, arising from its density and bulk conjunctly.
Aer densitate duplicata, in spatio etiam duplicato, sit quadruplus; in triplicato sextuplus. Idem intelige de nive & pulveribus per compressionem vel liquesactionem condensatis. Et par eft ratio corporum omnium, quæ per caufas quascunque diversimode condensantur. Medii interea, si quod fuerit, interstitia partium libere pervadentis, hic nullam rationem habeo. Hanc autem quantitatem sub nomine corporis vel masse in sequentibus passim intelligo. Innotescit ea per corporis cujusque pondus: Nam ponderi proportionalem esse reperi per experimenta pendulorum accuratissime instituta, uti posthac docebitur.Thus air of double density, in a double space, is quadruple in quantity; in a triple space, sextuple in quantity. The same thing is to be understood of snow, and fine dust or powders, that are condensed by compression or liquefaction; and of all bodies that are by any caused whatever differently condensed. I have no regard in this place to a medium, if any such there is, that freely pervades the interstices between the parts of bodies. It is this quantity that I mean hereafter everywhere under the name of body or mass. And the same is known by the weight of each body; for it is proportional to the weight, as I have found by experiments on pendulums, very accurately made, which shall be shewn hereafter.
Definitio. II.Definition II.
Quantitas motus est mensura ejusdem orta ex velocitate et quantite materiæ conjunctim.The quantity of motion is the measure of the same, arising from the velocity and the quantity of matter conjunctly.
Motus totius est summa motuum in partibus singulis; ideoque in corpore duplo majore, æ quali cum velocitate, duplus est, & dupla cum velocitate quadruplus.The motion of the whole is the sum of the motions of all the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity it is quadruple.
So what is mass?
pick an object to be the standard unit mass
push mass with reproducible force (or use the principle of action-reaction)
measure its acceleration
push an unknown mass with the same force
measure new acceleration
mass is inversely proportional to acceleration
Mass…
Mass is a measure of resistance to acceleration.
(More generally, mass is a measure of resistance to change.)
Mass is a scalar quantity associated with matter.
When a system is composed of several objects it is the total mass that matters.
The SI unit of mass is the kilogram [kg].
Mass of selected objects| mass(kg) | object |
--- |
| ~10 53 | observable universe |
| ~10 42 | Milky Way |
| 8 | ×10 40 | largest black hole (S5 0014+81) |
| 6 | ×10 32 | most massive star (R136a1) |
| 2 | ×10 31 | smallest black hole (XTE J1650−500) |
| 2.8~6 | ×10 30 | neutron star |
| 1.99 | ×10 30 | Sun |
| 1.90 | ×10 27 | Jupiter |
| 5.97 | ×10 24 | Earth |
| 6.42 | ×10 23 | Mars |
| 7.35 | ×10 22 | Moon |
| 1.31 | ×10 22 | Pluto |
| 1.35 | ×10 21 | Earth's hydrosphere |
| 5.14 | ×10 18 | Earth's atmosphere |
| 1.84 | ×10 15 | Earth's biosphere |
| ~150,000 | blue whale |
| ~5,000 | African elephant |
| ~1,750 | passenger car |
| 635 | world's heaviest man |
| 104 | the author |
| 7.72 | world's smallest woman |
| 3~7 | bowling ball |
| | one pound mass |
| 0.16 | billiard ball |
| ~3 | ×10−6 | snowflake |
| 2.18 | ×10−8 | Planck mass |
| ~10−12 | bacterium |
| ~10−15 | virus |
| 3.9529264 | ×10−25 | uranium 238 atom |
| 3.0762534 | ×10−25 | top quark |
| 1.6749275 | ×10−27 | neutron |
| 1.6735328 | ×10−27 | hydrogen 1 atom |
| 1.6726219 | ×10−27 | proton |
| 9.1093837 | ×10−31 | electron |
| <2.0 | ×10−36 | neutrino (upper limit) |
| <1.8 | ×10−54 | photon (upper limit) |
So what is a force?
Force…
A force is an interaction that causes acceleration. More generally, a force is an interaction that causes a change.
Force is a vector quantity associated with an interaction.
When several forces act on a system it is the net, external force that matters. Since force is a vector quantity, use geometry instead of arithmetic when combining forces.
For a force to accelerate an object it must come from outside it. External force. Can't pull yourself up by your own bootstraps. Anyone who says you can is engaging in hyperbole.
The SI unit of force is the newton [N=kg m/s 2]. Rule of thumb: one newton is approximately equal to a quarter pound
Selected forces (smallest to largest)| force(N) | device, event, phenomenon, process |
--- |
| 10−14 | langevin forces of brownian motion |
| 10−11 | molecular motors consuming ATP |
| 10−10 | breaking noncovalent bonds (denaturing proteins) |
| 10−0 9 | breaking covalent bonds |
| 10−0 6 | adhesive force of bacteria |
| 0.032 | threshold of touch sensation; index, middle, pinky |
| 0.044 | threshold of touch sensation; ring finger |
| 4.436–4.460 | weight of 1 lb on Earth |
| 4.44822 | weight of 1 lb in standard gravity |
| 9.780–9.832 | weight of 1 kg on Earth |
| 9.80665 | weight of 1 kg in standard gravity |
| 32 | French press coffee plunger |
| 256 | average dog bite |
| 1020 | weight of the author |
| 2,200 | peak foot force, 75 kg human, running |
| 140,000 | peak foot force, 10,000 kg asian elephant, running |
The concept of inertia comes in many forms.
Analogous applications of Newton's second law of motion| | cause of change | = | resistance to change | × | rate of change of… | |
--- --- ---
| newton's second law | force | | mass | | velocity | F=m dv dt |
| rotational dynamics | torque | | moment of inertia | | angular velocity | τ=I dω dt |
| newtonian fluids | shearing stress | | viscosity | | shear | Fx=η dx/dz A dt |
| thermal conduction | temperature gradient | | r-factor | | heat | ∆T=R dq dt |
| ohm's law | potential difference | | electrical resistance | | charge | V=R dq dt |
| faraday's law | potential difference | | inductance | | current | V=L dI dt |
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Energy and Entropy
Absolute Zero
Waves & Optics
Wave Phenomena
The Nature of Waves
Periodic Waves
Interference and Superposition
Interfaces and Barriers
Sound
The Nature of Sound
Intensity
Doppler Effect (Sound)
Shock Waves
Diffraction and Interference (Sound)
Standing Waves
Beats
Music and Noise
Physical Optics
The Nature of Light
Polarization
Doppler Effect (Light)
Cerenkov Radiation
Diffraction and Interference (Light)
Thin Film Interference
Color
Geometric Optics
Reflection
Refraction
Spherical Mirrors
Spherical Lenses
Aberration
Electricity & Magnetism
Electrostatics
Electric Charge
Coulomb's Law
Electric Field
Electric Potential
Gauss's Law
Conductors
Electrostatic Applications
Capacitors
Dielectrics
Batteries
Electric Current
Electric Current
Electric Resistance
Electric Power
DC Circuits
Resistors in Circuits
Batteries in Circuits
Capacitors in Circuits
Kirchhoff's Rules
Magnetostatics
Magnetism
Electromagnetism
Ampère's Law
Electromagnetic Force
Magnetodynamics
Electromagnetic Induction
Faraday's Law
Lenz' Law
Inductance
AC Circuits
RC Circuits
RL Circuits
LC Circuits
Alternating Current
Impedance
Electromagnetic Waves
Maxwell's Equations
Electromagnetic Waves
Electromagnetic Spectrum
Modern Physics
Relativity
Space-Time
Mass-Energy
General Relativity
Quanta
Blackbody Radiation
Photoelectric Effect
X-rays
Antimatter
Wave Mechanics
Matter Waves
Atomic Models
Semiconductors
Condensed Matter
Nuclear Physics
Isotopes
Radioactive Decay
Half Life
Binding Energy
Fission
Fusion
Nucleosynthesis
Nuclear Weapons
Radiobiology
Particle Physics
The Standard Model
Quantum Electrodynamics
Quantum Chromodynamics
Quantum Flavordynamics
Beyond the Standard Model
Foundations
Units
International System of Units
Gaussian System of Units
British-American System of Units
Miscellaneous Units
Time
Unit Conversion
Measurement
Significant Digits
Order of Magnitude
Graphs
Graphical Representation of Data
Linear Regression
Curve Fitting
Calculus
Vectors
Trigonometry
Vector Addition and Subtraction
Vector Resolution and Components
Vector Multiplication
Reference
Special Symbols
Frequently Used Equations
Physical Constants
Astronomical Data
Periodic Table of the Elements
People in Physics
Back matter
Preface
About this book
Contact the author
glennelert.us
Behance
Bluesky
Google Scholar
Instagram
Threads
YouTube
Affiliated websites
hypertextbook.com
midwoodscience.org
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